MCQ Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 6

If 5 men can cane 5 chairs in 5 hours then 1 man shall cane 1 chair in how many hours

0%
1 hour
0%
15 hours
0%
5 hours
0%
10 hours

Explanation

The time taken by 5 men to cane 5 chairs=5hours
If one man has to cane all the 5 ,then it is

$=5\times 5=25 hours$
If same man has to cane one chair it is=

$\frac{25}{5}=5 hours$

$A$ completes a work in

$12$ days.

$B$ completes the same work in

$15$ days.

$A$ started working alone and after

$3$ days

$B$ joined him. How many days will they now take together to complete the remaining work?

0%
$6$
0%
$8$
0%
$5$
0%
$4$
0%
None of these

Explanation

Let us Find the Work Rate

$\left(=\dfrac{W}{Days/Time}\right)$

Now, Let us Assume

$x$ be the Number of Extra Days.

$(i)$ for

$A=\dfrac{W}{12}$

$(ii)$ for

$B=\dfrac{W}{15}$

$\Rightarrow$

$\dfrac{W}{12}(3+x)+\dfrac{W}{15}x=W$

$\Rightarrow$

$\dfrac{3+x}{12}+\dfrac{x}{15}=1$

$\Rightarrow$

$x=5$

Therefore Answer is

$(C)$

12 men and 16 boys can do a piece of work in 5 days; 13 men and 24 boys can do it in 4 days. The ratio of the daily work done by a man to that of a boy is

0%
2 : 1
0%
3 : 1
0%
3 :2
0%
5 : 4

Explanation

Let 1 man's 1 days' work = x and 1 boy's 1 days' work = y

Then 12 x + 16y =

$\displaystyle \frac {1}{5}.............(i)$

13x + 24y =

$\displaystyle \frac {1}{4}.............(ii)$

Multiplying (i) by 3 and (ii) by 2 we get

36 x + 48y =

$\displaystyle \frac {3}{5}.............(iii)$

26x + 48y =

$\displaystyle \frac {1}{2}.............(iv)$

Subtracting eqn (iv) from eqn (iii) we get

$\displaystyle 10x=\frac{3}{5}-\frac{1}{2}=\frac{6-5}{10}=\frac{1}{10}\Rightarrow x=\frac{1}{10}$

Putting x =

$\displaystyle \frac{1}{100}$ in (i) we get

$\displaystyle \frac{12}{100}+16y=\frac{1}{5}\Rightarrow 16y=\frac{1}{5}-\frac{12}{100}=\frac{20-12}{100}=\frac{8}{100}\Rightarrow y=\frac{1}{200}$

$\therefore$ Required ratio =

$\displaystyle x:y=\frac{1}{100}:\frac{1}{200}=2:1$

A takes 5 hours more time than that taken by B to complete a work. If working together they can complete a work in 6 hours, then the number of hours , then the number of hours that a takes to complete the work individually is

0%
15
0%
12
0%
10
0%
9

Explanation

Let B take x hours to complete a work Then A shall take (x + 5) hours to complete the same work
B's 1 hours work =

$\displaystyle \frac{1}{x}$ A's 1 hours' work =

$\displaystyle \frac{1}{x+5}$ Given

$\displaystyle \frac{1}{x}+\frac {1}{(x+5)}=\frac {1}{6}$

$\displaystyle \Rightarrow \frac{x+5+x}{x(x+5)}=\frac{1}{6}\Rightarrow \frac{2x+5}{x^{2}+5x}=\frac{1}{6}$

$\displaystyle \Rightarrow 12x+30=x^{2}+5x\Rightarrow x^{2}-7x-30=0$

$\displaystyle \Rightarrow x^{2}-10x+3x-30=0\Rightarrow x(x-10)+3(x-10)=0$

$\displaystyle x=10\: \: or\: \: -3$ Neglecting negative value x = 10

A man, a woman and a boy can complete a job in 3, 4 and 12 days respectively. How many boys must assist 1 man and I women to complete the job in

$\displaystyle \frac{1}{4}$ of a day ?

0%
1
0%
4
0%
19
0%
41

Explanation

1 man's 1 days work =

$\displaystyle \frac{1}{3}$ 1 woman's 1 days' work =

$\displaystyle \frac{1}{4}$
1 boy's 1 days' work =

$\displaystyle \frac{1}{12}$ Let the number of boys to assist be x Then

$\displaystyle \frac{1}{3}+\frac{1}{4}+\frac{x}{12}=\frac{1}{1/4}$

$\displaystyle \Rightarrow \frac{4+3+x}{12}=4\Rightarrow 7+x=48\Rightarrow x=41$ boys

A job can be completed by 12 men in 12 days. How many extra days will be needed to complete the job . if 6 men. leave after working for 6 days?

