MCQ Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 8

If the day before yesterday was Thursday when will be Sunday?

0%
Today
0%
Two days after today
0%
Tomorrow
0%
Day after tomorrow

In each of the following questions, select the one which is different from the other three.

0%
Watch
0%
Clock
0%
Bangle
0%
Bracelet

If day before yesterday was Friday, what will be the third day after the day - after - tomorrow ?

0%
Thursday
0%
Friday
0%
Saturday
0%
Sunday

Explanation

The day before yesterday was Friday.
Therefore, today is Sunday.
The day after tomorrow will be Tuesday

$\text{Tuesday + 3 = Saturday}$

A is 3 times more efficient than B. Hence, he takes 60 days less in painting a room. In how many days, work will be completed, if A & B both work together ?

0%
$30$ days
0%
$45$ days
0%
$\displaystyle 22\frac { 1 }{ 2 }$ days
0%
$\displaystyle 17\frac { 1 }{ 2 }$ days

Explanation

A : B = x : 3x
Given that 3x - x = 60 = x = 30
A's efficiency = 30 and B's efficiency = 3 x 30 = 90
When A and B work together, they can complete the work in

$\displaystyle \dfrac { 30\times 90 }{ 120 } =22\dfrac { 1 }{ 2 }$ days

$x$ men can do a piece of work in

$y$ days, in how many days will

$z$ men do the same work?

0%
$\dfrac {xz}{y}$
0%
$\dfrac {xy}{z}$
0%
$\dfrac {yz}{x}$
0%
$xyz$

Explanation

x men can do the work in y days.
1 man can do the work in xy days.
z men can do the work in

$\frac {xy}{z}$ days.

The arithinetic mean of

$5, 6, 8, 9, 12, 13, 17$ is

0%
$20$
0%
$15$
0%
$10$
0%
$25$

Explanation

Mean

$= \dfrac { \text{Sum of observations}}{\text{Total number of observations}} = \dfrac {5 + 6 + 8 + 9 + 12 + 13 + 17}{7} = \dfrac {70}{7} = 10$

If 4 days before today it was Monday, what day will it fall on after 3 days ?

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

Find the day of the week on

$\text{16 January, 1969}$

0%
Thursday
0%
Monday
0%
Tuesday
0%
Friday

Explanation

$\Rightarrow\;1600$ years have

$0$ odd days .....(i)

$300$ years have

$1$ odd day .....(ii)

$68$ years have

$17$ leap years and

$51$ ordinary years

$=(17\times2+51\times1)$

$=85\,\text{odd days}=1\text{odd day}$ ......(iii)

$\text{16 January}$ has

$2$ odd days .........(iv)
Adding odd days,

$0+1+1+2=4\,\text{odd days}=\text{Thursday}$ .

If the day before yesterday was Sunday, what day will fall on the day after tomorrow ?

0%
Thursday.
0%
Monday
0%
Friday
0%
Wednesday

Explanation

$\Rightarrow$ Today it is Tuesday and day after tmorrow will be Thursday.

A person born on a Monday in July 1921 retires in 1981 on the day he attains the age of He retires on

0%
Monday
0%
Wednesday
0%
Friday
0%
Saturday

Explanation

As we know that date jumps by one day when year is changed and date jumps by

$5$ days in a four year time gap because it include a leap year.

Since there will be

$15$ leap years between

$1921$ to

$1981$ .

So, he retires in

$1981$ on the day

$=\dfrac{5\times 15}{7}\Rightarrow 5(Remainder)$

Hence, it will be

$5^{th}$ day of week that is

$Friday.$

Pipe A can fill a tank in 4 hours and pipe b can fill it in 6 hours . If they are opened at alternate hours and if pipe A is opened first, in how many hours will the tank be filled ?

