MCQ Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 3

An application was received by inward clerk in the afternoon of the week day. Next day he forwarded it to the table of the senior clerk, who was on leave that day. The senior clerk next day evening put up the application to the desk officer. Desk officer studied the application and disposed off the matter on the same day i.e. Friday. Which day was the application received by the inward clerk ?

0%
Monday
0%
Tuesday
0%
Wednesday
0%
Earlier week`s Saturday
0%
None of these

$A$ and

$B$ can complete a work in

$15$ days and

$10$ days respectively. They started doing the work together but after

$2$ days

$B$ had to leave and

$A$ alone completed the remaining work. The whole work was completed in:

0%
$8$ days
0%
$10$ days
0%
$12$ days
0%
$15$ days

Explanation

$(A+B)$ 's

$1$ day's work

$\left( \cfrac { 1 }{ 15 } +\cfrac { 1 }{ 10 } \right) =\cfrac { 1 }{ 6 }$
Work done by

$A$ and

$B$ in

$2$ days

$=\left( \cfrac { 1 }{ 6 } \times 2 \right) =\cfrac { 1 }{ 3 }$
Remaining work

$=\left( 1-\cfrac { 1 }{ 3 } \right) =\cfrac { 2 }{ 3 }$
Now,

$\cfrac{1}{15}$ work done by

$A$ in

$1$ day.

$\therefore$

$\cfrac{2}{3}$ work will be done by

$A$ in

$\left( 15\times \cfrac { 2 }{ 3 } \right) \quad =10$ days.
Hence, the total time taken

$=(10+2)=12$ days.

$2, 10, m, 12, 4$
A group of

$5$ integers is shown above. If the average (arithmetic mean) of the numbers is equal to

$m$ , find the value of

$m$ .

0%
$7$
0%
$8$
0%
$9$
0%
$10$
0%
$11$

Explanation

We know the average of a group of numbers is the sum of the numbers divided by the number of numbers, we can make an equation:

$\Rightarrow \dfrac { 2+10+m+12+4 }{ 5 } =m$

$\Rightarrow 28+m=5m$

$\Rightarrow m-5m=-28$

$\Rightarrow -4m=-28$

$\Rightarrow m=7$

$10$ pipes fill a tank in

$3$ hours. How many pipes are required to fill it in

$2$ hours?

0%
$15$
0%
$25$
0%
$30$
0%
$18$

Explanation

This is an example is of inverse proportion

No. of pipes Time taken

$10$

$3$ hours

$x$

$2$ hours

By inverse proportion,

$10\times 3 = x\times 2$

$\therefore x = \dfrac{10\times3}{2}$

$\therefore x = 15$

Hence the answer is option A

What will be the day of the week 15th August, 2010?

0%
Sunday
0%
Monday
0%
Tuesday
0%
Friday

Explanation

15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days 4 odd days.
Jan. Feb. March April May June July Aug.
(31+28+31+30+31+30+31+15) = 227 days.
$$\displaystyle \therefore $$ 227 days = (32 weeks + 3 days) 3 odd days.
Total number of odd days = (0 + 0 + 4 + 3) = 7

$\displaystyle \equiv$ 0 odd days.
Given day is Sunday.

How much time will the leak take to empty the full cistern?
I. The cistern is normally filled in

$9$ hours.
II. It takes one hour more than the usual time to fill the cistern because of la leak in the bottom.

0%
I alone sufficient while II alone not sufficient to answer
0%
II alone sufficient while I alone not sufficient to answer
0%
Either I or II alone sufficient to answer
0%
Both I and II are not sufficient to answer
0%
Both I and II are sufficient to answer

Explanation

I. Time taken to fill the cistern without leak

$=9$ hours.
Part of cistern filled without leak in

$1$ hour

$=\displaystyle\frac{1}{9}$
II. Time taken to fill the cistern in presence of leak

$=10$ hours.
Net filling in

$1$ hour

$=\displaystyle\frac{1}{10}$
Work done by leak in

$1$ hour

$=\left(\displaystyle\frac{1}{9}-\frac{1}{10}\right)=\displaystyle\frac{1}{90}$

$\therefore$ Leak will empty the full cistern in

$90$ hours.
Clearly, both I and II are necessary to answer the question.