0%
10 days
0%
12 days
0%
8 days
0%
24 days

Explanation

12 men can complete in 12 days 1 work 1 man can complete in 1 day

$\displaystyle \frac{1}{12\times 12}$ part of the work

$\therefore$ men can complete in 6 days

$\displaystyle \frac{6\times 6}{12\times 12}$ part of the work =

$\displaystyle \frac{1}{4}th$ of the work Number of remaining men = 6
Remaining work =

$\displaystyle \frac{3}{4}$

$\because$ 12 men can complete 1 work in 12 days
6 men can complete

$\displaystyle \frac{3}{4}$ work in

$\displaystyle \frac{12\times 12\times 3}{6\times 4}=18$ days

$\therefore$ Number of extra days = 18 - 6 = 12 days

$A$ and

$B$ can do a piece of work in

$20$ days and

$12$ days respectively.

$A$ started the work alone and then after

$4$ days.

$B$ joined him till the completion of the work. How long did the work last.?

0%
$10$ days
0%
$20$ days
0%
$15$ days
0%
$6$ days

Explanation

$A$ 's one days' work

$=$

$\displaystyle \frac{1}{20}$ .

$B$ 's one days' work

$=$

$\displaystyle \frac{1}{12}$ .

Work done by

$A$ in

$4$ days

$=$

$\displaystyle \frac{4}{20}=\frac {1}{5}$ .
Remaining work

$=$

$\displaystyle 1-\frac{1}{5}=\frac {4}{5}$

$(A + B)$ 's

$1$ days' work

$=$

$\displaystyle \frac{1}{20}+\frac{1}{12}=\frac{2}{15}$

$\therefore$ Number of days to do the remaining work

$=$

$\displaystyle \frac{4}{5}\div\frac{2}{12}=\frac{4}{5}\times \frac {15}{2}=6$ days

$\therefore$ Total number of days taken in the completion of work

$= 6 + 4 = 10$ days.

A contractor undertakes to complete a road 360 m long in 120 days and employs 30 men for the work. After 60 days he finds that only 120 m length
of the road has been made . How many more men should he emply so that the work many be complete in time ?

0%
20
0%
30
0%
15
0%
45

Explanation

Let the number of extra men employed be x 30 men in 60 days complete 120 m long road
1 man in 1 day will complete

$\displaystyle \frac{120}{30\times 60}$ m long road Also (30 + x) men in 60 days complete 240 m long road

$\therefore$ 1 man in 1 day will complete

$\displaystyle \frac{240}{(30+x)\times 60}m$ long road

$\displaystyle \Rightarrow \frac{120}{30\times 60}=\frac{240}{(30+x)\times 60}\Rightarrow \frac{1}{15}=\frac{4}{(30+x)}$

$\Rightarrow$ 30 + x = 60

$\Rightarrow$ x = 30

$\therefore$ 30 more men have to be employed

Three men , four women and six children can compete a work in seven days. A women does double the work a man does and a child does half the work a man does . How many women alone can complete the work in 7 days ?

0%
7
0%
8
0%
14
0%
12

Explanation

Let 1 woman's 1 days' work = x Then 1 man's 1 days' work =

$\displaystyle \frac{x}{2}$ and 1 child's 1 days' work =

$\displaystyle \frac{x}{4}$
Given (3 men's+ 4 women's + 6 children's) 1 day's work =

$\displaystyle \frac{1}{7}$

$\displaystyle \Rightarrow \frac{3x}{2}+4x+\frac{6x}{4}=\frac{1}{7}\Rightarrow \frac{6x+16x+6x}{4}=\frac{1}{7}$

$\displaystyle \Rightarrow \frac{28x}{4}=\frac{1}{7}\Rightarrow x=\frac{1}{49}$

$\therefore$ 1 women can alone complete the work in 49 days To complete the work in 7 days
number of women required =

$\displaystyle \frac{49}{7}=7$

8 men working for 9 hours a day can complete a piece of work in 20 days. In how many days can 7 men working for 10 hours a day complete the same piece of work?

