MCQ Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 1

If x men working x hours a day for x days can produce x articles, then the number of articles produced by y men working for y hours a day for y days is _____________________.

0%
$\displaystyle\frac{y^2}{x^3}$
0%
$\displaystyle\frac{x^2}{y^3}$
0%
$\displaystyle\frac{y^3}{x^2}$
0%
$\displaystyle\frac{x^3}{y^2}$

The greatest number that can be formed by the digits

$7,0,9,8,6,3$

0%
$9,87,360$
0%
$9,87,063$
0%
$9,87,630$
0%
$9,87,603$

Explanation

The greatest number that can be formed by the digits

$7,0,9,8,6,3$ is

$9 8 7 6 3 0$

To achieve this arrange the given numbers in descending order.

For a group of

$50$ male workers , the mean and S.D. of their daily wages are Rs.

$630$ and Rs.

$90$ respectively. For a group of

$40$ female workers, these are Rs.

$540$ and Rs.

$60$ respectively.The S.D. of these

$90$ workers is _____________.

0%
$60$
0%
$70$
0%
$80$
0%
$90$

If 20 pumps of equal capacity can fill a tank in 7 days, then how many more such pumps will be required to fill the tank in 5 days?

0%
28
0%
16
0%
8
0%
7

$\displaystyle\sum^m_{k=1}(k^2+1)k!=1999(2000!)$ , then m is?

0%
$1999$
0%
$2000$
0%
$2001$
0%
$2002$

Ramu scored

$2$ times more than the rahim scored. In an examination and shashi scored

$30$ marks more than ram in total they scored

$240$ marks. What was rahim score?

0%
$42$
0%
$20$
0%
$30$
0%
$35$

$8:00\ a.m$ ., Uday started from city

$A$ and travelled towards city

$B$ at

$50\ kmph$ and at

$9:30\ a.m$ Vikram started from city

$A$ and travelled to city

$B$ at

$60\ kmph$ . At what would Vikram overtake Uday?

0%
$3:30\ p.m.$
0%
$4:00\ p.m.$
0%
$4:30\ p.m.$
0%
$5:00\ p.m.$

The number of 5 letter words formed using letters of word "CALCULUS" is

0%
$280$
0%
$15$
0%
$1110$
0%
$56$

A person is standing on the top of a tower of height

$15(\sqrt{3} +1)$ and observing a car coming towards the tower . He observed that angle of depression changes from

$30^0$ to

$45^0$ in

$3$ second. What is the speed of the car in km/hr?

0%
$36$ km/hr.
0%
$72$ km/hr.
0%
$18$ km/hr.
0%
$30$ km/hr.

$n\ \in\ N$ &

$n$ is even, then

$\dfrac {1}{1\ .\ (n-1)\ !}+\dfrac {1}{3\ !(n-3)\ !}+\dfrac {1}{5\ !\ (n-5)\ !}+....+\dfrac {1}{(n-1)\ !\ 1\ !}=$

0%
$2^{n}$
0%
$\dfrac {2^{n-1}}{n\ !}$
0%
$2^{n}n\ !$
0%
$none\ of\ these$

The number of words which can be made out of the letters of the word

$'MOBILE'$ when consonants always occupy odd places is _______.

0%
$20$
0%
$36$
0%
$30$
0%
$720$

Mean of

$41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 49, 42, 52, 60 \ is\ 54.8$

0%
True
0%
False
0%
Neither
0%
None

Explanation

$Mean$

$=$

$\dfrac{sum of all numbers}{no. of numbers}$

$\therefore$

$mean$

$=$

$\dfrac{41+ 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 49 + 42 + 52 + 60}{15}$

$=$

$55.4$

If the day before yesterday was Thursday, when will Sunday be?

0%
Tomorrow
0%
Day after tomorrow
0%
Today
0%
Two days after today

Seven people are seated in a circle. How many relative arrangements are possible?

0%
$7!$
0%
$6!$
0%
$^7P_6$
0%
$^7C_6$

Explanation

The number of circular arrangement for

$n$ objects is

$(n-1)!.$
There would be

$(7 - 1)! = 6!$ arrangements possible.

A work can be completed by

$40$ workers in

$40$ days. If

$5$ workers leave every

$10$ days, in how many days work will be completed?

0%
$55.66$
0%
$56.44$
0%
$56.66$
0%
$54.66$

Explanation

A work can be completed by

$40$ workers in

$40$ days.

