Whole Numbers - Class 6 Maths - Extra Questions

Every natural number is a whole number, if true then enter $$1$$ and if false then enter $$0$$

Natural Numbers are $$ 1, 2, 3....$$ and so on.

Whole numbers are $$ 0, 1, 2, 3... $$ and so on.

Hence, every natural number is a whole number.


Predecessor of $$801010$$ is _____

Predecessor $$801010 $$
$$801010- 1 = 801009$$


Find the additive inverses of the following integers:
$$6, 9, 123, -76, -85, 1000$$

Two number are called additive inverse of each other if sum of them is $$0$$
So additive inverse of any number $$a$$ is $$-a$$
Hence additive inverse of $$6,9,123,-76,-85,1000$$ are $$-6,-9,-123,76,85,-1000$$ respectively 


Find the integer $$m$$ in the following:
$$m + 25 = 15$$

We have, $$m+25=15$$
$$\Rightarrow m+25-25=15-25$$, subtract $$25$$ from each sides
$$\Rightarrow m = -10$$, simplify 
Hence required solution is $$m=-10$$


How many rational numbers are there strictly between $$0$$ and $$1$$ such that the denominator of the rational number is $$80$$?

We know that, $$\dfrac{0}{80}=0$$ and $$\dfrac{80}{80}=1$$
So, the numbers between $$0$$ and $$1$$ having denominator $$80$$ are $$\dfrac{1}{80}, \dfrac{2}{80}, \dfrac{3}{80}, \dfrac{4}{80}, ... \dfrac{78}{80}, \dfrac{79}{80}$$
Therefore, there are $$79$$ such positive rational numbers.


Write the additive inverse of : $$\dfrac{2}{7}$$

The additive inverse of a number is the number, when added to the given number, gives a sum of zero.
So, the additive inverse of $$\displaystyle \frac{2}{7}$$ is  $$\displaystyle -\frac{2}{7}$$ 


Restate the following statements with appropriate conditions, so that they become true statements.
For any $$x, x^3\geq 0$$.

If $$x\geq 0$$ then  $$x^{3}\geq 0$$ the statement  becomes true
this can be verified as follows,
 If  $$x=0$$ then  $$x^{3}= 0$$ 
If $$x=1$$Then  $$x^{3}=1$$
If  $$x=2$$  then $$x^{3}=8$$
hence,$$x^3>0$$ only if $$x>0$$


Write down $$15$$ rational numbers which are equivalent to $$\dfrac{11}{5}$$ and the numerator not exceeding $$180$$

We know that  A number will be unchanged if you multiply and divide it by same non- zero number.
So multiply and divide given number by 2 to 16
Hence 15 rational numbers which are equivalent to $$\dfrac{11}{5}$$ and denominator  not exceeding $$180$$
are $$\dfrac{11\times 2}{5\times 2}$$, $$\dfrac{11\times 3}{5\times 3}$$, $$\dfrac{11\times 4}{5\times 4}$$,$$\dfrac{11\times 5}{5\times 5}$$,$$\dfrac{11\times 6}{5\times 6}$$,$$\dfrac{11\times 7}{5\times 7}$$,$$\dfrac{11\times 7}{5\times 7}$$,$$\dfrac{11\times 8}{5\times 8}$$$$\dfrac{11\times 9}{5\times 9}$$,$$\dfrac{11\times 10}{5\times 11}$$,$$\dfrac{11\times 12}{5\times 12}$$,$$\dfrac{11\times 13}{5\times 13}$$,$$\dfrac{11\times 14}{5\times 14}$$,$$\dfrac{11\times 15}{5\times 15}$$,$$\dfrac{11\times 16}{5\times 16}$$

or  $$\dfrac{22}{10}$$, $$\dfrac{33}{15}$$, $$\dfrac{44}{20}$$, $$\dfrac{55}{25}$$, $$\dfrac{66}{30}$$, $$\dfrac{77}{35}$$, $$\dfrac{88}{40}$$, $$\dfrac{99}{45}$$, $$\dfrac{110}{50}$$, $$\dfrac{121}{55}$$, $$\dfrac{132}{60}$$, $$\dfrac{143}{65}$$, $$\dfrac{154}{70}$$, $$\dfrac{165}{75}$$, $$\dfrac{177}{80}$$


Earlier you have studied decimals $$0.9, 0.8$$. Can you write these as rational numbers?

Yes, we can write decimals like $$0.9, 0.8$$ as rational numbers.
Such as $$0.9 = \dfrac{0.9\times 10}{10}=\dfrac{9}{10}$$ and $$0.8 = \dfrac{0.8\times 10}{10}=\dfrac{8}{10}$$.


Name the property involved in the given example: $$\dfrac{2}{5} + \dfrac{1}{3} = \dfrac{1}{3} + \dfrac{2}{5}$$

The property used here is called as Associative Property.
If $$A$$ and $$B$$ are the two rational numbers which are to be added, then associative property states that: $$A + B = B + A$$


How many trees are on the right side of the house?
695079_80a273c04568498ebfb788de82574d10.PNG

The number of trees on the right side of first house is 1 
whereas the number of trees on the right side of second house is 3


Find the sum by suitable rearrangement:
$$238 + 695 + 162 $$ 

$$238+695+162$$
$$=230 + 8 + 690 + 5 + 160 + 2$$
$$=(230+160)+690+(2+8)+5$$
$$=390+(690+10)+5$$
$$=(390+700)+5$$
$$=(1090+5)$$
$$=\boxed{1095}$$


Join the points A, B, C, D, D', C', B' and A in order, so as to form a closed figure. Write down the equation of the line of symmetry of the figure formed.


So the line of symmetry is $$x=0$$ as it can be seen from the figure

750437_728504_ans_b243012514404ff2a3b6aa4d5f30b831.png


Reflect points B, C, D on the y-axis and write down their coordinates. Name the images as B', C', D' respectively.

The graph is shown in the figure.
750248_728503_ans_9a19e88265454d9e9ee9be3f92e983ec.png


Find the sum  $$ 154 + 197 + 46 + 203$$

We have,
$$154+197+46+203$$
$$=(154+46)+(197+203)$$ ....... appplying associative property
$$=200+400$$
$$=600$$

Hence, this is the answer.


Verify whether commutative property is satisfied for addition, subtraction, multiplication and division of the following pairs of rational numbers.
(i) $$4$$ and $$\dfrac {2}{5}$$
(ii) $$\dfrac {-3}{7}$$ and $$\dfrac {-2}{7}$$

(i)Addition
$$4+\dfrac{2}{5}=\dfrac{20+2}{5}=\dfrac{22}{5}$$
$$\dfrac{2}{5}+4=\dfrac{2+20}{5}=\dfrac{22}{5}$$
Commutative property is satisfied.
Subtraction
$$4-\dfrac{2}{5}=\dfrac{20-2}{5}=\dfrac{18}{5}$$
$$\dfrac{2}{5}-4=\dfrac{2-20}{5}=\dfrac{-18}{5}$$
Commutative property is not satisfied.
Multiplication
$$\dfrac{2}{5}\times 4=\dfrac{8}{5}$$
$$4\times \dfrac{2}{5}=\dfrac{8}{5}$$
Commutative property is satisfied.
Division
$$4\div \dfrac{2}{5}=4\times \dfrac{5}{2}=\dfrac{20}{2}$$
$$\dfrac{2}{5}\div 4=\dfrac{2}{5}\times \dfrac{1}{4}=\dfrac{2}{20}$$
Commutative property is not satisfied.
(ii)Addition
$$\dfrac{-3}{7}+\left ( \dfrac{-2}{7} \right )=\dfrac{-3-2}{7}=\dfrac{-5}{7}$$
$$\dfrac{-2}{7}+\left ( \dfrac{-3}{7} \right )=\dfrac{-2-3}{7}=\dfrac{-5}{7}$$
Commutative property is satisfied.
Subtraction
$$\dfrac{-3}{7}-\left ( \dfrac{-2}{7} \right )=\dfrac{-3+2}{7}=\dfrac{-1}{7}$$
$$\dfrac{-2}{7}-\left ( \dfrac{-3}{7} \right )=\dfrac{-2+3}{7}=\dfrac{1}{7}$$
Commutative property is not satisfied.
Multiplication
$$\dfrac{-2}{7}\times \dfrac{-3}{7}=\dfrac{6}{7}$$
$$\dfrac{-3}{7}\times \dfrac{-2}{7}=\dfrac{6}{7}$$
Commutative property is satisfied.
Division
$$\dfrac{-3}{7}\div \dfrac{-2}{7}=\dfrac{-3}{7}\times \dfrac{7}{-2}=\dfrac{3}{2}$$
$$\dfrac{-2}{7}\div \dfrac{-3}{7}=\dfrac{-2}{7}\times \dfrac{7}{-3}=\dfrac{2}{3}$$
Commutative property is not satisfied.


Find the square root:
(1) $$\displaystyle \frac{36}{49}$$
(2) $$\displaystyle \frac{484}{625}$$
(3) $$\displaystyle 5\frac{19}{25}$$
(4) 72.25
(5) 39.69

(1) $$\displaystyle \frac{6}{7}$$
(2) $$\displaystyle \frac{22}{25}$$
(3) $$\displaystyle \frac{12}{5} = 2\frac{2}{5}$$
(4) 8.5
(5) 6.3


Find the integer m in the following:
$$m + 28 = -49$$

We have, $$m+28=-49$$
$$\Rightarrow m+28-28=-49-28$$,  subtract $$28$$ from each sides
$$\Rightarrow m = -77$$, simplify 
Hence required solution is $$m=-77$$


Write the smallest natural number. Can you write the largest natural number?

The smallest natural number is $$1$$, not $$0$$ because $$0$$ was invented by Aryabhatta.
The largest natural number cannot be written because natural number goes up to infinite.


Is it right to say that the median of first 'n' natural numbers is $$\cfrac { n+1 }{ 2 } $$? Justify?

No, because if n is even, the number $$\left( \cfrac { n+1 }{ 2 }  \right) $$ will not be a natural number.


Write all the primes less than $$25$$, to which number set the square roots of these prime numbers belongs to?

The prime numbers less than $$25$$ are $$2,3,5,7,11,13,17,19,23$$
The square roots of prime numbers are Irrational numbers.


Write the following numbers in word:
$$(i) 5,27,864 $$                             $$ (ii) 95,432$$

$$1)5,27,864=$$ five lakhs twenty seven thousand eight hundred sixty four
$$2)95,432=$$ ninty five thousand four hundred thirty two


How are the quantum numbers $$n$$, $$1$$ and $$m_{1}$$ arrived at? Explain the significance of these quantum numbers.?

Quantum numbers are a way to describe the discrete states that a particle can hold. These can then be used for solutions to the Schrodinger equation(above).

Principal Quantum Number (n)—this designates the size of the orbital, also called the shell. The larger the value for $$n$$, the greater the average distance of an electron in the orbital from the nucleus and therefore the larger the orbital. (I could always remember this trend by relating it back to Coulomb’s Law of Attractive Forces.) From the equation that defines the line spectrum of Hydrogen, we can also know that the value for n represents energy levels.

Angular Momentum Quantum Number (l)—this describes the shape of the orbital, but again, you can relate it back to the periodic table. The values of $$l$$ are integers that depend on the value of the principal quantum number, $$n$$. For any given value of $$n$$, the possible range of values for $$l$$ go from $$0$$ to $$n - 1$$. If $$n = 1$$, there’s only one possible value of $$l$$; that is, $$0$$ ($$n - 1$$ where $$n = 1$$). If $$n = 2$$, there’s two values for $$l$$ : $$0$$ and $$1$$. If $$n = 3$$, $$l$$ has three values: $$0$$, $$1$$, and $$2$$.

Magnetic Quantum Number (ml)—this describes the orientation of the orbital in space, or the direction along the axes in which it faces. Within a sub-shell, the value of this depends on the value of $$l$$. For a certain value of $$l$$, there are $$(2l + 1)$$ integral values, or $$-l, … 0, … +l$$

982990_1057706_ans_b1899061f08c41d3844ce0839733ca7e.JPG


Between which two numbers in the tenths place on the number line does each of the given number lies
$$0.17$$, $$0.54$$, $$0.61$$, $$0.78$$

$$i)0.10<0.17<0.20$$
$$ii)0.50<0.54<0.60$$
$$iii)0.50<0.61<0.60$$
$$iv)0.70<0.78<0.80$$

Hence, all the values lie between $$0.1\ to\ 0.8$$


Change these numbers into special numbers
$$(i)\ 28\ \\ (ii)\ 132$$

Solution:-
(i) $$28$$
Given no. $$= 28$$
Turning the no. from back to front, i.e., $$82$$
Adding both of them, i.e.,
$$82 + 28 = 110$$
Since $$110$$ is not a special no.
Again turning from back to end, i.e. $$011$$
Now again adding both, i.e.,
$$110 + 011 = 121$$
We got $$121$$ which is the required special no.
(ii) $$132$$
Given no. $$= 132$$
Turning the no. from back to front, i.e., $$231$$
Adding both of them, i.e.,
$$132 + 231 = 363$$
We got $$363$$ which is the required special no.


Find the following product.
$$2dc^{2}\times -6dc$$

$$2dc^2\times (-6dc)$$
$$=2\times (-6)\times d\times d\times c^2\times c$$
$$=-12\times d^{1+1}\times c^{2+1}$$
$$=-12\cdot d^2\cdot c^3$$.

1195547_1242425_ans_4905b8acb2cf499d82dde4c8a374e5ac.jpg


Write down three consecutive predecessors of $$7510001$$.  

three consecutive predecessors of $$7510001$$ are  
 $$7509998,7509999,7510000$$


Write the following numbers in Hindu-Arabic system.
(i) $$1256345$$          (ii) $$12027896$$

(i) $$12,56,345$$
Twelve lakh fifty six thousand three hundred forty five.
(ii) $$1,20,27,896$$
One crore twenty lakh twenty seven thousand eight hundred ninety six.


