MCQ Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 10

In how many way can

$12$ gentlemen sit around a round table so that three specified gentlemen are always together.

0%
$9!$
0%
$10!$
0%
$3! 10!$
0%
$3! 9!$

Explanation

Let us consider the

$3$ specific gentlemen as a single person that occupies three times the space.

So, now we have

$10$ persons who can be made to sit in

$9!$ ways and in the space for

$3$ gentlemen we can permute them in

$3!$ ways.

So, total ways are

$3!\, 9!$ .

The maximum number of intersection points of n circles and n straight lines , among themselves is 80.The value of n is

0%
$7$
0%
$6$
0%
$5$
0%
$8$

A is thrice as good a workman as B and takes 10 days less to do a piece of work than B takes. B can do the work in _______.

0%
$12$ days
0%
$15$ days
0%
$20$ days
0%
$30$ days

Explanation

Given, Work Efficiency ratio of A:B is 3:1

Let B takes

$x days$ to do the work. So according to the question, A takes

$x-10\: days$ to do the same work. Time ratio and efficiency ratio are inversely proportional. So equating the ratios,

$\dfrac{x-10}{x}=\dfrac{1}{3}$

$\implies x=15$

Therefore, B can finish the work alone in

$15\: days$

When

$n!+1$ is divided by any natural number between

$2$ and

$n$ then remainder obtained is

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

$\alpha ,\beta , \gamma$ are three consecutive integers. If these integers are raised to first, second and third positive powers respectively, and added then they form a perfect square, the square root of which is equal to the sum of these integers. Also,

$\alpha < \beta < \gamma$ . Then,

$\gamma$ is equals to:

0%
$3$
0%
$14$
0%
$5$
0%
$11$

Explanation

Let the numbers be

$n-1,n,n+1$

As per the given information, we have

$(n-1)+{ n }^{ 2 }+{ (n+1) }^{ 3 }={ (n-1+n+n+1) }^{ 2 }$

$\Rightarrow { n }^{ 3 }-5{ n }^{ 2 }+4n=0$

$\Rightarrow n=0,1,4$

For

$n=0,$ the numbers are:

$-1,0,1$ this is out as all the numbers should be positive

For

$n=1,$ the numbers are:

$0,1,2$ this is out as all the numbers should be positive (‘0’ can’t be taken as positive)

For

$n=4,$ the numbers are:

$3,4,5$

We have got largest number which is

$5$

Hence, the value of

$\gamma$ is

$5$ .

X was born on March 6,The same year Independence Day was celebrated on Friday.Find out the birthday of X.

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

Explanation

(a) Independence Day was celebrated after [25 (March) + 30 (April) + 31 (May) + 30 (June) + 31 (July) + 15 (August)] days
= 162 days = 23 x 7 + 1
= 23 weeks + 1 days

Number off odd days = 1

$\therefore$ Hence,X was born on Thrusday.

$150$ workers were engaged to finish a piece of work in a certain number of days.

$4$ workers dropped the second day,

$4$ more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is

0%
$15$
0%
$20$
0%
$25$
0%
$30$

Explanation

$\textbf{Step-1: Forming a linear equation using the concept of Arithmetic Progression}$

$\text{Let the work be finished in}$

$n$

$\text{days}$

$\text{The total number of workers who worked during these days will be the sum of the following A.P.}$

$150+146+142+.......n$

$\text{The sum of }n\text{ terms in an A.P. is given by:}$

$S_n=\dfrac{n}{2}[2a+(n-1)d]$

$\text{Hence, we get}$

$S_n=n(152-2n)$

$...(i)$

$\textbf{Step-2: Finding the number of days}$

$\text{If the workers hadn't dropped, then the number of days required would be}$

$(n-8)$

$\text{So, the sum would be}$

$150(n-8)$

$...(ii)$

$\text{Equating}$

$(i)$

$\text{and}$

$(ii)$

$n(152-2n)=150(n-8)$

$n^2-n-600=0$

$n^2-25n+24n-600=0$

$(n-25)(n+24)=0$

$\textbf{As the number of days can't be negative, we neglect}$

$(n+24)=0$

$n=25$

$\textbf{Hence, number of days in which the work was completed is}$

$\mathbf{25}$

How many times from 4 pm to 10 pm, the hands of a clock are at right angles?