Number of days required to complete the work by one worker

$=40\times 40=1600$ mandays(days per worker)

Day----mandays----total mandays

10 400 400

20 350 750

30 300 1050

40 250 1300

50 200 1500

After 50 days:

1500 man-hrs have been accumulated, 15 men are on the job.

Then, 100 more man-days required

$15 d = 100$

$\therefore$

$d = 100/15=6.66$

$\therefore$ Total no. of days required

$=50+6.66=56.66$ days.

Hence, option C is correct.

On what day of the week India will celebrate its Republic Day on 26th January, 2015 ?

0%
Sunday
0%
Monday
0%
Tuesday
0%
Wednesday

Monday, Tuesday etc. are numbered as 1,2 and so on. If 8th December 1994 was Monday, what would be the number of the day 8th January 1995 ?

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

$A$ can do a piece of work in

$24$ days. If

$B$ is

$60\%$ more efficient than

$A$ , then the number of days required by

$B$ to do the twice as large as the

$A's$ work is -

0%
$24$
0%
$36$
0%
$15$
0%
$30$

Explanation

A can do a work say

$w$ in

$24$ days

$A\rightarrow \dfrac { w }{ 24 }$ (work done by A in one day)

$B$ is

$60$ % more efficient

$B\rightarrow 1.6\times \dfrac { w }{ 24 }$ (work done by B in one day)

Now B has to do work

$2w$

For one day B can do

$B\rightarrow 1.6\times \dfrac { w }{ 24 }$

Say B takes

$x$ days to complete

$2w$ amount of work.

$1.6\times \dfrac { w }{ 24 } \times x=2w$

$x=\dfrac { 48 }{ 1.6 } =30$

$20$ people can build a wall in

$7$ days, then the number of persons required to complete the work in

$35$ days is:

0%
$4$
0%
$8$
0%
$12$
0%
$20$

Explanation

The number of people working and length of the wall are in inverse proportion.

Work done by

$20$ people in

$7$ days=

$20\times7$
Let,the number of people required to complete the same work in

$35$ days be

$x$
Now,

$20\times7=35x$

$=>\cfrac{140}{35}=x$

$=>x=4$

Choose the correct option for the following.

$n!=n(n-1)(n-2).....3.2.1$

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

Factorial : the product of an integer and all integer less than that

$\therefore n!=n\times (n-1)\times (n-2)............3\times 2\times 1$

$\therefore$ The given statement

$n!=n\times (n-1)\times (n-2)............3\times 2\times 1$ is True

Factorial of negative numbers is always greater than 1.

0%
True
0%
False
0%
Either
0%
Neither

The base of the right pyramid is a square of side 16 cm and height 15 cm. Its volume

$(cm^{3})$ will be

0%
$3840$
0%
$1920$
0%
$1280$
0%
$960$

Explanation

Area of the base

$=$

$(16 \times 16) cm^2$
Volume of the pyramid

$=$

$\frac{1}{3} \times B h$
Volume of the pyramid

$=$

$\frac{1}{3}\left ( 16\times 16 \right )\times 15$

$= 1280 cm^2$

If the fifth day of month is Friday, which of the following will be the seventh day from 10

$^{th}$ of that month?

0%
Wednesday
0%
Thursday
0%
Friday
0%
Saturday

At what angle are the hands of clock inclined at 30 minutes past 6?

0%
$7 \frac{1^{\circ}}{2}$
0%
$11 \frac{1^{\circ}}{2}$
0%
$15^{\circ}$
0%
$23^{\circ}$

Explanation

$30 \times 6 - 11/2 \times 30 = 15^{\circ}$

If 21

$^{st}$ November falls five days before Wednesday, then what will be the day on 25

$^{th}$ December?

0%
Monday
0%
Tuesday
0%
Wednesday
0%
Thursday

Value of 0! is always 1.

0%
True
0%
False
0%
Either
0%
Neither

Explanation

we know

$1!=1$

Also

$n!=n\times (n-1)\times (n-2)........3\times 2\times 1\\ n!=n\times (n-1)!\\ 1!=1(1-1)!\\ 1=1(0)!\\ 0!=1$

$0!$ is always

$1$

Hence its true that

$0!$ is always

$1$

A train

$140$ m long passes a telegraph post in

$14$ sec. Find its speed in km/hr .

