Ratio And Proportion - Class 6 Maths - Extra Questions

The ratio in the simplest form of 5 minutes to 15 minutes is $$1:4$$
If true then enter $$1$$ and if false then enter $$0$$

$$5$$ minutes to $$15$$ minutes
$$=\frac{5 \quad minutes}{15\quad minutes}$$
$$=\frac{1}{3}$$


The ratio in the simplest form of 8 months to $$2$$ year is $$1:3$$
If true then enter $$1$$ and if false then enter $$0$$

$$8$$ months to $$2$$ years
$$=\frac{8 \quad months}{2\quad years}$$
$$=\frac{8 \quad months}{2\times 12\quad months}$$
$$=\frac{1}{3}$$



The ratio in the simplest form of 16 years to 18 years is 8:10
If true then enter 1 and if false then enter 0

$$16$$ years to $$18$$ years
$$=\frac{16 \quad years}{18\quad years}$$
$$=\frac{8}{9}$$


A shop sells 60 pastries, 45 sandwiches, and 80 chocolate cakes. Write down the following ratio in the simplest form.
Number of chocolate cakes to the number of pastries is 4:5
If true then enter $$1$$ and if false then enter $$0$$

A shop sells
                  pastries$$=60$$ 
                  sandwiches$$=45$$ 
                  chocolate cakes$$=80$$
 
Number of chocolate cakes to the number of pastries $$=80:60$$
                                                                                           $$=4:3$$ 


A shop sells 60 pastries, 45 sandwiches and 80 chocolate cakes. Write down the following ratio in the simplest form.
Number of sandwiches to the number of chocolate cakes.

A shop sells 60 pastries, 45 sandwiches, and 80 chocolate cakes.

Number of sandwiches to the number of chocolate cakes $$=45:80=9:16$$


A shop sells 60 pastries, 45 sandwiches and 80 chocolate cakes. Write down the following ratio in the simplest form.
Number of pastries to the number of sandwiches.

A shop sells 60 pastries, 45 sandwiches and 80 chocolate cakes.

Number of pastries to the number of sandwiches $$=60:45=4:3$$


Find the ratio of the given quantities: A pot containing 8 liters of milk while another containing 750 milliliters of milk.


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Express the following ratio in their simplest form. $$2:3$$.

As, 2 and 3 does not have anything in common .
So, no further simplest form of 2:3 is possible


Geetha earns Rs.$$12000$$ a month, out of which she saves Rs.$$3000$$. Find the ratio of her expenditure to savings.

money earned, M.E.=Rs 12000
money saved, M.S.=Rs 3000
expenditure,E=M.E-M.S.
=12000-3000
=Rs 9000
Ratio=E:M.S.
=9000:3000
$$\begin{array}{c}\frac{{9000}}{{3000}} = \frac{{3 \times 3000}}{{3000}}\\ = \frac{3}{1}\\ = 3:1\end{array}$$



Solve the following.
A milk man adds $$250$$ml of water to $$1$$ litre of milk. Find the ratio of water to milk.

Amount of water=250ml
Amount of milk=1l=1000ml
ratio=amount of water:amount of milk
=250:1000
=1:4


Find the ratio of the given quantities: Speed of a cycle is 15 km/h and speed of the scooter is 30 km/h.


1688067_565058_ans_226bb22ceb4d4f3695e2476e079fa8b4.PNG


Express the following ratio in their simplest form. $$20:60$$.

20:60 can be written as,
$$\begin{array}{c}\dfrac{{20}}{{60}} = \dfrac{{20}}{{20 \times 3}}\\ = \dfrac{1}{3}\end{array}$$

Thus the simplest form will be 1:3


Ramu spent $$\displaystyle\frac{2}{5}$$th of his money on a story book. Find the ratio of money spent to the money with him at the beginning.

Let Ramu has Rs.$$x$$ at the beginning.

Thus spends Rs. $$\dfrac{2x}{5}$$ on story books.

Ratio of his spending with the money at beginning is $$\dfrac{2x}{5}:x$$

On further simplifying it is clear that the required ratio is $$2:5$$.


A point X on $$AB=14$$cm divides it in the ratio $$3:4$$. Find the length of AX and XB.


1348837_701705_ans_81feb0e4651c41c0ae08981069647985.jpg


If the cost of 7 meters of cloth is Rs.294, find the cost of 5m of cloth.

If the cost of $$7m$$ cloth is $$Rs.294$$
Then cost of $$1m$$ cloth is $$\cfrac{294}{7}=Rs.42$$
If the cost of $$5m$$ cloth is $$Rs.42\times 5=Rs.210$$.


Geetha earns Rs.$$12000$$ a month, out of which she saves Rs.$$3000$$. Find the ratio of her savings to her income.

money earned, E=Rs 12000
money saved, S=Rs 3000
Ratio=S:E
=3000:12000
$$\begin{array}{l} = \dfrac{{3000}}{{12000}}\\ = \dfrac{{3000}}{{4 \times 3000}}\\ = \dfrac{1}{4}\\ = 1:4\end{array}$$



A rectangle measures $$40\ cm$$ at its length and $$20\ cm$$ at its width. Find the ratio of the length to the width.

Given:
Length of the rectangle $$=40\ cm$$
Width of the rectangle $$=20\ cm$$

Therefore,
Required ratio $$=\dfrac{40}{20}=\dfrac{2}{1}=2:1$$


 
710639_8380b9fba90e4d1589dc494c5268455f.jpg

Factorization of the following.
$$\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{3}{4}$$


Geetha earns Rs.$$12000$$ a month, out of which she saves Rs.$$3000$$. Find the ratio of her expenditure to her income.

Money earned,$$ E=\text{Rs.} 12000$$
Money saved, $$S=\text{Rs.} 3000$$
Expenditure, $$E_x=E-S$$
                             $$=12000-3000$$
                             $$=\text{Rs.}\ 9000$$
Ratio$$=E_x:E$$
         $$=9000:12000$$ = $$\dfrac{{9000}}{{12000}}$$
         $$ = \dfrac{{3 \times 3000}}{{4 \times 3000}} = \dfrac{3}{4}\\ = 3:4$$



The steps for comparison of ratios is given in jumbled form as follows:
Compare the like fractions.
The fraction with greater numerator will be greater.
Convert the fractions to like fractions. (By equating denominator) 
Convert the given ratios in fractions.
Arrange these steps in the ascending order and state which will be the 4th step.

The steps for comparison of ratios is given in ascending order.
1. Convert the given ratios in fractions.
2. Convert the fractions to like fractions. (By equating denominator) 
3. Compare the like fractions.
4. The fraction with greater numerator will be greater.


The steps for comparison of ratios is given in jumbled form as follows:
Compare the like fractions.
The fraction with greater numerator will be greater.
Convert the fractions to like fractions. (By equating denominator) 
Convert the given ratios to fractions.
Arrange these steps in the ascending order and state which will be the 1st first.

The steps for comparison of ratios are given in ascending order.
1. Convert the given ratios to fractions.
2. Convert the fractions to like fractions.    ...(By equating denominator) 
3. Compare the like fractions.
4. The fraction with greater numerator will be greater.


How many $$cm$$ in $$28\ m$$?  $$(1\ m=100\ cm)$$

$$ \displaystyle 28 \,m = 28 \times 100 = 2800 \,cm. $$ as $$1\ m=100\ cm$$


What is the ratio of 2 kg. 700 g. to 900 g.?

The required problem is
$$\begin{array}{l} 2\ kg\ 700\ gm=2700\ gm \\ Ratio=\dfrac { { 2700 } }{ { 900 } }  \\ =\dfrac 3 1 \end{array}\\$$
$$ = 3:1$$


Express the following as ratio in simplest form.
$$4$$ kg to $$250$$ gm.

4 kg to 250 gm.
$$\dfrac{4kg}{250g}= \dfrac{4000g}{250g}= \dfrac{16}{1}= 16:1$$
$$\therefore $$ ratio = 16:1.

1113952_1142749_ans_728d4d0031e04c66b9c6576a4e61a08b.jpg


Divide $$\dfrac {+3}{-6}$$ by $$\dfrac {-9}{12}$$

We have to divede $$\dfrac{+3}{-6}$$ by $$\dfrac{-9}{12}$$
$$\Rightarrow$$  $$\dfrac{\dfrac{3}{-6}}{\dfrac{-9}{12}}$$

$$\Rightarrow$$  $$\left(-\dfrac{3}{6}\right)\times \left(-\dfrac{12}{9}\right)$$

$$\Rightarrow$$  $$\dfrac{3\times 12}{6\times 9}$$

$$\Rightarrow$$  $$\dfrac{2}{3}$$


Find the ratio of Rs. $$5$$ to $$50$$ paise.

$${\textbf{Step  - 1: Converting rupees to paise}}$$

                  $${\text{We know that, 1 Rupees  =  100 paise}}$$

                  $$\therefore {\text{ 5 Rupees  =  }}\dfrac{{{\text{5 }} \times {\text{ 100}}}}{{\text{1}}}{\text{ paise}}$$

                  $$ \Rightarrow {\text{ 5 Rupees  =  500 paise}}$$

$${\textbf{Step  - 2: Finding ratio}}$$

                  $${\text{Ratio of 5 Rupees}}{\text{ to 50 paise ,}}$$

                  $$ \Rightarrow {\text{ }}\dfrac{{{\text{5 Rupees}}}}{{{\text{50 paise}}}}$$

                  $$ \Rightarrow {\text{ }}\dfrac{{{\text{500 paise}}}}{{{\text{50 paise}}}}$$

                  $$ \Rightarrow {\text{ 10}}$$

$${\textbf{Thus, the ratio of 50 paise to 5 Rupees is 10:1}}$$



$$28%$$ expressed as a ratio is

$$28=$$ $$\cfrac{28}{1}$$
$$\Rightarrow 28=28:1$$


Find the ratio of $$2\ m$$ to $$10\ m$$.

Consider the given question.

$$ 2\,\,to\,\,10 $$

$$ a:b=2:10 $$

$$ a:b=1:5 $$

Hence, this is the answer.



Find the reduced form of the ratio of the first quantity to the second quantity.
(1) $$Rs. 14,\ Rs. 12.40$$.

The ratio of $$Rs. 14$$ to $$Rs.12 .40 $$

$$ = 14 : 12.40$$

$$ = 1400 : 1240 $$

$$ = 35 : 31$$ (Ans)                        ($$\because$$ HCF of $$1400$$ and $$1240=40$$)


Find the ratio of $$30$$ days to $$36$$ hours.

$$30\text{days}:36\text{hours }\\30\times 24:36\\20:1$$


Sushma bought a saree from the mall. She paid $$Rs 2640$$ for it. When she came home she measured the length of the saree and found that the saree was half metre less in length then what the shopkeeper had told her. She found that she is at a loss of $$Rs 220$$ because of this. What is the length of the saree she brought home?

Let the length of the sari be $$x\,m$$
cost of $$\frac{1}{2}m$$ = Rs $$220$$
then cost of $$1m=2$$(Rs220)=Rs$$440$$
Cost of the sari $$=x$$(Rs440)=Rs$$440x$$
But given that the cost of sari $$=$$Rs$$2640$$
Implies that:
$$440x$$ = $$2640$$
Implies that
$$x = \dfrac{{2640}}{{440}}$$
$$x=6$$.


$$25p$$ is _________ part of a rupee

25P =x part of 100P (1 Rs = 100P)
$$ \therefore x = \dfrac{25}{100}$$
$$ \therefore  x = \dfrac{1}{4}$$


Divide $$104$$ pencils among three children in the ratio $$\dfrac{1}{2}\,,\dfrac{1}{3},\,\dfrac{1}{4}.$$

Given that there are $$104$$ pencils that are to be divided among three children in the ratio $$ \dfrac{1}{2} , \dfrac{1}{3}, \dfrac{1}{4}$$.


Sum of ratio $$=\dfrac{1}{2}+ \dfrac{1}{3}+ \dfrac{1}{4} = \dfrac{6+4+3}{12}$$

$$= \dfrac{13}{12}$$

Hence, first child will get,
$$104 \times \dfrac{\frac{1}{2}}{\frac{13}{12}}$$

$$=\dfrac{104 \times 6}{13}$$

$$\Rightarrow \dfrac{624}{13}=48$$

Second child will get,
$$104 \times \dfrac{\frac{1}{3}}{\frac{13}{12}}$$

$$=\dfrac{104 \times 4}{13}$$

$$=\dfrac{416}{13}=32$$

And third child will get,
$$104 \times \dfrac{\frac{1}{4}}{\frac{13}{12}}$$

$$=\dfrac{104 \times 3}{13}$$

$$=\dfrac{312}{13}=24$$

Hence, on dividing $$104$$ pencils among three children in the ratio $$ \dfrac{1}{2} , \dfrac{1}{3}, \dfrac{1}{4}$$, they will get, $$48,\ 32$$ and $$24$$ pencils respectively.


Express the following ratio in the simplest form:
One century to one decade.

One century to one dacade = $${100}:{10}=10:1$$



Find $$x$$ in the following:
$$x:5=10:50$$

According to question,

$$x:5=10:50$$

$$\implies x:5=1:5$$

Hence , $$x=1$$


Find $$x$$ in the following:
$$12:30=8:x$$

According to question,

$$12:30=8:x$$

$$\implies 2:5=8:x$$

$$\implies 8:20=8:x$$

Hence, $$x=20$$


Share Rs. $$117$$ between Mr. Kohli, Mr.Dubey and Mr.Shukla in the ratio $$2:4:7$$

$$2x,4x,7x \rightarrow $$ shares
$$2x+4x+7x=117$$
$$13x=117$$
$$x=9$$
Mr. Kohli $$= 2x= 18 Rs.$$
Mr. Dubey $$= 4x= 36 Rs.$$
Mr. Shukla $$= 7x = 63 Rs.$$


Divide $$1$$ hour in the ratio $$2:3$$

We have to divide $$1$$ hour in the ratio $$2  : 3 $$ 
Let the parts be $$2x,3x$$.
$$\Rightarrow 2x+3x=1$$ hour
$$\Rightarrow 5x=60$$ min
$$\Rightarrow x= 12$$ min
$$\therefore$$ The parts are $$2x=24$$ min and  $$3x=36$$ min.


Divide Rs. $$225$$ between Kanchan and Riya in the ratio $$7:8$$.

Let Kachan's part be 7x and Riya's part be 8x
As per question 
$$7x+8x=225$$
$$15x=225$$
$$x=15$$
$$7x=7\times15=105$$
$$8x=8\times15=120$$
Therefore, Kanchan's part is Rs.105 and Riya's part is Rs.120.


Express the following situations using ratios.
$$9$$ rose plants to $$1$$ sunflower plant

Ratio between $$9$$ rose plants to $$1$$ sunflower plant $$= 9 : 1$$


Express the following situations using ratios.
$$8$$ coins to $$3$$ coins

Ratio between $$8$$ coins to $$3$$ coins $$= 8 : 3 $$


Two numbers are in the ratio $$7 : 2$$. If their sum is $$54$$, find the numbers.

Given that,

Ratio of two numbers $$ = 7 : 2 $$

Let, the numbers are $$7x, 2x $$ 

$$\because 7x + 2x = 54$$

$$ 9x = 54 $$

$$ x = \dfrac{54}{9} = 6 $$

$$7x=7\times6=42$$


$$2x=2\times6=12$$

$$\therefore $$ The numbers are $$42, 12 $$ respectively.


Find $$a : c$$ when 
$$(1) a : b = 2 : 5$$ and $$b : c = 4 : 7$$ 

$$(2) a : b = 1 \dfrac { 1 } { 2 } : 1 \dfrac { 2 } { 3 }$$ and $$b : c = 1 \dfrac { 2 } { 3 } : 5$$ .

Given:
(1) $$\dfrac{a}{b} = \dfrac{2}{5}, \dfrac{b}{c} = \dfrac{4}{7}$$

$$\dfrac{a}{c} = \dfrac{a}{b} \times \dfrac{b}{c} = \dfrac{2}{5}\times \dfrac{4}{7} = \dfrac{8}{35}$$

$$a:c=8:35$$

(2) $$\dfrac{a}{b} = \dfrac{1\dfrac{1}{2}}{1\dfrac{2}{3}},\quad  \dfrac{b}{c} = \dfrac{1\dfrac{2}{3}}{5}$$

     $$\dfrac{a}{b} = \dfrac{\dfrac{3}{2}}{\dfrac{5}{3}}, \quad \dfrac{b}{c} = \dfrac{\dfrac{5}{3}}{5}$$

     $$\dfrac{a}{b} = \dfrac{9}{10}, \quad \dfrac{b}{c} = \dfrac{1}{3}$$

$$\dfrac{a}{c} = \dfrac{a}{b} \times \dfrac{b}{c} = \dfrac{9}{10} \times \dfrac{1}{3} = \dfrac{3}{10}$$

$$a:c=3:10$$


Express the following situations using ratios.
$$6$$ green balls to $$7$$ red balls

Ratio between $$6$$ green balls to $$7$$ red balls $$= 6 : 7 $$


In a camp $$95kg$$ food is sufficient from $$5$$ percent. Find the quantity of food required for $$23$$ persons.

For $$5$$ persons, food required $$=95$$ kg

For $$1$$ person, food required $$=\dfrac{95}{5}=19$$ kg

For $$23$$ persons, food required$$=19\times 23$$
                                                   $$=437$$ kg.


There are $$2500$$ student in a school. Out of them $$1100$$ are girls and the rest  are boys. Find the ratio of :
number of girls to number of students.

Total no. of students$$=2500$$
No. of girls$$=1100$$
No. of boys$$=2500-1100=1400$$
ratio of girls to total no. of students $$=\cfrac { 1100 }{ 2500 } =\cfrac { 11 }{ 25 } $$


There are $$2500$$ student in a school. Out of them $$1100$$ are girls and the rest  are boys. Find the ratio of :
number of boys to number of girls.

Total strength$$=2500$$
No. of girls$$=1100$$
No. of boys$$=2500-1100=1400$$
Ratio of boys to girls$$=\cfrac { 1400 }{ 1100 } =\cfrac { 14 }{ 11 } $$


Which ratio is greater?
$$7 : 8$$ or $$2 : 3$$.

In two or more ratios, wherever the numerator and denominator differs by one, 

the ratio with higher numerator is always the greater one.

$$7:8=\dfrac{7}{8}=0.875$$

$$2:3=\dfrac{2}{3}=0.67$$

$$\therefore \dfrac{7}{8}>\dfrac{2}{3}$$


Following ratios as fractions and write them in decimal
$$15:10$$

$$\dfrac { 15 }{ 10 } =\dfrac { 5\times 3 }{ 5\times 2 } =\dfrac { 3 }{ 2 } $$
and $$\dfrac { 15 }{ 10 } $$ in decimal form is $$1.5$$.


Which ratio is greater?
$$11 : 21$$ or $$19 : 28$$.

$$11:21=\dfrac{11}{21}=0.5238$$

$$19:28=\dfrac{19}{28}=0.6785$$

$$\therefore \dfrac{19}{28}>\dfrac{11}{21}$$


A pencil box weighs $$225g$$. How much do eight such pencil boxes weigh?

Weight of 1 pencil box $$=225\ g$$
Weight of 8 pencil box $$=8\times225=1800\ g$$
Hence weight of 8 such pencil boxes is $$1800$$ g.


If $$7kg$$ onions cost $$140$$ rupees, how much most we pay for $$12kg$$ onions.

Cost of 7 kg onions $$=140\ rupees$$
Cost of 1 kg onions $$=\dfrac{140}{7}=20\ rupees$$
then,
Cost of 12 kg onions $$=20\times12=240\ rupees$$


Thomas and Manoj started a business together. Thomas invested RS. $$60000$$ and Manoj Rs. $$40000$$
Write the ratio of the amounts in the smallest form which Thomas and Manoj invested.

Ratio will be given as 
$$Ratio =\dfrac{60000}{40000}=\dfrac{6}{4}=\dfrac{3}{2}$$
So 3:2 is the ratio of amount in the smallest form.


The length and breadth of a rectangle are $$25$$ cm and $$15$$ cm.
Find the ratio of its length to its breadth.

Ratio of length to breadth $$=\cfrac{25}{15}=\cfrac{5}{3}=5:3$$



Find the radio of :
(d) 30 days oto 36 hours 

$$30$$ days to $$36$$ hrs

$$\Rightarrow \dfrac{30\times 24 hrs}{36 hrs}$$

$$\Rightarrow \dfrac{5}{6}\times 24$$

$$\Rightarrow \dfrac{20}{1}$$.


$$21{m^2} \div 7m$$

$$21m^2\div 7m$$
$$=\dfrac{21m^2}{7m}=3m$$.


Find the ratio of :
(b) 15 kg to 210 g

$$15$$ kg $$=15\times 1000=15000$$g
 So ratio will be $$\dfrac{15000}{210}=\dfrac{500}{7}$$ 


Find the ratio of :
(a)INR 5 to 50 paise 

$$INR\,\,5=500\,\,paise$$

$$\dfrac{INR \,5}{50 \,paise}=\dfrac{500}{50}=\dfrac{10}{1}$$


Find the ratio of :
(c)9m to 27 cm 

$$9m$$ to $$27cm$$

$$\Rightarrow \dfrac{9m}{27cm}$$

$$\Rightarrow \dfrac{9\times 100cm}{27cm}$$

$$\Rightarrow \dfrac{100}{3}$$.