0%
$5$
0%
$\displaystyle 5\dfrac{2}{3}$
0%
$4$
0%
$\displaystyle 4\dfrac{2}{3}$

Explanation

Let the unit of work be 12 LCM of 4 and 6

$\therefore$ work done by A in one hour is

$3\, unit$
and work done by B in one hour is

$2\, unit$
Now it starts with A and then B (alternate manner)
ie

$3+2+3+2=10$
10 unit work is done in 4 hours and now 2 unit work is left
which is filled by A only andA takes

$\dfrac{2}{3}$ hours to fill remaining 2 unit

$\therefore$ total time is

$4+\dfrac{2}{3}=4\dfrac{2}{3}$

An electric pump can fill a tank in

$3$ hours. Because of a leak in the tank, it took

$\displaystyle3\frac{1}{2}$ hours to fill the tank. The leak can drain all the water off the tank in

0%
$12$ hours
0%
$\displaystyle10\frac{1}{2}$ hours
0%
$\displaystyle6\frac{1}{2}$ hours
0%
$21$ hours

Explanation

Let the leak be able to empty the tank in

$x$ hours.
Then, amount filled by the pump in one hour =

$\dfrac{1}{3}$
Amount emptied by the leak =

$\dfrac{1}{x}$
Amount actually filled =

$\dfrac{1}{3.5}$ =

$\dfrac{2}{7}$
i.e

$\dfrac{1}{3}$ -

$\dfrac{1}{x}$ =

$\dfrac{2}{7}$

$\dfrac{1}{x}$ =

$\dfrac{1}{3}$ -

$\dfrac{2}{7}$
=

$\dfrac{7}{21}$ -

$\dfrac{6}{21}$ =

$\dfrac{1}{21}$
Thus, x = 21 hours

If p persons working p hours a day for each of p days produce p units of works, then the units of work produced by q persons working q hours a day for each of q days is

0%
$\displaystyle\frac{q^3}{p^2}$
0%
$\displaystyle\frac{q^2}{p^3}$
0%
$\displaystyle\frac{p^2}{q^3}$
0%
$\displaystyle\frac{p^3}{q^2}$

Explanation

$\because$ Persons working p hours a day in p days produce p units of work

$\therefore$ 1 person working 1 hours a day in 1 day produces

$\displaystyle=\frac{p}{p\times p\times p}$

$\therefore$ q persons working q hours a day in q days produce

$\displaystyle=\frac{p\times q\times q\times q}{p^3}=\frac{q^3}{p^2}$

What day of the week will 1st January 2008 be given that 1st January 2000 is a Saturday?

0%
Sunday
0%
Wednesday
0%
Tuesday
0%
None of these

Explanation

Since there will be

$2$ leap years between January

$1^{st},$

$2000$ and January

$1^{st},2008$ and

$6$ simple years

Hence, the date

$1^{st}$ January,

$2008$ have been jumped

$\dfrac{(4+6)}{7}=3(Remainder)$ days since January

$1^{st},2000$ .

So, It will be

$Tuesday$ on January

$1^{st},2000$ .

Mohan was born on February 29 in the year 1960 How may birthdays will be celebrate upto 1976 February ?

0%
16
0%
8
0%
4
0%
5
0%
10

At what time between

$2$ and

$3$ O'clock will the hands of a clock be together?

0%
$10 \cfrac{10}{11}$ min. past $2$
0%
$11 \cfrac{10}{11}$ min. past $2$
0%
$13 \cfrac{10}{11}$ min. past $2$
0%
$7 \cfrac{10}{11}$ min. past $2$

Explanation

If the time is

$2$ hours and

$x$ minutes after

$2$ , when the two hands of the clock are same, then the angle formed by minute hand is:

$\cfrac{x}{60} \times 360 = 6x$
The hour hand moves

$\cfrac{1}{2}$ degree each minute, then angle traced by hour hand is =

$60 + \cfrac{x}{2}$
Hence,

$60 + \cfrac{x}{2} = 6x$

$60 = \cfrac{11x}{2}$

$x = \cfrac{120}{11}$

$x = 10 \cfrac{10}{11}$ min. past

$2$

If the first day of the year 1990 was a Monday what day of the week was 1st January 1998?

0%
Thursday
0%
Tuesday
0%
Wednesday
0%
Friday

Explanation

IN the leap year odd no. of day in the year

$=2$

Leap year has

$29$ day in feb.

The Leap in that divided by

$4$

So,

$1990=1$

$1991=1$

$1992=2$

$1993=1$

$1994=1$

$1995=1$

$1996=2$

$1997=1$

$1998=1$

$(1+1+2+1+1+1+2+1+1)=11$

The odd no. of the year

$=$ No. of day in year

$7$

$11$ /

$7$ =

$4$

Remainder=

$4$

So,

$4$ =TUESDAY

Hence, this is the answer.