0%
$\displaystyle 20\frac{1}{2}$ days
0%
$\displaystyle 20\frac{3}{5}$ days
0%
21 days
0%
$\displaystyle 20\frac{4}{7}$ days

Explanation

8 men working for 9 hours a day can complete a piece of work in 20 days
8 men working for 1 hour a day can complete the work in (20 x 9) days (Less hours, More days)
1 man working for 1 hour a day can complete the work in (20 x 9 x 8) days (Less men, more days)

1 man working for 10 hours a day can complete the work in

$\displaystyle \left ( \frac{20\times 9\times 8}{10} \right )$ days

7 men working for 10 hours a day can complete the work in

$\displaystyle \left ( \frac{20\times 9\times 8}{7\times 10} \right )$ days =

$\displaystyle \frac{144}{7}$ days =

$\displaystyle 20\frac{4}{7}$ days

A wall of 100 metres can be built by 7 men or 10 women in 10 days. How many days will 14 men and 20 women take to build a wall of 600 metres?

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

Explanation

According to question 10 women = 7 men

$\therefore$ 20 women = 14 men
14 men + 20 women = 14 men + 14 men = 28 men
A wall of 100 m is built by 7 men in 10 days A wall of 100 m is built by 1 man in (10 x 7) days
A wall of 1 m is built by 1 man in

$\displaystyle \frac{10\times 7}{100}$ days (less length, less days)
A wall of 600 m is built by 1 man in

$\displaystyle \frac{10\times 7\times 600}{100}$ days (more length, more days)
A wall of 600 m is built by 28 men in

$\displaystyle \frac{10\times 7\times 600}{100\times 28}$ days = 15 days

A tank can be filled by two pipes A and B separately in 3 hours and 3 hours 45 minutes respectively, A third pipe C can empty the full tank in 1 hour, when the tank was exactly half-filled, all the three pipes are opened. The tank will become empty in

0%
1 hours 15 min
0%
2 hours 30 min
0%
3 hours 15 min
0%
4 hours 10 min

Explanation

Part of the tank emptied in 1 hour = 1 -

$\displaystyle \left ( \frac{1}{3}+\frac{4}{15} \right )=1-\frac{9}{15}=\frac{6}{15}=\frac{2}{5}$

$\displaystyle \left ( \because 3 \: hrs\: 45\: min=3\frac{45}{60}=\frac{15}{4}hrs \right )$
Time taken to empty the full tank =

$\displaystyle \frac{1}{\frac{2}{5}}=\frac{5}{2}hrs$

$\displaystyle \therefore$ Time taken to empty the exactly half filled tank =

$\displaystyle \frac{1}{2}\times \frac{5}{2}hours=\frac{5}{4}hours=1\: hr\: 15\: min$

A tap can fill a bath in

$20$ minutes and another tap can fill it in

$30$ minutes. Amit opens both the taps simultaneously. When the bath should have been full, he finds that the waste pipe was open. He then closes the waste pipe and in another

$4$ minutes the bath is full. In what time would the waste pipe empty it ?

0%
$35$ min
0%
$38$ min
0%
$36$ min
0%
$39$ min

Explanation

A work done in

$1$ minute by both the taps

$=\left (\dfrac {1}{20}+\dfrac {1}{30}\right)=\dfrac {5}{60}=\dfrac {1}{12}$

Both the taps can fill the bath in

$12$ minutes. Clearly the waste pipe empties as much the bath in

$12$ minutes as the other two two taps together can fill in

$4$ minutes.

Work done by two taps in

$4$ minutes

$=\dfrac {1}{12}\times 4=\dfrac {1}{3}$ .

$12$ minutes, the waste pipe empties

$=\dfrac {1}{3}$ part.

$1$ minute, the waste pipe empties

$=\dfrac {1}{3}\times \dfrac {1}{12}=\dfrac {1}{36}$ part.

Thus the waste pipe can empty the bath in

$36$ minutes.

A swimming pool is fitted with three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time as that taken by the third pipe alone. The second pipe fills the pool

$5\ hours$ faster than the first pipe and

$4\ hours$ slower than the tired pipe. Find the time required by the third pipe to fill the pool.

0%
$10\ hours$
0%
$6\ hours$
0%
$16\ hours$
0%
$5\ hours$

Explanation

Let the time required to fill the three pipes be

$a,b$ and

$c$ hours.

$1$ hour, they respectively fill

$\dfrac {1}{a}, \dfrac {1}{b}$ and

$\dfrac {1}{c}$ part of the pool.

