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CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 9 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Numerical Applications
Quiz 9
If $$6$$ painters can complete $$9$$ drawing in $$5$$ hours then
How many painters will make $$18$$ drawing in $$5$$ hours?
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0%
$$11$$
0%
$$12$$
0%
$$13$$
0%
$$14$$
Explanation
Let number of painters required for painting 18 drawing is $$x$$
Now since no. of painters is directly proportion to no. of drawing
$$\dfrac {6}{9} = \dfrac {x}{18} \Rightarrow x = \dfrac {6\times 18}{9} = 12$$
A single frame of $$35$$ mm film is about three-quarters of an inch long. A film reel holds up to $$1000$$ feet of film. Find the number of reels required for a $$2:47:00$$ ($$2$$ hr $$47$$ min) film shot at $$24$$ frames per second.
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0%
13
0%
14
0%
15
0%
16
Explanation
There are $$12$$ inches in one foot, so a reel is $$12 \times 1,000 = 12,000$$ inches long. Set up a proportion to determine how many frames per reel: $$=$$ $$\dfrac{1 frame }{\dfrac{3}{4}}=12000$$
$$\dfrac{3}{4}x=12000$$
Cross-multiply to get
$$\dfrac{3}{4}x=12000$$
.
Divide both sides by
$$\dfrac{3}{4}$$
to get $$x = 16,000$$ frames per reel. Next, find the number of frames the film requires.
Convert the time to seconds.
There are $$60$$ minutes in an hour, so $$2$$ hours and $$47$$ minutes is equal to $$(2\times 60)+47=167$$ minutes.
There are $$60$$ seconds in a minute, so there are $$60\times 167=10020$$ seconds in this film.
If each second consists of $$24$$ frames, then there are $$24\times 10020=240480$$ frames in this film.
To determine the number of reels, divide by the number of frames per reel: $$= 15.03$$ reels. Because $$15$$ reels does not hold quite enough frames, the film requires $$16$$ reels, which is (D).
The number of ways four boys can be seated around a round table in four chairs of different colours is:
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0%
$$24$$
0%
$$12$$
0%
$$23$$
0%
$$64$$
Explanation
Since the chair are of different colours, so you can treat it as linear permutation
So number of ways will be $$4!=24$$
A scanner can print at a rate of $$5$$ pages per minute. Calculate the number of hours will it take to print $$300$$ pages?
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0%
$$0.5$$
0%
$$1$$
0%
$$1.5$$
0%
$$3$$
Explanation
Scanner print $$5$$ pages in $$1$$ minute.
$$\therefore$$ number of hours required to print $$300$$ pages would be $$=$$ $$\dfrac{300}{5}=60$$ minute $$=1$$ hour.
If $$7$$ pipes fill a tank in $$8$$ hours, then $$4$$ pipes will fill it in ......... hours
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0%
$$14$$
0%
$$16$$
0%
$$18$$
0%
$$20$$
Explanation
No of pipes is inversely proportional to the taken to fill the tank
$$\therefore 7\times 8 = k = 4\times x \Rightarrow 56 = 4\times x \Rightarrow x = 14$$
If $$4$$ persons do a work in $$8$$ days then how many days it will take if $$8$$ persons do the same work?
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0%
$$4$$
0%
$$16$$
0%
$$8$$
0%
$$32$$
Explanation
The no of persons (p) is inversely proportional to the no. of days (d). Hence $$p\propto \dfrac {1}{d}$$
$$\Rightarrow p_{1}d_{1} = p_{2}d_{2}$$
Here, $$p_{1} = 4, d_{1} = 8, p_{2} = 8, d_{2}$$ is unknown.
$$\Rightarrow 4\times 8 = 8d_{2}$$
$$\Rightarrow d_{2} = 4$$.
Find the no. of pipes required to fill the tank in $$16$$ hours.
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0%
$$\dfrac {5}{2}$$
0%
$$\dfrac {7}{2}$$
0%
$$7$$
0%
$$\dfrac {9}{2}$$
Explanation
$$7\times 8 = x \times 16 \Rightarrow x =\dfrac {7\times 8}{16} = \dfrac {7}{2}$$
A coastal geologist estimates that a certain countrys beaches are eroding at a rate of $$1.5$$ feet per year.