List first $$n$$ natural numbers

The natural (or counting) numbers are$$1,2,3,4,5.....$$ etc.
There are infinitely many natural numbers.
Hence the first $$5$$ natural numebrs are $$1,2,3,4,5$$


Use distributive law to find the value of 
$$1063 \times 128 - 1063 \times 28$$.

Distributive law-
$$a(b+c)=ab+ac$$

So,
$$1063\times 128-1063\times 28=1063\;(128-28)$$

$$=1063\times 100$$
$$=106300$$


Add $$8+(-5)$$ on number line.

REF.Image
$$\therefore 8+(-5)=3$$

1089444_1143717_ans_c662fbe223d84c31bac5080db7d1a7ed.png



Write first ten Natural Numbers.

First ten natural numbers are ... 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.


Write first 10 Whole Numbers.

Whole numbers start with $$0$$.
First ten whole numbers are :$$ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.$$


Find how many three digits natural numbers are divisible by $$5$$.

Three digit number are $$100,105.........995$$
a$$= 100$$                            (a= first term)
d$$= 105-100=5$$         (d=common difference)
$$\begin{array}{l} \Rightarrow { a_{ n } }=a+(n-1)d \\ \Rightarrow 995=100+(n-1)5 \\ \Rightarrow 995-100=(n-1)5 \\ \Rightarrow \frac { { 895 } }{ 5 } =n-1 \\ \Rightarrow 179=n-1 \\ \Rightarrow n=179+1 \\ \Rightarrow n=180 \end{array}$$
Hence,
$$180$$ three digit natural number are divisible by 5.



Find the sum of the smallest five-digit and six-digit numbers?

Smallest $$6$$ digit number is $$100000$$

largest $$6$$ digit number is $$999999$$

sum is $$100000 + 999999 = 1099999$$

1194415_1294986_ans_6289feb644a246d3842575d2d258e0ea.jpg


If $$x - \frac{1}{x} = 3$$ then find the value of $$x ^ { 3 } - \dfrac { 1 } { x ^ { 3 } }$$

$$\begin{array}{l} We\, have \\ x-\frac { 1 }{ x } =3 \\ Now, \\ { \left( { x-\frac { 1 }{ x }  } \right) ^{ 3 } }={ x^{ 3 } }-\frac { 1 }{ { { x^{ 3 } } } } -3\times x\times \frac { 1 }{ x } \left( { x-\frac { 1 }{ x }  } \right)  \\ { \left( 3 \right) ^{ 3 } }={ x^{ 3 } }-\frac { 1 }{ { { x^{ 3 } } } } -3\left( 3 \right)  \\ { x^{ 3 } }-\frac { 1 }{ { { x^{ 3 } } } } ={ \left( 3 \right) ^{ 3 } }+3\times 3 \\ =27+6 \\ =36 \\ Hence, \\ { x^{ 3 } }-\frac { 1 }{ { { x^{ 3 } } } } =36. \\  \end{array}$$


Find the ratio of:
$$15\ kg$$ to $$210\ g$$

$$\dfrac { 15kg }{ 210g } =\dfrac { 15\times 1000 }{ 210\times 1 } =\underline { 500:7 } $$

Hence, this is the answer.


Separate the periods of the numeral  $$'93574862'$$  by commas and write it in words.

$$93574862 \rightarrow 9, 35, 74,862$$
In words - Nine crore thirty five Lakh seventy four thousand eight hundred sixty two.


From the following pairs of numbers, find the reduced form of ratio of first number second number.
$$52,78$$


$$\frac{52}{78}\times \frac{13}{13}=\frac{4}{6}=\frac{2}{3}$$





































































































































































































1188303_1344241_ans_c5e9429959c04d2db5e573100773cc97.jpg


Find the reduced form of the ratio of the first quantity to second quantity.
$$5\ litre,\ 2500\ ml$$

$$\frac{5litre}{2500ml}=\frac{5\times 1000ml}{2500ml}=\frac{50}{25}=2$$
1188307_1344270_ans_2a837ba9ebac46308b88dfa7ccf9e6cb.jpg


Find the coordinates of the points $$A,B,C,D$$ from the following graph.
1253714_09ef4ce6c76f493f984a8e672593e3ec.png

A = (2,4)
B = (2,-5)
C = (-2,1)
D = (-3,-5)


Express the following presentages as ratios in the reduced form.
$$44:100$$

$$\frac{44}{100}\times \frac{4}{4}=\frac{(\frac{44}{4})}{(\frac{100}{4})}=\frac{11}{25}$$
1188319_1344310_ans_1e9afa1b1d174d92b2199b4e4ee5c733.jpg


Answer the following question:
The counting numbers are also called ________

The counting numbers are $${Z}^{+}$$, i.e., $$\left\{ 1, 2, 3, ....., \infty \right\}$$ and thus are also known as Natural numbers.


2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Let number of rows = x
Each row contain plants = x
Total plants = 2025
Therefore,
$$x^{2} = 2025$$
$$x=45$$.


Solve
$$\left(\dfrac {5}{3}\right)^{3/2}$$

We have,
$${ \left( \dfrac { 5 }{ 3 }  \right)  }^{ 3/2 }$$
$$=\sqrt { \dfrac { { 5 }^{ 3 } }{ { 3 }^{ 3 } }  } $$
$$=\sqrt { \dfrac { 125 }{ 27 }  } $$

Hence, this is the answer.


Fill in the blanks:
The smallest $$7-$$digits number in the system_____

The smallest 7 - digit number in the system is 1000000.
1212896_1365749_ans_36f465e8b29440239a3d286747a4bfee.JPG


$$B = \{ x | 5 x - 2 = 0 , x \in N \}$$


1320169_1535218_ans_6314648decf1441da9eb96f67d383b95.jpeg


$$(i)$$Which is the smallest whole number?
$$(ii)$$Which is the smallest natural number ?

$$(i)$$Smallest whole number $$= 0$$
$$(ii)$$Smallest natural number $$= 1$$


Rajni is 6th from either end of a row of girls. How many girls are there in that row ?

Rajni is in $$6th$$ place.
So, $$n$$  $$= 6$$(From front) $$+6$$(From back) $$-1$$(Rajni is counted twice)$$=11$$


Solve:
$$\dfrac {1}{7}+\dfrac {3}{14}$$

We have,
$$\dfrac { 1 }{ 7 } +\dfrac { 3 }{ 14 }$$
$$ \Rightarrow \dfrac { 2 }{ 14 } +\dfrac { 3 }{ 14 } \Rightarrow \boxed { \dfrac { 5 }{ 14 }  } $$

Hence, this is the answer.


Show that the number $$4^{n}$$, when n is a natural number cannot end with the digit zero 

If any number ends with zero, then it must be divisible by 5
Hence, its prime factorization must contain 5
But we know,
$$4=2\times{2}$$
So, there is no natural number $$n$$ for which $$4^n$$ can end with zero


Is there any natural number which has no successor? 

Every natural number has successor.
There are infinite natural numbers & hence there is no last natural number.
Hence, every natural number has successor.


How many whole numbers are there between $$32$$ and $$53$$?

Whole numbers between $$32$$ and $$53$$ is $$20$$ $$(53-32-1=20)$$
$$(33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52)$$


A student multiplied $$7236$$ by $$65$$ instead of multiplying by $$56$$. By how much was his answer greater than the correct answer?

(Hint: Do you need to do both the multiplications?)
Difference between $$65$$ and $$56=9$$
Difference in the answer$$=7236\times 9$$
$$7236$$
$$\times 9$$
_________
$$65124$$
_________
Therefore, His answer was greater than the correct answer by $$65,124$$.


Is there any natural number that has no predecessor?


$$\textbf{Step 1: Checking for the smallest natural number.}$$
                $$\text{The range of natural numbers is from 1 to infinity}$$
                $$\therefore$$ $$ \text{The smallest natural number is 1.}$$

$$\textbf{Step 2: Checking for the predecessor of 1}$$
                $$\text{The predecessor of 1 i.e. } 1 - 1 = 0$$
                $$\text{But 0 is a whole number and not a natural number}$$
                $$\therefore \text{1 is a Natural number that has no predecessor.}$$

$$\textbf{Yes, there is a natural number that has no predecessor, which is 1.}$$


Are all natural numbers also whole numbers?

$$\textbf{Step -1: Comparing natural numbers and whole numbers.}$$
                $$\text{Whole numbers include all positive integers and 0}$$
                $$\text{Natural numbers include all positive integers}$$
                $$\therefore\text{All natural numbers are whole numbers}$$
$$\textbf{Hence , all natural numbers are also whole numbers.}$$


Which is the greatest whole number?

So, $$0, 1, 2, 3,….,10$$ are called whole numbers. So, 1 is the smallest natural number and 0 is the smallest whole number. But there is no largest whole number or natural number because each number has its successor.
There is no largest whole number because each number has its successor.


Are all whole numbers also natural numbers?

$$\textbf{Step -1: Comparing natural numbers and whole numbers.}$$
                $$\text{Whole numbers include all positive integers and 0}$$
                $$\text{Natural numbers include all positive integers}$$
                $$\therefore\text{All whole numbers aren't natural numbers because 0 is a whole number but not a natural number.}$$
$$\textbf{Hence , all whole numbers are not natural numbers.}$$


Which is the smallest whole number?

Step 1: To find
(1) Smallest whole number

Step 2: Smallest whole number
Numbers $$0, 1, 2, 3,….,10,......$$ are called whole numbers. Therefore, $$0$$ is the smallest whole number. 


Find the product using suitable properties.
$$738\times 103$$.

$$738\times 103=738\times (100+3)$$
$$=738\times 100+738\times 3$$  $$(Distributive$$ $$property)$$
$$=73800+2214$$
$$=76014$$.


Find the value of the following.
$$81265\times 169-81265\times 69$$.

$$81265\times 169-81265\times 69$$
$$=81265\times (169-69)$$
$$=81265\times 100$$
$$=8126500$$.


Find the value of the following.
$$54279\times 92+8\times 54279$$.

$$54279\times 92+8\times 54279$$
$$=54279\times 92+54279\times 8$$
$$=54279\times (92+8)$$
$$=54279\times 100$$
$$=5427900$$.


Find $$15\times 68$$ using distributive property.

$$15\times 68=15\times (60+8)$$
$$=15\times 60+15\times 8$$  $$(Distributive\space property)$$
$$=900+120$$
$$=1020$$.


Find the value of the following.
$$3845\times 5\times 782+769\times 25\times 218$$.

$$3845\times 782+769\times 25\times 218$$
$$=3845\times 5\times 782+769\times 5\times 5\times 218$$
$$=3845\times 5\times 782+3845\times 5\times 218$$
$$=3845\times 5\times (782+218)$$
$$=19225\times 1000=19225000$$.


Find $$17\times 23$$ using distributive property.

$$17\times 23=17\times (20+3)$$
$$=17\times 20+17\times 3$$(Distributive property)
$$=340+51$$
$$=391$$.


Find the product using suitable properties.
$$854\times 102$$.

Step 1: To find
(1) Product of $$854$$ and $$102$$ using suitable properties.

Step 2: Product
$$854\times 102=854\times (100+2)$$
                   $$=854\times 100+854\times 2$$          (using Distributive property)
                   $$=85400+1708$$
                   $$=87108$$.


Find the value of the following.
$$297\times 17+297\times 3$$.

$$297\times 17+297\times 3$$
$$=297\times (17+3)$$
$$=297\times 20$$
$$=5940$$.


Find $$69\times 78+22\times 69$$ using distributive property.

$$69\times 78+22\times 69$$
$$=69\times (70+8)+22\times (60+9)$$ $$(Distributive$$ $$property)$$
$$=69\times 70+69\times 8+22\times 60+22\times 9$$
$$=4830+552+1320+198$$
$$=6900$$.


Find the product using suitable properties.
$$258\times 1008$$.

$$258\times 1008=258\times (1000+8)$$
$$=258\times 1000+258\times 8$$  $$(Distributive$$ $$property)$$
$$=258000+2064$$
$$=260064$$.


Find using distributive properly.
$$4275\times 125$$.

$$4275\times 125=(4000+200+100-25)\times 125$$
$$=4000\times 125+200\times 125+100\times 125-25\times 125$$
$$=500000+25000+12500-3125$$
$$=534375$$.


How many whole numbers are there between $$1032$$ and $$1209$$?

Number of whole numbers between two whole numbers $$a$$ and $$b$$ is given by $$b-a-1$$.
To find the whole numbers between $$1032$$ and $$1209$$
$$=(1209 - 1032) - 1 \\= 177 - 1$$
$$ = 176$$
Hence $$176$$ whole numbers lie between $$1032$$ and $$1209$$


Find using distributive properly.
$$5437\times 1001$$.

$$5437\times 1001$$
$$=5437\times (1000+1)$$
$$=5437\times 1000+5437\times 1$$
$$=5437000+5437=5442437$$.


Find using distributive properly.
$$728\times 101$$.

$$728\times 101$$
$$=728\times$$$$(100+1)$$
$$=728\times 100+728\times 1$$
$$=72800+728=73528$$.


Add the following numbers and check by reversing the order of the addends:
$$2359 + 548$$

$$2359 + 548$$
$$2359 + 548 = 2907$$
Reversing the order of the addends, we get
$$548 + 2359 = 2907$$
$$\therefore 2359 + 548 = 548 + 2359$$


Write three consecutive whole numbers occuring just before $$320001.$$

The three consecutive whole numbers occuring just before $$320001$$ are:
$$320001 - 1 =  320000 - 1 = 319999 - 1 = 319998$$
These are : $$320000, 319999, 319998$$


 Determine the sums given below using suitable rearrangement.
$$1 + 2 + 3 + 4 + 96 + 97 + 98 + 99$$

$$1 + 2 + 3 + 4 + 96 + 97 + 98 + 99$$
Using associative property of addition
$$(1 + 2 + 3 + 4) + (96 + 97 + 98 + 99) = 10 + 390 = 400$$
1+2+3+4+96+97+98+99


Find the product using suitable properties.
$$1005\times 168$$.

$$1005\times 168=(1000+5)\times 168$$
$$=1000\times 168+5\times 168$$  $$(Distributive$$ $$property)$$
$$=168000+840=168840$$.