Divide $$84$$ into two parts such that $$\left(\dfrac{4}{17}\right)^{th}$$ of one part is equal to other part. find the product of both the parts.

Given, $$84$$ is divided into two parts
Let one part be $$'a'$$
Then other part$$=\dfrac{4}{17}\times a$$ [given]
Sum of part$$=84$$
$$a+\dfrac{4}{17}a=84$$
$$\dfrac{21a}{17}=84$$
$$a=68$$
Product of both parts$$=a\times \left(\dfrac{4}{17}a\right)$$
$$=\dfrac{4}{17}\times a^2\\$$
$$=\dfrac{4}{17}\times 68\times 68\\$$
$$=1088$$.


Compare which ratio is greater? 

$$7 : 16$$ or $$13 : 24$$

$$7:16=\dfrac{7}{16}$$

           $$=0.4375$$

Now, $$13:24=\dfrac{13}{24}$$

                       $$=0.5416$$

$$\Rightarrow 0.5416>0.4375$$

$$\therefore\ \  \dfrac{13}{24}>\dfrac{7}{16}$$


Find the ratio of the following :
$$55\ \text{kms}$$ to $$121\ \text{kms}$$

$$55$$ kms to $$121$$ kms
$$\cfrac{55}{121}=5:11$$


Solve
$$3 : x : 18 : 6$$


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In the following figure, express the ratio of shaded parts to unshaded parts in the simplest term.
1303428_d5d4bdbeac4a49018dacb34186a5e942.png

In the given figure 
Total no. of column $$=8$$
No. is column shaded $$=4$$
$${\mathop{\rm Re}\nolimits} quired\,\,ratio = \frac{{No.\,of\,shaded\,column}}{{Total\,no.\,of\,column}}$$
$$ = \frac{4}{{8}} = \frac{1}{2}$$
So, $${\mathop{\rm Re}\nolimits} quired\,\,ratio = 1:2$$
Which is the required answer.

1203537_1303428_ans_f8ba16bfd1ca4f02bae36ec0b0fceb31.JPG


Identify the number in the box:
(i) $$\dfrac{16}{24}=\dfrac{4}{\square }$$

(ii) $$\dfrac{28}{36}=\dfrac{-7}{\square }$$

(iii)$$\dfrac{18}{54}=\dfrac{\square}{6 }$$

(iv) $$\dfrac{-21}{36}=\dfrac{\square}{4}$$

(i) $$\dfrac{16}{24}=\dfrac{4}{6}$$ $$\Rightarrow \dfrac{16\div 4}{24\div 4}=\dfrac{4}{6}$$
(ii) $$\dfrac{28}{36}=-\dfrac{7}{9}$$ $$\Rightarrow \dfrac{28\div(-4)}{36\div (-4)}=\dfrac{-7}{9}$$
(iii) $$\dfrac{18}{54}=\dfrac{2}{6}$$ $$\Rightarrow \dfrac{18\div 9}{54\div 9}=\dfrac{2}{6}$$
(iv) $$\dfrac{-21}{36}=\dfrac{-7}{4}$$ $$\Rightarrow \dfrac{-24\div 9}{36\div 9}=\dfrac{\dfrac{-7}{3}}{4}$$.

1164043_1282527_ans_f4ff21d9a1fc42c7aa6ba93e063156e2.jpg


The value of $$x$$ in $$x:5::2:10$$

To check this, we need to know whether$$\displaystyle \frac{x}{5}= \frac{2}{10}$$

$$\displaystyle \frac{x}{5}= \frac{2}{5\times2}$$
$$\displaystyle \frac{x}{1}= \frac{1}{1}$$
Therefore, $$ x = 1$$


Find the ratio of the following 
$$20\ cm$$ to $$5\ m$$

Converting into same units.
We know that,
$$1m=100cm$$
$$ \therefore$$ $$5m=500cm$$
Thus the required ratio is,
$$20cm:500cm$$
$$1cm:25cm$$
Then,
we get the required ratio as $$1:25$$


The simplest from of $$3:15$$ is                      


1337990_1271318_ans_171d3f295f464fae983e3740c40676c8.PNG


In the following figure, express the ratio of shaded parts to unshaded parts in the simplest term.
1303425_f12be0d109854c6ab9fcf1ceae66d3b7.png

In the given figure 
Total no. of column $$=16$$
No. is column shaded $$=4$$
$${\mathop{\rm Re}\nolimits} quired\,\,ratio = \frac{{No.\,of\,shaded\,column}}{{Total\,no.\,of\,column}}$$
$$ = \frac{4}{{16}} = \frac{1}{4}$$
So, $${\mathop{\rm Re}\nolimits} quired\,\,ratio = 1:4$$
Which is the required answer.

1203520_1303425_ans_99dc80b29b634290b9e87ea3b1a9c8e7.JPG


Find the ratio of the following 
$$360\ gm$$ to $$18\ kg$$

We have,
$$ = 360gm:18kg$$
$$ = 360gm/18kg$$
We know that,
$$1kg=1000gm$$ Then,
$$18kg=18000gm$$
$$ = 360gm/18000kg$$
$$=1gm:50gm$$
Then,
We get ratio $$1:50$$


In the following figure, express the ratio of shaded parts to unshaded parts in the simplest term.
1303426_f0490050bc514021ab46acd00b351877.png

In the given figure,
Total no. of columns $$=9$$
Number of columns shaded $$=3$$

$$\text{Required ratio}= \dfrac{\text{Number of shaded columns}}{\text{Total number of columns}}$$
$$ = \dfrac{3}{{9}}$$
$$ = \dfrac{1}{3}$$

1203528_1303426_ans_7cdc1f9644504c0aa93449757d1c7760.png


Write the following ratios in ascending order $$(13:10), (24:25), (16:20)$$

$$\dfrac{13}{10}$$;           $$\dfrac{24}{25}$$;           $$\dfrac{16}{20}$$
$$(> 1)$$       $$(< 1)$$         $$(<< 1)$$
$$\therefore$$ comparing these:-
$$\dfrac{16}{20}< \dfrac{24}{25} < \dfrac{13}{10}$$.


Express the following ratios in the simplest form :
$$80 : 90$$
$$200 : 500$$
$$225 : 500$$
$$121 : 1331$$

$$\left( A \right)\,80:90 \Rightarrow 8:9$$
$$\left( B \right)\,200:500 \Rightarrow 2:5$$
$$\left( C \right)\,225:500 \Rightarrow 9:20$$
$$\left( D \right)\,121:1331 \Rightarrow 1:11$$


Find the ratio of the following
$$25$$ to $$150$$

$$25 : 150$$
$$25 = {5}^{2}$$
$$150 = {5}^{2} \times 6$$
Therefore,
$$25 : 150 = \cfrac{25}{150} = \cfrac{{5}^{2} \times 1}{{5}^{2} \times 6} = \cfrac{1}{6}$$
$$\therefore 25 : 150 = 1 : 6$$


Write the ratio for each of the following :
$$1.5\ km$$ to $$500\ m$$

We know that $$1 \; km = 1000 \; m$$
$$\Rightarrow 1.5 \; km = 1.5 \times 1000 = 1500 \; m$$
Therefore,
$$\cfrac{1.5 \; km}{500 \; m} = \cfrac{1500 \; m}{500 \; m} = \cfrac{500 \times 3}{500 \times 1} = \cfrac{3}{1} = 3 : 1$$


Write the ratio for each of the following :
$$1$$ year to $$4$$ months

No. of months in $$1$$ year $$= 12$$
Therefore,
$$\cfrac{1 \text{ year}}{4 \text{ months}}$$
$$= \cfrac{12 \text{ months}}{4 \text{ months}}$$
$$ = \cfrac{4 \times 3}{4 \times 1}$$
$$= \cfrac{3}{1} = 3 : 1$$


Find the ratio of the following
$$72$$ to $$36$$

$$72 : 36$$
$$72 = {6}^{2} \times 2$$
$$36 = {6}^{2}$$
Therefore,
$$72 : 36 = \cfrac{72}{36} = \cfrac{{6}^{2} \times 2}{{6}^{2} \times 1} = \cfrac{2}{1}$$
$$\therefore 72 : 36 = 2 : 1$$


In Mathematics Sunitha scored $$75$$ marks and Vinitha scored $$25$$ marks. Find the ratio of Vinitha's marks to Sunita's marks

Given:-
Sunitha's marks $$\left( {m}_{1} \right) = 75$$
Vinitha's marks $$\left( {m}_{2} \right) = 25$$
Therefore,
$$\cfrac{{m}_{2}}{{m}_{1}} = \cfrac{25}{75} = \cfrac{25 \times 1}{25 \times 3} = \cfrac{1}{3}$$
$$\Rightarrow {m}_{2} : {m}_{1} = 1 : 3$$


$$\dfrac{{{{\text{R}}_{\text{1}}}^{\text{2}}}}{{{{\text{R}}_{\text{2}}}^{\text{2}}}}\;{\text{ = }}\;\dfrac{{\text{9}}}{{{\text{576}}}}, \dfrac{{{{\text{R}}_{\text{1}}}}}{{{{\text{R}}_{\text{2}}}}}\;{\text{ = }}\;{\text{?}}$$

We have,
$$\dfrac{R^2_1}{R^2_2}=\dfrac{9}{576}$$

$$\dfrac{R^2_1}{R^2_2}=\dfrac{3^2}{24^2}$$

$$\dfrac{R_1}{R_2}=\dfrac{3}{24}$$

Hence, this is the answer.


Write the ratio for each of the following :
$$60$$ paise to $$Rs.\ 2$$

$$Rs. 1 = 100 \text{ paise}$$
$$Rs. 2 = 2 \times 100 = 200 \text{ paise}$$
Therefore,
$$\cfrac{60 \text{ paise}}{2 \text{ Rs.}} = \cfrac{60 \text{ paise}}{200 \text{ paise}} = \cfrac{3 \times 20}{10 \times 20} = \cfrac{3}{10}$$


Express the following 
$$125:175$$

$$\cfrac{125}{175} = \cfrac{25 \times 5}{25 \times 7} = \cfrac{5}{7}$$
$$\therefore 125 : 175 = 5 : 7$$


If $$\dfrac{a}{2b}=\dfrac{7}{4}$$ find the value of ratio $$\dfrac{a+2b}{a-2b}$$.

$$\cfrac{a}{2b} = \cfrac{7}{4} \quad \left( \text{Given} \right)\\$$
$$\cfrac{a+2b}{a-2b}\\$$
$$= \cfrac{\cfrac{a}{2b} + 1}{\cfrac{a}{2b} - 1}\\$$
$$= \cfrac{\cfrac{7}{4} + 1}{\cfrac{7}{4} - 1}\\$$
$$= \cfrac{7 + 4}{7 - 4} = \cfrac{11}{3}$$


Divide $$1500$$ among $$A,\,B,\,C$$ in the ratio $$3:5:2$$


We have, sum of the terms of the ratio$$=3+5+2=10$$

$$A$$'s share$$=\dfrac{3}{10}\times 1500=450$$

$$B$$'s share$$=\dfrac{5}{10}\times 1500=750$$

$$C$$'s share$$=\dfrac{2}{10}\times 1500=300$$


Find the ratio of the following 
$$24$$ to $$84$$

24 to 84
$$=\frac{24}{84} $$
$$=\frac{2}{7}$$
$$=2:7$$


Solve:
If $$a:b = 5:3,$$ then find $$(5a + 8b):(6a - 7b).$$

We have,
$$\dfrac{a}{b}=\dfrac{5}{3}$$

Since,
$$\dfrac{5a+8b}{6a-7b}$$

Therefore,
$$\Rightarrow \dfrac{5\dfrac{a}{b}+8}{6\dfrac{a}{b}-7}$$

$$\Rightarrow \dfrac{5\dfrac{5}{3}+8}{6\dfrac{5}{3}-7}$$

$$\Rightarrow \dfrac{\dfrac{25}{3}+8}{\dfrac{30}{3}-7}$$

$$\Rightarrow \dfrac{25+24}{30-21}$$

$$\Rightarrow \dfrac{49}{9}$$

Hence, this is the answer.


Find the ratio of the following 
$$30$$ minutes to $$1.5$$ hours

Ratio of 30 minutes to 1:5 hours
$$=\frac{30}{1.5\times 60}=\frac{1}{3}$$
ratio = 1:3

1212948_1365794_ans_156693aa4cc3490b819e007609eda456.JPG


Find the ratio of the first quantity to second quantity
(i) Rs $$700$$, Rs $$308$$
(ii) $$5$$ liter, $$2500$$ ml

$$(i) 700:308 = 25:11$$

$$(ii) 5000:2500 = 2:1$$


If $$\left(4x+5\right):\left(3x+11\right)=13:17,$$ find the value of $$x$$

Given :$$\left(4x+5\right):\left(3x+11\right)=13:17$$

$$\Rightarrow \dfrac{4x+5}{3x+11}=\dfrac{13}{17}$$

$$\Rightarrow \,17\left(4x+5\right)=13\left(3x+11\right)$$

$$\Rightarrow \,68x+85=39x+143$$

$$\Rightarrow \,68x-39x=143-85$$

$$\Rightarrow \,29x=58$$

$$\Rightarrow x=\dfrac{58}{29}=2$$

$$\therefore x=2$$


Express the ratio $$45:108$$ in its simplest form?

Factors of $$45=3\times 3\times 5$$

Factors of $$108=2\times 2\times 3\times 3\times 3$$

So, H.C.F of $$45$$ and $$108$$ is $$3\times 3=9$$

$$\therefore 45:108=\dfrac{45}{108}=\dfrac{45\div 9}{108\div 9}=\dfrac{5}{12}=5:12$$


Find smallest fraction
$$(1)$$ $$50$$ girls $$:80$$ girls
$$(ii)$$ $$55$$ girls $$:88$$ girls

$$Part\ (1)$$
$$50$$ girls $$:80$$ girls

$$\Rightarrow \dfrac{50}{80}$$

$$\Rightarrow \dfrac{5\times 10}{8\times 10}$$

$$\Rightarrow \dfrac{5}{8}$$

$$Part\ (ii)$$ 
$$55$$ girls $$:88$$ girls

$$\Rightarrow \dfrac{55}{88}$$

$$\Rightarrow \dfrac{5\times 11}{8\times 11}$$

$$\Rightarrow \dfrac{5}{8}$$

Hence, this is the answer.


Snehal has a red ribbon that is $$80$$ cm long and a blue ribbon, $$220 \,\,m$$ long. What is the ratio of the length of the red ribbon to that of the blue ribbon?

Red Ribbon : Blue Ribbon = $$80 : 220 = 4:11$$


express the following as ratios
$$(1) 6\ kg\ to\ 10kg$$
$$(2)50\ km\ to\ 70km$$
$$(3)250ml\ to\ 50ml$$

$$1) 6\ kg:10\ kg =\dfrac{6}{10}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3:5$$

$$2) 50\ km:70\ kn =\dfrac{50}{70}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5:7$$

$$3) 250\ ml:50\ ml =\dfrac{250}{50}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5:1$$


Find the ratio of the following
$$30\ minutes$$ to $$45\ minutes$$

$$30$$minutes$$:45$$minutes
$$=30:45$$
$$=2:3$$
The ratio is $$2:3$$


Find the ratio of the following
$$81$$ to $$108$$

$$81:108$$
$$=9\times 9:12\times 9$$
$$=9:12$$
$$=3\times 3:4\times 3$$
$$=3:4$$


Find the ratio of $$81:108$$

$$81:108$$
$$=9\times 9:12\times 9$$
$$=9:12$$
$$=3\times 3:3\times 4$$
$$=3:4$$
$$\therefore$$ the ratio is $$3:4$$


Are the rations 45g:60g and 36kg:48kg in proportion?

$$45g:60g$$=$$3g:4g$$
And $$36kg:48kg$$=$$36000g:48000g$$$$=2g:4g$$
Since both the rations are not equivalent.
Therefore,$$45g:60g$$ and $$36kg:48kg$$ are not in proportion.


Find each of the following ratios in the simplest form.
$$4\ kg\ to\ 750\ g$$ 

$$4$$kg$$:750$$gms
$$=4\times 1000$$gms$$:750$$gms since $$1$$kg$$=1000$$gms
$$=4000:750$$
$$=400:75$$
$$=25\times 8:25\times 3$$
$$=8:3$$
$$\therefore$$ the simplest form of $$4$$kg$$:750$$gms is $$8:3$$


Find each of the following ratios in the simplest form.
$$1.8\ kg\ to\ 6\ kg$$ 

$$1.8$$kg$$:6$$kg
$$=1.8:6$$
$$=1.8\times 10:6\times 10$$
$$=18:60$$
$$=6\times 3:6\times 10$$
$$=3:10$$
$$\therefore$$ the simplest form of $$1.8$$kg$$:6$$kg is $$3:10$$


Find the ratio of the following
$$30\ minutes$$ to $$1.5\ minutes$$

$$30$$minutes$$:1.5$$minutes
$$=30:1.5$$
$$=30\times 10:1.5\times 10$$
$$=300:15$$
$$=20:1$$


Let's suppose you earn $$200$$ a week.Your house rent is $$40$$ weekly.What is the ratio of your rent to your income?

Ratio of rent to the income$$=\dfrac{Rent}{Income}=\dfrac{40}{200}=\dfrac{40\times 1}{40\times 5}=\dfrac{1}{5}$$
or the ratio is $$1:5$$


Find the ratio of the following
$$33\ km$$ to $$121\ km$$

$$33$$km$$:121$$km
$$=33:121$$
$$=3\times 11:11\times 11$$
$$=3:11$$
$$\therefore$$ the ratio is $$3:11$$


Find the ratio of $$20\ \text{cm}$$ to $$3\ \text{m}$$. 

$$20\ \text{cm}:3\ \text{m}=\dfrac{20\ \text{cm}}{3\ \text{m}}=\dfrac{20}{300}=\dfrac{1}{15}$$  $$[\because 1\text{ m}=100\text{ cm}]$$


In a school, there were $$73$$ holidays in one year. What is the ratio of the number of holidays to the number of days in one year?

The answer is
$$\dfrac{365}{73}$$  =  $$5$$ : $$1 $$


Rs 120 is to be divided between Hari and Gopi in the ratio 5 :How much does each get ?

Given
Total amount = Rs $$120$$
Amount divided between Hari and Gopi in the ratio $$ = 5: 3 $$
Hence,
The sum of ratio is as follows:
Sum of ratio $$ = 5 + 3 $$
$$ = 8 $$
Hence, Hari’s and Gopi’s share will be as follows:
Hari’s share $$ = (120 \times  5) / 8 $$
= Rs $$75$$
Gopi’s share $$ = (120 \times  3) / 8 $$
= Rs $$45$$
Therefore, Hari get Rs $$75$$ and Gopi get Rs $$45$$


Find the ratio of Rs. 17 to 34 paise.

As we know that
$$1\ Rs=100\ paise$$
$$\mathrm{Rs. } 17=17\times 100\ \mathrm{paise}=1700\ \mathrm{paise}$$
Ratio of $$Rs.17$$ to $$34$$ paise is $$1700\ \mathrm{paise}:34\ \mathrm{paise}=\dfrac{1700}{34}=50$$


Find the ratio of the following
$$40\ cm$$ to $$1.5\ m$$

$$40$$cm$$:1.5$$m
$$=40$$cm$$:1.5\times 100$$cm since $$1$$m$$=100$$cm
$$=40:150$$
$$=4:15$$
$$\therefore$$ the ratio is $$4:15$$ 


Convert  1223 mg into grams.


1313483_1487387_ans_ab81b8d3850342a0a62dd9e64d9bc2a8.JPG


Find the ratio in the simplest form.
$$35$$ minutes to $$45$$ minutes.

$$35$$ minutes: $$45$$ minutes = $$\dfrac{35}{45}$$

                                         = $$\dfrac{7}{9}$$

                                         =$$7:9$$


Express the following ratios in simplest form:
$$81:108$$

$$81:108=\dfrac{81}{108}=\dfrac{9}{12}=\dfrac{3}{4}=3:4$$


Express the following ratios in simplest form:
$$40:20$$

$$40:20=\dfrac{40}{20}=\dfrac{2}{1}=2:1$$


$$\cfrac{7}{8}=\cfrac{x}{176}$$
Find x

$$\cfrac{7}{8}=\cfrac{x}{176}$$
$$x=\dfrac{7 \times 176}{8}$$
$$x=154$$


Without using your calculator, work out $$2\dfrac {7}{9}\div \dfrac {5}{6}$$.
Give your answer as a fraction in its lowest terms.
You must show each step of your working. 

$$2\dfrac 79\div \dfrac 56\\\dfrac {25}9\times \dfrac 65\\5\times \dfrac 23=\dfrac {10}3$$


Express the following ratios in simplest form:
$$20:40$$

$$20:40=\dfrac{20}{40}=\dfrac{1}{2}=1:2$$


Check whether the given ratios are equal, i.e., they are in proportion. If yes, then write them in the proper form.
Rs. $$10$$ to Rs. $$15$$ and $$4$$ to $$6$$.

$$10 \ to \ 15 \ and \ 4 \ to \ 6$$
$$\dfrac{10}{15} and \dfrac{4}{6}$$
$$\dfrac{2}{3} and \dfrac{2}{3}$$
Given ratios are equal



Find the ratio of number of desks and chairs in your classroom.

