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

$A$ and

$B$ can do a work in

$10$ days,

$B$ and

$C$ in

$15$ days and

$A$ and

$C$ in

$12$ days. In how many days the work will be finished by

$A, B$ and

$C$ working together.

0%
$6$
0%
$7$
0%
$8$
0%
$9$

Explanation

According to the question-

$\cfrac{1}{A} + \cfrac{1}{B} = \cfrac{1}{10} ..... \left( 1 \right)$

$\cfrac{1}{B} + \cfrac{1}{C} = \cfrac{1}{15} ..... \left( 2 \right)$

$\cfrac{1}{C} + \cfrac{1}{A} = \cfrac{1}{12} ..... \left( 3 \right)$

Adding equation

$\left( 1 \right), \left( 2 \right) \& \left( 3 \right)$ , we have

$\left( \cfrac{1}{A} + \cfrac{1}{B} \right) \left( \cfrac{1}{B} + \cfrac{1}{C} \right) + \left( \cfrac{1}{C} + \cfrac{1}{A} \right) = \cfrac{1}{10} + \cfrac{1}{15} + \cfrac{1}{12}$

$\Rightarrow 2 \left( \cfrac{1}{A} + \cfrac{1}{B} + \cfrac{1}{C} \right) = \cfrac{6 + 4 + 5}{60}$

$\Rightarrow \cfrac{1}{A} + \cfrac{1}{B} + \cfrac{1}{C} = \cfrac{15}{2 \times 60} = \cfrac{1}{8}$

Hence the work will be finished in

$8$ days by

$A, \; B$ and

$C$ working together.

The mean of squares of consecutive odd numbers from 1 to 13 is

0%
70
0%
65
0%
75
0%
66

In a project, a team of

$400$ labourers were assumed to complete a job in

$18$ months. Later on, it was required to complete the job in

$10$ months. The number of extra labourers are employed to complete the work is

0%
$240$
0%
$320$
0%
$360$
0%
$420$

$\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{x}{8!}$ , then

$x=?$

0%
$32$
0%
$48$
0%
$56$
0%
$64$

Explanation

$\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{x}{8!}\Rightarrow \dfrac{8\times 7}{8\times 7\times (6!)}+\dfrac{8}{8\times (7!)}=\dfrac{x}{8!}$

$\Rightarrow \dfrac{56}{8!}+\dfrac{8}{8!}=\dfrac{x}{8!}\Rightarrow x=56+8=64$ .

The exponent of 3 in

$100!$ is

0%
$12$
0%
$24$
0%
$48$
0%
$96$

The letters of word 'ZENITH' are written in all positive ways. If all these words are written in the order of a dictionary, then the rank of the word 'ZENITH' is

0%
$716$
0%
$692$
0%
$698$
0%
$616$

Explanation

The total number of words is

$6! = 720$ . Let us write the letters of word ZENITH alphabetically, i.e, EHINTZ.

For ZENITH word starts with	Word starting with	Number of words
$Z$	$E$	$5!$
	$H$	$5!$
	$I$	$5!$
	$N$	$5!$
	$T$	$5!$
ZEN	ZEH	$3!$
	ZEI	$3!$
ZENI	ZENH	$2$
ZENIT	ZENIH	$1$
	Total number of words before ZERNITH	$615$

Hence, there are

$615$ words before ZENITH, so the rank of ZENITH is

$616$ .

There are n seats round a table marked

$1, 2, 3,...,n.$ The number of ways in which

$m(<n)$ persons can take seats is

0%
$^nP_m$
0%
$^nC_m\times(m-1)!$
0%
$^nC_{m!}$
0%
$^{n-1}P_{m-1}$

$8$ men can finish a piece of work in

$40$ days. If

$2$ more men join them, the work will be completed in

0%
$30$ days
0%
$32$ days
0%
$36$ days
0%
$25$ days

Explanation

$(b) 32$ days
Because,

$8$ men can finish a piece of work in

$=40$ days

$1$ men can finish a piece of work in

$=(8\times 40)$
If

$2$ more men join them, i.e.

$10$ men can finish a work in

$=(8\times 40)/10$

$=32$ days

$(4/5)$ of a cistern is filled in

$1$ minutes, how much more time will required to fill the rest of it?

0%
$20$ seconds
0%
$15$ seconds
0%
$12$ seconds
0%
$10$ seconds

Four alternative options are given for each of the following statement . Select the correct option .
The mean of score

$10 , 15 , 12 , 15 , 15$ is

0%
$15$
0%
$13$
0%
$13 . 4$
0%
$14 . 3$

Four alternative options are given for each of the following statement . Select the correct option .
The mean of first three multiples of

$5$ is

0%
$5$
0%
$10$
0%
$15$
0%
$30$

A student got 85, 87 and 83 marks in Mathematics, Physics and Chemistry respectively. Mean of obtained marks.

