MCQ Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 5

If February, 1, 1996 is Wednesday, what day is March 3, 1996?

0%
Monday
0%
Sunday
0%
Saturday
0%
Friday

$12$ 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.
Hence, option

$D$ is the correct answer.

A clock strikes the number of times of the hour. How many strikes does it make in one day?

0%
$165$ times
0%
$156$ times
0%
$78$ times
0%
$87$ times

Explanation

For the first 12 hours of the day,the clock will strike

$1+2+3+......+12 = \cfrac{12}{2}(1+12) =78$ times
For the next

$12$ hours, there will be another 78 times, so in one day,the clock willl strike

$156$ times

If the mean of

$7$ ,

$9$ ,

$11$ ,

$13$ ,

$x$ ,

$21$ is

$13$ , find the value of

$x$ .

0%
$17$
0%
$71$
0%
$15$
0%
$20$

A fort had provisions for

$450$ soldiers for

$40$ days. After

$10$ days,

$90$ more soldiers came to fort. Find for how many days will the remaining provisions last, if consumed at the same rate ?

0%
$15$ Days
0%
$25$ Days
0%
$35$ Days
0%
$5$ Days

Explanation

since

$450$ soldiers had provision for

$40$ days in a fort.

After

$10$ days:

$450$ soldiers will remain with

$30$ days of provision

therefore for

$540$ soldiers provision will decrease to

$x$ days

therefore By inverse proportion,

$x= \dfrac{450 * 30}{540}$

$\therefore x=25$ days

Answer is

$25$ days

Twelve men can do a job in

$8$ days. Six days after they start

$4$ more men join them. How many more days will it take to do the job?

0%
$2.5$ days
0%
$3.5$ days
0%
$1.5$ days
0%
$6$ days

Explanation

Let

$x$ be the units of works per man day.

$\therefore$ Total work

$=12 \times 8=96$ units of works.

After 6 days,

$12 \times 6=7$ units of work will be completed.

Hence, the remaining work

$=96-72=24$

Now, more days required to complete the work by 16 men will be

$=\dfrac{24}{16}=1.5$ days.

A tea party is arranged for

$16$ people along two sides of a large table with

$8$ chairs on each side. Four men to sit on one particular side and two on the other side. In how many ways can they be seated?

0%
$\displaystyle\frac{8!\space8!\space10!}{(4!)^2\space6!}$
0%
$\displaystyle\frac{8!\space8!\space10!}{6!}$
0%
$\displaystyle\frac{8!\space8!\space10!}{4!!}$
0%
$\displaystyle\frac{8!\space8!\space10!}{4!\space6!}$

Explanation

There are

$8$ chairs on each side of the table. Let sides be represented by

$A$ and

$B$ . Let four persons sit on side

$A$ , then number of ways arranging

$4$ persons on

$8$ chairs on side

$A = \space ^8P_4$ and then two persons sit on side

$B$ , then the number of ways arranging

$2$ persons on

$8$ chairs on side

$B = \space ^8P_2$ and arranging the remaining

$10$ persons in remaining

$10$ chairs in

$10!$ ways.
Hence the total number of ways in which the persons can be arranged

$\quad = \space ^8P_4\times\space^8P_2\times\space10!$

$\quad = \displaystyle\frac{8!\space8!\space10!}{4!\space6!}$

Shrikant reads the following number of pages of a book daily. Find the mean.

$34, 43, 29, 51, 36, 41, 38, 35, 34, 36, 34, 38, 40, 36, 35,$

$35, 34, 35, 36, 34, 38, 38, 35, 36, 34, 38, 36, 34, 35, 36$

0%
$35.47$
0%
$35.67$
0%
$36$
0%
$36.47$

Explanation

The data numbers are:

$34, 43, 29, 51, 36, 41, 38, 35, 34, 36, 34, 38, 40, 36, 35, 35, 34, 35, 36, 34, 38, 38, 35, 36, 34, 38, 36, 34, 35, 36$
Mean =

$\cfrac{\text{Sum}}{\text{Total Numbers}}$
Sum of Numbers

$= 34 + 43 + 29 + 51 + 36 + 41 + 38 + 35 + 34 + 36 + 34 + 38 + 40 + 36 + 35 + 35$

$+ 34 + 35 + 36 + 34 + 38 + 38 + 35 + 36 + 34 + 38 + 36 + 34 + 35 + 36$
Sum

$= 1094$
Total Numbers

$= 30$
Mean

$= \cfrac{1094}{30}$
Mean

$= 36.47$

The mean of

$3, a+ 2, 8, 12, 2a -1$ and

$6$ is

$7$ , find the value of

$a$ .

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

Explanation

Given, mean is

$7$ .

Therefore, mean

$=\dfrac {3+a+2+8+12+2a-1+6}{6}$

$\Rightarrow 7=\dfrac {30+3a}{6}$

$\Rightarrow 42=30+3a$

$\Rightarrow 3a=12$

$\Rightarrow a=4$

$20$ th September in a year is Wednesday, the number of Fridays in the same month is

0%
$6$
0%
$5$
0%
$4$
0%
$3$

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

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

Explanation

$26$ January

$2015$

$=$ (

$2014$ years

$+$ period from

$1.1.2015 to 26.1.2015$ )
Odd days in

$1600$ years

$= 0$
Odd days in

$400$ years

$= 0$

$14$ years =(

$11$ ordinary years +

$3$ leap years )=

$11\times 1+3\times 2=17$ odd days
January

$=26$ days
Therefore

$26$ days

$=3$ weeks

$+ 5$ days

$= 5$ odd days
Total number of odd days

$= 17+5=22$
if we divide it by

$7$ then =

$1$ odd day
Therefore, given day is Monday.