$\therefore$ Correct answer is (E).

How many worker are required for completing the construction work in

$10$ days?

$I$ .

$20$ % of the work can be completed by

$8$ workers in

$8$ days.

$II$ .

$20$ workers can complete the work in

$16$ days.

$III$ . One-eighth of the work can be completed by

$8$ workers in

$5$ days.

0%
$I$ only
0%
$II$ and $III$ only
0%
$III$ only
0%
$I$ and $III$ only
0%
Any one of the three

Explanation

$I$ .

$\cfrac{20}{100}$ work can be completed by

$(8\times 8)$ workers in

$1$ day.

$\Rightarrow$ Whole work can be completed by

$(8\times 8\times 5)$ workers in

$1$ day.

$=\cfrac{8\times 8\times 5}{10}$ workers in

$10$ days

$=32$ workers in

$10$ days.

$II$ .

$(20\times 16)$ workers can finish it in

$1$ day.

$\Rightarrow$

$\cfrac{(20\times 16)}{10}$ workers can finish it in

$10$ days.

$\Rightarrow$

$32$ workers can finish it in

$10$ days.

$III$ .

$\cfrac{1}{8}$ work can be completed by

$(8\times 5)$ workers in

$1$ day.

$\Rightarrow$ Whole work can be completed by

$(8\times 5\times 8)$ workers in

$1$ day.

$=\cfrac{8\times 5\times 8}{10}$ workers in

$10$ days

$=32$ workers in

$10$ days.

$\therefore$ Any one of the three gives the answer.

$\therefore$ Correct answer is (E).

Two pipes can fill a tank in

$20$ and

$24$ minutes respectively and a waste pipe can empty

$3$ gallons per minute. All the three pipes working together can fill the tank in

$15$ minutes. The capacity of the tank is.

0%
$60$ gallons
0%
$100$ gallons
0%
$120$ gallons
0%
$180$ gallons

Two pipes A and B together can fill a cistern in

$4$ hours. Had they been opened separately, then B would have taken

$6$ hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?

0%
$1$ hour
0%
$2$ hours
0%
$6$ hours
0%
$8$ hours

Explanation

Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fit it in

$(x+6)$ hours.

$\therefore \displaystyle\frac{1}{x}+\frac{1}{(x+6)}=\frac{1}{4}$

$\Rightarrow \displaystyle\frac{x+6+x}{x(x+6)}=\frac{1}{4}$

$\Rightarrow x^2-2x-24=0$

$\Rightarrow (x-6)(x+4)=0$

$\Rightarrow x=6$ . [neglecting the negative value of x]

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 will be able to fill the tank in.

0%
$81$ min.
0%
$108$ min.
0%
$144$ min.
0%
$192$ min.

Explanation

Let the slower pipe alone fill the tank in x minutes.
Then, faster pipe will fill it in

$\displaystyle\frac{x}{3}$ minuts.

$\therefore \displaystyle\frac{1}{x}+\frac{3}{x}=\frac{1}{36}$

$\Rightarrow \displaystyle\frac{4}{x}=\frac{1}{36}$

$\Rightarrow x=144$ min.

Twenty women can do a work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?

0%
$3:4$
0%
$4:3$
0%
$5:3$
0%
Data inadequate

Explanation

$(20\times 16)$ women can complete the work in

$1$ day.

$\therefore$

$1$ women's

$1$ day's work

$=\cfrac{1}{320}$ .

$(16\times 15)$ men can complete the work in

$1$ day.