Find using distributive property:
$$824\times 25$$

To find:  $$824\times 25$$

Now,
$$824\times 25$$
$$=(800+20+4)\times 25$$
$$=800\times 25+20\times 25+4\times 25$$
$$=20000+500+100$$
$$=20000+600$$
$$=20600$$

Hence, the answer is $$20600$$ .


Add the following numbers and check by reversing the order of the addends:
$$19753 + 2867$$

$$19753 + 2867$$
$$19753 + 2867 = 22620$$
Reversing the order of the addends,
$$2867 + 19753 = 22620$$
$$\therefore 19753 + 2867 = 2867 + 19753$$


Natural number are closed under ____ and under ____.

Natural number are closed under Addition and under multiplication


$$32 \times (27 \times 19)=(32 \times $$ ______$$) \times 19$$

Let, the member is $$K$$
$$32 \times (27 \times 19)=(32 \times K) \times 19$$
$$K=\dfrac{\not {32} \times 27 \times \not {19}}{ \not {32} \times \not {19}}=27$$


Smallest natural number is ..

Smallest natural number is $$1$$.


Fill in the blanks:
The additive identity in whole numbers is............

$$7750 \div 17$$
On dividing $$7750$$ by $$17$$, we get
$$Quotient = 455 and Remainder = 15$$
Check by division algorithm:
$$Divident = Divisor \times Quotient + Remainder$$
$$= 17 \times 455 + 15 = 7750$$


Using suitable properties, evaluated the following:
$$724\times (-56)+(-724)\times 44 $$

$$724\times (-56)+(-724)\times 44 $$
Using distributive law of multiplication
$$=724\times (-56-44)$$
$$=724\times 100$$
$$=72400$$


Can a rectangular number be a square number ?

$$1 \times 9 + 1 = 10$$
$$12 \times 9 + 2 = 110$$
$$123 \times 9 + 3 = 1110$$
$$1234 \times 9 + 4 = 11110$$
$$12345 \times 9 + 5 = 111110$$


Fill in the blanks:
$$(-10)\times ((-15)\times 33)=$$______ $$\times (-15)+(-10)\times 33$$

$$(-10)\times ((-15)\times 33)=$$______ $$\times (-15)+(-10)\times 33$$
$$=(-10)\times \left\{(-15)+33\right\}$$
$$=(-10)\times (-15)+(-10)\times 33$$ (Distributive law of multiplication)


Find the rule which gives the number of matchsticks required to make the following matchsticks patterns.
Use a variable to write the rule.
A pattern of letter $$Z$$ as

From the above figure, it can be observed that the letter Z will require 3 matchsticks


Therefore,


The pattern is $$3n$$


1718743_1787032_ans_c94b4b2907f54cf5babc59040c0ead7f.PNG


$$54 + 234 = 234 + $$

$$54 + 234 = 234 + 54$$


Largest natural number is

Largest natural number is not possible to imagine as it is infinite


Smallest whole number is

Smallest whole number is $$0$$.


Largest whole number is ..

Largest whole number is not possible to imagine as it is infinite


All natural numbers are

All natural numbers are whole numbers.


$$332 + 497 = + 332$$

$$332 + 497 = 497 + 332$$


Represent the following whole numbers on a number line:
$$0, 3, 5, 8, 10$$


The given whole numbers $$0, 3, 5, 8, 10$$ can be represented on a number line as shown below:

1732635_1818632_ans_16a59727b11041bcb51f6dc40bc7eb6f.png


Is $$a + b + c = a + (b + c) = (b + a) + c$$ ?

Yes, for any set of three whole numbers if the sum of any two whole numbers is added to the third whole number, then whatever be the order, the sum will remain the same.


$$a + (b + c) = (a + ) + c$$

$$a + (b + c) = (a + b) + c$$


By re-arranging the given numbers, evaluate: $$162 + 253 +338 + 47$$

$$162 + 253 + 338 + 47$$
$$= (162 + 338) + (253 + 47)$$ (By associative law)
$$= 500 + 300$$
$$= 800$$.


Which property of addition is satisfied by:  $$8 + 7 = 15$$

$$8 + 7 = 15$$
The property used as closure property satisfied by $$15$$.


By re-arranging the given numbers, evaluate: $$237 + 308 + 163$$

$$237 + 308 + 163$$
$$= (237 + 163) + 308$$ (By associative law)
$$= 400 + 308$$
$$= 708$$.


Which property of addition is satisfied by:  $$3+ (5 + 4) = (3 + 5)+ 4$$

$$3 + (5 + 4) = (3 + 5) + 4$$,
$$3 + (5 + 4) = 3 + 9 = 12$$, and
$$(3 + 5) + 4 = 8 + 4 = 12$$
The property used as Associative law of addition satisfied by $$12$$.


Evaluate:  $$355 \times 18$$.

$$355 \times 18$$
This can be written as,
$$= (300 + 50 + 5) \times 18$$
$$= (300 \times 18) + (50 \times 18) + (5 \times 18)$$
$$= 5400 + 900 + 90$$
$$= 6390$$.


Evaluate:  $$6214 \times 12$$.

$$6214 \times 12$$
This can be written as,
$$= (6000 + 200 + 10 + 4) \times 12$$
$$= (6000 \times 12) + (200 \times 12) + (10 \times 12) + (4 \times 12)$$
$$= 72000 + 2400 + 120 + 48$$
$$= 74568$$.


Evaluate:  $$15 \times 49372$$.

$$15 \times 49372$$
This can be written as,
$$= 15 \times (40000 + 9000 + 300 + 70 + 2)$$
$$= (15 \times 40000) + (15 \times 9000) + (15 \times 300) + (15 \times 70) + (15 \times 2)$$
$$= 600000 + 135000 + 4500 + 1050 + 30$$
$$= 740580$$.


Let $$N$$ denote the set of all natural numbers and $$R$$ be the relation on $$N\times N$$ defined by $$(a, b) R (c, d)\Longleftrightarrow  ad (b+c)=bc(a+d)$$. Check weather $$R$$ is an equivalence relation.

Let R be defined on $$N\times N$$ as 
$$(a, b) R (c, d)\Longleftrightarrow  ad (b+c)=bc(a+d)$$.     ....(1)

Reflexivity:
We can write $$ab(b+a)= ba(a+b)$$ for all $$a, b\in N$$
Since, sum and product of natural numbers obeys commutative property
Hence, by def (1), we can write 
$$(a,b) R (a,b)$$   for all $$(a,b) \in N\times N$$
Hence, $$R$$ is reflexive.

Symmmetry :
Let $$(a,b) R (c,d)$$ 
$$\Rightarrow ad(b+c)=bc(a+d)$$
$$\Rightarrow da(c+b)=cb(d+a)$$    (Since, sum and product of natural numbers obeys commutative property)
or $$cb(d+a)=da(c+b)$$
$$\Rightarrow (c,d) R(a,b)$$
Hence, $$R$$ is symmetric 

Transitivity :
Let $$(a,b) ,(c,d) , (e,f) \in N\times N$$
Let $$(a,b)R (c,d)$$ and $$(c,d)R(e,f)$$
$$ad(b+c)=bc(a+d)$$ and $$cf(d+e)=de(c+f)$$
$$\Rightarrow \dfrac{ab}{a-b} =\dfrac{cd}{c-d}$$ and $$\dfrac{cd}{c-d}=\dfrac{ef}{e-f}$$
$$\Rightarrow \dfrac{ab}{a-b}=\dfrac{ef}{e-f}$$
$$\Rightarrow af(b+e) =be(a+f)$$
$$\Rightarrow (a,b)R(e,f)$$
Hence ,$$R$$ is transitive

$$\therefore R$$ is Equivalence Relation.


What are the properties of $$R$$ used in the following? $$0+0=0$$ 

We know that $$0$$ is the identity element under addition for the real numbers, since if $$a$$ is any real number then $$a + 0 = 0 + a = a$$

Now, since it is given that $$0+0=0$$, therefore, $$0$$ is the identity element.

Hence, the identity element property of addition is used in $$0+0=0$$.


Identify the property in the following statements:
$$8\cdot (6 + 5) = (8\cdot 6) + (8\cdot 5)$$

The distributive property is used to multiply a single term and two or more terms inside the parenthesis 
As 8 is distributed inside parenthesis with 6 and 5.
Fact: Distributive property $$a(b+c)=ab+ac$$
Hence $$8\cdot (6+5)=(8\cdot 6)+(8\cdot 5)$$ is distributive property 


Is $$\dfrac {3}{-2}$$ a rational number? If so, how do you write it in a form conforming to the definition of a rational number (that is, the denominator as a positive integer)?

$$\dfrac {3}{-2}$$ is not a rational number because in a rational number, denominator is always a positive integer and numerator can both be negative or positive. 

So, to make the fraction $$\dfrac {3}{-2}$$ a rational number, simply multiply the numerator and denominator by $$-1$$ as shown below:

$$\dfrac { 3\times -1 }{ -2\times -1 } =\dfrac { -3 }{ 2 }$$ 

Hence, $$\dfrac { -3 }{ 2 }$$ is a rational number.


Give an example to show that subtraction on the set of real numbers is not commutative.

Let the two numbers be $$4$$ , $$5$$
Here $$4-5 \ne 5-4$$
Which implies they are not commutative
Therefore, subtraction on set of real numbers is not commutative.


Write the additive inverse of : $$\dfrac{-11}{5}$$

The additive inverse of a number is that number, when added to the given number, gives a sum of zero.
So, the additive inverse of $$\displaystyle -\frac{11}{5}$$  is  $$\displaystyle \frac{11}{5}$$, as $$\displaystyle -\frac{11}{5}\, +\, \displaystyle \frac{11}{5}\, =\, 0$$ 


Find an irrational number between $$3$$ and $$4$$

$$\textbf{Step - 1 : Find an irrational number between 3 and 4.}$$
                    $$\text{We know,}{ \sqrt {ab} } \text{ is an irrational number between a and b.}$$
                    $$\text{Let, }a= 3 \text{ and } b= 4 $$
                    $$\therefore\sqrt {ab} = \sqrt {3\times 4}$$
                                 $$= 2\sqrt 3$$

$$\textbf{Hence, an irrational number between 3 and 4 is }\bf{2\sqrt 3.}$$


Write down ten rational numbers which are equivalent to $$\dfrac{5}{7}$$ and the denominator not exceeding $$80$$

Firstly, we multiply the numerator and denominator of the given fraction $$\dfrac {5}{7}$$ by $$2$$ and then we clearly see that the denominator is less than $$80$$

$$\dfrac { 5\times 2 }{ 7\times 2 } =\dfrac { 10 }{ 14 }$$

Now, we multiply the numerator and denominator of the given fraction $$\dfrac {5}{7}$$ by $$3$$ and then again the denominator is less than $$80$$

$$\\ \dfrac { 5\times 3 }{ 7\times 3 } =\dfrac { 15 }{ 21 }$$

Similarly, we get  the remaining eight rational numbers that are equivalent to $$\dfrac {5}{7}$$ and the denominator does not exceeds $$80$$ as shown below:

$$\\ \dfrac { 5\times 4 }{ 7\times 4 } =\dfrac { 20 }{ 28 }$$ 

$$\\ \dfrac { 5\times 5 }{ 7\times 5 } =\dfrac { 25 }{ 35 }$$ 

$$\\ \dfrac { 5\times 6 }{ 7\times 6 } =\dfrac { 30 }{ 42 }$$ 

$$\\ \dfrac { 5\times 7 }{ 7\times 7 } =\dfrac { 35 }{ 49 }$$ 

$$\\ \dfrac { 5\times 8 }{ 7\times 8 } =\dfrac { 40 }{ 56 }$$ 

$$\\ \dfrac { 5\times 9 }{ 7\times 9 } =\dfrac { 45 }{ 63 }$$ 

$$\\ \dfrac { 5\times 10 }{ 7\times 10 } =\dfrac { 50 }{ 70 }$$ 

$$\\ \dfrac { 5\times 11 }{ 7\times 11 } =\dfrac { 55 }{ 77 }$$ 


How many rational numbers are there strictly between $$0$$ and $$1$$ with the property that the sum of the numerator and denominator is $$70$$?

The rational numbers between $$0$$ and $$1$$ whose the sum of the numerator and denominator is $$70$$ are as follows:

$$\dfrac { 1 }{ 69 } =0.014$$ 

$$\dfrac { 2 }{ 68 } =0.029$$

$$\dfrac { 3 }{ 67 } =0.044$$

$$\dfrac { 4 }{ 66 } =0.0.06$$
$$\vdots$$
$$\vdots$$
$$\vdots$$
$$\vdots$$ 
$$\dfrac { 33 }{ 37 } =0.89$$

$$\dfrac { 34 }{ 36 } =0.94$$

Hence, there are a total of $$34$$ positive rational numbers  between $$0$$ and $$1$$ such that the sum of the numerator and denominator is $$70$$.


Name the property indicated in the following:
$$5 + 0 = 0 + 5 = 5$$

We know that for the set of real numbers, zero is said to be additive identity. It means when any real number is added to zero or zero is added to any real number, the result is that real number itself.

Here, a real number $$5$$ is added to $$0$$ or $$0$$ is added to the real number $$5$$, the result is that real number itself that is $$5$$.

Hence, $$0$$ is the additive identity in the given expression $$5+0=0+5=5$$.


Name the property involved in the given question: $$\dfrac{8}{5} + 0 = \dfrac{8}{5} = 0 + \dfrac{8}{5}$$

Its Commutative Property


If $$a = \dfrac {\sqrt {3} - 1}{ \sqrt {3} + 1}, b = \dfrac {\sqrt {3} + 1}{ \sqrt {3} - 1}$$, then find the value of $$a + b + ab$$.

$$a+b+{a}{b}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}.\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$$
$$=\dfrac{{(\sqrt{3}-1)}^2+{(\sqrt{3}+1)}^2}{(\sqrt{3}+1)(\sqrt{3}-1)}+1$$
$$=2\ast\dfrac{3+1}{3-1}+1$$
$$=5$$


Show that the sum of the first $$n$$ natural numbers is a perfect square, if $$n$$ is equal to $${ k }^{ 2 }$$ or $${ { k }^{ ' } }^{ 2 }-1$$, where $$k$$ is the numerator of an odd and $${ k }^{ ' }$$ the numerator of an even convergent to $$\sqrt { 2 } $$.