0%
10
0%
11
0%
9
0%
6

Choose the correct answer from the alternatives given :
A and B together can complete a work in 12 days. A alone can complete it in 20 days. If B now does the work only for half a day daily, then the number of days required to complete the work, by A and B together, is

0%
14
0%
16
0%
18
0%
15

Explanation

Let the work be 60 units. [LCM of 12 and 20]
A and B i.e. A + B do =

$\displaystyle \frac{60}{12}$ = 5 units/day
A does =

$\displaystyle \frac{60}{20}$ = 3 units/day
Hence, B does 5 3 = 2 units/day
If B works for half a day, so B does

$\displaystyle \frac{2}{2}$ = 1 unit / day
A and B together now do 3 + 1 = 4 units/day
Hence, will complete the work in

$\displaystyle \frac{60}{4}$ = 15 days.

A telephone number

$d_1d_2d_3d_4d_5d_6d_7$ is called memorable if the prefix sequence

$d_1d_2d_3$ is exactly the same as either of the sequence

$d_4d_5d_6$ or

$d_5d_6d_7$ (or possibly both). If each

$d_1\epsilon\{x|0\leq x\leq 9, x\epsilon W\}$ , then number of distinct memorable telephone number is(are).

0%
$19810$
0%
$19,910$
0%
$19,990$
0%
$20,000$

The number of ways in which

$6$ rings can be worn on the four fingers of one hand is

0%
$4^{6}$
0%
$^{6}C_{4}$
0%
$6^{4}$
0%
None of these

Explanation

Each ring can be worn on

$4$ fingers, means each have

$4$ possibilities.

$\therefore$ Total ways

$=4\times4\times4\times4\times4\times4=4^{6}$

Hence,

$(A)$

A supplies

$20$ men who work for

$8$ hrs. a day working for

$6$ days.

$B$ supplies

$15$ men working at

$9$ hrs. a day for

$7$ days and

$C$ supplies

$10$ men working

$6$ hrs. a day for

$8$ days to do a certain job. If Rs.

$636$ is paid for all labour, what is

$C's$ share?

0%
$130$
0%
$125$
0%
$128$
0%
$135$

Explanation

Man hours supplied by

$A = 20\times 8\times 6$
Man hours supplied by

$B = 15\times 7\times 9$
Man hours supplied by

$C = 10\times 6\times 8$

$\therefore$ Ratio of their shares

$= (20\times 8\times 6) : (15\times 7\times 9) : (10\times 6\times 8)$

$= 64 : 63 : 32$

$\therefore C's\ share = \dfrac {32}{159}\times 636 = Rs. 128$ .

A completes half as much work as B and C completes half as much work as A and B together, in the same time. If C alone can completes the work in 40 days, all of them can together finish the work in

0%
13 $\dfrac{1}{3}$ days
0%
14 $\dfrac{1}{3}$ days
0%
20 days
0%
30 days

Explanation

Given that,

$\displaystyle A \, = \, \dfrac{1}{2} \, \times \, B \, \Rightarrow \, \frac{A}{B}= \, \dfrac{1}{2}$
A : B = 1 : 2 or B : A = 2 : l

Also, given that

$C= \, \dfrac{1}{2} \, \times \, (A \, + \, B)$

$C= \, \dfrac{1}{2} \, \times \, (A \, + \, 2A)$
C =

$\dfrac{3}{2}$ A
C : A = 3 : 2 or A : C = 2 : 3
From equation (1) & (2), we get
B : A: C = 4 : 2 : 3 or A : B : C = 2 : 4 :3
Time taken to do the work

$\displaystyle \, \frac{1}{work \, done}$

Hence

$\displaystyle t_A \, : \, t_B \, : \, t_C \, = \, \frac{1}{2} \, : \, \frac{1}{4} \, : \, \frac{1}{3}$ or 6 : 3 :4
Now,

$t_C$ = 40

$t_A$ = 60

$t_B$ = 30
Let the work be 120. [LCM of 40, 60, 30]
A does

$\displaystyle A does \, = \, \frac{120}{60} \, = \, 2$ units/day

B does

$\displaystyle B does \, = \, \frac{120}{30} \, = \, 4$ units/day

C does

$\displaystyle C does \, = \, \frac{120}{40} \, = \, 3$ units/day
Together they do the work in =

$\displaystyle \frac{120}{2 \, + \, 4 \, + \, 3} \, = \, \frac{120}{9} \, = \, \frac{40}{3} \, = \, 13 \, \frac{1}{3}$ days

A tank

$15$ m long,

$10$ m wide and

$6$ m deep open at the top. If the width of the sheet is

$2$ m, then cost of iron sheet at the rate of Rs.