According to the geologists estimate, how long will it take, in years, for the countrys beaches to erode
by $$21$$ feet?
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0%
$$14$$ years
0%
$$12$$ years
0%
$$8$$ years
0%
$$15$$ years
Explanation
If the beaches are eroding at the rate of 1.5 feet per year, then for eroding by 21 feet, it will take $$ \frac {21}{1.5} = 14 $$ years
An 11.2 gb image has been taken of the surface of Jupiter by a camera . A tracking station in Earth can receive data from the spacecraft at a data rate of $$3$$ megabits per second for a maximum of $$11$$ hours each day. If $$1$$ gb equals $$1,024$$ mb, find the maximum number of images that the tracking station can receive from the camera each day. (gb:
gigabits, mb: megabits)
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0%
$$3$$
0%
$$10$$
0%
$$56$$
0%
$$144$$
Explanation
Given: A tracking station in earth can received data from the spacecraft at a data rate $$3$$ megabits per second for a maximum of $$11$$ hour each day.
Then total data received per day $$=$$ $$3\times 11\times 3600$$ megabits
Then total data received per day $$=$$
$$\dfrac{118800}{1024}=116.01$$ gb
Then tricking station received images by camera $$=$$
$$\dfrac{116.01}{11.02}=10.35$$
Then camera received image $$=10$$.
Jennilina's hobby is to paint the rooms at her college. At top speed, she could paint $$5$$ identical rooms during one $$6$$-hour shift. Find the time it took her to paint each room.
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0%
$$50$$ minutes
0%
$$1$$ hour and $$10$$ minutes
0%
$$1$$ hour and $$12$$ minutes
0%
$$1$$ hour and $$15$$ minutes
0%
$$1$$ hour and $$20$$ minutes
Explanation
There are $$60$$ minuet in $$1$$ hour
So in $$6$$ hours there are $$=$$ $$6\times 60=360$$ minutes
So Jennilina takes $$360$$ minutes to paint $$5$$ rooms.
So she paint one room in $$=$$ $$\dfrac{360}{5}=72$$ minutes.
$$72$$ minutes is equal to $$1$$ hour and $$12$$ minutes.
A square pyramid is inscribed in a cube of total surface area of $$24$$ square cm such that the base of the pyramid is the same as the base of the cube. What is the volume of the pyramid?
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0%
$$\dfrac 13$$
0%
$$\dfrac 83$$
0%
$$6$$
0%
$$4$$
0%
$$8$$
Explanation
Given that total surface area of cube is $$24$$ and cube has $$6$$ faces. so the area of each face is $$\dfrac {24}6 = 4$$
therefore the length of side of each face and the height of cube is $$2$$
So the volume of pyramid is $$\dfrac {lwh}3 = 2 \times 2 \times \dfrac 23 = \dfrac 83$$
Computer K can perform $$x$$ calculations in $$y$$ seconds and computer L can perform $$r$$ calculations in $$s$$ minutes. Find the number of calculations they can perform in $$t$$ minutes by working together simultaneously.
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0%
$$t\left (\dfrac {x}{60y} + \dfrac {r}{s}\right )$$
0%
$$t\left (\dfrac {60x}{y} + \dfrac {r}{s}\right )$$
0%
$$t\left (\dfrac {x}{y} + \dfrac {r}{s}\right )$$
0%
$$t\left (\dfrac {x}{y} + \dfrac {60r}{s}\right )$$
0%
$$60t\left (\dfrac {x}{y} + \dfrac {r}{s}\right )$$
Explanation
Given, computer $$K$$ can perform $$x$$ calculations in $$y$$ seconds and computer $$L$$ can perform $$r$$ calculations in s minutes.
Then computer $$K$$ can do one calculation $$=$$ $$\dfrac{60x}{y}$$ minutes
And
computer $$L$$ can do one calculation $$=$$ $$\dfrac{r}{s}$$ minutes
Then both computer
calculation
in $$t$$ minutes $$=$$
$$\left ( \dfrac{60x}{y}+\dfrac{r}{s} \right )t$$
Vasu can run $$4$$ Km in $$48$$ minutes. If Vikas can run twice as fast as Vasu, how many minutes does it take Vikas to run $$6$$ Km?