In my class room
No. of desks are $$30$$
No. of chairs are also $$30$$
Ratio = $$30$$ / $$30$$
=$$1$$


Check whether the given ratios are equal, i.e., they are in proportion. If yes, then write them in the proper form.
$$4:12$$ and $$9:27$$.

$$4:12 \ and \ 9:27$$
$$\dfrac{4}{12} and \dfrac{9}{27}$$

$$\dfrac{1}{3} and \dfrac{1}{3}$$ 


Fill in the blanks:
In a ratio, the first term is also called...... and second term is also called ..... .

In a ratio, the first term is also called antecedent and second term is also called consequent.


Find the ratio of each of the following in simplest form:
$$2.1\ m$$ to $$1.2\ m$$

$$2.1\ m: 1.2\ m=\dfrac{2.1}{1.2}=\dfrac{21}{12}\times \dfrac{10}{10}=\dfrac 74=7:4$$


From the given, find the ratio of
Number of circle to the number of remaining figures inside the rectangle.
1787099_41ec789e04364d4493f26b8c7a7b4496.png

Number of triangle $$=3$$
Number of rectangle $$=2$$
Number of circle $$=2$$
$$2:5$$


Find the ratio of each of the following in simplest form:
$$15\ mm$$ to $$2\ cm$$

$$15\ mm: 20\ cm$$
$$10\ mm: 1\ cm$$
$$15\ mm=\dfrac {15}{10}\ cm$$
$$\dfrac{\dfrac{15}{10}}{\dfrac 21}=\dfrac {15}{10}\times \dfrac12=\dfrac 34=3:4$$


From the given, find the ratio of
Number of squares to the number of all the figure inside the rectangle.
1787094_6bf668d635be4439bd1f508773adc7bc.png

Number of triangle $$=3$$
Number of square $$=2$$
Number of circle $$=2$$
$$2:7$$


From the given, find the ratio of
Number of triangles to the number of circle inside the rectangle.
1787065_bbb4b27335af47869dc9bdf531b86164.png

Number of triangle $$=3$$

Number of circle $$=2$$
ratio $$=3:2$$


Find the ratio of the following in simplest form:
$$3.5\ kg$$ to $$250\ gm$$

$$3.5\ Kg: 250\ gm$$ 

$$=3.5\times 1000\ gm: 250\ gm$$

$$=3500\ gm: 250\ gm$$

$$=\dfrac{3500}{250}$$

$$=\dfrac{14}{1}$$

$$=14:1$$

Hence, the ratio is $$14:1$$


Express the following ratios in simplest form:
$$98:63$$

$$98:63=\dfrac{98}{63}=\dfrac{14}{9}=14:9$$


Find the ratio of each of the following in simplest form:
$$91\ cm$$ to $$1.04\ m$$

$$91\ cm: 1.04\ cm$$ or $$1.04\times 100$$ or $$104\ cm$$
$$91\ cm: 104\ cm=\dfrac{91}{104}=7:8$$


Find the ratio of each of the following in simplest form:
$$1\ \mathrm{minute}$$ to $$15\ \mathrm{seconds}$$

$$1$$ minute : $$15$$ seconds
$$60$$ seconds $$=1$$ minute 
$$15$$ seconds  $$ =\dfrac{15}{60}$$ min $$=\dfrac 14$$ min
The ratio becomes- $$1$$ min : $$\dfrac 14$$ min 
$$\dfrac{1}{\dfrac 14}=\dfrac{1\times 4}{1}=\dfrac 41=4:1$$


$$20m:70m =8:\_\_\_$$

Let,
$$ 20m:70m = 8: x$$

$$\Rightarrow \dfrac{20}{70} = \dfrac{8}{x}\\ \Rightarrow 20 \times x=8\times 70\qquad\text{ [ On cross multiplication]}\\ \Rightarrow x =\dfrac{8\times 70}{20}\\ \therefore x =28$$

Hence, $$20m:70m = 8:28$$


Fill in the blanks:
$$90\ cm:1.5\ m=$$ ........

$$90\ cm:1.5\ m=3.5$$


Fill in the blanks:
The simplest form of the ratio $$8:12$$ is .......

The simplest form of the ratio $$8:12$$ is $$2:3$$


Is this pair of ratios equal? why?
$$\dfrac{2}{3}, \dfrac{4}{6}$$

Given ratios are : $$\dfrac{2}{3}$$ and $$\dfrac{4}{6}$$

Here, $$\dfrac46=\dfrac{4\div2}{6\div2} = \dfrac{2}{3}$$
$$\Rightarrow \dfrac{2}{3}=\dfrac{4}{6}$$

Hence, the two ratios are equal.


A profit of Rs $$2, 500$$ is to be shared among three persons in the ratio $$6: 9: 10$$. How much does each person get?

Given 
Total profit $$ = Rs. 2,500 $$ 
Ratio $$ = 6 : 9 : 10 $$ 
Sum of ratio $$ = 6 + 9 + 10 $$ 
$$ = 25 $$ 
Share of first person $$ = \dfrac{6}{25} \times 2500 $$ 
$$ = 6 \times 100 $$ 
$$ = 600 $$ 
Share of second person $$ = \dfrac{9}{25} \times 2500 $$ 
$$ = 9 \times 100 $$ 
$$ = 900 $$ 
Share of third person $$ = \dfrac{10}{25} \times 2500 $$ 
$$ = 10 \times 100 $$ 
$$ = 1000 $$


How will Rs $$ 31500 $$ be shared between A, B and C; if A gets the double of what B gets, and B gets the double of what C gets?

Given 
Amount $$ = Rs. 31500 $$ 
Let the share of C $$ = 1 $$ 
Share of B = double of C 
$$ = 2 \times 1 $$ 
$$ = 2 $$ 
Share of A = double of B 
$$ = 2 \times 2 $$ 
$$ = 4 $$ 
Therefore , given ratio will be 
A : B : C $$ = 4 : 2 : 1 $$ 
Sum of ratios $$ = 4 + 2 + 1 $$ 
$$ = 7 $$ 
A's share $$ = \dfrac{4}{7} \times 31500 $$ 
$$ = 4 \times 4500 $$ 
$$ = Rs. 18000 $$ 
B's share $$ = \dfrac{2}{7} \times 31500 $$ 
$$ = 2 \times 4500 $$ 
$$ = Rs. 9000 $$ 
C's share $$ = \dfrac{1}{7} \times 31500 $$ 
$$ = 1 \times 4500 $$ 
$$ = Rs. 4500 $$


Divide Rs $$ 10,400 $$ among A , B and C in the ratio $$ \dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4} $$ 

Given 
Amount $$ = Rs\ 10400 $$ 
Amount to be divided into ratio $$ =  \dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4} $$  
The L.C.M of $$ 2 , 3 $$ and $$ 4 $$ is $$ 12 $$ 
Hence , given ratio will be as given below 
$$ \dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4} = \dfrac{1}{2} \times 12 : \dfrac{1}{3} \times 12 : \dfrac{1}{4} \times 12 $$ 
$$ = 6 : 4 : 3 $$ 
A's part $$ = \dfrac{6}{13} \times 10400 $$ 
We get 
$$ = 6 \times 800 $$ 
$$ = 4800 $$ 
B's part $$ = \dfrac{4}{13} \times 10400 $$ 
We get 
$$ = 4 \times 800 $$ 
$$ = 3200 $$ 
C's part $$ = \dfrac{3}{13} \times 10400 $$ 
We get 
$$ = 3 \times 800 $$ 
$$ = 2400 $$


Is the ratio of $$6\, g $$ and $$ 15 \,g $$ same as the ratio of $$ 36\,cm $$ and $$90\,cm $$ ? 

Ratio of $$ 6\,g $$ and $$ 15 \,g = 6 \,g : 15\, g $$ 
On calculating further, we get
$$ = \dfrac{6}{15} $$ 

$$ = \dfrac{2}{5} $$ 

$$ = 2: 5 $$
Now, the ratio of $$ 36 \,cm$$  and $$ 90\,cm = 36\,cm : 90\,cm $$ 
By calculating further, we get
$$ = \dfrac{36}{90} $$ 

$$ = \dfrac{18}{45}  $$ 

$$ = \dfrac{6}{15} $$ 

$$ = \dfrac{2}{5} $$ 

$$ = 2: 5 $$
Since both the ratios $$ = 2: 5 $$
Hence, the ratios are same in both the cases


If two ratios are equal, then they are in              


For example:
If a : b = c : d, then a, b, c, d are in proportion.
$$\textbf{If two ratios are equal, then they are in proportion.}$$


To find the ratio of two quantities, they must be expressed in                units.


For example:
To express the ratio of $$1$$ metre : $$200$$ cm, both the quantities have to expressed in same units.
Ratio of $$1$$ metre : $$200$$ cm $$= 100$$ cm : $$200$$ cm
                                                          $$= 1 : 2$$
$$\therefore \textbf{To find the ratio of two quantities, they must be expressed in same units}$$.


Divide $$ 81 $$ into three parts in the ratio $$ 2 : 3 : 4 $$

Given 
Number $$ = 81 $$ 
Ratio $$ = 2 : 3 : 4 $$ 
Hence , the sum of ratio is calculated as follows 
Sum of ratios $$ = 2 + 3 + 4 $$ 
$$ = 9 $$ 
First divide $$ = \dfrac{2}{9} \times 81 $$ 
$$ = 18 $$ 
Second divide $$ = \dfrac{3}{9} \times 81 $$ 
$$ = 27 $$ 
Third divide $$ = \dfrac{4}{9} \times 81 $$ 
$$ = 36 $$ 
Therefore , $$ 81 $$ can be divided into $$ 18 , 27 $$ and $$ 36 $$ in the ratio $$ 2 : 3 : 4 $$


Is the ratio of $$15\, kg $$ and $$ 35 \,kg $$ same as the ratio of $$6$$ years and $$14$$ years ?

Ratio of $$ 15 \,kg $$ and $$ 35 \,kg = 15 \,kg : 35\, kg $$ 
We get
$$ \dfrac{15\,kg}{35 \, kg} $$ 
On simplification, we get
$$ = \dfrac{3 \, kg}{7\, kg} $$ 
$$ = 3: 7$$ 
Now, the ratio of $$6$$ years and $$14$$ years = $$6$$ years : $$14$$ years
We get
$$ = \dfrac{6 {\text{years}}}{14 \, {\text{years}}} $$ 
On simplification, we get
$$ = \dfrac{3 {\text{years}}}{7 \, {\text{years}}} $$ 
$$ = 3: 7 $$
Since both the ratio $$ = 3: 7 $$
Hence, the ratios are same in both cases


 Mr. Gupta divides Rs $$ 81000$$ among his three children Ashok, Mohit and Geeta in such a way that Ashok gets four times what Mohit gets and Mohit gets $$2.5$$ times what Geeta gets. Find the share of each of them.

Given
Amount = Rs $$ 81000 $$ 
Let the share of Geeta $$ = 1 $$
Share of Mohit is $$2.5$$ times of Geeta
Hence, share of Mohit becomes $$ = 2.5 $$
Share of Ashok is $$4$$ times of Mohit
Hence, share of Ashok becomes $$ = 4 \times 2.5 $$
$$ = 10 $$ 
Ratio $$ = 1: 2.5: 10 $$ 
$$ = 1 \times 2 : 2.5 \times 2 : 10 \times 2 $$ 
$$ = 2 : 5 : 20 $$ 
Thus, sum of ratios $$ = 2 + 5 + 20 $$ 
$$ = 27 $$ 
Share of Geeta $$ = \dfrac{2}{27} \times 81000 $$ 
$$ = 2 \times 3000 $$ 
$$ = Rs. 6000 $$ 
Share of Mohit $$ = \dfrac{5}{27} \times 81000 $$ 
$$ = 5 \times 3000 $$ 
$$ = Rs. 15000 $$ 
Share of Ashok $$ = \dfrac{20}{27} \times 81000 $$ 
$$ = 20 \times 3000 $$ 
$$ = Rs. 60000 $$


Which ratio is smaller :
$$ \dfrac{9}{17} $$ or $$ \dfrac{8}{15} $$

Given 
$$ \dfrac{9}{17} $$ or $$ \dfrac{8}{15} $$
The above expression can be written as follows :
$$ \dfrac{9}{17} $$ or $$ \dfrac{8}{15} $$
$$ = 9 \times 15 $$ or $$ 8 \times 17 $$ 
$$ = 135 $$ or $$ 136 $$ 
We know that , $$135 $$ is smaller than $$ 136 $$ 
Therefore $$ \dfrac{9}{17} $$ is smaller


Which ratio is greater :
$$ \dfrac{3}{7} $$ or $$ \dfrac{6}{13}  $$

$$ \dfrac{3}{7} $$ or $$ \dfrac{6}{13}  $$
The above expression can be written as follows :
$$ \dfrac{3}{7} $$ or $$ \dfrac{6}{13}  $$
$$ = 3 \times 13 $$ or $$ 6 \times 7 $$ 
$$ = 39 $$ or $$ 42 $$ 
We know that $$ 42 $$ is greater than $$ 39 $$ 
Therefore $$ \dfrac{6}{13}  $$ is greater


Which ratio is smaller :
$$ \dfrac{7}{15} $$ or $$ \dfrac{15}{32} $$

Given 
The above expression can be written as follows :
$$ \dfrac{7}{15} $$ or $$ \dfrac{15}{32} $$ 
$$ = 7 \times 32 $$ or $$ 15 \times 15 $$ 
$$ = 224 $$ or $$ 225 $$ 
We know that , $$ 224 $$ is smaller than $$ 225 $$
Therefore $$ \dfrac{7}{15} $$ is smaller


Which ratio is greater :
$$ \dfrac{8}{15} $$ or $$ \dfrac{5}{9} $$

Given 
$$ \dfrac{8}{15} $$ or $$ \dfrac{5}{9} $$ 
The above expression can be written as follows :
$$ \dfrac{8}{15} $$ or $$ \dfrac{5}{9} $$ 
$$ = 8 \times 9 $$ or $$ 15 \times 5 $$
$$ = 72 $$ or $$ 75 $$ 
We know that $$ 75 $$ is greater than $$ 72 $$ 
Therefore $$ \dfrac{5}{9} $$ is greater


Divide Rs.1162 among A, B, C in the ratio $$35 : 28 : 20.$$

Sum of ratio terms $$= (35+28+20) = 83$$

A's share = Rs.$$\displaystyle \left ( 1162\times \frac{35}{83} \right )=Rs.490$$

B's share = Rs.$$\displaystyle \left ( 1162\times \frac{28}{83} \right )=Rs.392$$

C's share = Rs.$$\displaystyle \left ( 1162\times \frac{20}{83} \right )=Rs.280$$


Are the ratio $$1 : 2$$ and $$2 : 3$$ equivalent ?

To check this, we need to know whether$$\displaystyle \frac{1}{2}= \frac{2}{3}$$
We have, $$\displaystyle \frac{1}{2}=\frac{1\times 3}{2\times 3}$$$$\displaystyle =\frac{3}{6}$$
and $$\cfrac{2}{3}=\cfrac{2\times 2}{3\times 2}=\cfrac{4}{6}$$
We find that$$\displaystyle \frac{3}{6}< \frac{4}{6},$$ which means that $$\displaystyle \frac{1}{2}< \frac{2}{3}.$$
Therefore, The ratio $$1 : 2$$ is not equivalent to the ratio $$2 : 3 $$


Arrange the ratios $$ 7 : 20, 13 : 25, 17 : 30$$ and $$11 : 15$$ in decreasing order.

To arrange the ratios, we need to find their fractional values
$$7:20=\dfrac{7}{20}=0.35$$

$$13:25=\dfrac{13}{25}=0.52$$

$$17:30=\dfrac{17}{30}=0.57$$

$$11:15=\dfrac{11}{15}=0.73$$
Arranging these numbers in decreasing order we get
$$0.73, 0.57, 0.52, 0.35$$
$$11:15, 17:30, 13:25, 7:20$$


If the cost of a pair of scissor is $$Rs. 40$$ and cost of a pen is $$Rs. 10$$, then the ratio of cost of the pair of scissor and the cost of a pen is $$x:y$$, Find $$x+y$$.


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The speed of a Garden-Snail is $$50$$ meters per hour and that of the Cheetah is $$120$$ kilometers per hour. Find the ratio of the speeds.

The speed of snail is 50 meter per hour or 0.05 km per hour.

The ratio of the speed of snail and cheetah is,

$$120:0.05$$

$$2400:1$$

Therefore, for each 2400 meter covered by cheetah, the garden-snail would have covered 1 meter.



In the given figure, ABC is a triangle. Write all possible ratios by taking measures of sides pair wise.
565067_b4d17cbc2e9a4633aa09507be02434c3.png

$$AB:BC = 8cm:6cm = 4:3$$
$$BC:CA = 6cm:10cm = 3:5$$
$$CA:AB = 10cm:8cm = 5:4$$

1175586_565067_ans_8fa333c0c0de4ad290295a72ae5af65d.png


If $$y(3x-y):x(4x+y)=5:12$$, then find the ratio $$(x^2+y^2):(x+y)^2$$.

$$y(3x-y):x(4x+y)=5:12$$
$$\Rightarrow \dfrac{y(3x-y)}{x(4x+y)}=\dfrac{5}{12}$$
$$12y(3x-y)=5x(4x+y)$$
$$\Rightarrow 36xy-12y^{2}=20x^{2}+5xy$$
$$36xy-12y^{2}-20x^{2}-5xy=20x^{2}+5xy-20x^{2}-5xy$$
$$\Rightarrow -20x^{2}+31xy-12y^{2}=0$$
$$\Rightarrow 20x^{2}-16xy-15xy+12y^{2}=0$$
$$\Rightarrow 4x(5x-4y)-3y(5x-4y)=0$$
$$\Rightarrow (4x-3y)(5x-4y)=0$$
Then if $$4x-3y=0\Rightarrow 4x=3y\Rightarrow x=\dfrac{3}{4}y$$.................(1)
And $$5x-4y=0\Rightarrow 5x=4y\Rightarrow x=\dfrac{4}{5}y$$.......................(2)
Then If  $$x=\dfrac{3}{4}y$$
$$\therefore x^{2}+y^{2}=\left ( \dfrac{3}{4}y \right )^{2}+y^{2}=\dfrac{9y^{2}}{16}+y^{2}=\dfrac{9y^{2}+16y^{2}}{16}=\dfrac{25y^{2}}{16}$$.......................(4)
$$\therefore (x+y)^{2}=\left ( \dfrac{3}{4}y+y \right )^{2}=\left ( \dfrac{3y+4y}{4} \right )^{2}=\left ( \dfrac{7y}{4} \right )^{2}=\dfrac{49y^{2}}{16}$$..................(5)
If $$x=\dfrac{5}{4}y$$
$$\therefore x^{2}+y^{2}=\left ( \dfrac{5}{4}y \right )^{2}+y^{2}=\dfrac{25}{16}y^{2}+y^{2}=\dfrac{25y^{2}+16y^{2}}{16}=\dfrac{41}{16}y^{2}$$.................(6)
$$\therefore (x+y)^{2}=\left ( \dfrac{5}{4}y +y\right )^{2}=\left ( \dfrac{5y+4y}{4} \right )^{2}=\left ( \dfrac{9}{4}y \right )^{2}=\dfrac{81}{16}y^{2}$$.....................(7)
As per (4) and (5) 
$$x^{2}+y^{2}:(x+y)^{2}=\dfrac{25y^{2}}{16} :\dfrac{49y^{2}}{16}=25 :49$$
As per (6) and (7)
$$x^{2}+y^{2}:(x+y)^{2}=\dfrac{41y^{2}}{16} :\dfrac{81y^{2}}{16}=41 :81$$


The diameter of the Sun is $$1.4\times {10}^{9}$$ meters and that of the Earth is about $$1.2768\times {10}^{7}$$ meters. Find the approximate ratio of the diameter of the Sun to that of the Earth.

Diameter of the sun is $$1.4\times 10^{9}$$ meters
Diameter of the earth is $$1.2768\times 10^{7}$$ meters
Ratio of diameter of sun to that of earth is given by 
$$\dfrac{1.4\times 10^{9}}{1.2768\times 10^{7}}$$

$$=\dfrac{(14\times 0.1)\times 10^{9}}{(14\times 0.0912)\times 10^{7}}$$

$$=\dfrac{14^{1-1}\times 0.1\times 10^{9-7}}{(0.0912)}$$ .......... [Using quotient rule of exponents]

$$=\dfrac{(0.1)\times 10^{2}}{(0.0912)}$$

$$=\dfrac{(10)^{-1}\times 10^{2}}{(0.0912)}$$

$$=\dfrac{(10)^{2-1}}{(0.0912)}$$ ........ [Using product rule of exponents]

$$=\dfrac{10}{(0.0912)}$$

$$=109.6$$

Hence, the answer is $$109.6$$.


A partially filled pool contains 600 gallons of water. A hose is turned on, and water flows into the pool at the rate of 8 gallons per minute. How many gallons of water will be in the pool after 70 minutes?