0%
86
0%
84
0%
85
0%
85.5

Explanation

Given marks:

$85,\ 87,\ 83$

Mean of marks

$= \dfrac{\mathrm{Sum\ of\ marks}}{\mathrm{No.\ of\ subjects}}$

$= \dfrac{85+87+83}{3}$

$= \dfrac{255}{3}$

$= 85$

53, 75, 42, 70 are marks obtained by 4 students is statistics. Arithmetic mean of their marks is:

0%
42
0%
64
0%
60
0%
56

Explanation

Given marks = 53, 75, 42, 70
Arithmetic mean

$= \frac{Sum\ of\ observations}{No.\ of\ observations}$

$= \frac{53+75+42+70}{4}$

$= \frac{240}{4} = 60$

The average of statistical data is called :

0%
Arithmetic mean
0%
Median
0%
Mode
0%
Frequency

Explanation

The average of statistical data is called

$Mean$ or

$Arithmatic\ mean$

Hence,

$Op-A$ is correct.

_________ represent the average of values.

0%
Mean
0%
Mode
0%
Median
0%
Range

Which measure of central tendency takes into account the magnitude of scores?

0%
Mean
0%
Mode
0%
Range
0%
Median

$A$ and

$B$ can do a piece of work in

$5$ days,

$B$ and

$C$ can do it in

$7$ days.

$A$ and

$C$ can do it in

$4$ days. Who among these will take the least time if put to do it alone ?

0%
$A$
0%
$B$
0%
$C$
0%
Data inadequate

Explanation

$Let \, \,the \, \,work \, \,be \, \,W.$

$A \, \, and\, \, B\, \, can\, \, do\, \, a \, \, piece \, \,of\, \, work\, \, in \, \, 5 \, \,days$

$\therefore \, \,5A+5B=W \, \,..........(1)$

$B \, \, and \, \, C \, \, can \, \,do \, \, it \, \, in \, \,7 \, \, days$

$\therefore, \, \,7B+7C=W \, \,......(2)$

$A \, \, and \, \, C \, \, can \, \,do \, \, it \, \, in \, \,4 \, \, days$

$\therefore, \, \,4A+4C=W \, \,.......(3)$

$Solving \, \, (1) \, \,(2) \, \,and \, \,(3) \, \,We \, \,get,$

$Speed \, \, of \, \, A=\dfrac{43}{280} \, \, work/day$

$Speed \, \, of \, \, B=\dfrac{13}{280} \, \, work/day$

$Speed \, \, of \, \, C=\dfrac{27}{280} \, \, work/day$

$Since \, \,speed \, \,of \, \,A \, \,is \, \,maximum \, \,among \, \,all \, \,three, \,$

$Hence \, \,A \, \, will \, \,take \, \,least \, \,time \, \,to \, \,complete \, \,the \, \,work.$

A and B can do a job in

$12$ days, B and C can do it in

$16$ days. After A has worked for

$5$ days and B has worked for

$7$ days, C can finish the rest in

$13$ days. In how many days, can C do the work alone?

0%
$16$ days
0%
$24$ days
0%
$36$ days
0%
$48$ days

Explanation

$A$ and

$B$ can do a job in

$12$ days.

$B$ and

$C$ can do a job in

$16$ days.

$A$ worked for

$5$ days and

$B$ worked for

$7$ days.

Part of work done by

$A$ and

$B$ in

$5$ days=

$\dfrac{5}{12}$

$\therefore$ Part of work left after

$A$ left=

$1-\dfrac{5}{12}=\dfrac{12-5}{12}=\dfrac{7}{12}$

After

$A$ left work

$B$ worked for

$2$ more days.

Part of work done by

$B$ and

$C$ in

$2$ days=

$\dfrac{2}{16}=\dfrac{1}{8}$

$\therefore$ part of work left after

$B$ left=

$\dfrac{ 7 }{ 12 } -\dfrac{ 1 }{ 8 } =\dfrac{ 14-3 }{ 24 } =\dfrac{ 11 }{ 24 }$

Now

$\dfrac{ 11 }{ 24 }$ part o work

$C$ has to complete in

$11$ days

Since

$C$ already had worked for

$2$ dayswith

$B$

(part)complete work is done by

$C$ in

$\dfrac { 11 }{ \dfrac { 11 }{ 24 } }$

$24$ days.

$3$ men and

$4$ boys can do a piece of work in

$14$ days, while

$4$ men and

$6$ boys do the same work in

$10$ days. How long would it take one man to finish the same work ?

