MCQ Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 12

If four men working at the same rate can do

$2/3$ of a job in

$40$ minutes it take one man working at this rate to do

$2/5$ of the job takes how many minutes?

0%
$80$
0%
$88$
0%
$92$
0%
$96$
0%
$112$

A library has

$'a'$ copies of one book,

$'b'$ copies of each of two books,

$'c'$ copies of each of three books, and single copy each of

$'d'$ . The total number of ways in which these books can be arranged in a row is

0%
$\dfrac{(a+b+c+d)!}{a!b!c!}$
0%
$\dfrac{(a+2b+3c+d)!}{a!(b!)^{2}(c!)^{3}}$
0%
$\dfrac{(a+2b+3c+d)!}{a!b!c!}$
0%
none of these

$\text{The number of ways dividing 52 cards amongst four players equally, are}$

0%
$\dfrac {52!}{(13!)^{4}}$
0%
$\dfrac {52!}{(13!)^{4}4!}$
0%
$\dfrac {52!}{(12!)^{4}4!}$
0%
$\text{None of these}$

Explanation

$\textbf{Hint:-Number of ways in which 3n diiferent things can be divided equally three people }$

$=\large\bf\frac{(3n)!}{n!n!n!}$

$\textbf{Step:-1 Find the total number of ways }$

$\text{Here 52 cards are distributed among four players equally in an order so each gets 13 cards}$

$\text{Total number of ways dividing 52 cards amongst four players equally,are=}$

$^{52}C_{13}\ \times \ ^{39}C_{13}\ \times$

$\ ^{26}C_{13}\ \times \ ^{13}C_{13}$

$=\large\frac{52!}{39!13!}$

$\times\large\frac{39!}{26!13!}$

$\times\large\frac{26!}{13!13!}$

$\times\large\frac{13!}{0!13!}$

$=\large\frac{52!}{13!\times13!\times13!\times13!}$

$=\large\frac{52!}{{13!}^4}$

$\textbf{Heence the number of ways the cards can be divided equally amongst number of players=}$

$\large\bf\frac{52!}{{13!}^4}$

If a pipe can fill a cistern in 8 hours and another pipe can empty the cistern in 10 hours, then what part of the cistern will be filled in 1 hour, if both pipes are opened simultancously?

0%
$\frac { 1 }{ 40 }$
0%
$\frac { 1 }{ 30 }$
0%
$\frac { 1 }{ 35 }$
0%
$\frac { 1 }{ 8 }$

A pipe can fill a cistern in

$12$ min and another pipe can fill it in

$15$ min, but a third pipe can empty it in

$6$ min. The first two pipes are kept open for

$5$ min in the beginning and then the third pipe is also opened. Time taken to empty the cistern is

0%
30 min
0%
33 min
0%
$37\frac{1}{2}\min$
0%
45 min

A street-light is hung

$20\ ft$ , above the ground. An object falls freely under the gravity, starting from rest at the same height as the lamp and at a horizontal distance of

$5\ ft$ . from it. When the object has fallen through

$16\ ft$ , the speed of the shadow of the object on the ground is

0%
$12\ ft./sec$
0%
$11\ ft./sec$
0%
$10\ ft./sec$
0%
$12.5\ ft./sec$

Four of the six numbers 1867, 1993, 2019, 2025, 2109 and 2121 have a mean ofWhat is the mean of the other two numbers.

0%
$1994$
0%
$2006$
0%
$2022$
0%
$2051$

Two pipes fill a tank in 20 minutes and 60 minutes. A third pipe empties the same tank in 40 minutes. In how much time will the tank be filled if all three pipes are opened together?

0%
6 min
0%
24 min
0%
30 min
0%
40 min

If a pipe fills a tank in 20 minutes and a pipe empties the same tank in 60 minutes. Then in how much time the tank will be filled completely if both the pipes are opened together?

0%
10 minutes
0%
70 minutes
0%
40 minutes
0%
30 minutes

The mean of series

$x_{1},x_{2},x_{3},........x_{n} \ is \bar{x},$ then mean of the series

$x_{1},+2i, i=1,2,3,....................n$ will be

0%
$\bar{x}+n$
0%
$\bar{x}+n+1$
0%
$\bar{x}+2$
0%
$\bar{x}+2n$

$16$ men can build a wall of

$52$ m long in

$25$ days working for

$8$ hours a day, in how many days can

$64$ men build a similar wall of

$260$ m long working for

$10$ hours a day?