In $$ 70 $$ minutes, gallons of water flowing into the pool $$ = 70 \times 8 = 560 $$

So total water $$ = 600 + 560 = 1160 $$ gallons


In a school, number of boys are $$30$$ and number of girls are $$20$$, then the ratio of number of boys and girls is $$x:y$$. Find $$x+y$$

Number of boys $$=$$ $$30$$
Number of girls $$=$$ $$20$$

Ratio of number of boys to the girls is,
$$x:y=30:20$$
         $$=3:2$$
$$\Rightarrow x=3$$ and $$y=2$$

$$\therefore x + y=5$$


Weight of Pratik is $$40\ kg$$ and weight of his father is $$50\ kg$$. If the ratio of weight of Pratik and his father is $$x:y$$, then find $$x+y$$.


2015686_710845_ans_760027a1e69b447b8d251fbb7c56df05.jpeg


If the price of a scooter is $$Rs. 45000$$ and price of a bicycle is $$Rs. 5000$$, then the ratio of price of bicycle and a scooter is $$x:y$$. Find $$x+y$$


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In a cloth-shop there are $$150$$ metre silk clothes and $$200$$ metre cotton clothes, then the ratio of silk clothes to cotton clothes is $$x:y$$. Find $$x+y$$


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If cost of a belt is $$Rs. 60$$ and cost of shoes is $$Rs. 150$$, then the ratio of cost of the belt and shoes is $$x:y$$. Find $$x+y$$


2015687_710848_ans_cee103ef05024af68b04a42777da48e9.jpeg


If Mayank is $$8$$ years old and his sister Shilpa is $$12$$ years old, then the ratio of their ages is $$x:y$$. Find $$x+y$$


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An alloy consists of two metals taken in the ratio $$1: 2$$ and another alloy consists of the same metals taken in the ratio $$2 : 3$$. How many parts of the two alloys must be taken to obtain a new alloy consisting of the same metals which are in the ratio $$\dfrac {17}{27}$$?

Let say alloy 1 consists of metal $$m_1$$ and $$m_2$$ in ratio $$1:2$$ and alloy 2 consists of metal $$m_1$$ and $$m_2$$ in ratio $$2:3$$. 
Now, combining both the alloys to form a new alloy. 
So, according to given information $$1x+2y=17 $$ and $$2x+3y=27$$
Where, $$x, y$$ represents metal $$m_1$$ and $$m_2$$
$$\implies 2(17-2y)+3y=27\implies y=7$$
By replacing $$y=7$$, we get $$x=3$$
$$\implies $$ we need to take first alloy $$3$$ parts and second alloy $$7$$ parts 


In two alloys copper and zinc are related as $$4 :1$$ and $$1 : 3$$. After alloying together $$10$$ kg of the first alloy, $$16$$ kg of the second and several kg of pure copper, an alloy was obtained in the ratio of copper to zinc was $$3 : 2$$. Find the weight of the new alloy.

Copper $${ 1 }^{ st }$$ Alloy of $$10kg$$ $$=\dfrac { 4 }{ 5 } \times 10$$
                                       $$=8kg$$
Zinc in $${ 1 }^{ st }$$ Alloy of $$10kg=\dfrac{1}{5}\times 10$$
                                      $$=2kg$$
Copper in $${ 2 }^{ nd }$$ Alloy of $$16kg$$ $$=\dfrac { 1 }{ 4 } \times 16$$
                                            $$=4kg$$
Zinc in $${ 2 }^{ nd }$$ Alloy of $$16kg$$ $$=\dfrac { 3 }{ 4 } \times 16$$
                                       $$=12kg$$
Total Copper $$=12kg$$
Total Zinc $$=14kg$$
In the new alloy let $$x\ kg$$ of copper was added as extra.
$$\therefore $$  Total copper in new alloy $$=12+x$$
$$\therefore $$  New alloy copper to zinc ratio is,
$$\dfrac { 12+x }{ 14 } =\dfrac { 3 }{ 2 } $$
$$24+2x=42$$
$$2x=18$$
$$x=9kg$$
Thus, 
Weight of the new alloy $$=(9+16+10)\ kg$$
                                      $$=35\ kg$$


If $$\dfrac{7m + 2n}{7m - 2n} = \dfrac{5}{3}$$ use properties of proportion to find $$m:n$$ and hence find $$\dfrac{m^2-n^2}{m^2+n^2}$$.

Given
$$\dfrac{7m + 2n}{7m - 2n} = \dfrac{5}{3}$$

By cross multiplying numerator and denominator,

$$21m+6n=35m-10n$$

or, $$14m=16n$$

$$\dfrac{m}{n}=\dfrac{16}{14}=\dfrac{8}{7}$$.......$$(1)$$

Now,
$$\dfrac{m^2-n^2}{m^2+n^2}$$

$$=\dfrac{\left(\dfrac{m}{n}\right)^2-1}{\left(\dfrac{m}{n}\right)^2+1}$$

$$=\dfrac{8^2-7^2}{8^2+7^2}$$ [Using $$(1)$$]

$$=\dfrac{15}{113}$$


Reena cuts $$46\ m$$ of cloth into pieces of $$1.15\ m$$ each. How many pieces did she get

Number of pieces

$$=\dfrac{46}{1.15}=\dfrac {4600}{115}=40$$


Find the ratio of the following in the simplest form :
1.48 to 60
2.80 paise to INR 40
3.45 minutes to 3 hours
4.4 min to 45 sec

Part $$(A)$$
$$48 $$ to $$60$$
Ratio $$=\dfrac{48}{60}$$

Ratio $$=\dfrac{4\times 12}{5\times 12}$$

Ratio $$=\dfrac{4}{5}$$

Part $$(B)$$
$$80 $$ paise to INR $$40$$
Ratio $$=\dfrac{80}{40\times 100}$$

Ratio $$=\dfrac{1}{50}$$

Part $$(C)$$
$$45 $$ minutes to $$3$$ hours
Ratio $$=\dfrac{45}{3\times 60}$$

Ratio $$=\dfrac{1}{4}$$

Part $$(D)$$
$$4 $$ minutes to $$45$$ seconds
Ratio $$=\dfrac{4\times 60}{45}$$

Ratio $$=\dfrac{16}{3}$$

Hence, this is the answer.


An alloy of zinc and copper weigh $$12\dfrac{1}{2}\ kg$$. If, in the alloy, the ratio of zinc and copper is $$1 : 4$$, find the weight of copper in it.

Weight of an alloy of zinc and copper $$= 12\dfrac{1}{2} kg $$

The ratio of zinc and copper in the alloy $$ = 1 : 4$$

The weight of copper in alloy $$ = \dfrac{4}{5} \ of \  12\dfrac{1}{2}kg$$

$$ = \dfrac{4}{5} \times \dfrac{25}{2}$$

$$= 10 kg$$ (Ans)


A bag containing $$30$$ apples weighs $$6\,kg$$. How much will $$1080$$ such apples weigh?

weigh of $$30$$ apples$$=6\,kg$$
weigh of $$1$$ apples$$=\dfrac {6}{30}\,kg$$
weigh of $$1080$$ apples$$=\dfrac {1080 \times 6}{30}=216\,kg$$


Rohit earns $$Rs. 15300$$ and saves $$Rs. 1224$$ per month. Find the ratio of
(i) his income and savings
(ii) his income and expenditure
(iii) his expenditure and savings.

(i) $$\dfrac{Income}{Savings}$$  $$=\dfrac{15300}{1224}$$ $$= 12.5$$

(ii) $$\dfrac{Income}{Expenditure}$$  $$= \dfrac{Income}{Income-savings}=\dfrac{15300}{15300-1224}$$ $$= 1.087$$

(iii) $$\dfrac{Expenditure}{Savings}$$  $$=\dfrac{Income-savings}{Savings}=\dfrac{15300 - 1224}{1224}$$ $$= 11.5$$


Total population of a district is 3,540 out of which 2,065 are males. Find the ratio of males to females.

Given, total population of a district is $$3,540$$ out of which $$2,065$$ are males. 
To find out: The ratio of males to females.

Numbers of females $$=(3,540-2.065)$$

$$=1,475$$

$$\therefore \ $$ The ratio of males to females $$=\dfrac{2,065}{1,475}$$

$$=\dfrac{7}{5}$$

Hence, the ratio of males to females is $$7:5$$.


If $$a, b, c$$ are all non-zero and $$a + b + c = 0$$, prove that $$\dfrac{a^2}{bc} + \dfrac{b^2}{ac} + \dfrac{c^2}{ab} = 3$$

Given, 

$$a+b+c=0$$

or, $$a+b=-c$$

cubing both sides we get,

$$(a+b)^3=-c^3$$

$$a^3+b^3+3ab(a+b)=-c^3$$     [$$\because(a+b)^3=a^3+3a^2b+3ab^2+b^3$$]

 $$a^3+b^3-3abc=-c^3$$ [Since $$a+b=-c$$]

 $$a^3+b^3+c^3=3abc$$.......(1).

Now,

$$\dfrac{a^2}{bc} + \dfrac{b^2}{ac} + \dfrac{c^2}{ab}$$

$$=\dfrac{a^3+b^3+c^3}{abc} $$

$$=\dfrac{3abc}{abc} $$ [Using (1)]

$$=3$$.


The weights of five cartons of books are in the ratio $$2:3:4:5:6.$$ If their total weight is $$400\ kg$$, find the weight of each carton.

Let the weight of each carton be $$2x,3x,4x,5x,6x$$ respectively. Since the total weight is given as 400 kg, then,

$$2x + 3x + 4x + 5x + 6x = 400$$

$$20x = 400$$

$$x = 20\;{\rm{kg}}$$

Therefore, 

Weight of the first carton is $$2*20 = 40\;{\rm{kg}}$$.

Weight of the second carton is $$3*20 = 60\;{\rm{kg}}$$.

Weight of the third carton is $$4*20 = 80\;{\rm{kg}}$$.

Weight of the fourth carton is $$5*20 = 100\;{\rm{kg}}$$.

Weight of the fifth carton is $$6*20 = 120\;{\rm{kg}}$$.



Two numbers are in the ratio 4 : 5. If 30 is subtracted from each of the numbers, the ratio becomes 1:2. Find the numbers.

Let $$x$$ and $$y$$ be the numbers.

Given,$$x:y=4:5$$ or $$\dfrac{x}{y}=\dfrac{4}{5}$$
$$x=\dfrac{4y}{5}$$       $$\dots (1)$$

$$x-30:y-30=1:2$$ or $$\dfrac{x-30}{y-30}=\dfrac{1}{2}$$

$$2(x-30)=y-30$$

$$x=\dfrac{y+30}{2}$$ $$\dots (2)$$

From $$(1)$$ and $$(2)$$:

$$\dfrac{4y}{5}=\dfrac{y+30}{2}$$

$$8y=5y+150$$

$$3y=150 \implies y=50$$

Substituting this value of $$y$$ in $$(1)$$:

$$x=\dfrac{4y}{5}=\dfrac{4(50)}{5}=40$$

$$\therefore 40$$ and $$50$$ are the numbers.


The length of a line segment $$AB$$ is $$38\ cm$$. A point $$X$$ on it divides it the ratio $$9:10$$. Find the length of the line segments $$AX$$ and $$XB$$
1047912_de84caf286344c3ba9b2ed2b44651c9f.png

Let the length of $$AX$$ and $$XB$$ be $$9k$$ and $$10k$$ respectively.
Then  $$9k + 10k = 38$$
$$ \Rightarrow 19k = 38$$
$$ \Rightarrow k = 2$$
So, Length of $$AX=9k$$ $$=$$ $$9 \times 2 $$ $$=18$$ $$cm$$
Length of $$XB=10k=10\times2=20$$ $$cm$$


A school has 630 students. The ratio of the number of boys to the number of girls is 3:The ratio changes to 7:5 after the admission of 90 new students. Find the number of newly admitted boys.

Let the number of boys be $$3x$$

Then the number of girls $$ =2x$$

$$\therefore 3x+2x=630$$

$$\Rightarrow 5x=630 \therefore x=126$$

$$\Rightarrow$$ number of boys $$ =3x =3\times 126=378$$

And number of girls $$ =2x=2\times 126 =252$$

After admission of $$90$$ students we have total number of students $$ =630+90=720$$

Now, let the number of boys be $$7x$$

Then, the number of girls $$=5x$$

$$\therefore 7x+5x=720$$

$$\Rightarrow 12x=720$$

$$\therefore x=60$$

$$\Rightarrow $$ number of boys $$=7x=7\times 60 =420$$

And number of girls $$ =5x =5\times 60 =300$$

$$\therefore$$ Number of newly admitted boys $$ = 420-378 =42$$



Write the following ratios in decreasing order $$4:7,5:14,3:14,1:3$$ ?


$$0.21,0.33,0.35,0.57$$

$$3:14,1:3,5:14,4:7$$


Divide Rs.$$1650$$ among $$A,B$$, and $$C$$ in the ratio $$3:2:1$$.

Let A=3, B=2. C=1

In A$$=\dfrac { 16504 }{ 3 } =Rs.5501$$

In B$$=\dfrac { 16504 }{ 2 } =Rs.8252$$

In C$$=\dfrac { 16504 }{ 1 } =Rs.16504$$


The interior angles of a pentagon are in the ratio $$4:8:6:4:5$$ Find each angle of the pentagon.

Let the interior angles of the pentagon be $$4x,\, 8x,\,  6x,\,  4x,\,  5x$$.  
Their sum $$= 4x + 8x + 6x + 4x + 5x = 27x$$
Sum of interior angles of a polygon$$=\left(n-2\right){180}^{\circ}=\left(5-2\right){180}^{\circ}={540}^{\circ}$$
$$\therefore 27x={540}^{\circ}$$
$$\Rightarrow \dfrac{{540}^{\circ}}{27}={20}^{\circ}$$
$$\therefore$$ Angles are 
$$4x=4\times{20}^{\circ}={80}^{\circ}$$
$$8x=8\times{20}^{\circ}={160}^{\circ}$$  
$$6x=6\times{20}^{\circ}={120}^{\circ}$$
$$4x=4\times{20}^{\circ}={80}^{\circ}$$  
$$5x=5\times{20}^{\circ}={100}^{\circ}$$


Divided $$120$$ into two parts such that $$\dfrac{2}{3}$$ of one part is equal to $$\dfrac{2}{3}$$ of the other part.

Let one part be $$x$$

other part be $$120-x$$

$$\therefore \displaystyle \frac{2}{3}x=\frac{2}{3}(120-x)$$

$$\therefore x+x=120$$

$$\therefore x=60$$

$$\therefore $$ Two part are 60 and 60


A taxi-charges $$Rs\ 6.40\ per\ km$$. How much will he charge for journey of $$16.5\ km$$?

$$Taxi\quad charge\quad per\quad km=Rs.6.40$$

$$Total\quad journey=16.5kms\\ $$

$$Charge=16.5\times 6.40\\ $$

$$=Rs.105.60$$


Find the ratio of $$9$$ m to $$27$$ cm.


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Find the ratio of $$15$$ kg to $$210$$ g.


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A scooter requires $$2$$ litres of petrol to cover $$80$$km. How many litres of petrol is required to cover $$1$$ km?

Given a scooter requires 2 liters of petrol to cover 80 km.
So, by unitary method, 1 liter of petrol covers 40 km
$$ \Rightarrow $$  1 km require $$ \dfrac{1}{40}$$ liters
= 0.025 liters of petrol


Simplfy:
$$\cfrac{1200}{480}$$

$$\frac { 1200 }{ 480 } =2.5$$
Since  480$$\times2.5$$=1200



If $$\dfrac{3a + 7b}{3a - 7b} = \dfrac{4}{3}$$ then find the value of the ratio $$\dfrac{3a^2 - 7b^2}{3a^2 + 7b^2}$$.

Given $$\dfrac{3a+7b}{3a-7b}=\dfrac{4}{3}$$

$$\dfrac{3a^2-7b^2}{3a^2+7b^2}=?$$

$$\dfrac{3a+7b}{3a-7b}=\dfrac{4}{3}$$

$$\dfrac{a(3+7\dfrac{b}{a})}{a(3-7\dfrac{b}{a})}$$ $$=\dfrac{4}{3}$$

$$9+21\left(\dfrac{b}{a}\right)=12-28$$ $$\left(\dfrac{b}{a}\right)$$

$$\Rightarrow 49\left(\dfrac{b}{a}\right)=3$$

$$\Rightarrow \left(\dfrac{b}{a}\right)=\dfrac{3}{49}$$

$$\Rightarrow \dfrac{b^2}{a^2}=\dfrac{9}{49\times 49}$$

$$\dfrac{3a^2-7b^2}{3a^2+7b^2}=\dfrac{a^2\left(3-7\dfrac{b^2}{a^2}\right)}{a^2\left(3+7\dfrac{b^2}{a^2}\right)}=\dfrac{3-7\left(\dfrac{9}{49\times 49}\right)}{63+7\times \left(\dfrac{9}{49\times 49}\right)}$$

$$=\dfrac{3\left[1-\dfrac{3}{7\times 49}\right]}{3\left[1+\dfrac{3}{7\times 49}\right]}$$

$$=\dfrac{(7\times 49-3)}{(7\times 49 +3)}$$

$$=\dfrac{343-3}{343+3}$$

$$\Rightarrow \dfrac{340}{346}$$

$$\Rightarrow \dfrac{170}{173}$$

$$\dfrac{3a^2-7b^2}{3a^2+7b^2}=\dfrac{170}{173}$$.


In what ratio must tea at Rs. $$62$$ per kg be mixed with tea at Rs. $$72$$ per kg so that the mixture must. be worth Rs. $$64.50$$ per kg?

Let the quantity of the tea that is $$Rs.\ 62$$ per kg be $$x$$ kg and that of tea that is $$Rs.\ 72$$ per kg be $$y$$ kg.

Hence, the required ratio is $$x:y$$.

Price of $$x$$ kg tea $$=Rs.\ 62x$$
Price of $$y$$ kg tea $$=Rs.\ 72y$$
If we mix both varieties of tea, the quantity of the mixture will be $$x+y$$.

Now, this mixture is $$Rs.\ 64.50$$ per kg.
Hence, the total price of mixture $$=Rs.\ 64.50(x+y)$$

So, $$62x+72y=64.50(x+y)$$

$$\Rightarrow 62x+72y=64.5x+64.5y$$

$$\Rightarrow 64.5x-62x=72y-64.5y$$

$$\Rightarrow 2.5x=7.5y$$

$$\Rightarrow \dfrac xy=\dfrac{7.5}{2.5}$$

$$\Rightarrow \dfrac xy=3$$

So, $$x:y=3:1$$

Hence, the required ratio is $$3:1$$.


Annual income of A and B is in the ratio of $$4 : 3$$ and this annual expenses bear a ratio of $$3 : 2$$. If each of them saves $$Rs. 600$$ at the end of the year, the annual income of A is,

Let their income be $$4x$$ and $$3x$$.
Let their expenditure be $$3y$$ and $$2y$$

We know that
$$Income - Exp = Saving$$

So, 
$$4x - 3y = 600$$
$$3x - 2y = 600$$

On solving
$$8x - 6y = 1200$$
$$- 9x + 6y = 1800$$

Hence on adding these equations,
$$x = 600$$
$$y = 600$$

Income of $$A= 4\times 600= Rs2400$$

Hence, this is the answer.


Express the following as ratio in simplest form.
$$10$$ cm to $$10$$ m.

$$\begin{matrix} simplest\, form \\ 10cm\, to\, 10m \\ \Rightarrow \dfrac { { 10 } }{ { 10\times 100 } } =\dfrac { 1 }{ { 100 } }  \\  \end{matrix}$$


Express the following as ratio in simplest form.
$$45$$ minutes to $$4$$ hours.

$$\begin{matrix} 45\, \min  utes\, to\, 4\, hours \\ Simplest\, ratio=\dfrac { { 45 } }{ { 4\times 60 } }  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\dfrac { 3 }{ { 16 } }  \\  \end{matrix}$$


If angles of a triangle are in the ratio $$2:3:4$$. Find the value of each angle.

Let common ratio be $$2x, 3x$$ and $$4x$$
$$\rightarrow $$ let common ratio be $$x$$
$$\therefore \ $$ Angle are $$2x,\ 3x$$ and $$4x$$
$$\therefore \ 2x+3x4x=180^o$$
$$\therefore \ 9x=180^o$$
$$\therefore \ x=\dfrac {180^o}{9}=20^o$$
$$\therefore \ $$ Angle are $$40^o,\ 60^o,\ 80^o$$



Find the ratio of :
  1. Rs $$5$$ to $$50$$ paisa
  2. $$15kg$$ to $$210g$$
  3. $$9m$$ to $$27cm$$
  4. $$30$$ days to $$36$$ hours

(A)
Ratio of $$5\ Rs$$ to $$50\ Paisa$$
if $$1\ Rs=100\ Paisa$$
then $$5\ Rs=500\ Paisa$$
Ratio $$=\dfrac{500}{50}=\dfrac{10}{1}=10:1$$
Ratio of $$5\ Rs$$ to $$50\ Paisa$$ is $$10:1$$
(B)
Ratio of $$15\ kg$$ to $$210\ g$$
if $$1\ kg$$ is $$= 1000\ g$$.
then $$15\ kg=15000\ g$$
and Ratio $$=\dfrac{15000}{210}=\dfrac{500}{7}$$
Ratio of $$15\ kg$$ to $$210\ g$$ is $$500:7$$
(C)
Ratio of $$9\ m$$ to $$27\ cm$$
if $$1\ m=100\ cm$$
then $$9\ m=900\ cm$$
and $$Ratio $$=\dfrac{900}{100}=\dfrac{9}{1}$$
Ratio of $$9\ m$$ to $$27\ cm$$ is $$9:1$$
(D)
Ratio of $$30\ days$$ to $$36\ hour$$
$$1\ day=24\ hour$$
$$30\ days=720\ hour$$
and ratio $$=\dfrac{720}{36}=\dfrac{1}{20}$$
Ratio of $$30\ days$$ to $$36\ hour$$ is $$20:1$$ 



If the number of workers employed to construct the Taj Mahal were $$20,000$$ and the ratio of Turks:Persians:Indians was $$1:2:17$$, find the  number of Indians, Turks and Persians.