$5$ per meter is

0%
Rs. $550$
0%
Rs. $1050$
0%
Rs. $1125$
0%
Rs. $1150$

Explanation

Total surface Area (excluding top) of the tank

$= 450\ sq.m.$

$450\ sq. m = L\times 2m$ (i.e. area of sheet required)

$\therefore L = 225\ m$

$\therefore$ cost at rate

$Rs. 5/m$ is

$225\times 5 = Rs. 1125$ .

Choose the correct answer from the alternatives given.
A contractor took a contract for building

$12$ kilometre road in

$15$ days and employed

$100$ labours on the work. After

$9$ days he found that only

$5$ kilometre road had been constructed. How many more labours should be employed to ensure that the work may be completed with in the given time?

0%
$120$
0%
$90$
0%
$110$
0%
$100$

Explanation

$100$ labours worked for

$9$ days,
Hence

$900$ man days =

$5$ km

$1$ km =

$\dfrac{900}{5}$ man days.
Now

$7$ km of the road needs to be built
Then

$7km$ =

$\dfrac{900}{5} \times 7 = 180 \times 7$ =

$1260$ man days.
Now thus

$1260$ man days of work needs to he completed in

$6$ days.
Let number of more workers required be x.

$1260=(100 + x)$

$\times$

$6$

$210 = 100 + x \Rightarrow x= 110$
Hence, number of more workers required are

$110$ .

If m men can do a job in d days, then m + r men can do the job in:

0%
d + r days
0%
d - r days
0%
$\frac{md}{m + r}$ days
0%
$\frac{d}{m+r}$ days

$20$ men working

$8$ hours per day can complete a piece of work in

$21$ days. How many hours per day must

$48$ men work to complete the same job in

$7$ days

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

Explanation

We know, Work = Men

$×$ hr/day×days

Work is same in both cases

Let

$48$ men will complete the same work in

$x$ hr/day

$20×8$ hr/day

$×21=48×x$ hr/day

$×7$

$48x=\dfrac{(20×8×21)}7$

$48x=20×8×3$

$x=\dfrac{(20×8×3)}{48}$

$x=\dfrac{480}{48}$

$x=10$

$48$ men must work

$10$ hr/day to complete the same work in

$7$ days

Khilona earned scores of $97$ , $73$ and $88$ respectively in her first three examinations. If she scored $80$ in the fourth examination, then her average score will be

0%
increased by $1$
0%
increased by $1.5$
0%
decreased by $1$
0%
decreased by $1.5$

Explanation

When earned scores are

$97,73,88$ then

Average

$=\dfrac{97+73+88}{3}\Rightarrow 86.$

When

$80$ is added in the score list then

Average

$=\dfrac{97+73+88+80}{4}\Rightarrow 84.5$

Hence, the average is decreased by

$1.5.$

$\dfrac { ^{ n }{ P }_{ r-1 } }{ a } =\dfrac { ^{ n }{ P }_{ r } }{ b } =\dfrac { ^{ n }{ P }_{ r+1 } }{ c }$ , then which of the following holds good:

0%
$c^{2}=a(b+c)$
0%
$a^{2}=c(a+b)$
0%
$b^{2}=c(a-b)$
0%
$\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}=1$

Explanation

$\dfrac{^{n} {P}_{r-1}}{a} = \dfrac{^{n}{P}_{r}}{b} = \dfrac{^{n}{P}_{r+1}}{c}$

$\implies \dfrac{n!}{(n - r+ 1)!a} = \dfrac{n!}{(n-r)!b} = \dfrac{n!}{(n-r-1)!c}$

$\implies \dfrac{n!}{(n - r+ 1)!a} = \dfrac{n!}{(n-r)(n-r+1)!b} = \dfrac{n!}{(n-r-1)(n-r)(n-r+1)!c}$

$\implies \dfrac{1}{a} = \dfrac{1}{(n-r)b} = \dfrac{1}{(n-r-1)(n-r)c}$

$\implies n-r = \dfrac{a}{b}$ and

$n-r-1 = \dfrac{b}{c}$

$\implies n-r - 1 = \dfrac{a}{b} - 1$

$\implies \dfrac{b}{c} = \dfrac{a}{b} - 1$

$\implies \dfrac{b}{c} - \dfrac{a}{b} = - 1$

$\implies b^2 - ac = - bc$

$\implies b^2 = ac - bc$

$\therefore b^2 = c(a - b)$ (Ans)