$\therefore$

$1$ man's

$1$ day's work

$=\cfrac{1}{240}$
So, required ratio

$=\cfrac{1}{240}:\cfrac{1}{320}$

$=\cfrac{1}{3}:\cfrac{1}{4}$

$=4:3$ (cross multiplied)

$X$ can do a piece work in

$40$ days. He works at it for

$8$ days and then

$Y$ finished it in

$16$ days. How long will they together take to complete the work?

0%
$13\cfrac{1}{3}$ days
0%
$15$ days
0%
$20$ days
0%
$26$ days

Explanation

Work done by

$X$ in

$8$ days

$=\left( \cfrac { 1 }{ 40 } \times 8 \right) =\cfrac { 1 }{ 5 }$ .
Remaining work

$=\left( 1-\cfrac { 1 }{ 5 } \right) =\cfrac { 4 }{ 5 }$ .
Now,

$\cfrac{4}{5}$ work done by

$Y$ in

$16$ days.
Whole work will be done by

$Y$ in

$\left( 16\times \cfrac { 5 }{ 4 } \right) =20$ days.

$\therefore$

$X$ 's

$1$ day's work

$=\cfrac{1}{40}$ ,

$Y$ 's

$1$ day's work

$=\cfrac{1}{20}$ .

$(X+Y$ )'s

$1$ day's work

$=\left( \cfrac { 1 }{ 40 } +\cfrac { 1 }{ 20 } \right) =\cfrac { 3 }{ 40 }$ .
Hence

$X$ and

$Y$ will together complete the work in

$\left( \cfrac { 40 }{ 3 } \right) =13\cfrac{1}{13}$ days.

If the third day of a month is Tuesday, which of the following would be the 4th day before the 27th day of that month?

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

Explanation

Dates on which Tuesday fall are 3,10,17, and 24
Hence , the 27th day of the month will be (Tuesday + 3 days=)Friday

$\therefore$ required day of the month = Friday - 4 = Monday

How many men will be required to plough

$100$ acres of land is

$10$ days if

$10$ men require

$8$ days to plough

$20$ acres of land?

0%
$30$
0%
$40$
0%
$50$
0%
$60$

Explanation

Area	No. of men	No. of days
$100$	$x$	$10$
$20$	$10$	$8$

More men more area

$\therefore$ direct variation
More men less no. of days

$\therefore$ inverse variation

$100 : 20 : : x : 10$

$8 : 10$

$100\times 8\times 10 = x \times 20\times 10$

$\Rightarrow x = \dfrac {8000}{200} = 40$

Nitin correctly remembers that Nidhi's birthday is before Friday but after Tuesday. Dheeraj correctly remembers that Nidhi's birthday is after Wednesday but before Saturday. On which of the following days does Nidhi's birthday definitely fall?

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

The basic unit of speed is:

0%
km $/$ min
0%
m $/$ min
0%
km $/$ h
0%
m $/$ s

Explanation

We know that the basic unit of speed is

$=meter\ per\ second$

Hence, this is the answer.

A bag contains Rs.

$112$ in the form of

$1$ -rupee,

$50$ -paise and

$10$ -paise coins in the ratio

$3 : 8 : 10$ . What is the number of

$50$ -paise coins?

0%
$112$
0%
$128$
0%
$96$
0%
$24$

Explanation

Here bag contains Rs.

$112$ in the form of

$1$ -rupee,

$50$ -paise and

$10$ -paise.

Coin ratio is

$3 : 8 : 10$ .
Value ratio

$= 3 \times 1: 8\times \dfrac {1}{2} : 10 \times \dfrac {1}{10}$

$= 3 : 4 : 1$
Number of

$50$ paise coins

$= \dfrac {4}{8} \times$ Rs.

$112 =$ Rs.

$56$
Therefore, number of

$50$ paise coins

$= 2\times 56 = 112$ .

$15$ men can do a piece of work in

$20$ days. In how many days can

$25$ men finish the same work?