Let $$Turks:Persians:Indians=x:2x:17x$$ 
Number of workers$$=x+2x+17x=20,000$$
$$\Rightarrow\,20x=20,000$$
$$\Rightarrow\,x=\dfrac{20000}{20}=1000$$
$$\therefore\,turks=1000$$ workers
Persians$$=2x=2\times 1000=2000$$ workers
Indians$$=17x=17\times 1000=17,000$$ workers.


Express the following as ratio in simplest form.
$$\text{Rs. } $$ $$3.50$$ to $$75$$ paise.

First of all, make the units same
so, $$\text{Rs } 3.50=350$$ paise
$$\therefore$$ Simplest ratio is $$\dfrac{350}{75}=\dfrac{14}{3}$$
$$\implies 14:3$$



Express the following as ratio in simplest form.
$$35$$ days to $$3$$ weeks.

$$\begin{matrix} 35\, days\, to\, 3\, weeks\,  \\ simplest\, ratio\,  \\ \Rightarrow \dfrac { { 35 } }{ { 3\times 7 } }  \\ \Rightarrow \dfrac { 5 }{ 3 }  \\  \end{matrix}$$


$$p+q =29$$
$$\frac{1}{9}+\frac{1}{5}=\frac{2}{p }$$ Prove that p:q =$$?$$

We have,

$$ p+q=29\,\,......\,\,\left( 1 \right) $$

$$ \dfrac{1}{9}+\dfrac{1}{5}=\dfrac{2}{p}\,\,......\,\,\left( 2 \right) $$

By equation (2) to and we get,

$$ \dfrac{5+9}{45}=\dfrac{2}{p} $$

$$ 14p=90 $$

$$ p=\dfrac{90}{14} $$

$$ p=\dfrac{45}{7} $$

Put the value of P in equation (1) and we get,

$$ p+q=29 $$

$$ \dfrac{45}{7}+q=29 $$

$$ q=29-\dfrac{45}{7} $$

$$ q=\dfrac{203-45}{7} $$

$$ q=\dfrac{158}{7} $$

The ratio of $$p:q=45:158$$

Hence, this is the answer.


If $$\dfrac {3} {4}$$ of the salary of a man per month is $$Rs. 600$$. What is the salary of him per month?

Let the salary be $$x$$.

According to the question,
$$\dfrac{3}{4}\times x=600$$

$$\dfrac{1}{4}\times x=200$$

$$x=800$$

Hence, this is the answer.


Shubham's age today is $$12$$ years and his father's age is $$42$$ years. Shubham's mother is younger than his father by $$6$$ years. Find the following ratios.
(1) Ratio of Shubham's age today to his mother's age today.
(2) Ratio of Shubham's mother age today to his father's age today.
(3) The ratio of Shubham's age to his mother's age when Shubham was $$10$$ years old.

Shubham's  age $$=12\ \text{years}$$
Father's age $$=42\ \text{years}$$
Mother is younger than father by 6 years therefore mother's age will be $$(42-6)=36\ \text{years}$$
1) Ratio $$=12:36=1:3$$
2) Ratio $$=36:42=6:7$$
As shubham's age was 10 years that means his mothers age was 34 years
3) Ratio $$10:34=5:17$$


Express the following ratios in their simplest form:
$$1m5cm$$ : $$63cm$$

Converting into the same units,

$$1m = 100 cm$$ 

$$1m 5cm = 100 cm + 5 cm = 105 cm$$

$$1m 5cm : 63 cm = 105 cm : 63 cm $$

$$= 7 × 5 × 3 : 7 × 3 × 3 $$

$$= 5 : 3$$






The weight of  $$72$$  books is  $$9kg$$ . What is the weight of  $$40$$  such books?


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Write the ratio of radius to the diameter of a circle.

Let diameter $$=d$$
Then radius $$=r= d/2$$
Ratio $$=\dfrac{radius }{ diameter }= \dfrac{d/2}{d}= \dfrac{1}{2}$$


Find the ratio of:
$$9m$$ to $$27cm$$

9 m to 27 cm
$$\Rightarrow 9\times100\ cm\ to\ 27\ cm\quad [\because 1\ m=100\ cm]$$
$$\Rightarrow 900\ cm\ to\ 27\ cm$$
Thus, ratio is $$=\dfrac{900}{27}=\dfrac{100}{3}=100:3$$


Divide Rs. $$150$$ between Abhishek and Akhil in the ratio $$3:7$$.

Let Abhishek's part is $$3x$$and Akhil's part is $$7x$$
ATQ,
$$3x+7x=150$$
$$10x=150$$
$$x=15$$
So 
Abhishek's part $$=3\times15=Rs.45$$

Akhil's part $$=7\times15=Rs.105$$


Solve the following question:
When $$x=2011$$, the value of $$\cfrac { 1 }{ 1+\cfrac { 1 }{ 1+\cfrac { 1 }{ 1+\cfrac { 1 }{ x }  }  }  } $$ is ______

$$\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}}$$ $$=\dfrac{1}{1+\dfrac{1}{1+\dfrac{x}{x+1}}}=$$ $$\dfrac{1}{1+\dfrac{1}{\left(\dfrac{2x+1}{x+1}\right)}}$$
$$\dfrac{1}{1+\dfrac{x+1}{2x+1}}=\dfrac{2x+1}{3x+2}=\dfrac{2(2011)+1}{3(2011)+2}=\dfrac{4022+1}{6033+2}=\dfrac{4023}{6035}$$.

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In a JRC one day camp, $$150\ gm$$ of rice and $$15\ ml$$ oil are needed for a student. If there are $$40$$ students to attend the camp how much of rice and oil are needed?

For one student, rice needed $$=150\ gm$$
For $$40$$ students, rice needed $$=150\times 40\ gm$$
                                                  $$=6000\ gm$$
                                                  $$=6\ kg$$

For one student, oil needed $$=15\ ml$$
For $$40$$ students, oil needed $$=15\times 40\ ml$$
                                                $$=600\ ml$$

Hence, the total rice needed is $$6\ kg$$ and the total oil needed is $$600\ ml$$.


If the cost of $$50$$ pencils is Rs. $$20$$, then the pencils can be bought for Rs. $$32$$ is ______

Number of pencils in Rs.20 $$=50$$
Number of pencils in Rs.1 $$=\dfrac{50}{20}=2.5$$
Number of pencils in Rs.32 $$=2.5\times32=80$$
Therefore 80 pencils can be bought from Rs.32


Ashu's mother wants to buy $$1kg$$ pastries. Each pastry weighs $$50g$$. How many pastries should she buy?

Total weight of pastries $$=1\ kg=1000\ g$$
Weight of 1 pastry $$=50\ g$$
Number of pastries $$=\dfrac{Total\ weight}{Weight\ of\ 1\ pastry}$$
                                 $$=\dfrac{1000}{50}$$
                                 $$=20$$
Therefore Ashu's mother buy $$20$$ pastries.


Find the ratio $$4$$m to $$24$$ cm.

Find the ratio of $$4m$$ to $$24cm$$
Now $$4m=4\times 100cm=400cm$$
$$\therefore$$ the ratio will be $$=400:24$$
                               $$=\dfrac{400}{24}$$

                               $$=\dfrac{4\times 100}{4\times 6}$$

                               $$=\dfrac{100}{6}$$

                               $$=\dfrac{50\times 2}{3\times 2}$$

                               $$=\dfrac{50}{3}$$

Hence, the answer is $$\dfrac{50}{3}.$$


Solve:
$$7 \mathrm { kg } : \ldots = 14 \mathrm { m } : 6 \mathrm { m }$$
$$\dots . .63 \mathrm { L }= 25 : : 35$$

1)
$$7:x=14:6$$.

$$\dfrac{7}{x}=\dfrac{14}{6}$$

$$\Rightarrow x=\dfrac{7\times 6}{14}=3$$

$$\therefore x=3$$

2)
$$x:63=25:35$$

$$\dfrac{x}{63}=\dfrac{25}{35}$$

$$\Rightarrow x=\dfrac{5}{7}\times 63$$

$$\therefore x=45$$


Express the following ratio in the simplest form
$$36:90$$

$$36:90$$  ratio
$$\dfrac{36}{90}=\dfrac{12}{30}=\dfrac{4}{10}=\dfrac{2}{5}$$
Simplest form of $$36:90$$ is $$2:5$$


a : b = 2 : 3 , b : c = 4 : 5 , c : d = 6 : 7 . Find a : b : c : d.

Given $$a:b=2:3 , b:c=4:5 , c:d=6:7$$
$$a:b=2\times 8:3\times 8=16:24\\ b:c=4\times 6:5\times 6=24:30\\ c:d=6\times 5:7\times 5=30:35\\ so\quad a:b:c:d=16:24:30:35$$


The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas. 

Surface area of sphere $$= \pi d^2 = 4\pi r^2$$

Surface area of moon = $$\pi d_1^2$$

Surface area of earth = $$\pi d_2^2$$

Given: $$\dfrac{d_1}{d_2}=\dfrac{1}{4}$$

Ratio of surface areas $$= \dfrac{\pi d_1^2}{\pi d_2^2}= (\dfrac{1}{4})^2$$

                                      $$=\dfrac{1}{16}$$




There are $$32$$ students in a class of which $$12$$ are girls. Find:
(i) the ratio of number of boys to number of girls
(ii) the ratio of number of boys to total number of students.

Boys$$=32-12=20$$
(i) $$\dfrac{20}{12}=\dfrac{5}{3}$$
(ii) $$\dfrac{20}{32}=\dfrac{5}{8}$$.


Express the following ratio in the simplest form
$$680:634$$

$$680 : 634$$
$$\dfrac{680}{634} =\dfrac{340}{317}$$
Simplest form of $$680 : 634$$ is $$340 : 317$$



If $$A:B=3:4$$ and $$B:C=6:7$$, find:
$$(i)$$ $$A:B:C$$
$$(ii)$$ $$A:C$$

$$A:B=3:4\Rightarrow\cfrac{A}{B}=\cfrac{3}{4}\\ \Rightarrow \cfrac{A}{3}=\cfrac{B}{4}\Rightarrow \cfrac{A}{9}=\cfrac{B}{12}$$
And $$B:C=6:7\Rightarrow \cfrac{B}{C}=\cfrac{6}{7}\\ \Rightarrow \cfrac{B}{6}=\cfrac{C}{7}\Rightarrow \cfrac{B}{12}=\cfrac{C}{14}$$
$$ \Rightarrow \cfrac{A}{9}=\cfrac{B}{12}=\cfrac{C}{14}\\ (i)\therefore A:B:C=9:12:14\\(ii) also\quad \cfrac{A}{9}=\cfrac{C}{14}\Rightarrow \cfrac{A}{C}=\cfrac{9}{14}\Rightarrow A:C=9:14$$


Divide $$Rs. 7400$$ among three people $$A, B,$$ and $$C$$ in the ratio $$3:5:12$$

$$ \rightarrow $$ Total divisions = 20 
one division $$ = \frac{7400}{20} = 370 $$
$$ \Rightarrow $$ A get $$ 3\times 370 = 1110$$
$$ \Rightarrow $$ B get $$ 5\times 370 = 1850 $$
$$ \Rightarrow $$ C get $$ 12\times 370 = 4440 $$

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There are $$2500$$ student in a school. Out of them $$1100$$ are girls and the rest  are boys. Find the ratio of :
number of student to number of boys.

Total no. of students$$=2500$$
No. of girls$$=1100$$
No. of boys$$=2500-1100=1400$$
ratio of girls to total no. of students $$=\cfrac { 2500 }{ 1400 } =\cfrac { 25 }{ 14 } $$


 What is the ratio of Quarter past 6 and 10 minutes to 7?


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If $$2$$ score strawberries can be bought for $$Rs\ 278.50$$, what is the price of $$1$$ dozen strawberries?

We know 
$$1 $$dozen means $$12$$
$$1$$ score = $$20$$
Thus $$2$$ scores =$$ 40$$
Now $$40$$ strawberries cost $$Rs  278.5$$
then cost of one strawberry is $$Rs. 6.9625$$
So cost of $$1$$ dozen is $$12 * 6.9625 = Rs. 83.55$$


The cost of dozen eggs is $$60\ Rs$$ find the cost of $$4$$ eggs.

$$1\ \text{dozen} = 12\ \text{items}$$
Given, cost of $$12$$ eggs is $$60Rs.$$
The cost of $$1$$ egg is $$=\dfrac{60}{12}=5\ Rs.$$
The cost of $$4$$ eggs is $$5\times 4=20\ Rs.$$


Find the ratio of :
$$(a)\,\,Rs\,5$$ to $$50$$ paise
$$(b)\,\,15\,kg$$ to $$210\,g$$
$$(c)\,\,9\,m$$ to $$27\,cm$$
$$(d)\,\,30$$ days to $$36$$ hours.

a) $$Rs.5$$ to $$50$$ paise

$$Rs.5 = 5 \times 100$$ paise

         $$= 500 $$ paise

Ratio $$= \dfrac{500}{50} = \dfrac{10}{1}$$

          $$ = 10 : 1$$

b) $$15 kg$$ to $$210$$ gm

$$15 kg = 15 \times 1000 $$ gm

         $$= 15000 \, gm$$

Ratio $$= \dfrac{15000}{210} = \dfrac{500}{7}$$

          $$= 500 : 7$$

c) $$9m$$ to $$27cm$$

$$9 m = 9 \times 100 \, cm$$

       $$= 900 \, cm$$

Ratio $$= \dfrac{900}{27} = \dfrac{100}{3}$$

         $$= 100 : 3$$

d) $$30 $$ days to $$36$$ hours

$$\Rightarrow  30$$ days $$ = 30 \times 24$$ hrs

                  $$= 720 \, hrs$$

Ratio $$= \dfrac{720}{36} = \dfrac{20}{1}$$

         $$= 20 : 1$$


Divide $$268$$ into two parts such that their ratio is $$1:3$$.

Let the two parts be: x & y

$$\therefore \dfrac{x}{y}=\dfrac{1}{3}$$

$$\Rightarrow y=3x$$

$$\therefore x+y = x+3x=268$$

$$4x=268$$

$$x=\dfrac{268}{4}=67$$

$$y=3\times 67=201$$

$$\therefore x=67, y=201$$.


Give three equivalent ratios of 4 : 9.

Ratio $$4 : 9$$ is also $$\dfrac {4}{9}$$. This ratio is in reduced form. 

Equivalent ratios can be found by multiplying both parts of the ratio by same constant.

$$\dfrac {4}{9}\times \dfrac {2}{2} = \dfrac {8}{18}$$

$$\dfrac {4}{9}\times \dfrac {3}{3} = \dfrac {12}{27}$$

$$\dfrac {4}{9}\times \dfrac {4}{4} = \dfrac {16}{36}$$


If $$A:B=3:4$$ and $$B:C=8:9$$, then find $$A:C$$.

We have,

$$ A:B=3:4 $$

$$ \dfrac{A}{B}=\dfrac{3}{4}\,\,......\,\,\left( 1 \right) $$

On multiplying by 8 in equation ( 1)

$$\dfrac{A}{B}=\dfrac{3\times 8}{4\times 8}=\dfrac{24}{32}$$ …… ( 2)

So,

$$\dfrac{B}{C}=\dfrac{8}{9}\,\,......\,\,\left( 3 \right)$$

On multiplying by 4 in equation ( 3)

$$\dfrac{B}{C}=\dfrac{8\times 4}{9\times 4}\,\,=\dfrac{32}{36}\,\,......\,\,\,\left( 4 \right)$$

From equation ( 2) and ( 4) to, and we get,

$$ \dfrac{A}{C}=\dfrac{24}{36}=\dfrac{2}{3} $$

$$ A:C=2:3 $$

Hence, this is the answer.



Find the ratio of the following 
$$3,000\ mL$$ to $$15$$ liters

Given ratio,
$$3,000$$ mL to $$15$$ liters
$$15$$ litre=$$15 \times1000=15000$$mL   $$\left[ {\because 1\,\,litre = 1000mL} \right]$$
Now, ratio of $$3000$$mL to $$15000$$mL
$$=\dfrac{3000}{15000} =\dfrac{1}{5}$$
$$=1:5$$


If  $$x : y = 4 : 7 ,$$  find the value of  ($$ 3 x + 2 y$$ ) : ( $$5 x + y$$ )

Given, $$x:y=4:7\implies \dfrac{x}{y}=\dfrac{4}{7}$$
$$\dfrac{3{x}+2{y}}{5{x}+y}=\dfrac{3\bigg(\dfrac{x}{y}\bigg)+2}{5\bigg(\dfrac{x}{y}\bigg)+1}=\dfrac{\dfrac{12}{7}+2}{\dfrac{20}{7}+1}=\dfrac{26}{27}$$


If $$x : 95 = 51:85$$, the 
value of x 

$$\dfrac{x}{95}=\dfrac{51}{85}$$
$$\Rightarrow x=\dfrac{51\times 95}{85}$$
$$=57$$.


Write each ratio in its simplest form.
$$36\ minutes$$ to an hour

$$\dfrac { { 36 } }{ { 60 } } =\dfrac { 3 }{ 5 } $$


The length of the Indian National Flag is $$90\ cm$$ and the breadth is $$60\ cm$$. Find the ratio of its length to breadth.

Given:-
Length of National Flag $$\left( l \right) = 90 \; cm$$
Breadth of National Flag $$\left( b \right) = 60 \; cm$$

Therefore,
$$\cfrac{l}{b} = \cfrac{90}{60} $$
     $$= \cfrac{3 \times 30}{2 \times 30} $$
     $$= \cfrac{3}{2}$$
$$\Rightarrow l : b = 3 : 2$$


A scooter consumer $$28$$ litres of petrol in covering $$2100\ km$$. How much petrol will be needed to cover a distance of $$3600\ km$$?

Using unitary method :- 
2100 km  $$ \rightarrow  $$ 28 litres
 1 km     $$ \rightarrow  \frac{28}{2100} litres $$
 3600 km $$\rightarrow  \frac{28}{2100}\times 3600 $$ litres
    = 48 litres 

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The ratio of expenditure to the income of a person is 7 :What amount will he save for month if his monthly salary is Rs. 64,000?

Let the expenditure and income be $$7x$$ and $$8x$$ respectively.

Monthly salary $$= Rs. 64000 \quad \left( \text{Given} \right)$$

$$8x = 64000$$

$$ = \cfrac{64000}{8} = 8000$$

Expenditure $$= 7x = 7 \times 8000 = 56000$$

Therefore,

Saving $$=$$ Income $$-$$ Expenditure

Saving $$= 64000 - 56000 = Rs. 8000$$

Hence the saving of the person $$Rs. 8000$$.


Two angles of a triangle are $$54^{o}$$ & $$72^{o}$$. Find the ratio of third angle to the sum of first two angles.

Given two angles are $$54^{\circ}$$ and $$72^{\circ}$$
We know that sum of angles of a triangle is $$180^{\circ}$$
Let the third angle be $$\theta$$
$$54^{\circ}+72^{\circ}+\theta =180$$
                      $$\theta =180^{\circ}-126^{\circ}$$
                         $$=54^{\circ}$$
Sum of first two angles $$=54^{\circ}+72^{\circ}$$
                                       $$=126^{\circ}$$
Ratio of third angle to the sum of first two angles $$=\cfrac { 54 }{ 126 } =\cfrac { 18 }{ 42 } $$
                                                                                  $$=\cfrac { 3 }{ 7 } $$


Write the ratio ($$125\ ml$$ to $$2\ l$$) in its simplest form.

$$\begin{array}{l} 125\, \, ml\, \, to\, \, 2l \\ Ratio=\, \, \dfrac { { 125\, \, ml } }{ { 2l } } =\dfrac { { 125\, \, ml } }{ { 2000\, \, ml } } =\dfrac { 1 }{ { 16 } } =\, 1:16 \end{array}$$


The ratio of copper & zinc in a alloy is $$9:5$$. If the weight of copper in the alloy is $$48$$, gram, find the weight of zinc in the alloy.

Ratio of copper and zinc in alloy$$=9:5$$
Weight of copper$$=48ℊ$$
Let weight of zinc$$=x$$
$$\cfrac { 9 }{ 5 } =\cfrac { 48 }{ x } $$
$$9x=5\times 48$$
$$x=\cfrac { 240 }{ 9 } $$
$$x=25.5ℊ$$
$$\therefore $$Weight of zincin the alloy is $$25.5ℊ$$


Neeraj has thrice the amount of money as compared to his brother. If they together have $$Rs\ 1100$$, how much money does each have?