Seven girls are to dance in a circle. In how many different ways can they stand on the circumference of the circle?

0%
$120$
0%
$640$
0%
$720$
0%
$840$

Explanation

Since circular permutations of

$n$ objects is

$(n-1)!$
Hence, the number of ways is

$\quad (7-1)!=6!=720$

$45$ men working

$8$ hours a day can finish a project within a given time. Five men are unable to report for the work. How long should the rest work a day to finish the project on time.

0%
$\;10\:hrs$
0%
$\;12\:hrs$
0%
$\;9\:hrs$
0%
$\;8\:hrs$

Explanation

Total Work Hours are

$\Rightarrow 45\times 8$

When

$5$ Do not Report ,then Only

$45-5=40$ men will remain

So Now, Let us Assume the No of Hours be

$H$ they now have to work to finish the project

$\Rightarrow 40\times H=45\times 8$

$\Rightarrow H=9$

Therefoe Answer is

$(C)$

$16$ men can complete a piece of work in

$8$ days. In how many days can

$12$ men complete the same piece of work?

0%
$11\dfrac{1}{5}$ days
0%
$10\dfrac{2}{3}$ days
0%
$10\dfrac{1}{7}$ days
0%
$11\dfrac{5}{2}$ days

Explanation

$16$ men can complete a piece of work in

$8$ days

$\therefore\,1$ man can complete the work in

$(16\times8)$ days (Less men more days so multiply)

$\therefore\,12$ men can complete the work in

$\displaystyle\frac{(16\times8)}{12}$ days (More men less days, so divide)

$=\displaystyle\frac{32}{3}$ days

$=10\displaystyle\frac{2}{3}$ days.

$A$ and

$B$ together can do a piece of work in 8 days.

$B$ alone can do it in 12 days.

$B$ alone works at it for 4 days. In how many more days after that could

$A$ alone complete it.

0%
15 days
0%
18 days
0%
16 days
0%
20 days

Explanation

$A$ alone can do

$\displaystyle\frac { 1 }{ 8 } - \displaystyle\frac { 1 }{ 12 } = \displaystyle\frac { 3-2 }{ 24 } = \displaystyle\frac { 1 }{ 24 }$ of the work in 1 day, i.e.,

$A$ alone can do the same work in 24 days.

$B$ 's 1 day's work

$= \displaystyle\frac { 1 }{ 12 }$

$\Rightarrow B$ 's 4 day's work

$=\displaystyle\frac { 4 }{ 12 } = \displaystyle\frac { 1 }{ 3 }$

$\therefore$ Remaining

$\displaystyle\frac { 2 }{ 3 }$ of the work is done by

$A$ alone in

$\displaystyle\frac { 2 }{ 3 } \times 24 = 16$ days.

In a Zonal athletic long jump meet the distances jumped by

$10$ atheletes are:

$205\:cm, 200\:cm, 275\:cm, 260\:cm, 259\:cm, 199\:cm, 252\:cm, 239\:cm, 228\:cm$ and

$281\:cm$ . Find the arithmetic mean of the jumps.

0%
$212.4\:cm$
0%
$222.5\:cm$
0%
$230.8\:cm$
0%
$239.8\:cm$

Explanation

Distances jumped by

$10$ athletes (in

$cm$ ):

$205, 200, 275, 260, 259, 199, 252, 239, 228$ and

$281$
Mean

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

$= \cfrac{205 +200 +275 +260 +259 +199 +252 +239 + 228 + 281}{10}$
Mean

$= \cfrac{2398}{10}$
Mean

$= 239.8$

The ages of

$5$ children are

$13, 15, 11, 9$ and

$8$ years respectively. The average age is

0%
$10.2$
0%
$12.2$
0%
$11.2$
0%
$11$

Explanation

Average is the sum of the data values divided by the total number of values.

Given that,

$13, 15, 11, 9, 8$ are the ages of

$5$ children.

Therefore, average age

$=\dfrac{13+15+11+9+8}{5}=\dfrac{56}{5}=11.2$

The average of first

$4$ prime numbers is?

0%
$4$
0%
$4.25$
0%
$5$
0%
$6.25$

Explanation

Average is the sum of the data values divided by the total values.

The first

$4$ prime numbers are

$2,3,5,7$

Therefore, average of first

$4$ prime numbers

$=\dfrac{2+3+5+7}{4}=\dfrac{17}{4}=4.25$

A can finish a work in

$18$ days and

$B$ can do the same work in half the time taken by

$A.$ Then working together what part of the same work can they finish in a day?

0%
$\displaystyle \frac{1}{6}$
0%
$\displaystyle \frac{1}{9}$
0%
$\displaystyle \frac{2}{5}$
0%
$\displaystyle \frac{2}{7}$

Explanation

A can finish a work in

$18$ days.
B can finish a work in

$9$ days.
So In one day A can finish

$\dfrac{1}{18}$ work.

and In one day B can finish

$\dfrac{1}{9}$ work.
So, working together the work can they finish in a day.

$W=\dfrac { 1 }{ 18 } +\dfrac { 1 }{ 9 } =\dfrac { 3 }{ 18 } =\dfrac { 1 }{ 6 }$
Answer is (A)

$\displaystyle, \frac{1}{6}$ .

A man and a boy can do a piece of work in 24 days. If the man works alone for last 4 days, it is complete in 5 days. How long would the boy take to do it alone?