0%
$12$
0%
$15$
0%
$2$
0%
$20$

Explanation

Using direct proportion method

15 men do work in 20 days

Therefore 25 men do work in

$15 \ast \dfrac{20}{25} = 12$

Two pipes A and B can fill a tank in

$4$ hours and

$5$ hours respectively. If they are turned up alternatively for one hour each, the time taken to fill the tank is

0%
$2\ hrs\ 15\ min$
0%
$4\ hrs\ 24\ min$
0%
$5\ hrs$
0%
$3\ hrs$

Explanation

Tank filled in

$2\ hours = \dfrac {1}{4} + \dfrac {1}{5} = \dfrac {9}{20}$
Tank filled in

$4\ hours = 2\times \dfrac {9}{20} = \dfrac {9}{10}$ .
Now

$\dfrac {1}{10}th$ of the tank filled by A in

$\left (\dfrac {1}{10}\times 4\times 60\right )min$ . i.e.,

$24$ min. Total time

$= 4\ hrs. 24 min$ .

A can do a piece of work in

$20$ days B in

$15$ days and C in

$12$ days. How soon can the work be done if A is assisted by B on one day and by C on the next alternately?

0%
$14$ days
0%
$6$ days
0%
$8$ days
0%
$10$ days

Explanation

$(A + B)'s 1\ day$ work

$= \left (\dfrac {1}{20} + \dfrac {1}{15}\right )$ of the work, i.e.,

$\dfrac {7}{60}$

$(A + C)^{s} 1\ days$ work

$= \left (\dfrac {1}{20} + \dfrac {1}{12}\right )$ , i.e.

$\dfrac {8}{60}$ of the work.

$2$ days work done

$= \dfrac {7}{60} + \dfrac {8}{60} = \dfrac {1}{4}$ .
Since

$\dfrac {1}{4}$ of work is done in

$2$ days, so whole of the work is done in

$8$ days.

A pyramid whose base is a regular pentagon of area

$42\ {cm}^2$ and whose height is

$7$ cm. What is the volume (in

${cm}^3$ ) of the pyramid?

0%
$98$
0%
$105$
0%
$126$
0%
$147$

Explanation

Given : Area of base of pyramid

$=42\ cm^2$ , height

$=7\ cm$

We know that, Volume of pyramid

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

$=\dfrac{1}{3}\times 42\times 7$

$=98\ cm^3$

Hence, volume of pyramid is

$98\ cm^3$ .

A square pyramid can contain

$16\ m^3$ of water. The height of the pyramid is

$3\ m$ . Calculate the length of base of the square pyramid.

0%
$16\ m$
0%
$12\ m$
0%
$4\ m$
0%
$2\ m$

Explanation

Volume of square pyramid

$=16{ m }^{ 3 }$

$\Rightarrow \quad \dfrac { 1 }{ 3 } \times { a }^{ 2 }\times h=16{ m }^{ 3 }\Rightarrow \dfrac { 1 }{ 3 } \times { a }^{ 2 }\times 3=16\Rightarrow a=\sqrt { 16 } =4cm$

$\therefore$ Length of base

$= 4m$

The arithmatic mean of

$4,6,8$ is

0%
$4$
0%
$6$
0%
$8$
0%
$4.5$

Explanation

$\Rightarrow$ First adding numbers

$4,\, 6$ and

$8$

$=4+6+8=18$

$\Rightarrow$ We have number of terms

$3$

$\Rightarrow$

$Arithmetic\, mean=\dfrac{S}{N}=\dfrac{18}{3}=6$

The monthly average expenditure of a family with

$4$ members is Rs.

$2,800$ . Find the monthly average expenditure of a family with only

$3$ members.

0%

Rs. 2000
0%
Rs. 2300
0%
Rs. 2100
0%
Rs. 2800

Explanation

Here,

Expenditure of $4$ members $=Rs.\ 2800$

So, expenditure of $1$ member $=\dfrac{2800}{4}=Rs.\ 700$

Therefore,

Expenditure of $3$ members $=700\times 3=Rs.\ 2100$

Hence, this is the required result.

