MCQ Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 9

$6$ painters can complete

$9$ drawing in

$5$ hours then

How many painters will make

$18$ drawing in

$5$ hours?

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.

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:

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?

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.

$7$ pipes fill a tank in

$8$ hours, then

$4$ pipes will fill it in ......... hours

0%
$14$
0%
$16$
0%
$18$
0%
$20$

Explanation

No of pipes is inversely proportional to the taken to fill the tank

$\therefore 7\times 8 = k = 4\times x \Rightarrow 56 = 4\times x \Rightarrow x = 14$

$4$ persons do a work in

$8$ days then how many days it will take if

$8$ persons do the same work?

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.

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

$21$ feet?

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)

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.

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?

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.

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?

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?

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

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

$\left [ \dfrac{100}{7} \right ]+\left [ \dfrac{100}{7^{2}} \right ] = 14+2 = \boxed{16 =d}$

$\boxed{c = \dfrac{b}{2}}$ &

$\boxed{d = \dfrac{b}{3}}$

1108205_518574_ans_647006bd59e945a4ad68cddbe95cdad1.jpg

$20$ 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.

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?

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?

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?

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

$C$ ?

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?

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?

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.

1212122_1276265_ans_2c6d3617eb7a4c4087d2a6111f8235b6.JPG

$A, B$ 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?

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?

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?

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:

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?

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?

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?

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?

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 :

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.

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?

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.

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?

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

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?

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.

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.

0%
$cu. m$
0%
$cu. m$
0%
$cu. m$
0%
$cu. m$

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?

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?

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

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

$0$ .

$n$ 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 ?

0%
$120$
0%
$720$
0%
$360$
0%
$60$

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

0%
$8.5$
0%
$5.67$
0%
$7$
0%
$7.5$

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

0%
$2$
0%
$2.5$
0%
$3$
0%
$0$

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?

0%
$4$ hours
0%
$5$ hours
0%
$6$ hours
0%
$7$ 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

0%
Sunday
0%
Friday
0%
Thursday
0%
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?

0%
Tuesday
0%
Wednesday
0%
Thursday
0%
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?

0%
Thursday
0%
Friday
0%
Saturday
0%
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 _________.

0%
$18$ days
0%
$15$ days
0%
$12$ days
0%
$\displaystyle 7\frac{1}{2}$ 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

0%
$10!$
0%
$9!$
0%
$2(9!)$
0%
$\dfrac{9!}{2}$

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.

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.