0%
$35$
0%
$65$
0%
$40$
0%
$25$

Explanation

$\text { Solve:- on considering first situation } \\$

$\text { no. of men }\left(m_{1}\right)=16 \\$

$\text { no. of } \operatorname{days}\left(D_{1}\right)=25 \\$

$\text { no. of hours work daily }=8 \text { hours } \\$

$\text { - length of wall }=52 \mathrm{~m}$

and in second case by using similar

notation we get,

$M_{2} =64 \\ D_{2} =?$

no. of hour

$=10$ hours

length of wall

$=260 \mathrm{~m}$

We know that, no. of men

$\alpha$ work

no. of hours

$\alpha$ work

no. of days

$\alpha$ work

$\text { So, } \frac{M_{1} \times D_{1} \times H_{1}}{l_{1}}=\frac{M_{2} \times D_{2} \times H_{2}}{l_{2}} \\$

$\Rightarrow \frac{16 \times 25 \times 8}{52}=\frac{64 \times D_{2} \times 10}{260}$

$\quad \Rightarrow D_{2}=25$ Days

$H_{1}, H_{2}$ Denotes hours in first and second

case. Similarly

$l_{1}$ and

$l_{2}$ represent length

A and

$B$ can do a job together in

$7$ days

$A$ is

$1\dfrac { 3 } { 4 }$ times as efficient as

$B$ The same job can be done by

$A$ alone in

0%
$9 \dfrac { 1 } { 3 }days$
0%
$11 days$
0%
$12\dfrac { 1 } { 4 } days$
0%
$16\dfrac { 1 } { 2 } days$

Explanation

$A$ and

$B$ can do a job together in 7 days

$A$ is

$1 \frac{3}{4}$ times as efficient as

$B$

$\begin{aligned} \Rightarrow \quad A: B &=\mid \frac{3}{4}: 1 \\ &=7: 4 \\ A &=7 x \quad, \quad B=4 x \end{aligned}$

$\underline{A \cdot T \cdot Q}$

$\begin{array}{l} A+B=\frac{1}{7} \\ 7 x+4 x=\frac{1}{7} \\ \Rightarrow x=\frac{1}{77} \\ \therefore A=7 x=\frac{1}{11}=11 \text { days } \end{array}$

Hence, (B) is the correct option.

In a conference hall 10 executive are to be seated on 10 chairs placed left to right. Executive

$E_1$ wants to sit to the right of

$E_2$ Also,

$E_4$ wants to sit the left of

$E_1$ and

$E_5$ wants to sits to the left of ways

$E_4$ Number of ways the executives can be seated on the 10 chairs, is

0%
$\dfrac{10!}{4!}$
0%
$\dfrac{10!}{3}$
0%
$10!$
0%
$3 \times \dfrac{10!}{4!}$

Number of positive integers

$n ,$ less than

$17 ,$ for which

$n ! + ( n + 1 ) ! + ( n + 2 ) !$ is an integral multiple of

$49$ is

0%
$2$
0%
$5$
0%
$7$
0%
$9$

$4$ men and

$3$ women finish a job in

$6$ days. And

$5$ men and

$7$ women can do the same job in

$4$ days. How long will

$1$ man and

$1$ woman take to do the work?

0%
$22\dfrac { 2 }{ 7 } days$
0%
$25\dfrac { 1 }{ 2 } days$
0%
$5\dfrac { 1 }{ 7 } days$
0%
$12 \dfrac { 7 }{ 22 } days$

Explanation

(

$4$ men

$+3$ women)

$\times 6$ days

$=$ (

$5$ men

$+ 7$ women)

$\times 4$ days

$24$ men

$+18$ women

$=20$ men

$+28$ women

$4$ men days

$=10$ women days

$1$ men days

$=2.5$ women days

Total work

$=24$ men days

$+18$ women days

$=20$ men days

$+28$ women days

$=78$ women days

$1$ man

$+1$ women

$=2.5$ women

$+1$ women

$=3.5$

women can do

$\dfrac{78}{3.5}=22\left(\dfrac{2}{7}\right)$ days

$m$ and

$n$ are positive integers more than or equal to

$2 , m > n ,$ then

$( m n ) !$ is divisible by

0%
$( m ! ) ^ { n }$
0%
$( n ! ) ^ { m }$
0%
$( m + n ) !$
0%
$( m - n ) !$

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 12 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 12 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.