0%
$36$ km/hr
0%
$56$ km/hr
0%
$44$ km/hr
0%
$51$ km/hr

Explanation

Given, Distance

$=140$

$m= 0.14$

$km$ ,

Time

$=14 sec=\displaystyle \frac { 14 }{ 3600 } hr$

$\therefore$ Speed

$=\displaystyle \frac {Distance}{Time} = \frac { 0.14 }{ 14 } \times 3600$

$= 36$

$km/hr$ .

If 09/12/2001 happens to be Sunday,then 09/12/1971 would have been at

0%
Wednesday
0%
Tuesday
0%
Saturdays
0%
Thursday

The least count of a stop watch is generally :

0%
1 s
0%
1 min
0%
0.5 s
0%
1 h

Find the mean of:

$5, 15, 20, 8$ and

$12$ .

0%
$12$
0%
$11$
0%
$10$
0%
$9$

Explanation

The observations are:

$5, 15, 20, 8, 12$
Mean

$= \cfrac{\text{Sum}}{\text{Number of observations}}$
Mean

$= \cfrac{5 + 15 + 20 + 8 + 12}{5}$
mean

$= \cfrac{60}{5} = 12$

Find the mean of

$43, 51, 50, 57$ and

$54$

0%
$50$
0%
$42.5$
0%
$55$
0%
$57$

Explanation

The observations are :

$43, 51, 50, 57$ and

$54$
mean

$= \cfrac{\text{Sum}}{\text{Count of numbers}}$
mean

$= \cfrac{43 + 51 + 50 +57 +54}{5}$
mean

$= \cfrac{255}{5}$
mean

$= 51$

Find the mean of

$43, 51, 50, 57$ and

$54$ .

0%
$51$
0%
$57$
0%
$40$
0%
$43$

Explanation

Mean of

$43, 51, 50, 57, 54$
Mean

$= \cfrac{\text{Sum}}{\text{Number of observations}}$
Mean

$= \cfrac{43 + 51 + 50 + 57 + 54}{5} = \cfrac{255}{5} = 51$

Find the mean of the following data:

$18, 33, 30, 21$ and

$13$ .

0%
$18$
0%
$23$
0%
$25$
0%
$15$

Explanation

Given set of data is

$18,33,30,21,13$

Arranging the data in ascending order, we get

$13,18,21,30,33$

No of terms(numbers)

$=n=5$

Mean

$=\dfrac{\text{Sum of numbers}}{n}$

$=\dfrac{13+18+21+30+33}{5}=\dfrac{115}{5}=23$

Thus, Mean of given data is

$23$ .

Hence, option B is correct.

$12$ men can complete a piece of work in

$16$ days. How many days will

$4$ men take to complete the task?

0%
$60$ days
0%
$45$ days
0%
$54$ days
0%
$48$ days

Explanation

$1$ man will complete the job in

$16 \times 12$ days.
So,

$4$ will do in

$\displaystyle \frac{12 \times 16}{4} = 48$ days.

Find the mean of the observations

$8, 12, 16, 22, 10$ and

$4$ .

0%
$13$
0%
$15$
0%
$14$
0%
$12$

Explanation

The given observations are:

$8, 12, 16, 10, 22, 4$
Mean

$= \cfrac{\text{Sum}}{\text{Number of observations}}$
Mean

$= \cfrac{8+ 12 + 16 + 10 + 22 + 4}{6} = \cfrac{72}{6} = 12$

The mean of first six natural numbers is

0%
$5$
0%
$3.5$
0%
$4$
0%
None of these

Explanation

$\textbf{Step -1: Find the sum of first six natural numbers.}$

$\text{The first six natural numbers are }1,2,3,4,5,6$

$\text{Sum of these observations}=1+2+3+4+5+6$

$=21$

$\textbf{Step -2: Find the mean of first six natural numbers.}$

$\text{Here, number of observations}=6$

$\mathbf{\because\text{Mean}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}.}$

$\therefore\text{Mean of first six natural numbers}=\dfrac{21}{6}$

$=3.5$

$\textbf{Final Answer: Hence, the correct option is B.}$

$X$ is twice as good as a workman as

$Y$ .

$X$ finished a piece of work in 3 hours less than

$Y$ . In how many hours could they have finished that piece of work together?

0%
3
0%
2
0%
4
0%
5

Explanation

$X$ finished a work in 3 hours less than

$Y$ .

$X$ is twice as good as a workman as

$Y$ . Therefore,

$Y$ finished the same work in 6 hours.