The average maximum temperature for 7 days from the 12th September to 18th September is

$35^\circ C$ and that for 7 days from 13th to 19th September is

$34^\circ C$ . From this we can conclude that

0%
Maximum temperature on 19th is $1^\circ C$ less than that for 12th September
0%
Maximum temperature on 12th is $1^\circ C$ less than that for 19th September
0%
Maximum temperature on 19th is $7^\circ C$ less than that for 12th September
0%
None of the above

Explanation

The average maximum temp from 12th sept to 18th sept is 35 and The average maximum temp from 13th sept to 19th sept is 34

Then total temp from 12th sept to 18th sept =35

$\times$ 7=245

And total temp from 13th sept to 19th sept =34

$\times$ 7=238

Then diff=245-238=7

So Maximum temperature on 12th is7

$^{0}$ less than that for 19th September

7
77

A right pyramid on a regular hexagonal base is of height

$60$ m. Each side of the base is

$10$ m. The volume of the pyramid is

0%
$\displaystyle 4500\ \text{m}^{3}$
0%
$\displaystyle 5000\ \text{m}^{3}$
0%
$\displaystyle 5196\ \text{m}^{3}$
0%
$\displaystyle 6196\ \text{m}^{3}$

Explanation

Volume of the pyramid

$\displaystyle =\frac{1}{3}\times$ base area

$\displaystyle \times$ height

$\displaystyle =\frac{1}{3}\times \left ( \frac{3}{2}\sqrt{2}\times 10^{2} \right )\times 60m^{2}$ ,
since

$\displaystyle \sqrt{3}=1.732$
=

$\displaystyle =5196m^{3}$

A can do a piece of work in 40 days , He worked at it for 5 days then B finished it in 21 days. The number of days that A and B take together to finish the work are

0%
15 days
0%
14 days
0%
13 days
0%
10 days

Explanation

A completed one work in 40 days that means he do

$\dfrac{1}{40}$ unit work per day

Now Let B completed the same work in x days

According to question we have

$\dfrac{5}{40}+\dfrac{21}{x}=1$

$\dfrac{21}{x}=\dfrac{35}{40}$

$x=24$

Now let A and B together finish the same work in y days

So we have

$y\left(\dfrac{1}{40}+\dfrac{1}{24}\right)=1$

On solving, we have

$y=15$

The weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams. Their mean weight is

0%
60.5 gram
0%
60 gram
0%
59 gram
0%
62 gram

Explanation

Given weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams.

Then total weight of 9 apples =50+60+65+62+67+70+64+45+48=531

Then mean=

$\frac{539}{9}$ =59

How many times in a day the two hands of a clock are at

$\displaystyle 90^{\circ}$

0%
4
0%
12
0%
22
0%
44

Explanation

In 12 hours, they are at right angles 22 times.
$\therefore$ In 24 hours, they are at right angles 44 times.

Option D is correct.

$A$ is thrice as good a work man as

$B$ and takes

$60$ days less than

$B$ for doing a job. Find the time in which they can do it together.

0%
$15\dfrac{9}{3}$
0%
$22\dfrac{4}{9}$
0%
$15\dfrac{7}{4}$
0%
$22\dfrac{1}{2}$

Explanation

Ratio time taken by A and B=

$1:3$

The time difference is

$(3-1)=2$ days while B takes 3 days and A takes 1 day

If difference of time

$2$ days, B takes

$3$ days

If difference of time 60 days, B takes

$\dfrac{3\times 60}{2}= 90$ days

So A takes

$30$ days to do work

So A one day work=

$\cfrac{1}{30}$

And B one day work =

$\cfrac{1}{90}$

So A + B one day work=

$\cfrac{1}{30}+\cfrac{1}{90}=\cfrac{4}{90}=\cfrac{2}{45}$

So the work takes

$\cfrac{45}{2}=22\cfrac{1}{2}$ days

$A$ can finish a work in $24$ days, $B$ in $9$ days and $C$ in $12$ days. $B$ and $C$ start the work but are forced to leave after $3$ days. The remaining work was done by $A$ in:

0%
$25$ days
0%
$10$ days
0%
$16$ days
0%
$31$ days

Explanation

Work done by

$A$ in one day

$= \dfrac{1}{24}$

Work done by

$B$ in one day

$= \dfrac{1}{9}$

Work done by

$C$ in one day

$= \dfrac{1}{12}$

Work done by

$B$ and

$C$ in 3 days

$= 3(\dfrac{1}{9} + \dfrac {1}{12}) = \dfrac{21}{36}$

Time taken by

$A$ to finish remaining

$\dfrac{15}{36}^{th}$ work

$= \dfrac {15}{36} \times 24$

$=10$ days

$A$ and

$B$ together can complete a piece of work in

$15$ days and

$B$ alone in

$20$ days. In how many days a alone can complete the work ?