0%
120 days
0%
20 days
0%
80 days
0%
36 days

Explanation

(Man + boy)'s one day's work

$= \displaystyle\frac{1}{24}$
(Man + boy)'s 20 day's work

$= \displaystyle\frac{20}{24} = \displaystyle\frac{5}{6}$
Remaining

$\displaystyle\frac{1}{6}$ of the work is done by the man alone in 5 days, i.e., the man alone can do

$\displaystyle\frac{1}{30}$ of the work in 1 day.

$\therefore$ Boy's one day's work

$= \displaystyle\frac{1}{24} - \displaystyle\frac{1}{30} = \displaystyle\frac{1}{120}$ .

$\therefore$ Boy alone would take 120 days to complete the work.

$A$ and

$B$ together can complete a piece of work in 16 days while

$B$ and

$C$ together can complete the same work in 12 days and

$A$ and

$C$ together in 24 days. Find out the number of days that

$A$ will take to complete the work, working alone.

0%
36 days
0%
48 days
0%
96 days
0%
80 days

Explanation

1 day's work of

$\left( A + B \right) + \left( B + C \right) + \left( C + A \right) = \displaystyle\frac { 1 }{ 16 } + \displaystyle\frac { 1 }{ 12 } + \displaystyle\frac { 1 }{ 24 } = \displaystyle\frac { 9 }{ 48 }$

$\Rightarrow$ 1 day's work of

$\left( A + B + C \right) = \displaystyle\frac { 1 }{ 2 } \times \displaystyle\frac { 9 }{ 48 } = \displaystyle\frac { 9 }{ 96 }$

$\therefore$ 1 day's work of

$A$ alone

$= \displaystyle\frac { 9 }{ 96 } - \displaystyle\frac { 1 }{ 12 } = \displaystyle\frac { 1 }{ 96 }$

$\therefore A$ can do the whole work alone in

$96$ days.

The average of the four even numbers between

$31$ and

$39$ is

0%
$35$
0%
$21$
0%
$20$
0%
$19$

Explanation

Average is the sum of the data values divided by the total values.

The

$4$ even numbers between

$31$ and

$39$ are

$32,34,36,38$

Therefore, average of even numbers between

$31$ and

$39$

$=\dfrac{32+34+36+38}{4}=\dfrac{140}{4}=35$

A man and a boy received Rs.

$800$ as wages for

$5$ days for the work they did together. The man's efficiency in the work was three times that of the boy. What are the daily wages of the boy?

0%
Rs. $76$
0%
Rs. $56$
0%
Rs. $44$
0%
Rs. $40$

Explanation

5 day's wages for a man and a boy = Rs.

$800$

$\therefore$ 1 days' wage for a man and a boy = Rs.

$\displaystyle\dfrac{800}{5}$
= Rs.

$160$
Let the wage of the boy per day be Rs.

$x$ . Then, Wage of the man per day = Rs.

$3x$
Given

$3x + x = 160$

$\Rightarrow 4x = 160 \Rightarrow x = 40$

$A$ can do a piece of work in

$4$ hours,

$B$ and

$C$ together in

$3$ hours and

$A$ and

$C$ together in

$2$ hours. How long will

$B$ alone take to do it?

0%
$10$ hours
0%
$12$ hours
0%
$8$ hours
0%
$24$ hours

Explanation

$A$ 's 1 hour's work

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

$\left( B+C \right)$ 's 1 hour's work

$= \displaystyle\frac { 1 }{ 3 }$

$\left( A+C \right)$ 's 1 hour's work

$= \displaystyle\frac { 1 }{ 2 }$

$\therefore C$ 's 1 hour's work

$= \displaystyle\frac { 1 }{ 2 } - \displaystyle\frac { 1 }{ 4 } = \displaystyle\frac { 1 }{ 4 }$

$B$ 's 1 hour's work

$= \displaystyle\frac { 1 }{ 3 } - \displaystyle\frac { 1 }{ 4 } = \displaystyle\frac { 1 }{ 12 }$
Hence,

$B$ will complete the whole work in 12 days.

$10$ men and

$15$ women finish a work in

$6$ days. One man alone finishes that work in

$100$ days. In how many days will a women finish the work?

0%
$125$ days
0%
$150$ days
0%
$90$ days
0%
$225$ days

Explanation

1 man can finish the whole work in 100 days

$\therefore$ 1 man's 1 days' work

$= \displaystyle\frac { 1 }{ 100 }$

$\Rightarrow$ 10 men's 1 days' work

$= 10 \times \displaystyle\frac { 1 }{ 100 } = \displaystyle\frac { 1 }{ 10 }$
Let the time taken by one woman to finish the work be

$x$ days.
Then, 1 woman's 1 days' work

$= \displaystyle\frac { 1 }{ x }$

$\Rightarrow$ 15 women's 1 days' work

$= \displaystyle\frac { 15 }{ x }$
Given, (10 men's + 15 women's ) 1 days' work

$= \displaystyle\frac { 1 }{ 6 }$

$\Rightarrow \displaystyle\frac { 1 }{ 10 } + \displaystyle\frac { 15 }{ x } = \displaystyle\frac { 1 }{ 6 } \Rightarrow \displaystyle\frac { 15 }{ x } = \displaystyle\frac { 1 }{ 6 } - \displaystyle\frac { 1 }{ 10 } = \displaystyle\frac { 5-3 }{ 30 } = \displaystyle\frac { 2 }{ 30 } = \displaystyle\frac { 1 }{ 15 }$

$\Rightarrow x = 225$ days.

The time taken by 4 men to complete a job is double the time taken by 5 children to complete the same job. Each man is twice as fast as a woman. How long will 12 men, 10 children and 8 women take to complete a job, given that a child would finish the job in 20 days.