0%
$40$ days
0%
$20$ days
0%
$35$ days
0%
$70$ days

Explanation

Let one man's one day's work be

$x$

Let one boy's one day's work be

$y$

According to first condition,

$3x+4y = \dfrac{1}{14}$ .......Equation

$1$

According to second condition,

$4x+6Y = \dfrac{1}{10}$ .......Equation

$2$

Multiplying Equation

$1$ by

$4$ and Equation

$2$ by

$3$ ,

$12x + 16y = \dfrac{4}{14}$ .....Equation

$3$

$12x + 18y = \dfrac{3}{10}$ .....Equation

$4$

Now solving Equation

$3$ and

$4$ simultaneously by substracting Equation

$3$ by Equation

$4$ ,

we get,

$\therefore -2y = \dfrac{4}{14} -\dfrac{3}{10}$

$\therefore -2y = \dfrac{40-42}{140}$

$\therefore y = \dfrac{1}{140}$

Putting the value of y in equation

$1$ , we get,

$3x + \dfrac{4}{140} = \dfrac{1}{14}$

$\therefore x = \dfrac{1}{70}$

Hence one man's one day work is

$\dfrac{1}{70}$

$\therefore$ one man completes the whole work in

$70$ days.

Two men and

$5$ boys together can finish a piece of work in

$4$ days, while

$3$ men and

$6$ boys can finish in

$3$ days. Find the time taken :
(i) by one man alone to finish the work
(ii) by one boy alone to finish the work.

0%
$17$ days and $19$ days
0%
$14$ days and $22$ days
0%
$18$ days and $36$ days
0%
$17$ days and $13$ days

Explanation

Let the amount of work done by a Man alone in a day be

$M$ and by a Boy alone be

$B$

Given,

$2$ men and

$5$ boys together can finish a piece of work in

$4$ days

Amount of work they can do in one day

$2M + 5B$

$= \dfrac {1}{4}$ -- (1)

Similarly,

$3M + 6 B = \dfrac {1}{3}$ --- (2)

Solving equations

$(1)$ and

$(2)$ , we get

$B = \dfrac {1}{36}$

Substituting

$B$ in equation

$2$ , we get

$M = \dfrac {1}{18}$

Hence, one man alone will take

$18$ days to do the work and one boy will take

$36$ days.

A large tanker can be filled by two pipes A and B in

$60$ minutes and

$40$ minutes, respectively. How many minutes will it take to fill the empty tanker if only B is used in the first-half of the time and A and B are both used in the second-half of the time?

0%
$15$
0%
$20$
0%
$27.5$
0%
$30$

Explanation

Let total time consumed in filling the tank is

$x$ minutes

$\left( A+B \right) \prime s$ 1 minute work

$=\quad \dfrac { 1 }{ 60 } +\dfrac { 1 }{ 20 } =\dfrac { 2+3 }{ 120 } =\dfrac { 5 }{ 120 } =\dfrac { 1 }{ 24 }$

$\left( A+B \right)$ will take half of total time

$=\dfrac { x }{ 2 } \times \dfrac { 1 }{ 24 } \_ \_ \_ \_ \_ \_ \left( 1 \right)$

The rest half will be filled by B only in half of total time

$=\dfrac { x }{ 2 } \times \dfrac { 1 }{ 40 } \_ \_ \_ \_ \_ \_ \left( 2 \right)$

Adding (1) and (2)

$\dfrac { x }{ 48 } +\dfrac { x }{ 80 } =1\\ \Rightarrow \dfrac { 5x+3x }{ 240 } =1\\ \Rightarrow 8x=240\\ \Rightarrow x=30min$

$16$ men can complete a work in

$12$ days,

$24$ children can complete the same work in

$18$ days,

$12$ men and

$8$ children started working and after

$8$ days,

$3$ more children joined them. How many days will they now take to complete the remaining work ?

0%
$2$ days
0%
$4$ days
0%
$6$ days
0%
$8$ days

Explanation

One day work of 16 men is

$\dfrac { 1 }{ 12 }$ .

One day work of 24 children is

$\dfrac { 1 }{ 18 }$ .

One men's one day work is

$\dfrac { 1 }{ 12\times 16 }$ .

One child's one day work is

$\dfrac { 1 }{ 24\times 18 }$ .

One day work of 12 men and 8 children is

$\dfrac { 12 }{ 12\times 16 } +\dfrac { 8 }{ 24\times 18 }$ .

One day work =

$\dfrac { 35 }{ 54\times 8 }$ .

Work done in 8 days =

$8\times \dfrac { 35 }{ 54\times 8 } =\dfrac { 35 }{ 54 }$ .