0%
$30$ days
0%
$60$ days
0%
$15$ days
0%
$45$ days

$6$ men and

$8$ boys can do a piece of work in

$10$ days and

$26$ men and

$48$ boys can do same in

$2$ days, the time taken by

$15$ men and

$20$ boys to do the same type of work will be :

0%
$5$ days
0%
$4$ days
0%
$6$ days
0%
$7$ days

Explanation

Let

$1$ man's

$1$ day's work

$= x$ and

$1$ boy's

$1$ day's work

$= y$ .
Then,

$6x+8y=\dfrac {1}{10}$ and

$26x+\dfrac {481}{2}$

Solving these two equation, we get :

$x=\dfrac {1}{100}$ and

$y = \dfrac {1}{200}$
(

$15$ men

$+20$ boy)'s

$1$ day's work

$= \left (\dfrac {15}{100}+\dfrac {20}{200}\right) = \dfrac {1}{4}$
So,

$15$ men and

$20$ boys can do the work in

$4$ days.

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%
None of these

Explanation

$20$ women can complete the work in

$16$ days
So,

$1$ women can complete the work

$= (20 \times 16)$ days

$= 320$ days

$1$ women's one day work

$=\dfrac { 1}{320}$

$16$ men can complete the work in

$15$ days

$1$ man can complete the work

$= 15 \times 16 = 240$ days

$1$ man's one day work

$= \dfrac {1}{240}$
Ratio of capacity of man and woman,

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

$= 4:3$

2 men or 3 women can do a piece of work in 10 days. In how many days can 4 men and 3 women do the work together ?

0%
30 days
0%
3.33 days
0%
20 days
0%
5.5 days

A tap can fill a tank in

$16$ min and another empty it in

$8$ min. If the tank is already half full and both the taps are opened together, the tank will be

0%
Filled in $15$ min
0%
Filled in $8$ min
0%
Emptied in $12$ min
0%
Emptied in $8$ min

Explanation

Tap fill the tank in 1 minute

$=\frac{1}{16}$

Waste tap empty the tank

$=\frac{1}{8}$

If both are open then work done by both in 1 min

$=\frac{1}{8}-\frac{1}{16}=\frac{1}{16}$

Now full tank will be emptied by them in

$16$ min

Then half tank will be emptied by them in

$8$ min

$40$ men can complete a work in

$40$ days. They started working together. But

$5$ men kept on leaving the job every

$10th$ day. In how many days work will be completed ?

0%
$45$ days
0%
$50$ days
0%
$55.5$ days
0%
$56.66$ days
0%
None of these

Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to

$\displaystyle \frac { 1 }{ 2 }$ of capacity in 3 hours and a second inlet pipe fills the same empty tank to

$\displaystyle \frac { 2 }{ 3 }$ of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

0%
3.25
0%
3.6
0%
4.2
0%
4.4
0%
5.5

Explanation

The first pipe fills half tank in 3 hrs. So, it must take 6 hrs to fill the complete tank.

The second pipe takes 6 hrs to fill 2/3 of the tank. So it must take 9 hrs to fill the tank completely.

To fill the complete tank together they will need,

$\cfrac{1}{1/6 + 1/9}$ =

$\cfrac{18}{5}$ =

$3.6$ hrs.

$34$ men completed

$\dfrac{2}{5}$ th of a work in

$8$ days working

$9$ hours a day. How many more man should be engaged to finish the rest of the work in

$6$ days working

$9$ hours a day?

0%
$68$ men
0%
$45$ men
0%
$34$ men
0%
$28$ men
0%
None of these

Explanation

Work efficiency of $34$ men in $8$ days working $9$ hours a day is $\dfrac{2}{5}$ th

Hence, work efficiency of $x$ men to finish rest of the work in $6$ days working $9$ hours a day = $1 - \dfrac{2}{5} = \dfrac{3}{5}$

$\therefore$ By direct proportionality,

$\dfrac{34\times8\times9}{\dfrac{2}{5}} =\dfrac{x\times6\times9}{ \dfrac{3}{5}}$

$\therefore 34\times72\times\dfrac{3}{5} = x\times54\times\dfrac{2}{5}$

$\therefore x = \dfrac{34\times72\times3\times5}{54\times2\times5}$

$\therefore x = 68$

Hence extra men required = $68 - 34 = 34$ men

So, the answer is option C

A researcher plans to identify each participant in a

$174$ certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are

$12$ participants,and each participant is to receive a different code?