According to the condition,

$\dfrac {1}{a}+\dfrac {1}{b}=\dfrac {1}{c}$

$\Rightarrow \dfrac {b+a}{ab}=\dfrac {1}{c}$ ....(1)

$a-b=5$

$\Rightarrow a=b+5$ ....(2)

and

$b-c=4$

$\Rightarrow c=b-4$ ....(3)

Substituting the value of

$a$ and

$c$ in equation (1), we get

$\dfrac {b+(b+5)}{(b+5)\times b}=\dfrac {1}{b-4}$

After solving this, we get

$b=10, b=-2$

We need to neglect

$b=-2$ .

Thus

$b=10$ is correct.

Substituting this value in equations (2) and (3),

$a=10+5=15$ and

$c=10-4=6$

Therefore, time required for the first pipe

$=15$ hours,

time required for the second pipe

$=10$ hours

and time required for the third pipe

$=6$ hours.

A man covered a certain distance at some speed had he moved 3 km/hr faster he would have taken 40 minutes less If he had moved 2 km/hr slower he would have taken 40 minutes more The distance ( in km) is

0%
35
0%
$\displaystyle 36\frac{2}{3}$
0%
$\displaystyle 37\frac{1}{2}$
0%
40

Explanation

Let the distance = x km and usual rate = y km/hr Then

$\displaystyle \frac{x}{y}-\frac{x}{y+3}=\frac{40}{60}\Rightarrow \frac{x(y+3)-xy}{y(y+3)}=\frac{2}{3}\Rightarrow \frac{3x}{y(y+3)}=\frac{2}{3}$

$\displaystyle \Rightarrow 2y(y+3)=9x...............(i)$
and

$\displaystyle \frac{x}{y-2}-\frac{x}{y}=\frac{40}{60}\Rightarrow \frac{xy-x(y-2)}{y(y-2)}=\frac{2}{3}$

$\displaystyle \Rightarrow \frac{2x}{y(y-2)}=\frac{2}{3}$

$\displaystyle \Rightarrow y(y-2)=3x...............(ii)$
On dividing eqn (i) by eqn (ii) we get

$\displaystyle \frac{2(y+3)}{y-2}=3\Rightarrow 2y+6=3y-6\Rightarrow y=12$

$\therefore$ Putting the value of y in (i) we get 2 x 12 x 15 = 9x

$\Rightarrow$ x = 40 km/hr

A started a work and left after working for 2 days. Then B was called and he finished the work in 9 days. Had a left the work after working for 3 days, B would have finished the remaining work in 6 days. In how many days can each of them, working alone finish the whole work ?

0%
5 days , 8.5 days
0%
2.5 days, 7.5 days
0%
5 days, 15 days
0%
3 days, 15 days,

Explanation

Suppose A and B do the work in x and y days respectively Now work done by A in 2 days + work done by B in 9 days = 1

$\displaystyle \Rightarrow \frac{2}{x}+\frac{9}{y}=1$ Similarly

$\displaystyle \frac{3}{x}+\frac{6}{y}=1$ Let

$\displaystyle \frac{1}{x}=a$ and

$\displaystyle \frac{3}{y}=b$ Then the equations become
2a + 9b = 1 .......(i)
3a + 6b = 1 .......(ii)
Performing (ii) x 2 - (i) x 3 we get (6a + 12b) - (6a + 27b) = 2 - 3

$\displaystyle \Rightarrow -15b=-1\Rightarrow b=\frac{1}{15}\Rightarrow x=15$

$\therefore$ Putting the value of b in (i)

$\displaystyle 2a+9\times \frac{1}{15}=1\Rightarrow 2a=1-\frac{3}{5}=\frac{2}{5}\Rightarrow a=\frac{1}{5}\Rightarrow y=5$

A cistern has a leak which would empty it in 8 hours . A tap is turned on , which admits 6 liters a minutes into the cistern and it is now emptied in 12 hours, How many litres can the cistern hold?

0%
8000 litres
0%
8400 litres
0%
8640 litres
0%
8650 litres

Explanation

Suppose the cistern holds x liters of water.
The tap admits 6 liters of water in 1 minute, i.e., it admits 360 liters of water in 1 hour.

$\displaystyle \therefore$ x liters of water can be filled in

$\displaystyle \frac{x}{360}hours$
The leak would empty the tank in 8 hours.