Garlands are formed using 6 red roses and 6 yellow roses of different sizes. The number of arrangements in garland which have red roses and yellow roses come alternately is

0%
$5!\times 6!$
0%
$6! \times 6!$
0%
$\dfrac{5!}{2!}\times 6!$
0%
$2(6!\times 6!)$

Working 4 hours daily, Swati can embroid 5 sarees in 18 days. How many days will it take for her to embroid 10 sarees working 6 hours daily

0%
24 days
0%
6 days
0%
12 days
0%
20 days

Explanation

$\dfrac{4}{6}\times \dfrac{10}{5} \times 18 = n$

$n = 24$

$24$ days

1420471_1077278_ans_130c8f85e2644cbab6d2e6cdf367796f.png

Number of ways in which

$7$ green bottles and

$8$ blue bottles can be arranged in a row if exactly

$1$ pair of green bottles is side by side is (Assume all bottles to be a like except for the colour).

0%
$84$
0%
$360$
0%
$504$
0%
None of the above

Explanation

Consider the two green bottles as one entity. Let's call it

$GG$ .

We will also refer to the green bottle as:

$G$ , and to the blue bottle as:

$B$ .

So now we have

$14$ units to arrange ( 1 green pair bottle + the remaining 5 green bottles + the 8 blue bottles):

$GG, G, G, G, G, G, B, B, B, B, B, B, B, B$

Now place the 8 blue bottles in a row, such that between each bottle has a gap before and after it:

_B_B_B_B_B_B_B_B_

Now we need to take the six green units (the 1 pair and 5 bottles) and place them in six of the nine gaps.

Therefore, the solution is:

$^9C_6 = 84 ways$ .

Number of five-digit numbers divisible by 5 that can be formed from the digits

$0,1, 2, 3, 4, 5$ without repetition of digits are

0%
$240$
0%
$360$
0%
$148$
0%
$216$

Let

$5 < n_1 < n_2 < n_3 < n_4$ be integers such that

$n_1+n_2+n_3+n_4=35$ . The number of such distinct arrangements

$(n_1, n_2, n_3, n_4)$ .

0%
$^{38}C_3$
0%
$^8C_3$
0%
$5$
0%
$6$

Explanation

${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+.....+{ n }_{ k }=R$

for this arrangement,

No. of arrangement or distinct arrangements are

${ R+K-1 }_{ { C }_{ K-1 } }$

So, for

${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+{ n }_{ 4 }=35$

No. of arrangements

$={ 35+4-1 }_{ { C }_{ 4-1 } }={ 38 }_{ { C }_{ 3 } }$

$10$ Men begin to work together on a job, but after some days,

$4$ if them left the job. As a result the job which could have been completed in

$40$ days is completed in

$50$ days. How many days after the commencement of the work did the

$4$ men level?

0%
$25$
0%
$30$
0%
$10$
0%
$15$

Explanation

$10$ Men begin work

It could been done in

$40$ days.

Let

$4$ men left after x days.

Left days are

$40-x$ if they do not left and

$50-x$ if they left.

$(40-x) 10= (50-x)6$

$400-10 x=300-6x$

$1000=4x$

$x=25$ days.

The number of permutation of the letters of the word

$HINDUSTAN$ such that neither the pattern

$'HIN'$ nor

$'DUS'$ nor

$'TAN'$ appears, are :

0%
$166674$
0%
$169194$
0%
$166680$
0%
$181434$

Explanation

Total number of letters

$=9$ in which N is repeated twice

Therefore total number of permutation

$=\dfrac{9!}{2}$

The number of permutation in which HIN comes as block

$=9-3+1=7!$

The number of permutation in which DUS comes as block

$=7!$

The number of permutation in which TAN comes as block

$=\dfrac{7!}{2}$

The number of permutation in which HIN and DUS comes as block

$=5!$

The number of permutation in which DUS and TAN comes as block

$=5!$

The number of permutation in which HIN and TAN comes as block

$=5!$

Therefore the required number of permutations

$=\dfrac{9!}{2}-\left(7!+7!+\dfrac{7!}{2}-3\times5!+3!\right)$

$=\dfrac{362880}{2}-\left(5040+5040+\dfrac{5040}{2}-360+6\right)$

$=181440-\left(10080+2520-360+6\right)$

$=181440-12246$

$=169194$

A train of 320 m cross a platform in 24 seconds at the speed of 120 km/h. while a man cross same platform in 4 minute.What is the speed of man in m/s?.