0%
2 days
0%
$2\displaystyle\frac{1}{8}$ days
0%
4 days
0%
1 day

Explanation

Time taken by 5 children to complete the job

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

$= 4$ days

$\therefore$ Time taken by 4 men to complete the job

$= \left( 4 \times 2 \right)$ days

$= 8$ days
Time taken by 4 women to the complete the job

$= \left( 8 \times 2 \right)$ days

$= 16$ days
Therefore, 5 children's 1 day's work

$= \displaystyle\frac { 1 }{ 4 }$
4 men's 1 day's work

$= \displaystyle\frac { 1 }{ 8 }$
4 women's 1 day's work

$= \displaystyle\frac { 1 }{ 16 }$

$\therefore$ (12 men's + 10 children's + 8 women's )
1 day's work

$= 3 \times$ (4 men's 1 day's work)

$+ 2 \times$ (5 children's 1 day's work)

$+ 2 \times$ (4 women's 1 day's work)

$= 3 \times \displaystyle\frac { 1 }{ 8 } + 2\times \displaystyle\frac { 1 }{ 4 } + 2 \times \displaystyle\frac { 1 }{ 16 }$

$= \displaystyle\frac { 3 }{ 8 } + \displaystyle\frac { 1 }{ 2 } + \displaystyle\frac { 1 }{ 8 } = \displaystyle\frac { 8 }{ 8 } = 1$

$\therefore$ Required time

$= 1$ day.

$A$ can do a piece of work in 25 days and

$B$ in 20 days. They work together for 5 days and then

$A$ goes away. In how many days will

$B$ finish the remaining work?

0%
17 days
0%
11 days
0%
10 days
0%
None of these

Explanation

${\textbf{Step- 1: Calculating the work they do in 1 day}}$

${\text{A does the work in 25 days,}}$

$\therefore {\text{ The work A does in 1 day = }}\dfrac{{\text{1}}}{{{\text{25}}}}$

${\text{B does the work in 20 days,}}$

$\therefore {\text{ The work B does in 1 day = }}\dfrac{{\text{1}}}{{{\text{20}}}}$

${\text{A's and B's toghether work in 1 day = }}\dfrac{{\text{1}}}{{{\text{25}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{20}}}}{\text{ = }}\dfrac{{\text{9}}}{{{\text{100}}}}$

${\textbf{Step- 2: Solving further}}$

${\text{A's and B's work toghether in 5 days = 5 }} \times {\text{ }}\dfrac{{\text{9}}}{{{\text{100}}}}{\text{ = }}\dfrac{{{\text{45}}}}{{{\text{100}}}}$

${\text{Work remaining after 5 days = 1 - }}\dfrac{{{\text{45}}}}{{{\text{100}}}}{\text{ = }}\dfrac{{{\text{55}}}}{{{\text{100}}}}$

${\text{This work is done by B alone,}}$

$\therefore {\text{ Time taken = }}\dfrac{{{\text{Work remaining}}}}{{{\text{Work B does in 1 day}}}}$

$\Rightarrow {\text{ Time taken = }}\dfrac{{\dfrac{{{\text{55}}}}{{{\text{100}}}}}}{{\dfrac{{\text{1}}}{{{\text{20}}}}}}{\text{ = 11}}$

${\textbf{Hence option B is correct }}$

$A$ can cultivate

$\displaystyle\frac{2}{5}$ of a land in 6 days and

$B$ can cultivate

$\displaystyle\frac{1}{3}$ of the same land in 10 days. Working together,

$A$ and

$B$ can cultivate

$\displaystyle\frac{4}{5}$ of the land in:

0%
4 days
0%
5 days
0%
8 days
0%
10 days

Explanation

$A$ can cultivate the full part of the land in

$\displaystyle\frac { 6 \times 5 }{ 2 }$ days

$= 15$ days

$B$ can cultivate the full part of the land in

$\left( 10 \times 3 \right) = 30$ days

$\therefore$ Part of the land cultivated by

$\left( A + B \right)$ in 1 day

$= \left( \displaystyle\frac { 1 }{ 15 } + \displaystyle\frac { 1 }{ 30 } = \displaystyle\frac { 2 + 1 }{ 30 } = \displaystyle\frac { 3 }{ 30 } \right) = \displaystyle\frac { 1 }{ 10 }$

$\displaystyle\frac { 1 }{ 10 }$ part of the land is cultivated by

$\left( A + B \right)$ in 1 day

$\therefore \displaystyle\frac { 4 }{ 5 }$ part of the land will be cultivated by

$\left( A + B \right)$ in

$10 \times \displaystyle\frac { 4 }{ 5 }$ days

$= 8$ days.

$A$ and

$B$ can do a piece of work in 30 days. While

$B$ and

$C$ can do the same work in 24 days and

$C$ and

$A$ in 20 days. They all work together for 10 days before

$B$ and

$C$ leave. How many days more will

$A$ take to finish the work?

0%
18 days
0%
24 days
0%
30 days
0%
36 days

Explanation

$\left( A + B \right)$ 's one day's work

$= \displaystyle\frac { 1 }{ 30 }$

$\left( B + C \right)$ 's one day's work

$= \displaystyle\frac { 1 }{ 24 }$ ..........

$(1)$

$\left( C + A \right)$ 's one day's work

$= \displaystyle\frac { 1 }{ 20 }$

$\therefore \left( A + B + C \right)$ 's one day's work

$= \displaystyle\frac { 1 }{ 2 } \left[ \displaystyle\frac { 1 }{ 30 } + \displaystyle\frac { 1 }{ 24 } + \displaystyle\frac { 1 }{ 20 } \right] = \displaystyle\frac { 1 }{ 2 } \times \left[ \displaystyle\frac { 20 + 25 + 30 }{ 600 } \right] = \displaystyle\frac { 75 }{ 1200 } = \displaystyle\frac { 1 }{ 16 }$ ...........