The sum of lot '$$n$$' natural number is $$\cfrac { n(n+1) }{ 2 } $$ 
This expression is a perfect square when $$n={ k }^{ 2 }$$, provided that $$\cfrac { { k }^{ 2 }+1 }{ 2 } $$ is a perfect square$$={ x }^{ 2 }$$(say)
$$\Rightarrow { k }^{ 2 }-2{ x }^{ 2 }+1=0$$
$$\Rightarrow \sqrt { 2 } =1+\cfrac { 1 }{ 2+ } \cfrac { 1 }{ 2+ } \cfrac { 1 }{ 2+ } .....$$
The values of $$k$$ are the numerators of the odd convergents, since the no. of quotients in the period is odd.
Again, the expression $$\cfrac { n(n+1) }{ 2 } $$ is a perfect square when $$n+1={ k }^{ { ' }^{ 2 } }$$ provided that $$\cfrac { { k }^{ { ' }^{ 2 } }-1 }{ 2 } =a$$a perfect square$$={ x }^{ 2 }$$.
$$\Rightarrow { k }^{ { ' }^{ 2 } }-2{ x }^{ 2 }=1$$
In this case the roots of equations are the numerators of the even convergents.


Find the value of subtraction using number line:
$$8-7$$


1308770_702790_ans_874b8eaa493d4a04865956447ccef126.jpeg


Identify value of the following subtractions using number line:
$$8-5$$


To get the value of $$(8-5)$$,  we have to go $$5$$ units to the left from $$8$$ on the number line. Thus the required answer is $$3$$
1308784_702797_ans_76c7cb64190c4c8b89ec90af487d820c.png


Value of the subtraction of two numbers $$7$$ and $$3$$ using number line.


1694583_702791_ans_c76f53fff8dc481d99fb7ad39b0eb22b.jpeg


Decompose the number 20 into two terms such that their product is the greatest.

We can be written $$a(10+10)$$ and $$10\times 10=100$$
In this case product will be maximum
Other number may be
$$11+9=20$$
$$11\times 9=99$$
$$18+2=20$$
$$18\times 2=36$$
$$17+3=20$$
$$17\times 3=57$$
$$16+4=20$$
$$16\times 4=64$$
$$15+5=20$$
$$15\times 5=75$$
$$14+6=20$$
$$14\times 6=84$$
$$13+7=20$$
$$13\times 7=91$$
$$12+8=20$$
$$12\times 8=96$$
$$11+9=20$$
$$11\times 9=99$$
$$10+10=20$$
$$10\times 10=100$$
Hence by hit and trial method its product $$(10+10)$$ is answer.


Find the value of the following subtractions using number line:
$$2-1$$


1694586_702792_ans_ab68e8eb962d44beb4946a020348738c.jpeg


Given, $$A=\left\{ x:-1<x\le -5,x\in R \right\} $$ and $$B=\left\{ x:-4<x\le 3,x\in R \right\} $$
Represent on different number lines:
(i) $$A\cap B$$
(ii) $$A'\cap B$$
(iii) $$A-B$$

$$A=\{x:-1<x\leq 5,x\in R\}$$
$$B=\{x:-4<x\leq 3,x\in R\}$$

$$(i)$$  $$A\cap B$$=the numbers which are both in A and B

$$\implies A\cap B=\{x:-1<x\leq 3,x\in R\}$$

$$(ii)$$  $$A'\cap B$$=the numbers which are not in A but are in B

$$\implies A'\cap B=\{x:-4<x\leq -1,x\in R\}$$

$$(iii)$$   $$A-B=$$ the numbers which is in A but not in B

$$\implies A-B\{x:3<x\leq 5,x\in R\}$$

866766_875523_ans_9f78ffad1f044c1ebce0b79b2463f28f.png


Find the product $$8739 \times 102$$ using distributive property.

$$8739\times 102$$

$$=8739\times(100+2)$$

$$=8739\times100+8739\times2$$

$$=873900+17478$$

$$=891,378$$


In which quadrant does this points $$(5,3) , (-2, 4)$$ lie?

$$(5,3)$$ both $$x$$ and $$y$$ are positive.
so, $$(5,3)$$ is in $$I^{st}$$ quadrant.
$$(-2,4)\, x$$ is negative and $$y$$ is positive. 
$$(-2,4)$$ is in II quadrant.

1038137_1061282_ans_986c2c9816d34db3aecef1d1e4c4f6d0.PNG


Write natural numbers from $$30$$ to $$45$$. What fraction of these numbers is odd?

Natural numbers from $$30$$ to $$45$$ are $$30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45$$.
There are $$16$$ numbers and $$8$$ of them are odd ones.
So $$\dfrac{1}{2}$$ of the numbers are odd.


Using $$13587 \times 93=1263591$$, find $$13587 \times 94$$ (use distributive property)

It is given that $$13587 \times 93 = 1263591$$, then,

$$13587 \times 94 = 13587 \times \left( {93 + 1} \right)$$

$$ = 13587 \times 93 + 13587 \times 1$$

$$ = 1263591 + 13587$$

$$ = 1277178$$



A cement company earns a profit of Rs. $$8$$ per bag of white cement sold and a loss of Rs. $$5$$ per bag of grey cement sold.
The company sells $$3,000$$ bags of white cement and $$5,000$$ bags of grey cement in a month. What is its profit or loss?

We are given that a cement company earns a profit of $$Rs.\ 8$$ per bag of white cement sold.

now, company sells $$3, 000$$ bags of white cement in a month. So, 
Total earn by company is, $$3000\times 8=24,000\ Rs$$

Now, we are also given that a cement company loss of $$Rs.\ 5$$ per bag or grey cement sold.

now company sells $$5000$$ bags of grey cement in a month. So,
Total Loss by company is, $$5000\times 5=25,000\ Rs$$

Now, the lossing parts is greater than the earning part. So company having loss

i.e, $$24000-25000=-100 < 0$$

So, company having loss of $$Rs.100$$


Manika makes 8 gift packs containing chocolates for Diwali. She puts 12 vanilla chocolates. 14 milk chocolates and 8 nut chocolates in each pack. With the help of which property, can you calculate the total number of chocolates in 8 packs?

We have,

Using $$BODMAS$$ property,

$$ 12+14+8=34 $$

$$ \text{Total}\,\text{number}\,\text{of}\,\text{chocolates = 34}\times \text{8=262} $$

Hence, this is the answer.



Find the number of natural numbers less than $$1000$$ and divisible by $$5$$ can be formed with ten digit, each digit not occurring more than once in each number is.


1116288_1140174_ans_83eec67c9c3c4c94a18a77b498cd9e06.jpeg


Simplity $$2\times(71\times5)$$ by using associate property

According to Associative Property
$$2\times(71\times5)=(2\times71)\times5=(2\times5)\times71=10\times71=710$$


Find $$285 \times 98$$ using distributive property ?

$$285\times 98=285\times  (90+8)$$

               $$=25650+2280$$

               $$=27930$$


If $$a, b, c, d, e, f, g, h$$ are eight consecutive natural numbers, then prove that
$${a^2} + {d^2} + {f^2} + {g^2} = {b^2} + {c^2} + {e^2} + {h^2}$$

Solution -
Let numbers be $$ a,a+1,a+2,----a+7 $$
Their square sum $$ = a^{2}+(a+1)^{2}+(a+2)^{2}+----(a+7)^{2} $$
$$ = 8a^{2}+140+56a $$
Now LHS 
$$ a^{2}+d^{2}+f^{2}+g^{2} = a^{2}+(a+3)^{2}+(a+5)^{2}+(a+6)^{2}$$
$$ = 4a^{2}+70+28a $$
RHS, $$ b^{2}+c^{2}+e^{2}+h^{2} = (a+1)^{2}+(a+2)^{2}+(a+4)^{2}+(a+7)^{2} $$
$$ = 4a^{2}+70+28a $$
LHS = RHS 

1100699_1188172_ans_abcdacff3d9144b8b64aae7ce651cacf.jpg


A shopkeeper allows $$8$$% discount on his goods and still earns a profit of $$15$$%. If an article is sold for $$Rs.460$$, find
(i) the marked price 
(ii) the cost price.

$$SP=460Rs$$
 
$$Discount=8\%$$
 
$$MP=SP-Discount$$
 
$$MP=460-\dfrac{8}{100}\times 460$$
 
$$MP=460-36.8=423.3Rs$$
 
MP is $$15\%$$ higher than CP
 
$$MP=115\%$$ of $$CP$$
 
$$MP=\dfrac{115}{100}\times CP$$
 
$$MP\times \dfrac{100}{115}=CP$$
 
$$\dfrac{42320}{115}=CP$$
 
$$CP=368Rs$$



Whether natural number follow the commutative property under subtraction, Explain with one example?

No natural numbers do not follow commentative property under subtraction
Ex. $$5-2=3$$
$$2-5=-3$$
So $$3$$ and $$-3$$ are not equal.
So natural numbers do not follow commentative property under subtraction.

1165185_1194603_ans_39c836b5d1a9459d88feabb5ffe5b4bb.jpg


 solve : $$6 - 3 \dfrac { 1 } { 2 } - 2 \dfrac { 1 } { 5 }$$

$$6-3\dfrac { 1 }{ 2 } -2\dfrac { 1 }{ 5 } $$
$$=6-\dfrac { 7 }{ 2 } -\dfrac { 11 }{ 5 } $$
$$=\dfrac { 60-35-22 }{ 10 } $$
$$=\dfrac { 60-57 }{ 10 } $$
$$=\dfrac { 3 }{ 10 } $$


$$\left\{ {{{\left( {\frac{1}{3}} \right)}^{ - 3}} - {{\left( {\frac{1}{2}} \right)}^{ - 3}}} \right\} + {\left( {\frac{1}{4}} \right)^{ - 3}} = $$?

$$\left \{ (\dfrac{1}{3})^{-3}-(\dfrac{1}{2})^{-3} \right \}+(\dfrac{1}{4})^{-3}$$

$$\Rightarrow \left \{ 3^{3}-2^{3} \right \}+(4)^{3}$$

$$\Rightarrow 27-8+64$$

$$\Rightarrow 19+64=83$$


Find two consecutive odd natural numbers, the sum of whose squares is $$202$$. 

We have,

Let the two odd consecutive numbers $$x\,\text{and}\,x+2$$

So,

According to given question,

$$ {{x}^{2}}+{{\left( x+2 \right)}^{2}}=202 $$

$$ \Rightarrow {{x}^{2}}+{{x}^{2}}+4+4x=202 $$

$$ \Rightarrow 2{{x}^{2}}+4x=198 $$

$$ \Rightarrow {{x}^{2}}+2x=99 $$

$$ \Rightarrow {{x}^{2}}+2x-99=0 $$

$$ \Rightarrow {{x}^{2}}+\left( 11-9 \right)x-99=0 $$

$$ \Rightarrow {{x}^{2}}+11x-9x-99=0 $$

$$ \Rightarrow x\left( x+11 \right)-9\left( x+11 \right)=0 $$

$$ \Rightarrow \left( x+11 \right)x-9=0 $$


If,

$$ x+11=0 $$

$$ \Rightarrow x=-11 $$

If,

$$ x-9=0 $$

$$ x=9 $$

So,

$$x+2=9+2=11$$


Hence, this is the answer.



If abc-cba = 792, where abc is a 3-digit number, if a+b+c = 18 and a = 9c, what is the number?

We have, $$abc-cba=792$$
$$abc$$ is a $$3-digit$$ number.
Hence, $$abc=100a+10b+c$$
Similarly, $$cba=100c+10b+a$$
So, $$abc-bca=100a+10b+c-100c-10b-a$$
$$=99a-99c$$
$$=99(a-c)$$

So, $$99(a-c)=792$$
$$\Rightarrow a-c=8$$
Also, $$a=9c\quad [given]$$
$$\therefore\ 9c-c=8$$
$$\Rightarrow 8c=8$$
Hence, $$c=1$$

So, $$a=9(1)$$
Hence, $$a=9$$

Also, we are given, $$a+b+c=18$$
So, $$9+b+1=18$$
$$\therefore \ b=8$$

Hence, the number is $$981$$.


Find the ratio of the total number of animals with the farmer to the number of cows with him?

Let sheeps & cows be 8k and 3k respectively
$$\Rightarrow $$ Total animals = 11 k
$$\Rightarrow \frac{Total animals}{cows}=\frac{11k}{3k}=11:3$$

1181954_1293098_ans_af58d30d2f4a49538680cd90cdd404c2.jpg


Can the number $${4^n}$$, n being a natural number end with the digit 0? Give reasons.


1975312_1329690_ans_7ee32ada482e45d382f16b63dab1d18e.jpg


Use Euclid's Division Lemma to show that the square of any positive integer is either of the form  $$3 n$$  or  $$3 n + 1$$  for some integer  $$n.$$

$$\textbf{Step-1:  Assume the positive integers to be equal to some variables}$$ 
                 $$\textbf{& then apply the relevant concept of Euclid's Division Lemma.}$$

                 $$\text{As per Euclid's Division Lemma.}$$ 

                 $$\text{If}$$ $$a$$ $$\text{&}$$ $$b$$ $$\text{are}$$ $$2$$ $$+ve$$ $$\text{integers then}$$ $$a=bq+r$$ ($$\because$$   $$0\le r<b$$). 

                 $$\text{Let}$$ $$a=3,$$ $$\text{then r = 0,1,2}$$

                  $$\underline { r=0 } $$                            $$\underline { r=1 } $$                                      $$\underline { r=2 } $$
                  $$a=3q$$                    $$a=3q+1$$                                  $$a=3q+2$$
                 $${ a }^{ 2 }={ 9q }^{ 2 }$$              $${ a }^{ 2 }={ 9q }^{ 2 }+1+6q$$                      $${ a }^{ 2 }={ 9q }^{ 2 }+4+12q$$
                 $$=3\left( { 3q }^{ 2 } \right) $$            $$=3\left( { 3q }^{ 2 }+2q \right) +1$$                  $$=3\left( { 3q }^{ 2 }+4q+1 \right) +1$$
                 $${ a }^{ 2 }=3m$$              $${ a }^{ 2 }=3m+1$$                              $$=3m+1$$
                 $$\left\{ m={ 3q }^{ 2 } \right\} $$        $$\left\{ m={ 3q }^{ 2 }+2q \right\} $$                      $$\left\{ m={ 3q }^{ 2 }+4q+1 \right\} $$

$$\textbf{Hence, the square of any positive integer is either of the form 3m or 3m +1}$$


In a fraction, twice the numerator is $$2$$ more than the denominator.If $$3$$ is added to the numerator and to the denominator, the new fraction is $$\dfrac{2}{3}$$.Find the original fraction.