0%
$4$
0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

The number of single letter code possible are

$n$ and no of distinct codes

$={^n{C}_{2}}$

So,

${^n{C}_{2}}+n\ge 12$

$\dfrac{n(n-1)}{2}+n\ge 12$

$n(n+1)\ge 24$

$n$ min

$=5$ as least number is asked.

Hence the answer is

$5.$

Mr. Leelanand Kumar looks at the calendar for 2012, He finds that April 2012 has exactly four Mondays and four Fridays. 1st April 2012 would fall on ?

0%
Tuesday
0%
Wednesday
0%
Thursaday
0%
Saturday

Explanation

Let us start with

$1$ st April.

$1$ st is a Monday, then April will contain

$5$ Mondays. This contradicts the given condition. Hence,

$1$ st April is not a Monday.

Similarly,

$2$ nd April cannot be a Monday.

So,

$3$ rd April is a Monday which satisfies the condition of

$4$ Mondays in April. Hence, there will be

$4$ Fridays in April (i.e.

$7, 14, 21, 28$ )

Therefore,

$1$ st April is a Saturday.

$36$ workmen are employed to finish a certain work in

$48$ days but it is found that in

$24$ days , only

$\dfrac{2}{5}$ work is done. How many more men must be taken to finish the work in time?

0%
$16$
0%
$20$
0%
$18$
0%
None of these

Explanation

$\cfrac{2}{5} =\cfrac {(2 \times 100)}{ 5} = 40 \%$ work completed in

$24$ days
so,

$60 \%$ more is required to complete in

$24$ days
If

$36$ worker works for rest

$48$ days then work will be

$80 \%$ completed.
To complete

$20\ %$ more work within the time they need

$18$ workers more.

Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

0%
$\displaystyle \frac { x }{ x+y }$
0%
$\displaystyle \frac { y }{ x+y }$
0%
$\displaystyle \frac { xy }{ x+y }$
0%
$\displaystyle \frac { xy }{ x-y }$
0%
$\displaystyle \frac { xy }{ y-x }$

Explanation

Nails Produced per Hour by Machine

$A=\dfrac{800}{y}$

Nails Produced per Hour by Machine

$A$ and

$B$ together

$=\dfrac{800}{x}$

Let us assume Nails Produced per Hour by Machine

$B=\dfrac{800}{B}$

$\Rightarrow \dfrac{800}{x}=\dfrac{800}{y}+\dfrac{800}{B}$

Solving we get,

$\Rightarrow B=\dfrac{xy}{y-x}$

Hence, the answer is

$(E)$

$2$ men and

$5$ women can do

$\dfrac{1}{4}$ th of a job in

$3$ days . After

$3$ days ,

$1$ man joined team and they completed another

$\dfrac{ 1}{4}$ th of the job in

$2$ days . How many men can complete the job in

$4$ days ?

0%
$6$
0%
$24$
0%
$34$
0%
$12$

A , B and C together can complete a piece of work in

$10$ days. All the three started working at it together and after

$4$ days A left. Then B and C together completed the work in

$10$ more days. A alone could complete the work in :

0%
$15$ days
0%
$16$ days
0%
$25$ days
0%
$50$ days

Explanation

(A+B+C)'s one days's work $=\cfrac{1}{10}$

So (A+B+C)'s 4 day's work $=\cfrac{4}{10}=\cfrac{2}{5}$

Remaining part of the work $=1-\cfrac{2}{5}=\cfrac{3}{5}$

$\because \cfrac{3}{5}$ part of the work done by (B+C)= $10$ days

$\therefore$ 1 work done by (B+C) $=\cfrac{10}{\cfrac{3}{5}}=\cfrac{50}{3} Days$

$\therefore$ A alone work $=\cfrac{1}{10}-\cfrac{3}{50}=\cfrac{1}{25}$

Hence A alone complete the work in $25$ days.