Remaining work =

$1-\dfrac { 35 }{ 54 } =\dfrac { 19 }{ 54 }$ .

The remaining work is 12

$\times$ one men's one day work + 11

$\times$ one child's one day work.

Hence,

$12\times \dfrac { 1 }{ 12\times 16 } +11\times 24\times 18$

One day work multiplied by number of days to complete remaining work = remaining work

Therefore, the work is done in 4 days.

$\left( \dfrac { 1 }{ 16 } +\dfrac { 11 }{ 24 } \times 18 \right) x=\dfrac { 19 }{ 54 }$

$\Rightarrow$

$x=4$

Thus they take 4 days to complete the remaining work.

Hence option B is correct.

A can do a piece of work in

$7$ days of

$9$ hr each, and B can do it in

$6$ days of

$7$ hour. How long will they take to do it working together

$\displaystyle \frac{42}{5}$ hr a day ?

0%
$3$ days
0%
$4$ days
0%
$4.5$ days
0%
None

Explanation

Unit of work		Days	Hour		Total Work
2	A	7	9	63	126
3	B	6	7	42	(L.C>M of 63 and 46)

Time taken to complete the work by

$A$ and

$B$ =

$=\dfrac { 126 }{ \left( 2+3 \right) \times \dfrac { 42 }{ 5 } }$ days.

$=\dfrac { 126\times 5 }{ 5\times 42 }$ days.

$=3$ days.

$2$ men and

$3$ boys together can do a piece of work in

$8$ days. The same work is done in

$6$ days by

$3$ men and

$2$ boys together. How long would

$1$ boy alone or

$1$ man alone take to complete the work.

0%
One boy alone can do a piece of work in $130$ days

and one man alone can do a piece of work in $160$ days
0%
One boy alone can do a piece of work in $90$ days

and one man alone can do a piece of work in $70$ days
0%
One boy alone can do a piece of work in $120$ days

and one man alone can do a piece of work in $20$ days
0%
One boy alone can do a piece of work in $160$ days

and one man alone can do a piece of work in $12$ days

Explanation

Let

$1$ boy alone can do a piece of work in

$x$ days and

$1$ man alone can do a piece of work in

$y$ days.
So,

$1$ day work of

$1$ boy

$=\cfrac{1}{x}$
and

$1$ day work of

$1$ man

$=\cfrac{1}{y}$
ATP

$\cfrac{3}{x}+\cfrac{2}{y}=\cfrac{1}{8}$ .........(i)
and

$\cfrac{2}{x}+\cfrac{3}{y}=\cfrac{1}{6}$ .......(ii)
Multiplying equation (i) by

$2$ and equation (ii) by

$3$

$\cfrac{6}{x}+\cfrac{4}{y}=\cfrac{1}{4}$

$\cfrac{6}{x}+\cfrac{9}{y}=\cfrac{1}{2}$
------------------------------------------

$-\cfrac{5}{y}=-\cfrac{1}{4}$

$\Rightarrow$

$y=20$
substitute

$y=20$ in equation (i)

$\cfrac{3}{x}+\cfrac{3}{20}=\cfrac{1}{8}$

$\cfrac{3}{x}=\cfrac{1}{8}-\cfrac{1}{10}$

$\cfrac{3}{x}=\cfrac{5-4}{40}$

$\cfrac{3}{x}=\cfrac{1}{40}$

$\Rightarrow$

$x=120$
So, one boy alone can do a piece of work in

$120$ days
and one man alone can do a piece of work in

$20$ days.

A cabinet of ministers consists of 11 ministers, one minister being the chief minister. A meeting is to be held in a room having a round table and 11 chairsround it, one of them being meant for the chairman. The number of ways in which the ministers can take their chairs, the chief minister occupying the chairman's place, is

0%
$\displaystyle \frac{1}{2} (10!)$
0%
$\displaystyle 9!$
0%
$\displaystyle 10!$
0%
none of these

Explanation

These are circular permutations
so,

$=(n-1)!$ =

$10!$

Pooja and Ritu can do a piece of work in

$17\displaystyle \frac{1}{7}$ days. If one day work of Pooja be three fourth of one day work of Ritu; find in how many days each alone will do the work.