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0%
$$24$$
0%
$$30$$
0%
$$36$$
0%
$$48$$
Explanation
Vasu takes $$=\dfrac{48}{4}=12$$ minutes to run $$1$$ km.
Vikas can run twice as fast as Vasu, means Vikas takes $$6$$ min to run $$1$$ km.
Hence, he takes, $$= 6\times 6 = 36$$ min. to run $$6$$ km.
Three men paint a house in 20 days. How many days do 30 men take to do the same?
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0%
3 days
0%
2 days
0%
60 days
0%
12 days
Explanation
Using $$Work=men\times days$$ we can see that
$$Work=3\times 20=60$$
Since the work is same
$$60=30\times days$$
$$days=2$$
Therefore, Answer is $$(B)$$
If 100! = $$\displaystyle { 2 }^{ a }{ 3 }^{ b }{ 5 }^{ c }{ 7 }^{ d }...$$, then
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0%
$$a=97$$
0%
$$\displaystyle b=\frac { 1 }{ 2 } \left( a+1 \right) $$
0%
$$\displaystyle c=\frac { 1 }{ 2 } b$$
0%
$$\displaystyle d=\frac { 1 }{ 3 } b$$
Explanation
i) no. of 2's in 100! are
$$ \left [ \dfrac{100}{2} \right ]+\left [ \dfrac{100}{22} \right ]+\left [ \dfrac{100}{23} \right ]+\left [ \dfrac{100}{24} \right ]+\left [ \dfrac{100}{25} \right ]+\left [ \dfrac{100}{26} \right ] $$
$$ = 50+25+12+6+3+1 $$
$$ = 97 $$ ie $$ \boxed{a = 97} $$
ii) no of 3's 100! are
$$ \left [ \dfrac{100}{3} \right ]+\left [ \dfrac{100}{3^{2}} \right ]+\left [ \dfrac{100}{3^{3}} \right ]+\left [ \dfrac{100}{3^{4}} \right ] $$
$$ = 33+11+3+1 = \boxed{48 = 6} $$
iii) no of 5's in 100 ! are
$$ \left [ \dfrac{100}{5} \right ]+\left [ \dfrac{100}{5^{2}} \right ] = 20+4 = \boxed{24 = c} $$
iv) no of 7's in 100 ! are
$$ \left [ \dfrac{100}{7} \right ]+\left [ \dfrac{100}{7^{2}} \right ] = 14+2 = \boxed{16 =d} $$
$$ \boxed{c = \dfrac{b}{2}} $$ & $$ \boxed{d = \dfrac{b}{3}} $$
If $$20$$ men working together can finish a job in $$20$$ days, find the number of days taken by $$25$$ men of the same capacity to finish the job.
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0%
$$25$$
0%
$$20$$
0%
$$16$$
0%
$$12$$
Explanation
Using $$Work=men\times days$$ we get
$$Work=20\times 20$$
$$Work=400$$
Now, work is same and number of days are to be calculated
$$400=25\times days$$
$$days=16$$
Therefore, the answer is $$(C)$$.
Riya can do $$\displaystyle \frac{1}{12}$$ of a job in an hour, how many days will she take to complete the job?
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0%
$$12$$
0%
$$\displaystyle \frac{1}{12}$$
0%
$$\displaystyle \frac{1}{24}$$
0%
$$24$$
Explanation
Assuming Riya work 1 $$hour$$ per $$day$$
and it is given that Riya can do $$\dfrac{W}{12} $$ in an hour
So, she will take $$12$$ $$hours $$ to complete the work and
Thus, take $$12$$ $$days$$ to complete it as she works 1 $$hour$$ per $$day$$
Therefore Answer is $$(A)$$
Three pipes $$A, B$$ and $$C$$ can fill a tank from empty to full in $$30$$ minutes, $$20$$ minutes, and $$10$$ minutes respectively. When the tank is empty, all the three pipes are opened. $$A, B$$ and $$C$$ discharge chemical solutions $$P, Q$$ and $$R$$ respectively. What is the proportion of the solution $$R$$ in the liquid in the tank after $$3$$ minutes?