Let $$x$$ be the amount of money neeraj's brother has.
Amount of money Neraj has $$= 3x$$
According to the question-
$$x + 3x = 1100$$
$$x = \cfrac{1100}{4} = 275$$
Therefore,
Amount of money Neeraj's brother has $$= Rs. 275$$
Amount of money Neeraj has $$= 3 \times 275 = Rs. 825$$


Write each ratio in its simplest form.
$$3.75\ kg$$ to $$75\ g$$

We know that,
$$1kg=1000g$$
$$ = \dfrac{{3.75 \times 1000}}{{75}}$$
$$ = \dfrac{{3750}}{{75}}$$
$$=50:1$$


The distance between City A and City B on a map is given as 6 cm. If the scale represents 1 cm = 200 km, then find the actual distance between City A and City B, in metres.

Given that the distance between City A and City B is $$6 \ cm$$ and that $$1\ cm$$ on map represents $$200 \ km$$
Using unitary method,
If $$1\ cm\rightarrow 200 \ km$$
then, $$6\ cm\rightarrow 200\times 6\ km$$
$$\Rightarrow 1200\ km$$
We know that, $$1\ km=1000 \ m$$
$$\therefore \ 1200\ km=1200\times 1000\ m$$
$$\Rightarrow 1200000\ m$$
Hence, the actual distance between City A and City B is $$1200000\ m$$


One number is greater than thrice the other number by $$2$$ and $$4$$ times smaller number exceeds greater number by $$5$$.Find the numbers.

Let the greater number be $$x$$ and smaller number be $$y$$

Then, according to the given conditions, we have

$$x-3y=2,$$

$$4y-x=5$$

Thus, we have $$x-3y=2$$      ........$$(1)$$

$$-x+4y=5$$        .......$$(2)$$

Adding $$(1)$$ and $$(2)$$ we get

$$x-3y-x+4y=2+5$$

$$\Rightarrow y=7$$

Put $$y=7$$ in $$(1)$$ we get

$$x-3y=2$$

$$\Rightarrow x-3\times 7=2$$

$$\Rightarrow x-21=2$$

$$\Rightarrow x=2+21=23$$

$$\therefore$$ Greater number$$=23$$ and smaller number $$=7$$


If  $$\dfrac { 3 a + 7 b } { 3 a - 7 b } = \dfrac { 4 } { 3 }$$  then find the value of the ratio  $$\dfrac { 3 a ^ { 2 } - 7 b ^ { 2 } } { 3 a ^ { 2 } + 7 b ^ { 2 } }.$$


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Divide $$34$$ into two parts in such a way that $${\left(\dfrac{4}{7}\right)}^{th}$$ of one part is equal to $${\left(\dfrac{2}{5}\right)}^{th}$$ of the other.

Let First part$$=x$$

then second part$$=34-x$$

According to the problem,$${\left(\dfrac{4}{7}\right)}^{th}$$ of First part$$={\left(\dfrac{2}{5}\right)}^{th}$$ of the second part.

$$\Rightarrow \dfrac{4}{7}x=\dfrac{2}{5}\left(34-x\right)$$

$$\Rightarrow \dfrac{4x}{7}=\dfrac{68}{5}-\dfrac{2x}{5}$$

$$\Rightarrow \dfrac{4x}{7}+\dfrac{2x}{5}=\dfrac{68}{5}$$

$$\Rightarrow \dfrac{4x\times 5}{7\times 5}+\dfrac{2x\times 7}{5\times 7}=\dfrac{68}{5}$$

$$\Rightarrow \dfrac{20x}{35}+\dfrac{14x}{35}=\dfrac{68}{5}$$

$$\Rightarrow \dfrac{20x+14x}{35}=\dfrac{68}{5}$$

$$\Rightarrow x=14$$

Hence two parts are $$14$$ and $$34-14=20$$

$$\therefore$$ the first part is $$14$$ and second part is $$20$$


A aeroplane flies $$300\ km$$ in $$5$$ hours. How much distance will be covered if it travels for $$7$$ hours.

$$5$$ hours $$\rightarrow 300$$ km
$$1$$ hr $$\rightarrow \dfrac{300}{5}=60km$$       [unitary method]
$$7$$ hours $$=60\times 7=420$$ km.


Out of $$500$$ oranges in a box,$$\dfrac{3}{25}$$ are rotten and $$\dfrac{1}{5}$$ are kept for some guest.How many oranges are left in the box?

Number of oranges$$=500$$
Bad oranges$$=\dfrac{3}{25}$$ of $$500=\dfrac{3}{25}\times 500=60$$
Kept for guests$$=\dfrac{1}{5}$$ of $$500=\dfrac{1}{5}\times \,500=100$$
$$\therefore\,$$Number of oranges which can be used$$=500-60-100=500-160=340$$


Find the cost of $$8\ kg$$ of rice, if the cost of $$10\ kg$$ is $$Rs.\ 325$$.

$$9$$ kg$$ \rightarrow$$ Rs. $$325$$                       using unitary method
$$1$$kg $$\rightarrow 325/9$$                               
$$10$$kg$$\rightarrow $$ $$\frac{325}{9}\times 10$$
$$=$$ Rs. $$361.12$$.

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Find two equivalent ratios of the following
$$5:7$$

$$5:7$$
$$5\times 2:7\times 2=10:14$$
$$5\times 3:7\times 3=15:21$$
$$\therefore$$ the two equivalent ratios are $$10:14$$ and $$15:21$$


On a work place mathematics test of $$20$$ questions, you missed $$2$$ questions.What is the ratio of the number you answered correctly to the number you missed?

Total questions$$=20$$

The number missed$$=2$$

Number correct$$=20-2=18$$

$$\dfrac{Number\,correct}{Number\,missed}=\dfrac{18}{2}=\dfrac{9}{1}=9:1$$


Find each of the following ratios in the simplest form.
$$2.4\ km\  ,\ 900\ m$$ 

$$2.4\ Km:900\ m$$

$$=2.4\times 1000\ m$$$$:900\ m$$         ( $$1\ km$$$$=$$$$1000\ m$$)
$$= 2400:900$$
$$=24:9$$
$$=3\times 8:3\times 3$$
$$=8:3$$

$$\therefore$$ The simplest form of $$2.4\ km:900\ m$$ is $$8:3$$


Write each of the following ratios in the simplest form:
$$3\ weeks:30\ days$$

$$3$$weeks$$:30$$days
$$=3\times 7$$days$$:30$$days
$$=21$$days$$:30$$days
$$=3\times 7:3\times 10$$
$$=7:10$$
$$\therefore$$ the simplest ratio is $$7:10$$


Express each of the following ratios in the simplest form:
$$480:348$$

$$480:348$$
$$=4\times 120:4\times 87$$
$$=120:87$$
$$=3\times 40:3\times 29$$
$$=40:29$$


Find each of the following ratios in the simplest form.
$$84\ paise\ to\ Rs.3$$ 

$$84$$paise$$:3$$rupees
$$=84$$paise$$:3\times 100$$paise since $$1$$ rupees$$=100$$paise.
$$=84$$paise$$:300$$paise
$$=4\times 21$$paise$$:4\times 75$$paise
$$=21$$paise$$:75$$paise
$$=3\times 7$$paise$$:3\times 25$$paise
$$=7$$paise$$:25$$paise
$$\therefore$$ the simplest form of $$84$$paise$$:3$$rupees is $$7$$paise$$:25$$paise or $$7:25$$ 



Two numbers are in the ratio $$11:12$$ and their sum is $$460$$. Find the numbers.

Let the numbers be $$11x,12x$$ then their sum $$=11x+12x=460$$
$$\Rightarrow 23x=460$$
$$\therefore x=\dfrac{460}{23}=20$$
$$\therefore$$ the numbers are $$11x=11\times 20=220$$ and $$12x=12\times 20=240$$


Find each of the following ratios in the simplest form.
$$24\ to\ 56$$ 

$$24:56$$
$$=2\times 12:2\times 28$$
$$=12:28$$
$$=2\times 6:2\times 14$$
$$=6:14$$
$$=2\times 3:2\times 7$$
$$=3:7$$
$$\therefore$$ the simplest form of $$24:56=3:7$$


Express each of the following ratios in the simplest form:
$$36:90$$

$$36:90$$
$$=9\times 4:9\times 10$$
$$=4:10$$
$$=2\times 2:2\times 5$$
$$=2:5$$


Write each of the following ratios in the simplest form:
$$Rs.6.30:Rs.16.80$$

Rs$$6.30:$$Rs.$$16.80$$
$$=$$Rs$$6.30\times 100:$$Rs.$$16.80\times 100$$
$$=$$Rs$$630:$$Rs.$$1680$$
$$=$$Rs$$63:$$Rs.$$168$$
$$=3\times 21:3\times 56$$
$$=21:56$$
$$=7\times 3:7\times 8$$
$$= 3: 8$$
$$\therefore$$ the simplest form is $$3:8$$


Write each of the following ratios in the simplest form:
$$3\ m\ 5\ cm:35\ cm$$

$$3$$m$$5$$cm$$=3\times 100+5$$cm$$=300+5=305$$cm
$$3$$ m $$5$$ cm$$:35$$ cm
$$=305$$cm$$:35$$ cm
$$=305:35$$
$$=5\times 61:5\times 7$$
$$=61:7$$
$$\therefore$$ the simplest form is $$61:7$$


Find the ratio of the following:
98 to 63.

Required ratio is $$98:63$$

It can be simplified as:
$$=14\times 7:9\times 7$$

Dividing both sides by $$7$$, we get:
$$=14:9$$

Hence, the ratio of $$98$$ to $$63$$ is $$14:9$$.


Write each of the following ratios in the simplest form:
$$48\ min:2\ hours\ 40\ min$$

$$2$$hrs$$40$$mins$$=2\times 60+40$$mins$$=120+40=160$$mins
$$48$$ min$$:2$$ hours $$40$$ min
$$=48:160$$
$$=4\times 12:4\times 40$$
$$=12:40$$
$$=4\times 3:4\times 10$$
$$=3:10$$
$$\therefore$$ the simplest form is $$3:10$$


Divide 72 in the ratio $$2\frac { 1 }{ 2 } : 1\frac { 1 }{ 2 } $$


1896585_1441820_ans_3a82f6e8854840ba92ff5a2895c39f66.jpg


In a mixture of $$35$$ litres, the ratio of milk to water is $$4:1$$. If $$7$$ litres of water is added to the mixture, then find the ratio of milk and water in the resulting mixture.

Total mixture$$=35l$$

Amount of milk $$=4x$$

Amount of water $$=x$$

Therefore we have

$$x+4x=35l$$

$$\Rightarrow 5x=35l$$

$$\Rightarrow x=\dfrac{35}{5}l$$

$$x=7l$$

Amount of water $$=x=7l$$

Amount of milk $$=4x=4\times 7l=28l$$

If $$7$$ liters of water is added then the amount of water $$=14l$$

Amount of milk $$=28l$$

and the ratio of milk $$:$$ water will be

$$28l:14l$$

$$=2l:1l$$.


Write each of the following ratios in the simplest form:
$$1\ L\ 35\ mL:270\ mL$$

As we know that,
$$1\ L = 1000\ ml$$

So,
$$1\ L\ 35\ ml=1000\ ml+35\ ml$$
                   $$=1035\ ml$$

Now,
$$\begin{aligned}{}\frac{{1\ L\ 35\ ml}}{{270\ ml}} &= \frac{{1035\ ml}}{{270\ ml}}\\ &= \frac{{3 \times 3 \times 5 \times 23}}{{2 \times 3 \times 3 \times 3 \times 5}}\\ &= \frac{{23}}{{2 \times 3}}\\& = \frac{{23}}{6}\\ &= 23:6\end{aligned}$$


Write the following ratio in the simplest form:
$$4\ kg:2\ kg\ 500\ g$$

The given quantities are $$4\ kg$$ and $$2\ kg \ 500\ g$$
Converting the given quantities to same units by using the relation $$1\ kg=1000\ g$$
$$4\ kg=4\times 1000=4000\ g$$ 
$$2\ kg \ 500\ g=2000\ g+500\ g=2500\ g$$
Hence, $$4\ kg:2\ kg \ 500\ g=4000\ g:2500\ g$$
$$\Rightarrow 40:25$$
$$\Rightarrow 5\times 8:5\times 5$$
$$\Rightarrow 8:5$$
$$\therefore\ $$ The simplest form of the given ratio is $$8:5$$.



Medicine is packed in boxes, each weighing $$4$$kg$$500$$g.How many such boxes can be loaded in a van which cannot carry beyond $$800$$kg?

Weight of $$1$$ medicine box$$=4$$kg$$500$$g

$$=4$$kg$$+500$$g 

$$=4\times 1000$$g$$+500$$g since $$1$$kg$$=1000$$g

$$=4000+500=4500$$g

Total weight$$=800$$kg

$$=800\times 1000=8,00,000$$g since $$1$$kg$$=1000$$g

Number of boxes can be loaded$$=\dfrac{Total\,\, weight}{weight \,\, of \,\, 1 \,\, box}$$

$$=\dfrac{8,00,000}{4500}=177$$ Remainder is $$35$$ is left.We will ignore 
this

$$\therefore$$ Number of boxex$$=177$$


Find the number which bears the same ratio to $$\dfrac{7}{{33}}$$ that $$\dfrac{8}{{21}}$$ does to $$\dfrac{4}{{9}}$$.


       

$${\textbf{Step -1: Find the given ratio.}}$$

            $${\text{Given ratio of }}\dfrac{8}{{21}}{\text{ to }}\dfrac{4}{9}{\text{  = }}\dfrac{8}{{21}} \times \dfrac{9}{4} = \dfrac{6}{7} $$

$${\textbf{Step -2: Put the ratio according to the given data.}}$$

     $${\text{Let the required number to be }}\dfrac{x}{y}$$

            $$\text{Hence, we have}$$

            $$ \Rightarrow \left( {\dfrac{x}{y}} \right) \div {\text{ }}\left( {\dfrac{7}{{33}}} \right){\text{ }} = {\text{ }}\left( {\dfrac{6}{7}} \right)$$

            $$ \Rightarrow \left( {\dfrac{x}{y}} \right) \times \dfrac{{33}}{7} = \dfrac{6}{7} \\$$

            $$ \Rightarrow \dfrac{x}{y} = \dfrac{6}{7} \times \dfrac{7}{{33}}$$

            $$ \Rightarrow \dfrac{x}{y} = \dfrac{2}{{11}} $$

$${\textbf{ Hence, the required number is }} $$ $$ \mathbf{\dfrac{2}{{11}} .}$$



Reema has 24 notebooks and 18 books. Find the ratio of notebooks to books .

Given that, Reema has $$24$$ notebooks and $$18$$ books.
To find out: The ratio of notebooks to books

We know that,
Number of notebooks $$=24$$
and number of books $$=18$$

$$\therefore \ $$ The ratio of notebooks to books $$=24:18$$
Simplifying the ratio, we get:
$$4:3$$

Hence, the ratio of notebooks to books is $$4:3$$.


Find the ratio of Rs. $$5$$ to $$50$$ paise. 

$${\textbf{Step -1: Converting rupees to paise}.}$$

                 $${\text{We know that, 1 Rupee  =  100 paise}}$$

                 $$\therefore {\text{ 5 Rupees  =  }}\dfrac{{{\text{5 }} \times {\text{ 100}}}}{{\text{1}}}{\text{ paise}}$$

                 $$ \Rightarrow {\text{ 5 Rupees  =  500 paise}}$$     $$\quad \quad \text{......eqn (i)}$$

$${\textbf{Step -2: Substitute the above result and find the required ratio}.}$$

                 $${\text{Ratio of 5 Rupees}}{\text{ to 50 paise }}$$$$ = {\text{ }}\dfrac{{{\text{5 Rupees}}}}{{{\text{50 paise}}}}$$

                                                                      $$ = {\text{ }}\dfrac{{{\text{500 paise}}}}{{{\text{50 paise}}}}$$   $$\quad \quad \textbf{[From eqn (i)]}$$

                                                                      $$ = {\text{ 10:1}}$$

$${\textbf{Thus, the ratio of 5 Rupees to 50  paise is 10:1}.}$$



Find the ratio of the following.
$$30$$ minutes to $$1.5$$ hours.


1318476_1586124_ans_be93febf5fa3475696696d55050581ad.jpg


Ravi walks $$6$$km in an hour while Roshan walks $$4$$km in an hour. What is the ratio of the distance covered by Ravi to the distance covered by Roshan?

distance covered by Ravi in $$1$$ hr = $$6$$ km

distance covered by Roshan in $$1$$ hr = $$4$$ km

ratio of the distance covered by Ravi in $$1$$ hr to  distance covered by Roshan in $$1$$ hr = $$6$$ km : $$4$$ km

 =$$6$$ : $$4$$
 = $$3$$ : $$2$$


Saurabh takes $$15$$ minutes to reach school from his house and Sachin takes one hour to reach school from his house. Find the ratio of the time taken by Saurabh to the time taken by Sachin.

when you compare two quantities then you must make sure that both the quantities should be expressed in the same unit.

Time taken by Saurabh to reach school = $$15$$ minutes
Time taken by Sachin to reach school = $$1$$hour = $$60$$ minutes

Now, the ratio of time taken by Saurabh to that taken by Sachin = $$15$$ / $$60$$
= $$1$$ / $$4$$


Reena wants to prepare vegetable pulao. she needs $$300$$ grams of rice. If she has to feed $$4$$ people. How much of rice is needed if the same pulao is prepared for $$7$$ people?

Rice needed to feed $$4$$ people $$= 300 g$$

Rice needed to feed $$1$$ people $$= \dfrac{300}{4} = 75 g$$

Rice needed to feed $$7$$ people $$= 75 \times 7 = 525 g$$ of rice



Cost of a toffee is $$50$$ paise and cost of a chocolate is $$Rs.10$$ . Find the ratio of the cost of a toffee to the cost of a chocolate.

$$\textbf{Step-1: Naming the variables}$$
                $$\text{Let the cost of toffee and chocolate be respectively}$$ $$x,y.$$
$$\textbf{Step-2: Finding the ratio}$$ 
               $$\text{Cost of chocolate in paise is, }10\times100$$               
               $$\dfrac{x}{y}=\dfrac{50}{10\times100}=\dfrac{1}{20}$$

$$\textbf{Hence, the ratio of cost of toffee to a chocolate is}$$ $$\mathbf{1:20.}$$


Find the ratio of number of notebooks to the number of books in a bag that has 6 notebooks, 4 textbooks , 8 diaries . 

The ratio of number of the notebook to the number of books in my bag is $$\dfrac{6}{4}$$
$$=3:2$$


A $$20$$ litre mixture of milk and water contains milk and water in the ratio $$3 : 2$$. $$10$$ litres of the mixture is removed and replaced with pure milk and the operation is repeated once more.
At the end of the two removals and replacement, what is the ratio of milk and water in the resultant mixture?

Let $$W_{o}$$ be the amount of water in the mixture with pure milk$$ = 8~l$$
Let $$W_{r}$$ be the amount of water in the mixture after the replacement have taken place.
$$\Rightarrow \dfrac{W_{r}}{W_{o}}=(1-\dfrac{R}{M})^{n}$$
Where R is the amount of mixture replaced by milk in each of the steps, M is the total volume of the mixture and n is the number of times the cycle is repeated.
Hence, $$\dfrac{W_{r}}{W_{o}}=(\dfrac{1}{2})^{2}=\dfrac{1}{4}$$
$$\therefore W_{r}=\dfrac{W_{o}}{4}=\dfrac{8}{4}=2~l$$
Hence, the mixture will have $$18l$$ of milk and $$2l$$ of water.
$$\therefore $$ ratio of $$milk$$ and $$water = 18:2=9:1$$


The length of a pencil is 16$$ { cm }$$ and its diameter is $$6mm$$ .What is the ratio of the diameter of the pencil to that of its length.

Length of a pencil $$=16 \ cm=16 \ cm \times 
10=160\ mm$$
Diameter of the pencil $$6\ mm$$
Ratio of the diameter of the pencil to that 
of its length $$=6:160=3:80$$


Find the number of students above twelve years of age in your class. Then, find the ratio of number of students with age above twelve years and the remaining students.

Total number of students in my class =  $$50$$
Total number of students above $$12$$ years of age = $$30$$
Ratio of number of students with age above twelve years and the remaining students = $$30$$ : $$20$$                                                                                                                                                 = $$3$$ : $$2$$



Find the ratio of the following:
$$30$$ minutes to $$1.5$$ hours.

Converting hours into minutes, we get$$1.5\ hr$$  = $$90\ min$$ . (1 hour = 60 minutes)

Ratio = $$30\ min$$ : $$90\ min$$ = $$1$$ : $$3$$



See the figure and find the ratio of number of circles to all the figures inside the rectangle.
1644637_7c99246f59c34a308176020d8eb1154a.png

Number of triangles$$=3$$
Number of circles$$=2$$
Number of squares$$=2$$
Total number of figures$$=7$$
Ratio of the number of circles to all the figures in the rectangle$$=2/7$$.


Find the ratio of its length to its breadth of rectangle with $$6$$ as length and $$12$$ as breadth.