The diagram shows a square-based pyramid with base lengths

$6\ in$ and height

$8\ in$ . What is the volume (in in

$^3$ ) of the pyramid?

0%
$64$
0%
$96$
0%
$144$
0%
$288$

Explanation

Given : Base length

$=6\ inch$ , height

$=8\ inch$

We know that, Volume of pyramid

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

$(1)$

Area of square

$=side^2=6^2=36$

From

$(1)$ we get,

Volume

$=\dfrac{1}{3}\times 36\times 8$

$=96\ inch^3$

Hence, volume of pyramid is

$96\ inch^3$ .

Which of the following relations is correct ?

0%
Speed $=$ Distance Time
0%
Speed $=$ Distance $/$ Time
0%
Speed $-$ Time $/$ Distance
0%
Speed $=1/$ Distance Time

Explanation

We know that

$Speed =\dfrac{Distance}{Time}$

Hence, this is the answer.

Find the number of ways in which

$5$ persons

$A,B,C,D,E$ can be seated round a table such that A and B always sit together?

0%
$24$
0%
$12$
0%
$6$
0%
$36$

Explanation

Consider

$AB$ as single unit .

There are total 4 units and arrange them in a circular table.

It will be

$(4-1)!=3!$

and rest

$"AB"$ can also arrange themselves.

Total number of permutations

$=3! \times 2$

$=12$

The mean of a sample is:

0%
Always equal to the mean of the population
0%
Always smaller than the mean of the population
0%
Computed by summing the data values and dividing the sum by $(n - 1)$
0%
Computed by summing all the data values and dividing the sum by the number of items

Explanation

$Mean= \dfrac {Total\quad of\quad sample\quad values}{sample\quad size}$
Answer-Option

$D$ .

Arithmetic mean of

$2$ and

$8$ is

0%
$5$
0%
$10$
0%
$16$
0%
$3.2$

Explanation

${\textbf{Step -1: Stating the arithmetic mean of n general numbers.}}$

${\text{Let there be }}n{\text{ numbers }}{a_1}{\text{,}}{a_2}{\text{,}}{a_3}.....{a_n}{\text{. }}$

${\text{Arithmetic mean of the numbers = }}\dfrac{{{a_1} + {a_2} + {a_3}{\text{ + }}.....{\text{ + }}{a_n}{\text{.}}}}{n}$

${\textbf{Step -2: Finding arithmetic mean of 2 and 8.}}$

${\text{Here we have only two numbers - }}2,8$

${\text{Arithematic mean }}A = \dfrac{{2 + 8}}{2} = \dfrac{{10}}{2} = 5$

${\textbf{Hence, }}\ {\textbf{Arithematic mean of 8 annd 2 is 5.}}$

An insect crawls up

$5$ cm every second on a

$60$ cm vertical rod and then falls down

$2$ cm over the next second. How many seconds will it take to climb the rod?

0%
$20$ seconds
0%
$39$ seconds
0%
$60$ seconds
0%
$30$ seconds

Explanation

For every

$2$ seconds insect climbs

$5-2=3$ cm

Therefore in

$38$ seconds it will climb

$38\times 2=57$ cm

$39$ second it will climb up

$5$ cm more than previous one.

$57+5=62$ cm
So total

$39$ seconds would be required by insect to climb the rod.

A tank contains

$500$ litres of water. Due to a small hole in the tank, the quantity of water is decreasing at the rate of

$9$ litres every hour. What will be the quantity of water in the tank after

$10$ hours?

0%
$410$ litres
0%
$491$ litres
0%
$400$ litres
0%
$90$ litres

Explanation

Capacity of tank =

$500$ litres

Water is decreasing at the rate of

$9$ litres per hour
Total capacity of tank after

$10$ hours =

$500-10\times 9=410$ litres
Hence option A is the correct answer.