$(2)$

$\left( A + B + C \right)$ 's 10 day's work

$= \displaystyle\frac { 10 }{ 16 } = \displaystyle\frac { 5 }{ 8 }$

Remaining work,

$1-\dfrac{5}{8}=\dfrac{3}{8}$ (which is done by A)

From

$(1)$ and

$(2)$ ,

$A$ 's one day's work

$= \displaystyle\frac { 1 }{ 16 } - \displaystyle\frac { 1 }{ 24 } = \displaystyle\frac { 1 }{ 48 }$

$\therefore$ Remaining

$\displaystyle\frac { 3 }{ 8 }$ of the work is done by

$A$ alone in

$\displaystyle\frac { 3 }{ 8 } \times 48 = 18$ days.

$A, B$ and

$C$ can complete a work in

$10, 12\ and\ 15$ days respectively. They started the work together, but

$A$ left the work before 5 days of its completion.

$B$ also left the work

$2$ days after

$A$ left. In how many days was the work completed.

0%
$4$
0%
$5$
0%
$7$
0%
$8$

Explanation

Let the work be completed in

$x$ days.

$A$ 's 1 days' work

$= \displaystyle\frac { 1 }{ 10 }$

$\Rightarrow A$ 's

$\left( x-5 \right)$ days' work

$= \displaystyle\frac { \left( x-5 \right) }{ 10 }$

$B$ 's 1 days' work

$= \displaystyle\frac { 1 }{ 12 }$

$\Rightarrow B$ 's

$\left( x-3 \right)$ days' work

$= \displaystyle\frac { \left( x-3 \right) }{ 12 }$

$C$ 's 1 days' work

$= \displaystyle\frac { 1 }{ 15 }$

$\Rightarrow C$ 's

$x$ days' work

$= \displaystyle\frac { x }{ 15 }$

$\Rightarrow \displaystyle\frac { x-5 }{ 10 } + \displaystyle\frac { x-3 }{ 12 } + \displaystyle\frac { x }{ 15 } = 1$

$\Rightarrow \displaystyle\frac { 6\left( x-5 \right) + 5\left( x-3 \right) + 4x }{ 60 } = 1$

$\Rightarrow 6x - 30 + 5x - 15 + 4x = 60$

$\Rightarrow 15x = 60 + 45 = 105 \Rightarrow x = 7$

$\therefore$ The work was completed in

$7$ days.

$A$ is twice as fast as workman as

$B$ and together they finish a piece of work in 14 days. In how many days can

$A$ alone finish the work?

0%
$18$ days
0%
$21$ days
0%
$24$ days
0%
$27$ days

Explanation

$B$ finishes the work in

$x$ days, then

$A$ finishes the work in

$\displaystyle\dfrac { x }{ 2 }$ days.

$\Rightarrow B$ 's one day's work

$= \displaystyle\dfrac { 1 }{ x }$

and

$A$ 's one day's work

$= \displaystyle\dfrac { 2 }{ x }$

Given

$\displaystyle\dfrac { 1 }{ x } + \displaystyle\dfrac { 2 }{ x } = \displaystyle\dfrac { 1 }{ 14 } \Rightarrow \displaystyle\dfrac { 3 }{ x } = \displaystyle\dfrac { 1 }{ 14 }$

$\Rightarrow x = 14 \times 3 = 42$ days

$\therefore A$ alone can finish the work in

$\displaystyle\dfrac { 42 }{ 2 } = 21$ days.

$A$ can do a piece of work in

$20$ days and

$B$ in

$40$ days.

$A$ begins the work and there after they work alternately one on each day. On which day will the work be completed?

0%
$24$ th day
0%
$25$ th day
0%
$26$ th day
0%
$27$ th day

Explanation

Work done by

$\left( A+B \right)$ in 2 days

$= \left( \displaystyle\dfrac { 1 }{ 20 } + \displaystyle\dfrac { 1 }{ 40 } \right) = \displaystyle\dfrac { 3 }{ 40 }$

$\displaystyle\dfrac { 3 }{ 40 }$ of the work is done by

$\left( A+B \right)$ in 2 days

$\therefore$ Whole work is done by

$\left( A+B \right)$ in

$2 \times \displaystyle\dfrac { 40 }{ 3 }$ days

$= \displaystyle\dfrac { 80 }{ 3 }$ days

$= 26 \displaystyle\dfrac { 2 }{ 3 }$ days

$\therefore$ Work will be completed on 27th day.

Babu and Asha can do a job together in 7 days. Asha is

$1\displaystyle\frac{3}{4}$ times as efficient as Babu. The same job can be done by Asha alone in

0%
$\displaystyle\frac{49}{4}$ days
0%
$\displaystyle\frac{49}{3}$ days
0%
11 days
0%
$\displaystyle\frac{28}{3}$ days

Explanation

If Babu can do a job in

$x$ days, then Asha can do the same job in

$\displaystyle\frac{4x}{7}$ days.

$\therefore$ (Asha and Babu)'s one day's job

$= \displaystyle\frac{1}{x} + \displaystyle\frac{7}{4x}$
Also given, (Asha and Babu)'s one day's job

$= \displaystyle\frac { 1 }{ 7 }$

$\therefore \displaystyle\frac { 1 }{ x } + \displaystyle\frac { 7 }{ 4x } = \displaystyle\frac { 1 }{ 7 } \Rightarrow \displaystyle\frac { 4+7 }{ 4x } = \displaystyle\frac { 1 }{ 7 } \Rightarrow x = \displaystyle\frac { 77 }{ 4 }$

$\therefore$ Asha alone can do the job in

$\displaystyle\frac { 4 }{ 7 } \times \displaystyle\frac { 77 }{ 4 }$ days

$= 11$ days.