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0%
$$\displaystyle\frac{5}{11}$$
0%
$$\displaystyle\frac{6}{11}$$
0%
$$\displaystyle\frac{7}{11}$$
0%
$$\displaystyle\frac{8}{11}$$
Explanation
Rate of filling the tank for A: $$\dfrac{1}{30}$$
Rate of filling the tank for B:
$$\dfrac{1}{20}$$
Rate of filling the tank for C:
$$\dfrac{1}{10}$$
Part filled by $$(A+B+C)$$ in $$3$$ minutes $$3\displaystyle\left(\displaystyle\frac{1}{30}+\frac{1}{20}+\frac{1}{10}\right)$$ $$=\displaystyle\left( 3\times \frac{11}{60}\right)=\frac{11}{20}$$.
Part filled by C in $$3$$ minutes $$=\displaystyle\frac{3}{10}$$.
$$\therefore$$ Required ratio$$=\left(\displaystyle\frac{3}{10}\times \frac{20}{11}\right)=\displaystyle \frac{6}{11}$$.
What is the average(arithmetic mean) of all the multiples of ten from $$10$$ to $$190$$ inclusive?
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0%
$$90$$
0%
$$95$$
0%
$$100$$
0%
$$105$$
0%
$$110$$
$$A$$ alone can do a piece of work in $$6$$ days and $$B$$ alone in $$8$$ days. $$A$$ and $$B$$ undertook to do it for Rs.$$3200$$. With the help of $$C$$, they completed the work in $$3$$ days. How much is to be paid to $$C$$?
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0%
Rs.$$375$$
0%
Rs.$$400$$
0%
Rs.$$600$$
0%
Rs.$$800$$
Explanation
$$C$$'s $$1$$ day's work $$=\cfrac{1}{3}-\left( \cfrac { 1 }{ 6 } +\cfrac { 1 }{ 8 } \right) =\cfrac { 1 }{ 3 } -\cfrac { 7 }{ 24 } =\cfrac { 1 }{ 24 } $$.
$$A$$'s Wages: $$B$$'s wages: $$C$$'s wages $$=\cfrac{1}{6}:\cfrac{1}{8}:\cfrac{1}{24}=4:3:1$$.
$$\therefore$$ $$C$$'s share (for $$3$$ days)$$=$$ Rs. $$\left( 3\times \cfrac { 1 }{ 24 } \times 3200 \right) =$$ Rs.$$400$$.
A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?
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0%
Rs. $$7.98$$
0%
Rs. $$8$$
0%
Rs. $$8.50$$
0%
Rs. $$9$$
Explanation
Total quantity of petrol consumed in 3 years = $$\begin{pmatrix}\dfrac{400}{7.50}+\dfrac{4000}{8}+\dfrac{4000}{8.50}\end{pmatrix}$$ litres
= $$4000 \begin{pmatrix}\dfrac{2}{15}+\dfrac{1}{8}\dfrac{2}{17}\end{pmatrix}$$ litres
= $$\begin{pmatrix}\dfrac{76700}{51}\end{pmatrix}$$ litres
Total amount spent = Rs. (3 x 4000) = Rs. 12000.
$$\therefore$$ Average cost = Rs. $$\begin{pmatrix}\dfrac{12000\times 51}{76700}\end{pmatrix}$$ = Rs. $$\dfrac{6120}{767}$$ = Rs. $$7.98$$
$$18$$ men can reap a field in $$35$$ days. For reaping the same field in $$15$$ days, how many men are required?
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0%
$$42$$
0%
$$28$$
0%
$$32$$
0%
$$40$$
Explanation
18 Men $$\rightarrow $$ 35 days.
1 Men $$\rightarrow 35\times 18$$ days.
$$\therefore $$ Reaping in 15 days -
No. of men = $$\frac{35\times 18}{15}= 42$$ men.
$$A, B$$ and $$C$$ can do a piece of work in $$20,30$$ and $$60$$ respectively. In how many days can $$A$$ do the work if he is assisted by $$B$$ and $$C$$ on every third day?