0%
Pooja in $44$ days and Ritu in $8$ days
0%
Pooja in $80$ days and Ritu in $500$ days
0%
Pooja in $20$ days and Ritu in $120$ days
0%
Pooja in $40$ days and Ritu in $30$ days

Explanation

Let Pooja's one day work be

$x$

And Ritu's one day work be

$y$

Accprding to first condition,

Pooja and Ritu can do work in

$17 \dfrac{1}{7}$ days i.e

$\dfrac{120}{7}$ days

$\therefore x + y = \dfrac {1}{\dfrac{120}{7}}$

$\therefore x + y = \dfrac{7}{120}$ .........Equation

$1$

According to second condition,

$x = \dfrac{3}{4} y$

$\therefore 4x - 3y = 0$ .......... Equation

$2$

Multiplying eqution

$1$ by

$3$ and solving equation

$1$ and

$2$ simultaneosly, we get

$3x + 3y = \dfrac{7}{40}$

$4x - 3y = 0$

$\therefore 7x = \dfrac{7}{40}$

$\therefore x = \dfrac{1}{40}$

Putting the value of

$x$ in equation

$1$ , we get

$\dfrac{1}{40} + y = \dfrac{7}{120}$

$\therefore y = \dfrac{4}{120} = \dfrac{1}{30}$

$\therefore$ Pooja's one day work is

$x = \dfrac{1}{40}$

i.e. her whole work will be completed in

$40$ days

Similarly Pooja's work will be completed in

$30$ days

Answer : Pooja can do the work alone in

$40$ days and Ritu in

$30$ days

$A$ can do a work in

$15$ days and

$B$ can do the same work in

$20$ days, if

$A$ and

$B$ has done work for

$4$ days together, the fraction of work left?

0%
$\dfrac{8}{15}$
0%
$\dfrac{2}{15}$
0%
$\dfrac{1}{5}$
0%
$\dfrac{7}{13}$
0%
$\dfrac{16}{15}$

Explanation

${\textbf{Step -1: Find the work left after 4 days.}}$

${\text{Let, total work = 1}}$

${\text{A can do the work in 15 days}}$

${\text{Amount of work A can do in 1 day = }}\dfrac{1}{{15}}$

${\text{B can do the work in 20 days}}$

${\text{Amount of work B can do in 1 day = }}\dfrac{1}{{20}}$

${\text{Amount of work done by A and B together in 1 day = }}\dfrac{1}{{15}} + \dfrac{1}{{20}} = \dfrac{7}{{60}}$

${\text{Amount of work A and B together can do in 4 days = 4}} \times \dfrac{7}{{60}} = \dfrac{7}{{15}}$

${\text{Work left after 4 days = } 1 - }\dfrac{7}{{15}}$

$= \dfrac{8}{{15}}$

${\textbf{Hence, the correct option is A.}}$

4 men and 6 women working together can complete the work within 10 days.3 men and 7 women working together will complete the same work within 8 days. In how many days 10 women will complete this work?

0%
5
0%
8
0%
5.5
0%
None

Two pipes can fill an empty tank separately in

$24$ minutes and

$40$ minutes respectively and a third pipe can empty

$30$ gallons of water per minute. If all the three pipes are open, empty tanks becomes full in one hour. The capacity of tank (in gallons) is

0%
$500$
0%
$550$
0%
$600$
0%
$700$

Explanation

Work done by two pipes individually is

$\dfrac{1}{24}$ and

$\dfrac{1}{40}$ respectively

work done by all three pipe together is

$\dfrac{1}{60}$

Work done by third pipe to empty tha tank

$=\dfrac{1}{60} - (\dfrac{1}{24} + \dfrac{1}{40})$

$= \dfrac{1}{60} - \dfrac{8}{120}$

$= - \dfrac{1}{20}$ [-ve sign means emptying]

$\therefore$ Volume of

$\dfrac{1}{20}$ part

$= 30$ gallons

$\therefore$ Volume of whole tank

$= 30 \times 20 = 600$ gallons

Answer is

$600$ gallons.

A tap A can fill a cistern in

$8$ hours while tap B can fill it in

$4$ hours. In how much times will the cistern be filled if both A and B are opened together?

0%
$2hrs\;30min$
0%
$2hrs\;40min$
0%
$2hrs\;50min$
0%
$3hrs$

Explanation

Time taken by tap A to fill the cistern

$=8$ hours.
Work done by tap A in

$1$ hour

$=\dfrac{1}{8}$
Time taken by tap B to fill the cistern

$=4$ hours.
Work done by tap B in

$1$ hour

$=\dfrac{1}{4}$
Work done by

$\begin{pmatrix}A+B\end{pmatrix}$ in

$1$ hour

$=\dfrac{1}{8}+\dfrac{1}{4}=\dfrac{3}{8}$
Therefore, time taken by

$\begin{pmatrix}A+B\end{pmatrix}$ to fill the cistern

$=\dfrac{8}{3}$ hours

$=2hours\;40\;min$ .

In half of a 600 square meters field, a fanner sowed ragi, in one-third of the remaining, he sowed coconut and in one-fourth of the left over, he sowed groundnut. He left half of the remaining land uncultivated. In one-third of the land still left, he dug a well, utilising the remaining land for growing orchids. What fraction of the total land is he using for orchid cultivation?