$\displaystyle \therefore$ In one hour,

$\displaystyle \frac{1}{8}$ of the tank becomes empty.
If the tap and the leak both are allowed simultaneously, then the tank becomes empty in 12 hours, i.e.,

$\displaystyle \frac{1}{8}-\frac{360}{x}=\frac{1}{12}\Rightarrow \frac{360}{x}=\frac{1}{8}-\frac{1}{12}=\frac{3-2}{24}=\frac{1}{24}$

$\displaystyle \Rightarrow x=\left ( 360\times 24 \right )liters=8640\: liters$

Two pipes

$A$ and

$B$ can fill a cistern is

$12$ minutes and

$15$ minutes respectively, but a third pipe

$C$ can empty the full tank in

$6$ minutes.

$A$ and

$B$ are kept open for

$5$ minutes in the beginning and then

$C$ is also opened. In what time will the cistern be emptied ?

0%
$30$ min
0%
$33$ min
0%
$37.5$ min
0%
$45$ min

Explanation

Part filled in

$5$ min

$=5\times \left (\dfrac {1}{12}+\dfrac {1}{15}\right)$

$=5\times \dfrac {9}{60}=\dfrac {3}{4}$

Part emptied in

$1$ minute when all the pipes are opened

$=\dfrac {1}{6}-\left (\dfrac {1}{12}+\dfrac {1}{15}\right)=\dfrac {1}{60}$ .

Now,

$\dfrac {1}{60}$ part is emptied in

$1$ minute.

Therefore,

$\dfrac {3}{4}$ part will be emptied in

$60\times \dfrac {3}{4}=45$ minutes.

Two pipes can fill a tank with water in 15 and 12 hours respectively and the third pipe can empty it in hours. If the pipers are opened in order at 8,9 and 11 A. M. respectively, the thank will be emptied at

0%
11 : 40 a.m
0%
12 : 40 p.m
0%
1 : 40 P,,m
0%
2:40 p.m

Explanation

Let the tank be emptied in x hrs after 8 A.M.

$\displaystyle \therefore$ Tank filled by pipe in x hours =

$\displaystyle \frac{x}{15}$
Tank filled by second pipe in (x - 1) hours =

$\displaystyle \frac{\left ( x-1 \right )}{12}$
Tank emptied by third pipe in (x - 3) hours =

$\displaystyle \frac{\left ( x-3 \right )}{4}$

$\displaystyle \therefore \frac{x}{15}+\frac{\left ( x-1 \right )}{12}-\frac{\left ( x-3 \right )}{4}=0$

$\displaystyle \Rightarrow \frac{4x+5\left ( x-1 \right )-15\left ( x-3 \right )}{60}=0$

$\displaystyle 4x+5x-5-15x+45=0\Rightarrow -6x+40=0$

$\displaystyle \Rightarrow x=\frac{40}{6}=\frac{20}{3}=6\frac{2}{3}hrs=6\: hrs\: 40\: min$

$\displaystyle \therefore$ The tank will be emptied at 2 : 40 p.m.

A drum can be filled with oil with a supply pipe in 40 minutes. A gain the full drum cam be made empty by another pipe in 60 minutes, When

$\displaystyle\frac{1}{2}$ part of the drum was filled, the second pipe was. opened and it was stopped after 15 minutes. if the supply pipe is opened how, time required to fill the drum is

0%
$\displaystyle23\frac{1}{2}$ min
0%
$\displaystyle25\frac{2}{3}$ min
0%
$\displaystyle27\frac{1}{3}$ min
0%
$\displaystyle28\frac{2}{3}$ min

Explanation

Part of the drum that can be emptied in 15 min by the second pipe =

$\displaystyle 15\times \frac{1}{60}=\frac{1}{4}$
After 15 min, part of the drum filled with oil =

$\displaystyle \frac{2}{3}-\frac{1}{4}=\frac{8-3}{12}=\frac{5}{12}$
After 15 min, part of the drum empited =

$\displaystyle 1-\frac{5}{12}=\frac{7}{12}$
Given, the whole drum can be filled in 40 min

$\displaystyle \therefore \frac{7}{12}$ part of the drum can be filled in

$\displaystyle \left ( 40\times \frac{7}{12} \right )min=\frac{70}{3}min=23\frac{1}{3}min$

15 men can complete a work in 210 days, They started the work but at the end of 10 days, 15 additional men, with double efficiency, were inducted, How many days in all did they take to finish the work ?