$\therefore \left( X+Y \right)$ 's one hour's work

$= \displaystyle\frac { 1 }{ 3 } + \displaystyle\frac { 1 }{ 6 } = \displaystyle\frac { 3 }{ 6 } = \displaystyle\frac { 1 }{ 2 }$

$\therefore X$ and

$Y$ together can finish the whole work in 2 hours.

Sachin scored

$80$ and

$120$ runs respectively in the two innings of a cricket match. What would have been his average score in each innings?

0%
$80$
0%
$120$
0%
$200$
0%
$100$

Explanation

$\Rightarrow$ Runs scored by Sachin in fist innings

$=80$

$\Rightarrow$ Runs scored by Sachin in fist innings

$=120$

$\Rightarrow$ Total runs scored in two innings by Sachin

$=80+120=200$

$\Rightarrow$ Average scored of Sachin in each innings

$=\dfrac{200}{2}=100$

The average of

$33.5, 30.4, 25.6, 31.5\ and \ 29$ is:

0%
$28.5$
0%
$30$
0%
$29.7$
0%
$30.4$

Explanation

$\displaystyle = \frac{33.5+30.4+25.6+31.5+29}{5}$

$\displaystyle = \frac{150}{5}=30$

A car moves with a speed of

$80$ km/hr. What is the speed of the car in meters per second?

0%
$\displaystyle 22 \frac{2}{9}$ m/sec
0%
$8$ m/sec
0%
$\displaystyle 20 \frac{1}{9}$ m/sec
0%
$\displaystyle 8 \frac{2}{3}$ m/sec

Explanation

$Speed = 80 km/hr = \displaystyle \left (80 \times \frac{5}{18} \right ) m/sec = 22 \frac{2}{9} m/sec.$

The runs scored by Sachin in

$5$ test matches are

$140$ ,

$153$ ,

$148$ ,

$150$ and

$154$ respectively. Find his mean

0%
$150$
0%
$149$
0%
$148$
0%
$147$

Explanation

Runs score by Sachin in

$5$ test matches:

$140,153,148,150,154$
Mean of runs

$= \cfrac{\text{Total runs}}{\text{Number of matches}}$
Mean

$= \cfrac{140 + 153 + 148 + 150 + 154}{5}$
Mean

$= \cfrac{745}{5}$
Mean

$= 149$

In a cricket test match the scores of ten players are:

$85$ ,

$32$ ,

$0$ ,

$54$ ,

$29$ ,

$101$ ,

$73$ ,

$64$ ,

$29$ and

$36$ . Find the mean of the runs.

0%
$36.4$ runs
0%
$45.5$ runs
0%
$50.3$ runs
0%
$63.7$ runs

Explanation

Runs scored by

$10$ players

$= 85,32,0,54,29,101,73,64,29,36$
Mean

$= \cfrac{\text{Sum}}{\text{Number of players}}$
Mean

$= \cfrac{85 + 32 + 0 + 54 + 29 + 101 + 73 + 64 + 29 + 36}{10}$
Mean

$= \cfrac{503}{10}$
Mean

$= 50.3$

The average of

$0.3, 0.03$ and

$0.003$ is ____

0%
$0.3$
0%
$0.1$
0%
$0.111$
0%
none of these

Explanation

$\text{Average} = \displaystyle \frac{0.3+0.03+0.003}{3}$

$=\displaystyle \frac{0.333}{3}\\=0.111$

The average of

$5, 0, 6,$

$\displaystyle \frac{1}{4}$ and

$\displaystyle 8\frac{3}{4}$ is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$\displaystyle \frac{5+0+6+\frac{1}{4}+8\frac{3}{4}}{5}$
=

$\displaystyle \frac{20}{5}=4$

Find the mean of first ten odd natural numbers.

0%
$7$
0%
$10$
0%
$13$
0%
$18$

Explanation

${\textbf{Step - 1: Listing first 10 odd natural numbers}}$

${\text{First 10 odd natural numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19}}$

${\textbf{Step - 2: Calculating mean}}$

${\text{We know that, mean of n numbers = }}\dfrac{{{\text{Sum of n numbers}}}}{{\text{n}}}$

$\therefore {\text{ Mean of first 10 odd natural numbers,}}$

${\text{M = }}\dfrac{{{\text{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}}}}{{{\text{10}}}}$

$\Rightarrow {\text{ M = }}\dfrac{{{\text{100}}}}{{{\text{10}}}}{\text{ = 10}}$

${\textbf{Thus, the mean of first 10 odd natural numbers is 10}}$

Twenty four men can complete a work in sixteen days. Thirty two women can complete the same work in twenty four days. Sixteen men and sixteen women started worked for 12 days. How many more men are to be added to complete the remaining work in 2 days ?