It one machine would take

$4$ hours to complete a large production order and another machine

$3$ hours to complete the same order. How many hours would both machines take, working simultaneously at their respective constant rates, to complete the order?

0%
$\displaystyle \frac { 7 }{ 12 }$
0%
$\displaystyle 1\frac { 1 }{ 2 }$
0%
$\displaystyle 1\frac {5}{ 7 }$
0%
$\displaystyle 3\frac { 1 }{ 2 }$
0%
$\displaystyle 7$

Explanation

Order completed by machine

$M_1$ in 1 hour

$=\dfrac{W}{4}$

Similarly, Order completed by machine

$M_2$ in 1 hour

$=\dfrac{W}{3}$

$\Rightarrow \dfrac{W}{4}+\dfrac{W}{3}=\dfrac{7W}{12}$

This means that when both machines are working together work done per hour

$\Rightarrow \dfrac{W}{\dfrac{12}{7}}$

So, the work will be completed in

$\dfrac{12}{7}hrs$

$\dfrac{W}{\dfrac{12}{7}}\times \dfrac{12}{7}=W$

Therefore Answer is

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

$6$ oxen or

$8$ cows can gaze a field in

$28$ days. How long would

$9$ oxen and

$2$ cows take to graze the same field?

0%
$16$ days
0%
$20$ days
0%
$15$ days
0%
$25$ days

Explanation

$6$ oxen

$=8$ cows

$1$ oxen

$=\dfrac{8}{6}$ cows

$9$ oxen

$=\dfrac{8}{6}\times9$ cows

$=12$ cows

$\begin{pmatrix}9\text{oxen}+2\text{cows}\end{pmatrix}\equiv\begin{pmatrix}12\text{cows}+2\text{cows}\end{pmatrix}=14\text{cows}$
Now,

$8$ cows can graze the field in

$28$ days

$1$ cow can graze the field in

$\begin{pmatrix}28\times8\end{pmatrix}$ days

$14$ cows can graze the field in

$\dfrac{28\times8}{14}$ days

$=16$ days.

$6$ typist working

$5$ hours a day can type the manuscript of a book in

$16$ days. How many days will

$4$ typists take to do the same job, each working

$6$ hrs a day?

0%
$21$ days
0%
$17$ days
0%
$18$ days
0%
$20$ days

Explanation

$6$ typists working

$5$ hours a day can finish the job in

$16$ days

$6$ typists working

$1$ hour a day can finish it in

$\begin{pmatrix}16\times5\end{pmatrix}$ days

$1$ typist working

$1$ hour a day can finish it in

$\dfrac{16\times5\times6}{6}$ days

$1$ typist working

$6$ hours a day can finish it in

$6$ days

$4$ typists working

$6$ hours a day can finish it

$\dfrac{16\times5\times6}{6\times4}$ days

$=20$ days.
Hence,

$4$ typists working

$6$ hours a day can finish the job in

$20$ days.

$10$ workers can do a work in

$15$ hours then

$30$ workers can do it in _____ hours

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

Explanation

This question is based on Inverse proprtion

According to the given conditions,

No. of workers Time taken

$10$

$15$ hours

$30$

$x$ days

By inversion proportion,

$10\times15 = 30\times x$

$\therefore x = \dfrac{10\times15}{30}$

$\therefore x =5$

Answer is

$5$ hours

Two hoses are pouring water into an empty pool. Hose

$1$ alone would fill up the pool in

$6$ hours. Hose

$2$ alone would fill up the pool in

$4$ hours. How long would it take for both hoses to fill up two-thirds of the pool?

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

Explanation

If Hose

$1$ can fill the pool in

$6$ hours, its are

$\dfrac {1}{6}$ 'pool per hour,' or the fraction of the job it can do in one hour. Likewise, if Hose

$2$ can fill the pool in

$4$ hours, its rate is

$\dfrac {1}{4}$ pool per hour. Therefore, the combined rate is

$\dfrac {5}{12}$ pool per hour

$\left (\dfrac {1}{4} + \dfrac {1}{6} = \dfrac {5}{12}\right )$ :

$RT = W$

$(5/12)t = 2/3$

$t = \left (\dfrac {2}{3}\right )\left (\dfrac {12}{5}\right ) = \dfrac {8}{5} = 1\dfrac {3}{5}$ hours.