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0%
$$12$$ days
0%
$$15$$ days
0%
$$16$$ days
0%
$$18$$ days
Explanation
$$A$$'s $$1$$ day's work $$=\left( \cfrac { 1 }{ 20 } \times 2 \right) =\cfrac { 1 }{ 10 } $$
$$(A+B+C)$$'s $$1$$ day's work $$=\left( \cfrac { 1 }{ 20 } +\cfrac { 1 }{ 30 } +\cfrac { 1 }{ 60 } \right) =\cfrac { 6 }{ 60 } =\cfrac { 1 }{ 10 } $$
Work done in $$3$$ days $$=\left( \cfrac { 1 }{ 10 } +\cfrac { 1 }{ 10 } \right) =\cfrac { 1 }{ 5 } $$
Now, $$\cfrac{1}{5}$$ work is done in $$3$$ days.
$$\therefore$$ Whole work will be done in $$(3\times 5)=15$$ days.
The amount of time taken to paint a wall is inversely proportional to the number of painters working on the job. If it takes 3 painters 5 days to complete such a job, how many days longer will it take if there are only 2 painters working?
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0%
$$2.5$$
0%
$$2$$
0%
$$2.4$$
0%
None
In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?
Report Question
0%
1
0%
$$\dfrac{1}{40}$$
0%
40
0%
80
Explanation
Let the required number of days be x.
Less cows, More days (Indirect Proportion)
Less bags, Less days (Direct Proportion)
$$\left.\begin{matrix}Cow & 1 : 40 \\ Bags & 40 : 1 \end{matrix}\right\} :: 40 : x$$
$$\therefore 1 \times 40 \times x = 40 \times 1 \times 40$$
$$\Rightarrow x = 40$$
$$A$$ takes twice as much time as $$B$$ or thrice as much time as $$C$$ to finish a piece of work. Working together, they can finish the work in $$2$$ days. $$B$$ can do the work alone in:
Report Question
0%
$$4$$ days
0%
$$6$$ days
0%
$$8$$ days
0%
$$12$$ days
Explanation
Suppose $$A,B$$ and $$C$$ take, $$x,\cfrac{x}{2}$$ and $$\cfrac{x}{3}$$ days respectively finish the work.
Then, $$\left( \cfrac { 1 }{ x } +\cfrac { 2 }{ x } +\cfrac { 3 }{ x } \right) =\cfrac { 1 }{ 2 } $$
$$\Rightarrow$$ $$\cfrac{6}{x}=\cfrac{1}{2}$$
$$\Rightarrow$$ $$x=12$$
So, $$B$$ takes $$(12/2)=6$$ days to finish the work.
What is the average of four tenths and five thousandths?
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0%
$$25002$$
0%
$$2502$$
0%
$$0.225$$
0%
$$0.2025$$
0%
$$0.02025$$
A can copy $$75$$ pages in $$25$$ hours, A and B together copy $$135$$ pages in $$27$$ hours. In what time can B copy $$42$$ pages?
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0%
$$17\ hours$$
0%
$$27\ hours$$
0%
$$31\ hours$$
0%
$$21\ hours$$
In how many ways $$7$$ men and $$7$$ women can be seated around a round table such that no two women can sit together?
Report Question
0%
$$(7!)^2$$
0%
$$7!\times 6!$$
0%
$$(6!)^2$$
0%
$$7!$$
Explanation
Lets first place the men (M). * here indicates the linker of round table *M -M - M - M - M* which is in $$(7-1)!$$ ways = $$6!$$
So we have to place the women in between the men which is on the $$5$$ empty seats So $$7$$ women can sit on $$7$$ seats in $$(7)!$$ ways
So the answer is $$7!\times 6!$$
A can do a piece of work in $$6\dfrac {2}{3}$$ days and B in $$5$$ days. They work together for $$2$$ days and then A leaves B to finish the work alone. How long will B take to finish it?
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0%
$$\dfrac {3}{4}$$ days
0%
$$1\dfrac {1}{2}days$$
0%
$$3\ days$$
0%
$$7\ days$$
Two pipes $$A$$ and $$B$$ can fill a cistern in $$37\displaystyle\frac{1}{2}$$ minutes and $$45\ minutes $$ respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the $$B$$ is turned off after :
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0%
$$5\ min$$
0%
$$9\ min$$
0%
$$10\ min$$
0%
$$15\ min$$
Explanation
Let B be turned off after x minutes. Then,
Part filled by $$(A+B)$$ in x min. $$+$$ Part filled by A in $$(30-x)$$min. $$=1$$.
$$\therefore x\left(\displaystyle\frac{2}{75}+\dfrac{1}{45}\right)+(30-x).\displaystyle\dfrac{2}{75}=1$$
$$\Rightarrow \displaystyle\dfrac{11x}{225}+\dfrac{(60-2x)}{75}=1$$
$$\Rightarrow 11x+180-6x=225$$.
$$\Rightarrow x=9$$.