Length of rectangle = $$6$$
breadth of rectangle = $$12$$
ratio =$$6$$ : $$12$$
 = $$1$$: $$2$$


Check whether the given ratios are equal, i.e., they are in proportion. If yes, then write them in the proper form.
$$15:45$$ and $$5:25$$.

$$\dfrac{15}{45}=\dfrac{5}{25}$$
$$\dfrac{1}{3}=\dfrac{1}{5}$$
$$\dfrac{1}{3} not \ equal \ to \dfrac{1}{5}$$


Which is greater in the following:
$$\cfrac{2}{3}$$, $$\cfrac{5}{2}$$

$$\cfrac{2\times 2}{3\times 2}=\cfrac{4}{6}$$ and  $$\cfrac{5\times 3}{2\times 3}=\cfrac{15}{6}$$
Since $$\cfrac{4}{6}< \cfrac{15}{6}$$ Therefore $$\cfrac{2}{3}< \cfrac{5}{2}$$


Which is greater in the following:
$$\cfrac{-5}{6}$$, $$\cfrac{-4}{3}$$

$$\cfrac{-5\times 1}{6\times 1}=\cfrac{5}{6}$$ and  $$\cfrac{-4\times 2}{3\times 2}=\cfrac{-8}{12}$$
Since $$\cfrac{-5}{6}> \cfrac{-8}{6}$$ Therefore $$\cfrac{-5}{6}> \cfrac{-4}{3}$$


See the figure and find the ratio of number of triangles to the number of circles inside the rectangle.
1644632_51dfb9b5f0d94236bdc8acdf7b68efef.png

Number of triangles$$=3$$
Number of circles$$=2$$
Number of squares$$=2$$
Ratio of the number of triangles to the number of circles$$=3/2$$.


Find the ratio of number of doors and the number of windows in your classroom.

In my class room
No. of doors are $$2$$
No. of windows are $$4$$
Ratio $$=\dfrac{2}{4}$$
 $$=1$$ : $$2$$


Find the ratio of the following.
$$30$$ minutes to $$45$$ minutes.

=$$30$$ minutes to $$45$$ minutes
=$$30$$ minutes : $$45$$ minutes
 =$$2$$ minutes : $$3$$ minutes

So , The Ratio is $$2$$ : $$3$$


Express the given ratio in its simplest form.
$$324 : 144$$.

$$324:144=\dfrac{324}{144}$$

                  $$=\dfrac{36\times 9}{36\times 4}$$

                  $$ = \dfrac{9}{4}$$.

                  $$=9:4$$



Find the ratio of the given statement in the simplest form.
$$2$$ dozen to $$3$$ scores.

We know that,
$$1 \;\text{dozen} = 12$$
$$1 \;\text{score} = 20$$.

Thus,
$$2 \;\text{dozen} : 3 \;\text{scores}$$ = $$2\times12 : 3\times20$$ 

                                = $$24 : 60$$

                                = $$\dfrac{24}{60}$$

                                = $$\dfrac{2}{5}$$

                                = $$2:5$$.


Express the given ratio in its simplest form.
$$85 : 391$$.

$$85:391=\dfrac{85}{391}$$

              $$=\dfrac{5}{23}$$

              $$=5:23$$


Find the ratio of the following in the simplest form.
$$8$$ kg to $$400$$ gm.

We know that,
$$1$$kg = $$1000$$gm

so,

$$8$$kg:$$400$$gm = $$8000$$ gm : $$400$$ gm.

                       = $$\dfrac{8000}{400}$$

                       = $$\dfrac{20}{1}$$
  
                     = $$20:1$$


In $$10$$ days, the earth picks up $$ 2.6 \times 10^{8} $$ pounds of dust from the atmosphere. How much dust will it pick up in $$45$$ days? (in terms of multiples of $$10^8$$ pounds)

Given, in $$10 $$ days the earth picks up $$2.6\times 10^8$$ pounds of dust from the atmosphere.

So, in $$1$$ day the earth picks up $$\dfrac{2.6\times 10^8}{10}$$ pounds of dust from the atmosphere.

Then, in $$45$$ days the earth will pick up $$\dfrac{2.6\times 10^8\times 45}{10}=11.7\times 10^8$$ pounds of dust from the atmosphere.

So, in terms of multiple of $$10^8,$$ the final answer will be $$11.7.$$


Express the given ratio in the language of daily life.
In India, the ratio of the number of villages to that of cities is about $$2000:1$$.

Given,
Ratio of villages and cities in India $$= 2000:1$$
It means, number of villages in india is $$2000$$ times that of number of cities in India.  


Express the given ratio in its simplest form.
$$60:72$$.

$$60:72=\dfrac{60}{72}$$.
              
              $$=\dfrac{5}{6}$$.
              
              $$=5:6$$



Find the ratio of the given statement in the simplest form.
$$3$$ weeks to $$3$$ days.

We know that,
$$1 week = 7 days$$.
So, 
$$3 weeks : 3 days$$ = $$3 \times 7 days : 3 days$$
 
                             = $$21 days : 3 days$$

                               = $$\dfrac{21}{3}$$

                                =$$\dfrac{7}{1}$$

                                = $$7 : 1$$


Which is greater in the following:
$$\cfrac{-3}{4}$$, $$\cfrac{2}{-3}$$

$$\cfrac{-3\times 3}{4\times 3}=\cfrac{-9}{12}$$ and  $$\cfrac{2\times (-4)}{-3\times (-4)}=\cfrac{-8}{12}$$
Since $$\cfrac{-9}{12}< \cfrac{-8}{12}$$ Therefore $$\cfrac{-3}{4}< \cfrac {2}{-3}$$


Which is greater in the following:
$$\cfrac{-1}{4}$$, $$\cfrac{1}{4}$$

$$\cfrac{-1}{4}< \cfrac{1}{4}$$ Since positive number is always greater than negative number.


Find the ratio of $$25$$ paise to Rs. $$60$$.

We know that,
1 rupee = 100 paise.

So, 
25 paise : Rs.60 = 25 : 60 $$\times$$100

                              =$$ 25 : 6000$$

                              = $$\dfrac{25}{6000}$$.

                              =$$\dfrac{1}{240}$$

                              = $$1 : 240$$
                       


Find the ratio of $$10$$ litres to $$0.25$$ litre.

10 liters : 0.25 liters = $$10 \times 100: 0.25 \times 100$$.

                                  =$$ 1000 : 25 $$

                                   =$$\dfrac{1000}{25}$$.

                                   =$$\dfrac{40}{1}$$

                                   =$$ 40 : 1$$
                 
      


Avinash works as a lecturer and earns Rs. $$12,000$$ per month. His wife who is a doctor earns Rs. $$15,000$$ per month.

Find the ratio of Avinash's income to the income of his wife ?

Given,
Earnings of Avinash $$= Rs.12,000$$.

Earnings of Avinash's wife $$=Rs. 15,000$$.

So, 

The ratio of Avinash's income to his wife $$= 12000: 15000$$.

                                                                     $$= \dfrac{12000} {15000}$$

                                                                      $$= \dfrac{4} {5}$$

                                                                      $$= 4:5$$.


Avinash works as a lecturer and earns Rs. $$12,000$$ per month. His wife who is a doctor earns Rs. $$15,000$$ per month.

Find the ratio of Avinash's income to their total income.

Given,
Earnings of Avinash $$=Rs. 12,000$$.

Earnings of Avinash's wife $$=Rs. 15,000$$.

We know that,

Total of income of Avinash and Avinash's wife $$=27,000$$. 

Thus,
The ratio of  Avinash's income to their total income $$=12000:27000$$

                                                                                       $$= \dfrac{12000} {27000}$$

                                                                                       $$= \dfrac{4} {9}$$

                                                                                        $$=4:9$$


The number of boys and girls in a school are $$1168$$ and $$1095$$ respectively. Express the ratio of the number of boys to that of the girls in the simplest form.

Given,
Number of boys in the school = $$1168$$.

Number of girls in the school =$$1095$$.

Number of boys to Number of girls = $$1168 : 1095.$$

                                                            =$$\dfrac{1168}{1095}$$.

                                                             =$$\dfrac{16}{15}$$

                                                           = $$16 : 15$$.





Find the ratio of the following in the simplest form.
$$48$$ min to $$2$$ hours $$40$$ min.

We know that,

$$1 hour = 60 minutes$$.

So, $$2 hours \,  40 min = 2 \times 60 + 40 minutes$$.

                                 $$=160 minutes$$.

Thus,

$$48 minutes : 2 hours \,  40 minutes$$ = $$48 minutes : 160 minutes$$.

                                                        = $$\dfrac{48}{160}$$

                                                        = $$\dfrac{3}{10}$$
   
                                                       =$$3:10$$. 


Find the ratio of $$10$$ metres to $$25$$ cm.

We know that,
1 meter = 100 cm.
So, 
10 meters: 25 cm = 10$$\times $$100cm : 25cm.

                              = $$1000 : 25$$

                              = $$\dfrac{1000}{25}$$.

                              =  $$\dfrac{40}{1}$$.

                              = $$40:1$$



Find the ratio of the following in the simplest form.
$$3$$m $$5$$cm to $$35$$cm.

We know that,
1 meter = 100 cm.

3 m 5 cm = 3 $$ \times $$ 100 cm + 5 cm.
                 = 305 cm.

Thus, 
3 m 5 cm : 35 cm = 305 cm : 35 cm.

                              = $$ \dfrac{305}{35}$$.

                              = $$ \dfrac{61}{7}$$

                              = 61 : 7.


Out of the $$72$$ persons working in an office, $$28$$ are men and the remaining are women.

Find the ratio of the number of men to that of women.

Given, 
Total number of people working in the office $$=72$$

Number of men in the office $$ = 28$$.

Number of women in the office $$=72-28$$.
                                                      $$=44$$.

Thus,
The ratio of number of men to the women in the office $$=28:44$$.

                                                                                             $$=\dfrac{28}{44}$$. 
                                                                                                    
                                                                                             $$=\dfrac{7}{11}$$.

                                                                                              $$=7:11$$.


Find the ratio of $$3.2$$ metres to $$56$$ metres.

3.2 meters : 56 meters = 3.2 $$ \times 10 $$ : 56 $$ \times 10 $$

                                        = $$32 : 560$$

                                        = $$\dfrac{32}{560}$$

                                        = $$\dfrac{4}{70}$$.

                                        = $$4 : 70$$


Which ratio is larger in the given pair?
$$15 : 16$$ or $$24 : 25$$.

Given ratios, $$15:16$$ and $$24:25.$$

On writing equivalent ratios.
$$15:16=15 \times 25: 16 \times 25$$
             $$=375:400$$
And
$$24:25=24\times16 : 25\times16$$
             $$=384:400$$      

Now, on comparing the numerator of the ratios as the denominator is same,
$$375:400<384:400$$
$$\Rightarrow 15:16< 24:25$$                                        


Out of the $$72$$ persons working in an office, $$28$$ are men and the remaining are women.
Find the ratio of the number of men to the total number of persons.

Total = Men + Women

$$72=28+\text{women}$$

Ratio $$=\dfrac{\text{Men}}{\text{Total}}$$

          $$=\dfrac{28}{72}$$

          $$=\dfrac{7}{18}$$


Which ratio is larger in the following pairs?
$$4 : 7$$ or $$5 : 8$$.

Given ratios,
$$4:7$$ and $$5:8$$.
On writing equivalent ratios.
=>$$4\times 8: 7 \times 8$$   and $$5\times7 : 8\times7$$        
=>$$ 32:56< 35:56$$   
=>$$4:7$$ < $$5:8$$                                               


Which ratio is larger in the given pair?
$$3 : 4$$ or $$9 : 16$$.

Given ratios are :
$$3:4$$ and $$9:16$$.

On writing the ratios into equivalent fractions with common denominators, we get
$$\dfrac{3\times 4}{4\times 4}=\dfrac{12}{16}$$    and $$\dfrac{9}{16}$$  
      
It can be observed $$ 12 > 9$$ 
  
Therefore$$3:4 > 9:16$$.                                                


Which ratio is larger in the given pairs?
$$9 : 20$$ or $$8 : 13$$.

Given Ratios,
$$9:20$$ and $$8:13$$.
On writing Equivalent ratios.
$$9 \times 13 : 20\times 13$$ and $$8\times20 : 13\times20$$.
$$117 : 260$$ and $$160:260$$
$$117<160$$
$$117 : 260$$ < $$160:260$$
$$9:20$$ < $$8:13$$.



The length of a steel tape for measurements of buildings is $$10$$m and its width is $$2.4$$cm. What is the ratio of its length to width?

Given,
Length of a steel tape $$= 10$$ $$m$$
                                       $$= 10\times 100 $$ $$cm$$             [$$\because 1 m = 100 cm$$]
                                       $$= 1000 cm$$

Width of a steel tape  $$= 2.4 $$ $$cm$$

Thus,
Ratio of length of the steel tape to its width $$ =1000:2.4 $$   

                                                                         $$ =\dfrac{1000}{2.4}$$                                                                                              

                                                                         $$ =\dfrac{10000}{24}$$  

                                                                         $$ =\dfrac{1250}{3}$$ 

                                                                         $$ =1250:3$$


An office opens at $$9$$ a.m. and closes at $$5$$ p.m. with a lunch interval of $$30$$ minutes. What is the ratio of lunch interval to the total period in office?

Given,
An office opens at $$9$$ $$ a.m.$$ and closes at $$ 5$$ $$p.m $$.
Lunch duration $$= 30 $$min.
So,
Total period in the office$$ = 8 hrs$$.
                                         $$ = 480 \,mins$$     [$$\because 1 hr = 60$$ minutes ]
Ratio of lunch interval to the total period in the office $$ = 30 : 480 $$.
                                                                                          $$ = 1 : 16 $$.



A bullock-cart travels $$24$$km in $$3$$ hours and a train travels $$120$$ km in $$2$$ hours. Find the ratio of their speeds.

We know that,
$$Speed = \dfrac{distance}{time}$$.

Given,

Speed of bullock car $$=\dfrac{24}{3} km/hr$$.

                                     $$=8$$ $$km/hr$$   

Speed of train car $$=\dfrac{120}{2} km/hr$$.

                                $$=60$$ $$ km/hr$$

Ratio of speed of bullock cart to train $$ = 8 : 60$$
                                                               $$ = 2 : 15$$


Which ratio is larger in the given pairs?
$$1 : 2$$ or $$13 : 27$$.

Given Ratios are :
$$1:2$$ and $$13:27$$.

Writing them as fractions, we get : $$\dfrac{1}{2},\dfrac{13}{27}$$ 

Converting them to fractions with same denominators :
 $$\dfrac{1 \times 27}{2 \times 27},\dfrac{13\times 2}{27 \times 2}$$

$$=\dfrac{27}{54},\dfrac{26}{54}$$ 

It can be observed that $$27> 26$$

Hence $$1:2$$ is larger than $$13:27$$


Divide Rs. $$1350$$ between Ravish and Shikha in the ratio $$2:3$$ and find the amount each get.

Given Rs. $$1350\,$$ is divided between Ravish and shikha in the ratio $$2:3$$
Let
Ravish $$=2x$$
Shikha $$3x$$

$$2x+3x=1350$$

$$5x=1350$$

$$x=\dfrac{5\times 270}{5}$$

$$x=270$$

Ravish money $$=2\times 270$$

                        $$=540$$

Shikha money $$=3\times 270$$

                         $$=810$$


Fill in the blank.
Ratio of Rs. $$2$$ and $$50$$ paise and Rs. $$3$$ and $$75$$ paise is _______.

Given,
$$\text{Rs. }2 \text{ and } 50 \text{ paise } = 2(100) + 50$$
                                   $$= 250 \text{ paise}$$   [1 Rupee = 100 paise]

$$\text{Rs. }3 \text{ and } 75 \text{ paise } = 3(100) + 75$$
                                   $$= 375 \text{ paise}$$

$$\text{Ratio } = \cfrac{250}{375}$$
             $$=\cfrac{2}{3}$$


Two numbers are in the ratio $$7:11$$. If $$7$$ is added to each of the numbers, the ratio becomes $$2:3$$. Find the numbers.

Given $$'2'$$ no's are in the ratio $$7:11$$

$$'7'$$ is added to these no's then the ratio $$2:3$$

Let 1st no= $$7x$$

2nd no= $$11x$$

$$\dfrac{7x+7}{11x+7}=\dfrac{2}{3}$$

$$3(7x+7)=2(11x+7)$$

$$21x+21=22x+14$$

$$22x-21x=21-14$$

$$x=7$$

1st no $$=7\times 7=49$$

2nd no $$=11\times 7=77$$


Divide Rs. $$2000$$ among $$P,Q,R$$ in the ratio $$2:3:5$$ and find the amount each person receive.


1512294_1675417_ans_263d8748fe1e467c9aa594de630d7dbf.jpg


A sum of $$Rs.5500$$ is to be divided between Raman and Aman in the ratio $$2:3$$. How much will each get?

Raman's share$$=2x$$
Aman's share$$=3x$$

Total$$=2x+3x=5500 \Rightarrow 5x=5500 \Rightarrow x=1100$$

Raman's share$$=2x=2200$$
Aman's share$$=3x=3300$$



The ages of two persons are in the ratio of $$5:7$$. Eighteen years ago their ages were in the ratio $$8:13$$. Find their present ages.

The ages of $$'2'$$ persons are in the ratio $$5:7$$
1st person age = $$5x$$
2nd person age = $$7x$$
$$'18'$$ years ago their ratio $$8:13$$

$$\dfrac{5x-18}{7x-18}=\dfrac{8}{13}$$

$$13(5x-18)=8(7x-18)$$

$$65x-234=56 x-144$$

$$65x-56x=234-144$$

$$9x=90$$

$$x=10$$

1st person age = $$5\times 10=50\,yrs$$

2nd person age = $$7\times 10=70\,yrs$$


The sides of a triangle are in the ratio $$1:2:3$$. If the perimeter is $$36\ cm$$, find its sides.

Since we are given that the sides are in the ratio $$1:2:3$$
So, we assume the sides of the triangle to be $$1\times t,2\times t,3\times t=t,2t,3t$$

Now,
As we are given that the perimeter is $$36\ cm$$
So, $$t+2t+3t=36\ cm$$
$$\Rightarrow 6t=36 \ cm$$
$$\Rightarrow t=6\ cm$$

Hence, the sides of the triangle are $$6\ cm , 12\ cm,18\ cm$$ respectively.


Two numbers are in the ratio $$2:7$$. If the sum of the numbers is $$810$$, find the numbers.

Given $$'2'$$ no's are in the ratio $$2:7$$

Let
1st no= $$2x$$

2nd no= $$7x$$

Given sum $$=810$$

$$2x+7x=810$$

$$9x=810$$

$$x=\dfrac{9\times 90}{9}$$

$$x=90$$

1st no $$=2\times 90=180$$

2nd no $$=7\times 90=630$$


The ratio of monthly income to the savings of a family is $$7:2$$. If the savings be of Rs. $$500$$, find the income and expenditure.

Since the ratio of monthly income to savings of the family is $$7:2$$, we assume that the income be $$7t$$ and savings be $$2t$$

Now, we are given that the savings is $$=Rs\ 500$$
So, According to our assumption, $$2t=500$$
                                                     $$\Rightarrow t=250$$

Hence, the income of the family is $$=7\times 250=Rs\ 1750$$

And the expenditure is $$= Income-Savings$$
                                       $$=Rs\ 1750-Rs\ 500$$
                                       $$=Rs\ 1250$$


Divide Rupees $$3,450$$ among $$A, B$$ and $$C$$ in the ratio $$3:5:7$$.

Given $$A:B:C=3:5:7$$

Sum of the ratios $$=3+5+7$$
$$=15$$

$$\therefore$$ Share of $$A=\dfrac3{15}\times 3450$$
$$=\dfrac{10350}{15}$$
$$=$$ Rupees $$690$$
Share of $$B=\dfrac5{15}\times 3450$$
$$=\dfrac{17250}{15}$$
$$=$$ Rupees $$1150$$
Share of $$C=\dfrac7{15}\times 3450$$
$$=\dfrac{24150}{15}$$
$$=$$ Rupees $$1610$$


In class, one out of every six students fails. If there are $$42$$ students in the class, how many pass.


1512455_1675477_ans_f86db23c357f42f2aaa00f8bc6203b50.jpg


Divide Rs. $$351$$ into two parts such that one may be to the other as $$2:7$$ and find the two parts.

We are asked to divide $$Rs\ 351$$ in two parts in the ratio $$2:7$$
So, we assume that the parts are $$2t,7t$$ respectively
Now, we know that the sum to be divided is $$351$$, so
$$\Rightarrow 2t+7t=351$$
$$\Rightarrow 9t=351$$
$$\Rightarrow t=39$$

Hence, the parts are $$2\times 39, 7\times 39$$ respectively
$$i.e \ Rs\ 78$$ and $$Rs\ 273$$


Divide $$81$$ into two parts so that one may be a multiple of $$8$$ and the other of $$5$$.


1891589_1743678_ans_d9a18b18eb304ef39bef8a7a0ebc2607.jpeg


Divide Rupees $$1,575$$ between Kamal and Madhu in the ratio $$7:2$$.

Given
Kamal : Mdhu $$=7:2$$

Sum of ratio $$=7+2=9$$

$$\therefore$$ Kamal's share $$=\dfrac79 \times 1575=\dfrac{11025}9$$$$=$$ Rupees $$1225$$
And, Madhu's share $$=\dfrac29\times 1575=\dfrac{3150}9$$$$=$$ Rupees $$350$$


Give two equivalent ratios of $$6:8$$.