How many times do the hands of a clock coincide in a day?

0%
24
0%
20
0%
21
0%
22

Explanation

The hands of a clock coincide

$11$ times in every

$12$ hours . The hands overlap about every

$65$ minutes, not every

$60$ minutes. Thus the minute hand and the hour hand coincide

$22$ times in a day

A car travelling with

$\displaystyle \dfrac{5}{7}$ of its usual speed covers

$42$ km in

$1$ hour

$40$ min

$48$ sec. What is the usual speed of the car?

0%
$\displaystyle 17 \dfrac{6}{7}$ km/hr
0%
$25$ km/hr
0%
$30$ km/hr
0%
$35$ km/hr

Explanation

Let the usual speed of the car be x km/hr.

Time = 1 hour 40 min 48 sec

$= \displaystyle 1 + \dfrac{40}{60} + \dfrac{48}{3600} hrs$

$= 1 \displaystyle + \dfrac{2}{3} + \dfrac{1}{75} hrs$

$= \displaystyle \dfrac{75 + 50 + 1}{75} hrs$

$= \displaystyle \dfrac{126}{75} hrs$

Given,

$\displaystyle \dfrac{5}{7} x \times \dfrac{126}{75} = 42 \Rightarrow x = \dfrac{42 \times 7 \times 75}{5 \times 126}$

$= 35 km/hr$

A cistern has two taps which fill it in

$12$ minutes and

$15$ minutes respectively. There is also a waste pipe in the cistern. When all the pipes are opened, the empty cistern is full in

$20$ minutes. How long will the waste pipe take to empty the full cistern?

0%
$12$ minutes
0%
$10$ minutes
0%
$8$ minutes
0%
$16$ minutes

Explanation

Let the waste pipe take

$x$ minutes to empty the full cistern. Then,

$\left( \displaystyle\frac { 1 }{ 12 } + \displaystyle\frac { 1 }{ 15 } - \displaystyle\frac { 1 }{ x } \right) = \displaystyle\frac { 1 }{ 20 }$

$\Rightarrow \displaystyle\frac { 1 }{ x } = \displaystyle\frac { 1 }{ 12 } + \displaystyle\frac { 1 }{ 15 } - \displaystyle\frac { 1 }{ 20 } = \displaystyle\frac { 5+4-3 }{ 60 } = \displaystyle\frac { 6 }{ 60 } = \displaystyle\frac { 1 }{ 10 }$

$\therefore$ The waste pipe takes 10 minutes to empty the full cistern.

One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in

$36$ minutes, then the slower pipe alone would be able to fill the tank in

0%
$81$ minutes
0%
$108$ minutes
0%
$144$ minutes
0%
$192$ minutes

Explanation

Suppose the faster pipe

$A$ can fill the tank in

$x$ minutes. Then,
Slower pipe

$B$ will fill it in

$3x$ minutes.
In one minute both the pipes can fill

$\displaystyle\frac { 1 }{ x } + \displaystyle\frac { 1 }{ 3x } = \displaystyle\frac { 4 }{ 3x }$ of the tank.
Also, given that in one minute both the pipes can fill

$\displaystyle\frac { 1 }{ 36 }$ of the tank.

$\therefore \displaystyle\frac { 4 }{ 3x } = \displaystyle\frac { 1 }{ 36 } \Rightarrow x = 48$
The slower pipe will take

$3 \times 48 = 144$ minutes.

A pump can fill a tank with water in

$2$ hours. Because of a leak in the tank, it takes

$2\displaystyle\frac { 1 }{ 3 }$ hours to fill the tank. The leak can empty the filled tank in

0%
$2\displaystyle\frac { 1 }{ 3 }$ hours
0%
$7$ hours
0%
$8$ hours
0%
$14$ hours

Explanation

Part of the tank filled by the pump in 1 hour

$= \displaystyle\frac { 1 }{ 2 }$
Total part of the tank filled by the pump and leak in 1 hour

$= \displaystyle\frac { 1 }{ \displaystyle\frac { 7 }{ 3 } } = \displaystyle\frac { 3 }{ 7 }$

$\therefore$ Part of the tank emptied by the leak in 1 hour

$= \displaystyle\frac { 1 }{ 2 } - \displaystyle\frac { 3 }{ 7 } = \displaystyle\frac { 7-6 }{ 14 } = \displaystyle\frac { 1 }{ 14 }$

$\Rightarrow$ The leak will empty the tank in

$14$ hours.

The sum of five numbers is

$555$ . The average of first two numbers is

$75$ and the third number is

$115$ . What is the average of the last two numbers?

0%
$145$
0%
$290$
0%
$265$
0%
$150$

Explanation

Let the five numbers be

$A, B, C, D, E$
Given,

$A + B + C + D + E =555$

$\displaystyle \frac{A+B}{2} = 75 \Rightarrow A + B = 150$ and

$C = 115$

$\therefore 150 + 115 + D + E = 555$

$\Rightarrow D + E = 555 - 265 = 290$

$\therefore$ Required Average

$= \displaystyle \frac{D + E}{2} = \frac{290}{2} = 145$

How long will two pipes together take to fill a cistern which they can separately fill in

$20$ and

$25$ minutes?