0%
$2.4$
0%
$1.5$
0%
$1.6$
0%
$2.0$

Explanation

Speed of train

$=120\ kmph$

$=\dfrac{120\times5}{18}$

$=\dfrac{100}{3}\ m/sec$

Let 'x' be the length of platform

Now,

$\dfrac{Length\ of\ train\ and\ platform}{Time\ taken\ in\ crossing}=\dfrac{100}{3}$

$\Rightarrow\dfrac{320+x}{24}=\dfrac{100}{3}$

$\Rightarrow 320+x=\dfrac{24\times100}{3}$

$\Rightarrow 320+x=800$

$\Rightarrow x=480\ m$

$\therefore$ Man's speed

$=\dfrac{480}{4\times60}=2\ m/sec$

Let the eleven letters,

$A, B, ....K$ denote an artbitrary permutation of the integers

$(1,2,....11)$ , then

$(A-1)(B-2)(C-3)...(K-11)$ is

0%
Necessarily zero
0%
Always odd
0%
Always evem
0%
None of these

A, B, and C together can finish a place of work in 12 days. A and C together work twice as much as B, A and B together work thrice as much as C. In what time ( day ) could each do it separately

0%
144/5, 36, 48
0%
36, 48, 144/5
0%
48, 144/5, 36
0%
none of these

Explanation

Let A's 1 day work = a

B's 1 day work = b

C's 1 day work = c

$\therefore A+B+C = \frac{1}{12}$ ...(1)

$A+C = 2B$ ...(2)

$A+B = 3C$ ...(3)

From(2)

$A = 2B -C$ ...(4)

from (3)

$A = 3C -B$ ...(5)

comparing (4) and (5)

$2B-C-3C-B$

$3B = 4C$

$B = \frac{4}{3}C$ ...(6)

putting in (5)

$A = BC -\frac{4C}{3}$

$A = \frac{5C}{3}$ ...(7)

putting (6) and (7) in (1)

$\frac{5C}{3}+\frac{4C}{3}+C = \frac{1}{12}$

$\frac{12C}{3} = \frac{1}{12}$

$C = \frac{1}{48}$

$\therefore A = \frac{5}{3}\times \frac{1}{48} = \frac{5}{144}$

$B = \frac{4}{3} \times \frac{1}{48} = \frac{1}{36}$

$\therefore$ A finishes in

$\frac{144}{5}$ days

B finishes in 36 days

C finishes in 48 days

1114349_1200765_ans_5c286ab1b13641cfb81f9da0b76e8ef0.jpg

Pipe A can fill a tank in

$10h$ and pipe

$B$ can fill the same tank in

$12h$ . Both the pipes are opened to fill the tank and after

$3h$ pipe

$A$ is closed. Pipe

$B$ will fill the remaining part of the tank in

0%
$5h$
0%
$4h$
0%
$5h$ $24min$
0%
$3h$

Explanation

Let capacity of tank = V liters

Speed of A = V/10 lit/hour

B = V/12 lit/hour

In 3hrs both fill

$= 3(\frac{V}{10})+3(\frac{V}{12}) = 3V(\frac{11}{60})$

$= 33V/60$ Remaining

$27V/60$

B fills remaining

$= \frac{\frac{27V}{60}}{\frac{V}{12}} = \frac{27}{5} = 5hr.24\, min$

$\Rightarrow (c)$

1118219_1190651_ans_f8ffda6b13c14107913fa5b7bc092b43.jpg

$24$ men working at

$8$ hours per day can do a piece of work in

$15$ days. In how many days can

$20$ men working at

$9$ hours per day do the same work?

0%
$14$ days
0%
$16$ days
0%
$13$ days
0%
$17$ days

Explanation

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

${\text{Total 24 men work for 8 hours a day and takes a total of 15 days to complete it}}{\text{.}}$

${\text{So total work units}} = 24 \times 8 \times 15 = 2880$

${\text{Thus, for the first case, it's a total of 2880 work units}}{\text{.}}$

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

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

${\text{Number of work units}} = 20 \times 9 \times x = 180x$

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

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

$\Rightarrow 2880 = 180x$

$\Rightarrow x = \dfrac{{2880}}{{180}} = 16$

${\textbf{Thus, it will take 16 days for the 20 workers to complete the work}}{\textbf{. Thus, option B }}$ $\textbf{is correct.}$

A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before.How many days did it take him to finish the job?.