0%
1/36
0%
1/28
0%
1/16
0%
1/12

Explanation

$Lets\quad total\quad land\quad be\quad x,\\ Ragi\quad occupies=\dfrac { 1 }{ 2 } x,\\ Land\quad left=\dfrac { 1 }{ 2 } x,\\ Groundnut\quad occupies=\dfrac { 1 }{ 4 } \times \dfrac { 1 }{ 2 } x=\dfrac { 1 }{ 8 } x,\\ Land\quad left=\dfrac { 3 }{ 8 } x,\\ Coconut\quad occupies=\dfrac { 1 }{ 3 } \times \dfrac { 3 }{ 8 } x=\dfrac { 1 }{ 8 } x,\\ Land\quad left=\dfrac { 1 }{ 4 } x,\\ Land\quad uncultivated=\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 4 } x=\dfrac { 1 }{ 8 } x=land\quad left,\\ Orchid\quad occupies=\dfrac { 2 }{ 3 } \times \dfrac { 1 }{ 8 } x=\dfrac { 1 }{ 12 } x$

A man was engaged on a job for

$30$ days on the condition that he would get a wage of Rs.

$10$ for the day he works , but he have to pay a fine of Rs.

$2$ for each day of his absence. If he gets Rs.

$216$ at the end. He was absent for work for how many days ?

0%
$7$ days
0%
$8$ days
0%
$9$ days
0%
$10$ days
0%
None of these

Explanation

Let he work for

$x$ days then he was absent for

$(30-x)$ days.
Total paid

$=$ Rs.

$216$
Now, according to question.

$10 \times x - 2 \times (30-x) = 216$

$\Rightarrow 10x - 60 + 2x = 216$

$\Rightarrow 12x = 276$

$\Rightarrow x = 23$
Number of working days

$= 23$
Therefore, number of absent

$= 30 - x = 30 - 23 = 7$ days.

A monkey ate 100 idli's in 5 days, each day eating 6 more than the previous day. How many did it eat on the last day?

0%
23
0%
26
0%
35
0%
32

Explanation

$Let\quad the\quad number\quad of\quad idlis\quad eaten\quad on\quad first\quad day\quad be\quad x,\\ idlis\quad eaten\quad on\quad day\quad n=x+(n-1)*6\quad (given)\\ This\quad is\quad an\quad arithematic\quad progression,\quad with\quad difference\quad 6.\\ for\quad n=5,\quad sum=100.\\ sum=\frac { n({ a }_{ n }+{ a }_{ 1 }) }{ 2 } ,\quad { a }_{ n }{ =a }_{ 5 }=x+24,\quad { a }_{ 1 }=x\\ \frac { 5(x+x+24) }{ 2 } =100,\\ 10x=200-120,\\ x=8,\\ { a }_{ 5 }=idlis\quad eaten\quad on\quad day\quad 5=8+24=32$

$10$ hectare field is reaped by

$2$ men,

$3$ women and

$4$ children in

$10$ days. If a man, a woman and a child work in the ratio

$5 : 4 : 2,$ a

$16$ hectare field will be reaped by

$6$ men,

$4$ women and

$7$ children in:

0%
$5$ days
0%
$\displaystyle6\frac{1}{2}$ days
0%
$7$ days
0%
$8$ days

Explanation

Man:Woman:child =

$5:4:2$

Man:Woman =

$5:4$

$4$ men =

$5$ women

Man:Child =

$5:2$

$2$ men =

$5$ children

Woman:Child =

$4:2$

$4$ children =

$2$ women
Or

$1$ woman =

$2$ child

$2$ men,

$3$ women and

$4$ children =

$5 + 6 + 4$ children =

$15$ children

$15$ children reap

$10$ hectare field in

$10$ days

Now

$6$ men,

$4$ women and

$7$ children:

$3\times 5 + 4 \times 2 + 7$ children =

$15 + 15$ =

$30$ children
We need to find time taken by

$30$ children to reap

$16$ hectares

We know that

$15$ children reap

$10$ hectare field in

$10$ days

$15$ children will reap 1 hectare in

$\dfrac{10}{10} = 1$ day

$15$ children will reap

$16$ hectare in

$1 \times 16 = 16$ days

$30$ children will reap

$16$ hectare in

$\dfrac{16}{2} = 8$ days
i.e

$6$ men,

$4$ women and

$7$ children will reap

$16$ hectares in

$8$ days

If a pipe fills the tank in

$10$ minutes, how much minutes it will take to half fill the tank in we connect two pipes?