0%
$11\displaystyle\frac { 1 }{ 9 }$ min
0%
$10\displaystyle\frac { 1 }{ 9 }$ min
0%
$11\displaystyle\frac { 6 }{ 9 }$ min
0%
$11\displaystyle\frac { 2 }{ 9 }$ min

Explanation

Part of cistern filled by 1st pipe in 1 min

$= \displaystyle\frac { 1 }{ 20 }$

Part of cistern filled by 2nd pipe in 1 min

$= \displaystyle\frac { 1 }{ 25 }$

$\therefore$ Part of cistern filled by both pipes together in 1 min

$= \displaystyle\frac { 1 }{ 20 } + \displaystyle\frac { 1 }{ 25 } = \displaystyle\frac { 5+4 }{ 100 } = \displaystyle\frac { 9 }{ 100 }$

$\therefore$ The cistern will be filled by both the pipes together in

$\displaystyle\frac { 100 }{ 9 }$ min

$= 11\displaystyle\frac { 1 }{ 9 }$ min.

A man can reach a certain place in

$30$ hours. If he reduces his speed by

$\displaystyle \dfrac {1}{15}$ th, he goes

$10$ km less in that time. Find his speed in km/ hr?

0%
$9$ km/hr
0%
$2$ km/hr
0%
$7$ km/hr
0%
$5$ km/hr

Explanation

Let the man's speed be

$x$ km/hr.

Distance covered in

$30$ hours =

$30x$ km

New speed =

$\displaystyle[x- \dfrac {x}{15}] km/hr = \displaystyle \dfrac {14x}{15} km/hr$

Distance covered in 30 hours =

$\displaystyle \dfrac {14x}{15} \times 30 =$ 28x km Given,

$30x-28x = 10km$

$\Rightarrow$

$x = 5$ km/hr

A truck covers a distance of

$550$ metres in

$1$ minute whereas a bus covers a distance of

$33$ km in

$45$ minutes. What is the ratio of their speeds?

0%
$3 : 4$
0%
$2 : 7$
0%
$5 : 8$
0%
$3 : 8$

Explanation

Required ratio = Speed of truck : Speed of bus=

$\displaystyle \frac{550}{60}$ m/sec :

$\displaystyle \frac{33\times 1000} {45 \times 60}$ m/sec =

$\displaystyle \frac{55}{6}$ m/sec :

$\displaystyle \frac{110}{9}$ m/sec =

$\displaystyle \frac{55}{6} \times \displaystyle \frac {9}{110} = \displaystyle \frac{3}{4} = 3:4$

If Ajit can do

$\displaystyle \frac{1}{4}$ of the works in

$3$ days and Sujit can do

$\displaystyle \frac{1}{6}$ of the same work in

$3$ days, how much will Ajit get if both work together and are paid Rs.

$180$ in all?

0%
Rs. $120$
0%
Rs. $108$
0%
Rs. $60$
0%
Rs. $36$

Explanation

Ajit's

$3$ days' work

$=$

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

$\Rightarrow$ Ajit's

$1$ days' work

$=$

$\displaystyle \frac{1}{3}\times \frac{1}{4}=\frac{1}{12}$
Sujit's

$4$ dasys' work

$=$

$\displaystyle \frac{1}{6}$

$\Rightarrow$ Sujit's

$1$ days' work

$=$

$\displaystyle \frac{1}{4}\times \frac{1}{6}=\frac{1}{24}$

$\therefore$ They are paid in the ratio

$\displaystyle \frac{1}{12}:\frac{1}{24}=2:1$

$\therefore$ Ajit's share

$=$

$\displaystyle \frac{2}{3}\times$ Rs.

$180=$ Rs.

$120$

Ten women can complete a piece of work in

$15$ days. Six men can complete the same piece of work in

$10$ days. In How many days can

$5$ women and six men together complete the piece of work?

0%
$15$ days
0%
$7.5$ days
0%
$9$ days
0%
$12.5$ days

Explanation

$10$ women can complete a piece of work in

$15$ days.

$\therefore$

$1$ woman can complete in 1 day

$\displaystyle \frac{1}{10\times 15}=\frac{1}{150}$ par of work.

$6$ men can complete the same work in

$10$ days

$\therefore$

$1$ man can complete in

$1$ day

$\displaystyle \frac{1}{6\times 10}=\frac{1}{60}$ part of the work.

$\therefore$

$(6$ men's

$+ 5$ women's

$)$

$1$ days' work

$\displaystyle =\frac{6}{60}+\frac{5}{150}$

$=\dfrac{1}{10}+\dfrac{1}{30}$

$=\dfrac{3+1}{30}$

$=\dfrac{4}{30}=\dfrac{2}{15}$

$\therefore$

$6$ men and

$5$ women can complete the whole work in

$\displaystyle \frac{15}{2}=7.5$ days.

X can do a piece of work in

$8$ days and Y in

$12$ days. When Z joined them, they completed the work in

$3$ days. If the total money received for the work was

$Rs. 400,$ what would be the share of X, based upon his proportionate output?

0%
$Rs. 200$
0%
$Rs. 160$
0%
$Rs. 150$
0%
$Rs. 170$

Explanation

From Given data,
X's 1 day work $= \dfrac{1}{8}$
Y's 1 day work $= \dfrac{1}{12}$
IF X, Y and Z complete the work in $3$ days , then Z's $1$ day work = One day work of X, Y and Z combined - ( One day work of X + One day work of )

$=\dfrac{1}{3} - [ \dfrac{1}{8} + \dfrac{1}{12}]$

$= \dfrac{1}{3} -[\dfrac{3 + 2}{24}]$

$= \dfrac{1}{3} - \dfrac{5}{24} = \dfrac{8-5}{24} = \dfrac{3}{24}$
Now we have to find Z's share of the salary.
X's share: Y's share : Z's share $= \dfrac{1}{8}: \dfrac{1}{12} : \dfrac{3}{24}: 3:2:3$
Z share of wage from the total wage of $Rs 400 = \dfrac{1}{(3+2+3)} \times 400 = Rs 150.$

$A$ can do

$\displaystyle \frac{1}{5}$ th part of the work in

$12$ days,

$B$ can do

$60\%$ of the work in

$30$ days and

$C$ can do

$\displaystyle \frac{1}{3}$ rd of the work in

$15$ days, Who will complete the work first ?