0%
$11$
0%
$10$
0%
$12$
0%
$14$

Explanation

$\textbf{Step 1 : Find the no. of days}$

$\text{Here, a=5 and d=2}$

$\text{Let the carpenter finish the job in n days}$

$\text{Then,}$

$S_n=192$

$\Rightarrow 192=\dfrac{n}{2}[2a+(n-1)d]$

$\Rightarrow 192=\dfrac{n}{2}[2\times5+(n-1)2]$

$\Rightarrow 192=n[5+n-1]$

$\Rightarrow n^2+4n-192=0$

$\Rightarrow n^2-12n+16n-192=0$

$\Rightarrow n(n-12)+16(n-12)=0$

$\Rightarrow (n-12)(n+16)=0$

$\therefore n=12$

$\quad \quad \textbf{[Rejecting n = -16 as number of days cannot be negative]}$

$\textbf{Hence, he will finish job in 12 days}$

$(1+x+x^2)^n=\displaystyle\sum^{2n}_{r=0}a_rx^r$ , then

$a_0a_{2r}-a_1a_{2r+1}+a_2a_{2r+2}-....=?$

0%
$a_r$
0%
$a_{n-2r}$
0%
$a_{n+r}$
0%
$a_{2r}$

$150$ workers were engaged to finish piece of work in a certain number of day. Four workers dropped the second day, four more workers dropped the third day and so on, it take

$8$ more days to finish the work now. Then the number of days in which the work was completed is

0%
$29$ days
0%
$24$ day
0%
$25$ days
0%
None of these

Find the average of the following set of scores

$253,124,255,534,836,375,101,443,760$

0%
$427$
0%
$413$
0%
$141$
0%
$409$

$A$ can do a piece of work in

$15$ days.

$B$ is

$50\%$ more efficient than

$A$ .

$B$ can finish it in

0%
$10$ days
0%
$7 \dfrac { 1 } { 2 }$ days
0%
$12$ days
0%
$12 \frac { 1 } { 2 }$ days

Explanation

$A\rightarrow 15$ days

capacity of

$A$ is

$\dfrac{W}{15}$ where

$W$ is worke

$B$ is

$50\%$ more efficient than

$A$

$\dfrac{W}{15}+\dfrac{50}{100}(\dfrac{W}{15})=\dfrac{150}{100}(\dfrac{W}{15})$

$=\dfrac{3}{2}(\dfrac{W}{15}$ )

Let

$B$ take

$t$ days to complete

$W$ work

$\dfrac{3}{2}(\dfrac{W}{15})\times t=W$

$t=10\,days$

Six people are going to sit in a row on a bench.

$A$ and

$B$ are adjacent,

$C$ does not want to sit adjacent to

$D.E$ and

$F$ can sit anywhere. Number of ways in which these six people can be seated is

0%
$200$
0%
$144$
0%
$120$
0%
$56$

Explanation

A, B, C, D, E, F

Consider AB as group so we have AB, C, D, E, F.

We have totally

$5$

No. of ways

$(w_1)=5!\times 2$

$=240$

Let CD are adjacent now AB, CD, E, F

No. of ways

$(w_2)=4!2!2!$

$=96$

Total no. of ways

$W=w_1-w_2$

$=240-96$

$=144$ .

1195752_1268180_ans_f3f29513dac144278531110cb0dd18f2.jpg

Working together, pipes

$A$ and

$B$ can fill an empty tank in

$10\ hours$ . they worked together for

$4$ hours and then

$B$ stopped and

$A$ continued filling the tank till was full. It took a total of

$13\ hours$ to fill the tank. How long would it take

$A$ to fill the empty tank alone?

0%
$13\ hours$
0%
$15\ hours$
0%
$17\ hours$
0%
$18\ hours$

$200$ students take

$40$ days to complete the project, how much time will be taken by

$250$ students?

0%
$31$ days
0%
$32$ days
0%
$23$ days
0%
$21$ days

Explanation

$200$ students

$\rightarrow$

$40$ days.

Rate work of 1 student

$\Rightarrow (\frac{x}{40})\times \frac{1}{200}$

rate of

$250$ students

$\Rightarrow \frac{x\times 250}{40\times 200}=\frac{x}{32}$

$\therefore$ Time required =

$32$ hours.