Let the numerator be $$x$$
Denominator $$= 2x − 2$$
Therefore the fraction $$= \dfrac{x}{2x − 2}$$
Given: when $$3$$ is added to both the numerator and denominator, the fraction becomes $$\dfrac{2}{3}$$.
Hence $$\dfrac{\left(x + 3\right)}{\left(2x − 2 + 3\right)} = \dfrac{2}{3}$$ 
$$\dfrac{\left(x + 3\right)}{\left(2x + 1\right)}= \dfrac{2}{3}$$
$$\Rightarrow 3x + 9 = 4x + 2$$ 
$$\Rightarrow x = 7$$
$$\therefore$$ the fraction is $$\dfrac{7}{12}$$.


Solve:$$2\left(2x+3\right)-10\le \,6\left(x-2\right)$$ when $$x$$ is a natural number.

$$2\left(2x+3\right)-10\le \,6\left(x-2\right)$$
$$\Rightarrow\,4x+12-10\le\,6x-12$$
$$\Rightarrow\,2\left(2x+1\right)\le\,2\left(3x-4\right)$$
$$\Rightarrow\,2x+1\le\,3x-4$$
$$\Rightarrow\,2x-3x\le\,-4-1$$
$$\Rightarrow\,-x\le\,-5$$
$$\Rightarrow\,x\ge 5$$
$$\therefore\,x\in\left\{5,6,7,8,...\right\}$$


Introducing a man, a woman says, " His wife is the only daughter of my father". How is the man related to the woman ?

The only daughter of woman's father is the woman herself. Therefore, the man is the husband of the woman.


Solve:$$7-7x\ge 37$$ when $$x$$ is an natural number.

$$7-7x\ge 37$$
$$\Rightarrow\,7-7x-7\ge 37-7$$
$$\Rightarrow\,-7x\ge 30$$
$$\Rightarrow\,7x\le -30$$
$$\Rightarrow\,x\le\dfrac{-30}{7}$$
$$\therefore\,x\in\phi$$


Find using distributivity:
$$\left[\dfrac{7}{5}\times\left(\dfrac{-3}{12}\right)\right]+\left[\dfrac{7}{5}\times \dfrac{5}{12}\right]$$

$$\left[\dfrac{7}{5}\times\left(\dfrac{-3}{12}\right)\right]+\left[\dfrac{7}{5}\times \dfrac{5}{12}\right]=\dfrac{7}{5}\times\left\{\left(\dfrac{-3}{12}\right)+\dfrac{5}{12}\right\}$$

$$=\dfrac{7}{5}\times\left\{\dfrac{-3+5}{12}\right\}$$

$$=\dfrac{7}{5}\times\left\{\dfrac{2}{12}\right\}$$

$$=\dfrac{7}{5}\times\dfrac{2}{12}$$

$$=\dfrac{7}{5}\times\dfrac{1}{6}$$

$$=\dfrac{7}{30}$$


Show on the number line $$x, x-4,2x+1, 3x-2$$.

Given numbers are $$x,\ x-4,\ 2x+1,\ 3x-2$$.

The above numbers can be represented on the number line as shown in the above figure.

1228034_1344038_ans_5b6df28cb7d74e54899919179be24432.png


What is the distance between two points $$A$$ and $$B$$ representing respective numbers $$x$$ and $$y$$ on a number line.

If we have two points $$A$$ and $$B$$ on the number line such that point $$A$$ is $$x$$ and point $$B$$ is $$y$$ then the distance between the two is given by
$$d=x-y$$ when $$x-y$$ is positive.
   $$=y-x$$ when $$x-y$$ is negative. 


What is the total number of numbers less than $$1000$$ and divisible by $$5$$, which can be formed with digits $$0$$ to $$9$$, such that no digit gets repeated?

Condition $$1 :$$ Number $$< 1000$$.
Condition $$2$$ : Number divisible by $$5$$
Suppose it’s a $$1$$ digit number. Then only '5' falls under the above condition (1 way)
Suppose it’s a $$2$$ digit number:$$ __ $$ $$__$$
If second digit is $$0$$, no of ways to fill 1st digit $$\Rightarrow$$ $$1$$ to $$9$$. ($$9$$ ways)
If second digit is $$5$$, no of ways to fill 1st digit $$\Rightarrow$$ $$1,2,3,4,6,7,8,9$$. ($$8$$ ways)
Total ways : $$9 + 8 = 17$$
Suppose it’s a $$3$$ digit number: $$__$$ $$ __$$ $$ __$$
If last digit is $$0$$; $${1}^{st}$$ digit = $$9$$ ways, $${2}^{nd}$$ digit =$$8$$ ways, Total $$=8\times 9=72$$ ways
If last digit is $$5$$; $${1}^{st}$$ digit = $$8$$ ways, $${2}^{nd}$$ digit = $$8$$ ways, Total $$=8\times 8=64$$ ways
Total ways : $$72 + 64 = 136$$
Total no of numbers :$$ 1 + 17 + 136 = 154 $$ ways.


Solve
$$\dfrac {a^{3}-27}{5a^{2}-16a+3}\div \dfrac {a^{2}+3a+a}{25a^{2}-1}$$

$$\dfrac { { a }^{ 3 }-27 }{ { 5a }^{ 2 }-16a+3 } \div \dfrac { { a }^{ 2 }+3a+2 }{ { 25a }^{ 2 }-1 } $$
$$=\dfrac { { a }^{ 3 }-27 }{ { 5a }^{ 2 }-16a+3 } \times \dfrac { { 25a }^{ 2 }-1 }{ { a }^{ 2 }+3a+9 } $$
$$=\dfrac { \left( a-3 \right) \left( { a }^{ 2 }+3a+9 \right)  }{ \left( 5a-1 \right) \left( a-3 \right)  } \times \dfrac { \left( 5a+1 \right) \left( 5a-1 \right)  }{ \left( { a }^{ 2 }+3a+9 \right)  } $$
$$=(5a+1)$$


Using distributivity of multiplication over addition of whole numbers, find each of the following products: 
$$736 \times 103$$ 

$$\begin{aligned}{}736 \times 103 &= \left( {730 + 6} \right)\left( {100 + 3} \right)\\ &= 730\left( {100 + 3} \right) + 6\left( {100 + 3} \right)\\& = 730 \times 100 + 730 \times 3 + 6 \times 100 + 6 \times 3\\ &= 73000 + 2190 + 600 + 18\\& = 75808\end{aligned}$$


Does there exist a whole number a such that $$a + a = a$$?


1503134_1658535_ans_b062ef25b0b445eb8ff3b888c867013c.jpg


The value of question mark is :
1373081_e253d33202254f829f9b1ec4a9873b4a.png

Working in rows, the number in the centre equals the sum of the left and right hand numbers, divided by 3.
In the first row, middle value is $$=\cfrac{5+7}{3}=\cfrac{12}{3}=4$$
In the second row, middle value is $$=\cfrac{12+6}{3}=\cfrac{18}{3}=6$$
Hence, required value = $$\cfrac{7+2}{3}=\cfrac{9}{3}=3$$


Using Euler's formula find the unknown.
Faces?520
Vertices6?12
Edges129?

 Faces$$(F)$$ $$F=2+E-V=2+12-6=8$$$$5$$  $$20$$
 Vertices$$(V)$$ $$6$$ $$V=2+E-F=2+5-9=-2$$ $$12$$
 Edges$$(E)$$ $$12$$ $$9$$ $$E=V+F-2=12+20-2=30$$


Using distributivity of multiplication over addition of whole numbers, find each of the following products: 
$$ 258 \times 1008 $$


1671031_1658505_ans_90045313285745298923936c814ac00b.jpg


Which property of addition satisfies the statement $$\dfrac{1}{5}+0=0+\dfrac{1}{5}=\dfrac{1}{5}$$.

$$Additive \  identity$$

For every rational number $$x$$,
$$x+0=0+x=x$$

$$\therefore Additive \  identity$$ property of addition satisfies the statement  

$$\dfrac{1}{5}+0=0+\dfrac{1}{5}=\dfrac{1}{5}$$ 


Observe the pattern in the following and fill in the blanks: 
$$9 \times 9 +7 =88 $$
$$98 \times 9+6 =888 $$
$$987 \times 9 + 5 =8888 $$
$$9876 \times 9+4 =......$$
$$98765 \times 9 + 3 =....$$  
$$987654 \times 9 + 2 = ....$$
$$9876543 \times 9 +1 = ......$$ 


1681617_1658574_ans_cf62402d3af24215bd82ddb02f165185.jpg


What are the whole numbers which when multiplied with itself gives the same number?

$$=0, 1$$


Write the first four products of two numbers with difference 4 starting from In the following order: 1, 2, 3, 4, 5, 6, .....
 Identify the pattern in the products and write the next three products. 


1671098_1658569_ans_74df6fa9286b488990646547943ac29f.jpg


Study the following pattern:

$$ 1 = \dfrac {1 \times 2}{2}$$

$$ 1+2 = \dfrac {2 \times 3}{2}$$

$$ 1+2+3 = \dfrac {3 \times 4}{2}$$

$$ 1+2+3+4 = \dfrac {4 \times 5}{2}$$

By observing the above pattern, find 
$$ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 $$

Clearly the pattern being followed is :
$$1+2+3+...+n=\dfrac{n(n+1)}{2}$$

Now,
$$ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 $$

here, $$n=10$$

So, $$1+2+3+4+5+6+7+8+9+10=\dfrac{10\times11}{2}=55$$

Hence, the required value is $$55$$ .


Study the following pattern: 
$$1+3   =2 \times 2$$
$$1+ 3+5 = 3 \times 3$$
$$ 1+ 3+7 = 4 \times4$$
$$1+ 3 + 5 + 7 + 9 = 5 \times 5 $$
By observing the above pattern, find
(i) $$1+ 3 + 5 + 7+ 9 +11$$
(ii) $$ 1+3+5+7+9+11+13+15 $$
(iii) $$21+ 23 + 25 + + 51$$


1872509_1658582_ans_8e4e6a75e1ff4e138feb7eb73d0f9ac2.jpg


Determine the following sum by using suitable rearrangement:
$$2+3 +4+ 5 + 45+ 46 +47+48$$. 

given, $$2+3 +4+ 5 + 45+ 46 +47+48$$

$$=2+48+3+47+4+46+5+45$$

$$=50+50+50+50$$

$$= 200$$


Observe the following pattern and extend it to three more steps: 
$$6 \times 2 - 5 =7 $$
$$7 \times 3 - 12 =9 $$
$$8 \times 4 - 21 =11 $$
$$ 9 \times 5 - 32 =13 $$
$$... \times ... - ... = ...$$
$$... \times ... - ... = ...$$
$$... \times ... - ... = ...$$


1681635_1658578_ans_20da1b637c0540549bd27db6458b20aa.jpg


Write down the smallest whole number.

Whole Numbers are $$0, 1, 2, 3, 4, .... $$

So, the smallest whole numbers is $$0$$ .


Match the following: 
  List I  List II
 (i) $$625 \times 436 = 625 \times 400 + 625 \times 30 + 625 \times 6$$ (a) Commutativity under addition
 (ii) $$25 \times 69 x 8= 25 \times 8 \times 69$$ (b) Commutativity under multiplicatio
 (iii) $$ 60 + 19758 + 840 = 60 + 940 + 19758$$ (c) Distributivity of multiplication over addition


1877178_1658672_ans_2013a4be899e4c9fa2fe68895635be90.jpg


Write down, if possible, the largest natural number.

Natural numbers are infinite. Series of natural numbers doesn't have any specific end. So we can't find the largest natural number.


Write down, if possible, the largest whole number.

Whole numbers start from $$0$$ and go till $$\infty$$.
That is, the whole numbers are $$0,\ 1,\ 2,\ 3,\ ............\ \infty$$

Hence, it is not possible to find the largest whole number.


Write down three consecutive whole numbers just preceding $$8510001$$.

Consecutive numbers: numbers in series that follows a particular order like.. n,n+1,n+2...

Here we need the preceding terms, so our series will be in the form n-1,n-2,n-3

where n=8510001 (given numbers)

And the three consecutive number preceding 8510001 are :
8510000,8509999,8509998


Represent the following numbers on the number line:
$$2, 0, 3, 5, 7, 11, 15$$.

$$2, 0, 3, 5, 7, 11, 15$$ are shown on number line, with circles.
1672195_1659736_ans_8a08440db58749e09eb10bf5a1460879.png


Give arguments in support of the statement that there does not exist the largest natural number.

Assume the existence of a largest natural number $$p$$

Being the largest natural number, $$p$$ cannot be followed by a greater natural number.
i.e., $$p + 1 ≤ p$$
Now subtracting $$p$$ from both sides,
$$\Rightarrow 1 ≤ 0$$ , which is not true.

This shows that the assumption of $$p$$ leads to a contradiction.
Hence, there does not exist any largest natural number.


Write down the next three consecutive whole numbers starting from $$4009998$$.

Three consecutive whole numbers starting from 4009998.
$$4009998, 4009998+1, 4009998+2$$
$$4009998,4009999,4010000$$ are the $$3$$ consecutive whole numbers.