Abhinav celebrates his birthday just after

$4^{th}$ Tuesday of September, 20XX. The date after his birthday will be _____ .

0%
$26^{th}$ September
0%
$18^{th}$ September
0%
$25^{th}$ September
0%
$29^{th}$ September

Explanation

$4^{th}$ Tuesday of September 20XX falls on

$28^{th}$ September. So, the date just after it is

$29^{th}$ September.

There is leak in the bottom of a tank. This leak can empty full tank in 8 hours.When the tank is full, a tap is opened into the tank which admits 6 litres per hour and the tank is now emptied in 12 hours.What is the capacity of the tank?

0%
$28.8$ litres
0%
$36$ litres
0%
$144$ litres
0%
$145$ litres

Explanation

Working by the in let in 1 hour

$\dfrac{1}{8}-\dfrac{1}{12}=\dfrac{1}{24}$

$\therefore$ Volume of

$\dfrac{1}{24}$ part = 6 litres

$\therefore$ Volume of whole =

$24 \times 6$ litres

$=144$ litres.

Everyone in Biba's neighbourhood is permitted to swim at Barton Pool at sometime during each day that it is open. No children under the age of 6 are permitted to swim at Barton Pool between noon and 5 P.M. From 5 P.M. until closing, Barton Pool is reserved for adults only.
If all the sentences above are true, then which one of the following must be true?

0%
Few children under the age of 6 live in Biba's neighbourhood
0%
If Biba's next door neighbour has a child under the age of 6, then Barton Pool is open before noon
0%
If most children who swim in Barton Pool swim in the afternoon, then the pool is generally less crowded after 5 P.M.
0%
On day's when Barton Pool is open, at least some children swim there in the afternoon

Mr. X , the president of a club arrived in a meeting at 10 minutes to 12:30 hrs. Mr. X came earlier by 20 minutes than the other participating members in the meeting, who arrived late by 30 minutes. At what time was the meeting scheduled?

0%
12:20 hrs
0%
12:10 hrs
0%
12:50 hrs
0%
12:40 hrs

The Bubna dam has four inlets.Through the first three inlets,the dam can be filled in 12 minutes; through the second, the third and the fourth inlet,it can be filled in 15 minutes; and through the first and the first inlet, in 20 minutes.How much time will take all the four inlets to fill up the dam?

0%
$8\ min$
0%
$10\ min$
0%
$12\ min$
0%
None of these

Explanation

Let Pipes be A,B ,C and D

$\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=\dfrac{1}{12}$

$\dfrac{1}{B}+\dfrac{1}{C}+\dfrac{1}{D}=\dfrac{1}{15}$

$\dfrac{1}{A}+\dfrac{1}{D}=\dfrac{1}{20}$
Adding all equations

$2[\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}+\dfrac{1}{D}]=\dfrac{1}{12}+\dfrac{1}{15}+\dfrac{1}{20}=\dfrac{1}{5}$

$\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}+\dfrac{1}{D}=\dfrac{1}{10}min$
All the inlets together can fill in 10 min

Two taps A and B can fill a tank in

$10$ hrs and

$12$ hrs respectively. If both the taps are opened together, the tank will be full in

0%
$5\dfrac { 6 }{ 11 }$
0%
$5\dfrac { 3 }{ 11 }$
0%
$5\dfrac { 4 }{ 11 }$
0%
$5\dfrac { 5 }{ 11 }$

Explanation

$A$

$B$

$\Downarrow$

$10\, hr$

$12\, hr$

$= \dfrac{1}{10} + \dfrac{1}{12} = \dfrac{6+5}{60} = \dfrac{11}{60}$

$= \dfrac{60}{11} hrs$

$= 5 \dfrac{5}{11} hrs$

1109745_809670_ans_1a184c60600340a6addb8113f6c1aec1.jpg

A group of workers accepted to do a piece of work in

$25$ days.If

$6$ of them did not turn up for the work and the remaining workers did the work in

$40$ days, then the original number of workers was

0%
$22$
0%
$18$
0%
$20$
0%
$16$

Explanation

Let

$x$ no. of men can do work in 25 days
Using

$M_1D_1=M_2D_2$

$25x=(x-6)40$

$15x=240$

$X=16$

$5$ men can build a wall in

$12$ days. How many men can build it in

$10$ days?