Twelve identical machines, running continuously at the same constant rate, take

$8$ days to complete a shipment. How many additional machines, each running at the same constant rate, would be needed to reduce the time required to complete a shipment by

$2$ days?

0%
$2$
0%
$3$
0%
$4$
0%
$6$
0%
$9$

Explanation

$4$ additional machines: Let the work rate of

$1$ machine be

$r$ . Then the work rate of

$12$ machines is

$12r$ , and you can set up an RTW chart:

The shipment work is then

$96r$ . To figure out how many machines are needed to complete this work in

$8 - 2 = 6 days$ , set up another row and solve for the unknown rate:

Therefore, there are

$\dfrac {96r}{6} = 16r$ machines in total, or

$16 - 12 = 4$ additional machines.

$16$ workers can do a work in

$20$ hours then how many workers can do the work in

$10$ hours?

0%
$8$
0%
$16$
0%
$32$
0%
$24$

Explanation

This question is based on Inverse proprtion

According to the given conditions,

No. of workers Time taken

$16$

$20$ hours

$x$

$10$ hours

By inversion proportion,

$16\times20 = x\times10$

$\therefore x = \dfrac{16\times20}{10}$

$\therefore x =32$

Answer is

$32$ workers

A tap A can fill a cistern in

$4$ hours and the tap B can empty the full cistern in

$6$ hours. If both the taps are opened together in the empty cistern, in how much time will the cistern be filled up?

0%
$8hrs$
0%
$10hrs$
0%
$12hrs$
0%
$14hrs$

Explanation

Time taken by tap A to fill the cistern

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

$1$ hour

$=\dfrac{1}{4}$
Time taken by tap B to empty the full cistern

$=6\;hours$ .
Work done by tap B in

$1$ hour

$=-\dfrac{1}{6}$ (since, tap B empties the cistern).
Work done by

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

$1$ hour

$\dfrac{1}{4}-\dfrac{1}{16}=\dfrac{3-2}{12}=\dfrac{1}{12}$ part of the tank is filled.
Therefore, the tank will fill the cistern

$=12\;hours$ .

A and B together can complete a work in

$4$ days. A alone can finish the work in

$12$ days, In how many days B alone can finish the work?

0%
$5$
0%
$15$
0%
$6$
0%
$8$

Explanation

$\begin{pmatrix}A+B\end{pmatrix}$ "s

$1$ day work

$=\dfrac{1}{4}$
A"s

$1$ day work

$=\dfrac{1}{12}$
B"s

$1$ day work

$=\begin{bmatrix}\begin{pmatrix}\dfrac{1}{4}\end{pmatrix}-\begin{pmatrix}\dfrac{1}{12}\end{pmatrix}\end{bmatrix}=\dfrac{1}{6}$
So, B can complete the work in

$6$ days

A group of workers can do a piece of work in

$24$ days. However as

$7$ of them were absent it took

$30$ days to complete work. How many people actually worked on the job to complete it?

0%
$28$
0%
$35$
0%
$30$
0%
$42$

Explanation

Let the original number of workers in the group be "

$x$ "
therefore, actual number of workers

$=x-7$ .
We know that the number of man hours required to do the job is the same in both the cases.
Therefore,

$x\begin{pmatrix}24\end{pmatrix}=\begin{pmatrix}x-7\end{pmatrix}.30$

$\Longrightarrow 24x=30x-210$

$\Longrightarrow 6x=210$

$\Longrightarrow x=35$ .
Therefore, the actual number of workers who worked to complete the job

$=x-7=35-7=28$ .

What is the proportionality between the production of cakes & days and production of cakes & hours respectively?

0%
direct, direct respectively
0%
direct, inverse respectively
0%
inverse, direct respectively
0%
none of these

A cylindrical water tank has a diameter of 14 meters and a height of 20 meters. A water truck can fill TT cubic meters of the tank every minute. How long will it take the water truck to fill the water tank from empty to half full?

0%
$400$ minutes
0%
$375$ minutes
0%
$490$ minutes
0%
$380$ minutes

Explanation

490 minutes, or 8 hours and 10 minutes: First find the volume of the cylindrical tank:

$v= \pi r^2 \times h$

$=\pi(7)^2 \times 20$

$= 980\pi$

If the water truck can fill n cubic meters of the tank every minute, it will take 980 minutes to fill the tank completely; therefore, it will take

$980 \div 2 = 490$ minutes to fill the tank halfway.

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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