Fill in the blanks to make the following a true statement:
$$53501 + (574 + 799) = 574 + (53501 + ..........)$$

We know that, by associative property
$$a+(b+c)=b+(a+c)$$ 
$$53501 + (574 + 799) = 574 + (53501 + 799)$$


Add the following numbers and check by reversing the order of the addends:
$$16509 + 114$$

$$16509 + 114$$
$$16509 + 114 = 16623$$
Reversing the order of the addends, we get
$$114 + 16509 = 16623$$
$$\therefore 16509 + 114 = 114 + 16509$$


Find the sum: $$(1546 + 498) + 3589$$
Also, find the sum: $$1546 + (498 + 3589).$$ Are the two sums equal? State the property satisfied.

We have
$$(1546 + 498) + 3589 = 2044 + 3589 = 5633$$
$$1546 + (498 + 3589) = 1546 + 4087 = 5633$$
Yes the two sums are equal
Hence associative property of addition is satisfied.


Fill in the blanks to make the following a true statement:
$$1946 + ........... = 984 + 1946$$

We know commutative property of addition.
$$a+b=b+a$$
$$1946 + 984 = 984 +\underline{ 1946}$$


Fill in the blanks to make the following a true statement:
$$864 + 2006 = 2006 + .............$$

We know commutative property of addition.
$$a+b=b+a$$
$$864 + 2006 = 2006 + \underline{864}$$


 Determine the sums given below using suitable rearrangement.
$$15409 + 278 + 691 + 422$$

$$15409 + 278 + 691 + 422$$
Using associative property of addition
$$(15409 + 278) + (691 + 422) = 15687 + 1113 = 16800$$


 Determine the sums given below using suitable rearrangement.
$$1983 + 647 + 217 + 353$$

1983+647+217+3531983+647+217+353


State the property used in each of the following Statements:
$$(29 \times 36) \times 18 = 29 \times (36 \times 18)$$


1679522_1778122_ans_efb3f18847e2418f9e488a40042ba7dc.jpg


Fill in the blanks to make the following a true statement:
$$68 \times 95 = 68 \times 100 - 68 \times .......$$

From property we know that,
$$a \times b-a \times c=a \times (b-c)$$

So from given expression,
$$68 \times 95 = 68 \times 100 - 68 \times .......$$
$$68 \times 95 = 68 \times 100 - 68 \times 5$$


State the property used in each of the following Statements:
$$72 \times 98 + 72 \times 2 = 72 \times (98 + 2)$$


1679532_1778128_ans_0046cdf5268f474fa4ce382a315392f6.jpg


 Determine the sums given below using suitable rearrangement.
$$3259 + 10001 + 2641 + 9999$$

$$3259 + 10001 + 2641 + 9999$$
Using associative property of addition
$$(3529 + 10001) + (2641 + 9999) = 13260 + 12640 = 25900$$


State the property used in the following statement:
$$(16 \times 32)$$ is a whole number


1679515_1778119_ans_3c7599de1d0248fd85367bac870e62a5.jpg


State the property used in the following Statements:
$$19 \times 17 = 17 \times 19$$


1679230_1778117_ans_52289095521c467580a72a9c726899d4.jpg


State the property used in each of the following Statements:
$$1480 \times 1 = 1480$$


1679526_1778123_ans_cb6b0e67359e4511bebcd4c0aecf2290.jpg


State the property used in each of the following Statements:
$$1732 \times 0 = 0$$


1679531_1778124_ans_f7b628830d87404d8f3483ce74036085.jpg


Find the following products, using distributive laws:
$$245 \times 1008$$

Using distributive law of multiplication over addition
We get
$$245 \times 1008 = 245 \times (1000 + 8)$$
$$ = 245 \times 1000 + 245 \times 8$$
$$ = 245000 + 1960$$
$$ = 246960$$


Find the following products, using distributive laws:
$$996 \times 367$$

Using distributive law of multiplication over addition
We get
$$996 \times 367 = 367 \times (1000 - 4)$$
$$ = 367 \times 1000 - 367 \times 4$$
$$ = 367000 - 1468$$
$$ = 365532$$


Find the following products, using distributive laws:
$$472 \times 1097$$

Using distributive law of multiplication over addition
We get
$$472 \times 1097 = 472 \times (1000 + 97)$$
$$ = 472 \times 1000 + 472 \times 97$$
$$ = 472000 + 45784$$
$$ = 517784$$


Find the following products, using distributive laws:
$$947 \times 96$$

Using distributive law of multiplication over addition
We get
$$947 \times 96 = 947 \times (100 - 4)$$
$$ = 947 \times 100 - 947 \times 4$$
$$ = 94700 - 3788$$
$$ = 90912$$


Find the value of the following using various properties:
$$647 \times 13 + 647 \times 7$$

By using distributive property we get
$$647 \times 13 + 647 \times 7 = 647 \times (13 + 7)$$
$$ = 647 \times 20$$
$$ = 12940$$


Find the following products, using distributive laws:
$$740 \times 105$$

Using distributive law of multiplication over addition
We get
$$740 \times 105 = 740 \times (100 + 5)$$
$$ = 740 \times 100 + 740 \times 5$$
$$= 74000 + 3700$$
$$= 77700$$


Find the value of the following using various properties:
$$8759 \times 94 + 8759 \times 6$$

By using distributive property we get
$$8759 \times 94 + 8759 \times 6 =8759 \times (94 + 6)$$
$$ = 8759 \times 100$$
$$ = 875900$$


Find the value of the following using various properties:
$$7459 \times 999 + 7459$$

By using distributive property we get
$$7459 \times 999 + 7459 = 7459 \times (999 + 1)$$
$$ = 7459 \times 1000$$
$$ = 7459000$$


Find the value of the following using various properties:
$$569 \times 17 + 569 \times 13 + 569 \times 70$$

By using distributive property we get
$$569 \times 17 + 569 \times 13 + 569 \times 70 = 569 \times (17 + 13 + 70)$$
$$ = 569 \times 100$$
$$ = 56900$$


State the property used in each of the following Statements:
$$63 \times 126 - 63 \times 26 = 63 \times (126 - 26)$$


1679536_1778129_ans_c4c0751b70764a55a3c94e633ec02d8d.jpg


Find the following products, using distributive laws:
$$439 \times 997$$

Using distributive law of multiplication over addition
We get
$$439 \times 997 = 439 \times (1000 - 3)$$
$$ = 439 \times 1000 - 439 \times 3$$
$$= 439000 - 1317$$
$$= 437683$$


Find the product:
$$3709$$ $$\times 89$$

$$\ \ \ \ 3709$$
$$\ \ \ \times 89$$
$$----$$
$$\ \ 33381$$
$$296720$$
$$----$$
$$330101$$
$$----$$
Hence by multiplying by $$9$$ and $$80$$ we get $$3709 \times 89 = 330101$$


Find the following products, using distributive laws:
$$580 \times 64$$

Using distributive law of multiplication over addition
We get
$$580 \times 64 = 580 \times (60 + 4)$$
$$ = 580 \times 60 + 580 \times 4$$
$$ = 34800 + 2320$$
$$ = 37120$$


Find the products:
$$15208$$
$$\times 542$$

$$\ \ \ 15208$$
$$\ \ \times 542$$
$$----$$
$$\ \ \ \ 30416$$
$$\ \ 608320$$
$$7604000$$
$$----$$
$$8242736$$
$$----$$
Hence by multiplying by $$2, 40$$ and $$500$$ we get $$15208 \times 542 = 8242736$$


Find the following products, using distributive laws:
$$3576 \times 9$$

$$3576 \times 9$$ can be written as
$$3576 \times 9 = 3576 \times (10 - 1)$$
$$ = 3576 \times 10 - 3576 \times 1$$
$$ = 35760 - 3576$$
$$ = 32184$$


Find the products:
$$4617$$
$$\times 234$$

$$\ \ \ \ 4617$$
$$\ \ \times 234$$
$$----$$
$$\ \ 18468$$
$$138510$$
$$923400$$
$$----$$
$$1080378$$
$$----$$
Hence by multiplying by $$4, 30$$ and $$200$$ we get $$4617 \times 234 = 1080378$$


Find the following products, using distributive laws:
$$2437 \times 999$$

$$2437 \times 999$$ can be written as
$$2437 \times 999 = 2437 \times (1000 - 1)$$
$$ = 2437 \times 1000 - 2437 \times 1$$
$$ = 2437000 - 2437$$
$$ = 2434563$$


Find the following products, using distributive laws:
$$1553 \times 198$$

Using distributive law of multiplication over addition
We get
$$1553 \times 198 = 1553 \times (100 + 98)$$
$$ = 1553 \times 100 + 1553 \times 98$$
$$ = 155300 + 152194$$
$$ = 307494$$


Find the products:
$$458$$
$$\times 67$$

$$458$$
$$\times 67$$
$$----$$
  $$ 3206$$
$$2748 \times$$
$$----$$
$$30686$$
$$----$$
Hence by multiplying by $$7$$ and $$60$$ we get $$458 \times 67 = 30686$$


Find the following products, using distributive laws:
$$847 \times 99$$

$$847 \times 99$$ can be written as
$$847 \times 99 = 847 \times (100 - 1)$$
$$ = 847 \times 100 - 847 \times 1$$
$$ = 84700 - 847$$
$$ = 83853$$


Find the value of
$$(1573 \div 1573) - (1573 \div 1573)$$

By BODMAS rule first we are going to solve brackets.
$$(1573 \div 1573) = 1$$
$$1 - 1 = 0$$
$$\therefore$$ Value of $$(1573 \div 1573) - (1573 \div 1573) = 0$$

1689047_1778215_ans_9e49b03c5e3249fa9e09efcc6ab91225.png


Find the rule which gives the number of matchsticks required to make the following matchsticks patterns.
Use a variable to write the rule.
A pattern of letter $$T$$ as

From the above figure, it can be observed that the letter T will require 2 matchsticks


Therefore,


The pattern is $$2n$$



1718736_1787031_ans_7a8b4bcc0de047858bc6783c7859a8d4.PNG


How many times does the digit 3 occur at tenth's place in natural numbers from $$100$$ to $$1000$$ ?

$$90$$ times i.e. $$3, 13, 23, 33, 43, 53, 63, 73, 83, 93$$ and upto $$983, 993$$


Find the value of
$$32277 \div (648 - 39)$$

By BODMAS rule first we are going to solve brackets.
$$648 - 39 = 609$$
$$32277 \div 609$$
By actual division we have:
$$\therefore$$ Value of $$32277 \div (648 - 39) = 53$$

1689034_1778212_ans_1e12f98b97884cd4bce78fc8de3cd579.png


Find the value of
$$1507 - (625 \div 25)$$

According to BODMAS rule, first perform division.
By actual division we have:
$$(625 \div 25) = 25$$
$$\therefore$$ Value of $$1507 - (625 \div 25) = 1507 - 25 = 1482$$

1689019_1778211_ans_787191e8d2604eab84e7300a3f35c7f2.png


In each of the following pairs, which number is to the right of the other on the number line?
$$ 3 , 5 $$ 

As $$3<5$$
Hence, $$ 5 $$ is in the right of $$3$$ on number line


Find the rule which gives the number of matchsticks required to make the following matchsticks patterns.
Use a variable to write the rule.
A pattern of letter $$V$$ as

From the above figure, it can be observed that the letter V will require 2 matchsticks


Therefore,


The pattern is 2n


1718741_1787026_ans_1aa3f68d9da24c348f6e692859937dd3.PNG


______ is the successor of the largest $$3$$ digit number.

Largest 3 digit number =999
999+1=1000


If $$0$$ is subtracted from a whole number, then the result is the ____ itself.

let $$k$$ be the any whole number 
If we subtract $$0$$ from $$k$$
$$k-0=k$$,
then we get $$k$$ itself
So, If $$0$$ is subtracted from a whole number, then the result is the $$\mathrm{Number}$$ itself.


The smallest whole number is

Whole numbers start from $$0$$ and go up to infinity.
So, Zero is smallest whole number.


Whole number are closed under ______ and under ______.

Whole number are closed under Addition and under multiplication


Multiplication is distributive over ______ for whole numbers.

Multiplication is distributive addition over for whole numbers.


$$1001 \times 2002=1001 \times 1001 +1001 \times $$ _____

Let, the member is $$K$$
$$1001 \times 2002 =1001 \times (1001+K)$$
$$\dfrac{\not {1001} \times 2002}{ \not {1001}}=(1001+K)$$
$$2002=1001+K$$
$$K=2002-1001=1001$$


$$125+(68+17)=(125+$$_____$$)+17$$

Let, the member is $$K$$
$$125+(68+17)=(125+K)+17$$
$$125+85=125+K+17$$
$$K=125+85-125-17$$
$$K=85-17$$
$$K=68$$


$$786 \times 3+786 \times 7=$$ ______

Let, the member is $$K$$
$$786 \times 3+786 \times 7=K$$
$$K=786(3+7)=786 \times 10$$
$$K=7860$$


How many natural numbers are there between $$102$$ and $$112$$ ? What fraction of them are prime numbers?

The nature numbers from $$102$$ to $$112$$ are : $$103, 104,105, 106, 107, 108, 109, 110, 111$$
Total number of natural number $$= 9$$
Out of these prime numbers are : $$ 103 , 107, 109 $$
$$\therefore$$ Total number of these prime numbers $$= 3$$
Required fraction $$= \dfrac{3}{9}$$ 
                           $$= \dfrac{1}{3}$$


How many natural numbers are there between $$102 \,and \, 112$$ ? What fraction of them are prime numbers?

The nature numbers from $$102 \, to 112$$ are : $$103, 104,105, 106, 107, 108, 109, 110, 111$$
Total number of natural number $$= 9$$
Out of these prime numbers are : $$ 103 , 107, 109 = 3$$
$$\therefore$$ Total number of these prime numbers $$= 3$$
Required fraction $$= \dfrac{3}{9} = \dfrac{1}{3}$$


Work out following multiplication,
$$\dfrac { \begin{matrix} 123456789 \\ \times \ 9 \end{matrix} }{  } $$
Use the result to answer the following questions.
By what number should $$123456789 $$ be multiplied to get $$8888888808$$?

$$\dfrac { \begin{matrix} 123456789 \\ \times \ 9 \end{matrix} }{ 1111111101 } $$

$$8888888808=1111111101\times8$$
$$\Rightarrow 8888888808=123456789\times9\times8$$
$$\Rightarrow 8888888808=123456789\times72$$
By $$72$$ number should $$123456789 $$ be multiplied we get $$8888888808$$


$$a(b +c) =a \times$$ __________ $$+ a \times$$ __________ .