A pump can fill a tank with water in $$2\ hours$$. Because of a leak, it took $$2\displaystyle\frac{1}{3}\ hours$$ to fill the tank. The leak can drain all the water of the tank in.
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0%
$$\displaystyle 4\frac{1}{3}\ hrs$$
0%
$$\displaystyle 7\ hrs$$
0%
$$\displaystyle 8\ hrs$$
0%
$$\displaystyle 14\ hrs$$
Explanation
Rate of filling the Tank by Pump : $$\dfrac{1}{2} $$
Rate of leak while water is being pumped in the tank: $$\dfrac{3}{7}$$
Rate of leak in the tank: $$1$$ hour $$=\left(\displaystyle\frac{1}{2}-\frac{3}{7}\right)=\displaystyle\frac{1}{14}$$.
$$\therefore$$ Leak will empty the tank in $$14$$ hrs.
Which of the following is correct?
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0%
When the positions are numbered, then the circular permutations are treated as a linear arrangement.
0%
In linear arrangements, it does not make difference whether the positions are numbered are not.
0%
Both A and B
0%
None of these.
Explanation
By numbering position we are actually fixing staring point which makes it similar to linear permutation.so option A is correct
Linear permutation already has a staring point so it dose not need seats to be numbered so option B is correct
A cistern has a leak which would empty it in $$8$$ hours. A tap is turned on which admits $$6$$ litres a minute into the cistern and is now emptied in $$12$$ hours. How many litres does the cistern hold?
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0%
$$8640$$
0%
$$8500$$
0%
$$4320$$
0%
$$60\ hours$$
$$A$$ is $$50\%$$ as efficient as $$B$$. $$C$$ does half of the work done by $$A$$ and $$B$$ together. If $$C$$ alone does the work in $$40$$ days, then $$A, B$$ and $$C$$ together can do the work in
Report Question
0%
$$\displaystyle 15\frac{1}{3}$$ days
0%
$$15$$ days
0%
$$13$$ days
0%
$$\displaystyle 13\frac{1}{3}$$ days
Explanation
$$Work Rate=\dfrac{Work} {Days}$$
Let Work Rate of A be $$W_A $$ and Work rate of B be $$W_B$$
and it is Given That
$$\Rightarrow W_A=\dfrac{W_B}{2}$$
$$\Rightarrow \dfrac{W}{D_A}=\dfrac{W}{2D_B}$$
$$\Rightarrow D_B=\dfrac{D_A}{2}$$...............................................(1)
Now, Work Done By$$ A$$ and$$ B$$ Together
$$\Rightarrow \dfrac{W}{D_A}+\dfrac{W}{D_B}=\dfrac{3W}{2D_B}$$
Using Given Information,
$$\Rightarrow \dfrac{W}{D_C}=\dfrac{3W}{4D_B}$$
$$\Rightarrow \dfrac{W}{40}=\dfrac{3W}{4D_B}$$
$$\Rightarrow D_B=30$$
From Relation (1) $$\Rightarrow D_A=60$$
Now,When $$A,B$$ and $$C$$ work Together
$$\Rightarrow \dfrac{W}{30}+\dfrac{W}{60}+\dfrac{W}{40}=\dfrac{W}{\dfrac{120}{9}}$$
Therefore The Work will be completed in $$\dfrac{120}{9}days=13\dfrac{1}{3}days$$
Therefore Answer is $$(D)$$
In a class of $$100$$ students there are $$70$$ boys whose average marks in a subject are $$75$$. If the average marks of the complete class is $$72$$, then what is the average of the girls?
Report Question
0%
$$73$$
0%
$$65$$
0%
$$68$$
0%
$$74$$
Explanation
Number of boys $$= 70$$
Average marks of boys $$=75$$
Total marks of boys $$=70 \times 75=5250$$
Total marks of the class $$=72 \times 100 =7200$$
Total marks of girls $$=1950$$
Average of the girls $$=\dfrac{1950}{30}=65$$
The length of the base of a square pyramid is $$2\ cm$$ and the height is $$6\ cm$$. Calculate the volume.