0%
$6$
0%
$7$
0%
$8$
0%
$4$

Explanation

Let total work be W, and work done by 1 man in 1 day be w, then

$5\times w\times 12=W$

Also,

$x\times w\times 10=W$ ( where x is number of men required to build the wall in 10 days)

Equating both equations, we get,
$x\times w\times 10=5\times w\times 12$

$\Rightarrow x\quad = \dfrac { 60 }{ 10 } =6\quad men$

$3$ 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%
$4$ days
0%
$10$ days
0%
$15$ days
0%
$20$ days

Explanation

$3 men$

$\equiv 5 women$

$6 men + 5 women \equiv 15 women$

Now,

$5 women$ can do it in

$12$ days.

$1 women$ can do it in

$12$

$\times 5 = 60$ days.

$15 women$ can do it in

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

$12$ workers can complete a piece of work in

$10$ days. If the number of workers are reduced to

$\dfrac {1}{3}rd$ of the original number, then how many more days would be required to complete the same work?

0%
$3$
0%
$5$
0%
$15$
0%
$20$

If February 1, 1996 is a Thusrday, then what day is March 10, 1996 ?

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

Explanation

$\displaystyle \because$ 1996 is a leap year.

$\displaystyle \therefore$ Days from 1 Feb to 10 March

$=$ 39 days

$=$ 4 odd days.

$\displaystyle \therefore$ Thusrday + 3 odd days (

$\displaystyle \because$ 1 Feb is Thursday)

$=$ Sunday

A can do

$\dfrac 12$ of a piece of work in

$5$ days. B can do

$3S$ of the same work in

$9$ days and C can do

$\dfrac 23$ of that work in

$8$ days. In how many days can three of them together do the work.

0%
$3$ days
0%
$5$ days
0%
$4 \, \dfrac{1}{2}$ days
0%
$4$ days

Explanation

Given that,
A does

$\displaystyle\frac{1}{2}$ work in 5 days.
A does the whole work in 10 days.
B does

$\displaystyle\frac{3}{5}$ work in 9 days.
B does the whole work in 15 days.
C does

$\displaystyle\frac{2}{3}$ of work in 8 days.
C does the whole work in 12 days.
Let the total unit of work be 60 units. [LCM of (10, 15, 12)]
A does =

$\displaystyle\frac{60}{10}$ = 6 units/day
B does =

$\displaystyle\frac{60}{15}$ = 4 units/day
C does =

$\displaystyle\frac{60}{12}$ = 5 units/day
(A + B + C)s together per day work = 6 + 4 - 5 = 15 units/day
Days required to complete the work =

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

Choose the correct answer from the alternatives given :
A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely.

0%
3 hours 15 min.
0%
3 hours 45 min.
0%
4 hours.
0%
4 hours 15 min.

Explanation

Let the capacity of tank be

$6$ units.
Tank filled by A per hour =

$\displaystyle \frac{6}{6} = 1$ unit hr.

Required time for filling half the tank =

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

$= 3$ hours
Remaining volume left

$= 6 - 3 = 3$ units

Now according to the questions,

$\displaystyle \therefore$ Required time =

$\displaystyle \frac{3}{(1 \, + \, 3)} \, = \, \frac{3}{4} \, \times \, 60 \, = \, 45 \, min.$

Total time =

$3$ hrs +

$45$ min =

$3$ hrs

$45$ min.

Choose the correct answer from the alternatives given.
A can complete a work in m days and B can complete it in n days. How many days will it take to complete the work if both A and B work together?