1975361_1792711_ans_15203387e2b04d88a361f40342055518.jpg


By re-arranging the given numbers, evaluate: $$21 + 22 + 23 + 24 + 25 + 75 + 76 + 77 + 78 + 79$$

$$21 + 22 + 23 + 24 + 25 + 75 + 76 + 77 + 78 + 79$$
$$= (21 + 79) + (22 + 78) + (23 + 77) + (24 + 76) + (25 + 75)$$     (By associative law)
$$= 100 + 100 + 100 + 100 + 100$$
$$= 500$$.


A loading tempo can carry $$482$$ boxes of biscuits weighing $$15\ kg$$ each, whereas a van can carry $$518$$ boxes each of the same weight. Find the total weight that can be carried by both the vehicles.

$$\textbf{Finding weight that can be carried by each of the vehicles and finding total weight}$$
     Tempo can carry $$482$$ boxes weighing $$15\ kg$$ each.
     Therefore Total weight tempo can carry $$= 482 \times 15\ kg$$
                                                                        $$= 7230\ kg$$

      Van can carry $$518$$ boxes, weighing $$15\ kg$$ each.
      Therefore, total weight van can carry $$= 518 \times 15\ kg$$
                                                                     $$= 7770\ kg$$

Weight of total biscuits that can be carried by both vehicles $$= (7230 + 7770)\ kg$$
                                                                                                    $$= 15000\ kg$$

$$\textbf{Hence, the total weight that can be carried by both the vehicles is 15000kg}$$.


By re-arranging the given numbers, evaluate: $$1 + 2 + 3 + 4 + 596 + 597 + 598 + 599$$

$$1 + 2 + 3 + 4 + 596 + 597 + 598 + 599$$
$$= (1 + 599) + (2 + 598) + (3 + 597) + (4 + 596)$$      (By associative law)
$$= 600 + 600 + 600 + 600$$
$$= 2400$$.


Evaluate the following, using the distributive property.
$$(\mathbf{i})-39 \times 99$$
(ii) $$(-85) \times 43+43 \times(-15)$$
(iii) $$53 \times(-9)-(-109) \times 53$$
(iv) $$68 \times(-17)+(-68) \times 3$$

Distributive property is $$a\times(b+c)=a\times b+a\times c$$
Associative property is $$a\times b\pm a\times c=a\times(b\pm c)$$
using this in given examples.

$$(i)$$ 
$$-39 \times 99\\=-39 \times[100-1]$$
$$=-39\times100-(-39)\times 1\cdots\cdots[\text{distributive property}]\\=-3900+39=-3861$$

(ii)
$$(-85) \times 43+43 \times(-15)$$
$$=43 \times(-85)+43 \times(-15)\cdots[\because a\times b=b\times a]$$
$$=43 \times[-85-15]\cdots\cdots[\text{associative property}]$$
$$=43 \times(-100)=-4300$$

(iii)
$$53 \times(-9)-(-109) \times 53$$
$$53 \times(-9)-53\times(-109)\cdots[\because a\times b=b\times a]$$
$$=53 \times[(-9)-(-109)]\cdots\cdots[\text{associative property}]$$
$$=53 \times[-9+109]=53 \times 100=5300$$

$$(iv)$$
$$68 \times(-17)+(-68) \times 3\\=68 \times(-17)-68 \times3$$
$$=68 \times[(-17)-3]\cdots\cdots[\text{associative property}]$$
$$=68 \times(-20)$$
$$=-1360$$


$$5 10 = .$$, which is not a
$$\Rightarrow$$ Subtraction of .. is not closed.

$$5 – 10 = - 5$$, which is not a whole number.
$$\Rightarrow$$ Subtraction of whole numbers is not closed.


Which property of addition is satisfied by:  $$(15 – 12) \times 18 = 15 \times 18 – 12 \times 18$$

Given, $$(15 – 12) \times 18 = 15 \times 18 – 12 \times 18$$,

Now,
$$(15 – 12) \times 18 = 3 \times 18$$,
$$= 54$$,   and
$$15 \times 18 – 12 \times 18 = 270 – 216$$,
$$= 54$$.

The property used as Distributive over subtraction satisfied by $$54$$.


Which property of addition is satisfied by:  $$(7 + 6) \times 10 = 7 \times 10 + 6 \times 10$$

Given, $$(7 + 6) \times 10 = 7 \times 10 + 6 \times 10$$,

Now,
$$(7 + 6) \times 10 = 13 \times 10$$,
$$= 130$$,   and
$$7 \times 10 + 6 \times 10 = 70 + 60$$,
$$= 130$$.

The property used as Distributive property over addition satisfied by $$130$$.


$$32 \times 15 = 32 \times 6 + 32 \times 7 + 32 \times ……….$$

$$32 \times 15 =32 \times (6+7+2)= 32 \times 6 + 32 \times 7 + 32 \times 2$$, (By commutative law of multiplication).


$$12 * (9 6) = .. = .$$
$$12 * 9 12 * 6 = . = .$$
Is $$12 * (9 6) = 12 * 9 12 * 6= $$.
Is this type of result always true ?
Name the property used here ..

$$12 * (9 – 6) = 12 * 3 = 36$$.
$$12 * 9 – 12 * 6 = 108 – 72 = 36$$.
Is $$12 * (9 – 6) = 12 * 9 – 12 * 6$$ ? $$Yes$$.
Is this type of result always true? $$Yes$$.
Name the property used here Distributive property.


$$(16 8) * 24 = .. = .$$.
$$16 * 24 8 * 24 = .. .. = .$$.
Is $$(16 8) * 24 = 16 * 24 8*24$$ ? ..
Is the type of result always true? .
Name the property used here

$$(16 – 8) * 24 = 8 * 24 = 192$$.
$$16 * 24 – 8 * 24 = 384 – 192 = 192$$.
Is $$(16 – 8) * 24 = 16 * 24 – 8 * 24$$ ? $$Yes$$.
Is the type of result always true? $$Yes$$.
Name the property used here: Distributivity.


$$23 \times 56 = 20 \times 56 + ………… \times 56$$

$$23 \times 56 = 20 \times 56 + 3 \times 56$$, (By distributive law of multiplication).


Which property of addition is satisfied by:  $$8 \times (8 + 0) = 8 \times 8 + 8 \times0$$

Given, $$8 \times (8 + 0) = 8 \times 8 + 8 \times 0$$,

Now,
$$8 \times (8 + 0) = 8 \times 8$$,
$$= 64$$,  and

$$8 \times 8 + 8 \times 0 = 64 + 0$$,
$$= 64$$.
The property used as Distributivity of multiplication over addition satisfied by $$64$$.


Fill in the blank.
$$229 \times ………… = 578 \times 229$$

$$229 \times 578 = 578 \times 229$$, (By commutative law of multiplication).


Fill in the blank.
$$328 \times 573 = ……….. \times 328$$

$$328 \times 573 = 573 \times 328$$, (By commutative law of multiplication).


Evaluate using properties:  $$924 \times 988$$.

$$924 \times 988$$
$$= (900 + 20 + 4) \times 988$$
$$= 900 \times 988 + 20 \times 988 + 4 \times 988$$
$$= 889200 + 19760 + 3952$$
$$= 912912$$.


Use distributive law to evaluate: $$385 \times 1004$$.

$$385 \times 1004$$
$$= 385 \times (1000 + 4)$$
$$= 385 \times 1000 + 385 \times 4$$
$$= 385000 + 1540$$
$$= 386540$$.


$$98 * 273 75 * 273 = (..)* 273 = . * 273$$

$$98 * 273 – 75 * 273 = (98 – 75) * 273 = 23 * 273$$.


Use distributive law to evaluate: $$984 \times 102$$.

$$984 \times 102$$
$$= 984 \times (100 + 2)$$
$$= 984 \times 100 \times 984 \times 2$$
$$= 98400 + 1968$$
$$= 100, 368$$.


$$83 \times  54 + 83 \times 16 = 83 \times (……….) = 83 \times ………… = ……….$$

$$83 \times 54 + 83 \times 16 = 83 \times (54 + 16) = 83 \times 70 = 5810$$.


Use distributive law to evaluate: $$446 \times 10002$$.

$$446 \times 10002$$
$$= 446 \times (10000 + 2)$$
$$= 446 \times 10000 \times 446 \times 2$$
$$= 4460000 + 892$$
$$= 4460892$$.


Evaluate using properties:  $$3023 \times 723$$.

$$3023 \times 723$$
$$= (3000 + 20 + 3) \times 723$$
$$= 3000 \times 723 + 20 \times 723 + 3 \times 723$$
$$= 2169000 + 14460 + 2169$$
$$= 2185629$$.


Evaluate using properties:  $$548 \times 98$$.

$$548 \times 98$$
$$= (500 + 40 + 8) \times 98$$
$$= 500 \times 98 + 40 \times 98 + 8 \times 98$$
$$= 49000 + 3920 + 784$$
$$= 53704$$.


Evaluate using properties:  $$679 \times 8 + 679 \times 2$$.

$$679 \times 8 + 679 \times 2$$
$$= 679 \times (8 + 2)$$,   (using distributivity)
$$= 679 \times 10$$
$$= 6790$$.


Evaluate using properties:  $$284 \times 12 – 284 \times 2$$.

$$284 \times 12 – 284 \times 2$$
$$= 284 \times (12 – 2)$$,   (using distributivity)
$$= 284 \times 10$$
$$= 2840$$.


Evaluate using properties:  $$55873 \times 94 + 55873 \times 6$$.

$$55873 \times 94 + 55873 \times 6$$
$$= 55873 \times (94 + 6)$$,   (using distributivity)
$$= 55873 \times 100$$
$$= 5587300$$.


Evaluate using properties:  $$8324 \times 1945 – 8324 \times 945$$.

$$8324 \times 1.945 – 8324 \times 945$$
$$= 8324 \times (1945 – 945)$$,   (using distributive property)
$$= 8324 \times  1000$$
$$= 8324000$$.


Identify the property in the following statements.
8 X (6+5)= (8X6) + (8 X5)

Distributive property of multiplication over addition in integers.


Evaluate using properties:  $$3333 \times 987 + 13 \times 3333$$.

$$3333 \times 987 + 13 \times 3333$$
$$= 3333 \times (987 + 13)$$,   (using distributive property)
$$= 3333 \times 1000$$
$$= 3333000$$.


Fill in the blank.
$$(437 + 3) * (400 – 3) = 397 * ……….$$

$$(437 + 3) * (400 – 3)\\ = 440 * 397$$
$$= 397 * 440$$ by commutative property


$$66 + 44 + 22 = 11 * (..) = 11 * .$$

$$66 + 44 + 22 = 11 * (6 + 4 + 2) = 11 * 12 = 132$$


Evaluate using properties:  $$7984 \times 15 – 7984 \times 5$$.

$$7984 \times 15 – 7984 \times 5$$
$$= 7984 \times (15 – 5)$$,   (using distributive property)
$$= 7984 \times 10$$
$$= 79840$$.


Define a new addition e on (be set of all natural numbers by $$ m ^{\ast} n = m^{2} + n^{2} $$
(i) Is N closed under e ?
(ii) Is $$^{\ast} $$ commutative on N ?
(iii) Is $$^{\ast} $$ associative on N ?
(iv) Is there an identity element in N with respect $$ ^{\ast} $$ ?

(i) Let $$m = 2 $$ and $$ n = 5 $$
$$ m^{\ast} n = m^{2} + n^{2} $$ 
$$ 2 ^{\ast} 5 = 2^{2} + 5^{2} = 4 + 25 = 29 \in N $$ 
$$ \forall m , n \in N , m^{2} + n^{2} \in N $$ 
$$ \therefore \, ^{\ast} $$ is closed under $$N$$.

(ii) Let $$ m = 4 $$ and $$ n = 7 $$
$$m ^{\ast} n = m^{2} + n^{2} = 4^{2} + 7^{2} = 16 + 49 = 65 $$
$$ n ^{\ast}  m = n^{2} + m^{2} = 7^{2} + 4^{2} = 49 + 16 = 65 $$ 
$$ \therefore \, m ^{\ast}  n = n ^{\ast}  m \forall m , n \in N $$
Hence $$ ^{\ast}  $$ is commutative on $$N$$.

(iii) Let $$m = 2 , n = 3 $$ and $$p = 4 $$.
$$ m ^{\ast}  (n ^{\ast}  p) = 2 ^{\ast}  (3  ^{\ast}  4) $$ 
$$ = 2 ^{\ast}  (3^{2} + 4^{2}) = 2 ^{\ast}  (9 + 16) $$ 
$$ = 2 ^{\ast}  25 = 2^{2} + 25^{2} = 4 + 625 $$ 
$$ - 629 $$ ............(i) 
$$ (m ^{\ast}  n) ^{\ast}  p = (2 ^{\ast}  3) ^{\ast}  4 $$ 
$$ = (2^{2} + 3^{2}) ^{\ast}  4 = (4 ^{\ast}  9) ^{\ast}  4 $$
$$ = 13 ^{\ast}  4 = 13^{2} + 4^{2} $$
$$ = 169 + 16 = 185 $$ ...........(ii) 
From (i) and (ii) $$ m ^{\ast}  (n ^{\ast}  p) \neq (m ^{\ast}  n) ^{\ast}  p $$ 
$$ \therefore \, ^{\ast} $$ is not associative on $$N$$.

(iv) Let $$k$$ be the identity element then $$ m ^{\ast} k = m , m^{2} + k^{2} = m^{2} $$ which means 
$$ k^{2} = m^{2} - m^{2} = 0 $$
$$ k = 0 , 0 $$ does not belong to $$N$$.
There is no identity element $$N$$ with respect to $$ ^{\ast}  $$.