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0%
$$8\ cm^3$$
0%
$$6\ cm^3$$
0%
$$4\ cm^3$$
0%
$$2\ cm^3$$
Explanation
Volume of square pyramid $$=\dfrac { 1 }{ 3 } \times { a }^{ 2 }\times h=\dfrac { 1 }{ 3 } \times 2\times 2\times 6=8{ cm }^{ 3 }$$
A regular square pyramid is $$3$$ m height and the perimeter of its base is $$16$$ m. Find the volume of the pyramid.
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0%
12
12
$$cu. m$$
0%
14
14
$$cu. m$$
0%
16
16
$$cu. m$$
0%
18
18
$$cu. m$$
Explanation
Given, height of regular square pyramid is $$3$$ m and the perimeter of its base is $$16$$ m
Let the base side of pyramid is $$l$$ m
Then perimeter of base $$=4a=16$$
So, $$ a=4$$
Then volume of pyramid $$=$$ $$\dfrac{1}{3}l^{2}h=\dfrac{1}{3}\times (4)^{2}\times 3=16 $$ $$cu. m$$
How many seating arrangements are possible with $$8$$ people around a round table?
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0%
$$5040$$
0%
$$8!$$
0%
$$7!$$
0%
$$720$$
Explanation
To arrange $$n$$ people around a round table
Number of arrangements $$=(n-1)!$$
To arrange them $$8$$ around the round table.
Number of arrangements $$=(8-1)!=7!=5040$$
Twelve friends go out for a dinner to $$UTSAV$$ restaurant. There are two circular tables one with $$7$$ chairs and one with $$5$$ chairs. In how many ways can the group settle down themselves for dinner?
Report Question
0%
$$\dfrac{12!}{7!\times5!}$$
0%
$$\dfrac{12!}{35}$$
0%
$$12!$$
0%
$$12! 5! 7!$$
Explanation
Number of ways of selecting $$7$$ people for first round table $$={}^{12}C_5$$
Number of ways of selecting remaining $$5$$ people for second round table $$={}^{5}C_5$$
Arranging $$7$$ people on first round table $$=(7-1)!=6!$$
Arranging $$5$$ people on second round table $$=(5-1)!=4!$$
Hence the answer is
$$={}^{12}C_5\times {}^{5}C_5\times 6! \times 4!$$
$$=\dfrac{{12}!\times 5! \times 6! \times 4!}{7! \times 5!\times 5!\times 0!}=\dfrac{12!}{7 \times 5}=\dfrac{12!}{35}$$
Hence the correct answer is $$\dfrac{12!}{35}$$.
The Arithmetic mean of integers from $$-5$$ to $$5$$ is
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0%
$$3$$
0%
$$0$$
0%
$$25$$
0%
$$10$$
Explanation
All integers from $$-5$$ to $$5$$ are:
$$-5,\ -4,\ -3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3,\ 4,\ 5$$.
We know that, arithmetic mean $$=\dfrac{\text{Sum of all observations}}{\text{Number of observations}}$$
Hence, sum of all numbers $$=(-5)+(-4)+(-3)+(-2)+(-1)+ 0+ 1+2+3+4+5$$
$$=0$$
Total integers $$=11$$
Hence, arithmetic mean $$=\dfrac{0}{11}$$
$$=0$$.
Hence, the arithmetic mean of integers from $$-5$$ to $$5$$ is $$0$$.
If $$n$$ books can be arranged on an ordinary shelf in $$720$$ ways, then in how many ways an these books be arranged in a circular shelf ?
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$$120$$
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$$720$$
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$$360$$
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$$60$$
Explanation
$$n$$ books can be arranged in $$N!$$ ways
Given: $$n!=720=6!$$
Therefore, $$n=6$$
The number of ways books can be arranged in circular shelf $$=(n-1)!=5! =5\times 4\times 3\times 2\times 1=120$$
Hence the correct answer is $$120$$.