0%
$20$
0%
$10$
0%
$5$
0%
$2.5$

Explanation

No of pipe (P) is inversely proportional to time (t)

$\Rightarrow P \propto \dfrac {1}{t}$
But here it takes

$10$ minutes to fill the tank.
To half fill the tank the pipe will take

$\dfrac {10}{2} = 5\ minutes$ .
Now,

$P_{1} t_{1} = P_{2} t_{2}$

$P_{1} = 1, t_{1} = 5, P_{2} = 2$ , we have to find

$t_{2}$ .

$\Rightarrow t_{1} = \dfrac {P_{1}t_{1}}{P_{2}} = \dfrac {1\times 5}{2} = 2.5\ minutes$

$15$ men complete a task in

$3$ days, how many days will

$9$ men take to complete it?

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

Explanation

No. of men is inversely proportion to no. of days taken

$\therefore 15\times 3 = K = 9\times x$

$\Rightarrow 4s = 9\times x \Rightarrow x = 5$ .

The average bonus per employee was.
The chart shows the amount paid in bonuses to the employees of a certain firm.

Bonus paid to an employee $(\$)$	$50$	$100$	$150$	$200$
Number of employees	$7$	$37$	$4$	$2$

0%
$81$
0%
$91$
0%
$100$
0%
$101$
0%
$105$

Raman can do a work in

$5$ days, Jatin can do the same work in

$7$ days and Sachin can do the same work in

$9$ days. If they do the same work together and they are paid Rs.

$2860$ , then what is the share (in Rs.) of Raman?

0%
$1260$
0%
$700$
0%
$900$
0%
$870$

Explanation

Let the total work be

$315$ [L.C.M of

$(5,7,9)$ ] units. Raman's. Jatin's and Sachin's each day work is

$63,45$ and

$35$ units.
Hence, Ratio of share

$63:45:35$
Raman's share

$=\cfrac { 63 }{ 143 } \times 2860=Rs.1260\quad$

Consider a pyramid with a square base side of

$6$ inches and with a height of

$12$ inches, as shown below. If we cut off the top of the pyramid parallel to the base

$3$ inches from the tip, what is the volume of the remaining solid?

0%
$141.75$
0%
$140$
0%
$135.48$
0%
$144$
0%
$130$

Explanation

a pyramid of a square base side of 6 inches.

height =

$12$ inches

for smaller pyramid given :-

height =

$3$ inches

Base square of side =

$1.5$ inches

volume of larger pyramid =

$\dfrac{1}{3}$ Base area

$\times$ height

$= \dfrac{1}{3} (6)^2 \times 12$

$= 36 \times 4 = 144 \, inch^3$

Volume of smaller pyramid

$= \dfrac{1}{3}$ Base area

$\times$ height

$= \dfrac{1}{3} (1.5)^2 \times 3$

$= (1.5)^2 = 2.25 inch^3$

So, volume of remaining solid

= larger volume - smaller volume

$= 144 - 2.25$

$= 141.75 \, inch^3$

$6$ painters can complete

$9$ drawing in

$5$ hours then, what is relation between no. of painters and no. of drawings if no. of hours is constant?

0%
Direct
0%
Indirect
0%
Both
0%
None

Explanation

$\because$ More no. of painters less no. hours taken to complete same no. of drawings

$\therefore$ These are in inverse proportion.

What was the average number of babies that Dr. Jones delivered each year from 1995 to 1998?

0%
35
0%
40
0%
45
0%
50
0%
55

Each side of the base of a square pyramid is reduced by

$20%$ . By what percent must the height be increased so that the volume of the new pyramid is the same as the volume of the original pyramid?

0%
20
0%
40
0%
46.875
0%
56.25
0%
71.875

Explanation

Let

$a$ be the side of the square.

Length of side of square when reduced by

$20\% = a-\dfrac{20a}{100}=0.8a$

Let

$a_1=0.8a$

Volume of pyramid

$V=\dfrac { 1 }{ 3 } \times$ Area of base

$\times height=\dfrac{1}{3}A\times h$

Area of base with side

$a = { a }^{ 2 }$

${ a }^{ 2 }=0.8a$

${V}_{ 1 }=\dfrac { 1 }{ 3 } \times { \left( { a }_{1} \right) }^{ 2 }\times { h }_{ 1 }$

${V}_{ 1 }=V$ ....... [Given]

$\Rightarrow \dfrac { 1 }{ 3 } { a }^{ 2 }\times h=\dfrac { 1 }{ 3 } { \left( 0.8 \right) }^{ 2 }{ a }^{ 2 }\times { h }_{ 1 }$

$\therefore { h }_{ 1 }=1.5625h$

$\implies \dfrac { { h }_{ 1 }-h }{ h } =1.5625-1=56.25%$

$\therefore$ 'h' need to be increase by

$56.25$

A cistern can be filled by one tap in

$5$ hours and by another in

$4$ hours. How long will it take to fill if both the taps are opened simultaneously?