Evaluate:  $$9999 \times 8$$.

$$9999 \times 8$$
This can be written as,
$$= (9000 + 900 + 90 + 9) \times 8$$
$$= (9000 \times 8) + (900 \times 8) + (90 \times 8) + (9 \times 8)$$
$$= 72000 + 7200 + 720 + 72$$
$$= 79992$$.


Show that:  Division of whole numbers is not closed. Give two suitable examples.

Example:
$$5$$ and $$8$$ are whole numbers, but $$5 \div 8$$ is not a whole number.
Therefore, closure property does not exist for division of whole numbers.


Name the property indicated in the following:
$$\frac { 8 }{ 17 } +\frac { -8 }{ 17 } =0$$


1971806_1870608_ans_0a8804d3286a4a02a74c90ceae6dba5e.png


Name the property indicated in the following:
$$\frac { 8 }{ 9 } \times 1=\frac { 8 }{ 9 } $$

1 is the multiplication identity.


Check the commutative property of multiplication for the following pairs
$$\frac { -7 }{ 13 } ,\frac { 25 }{ 27 } $$

$$a\times b = b\times a$$

$$\dfrac { -7 }{ 13 } \times\dfrac { 25 }{ 27 } =\dfrac { -7\times25 }{ 13\times27 } =\dfrac { -175 }{ 351 }  ....(i)$$

$$\dfrac { 25 }{ 27 } \times\dfrac { -7 }{ 13 } =\dfrac { 25\times-7 }{ 27\times13 } =\dfrac { -175 }{ 351 }....(ii)$$

From (i) and (ii) $$\dfrac { -7 }{ 13 } \times\dfrac { 25 }{ 27 }=\dfrac { 25 }{ 27 } \times\dfrac { -7 }{ 13 }$$

LHS=RHS




Check the commutative property of addition for the following pairs.
$$\frac { -7 }{ 9 } ,\frac { -18 }{ 19 } $$

$$a+b=b+a$$

$$\dfrac { -7 }{ 9 } +\dfrac { -18 }{ 19 } =\quad \dfrac { -7\times19+(-18)9 }{ 9\times19 } \\ =\dfrac { -133-162 }{ 171 } =\dfrac { -295 }{ 171 } ....(i)$$

$$\dfrac { -18 }{ 19 } +\dfrac { -7 }{ 9 } =\quad \dfrac { -18\times9+(-7)19 }{ 9\times19 } \\ =\dfrac { -162-133 }{ 171 } =\dfrac { -295 }{ 171 } ....(ii)$$

From (i) and (ii) $$\dfrac { -7 }{ 9 } +\dfrac { -18 }{ 19 } = \dfrac { -18 }{ 19 } +\dfrac { -7 }{ 9 } $$

LHS=RHS




In the set of all rational numbers, give 2 examples each illustrating the following properties.
i. Associative
ii. Commutativity
 iii. Distributivity of multiplication over addition.


1972269_1871524_ans_db55400e577646118db7ebd51af55bde.gif


Simplify the following using distributive property
$$\frac { 5 }{ 12 } \times\left( \frac { 25 }{ 9 } +\frac { 32 }{ 5 }  \right) $$

$$\dfrac { 5 }{ 12 } \times\dfrac { 25 }{ 9 } +\dfrac { 5 }{ 12 } \times\dfrac { 32 }{ 5 } =\dfrac { 125 }{ 108 } +\dfrac { 160 }{ 60 } $$

$$=\dfrac { 125\times6+16\times108 }{ 648 } $$

$$=\dfrac { 750+1728 }{ 648 } $$

$$=\dfrac { 2478 }{ 648 } =\dfrac { 1239 }{ 324 } =\dfrac { 413 }{ 108 } $$




Check the commutative property of addition for the following pairs.
$$\frac { -8 }{ 13 } ,\frac { 23 }{ 27 } $$

$$a+b = b+a$$

$$\dfrac { -8 }{ 13 } +\dfrac { 23 }{ 27 } =\dfrac { -8\times27+23\times13 }{ 13\times27 } $$

$$=\dfrac { -216+299 }{ 351 } =\dfrac { 83 }{ 351 } ....(i)$$

$$\dfrac { 23 }{ 27 } +\dfrac { -8 }{ 13 } =\dfrac { 23\times13+(-8)\times27 }{ 27\times13 } $$

$$=\dfrac { 299-216 }{ 351 } =\dfrac { 83 }{ 351 } ....(ii)$$

From (i) and (ii) $$\dfrac { -8 }{ 13 } +\dfrac { 23 }{ 27 } = \dfrac { 23 }{ 27 } +\dfrac { -8 }{ 13 } $$

LHS=RHS




Simplify the following using distributive property
1) $$\frac { 2 }{ 5 } \times\left( \frac { 1 }{ 9 } +\frac { 2 }{ 5 }  \right) $$

 $$\dfrac { 2 }{ 5 } \times\left( \dfrac { 1 }{ 9 } +\dfrac { 2 }{ 5 }  \right) $$

$$=\dfrac { 2 }{ 5 } \times\dfrac { 1 }{ 9 } +\dfrac { 2 }{ 5 } \times\dfrac { 2 }{ 5 } =\dfrac { 2 }{ 45 } +\dfrac { 4 }{ 25 } $$

$$=\dfrac { 2\times25+4\times45 }{ 45\times25 } =\dfrac { 50+180 }{ 1125 } $$

$$=\dfrac { 230 }{ 1125 } =\dfrac { 46 }{ 225 } $$




Check the commutative property of multiplication for the following pairs
$$\frac { 22 }{ 45 } ,\frac { 3 }{ 4 } $$

$$a\times b = b\times a$$

$$\dfrac { 22 }{ 45 } \times\dfrac { 3 }{ 4 } =\dfrac { 22\times3 }{ 45\times4 } =\dfrac { 66 }{ 180 } =\dfrac { 11 }{ 30 } ....(i)$$

$$\dfrac { 3 }{ 4 } \times\dfrac { 22 }{ 45 } =\dfrac { 3\times22 }{ 4\times45 } =\dfrac { 66 }{ 180 } =\dfrac { 11 }{ 30 } ....(ii)$$

From (i) and (ii) $$\dfrac { 22 }{ 45 } \times\dfrac { 3 }{ 4 }=\dfrac { 3 }{ 4 } \times\dfrac { 22 }{ 45 }$$

LHS=RHS




Check the commutative property of multiplication for the following pairs
$$\frac { -8 }{ 9 } ,\frac { -17 }{ 19 } $$

$$a\times b = b\times a$$

$$\dfrac { -8 }{ 9 } \times\dfrac { -17 }{ 19 } =\dfrac { -8\times-17 }{ 9\times19 } =\dfrac { 136 }{ 171}  ....(i)$$

$$\dfrac { -17 }{ 19 } \times\dfrac { -8 }{ 9 } =\dfrac { -17\times 19 }{ 19\times9 } =\dfrac { 136 }{ 171 }....(ii)$$

From (i) and (ii) $$\dfrac { -8 }{ 9 } \times\dfrac { -17 }{ 19 }=\dfrac { -17 }{ 19 } \times\dfrac { -8 }{ 9 }$$

LHS=RHS




$$\left( \frac { 51 }{ 2 } +\frac { 7 }{ 6 }  \right) \div \frac { 3 }{ 5 } $$

$$\left( \dfrac { 51 }{ 2 } +\dfrac { 7 }{ 6 }  \right) \div \dfrac { 3 }{ 5 } $$

 $$=\left( \dfrac { 306\times14 }{ 12 }  \right) \times\dfrac { 5 }{ 3 } =\dfrac { 320 }{ 12 } \times\dfrac { 5 }{ 3 } $$

$$=\dfrac { 1600 }{ 36 } =\dfrac { 800 }{ 18 } =\dfrac { 400 }{ 9 } $$




(3) $$\frac { 8 }{ 9 } \times\left( \frac { 11 }{ 2 } +\frac { 2 }{ 9 }  \right) $$

$$\dfrac { 8 }{ 9 } \times\dfrac { 11 }{ 2 } +\dfrac { 8 }{ 9 } \times\dfrac { 2 }{ 9 } =\dfrac { 88 }{ 18 } $$

$$=\dfrac { 88\times81+16\times18 }{ 18\times81 } =\dfrac { 7128+288 }{ 1458 } $$

$$=\dfrac { 7416 }{ 1458 } =\dfrac { 3708 }{ 729 } =\dfrac { 412 }{ 81 } $$




(ii) $$\left( \frac { 22 }{ 7 } +\frac { 36 }{ 5 }  \right) X\frac { 6 }{ 7 }$$

$$\left( \dfrac { 22 }{ 7 } +\dfrac { 36 }{ 5 }  \right) \times\dfrac { 6 }{ 7 }$$

$$=\left( \dfrac { 22\times5+36\times7 }{ 7\times5 }  \right) \times\dfrac { 6 }{ 7 } $$

$$\dfrac { 110+252 }{ 35 } \times\dfrac { 6 }{ 7 } =\dfrac { 362 }{ 35 } \times\dfrac { 6 }{ 7 } =\dfrac { 2172 }{ 235 } $$




Prove that the fraction $$\displaystyle \frac{m(n+1)+1}{m(n+1)-n}$$ is irreducible for every positive integers $$m$$ and $$n$$.

Let us consider three cases as follows:

Case 1: When $$m$$ is even and $$n$$ is even, then $$m=2p$$ and $$n=2q$$. Therefore,

$$\dfrac { m(n+1)+1 }{ m(n+1)-n } =\dfrac { 2p(2q+1)+1 }{ 2p(2q+1)-2q } =\dfrac { 4pq+2p+1 }{ 4pq+2p-2q }$$ which is not reducible.

Case 2: When $$m$$ is even and $$n$$ is odd, then $$m=2p$$ and $$n=2q+1$$. Therefore,

$$\dfrac { m(n+1)+1 }{ m(n+1)-n } =\dfrac { 2p(2q+1+1)+1 }{ 2p(2q+1+1)-(2q+1
) } =\dfrac { 2p(2q+2)+1 }{ 2p(2q+2)-2q-1 } =\dfrac { 4pq+4p+1 }{ 4pq+4p-2q-1 }$$ which is not reducible.

Case 3: When $$m$$ is odd and $$n$$ is odd, then $$m=2p+1$$ and $$n=2q+1$$. Therefore,

$$\dfrac { m(n+1)+1 }{ m(n+1)-n } =\dfrac { (2p+1)(2q+1+1)+1 }{ (2p+1)(2q+1+1)-(2q+1) } =\dfrac { (2p+1)(2q+2)+1 }{ (2p+1)(2q+2)-2q-1 } =\dfrac { 4pq+4p+2q+2+1 }{ 4pq+4p+2q+2-2q-1 }$$
$$=\dfrac { 4pq+4p+2q+3 }{ 4pq+4q+1 }$$ which is not reducible.
 
Hence the fraction $$\dfrac { m(n+1)+1 }{ m(n+1)-n }$$ is irreducible for every positive integer $$m$$ and $$n$$.


Prove that any integer $$a \ge 7$$ is the sum of two relatively prime integers. (We say two integers $$m$$ and $$n$$ are relatively prime if their HCF (i.e., GCD) is $$1$$)

Let $$a$$ be any integer.

When $$a$$ is odd then $$a=p+q$$ where either $$p$$ is even or $$q$$ is even.

If $$p$$ is even then $$q$$ is odd and hence $$p$$ and $$q$$ are relatively prime to each other. Similarly, vice versa.

Therefore, $$GCD(p,q)=1$$

Hence, any integer $$n\ge 7$$ is the sum of two relatively prime integers.


Simplify the following using the rules for identities :
(i)  $$28\times 32$$
(ii) $$102\times 98$$

(i) $$28 \times 32 = 28 \times \left( {30 + 2} \right)$$

$$ = 840 + 56$$

$$ = 896$$

(ii) $$102 \times 98 = 102 \times \left( {100 - 2} \right)$$

$$ = 102 \times 100 - 102 \times 2$$

$$ = 10200 - 204$$

$$ = 9996$$



Is $$(8\div2)\div 4 = 8 \div(2\div 4)?$$ is there any associative property for division?Check.

According to BODMAS Rule 
$$(8\div2)\div4=(4)\div4=1\\8\div(2\div4)=8\div(0.5)=\dfrac{8}{0.5}=16$$

Hence, We can see that associative property does not hold for division


What is the least number of square(s) which must be shaded to make the given figure symmetric along the dotted line?
1236414_6bc91c4e492b408caefec02b722cc3bc.PNG


1193738_1236414_ans_4fe58c07551d42ad964d1865e05b8794.PNG


Sakharam was playing the field from 3 in the morning At 12:30 in the after he not stopped and started for home. He reached home at 9:How long was he playing the field? How long did it take him to reach home from the field .

Ploughing=from 3 to 12:30=9hr 30min
time taken to reach home= 12:30 to 9:30= 3hr


Add together $$5, 15$$ and $$25$$ and divide the sum by $$9$$. 


2039432_1285349_ans_f46d322d33c34e50a604789a01d76f29.jpg


The number of ways in which the number $$10800$$ can be resolved as a product of two factors.

Let we have a no. $$N$$ which we can factorize as
$$N=\alpha_1^{k_1}\alpha_2^{k_3}...\alpha_n^{k_n}$$
Where $$\alpha_1,\alpha_2,...,\alpha_n$$ are coprime of each other and $$k_1,k_2,...,k_n$$ are natural numbers
Now, if $$N$$ is not a perfect square then the no. of ways in which it can be resolved as a product of two factors is given as
$$\cfrac{(k_1+1)(k_2+2)....(k_n+1)}{2}$$
Here $$N=10800$$
It's prime factorization will be 
$$10800=2^4\times3^3\times5^2$$
Hence $$N$$ is not a perfect square
So, $$\cfrac{(4+1)\times(3+1)\times(2+1)}{2}$$
$$=30$$ ways.


Class 6 Maths Extra Questions