The Arithmetic mean of all the factors of $$24$$ is
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$$8.5$$
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$$5.67$$
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$$7$$
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$$7.5$$
Explanation
$$\Rightarrow$$ Factors of $$24$$ are $$1,\,2,\,3,\,4,\,6,\,8,\,12,\,24$$
So, the observations are $$1,2,3,4,6,8,12,24$$
$$\therefore$$ Number of observations $$=8$$
Arithmetic mean $$=\dfrac{\text{Sum of observations}}{\text{Number of observations}}$$
$$\therefore$$ Arithmetic mean $$=\dfrac{1+2+3+4+6+8+12+24}{8}$$
$$\therefore$$
Arithmetic mean $$=\dfrac{60}{8}$$
$$\therefore$$ Arithmetic mean $$=7.5$$
The mean of first $$5$$ whole numbers is
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$$2$$
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$$2.5$$
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$$3$$
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$$0$$
Two workers $$A$$ and $$B$$ are engaged to do a piece of work. Working alone, $$A$$ takes $$8$$ hours more to complete the work than if both worked together. On the other hand, working alone, $$B$$ would need $$4\displaystyle\frac{1}{2}$$ hours more to complete the work than if both worked together. How much time would they take to complete the job working together?
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$$4$$ hours
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$$5$$ hours
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$$6$$ hours
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$$7$$ hours
Explanation
Let the number of hours taken by A and B both to complete the work be $$x$$.
As per the given information, A takes $$x + 8$$ hours to complete when he works alone.
Similarly, B takes $$x + 4.5$$ hours to complete the work, working alone.
$$\therefore \cfrac{1}{x + 4.5} + \cfrac{1}{x + 8} = \cfrac{1}{x}$$
$$\Rightarrow (x + 8 + x + 4.5)x = (x + 4.5)(x + 8)$$
$$\Rightarrow 2x^2 + 12.5x = x^2 + 12.5x + 36$$
$$\Rightarrow x^2 = 36$$ or $$x = 6$$ hours
If on the 15th September 2008 was a Friday. Then which day will be on 15th September 2009
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Sunday
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Friday
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Thursday
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Saturday
There are exactly twelve Sundays in the period from January $$1$$ to March $$31$$ in a certain year. Then the day corresponding to February $$15$$ in that year is?
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Tuesday
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Wednesday
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Thursday
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Not possible to determine from the given data
Explanation
Total number of days from $$1$$ jan to $$31$$ feb $$=31+28+31=90$$
Dividing number of days with $$7$$, we get
$$\Rightarrow \dfrac{90}{7}=12\dfrac{6}{7}$$
Total number of sundays are $$12$$
$$\therefore 1st$$ jan must be monday.
$$15$$ feb is $$46th$$ day of the year,
number of weeks
$$\Rightarrow\Bigr(\dfrac{46}{7}=6\dfrac{4}{7}\Bigl)$$
Thus, 46th day is thursday (or $$4$$th day of the week).
If 15th August 2011 was Tuesday, then what day of the week was it on 17th September, 2011?
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Thursday
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Friday
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Saturday
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Sunday
Explanation
The answer is option $$D$$
Number of days from 15th Aug 2011 and 17th Sep 2011 = 16+17=33
Number of odd days in this =Remainder (33/7)=5
Since number of odd days is 5 and 15th Aug 2011 was Tuesday, counting five days from Tuesday, 17th Sep 2011 will be Sunday
A and B can do a piece of work in $$6$$ days and A alone can do it in $$9$$ days. The time take by B alone to do the work is _________.
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$$18$$ days
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$$15$$ days
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$$12$$ days
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$$\displaystyle 7\frac{1}{2}$$ days
Explanation
A and B can do a piece of work in $$6$$ days, so work done by A and B in one day will be $$\dfrac{1}{6}$$
A alone can do the same work in $$9$$ days, so work done by A in one day is $$\dfrac{1}{9}$$
work done by B in one day will be $$=\dfrac{1}{6}-\dfrac{1}{9}$$
$$=\dfrac{1}{18}$$
so, we can say that B will take $$18$$ days to complete the same work.
In how many ways a garland can be made from exactly $$10$$ flowers
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$$10!$$
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$$9!$$
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$$2(9!)$$
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$$\dfrac{9!}{2}$$
Explanation
Since $$10$$ things can be permuted along a circle in $$9!$$ ways.
But in a garland anticlockwise and clockwise directions are the same, so we have $$\dfrac{9!}{2}$$.
Hence, option D is correct.
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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