Given: $$6:8$$
$$6\times 2:8\times 2=12:16$$
$$6\times 3:8\times 3=18:24$$
$$\therefore$$ the two equivalent ratios are $$12:16$$ and $$18:24$$


In the ratio $$7:8$$, if the consequent is $$40$$, what is the antecedent?

Let the antecedent and consequent be $$7x$$ and $$8x$$ respectively.
Then, consequent $$=40\Rightarrow 8x=40\Rightarrow x=5$$. 
Therefore, antecedent $$=7x=7\times 5=35$$


The ratio of zinc and copper in an alloy is $$7:9$$. If the weight of the copper in the alloy is $$11.7kg$$, find the weight of the zinc in the alloy.


1512423_1675457_ans_41c1daf420324469a0063dcdfe2fba90.jpg


Student of colony decided to go to an old age home in their vicinity to wish  Happy New year and get blessing from old people.
They carried the following items with them:
Bouquets $$63$$, New Year Cards $$70$$ and chocolates bars $$140$$. Answer the following question:
What is the ratio of number of bouquets to the number of chocolate bars?

Given,
Number of bouquets $$=63$$
Number of chocolates $$=140$$

$$\therefore$$ Required ratio $$=63:140$$
     $$=\dfrac {63}{140}=\dfrac{63\div7}{140\div7}\\=9:20$$


Divide Rs. $$880$$ between Rajan and kamal in the ratio $$(1/5):(1/6)$$

From the question
Sum of the ratios $$(1/5):(1/6)$$
$$=(1/5)+(1/6)$$
$$=(6+5)/30$$
$$=11/30$$
Total share Rajan has $$=Rs.880\times [(1/5)/(11/30)]$$
$$=880\times [(1/5)\times (30/11)]$$
$$=880\times (6/11)$$
$$=Rs. 480$$
Total share kama has $$=Rs.880\times [(1/6)/(11/30)]$$
$$=880\times [(1/6)\times (30/11)]$$
$$=880\times (5/11)$$
$$=Rs. 400$$


Student of colony decided to go to an old age home in their vicinity to wish  Happy New year and get blessing from old people.
They carried the following items with them:
Bouquets $$63$$, New Year Cards $$70$$ and chocolates bars $$140$$. Answer the following question:
What is the ratio of number of cards to the number of sum of all items?

Given,
Total number of cards $$=70$$
Sum of all items $$=63+70+140=273$$

$$\therefore$$ Required ratio $$=70:273=10:39$$


Divide Rs. $$360$$ between kunal and Mohit in the ratio $$7:8$$

From the question
Sum of the ratios $$7:8=7+8=15$$
Total share kunal has $$=Rs.360\times (7/15)=Rs.168$$
Total share Mohit has $$=Rs.360\times (8/15)=Rs.192$$


Find the ratio of each of the following in simplest form:
$$60\ paise$$ to $$4\ rupees$$

$$60$$ paise : $$4$$ ruppes
$$1$$ rupees $$=100$$ paise
$$\therefore 60$$ paise 
$$\dfrac{60}{100}=Rs. \dfrac 35$$
$$\dfrac{3}{5}$$ rupees : $$4$$ rupees
$$\dfrac{\dfrac 35}{\dfrac 41}=\dfrac 35\times \dfrac14=\dfrac{3}{20} =3:20$$


Divide Rs. $$5600$$ between $$A,B$$ and $$C$$ in the ratio $$1:3:4$$

From the question
Sum of the ratios $$1:3:4=1+3+4=8$$
Total share $$A$$ has $$=Rs. 5600\times (1/8)=Rs.700$$
Total share $$B$$ has $$=Rs. 5600\times (3/8)=Rs.2100$$
Total share $$C$$ has $$=Rs. 5600\times (4/8)=Rs.2800$$


Divide $$Rs\ 6000$$ among Irfan, Nagma and Ishan in the ratio $$3:5:7$$

Given,
Total amount $$=Rs\ 6000$$
Ratio of Irfan, Nagma and Ishan's share $$=3:5:7$$

Thus, Sum of ratios $$=3+5+7=15$$

$$\therefore$$ Irfan's share $$=Rs\ \dfrac{6000\times 3}{15}=Rs\ 1200$$

$$\therefore$$ Nagma's share $$=Rs\ \dfrac{6000\times 5}{15}=Rs\ 2000$$

$$\therefore$$ Ishan's share $$=Rs\ \dfrac{6000\times 7}{15}=Rs\ 2800$$


The energy content of different foods are as follows 
FoodEnergy content per kg
Wheat
Rice
Potatoes (cooked )
Milk		3.2 Joules
5.3 Joules
3.7 Joules
3.0 Joules

Which food provides the least energy and which provides the maximum? 
Calculate the ratio of least energy to the maximum energy.

From the above table it is clear that milk provides the least energy and rice provides the maximum.
$$ \therefore $$ ratio of the least energy to the maximum energy 
$$\Rightarrow  \dfrac { the \quad least \quad energy}{ the \quad maximum \quad energy } = \dfrac  {3}{5.3} = \dfrac  { 3 \times 10}{ 5.3 \times  10} = \dfrac {30}{53} $$


Heights of Anshul, Ankita and Dhruv are $$1.04\ m, 1.30\ m$$ and $$91\ cm$$ respectively. Divide $$100$$ sweets among them in the ratio of their heights.

Height of Anshul $$=1.04\ m=(1.04 \times 100)\ cm=104\ cm$$
Height of Ankita $$=1.30\ m=(1.30 \times 100)\ cm=130\ cm$$
Height of Dhruv $$=91\ cm$$
Ratio of their heights $$=104\ cm:130\ cm:91\ cm$$
$$104:130:91$$
$$=8:10:7$$ 
We are to divide $$100$$ sweets in the ratio $$8:10:7$$.
Sum of the terms of the ratio $$=8+10+7=25$$
$$\therefore $$ Anshul will get sweets $$=\dfrac{8}{25}$$ of $$100$$
$$=\dfrac{8}{25}\times 100=\dfrac{800}{25}=32=32$$ sweets
Ankita will get $$=\dfrac{10}{25}$$ of $$100$$
$$=\dfrac{10}{25}\times 100=\dfrac{1000}{25}=40=40$$ sweets
Dhruv will get $$\dfrac{7}{25}$$ of $$100$$
$$=\dfrac{7}{25}\times 100=\dfrac{700}{25}=28=28$$ sweets


Which ratio is greater?
$$5:6\ or\ 6:7$$

$$5:6\ or\ 6:7$$
$$5:6=\dfrac{5}{6}$$ and $$6:7=\dfrac{6}{7}$$
COnverting them into equivalent fraction by taking L.C.M of $$6\ and\ 7=42$$
$$\dfrac{5}{6}=\dfrac{5 \times 7}{6 \times 7}=\dfrac{35}{42}$$ and

$$\dfrac{6}{7}=\dfrac{6 \times 6}{7 \times 6}=\dfrac{36}{42}$$

As $$36 > 35 \therefore \dfrac{36}{42} > \dfrac{35}{42}$$
$$\Rightarrow \dfrac{6}{7} > \dfrac{5}{6}$$
Hence $$6:7$$ is the greater ratio.


Divide the angles of a triangle in the ratio $$2:3:4$$.

Given ratio $$=2:3:4$$
Sum of the terms of ratio $$=2+3+4=9$$
Sum of angles of a triangle $$=180^{\circ}$$
Thus, we are to divide $$180^{\circ}$$ in the ratio $$2:3:4$$
$$\therefore  \angle 1 =\dfrac 29 \times 180^{\circ}$$
$$=\dfrac{360^\circ}{9}=40^\circ$$
$$ \angle 2=\dfrac 39 \times 180^\circ$$
$$=\dfrac{540^\circ}{9}=60^\circ$$
$$ \angle 3=\dfrac 49 \times 180^\circ$$
$$=\dfrac{720^\circ }{9}=80^\circ$$
$$\therefore  \angle 1=40^\circ,  \angle 2=60^\circ$$ and $$ \angle 3=80^\circ$$


A pole of height $$3$$ metres is struck by a speeding car and breaks into two piece such that the first piece is $$\dfrac{1}{2}$$ of the second. Find the length of both pieces.

Total height of pole $$=3\ metres$$
Let length of $$2nd$$ piece $$=x$$
Length of $$1st$$ piece $$=\dfrac{1}{2} x$$
Ratio of length of two parts $$=\dfrac{1}{2} x: 1 x$$
$$=\dfrac{1}{2} : \dfrac{1}{1}$$
Taking L.C.M. of $$2$$ and $$1$$
$$=\left(\dfrac{1}{2} \times 2\right) $$ $$:\left(\dfrac{1}{1} \times 2\right)$$
$$=1:2$$
Sum of terms of ratio $$=1+2=3$$
$$\therefore$$ Length of $$1st$$ part $$=\dfrac{1}{3}\ of\ 3\ m$$
$$=\left(\dfrac{1}{3} \times 3 \right)m=1\ m$$
Length of $$2nd$$ part $$=\dfrac{1}{3}\ of\ 3\ m$$
$$=\left(\dfrac{2}{3} \times 3 \right)m=2\ m$$
Length of $$1st$$ part $$=1\ m$$
Length of $$2nd$$ part $$=2\ m$$


Is this pair of ratios equal? why?
$$\dfrac{8}{4}, \dfrac{2}{1}$$

Given ratios are,  $$\dfrac{8}{4}$$ and $$\dfrac{2}{1}$$

Here we have, 
$$  \dfrac{8}{4} = \dfrac{8\div 4}{4 \div 4} = \dfrac{2}{1}$$

$$\therefore \dfrac{8}{4}=\dfrac{2}{1}$$

Hence, the given ratios are equal.


Which ratio is smaller?
 $$9:13$$ or $$7:11$$

The given ratios are: 
$$9:13=\dfrac{9}{13}$$ and $$7:11=\dfrac{7}{11}$$
 
L.C.M of $$13$$ and $$11=143$$

Converting them into like fraction

$$\dfrac{9}{13}=\dfrac{9\times 11}{13\times 11}=\dfrac{99}{143}$$

and 
$$\dfrac{7}{11}=\dfrac{7\times 13}{11\times 13}=\dfrac{91}{143}$$


As, $$91 < 99$$

$$\Rightarrow \dfrac{91}{143} < \dfrac{99}{143}$$

$$\Rightarrow \dfrac{7}{11} < \dfrac{9}{13}$$

Hence, $$ 7:11$$ is the smaller ratio.


Which ratio is greater?
$$13:24\ or\ 17:32$$

$$13:24\ or\ 17:32$$
$$13:24=\dfrac{13}{24}$$ and $$17:32=\dfrac{17}{32}$$
Converting them into equivalent fraction by taking L.C.M of $$24\ \text{and}\ 32=96$$
$$\dfrac{13}{24}=\dfrac{13 \times 4}{24 \times 4}=\dfrac{52}{96}$$ and

$$\dfrac{17}{32}=\dfrac{17 \times 3}{32 \times 3}=\dfrac{51}{96}$$

As $$52 > 51$$

$$ \therefore \dfrac{52}{96} > \dfrac{51}{96}$$

$$\Rightarrow \dfrac{13}{24} > \dfrac{17}{32}$$

$$ \therefore 13:24$$ is the greater ratio.


Which ratio is larger $$10:21$$ or $$21:93$$?

To compare the ratios, convert the ratios to common denominators by taking LCM:
$$\begin{aligned}&10: 21=\frac{10}{21} \\&21: 93=\frac{21}{93}\end{aligned}$$
LCM of $$21$$ and $$93=651$$
Make the denominators of the fractions $$651$$
$$\begin{aligned}&\frac{10}{21}=\frac{10 \times 31}{21 \times 31}=\frac{310}{651} \\&\frac{21}{93}=\frac{21 \times 7}{93 \times 7}=\frac{147}{651} \\&\text { Clearly } \frac{310}{651}>\frac{147}{651} \\&\therefore 10: 21>21: 93 \\&\therefore 10: 21 \textbf{ is larger than } 21: 93\end{aligned}$$


  720 is divided between x and y in the ratio 4:How many rupees will each get?

It is given that Total amount = ₹ 720
Ratio between x and y = 4: 5
We know thatThe sum of ratios = 4 + 5 = 9
So x’s share $$= \dfrac49$$ of $$₹\ 720 = ₹\ 320$$
Similarly y’s share $$= \dfrac59$$ of $$₹\ 720 = ₹\ 400$$


A line is divided into two parts in the ratio 2.5: 1.If the smaller one is 35.1 cm, find the length of the line.

It is given that Ratio between two parts of a line = 2.5: 1.3
Multiplying by $$10= 25: 13$$
Here the sum of ratios $$= 25 + 13 = 38$$
Length of smaller one $$= 35.1\ cm$$
So the length of the line $$= \dfrac{38}{13 }× 35.1$$
We get $$= 38 × 2.7\ cm= 102.6\ cm$$


Express each of the following ratios in its simplest form 
$$ 48 : 54 $$ 

$$ 48 : 54 $$
$$ 24 : 27 $$
$$ 8:9 $$


The population of a town is 180,000, out of which males are 1/3 of the whole population. Find the
number of females. Also, find the ratio of the number of females to the whole population.

It is given that Total population = 180000
So the population of males $$=\dfrac 13 \times 180000 = 60, 000$$
Similarly the population of females = 180000 – 60000 = 120000
Here the ratio of females to whole population = 120000: 180000 = 2: 3


The ratio between two quantities is 3:If the first is 810, find the second.

It is given that the ratio between two quantities = 3: 4
So the sum of ratio = 3 + 4 = 7
Here the second quantity $$=\dfrac{ (810 × 4)}3$$
We get = 270 × 4 = ₹ 1080


Divide $$Rs. 1290$$ into $$A, B$$ and $$C$$ such that $$A$$ is $$\dfrac{2}{5}$$ of $$B$$ and $$B: C = 4: 3$$.

Given that, 
$$B: C = 4: 3$$ 
so, $$\dfrac{B}{C} = \dfrac{4}{3}$$
 $$\Rightarrow C = \dfrac{3}{4} B.....(i)$$

And, $$A = \dfrac{2}{5} B.....(ii)$$

Also given that, total amount$$=Rs.1290$$
So,
$$A + B + C = 1290$$
 $$\Rightarrow\dfrac{2}{5} B + B +  \dfrac{3}{4} B = 1290$$

$$\Rightarrow \dfrac{4\times2B + 20B + 5\times3B}{ 20} = 1290$$

$$\Rightarrow \dfrac{8B + 20B + 15B}{ 20} = 1290$$

$$\Rightarrow 43B = 1290 \times 20$$

$$\Rightarrow B = \dfrac{1290 \times 20}{43}$$

$$\Rightarrow B = 600$$

So,
$$A = \dfrac{2}{5} B= \dfrac{2}{5}\times Rs.600 $$
                 $$=Rs.240$$
And,
$$C = \dfrac{3}{4} B= \dfrac{3}{4}\times Rs.600 $$
                 $$=Rs.450$$

Hence, $$A$$ gets $$Rs. 240$$, $$B$$ gets $$Rs. 600$$ and $$C$$ gets $$Rs. 450$$.


300 is divided between A and B in such a way that A gets half of B. Find:
the share of A and the share of B.

We know that
Sum of the ratios = 1 + 2 = 3
Share of A $$= \dfrac {(300 × 1)}3 = ₹ 100$$
Share of B $$= \dfrac{(300 × 2)}3 = ₹ 200$$


₹ $$300$$ is divided between $$A$$ and $$B$$ in such a way that $$A$$ gets half of $$B$$. Find:
the ratio between the shares of $$A$$ and $$B$$.

Amount divided between $$A$$ and $$B$$ = $$₹. 300$$
We know that $$A$$ gets half of $$B$$
So the ratio between the shares of $$A$$ and $$B$$ = $$\dfrac {1}{2}:1 = 1: 2$$


Compare the following pair of ratios:
$$\frac{5}{18},\frac{17}{121}$$

$$5*121=605$$
$$18*17=306$$
Now, $$605>306$$
$$\dfrac{5}{18}>\dfrac{17}{121}$$


In two alloys, the ratios of copper to zinc are $$5 : 2$$ and $$3 : 4$$ (by weight). How many kg of the first alloy and of the second alloy should be alloyed together to obtain $$28$$ kg of a new alloy with equal contents of copper and zinc?

Let $$x$$ be the amount of first alloy mixed
let $$y$$ be the amount of second alloy mixed
Copper in $${ 1 }^{ st }$$ alloy $$=\dfrac { 5 }{ 7 } x$$

Copper in $${ 2 }^{ nd }$$ alloy $$=\dfrac { 3 }{ 7 } x$$

Total copper $$=\dfrac { 5x+3y }{ 7 } $$

Zinc in $${ 1 }^{ st }$$ alloy $$=\dfrac { 2 }{ 7 } x$$

Zinc in $${ 2 }^{ nd }$$ alloy $$=\dfrac { 4 }{ 7 } x$$

Total Zinc $$=\dfrac { 2x+4y }{ 7 } $$
Both copper & zinc should be in $$\left| : \right| $$ ratio $$\dfrac { \dfrac { 5x+3y }{ 7 }  }{ \dfrac { 2x+4y }{ 7 }  } =1$$
$$5x+3y=2x+4y$$
$$3x=y$$ $$\longrightarrow (I)$$
we know both alloy should make $$28kg$$
$$x+y=28$$ $$\longrightarrow (II)$$
where $$y=3x$$
           $$x+3x=28$$
           $$x=7kg$$ $$\longrightarrow { 1 }^{ st }$$ alloy
           $$y=21kg$$ $$\longrightarrow { 2 }^{ nd }$$ alloy



If $$bc:ac:ab = 1:3:5,$$ then find $$ \dfrac{a}{{bc}}:\dfrac{b}{{ca}}$$.

Given:
$$bc:ac:ab= 1:3:5$$

Let say $$\rightarrow bc=x$$ ; $$ac=3x$$ ; $$ab=5x$$

If we multiply all then -

$$bc\times ac\times ab= (x)(3x)(5x)$$

$$a^{2}b^{2}c^{2}=15x^{3}$$

$$abc =x\sqrt{15x}$$ ___ (i)

$$\because bc=x$$; so, $$a(x)=x\sqrt{15x}\Rightarrow a=\sqrt{15x}$$

similarly $$\because ac=3x$$; 

$$abc= x\sqrt{15x}$$

$$b(ac)=x\sqrt{15x}$$

$$b(3x)=x\sqrt{15x}\Rightarrow b=\dfrac{1}{3}\sqrt{15x}$$

Ratio :
$$\Rightarrow \displaystyle \frac{a}{bc}:\frac{b}{ca}$$

$$\displaystyle \Rightarrow \frac{\sqrt{15x}}{x}:\frac{\frac{1}{3}\sqrt{15x}}{3x}\Rightarrow \frac{1}{1}:\frac{1}{3\times 3}\Rightarrow 1:\frac{1}{9}$$

$$\Rightarrow 9:1$$ Answer


Divide $$0.1$$ in the ratio $$1:1.$$

Given ratio=$$1:1$$.
So, $$0.1$$ is divided equally.
$$1^{st}$$ part$$=2^{nd}$$ part$$=\dfrac{1}{2}\times \dfrac{1}{10}=0.05$$ Ans



There are two alloys of platinum, silver and gold. The first alloy is known to contain $$40$$ per cent of platinum and the silver alloy $$26$$ per cent of copper. The percentage of gold is the same in both alloys. Having alloyed $$150$$ kg of the first alloy and $$250$$ kg of the second, we get a new alloy which contains $$30$$ per cent of gold. How many kg of platinum are there in the new alloy?

FIRST ALLOY
$${ 1 }^{ st }$$ Alloy $$150kg$$ is used.
$$40$$% is platinum in first alloy
$$\dfrac { 40 }{ 100 } \times 150=60kg$$ of PLATINUM
SECOND ALLOY
$${ 2 }^{ nd }$$ ALLOY $$250kg$$ is used
$$26$$% of silver $$=\dfrac { 26 }{ 100 } \times 250=65kg$$
New alloy $$=150+250=400kg$$ has $$30$$% gold $$=\dfrac { 30 }{ 100 } \times 400=120kg$$ gold
since both contain same % of gold $$x$$% of $$150$$ $$+$$ $$x$$% of $$250=120$$
$$\dfrac { x }{ 100 } \times 150+\dfrac { x }{ 100 } \times 250=120$$
$$1.5x+2.5x=120$$
$$x=30$$%
In second alloy $$100-(30+26)=44$$% of platinum is there
$$44$$% of $$250$$ $$=\dfrac { 44 }{ 100 } \times 250=110kg$$
In new Alloy Platinum is $$110+60kg$$ $$=170kg$$


Mr Sahal and his wife are both school teachers and earn Rs. 16800 and Rs. 10500 per month respectively.  Find the ratio of 
(i) Mr Sahai's income to his wife's income.
(ii) Mrs Sahal's income to her husband's income
(iii) Mr Sahai's income to the total income of the two.


1297867_1544493_ans_086f7ddc408b460484e6fcda8127d4f8.jpg


Find the ratio of the price of coffee to that of tea  when coffee rs  24 per 100 g and tea costs rs 80 per kilogram.


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Class 6 Maths Extra Questions