1212186_1286090_ans_b4d54db52b21400e83d112e26a2d5540.JPG

Moving along the

$x-$ axis there are two points with

$x=10+6t,x=3+t^{2}$ , the speed with which they are reaching from each other at the time of encounter is (

$x$ is in

$cm$ and

$t$ is in seconds.)

0%
$16\ cm/sec$
0%
$20\ cm/sec$
0%
$8\ cm/sec$
0%
$12\ cm/sec$

The number of seven letter words that can be formed by using the letters of the word

$SUCCESS$ that the two

$C$ are together but no two

$S$ are together is

0%
$24$
0%
$18$
0%
$54$
0%
none of these

Explanation

using the letters of the word

$SUCCESS$ that the two

$C$ are together but no two

$S$ are together

let two C's be 1 unit

$\therefore$ no. of ways

$=\frac{6!}{4!}=30$

We have to put 1 letter between one S

So, No of ways

$=2\times 3!=12$

No. of ways=30-12=18

4 men and 6 woman get Rs 1600 by doing a piece of work in 5 days. 3 men and 7 woman get Rs 1740 by doing the same work in 6 days. In how many days, 7 men and 6 woman can complete the same work getting Rs 3760?

0%
$6$ days
0%
$8$ days
0%
$10$ days
0%
$12$ days

Explanation

$5$ days ,(

$4$ men

$+6$ women) get

$=$ Rs

$1600$

$\therefore$ , In

$1$ day, (

$4$ men

$+6$ women) get

$=\dfrac { 1600 }{ 5 } =$ Rs.

$320$ ...

$(1)$

$1$ day, number of persons to get Re.

$1=\dfrac { 320 }{ 4men+6women }$ ....

$(2)$

Similarly, in second condition,

$1$ day, number of person to get Re.

$1$

$=\dfrac { 1740 }{ 6\times \left( 3men+7women \right) } =\dfrac { 290 }{ 3men+7women }$ ...

$(3)$

From Eqs.

$(2)$ and

$(3)$ We get

$\dfrac { 320 }{ 3men+7women } =\dfrac { 290 }{ 3men+7women }$

$96$ men

$+224$ women

$=116$ men

$+174$ women

$\Rightarrow 20men=50women$

$\Rightarrow \dfrac { man }{ woman } =\dfrac { 5 }{ 2 }$

$\therefore, 1$ woman

$=\dfrac { 2 }{ 5 }$ man

from Eq.

$(2)$ in

$1$ day,

$\left( 4men+6women \right) =\left( 4men+6\times \dfrac { 2 }{ 5 } m \right)$

$=\dfrac { 32 }{ 5 } men\quad get\quad Rs.320$

$1$ day ,

$1$ man get

$\dfrac { 320\times 5 }{ 32 } =Rs.50$

$\therefore$ In

$1$ day,

$1$ woman get

$\dfrac { 50\times 2 }{ 5 } =Rs.20$

$\therefore$ , in

$1$ day

$\left( 7men+6women \right) get\quad 7\times 50+6\times 20=Rs.470$

Let, required number of days

$=\dfrac { 3760 }{ 470 } =8$ days

Find

$x$ , if

$\dfrac {1}{4!}-\dfrac {1}{x}=\dfrac {1}{5!}$ .

0%
$5$
0%
$4$
0%
$30$
0%
$None$

$3$ men or

$5$ women can do a work in

$12$ days. How long will

$6$ men and

$5$ women take to finish the work

0%
$6 \,\,days$
0%
$5\,\, days$
0%
$4 \,\,days$
0%
$3 \,\,days$

Explanation

$3 \ men \equiv 5 \ women$

$1 man = \dfrac{5}{3}\times 6 =10 women$

$(6 \ men + 5 \ women ) \equiv 15 \ women$

$5$ women can do the work in

$12$ days.

$1$ woman can do it in

$12 \times 5$ days.

Therefore,

$15$ women can do it in

$\dfrac{12 \times 5}{15}=4$ days

Pipe

$A$ can fill one fourth of a tank in

$5$ hours and pipe

$B$ can empty the full tank in

$30$ hours.If both pipes are opened simultaneously, then in how many hours will the empty tank be filled.

0%
$10$
0%
$60$
0%
$25$
0%
$45$

Explanation

Pipe can do

$\frac{1}{4}$ work in 5 hours.