0%
$(m + n)$ days
0%
$(\dfrac{1}{m} \times \dfrac{1}{n})$ days
0%
$(\dfrac{m + n}{mn})$ days
0%
$(\dfrac{mn}{m + n})$ days

Explanation

Let the total work be mn units. [LCM of m and n]
A does =

$\dfrac{mn}{m}$ = n units/day

B does =

$\dfrac{mn}{n}$ = m units / day
A and B together do = (n + m) units/day
Hence, the work will get completed by them in

$\dfrac{mn}{m + n}$ days.

$A$ and

$B$ together can finish a work in

$30$ days. They worked on it for

$20$ days and then

$B$ left the work. The remaining work was done by

$A$ alone in

$20$ days more. In how many days can

$A$ alone finish the work?

0%
$48\ days$
0%
$50\ days$
0%
$54\ days$
0%
$60\ days$

Explanation

Let the total work be

$60$ units. [LCM of

$30$ and

$20$ ]

$A$ and

$B$ do

$\dfrac {60}{30} = 2\ units/ day$
As they worked for

$20$ days, they did

$20\times 2 = 40\ units$ of work.
Work left is

$60 - 40 = 20\ units$
Thus,

$20$ units is finished by

$A$ in

$20\ days$ .
Hence,

$A$ does

$\dfrac {20}{20} = 1\ unit/ day$ .
So,

$A$ alone can finish the work in

$\dfrac {60}{1} = 60\ days$ .

$A$ and

$B$ together can complete a work in

$3$ days. They start together but, after

$2$ days,

$B$ left the work. If the work is completed after

$2$ more days,

$B$ alone could do the work in

0%
$4$ days
0%
$6$ days
0%
$8$ days
0%
$10$ days

Explanation

Let the total unit of work be

$6$ units (LCM of

$2$ and

$3$ )

$(A + B)do = \dfrac {6}{3} = 2\ units/ day$

$(A + B)'s 2\ day$ work

$= 2\times 2 = 4\ units$
Work left

$= 6 - 4 = 2\ units$
This remaining work is done by

$A$ in

$2$ days.

$\therefore A's$ per day work

$= \dfrac {2}{2} = 1\ unit/ day$

$\therefore B's$ per unit work

$= (A + B)'s\ work - A's\ work$

$= 2 - 1 = 1\ unit/ day$

$B$ alone can complete the work in

$= \dfrac {6}{1} = 6\ days$ .

Choose the most appropriate option for the following question.
If

$27th$ March

$2011$ was Sunday, what was the day on

$27th$ June,

$2011$ ?

0%
Tuesday
0%
Saturday
0%
Monday
0%
Sunday.

Explanation

Here finding odd days

March (

$4$ days left)

$= 4$ odd days

April (full month)

$= 2$ odd days

May (full month)

$= 3$ odd days

June (till

$27$ )

$= 6$ odd days

Now total odd days

$= 4 + 2 + 3 + 6 = 15$ i.e.

$1$ odd day.

${27}^{th}$ June

$2011$ is

$1$ day after Sunday i.e. on Monday.

Which is the only month in a normal year that has

$28$ days?

0%
April
0%
February
0%
December
0%
June

Explanation

$\Rightarrow$ The only month in a normal year that has

$28$ days is

$February$ .

$\Rightarrow$ In a non-leap year only February contains exactly

$28$ days.

A can do a piece of work in

$9$ days. B is

$50\%$ more efficient than A. The number of days it takes B to do the same piece of work is:

0%
$13\frac{1}{2}$
0%
$4\frac{1}{2}$
0%
$6$
0%
$3$

Explanation

In one day A can do

$1/9$ of the job.

Since B is

$50\%$ more efficient than A

B can do

$\dfrac32.\dfrac 19=\dfrac16$ of the job in one day.

Therefore, B needs

$6$ days for the complete job.

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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