0%
$\dfrac{10}{9}\;hrs$
0%
$9\;hrs$
0%
$10\;hrs$
0%
$\dfrac{20}{9}\;hrs$

Explanation

Time taken by first tap to fill the cistern

$=5$ hours.
Work done by first tap in

$1$ hour

$=\dfrac{1}{5}$
Time taken by second tap to fill the cistern

$=4\;hours$ .
Work done by second tap in

$1$ hour

$=\dfrac{1}{4}$
Work done by (first tap + second tap) in

$1$ hour

$=\dfrac{1}{5}+\dfrac{1}{4}=\dfrac{9}{20}$
Therefore, time taken by (first tap + second tap) to fill the cistern

$=\dfrac{20}{9}\;hours$ .

What was the average number of children taking swim lessons from 1990 to 1995?

0%
250
0%
308
0%
385
0%
450
0%
1,850

In how many ways can

$12$ people be seated around a table?

0%
$11!$
0%
$12!$
0%
$\dfrac{12!}{2}$
0%
$1$

Explanation

Number of people

$=12$

When in circular arrangement

The possible number of arrangements of

$n$ objects around a table (circular)

$(n-1)!$

Therefore number of ways of arranging

$12$ people

$\Rightarrow (12-1)!$

$\Rightarrow 11!$ Ways

Therefore number of ways

$=11!$

How many necklaces of

$10$ beads each can be made from

$20$ beads of different colors?

0%
$19!$
0%
$\dfrac{19!}{10!}$
0%
$\dfrac{19!}{10!^2}$
0%
$20!$

Explanation

$10$ Beads

$\longrightarrow$ Each necklece

$20$ Beads total

Number of ways of arranging

$20$ Beads in circular Form

$=19!$

There should be

$2$ sets of necklace of

$10$ beads each

Number of ways

$=10!\times 10!$

Total ways

$=\cfrac { 19! }{ 10!\times 10! } \\ =\cfrac { 19! }{ { (10!) }^{ 2 } }$

How many litres of water flow out of a pipe having an area of cross-section of

$5$

${cm}^{2}$ in one minute, if the speed of water in the pipe is

$30cm/sec$ ?

0%
$90$ litres
0%
$15$ litres
0%
$9$ litres
0%
$1.5$ litres

It 10 men or 20 women or 40 children can do a piece of work in 7 months, then 5 men, 5 women and 5 children together can do half of the work in :

0%
6 months
0%
4 months
0%
5 months
0%
8 months

Explanation

Let total work be

$W$

Work done by

$1$ man in

$1$ month =

$\frac { W }{ 10*7 } = \frac { W }{ 70}$

Work done by

$1$ woman in

$1$ month =

$\frac { W }{ 20*7 } = \frac { W }{ 140}$

Work done by

$1$ child in

$1$ month =

$\frac { W }{ 40*7 } = \frac { W }{ 280}$

Total work done by 5 men, 5 women and 5 children in 1 month=

$\frac{5W}{70}+\frac{5W}{140}+\frac{5W}{280}= \frac{W}{8}$

Let number of months be

$n$

$\frac{W}{8}*n=\frac{W}{2}$

n=4

Two classrooms A and B having capacity of

$25$ and

$(n-25)$ seats respectively.

$A_n$ denotes the number of possible seating arrangements of room

$'A'$ , when 'n' students are to be seated in these rooms, starting from room

$'A'$ which is to be filled up to its capacity. If

$A_n-A_{n-1}=25!(^{49}C_{25})$ then 'n' equals:

0%
$50$
0%
$48$
0%
$49$
0%
$51$

Explanation

Given

$A_n=nC_{25} \cdot 25!$

$A_{n-1}={n-1}C_{25} \cdot 25!$

Hence

$nC_{25} \cdot 25! - (n-1)C_{25} \cdot 25!=25! 49C_{25}$

$\Rightarrow (n-1)C_{25}+(n-1)C_{24}-(n-1)C_{25}=49C_{24}$

$\Rightarrow n-1=49$

$\Rightarrow n=50$

'Environment Day (

$5$ th June)' falls on Thursday, then on which day 'Children's Day (

$14$ th Nov)' will fall in the same year?

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

Explanation

Environment day falls on

$5^{th} June$ and Childrens day falls on

$14^{th} Nov$

Number of days between these dates

$=25+31+31+30+31+14=162$

Number of odd days in

$162$ days is

$1$

$\therefore$ If Environment days fall on a

$Thursday$ , Children's day will fall on

$Friday$

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

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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