0%
16
0%
24
0%
36
0%
48

Explanation

1 man's 1 days' work =

$\displaystyle \frac{1}{24\times 16}$

$\therefore$ 16 men's 1 days' work =

$\displaystyle \frac{16}{24\times 16}= \frac{1}{24}$ 1 woman's 1 days' work =

$\displaystyle \frac{1}{32\times 24}$
16 women's 1 days' work =

$\displaystyle \frac{16}{32\times 24}=\frac{1}{48}$

$\therefore$ 12 days' work of (16 men + 16 women)

$\displaystyle =12\left ( \frac{1}{24}+\frac{1}{48} \right )=12\left ( \frac{2+1}{48} \right )=12\times \frac {3}{48}=\frac {3}{4}$
Remaining work =

$\displaystyle 1-\frac{3}{4}=\frac{1}{4}$ Now (16 men's + 16 women's) 2 days' work =

$\displaystyle 2\left ( \frac{1}{24}+\frac{1}{48} \right )=2\times \frac{1}{16}=\frac{1}{8}$

$\therefore$ Remaining work =

$\displaystyle \frac{1}{4}-\frac{1}{8}=\frac{1}{8}$

$\displaystyle \Rightarrow \frac{1}{384}$ work will be done in 2 days by

$\displaystyle \left ( 384\times \frac{1}{8}\times \frac{1}{2} \right )=24$ men

A contractor undertook to do a certain piece of work in

$9$ days. He employed certain number of men, but

$6$ of them being absent from the very first day, the rest could finish the work in

$15$ days. The number of men originally employed were

0%
$12$
0%
$15$
0%
$18$
0%
$24$

Explanation

Let there be

$x$ men at the begining less men, more days

$\therefore$

$15 : 9 :: x : (x - 6)$

$\Rightarrow$

$15(x - 6) = 9x$

$\Rightarrow$

$6x = 90$

$\Rightarrow$

$x = 15$

The average of eight numbers is

$38.4$ and the average of seven of them is

$39.2$ . What is the eighth number?

0%
$0.8$
0%
$32.8$
0%
$34.8$
0%
$33.8$

Explanation

Sum of eight numbers

$= 38.4 \times 8 = 307.2$

$\because$ Total sum

$=$ Average

$\times$ Number of items
Sum of seven numbers

$=39.2 \times 7 = 274.4$

$\therefore$ Eight number

$= 307.2 - 274.4 = 32.8$

Two pipes

$A$ and

$B$ can separately fill a cistern in

$60\ minutes$ and

$75\ minutes$ respectively. There is a third pipe at the bottom of the cistern to empty it , If all the three pipes are simultancously opened, then the cistern is full in

$50\ minutes$ , In how much time, the third pipe alone can empty the cistern ?

0%
$110\ minutes$
0%
$100\ minutes$
0%
$120\ minutes$
0%
$90\ minutes$

Explanation

Let the third pipe empty the whole cistern in x minutes.

$\displaystyle \therefore \frac{1}{60}+\frac{1}{75}-\frac{1}{x}=\frac{1}{50}$

$\displaystyle \Rightarrow \frac{1}{x}=\frac{1}{60}+\frac{1}{75}-\frac{1}{50}=\frac{5+4-6}{300}=\frac{3}{300}=\frac{1}{100}$
The third pipe can empty the cistern in 100 minute

A fort had provision of food for

$150$ men for

$45$ days. After

$10$ days,

$25$ men left the fort, The number of days for which the remaining food will last, is

0%
$29\displaystyle \frac{1}{5}$
0%
$37\displaystyle \frac{1}{4}$
0%
$42$
0%
$54$

Explanation

After

$10$ days:

$150$ men had food for

$35$ days.
Suppose

$125$ men had food for

$x$ days. Less men, More days; More men less days.

$\therefore$

$125 : 150 : : 35 : x$

$\Rightarrow\, 125\, \times\, x\, =\, 150\, \times\, 35$

$\Rightarrow\, x\, =\, \displaystyle \frac{150\, \times\, 35}{125}\, \Rightarrow\, x\, =\, 42$

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 1 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 1 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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