$\Rightarrow$ pipe can do

$\frac{1}{20}$ work in 1 hours.

when both are opened simultaneously,

they can do

$\frac{1}{20} \frac{-1}{30} = \frac{1}{60}$ work in 1 hours

So, when opened together,

it takes 60 hours to fill the tank.

1214734_1340754_ans_b6130ad7fe484d8591ce0a4c82a2c2d5.jpg

A takes

$5$ days less than

$B$ to complete the work. Both complete the work in

$6$ days. In how many days

$A$ complete the work?

0%
$10$
0%
$15$
0%
$20$
0%
$30$

Explanation

From given, we have,

$\dfrac{1}{x}+\dfrac{1}{x+5}=\dfrac{1}{6}$

$6\left(x+5\right)+6x=x\left(x+5\right)$

$x^2+5x-30=12x+30-30$

$x^2-7x-30=0$

$(x-10)(x+3)=0$

$\therefore x=-3,10$

Days can't be negative.

Therefore, the number of days A uses to complete the work is

$10$ days.

$A$ and

$B$ can complete a task in

$12$ days. However,

$A$ had to leave a few days before the task was complete and hence it took

$16$ days in all to complete task. If

$A$ alone could complete the work in

$21$ days, how many days before the work getting over did

$A$ leave?

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

Explanation

B can alone do the work

$=\dfrac{1}{\dfrac{1}{12}-\dfrac{1}{21}}$ days

$\implies \dfrac 1B = \dfrac{7-4}{84}=\dfrac{3}{84} = \dfrac1{28}$

$=28$ Days

Let

$A$ worked

$X$ days, So

Work done by

$A+$ Work done by

$B =1$

$\dfrac{X}{21} +\dfrac{16}{28} =1$

$\implies \dfrac{X}{21} = \dfrac{12}{28}$

$\implies X=9$

Hence,

$A$ worked for

$9$ days means

$A$ left work before

$( 16-9)= 7$ days of completion.

$A$ and

$B$ can do a work in

$10$ days.

$B$ and

$C$ can do the same work in

$15$ days.

$C$ and

$A$ can complete the same work in

$20$ days. In how many days can

$C$ alone complete the work ?

0%
$120$ days
0%
$118$ days
0%
$115$ days
0%
$110$ days

Explanation

In 10 days

$A$ and

$B$ does 1 work.

$\therefore$ In 1 day

$A$ and

$B$ will do

$\dfrac{1}{10}$ work -(i)

In 15 days

$B$ and

$C$ does 1 work

$\therefore$ In 1 day

$B$ and

$C$ will do

$\dfrac{1}{15}$ work -(ii)

In 20 days

$C$ and

$A$ does 1 work

$\therefore$ In 1 day

$C$ and

$A$ will do

$\dfrac{1}{20}$ work - (iii)

Adding equation(i),(ii)and (iii)

$In \quad 1 \quad$ day

$\quad 2(A+B+C)$ will

$do\left(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{20}\right)$

$In \quad 1 \quad$ day

$\quad 2(A+B+C)$ will

$d o$

$\dfrac{13}{60}$

work.

In 1 day

$(A+B+C)$ will

$do \quad \dfrac{13}{120}$ work

In 1 d a y

$(A+B)+(C)$ will

$do \dfrac{13}{120}$ work

$A+B$ does

$\dfrac{1}{10}$ work in 1 day

$\therefore$ work done by C in 1 day

$=\dfrac{13}{120}-\dfrac{1}{10}$

$=\dfrac{1}{120}$

$\dfrac{1}{120}$ work is done by

$C$ in 1 day.

$\therefore 1$ work is done by

$C$ in 120 days.

Answer: option

$(A)$

$39$ persons can repair a road in

$12$ days, working

$5$ hours a day. In how many days will

$30$ persons, working

$6$ hours a day, complete the work?

0%
$9$
0%
$12$
0%
$13$
0%
$10$

Explanation

Given,

Let the required number of days be

$x$ .

$\left.\begin{matrix} Persons & 30:39 \\ Working hours/day & 6:5 \end{matrix}\right\} ⋮⋮ 12:x$

$\Rightarrow 30 \times 6 \times x = 39 \times 5 \times 12$

$\therefore x= 13$

Hence the number of days.

The number of ways in which

$9$ persons can be divided into three equal groups, is

0%
$1680$
0%
$840$
0%
$560$
0%
$280$

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 10 - 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 10 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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