0%
$A$
0%
$B$
0%
$C$
0%
$A$ and $B$ both

Explanation

$A$ can do

$\dfrac {1}{5}th$ part of the work in

$12$ days.

$\Rightarrow$

$A$ can do the whole work in

$(12 \times 5)$ days

$= 60$ days.

$B$ can do

$60\%$ of the work in

$30$ days.

$\Rightarrow$ B can do the whole work in

$\displaystyle \left ( \frac{30\times 100}{60} \right )$ days

$= 50$ days.

$C$ can do

$\displaystyle \frac {1}{3}rd$ of the work in

$15$ days

$= 45$ days.

$\therefore$

$C$ will complete the work first.

$36$ men can complete a piece of work in

$18$ days. In how many days will

$27$ men complete the same work?

0%
$12$
0%
$18$
0%
$22$
0%
$24$

Explanation

${\textbf{Step - 1: Finding number of work units for the first case}}{\text{.}}$

${\text{Total 36 men work for a total of 18 days to complete it}}{\text{.}}$

${\text{So total work units}} = 36 \times 18 = 648$

${\text{Thus, for the first case ,it's a total of 648 work units}}{\text{.}}$

${\textbf{Step - 2: Finding number of days for the second case}}{\text{.}}$

${\text{Let it takes }}x{\text{ days for the workers to complete the task,so we can write}}$

${\text{Number of work units}} = 27 \times x = 27x$

${\text{Also, the work is same so the work units must be same}}$

${\text{So, equating the work units for the two cases, we get}}$

$648 = 27x$

$x = \dfrac{{648}}{{27}} = 24$

${\textbf{Hence option D is correct}}$

$72$ men can build a wall

$280$ m long in

$21$ days, how many men will take

$18$ days to build a similar type of wall of length

$100$ m?

0%
$30$
0%
$10$
0%
$18$
0%
$28$

Explanation

In $21$ days $280$ m wall build by $72$ man

Then in one day $280$ m wall build by man = $21\times 72$

Or one day one m wall build by man= $\dfrac{21\times 72}{280}$

Or one day 100 m wall build by man = $\dfrac{21\times 72\times 100}{280}$

Or 18 days 100 m wall build by man= $\dfrac{21\times 72\times 100}{280\times 18}=30$ man

Two taps can separately fill a cistern in

$10$ minutes and

$15$ minutes, respectively and when the waste pipe is open, they can together fill it in

$18$ minutes. The waste pipe can empty the full cistern in:

0%
$7$ min
0%
$9$ min
0%
$13$ min
0%
$23$ min

Explanation

Let us name the

$3$

$Taps$ be

$T_1,T_2,T_3$

where

$T_1,T_2$ are filling pipes and

$T_3$ is Waste Pipe.

Water Supplied by

$T_1$ in

$10$

$min=Volume$

$of$

$Cistern$

$or$

$V$

Water Supplied by

$T_1$ in

$1$

$min=\dfrac{V}{10}$

Similarly, Water Supplied by

$T_2$ in

$15$

$min=V$

Water Supplied by

$T_1$ in

$1$

$min=\dfrac{V}{15}$

Let us assume

$t$ be the time by

$T_3$ to

$empty$ Cistern.

Similarly, Water removed from

$T_3$ in

$1$

$min=\dfrac{V}{t}$

Now,it is given that they can together fill

$V$ in

$18$

$minutes$ (

$i.e.$ in

$1 min$ Water supplied

$=\dfrac{V}{18}$ )

$\Rightarrow \dfrac{V}{10}+\dfrac{V}{15}-\dfrac{V}{t}=\dfrac{V}{18}$

$\Rightarrow \dfrac{1}{10}+\dfrac{1}{15}-\dfrac{1}{t}=\dfrac{1}{18}$

Solving this Equation we get

$\Rightarrow t=9 min$

Therefore Answer is

$(B)$

A water tank is

$\displaystyle\frac{2}{5}$ th full. Pipe

$A$ can fill the tank in 10 minutes and pipe

$B$ can empty it in

$6$ minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?

0%
$6$ minutes to empty
0%
$6$ minutes to fill
0%
$9$ minutes to empty
0%
$9$ minutes to fill

Explanation

Pipe

$A$ in 1 minute fills

$\displaystyle\frac { 1 }{ 10 }$ part and pipe

$B$ in 1 minute empties

$\displaystyle\frac { 1 }{ 6 }$ part.

$\therefore$ Work done by pipe

$\left( A+B \right)$ in 1 min

$= \displaystyle\frac { 1 }{ 10 } - \displaystyle\frac { 1 }{ 6 } = \displaystyle\frac { 1 }{ 15 }$

$\Rightarrow \displaystyle\frac { 1 }{ 15 }$ parts gets emptied in 1 min

$\Rightarrow \displaystyle\frac { 2 }{ 5 }$ part is emptied in

$15 \times \displaystyle\frac { 2 }{ 5 }$ min

$= 6$ min.

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 5 - 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 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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