0%
$\displaystyle72\frac{1}{2}$ days
0%
$\displaystyle84\frac{3}{4}$ days
0%
$\displaystyle76\frac{2}{3}$
0%
70 days

Explanation

15 men's 1 days' work =

$\displaystyle \frac{1}{210}$

$\therefore$ 15 men's 10 days' work

$\displaystyle \frac{10}{210}=\frac {1}{21}$ Remaining work =

$\displaystyle 1-\frac{1}{21}=\frac{20}{21}$
The additional 15 men's 1 days' work =

$\displaystyle 1-\frac{2}{210}=\frac{1}{105}$

$\therefore$ These 30 men's 1 days' work =

$\displaystyle \left ( \frac{1}{210}+\frac{1}{105} \right )=\frac{1+2}{210}=\frac{3}{210}=\frac{1}{70}$

$\displaystyle \frac{1}{70}th$ of the work can be done in 1 day

$\therefore$

$\displaystyle \frac{20}{21}th$ of the work can be done in

$\displaystyle \left ( 70\times \frac{20}{21} \right )$ days

$\displaystyle =\frac{200}{3}$ days =

$\displaystyle 66\frac{2}{3}$ days

$\displaystyle \therefore$ Total no. of days =

$\displaystyle 66\frac{2}{3}$ days + 10 days =

$\displaystyle 76\frac{2}{3}$ days

A leak in the bottom of a tank can empty it in

$6$ hours. A pipe fills the tank at the rate of

$4$ liters per minute. When the tank is full, the inlet is opened but leak emptied the tank in

$8$ hours. What is the capacity of the tank?

0%
$5260$ L
0%
$5670$ L
0%
$5946$ L
0%
$5760$ L

Explanation

Part emptied by the leak in 1 hr

$=\dfrac{1}{6}$

Net part emptied by the leak and the inlet pipe in 1 hr

$=\dfrac{1}{8}$

Part fill by the inlet pipe in 1 hr

$=\dfrac{1}{6}-\dfrac{1}{8}=\dfrac{1}{24}$

i.e inlet pipe fills the tank in 24 hr

$=24\times 60=1440 min.$

According to the question inlet pipe fills water at the rate of

$4$ liters per minute.

Hence, water fills in

$1440$ min

$=1440\times4=5760 L$

$A$ and

$B$ can do a job in

$12$ days and

$B$ and

$C$ can do it in

$16$ days. After

$A$ has been working for

$5$ days and

$B$ for

$7$ days,

$C$ finishes rest of the work in

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

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

Explanation

A's 5 days work

$+$ B's 7 days work

$+$ C's 13 days work

$=$ 1

$\therefore$

$(A+B)'s$ 5 days work

$+$

$(B+C)'s$ 2 days work

$+$ C's 11 days work

$=$ 1

$\therefore$

$\dfrac{5}{12}+\dfrac{2}{16}$

$+$ C's 11 days work

$=1$

$\therefore$ C's 11 days work

$=$

$1-(\dfrac{5}{12}+\dfrac{2}{16}$ )

$=$

$\dfrac{11}{24}$

$\therefore$ C's 1 days work

$=$

$(\dfrac{11}{24} \times \dfrac{1}{11}$ )

$=\dfrac{1}{24}$

$\therefore$ C alone can finish the work in 24 days.

$A$ can complete a certain job in

$12$ days.

$B$ is

$60$ % more efficient than

$A. B$ can complete the work alone in:

0%
$6$ days
0%
$6.25$ days
0%
$7.2$ days
0%
$7.5$ days

Explanation

Ratio of times taken by A and B

$=$

$160:100$

$=$

$8:5$ .
Suppose B alone takes

$x$ days to do the job.
Then,

$8:5$

$: :12:x$

$\rightarrow$

$8x=5 \times 12$

$\rightarrow$

$x=7.5 \, days$

$A$ can do a piece of work in

$80$ days. He works for

$10$ days and then

$B$ alone finishes the remaining work in

$42$ days. The two together could complete the work in

0%
$24$ days
0%
$25$ days
0%
$30$ days
0%
$6$ days

Explanation

$A$ can do a piece of work in

$80$ days.

So, in

$1$ day,

$A$ finishes

$\dfrac {1}{80}$ of the work.

Working for

$10$ days,

$A$ completes

$\left (\dfrac {1}{8}\right)^{th}$ of the work.

Amount of work remaining is

$\dfrac {7}{8}$ .

$B$ alone finishes the remaining work in

$42$ days.

So, in

$1$ day,

$B$ finishes

$\dfrac {1}{42}$ of the work.

So, to finish the work,

$B$ alone needs

$48$ days.

$1$ day working together,

$A$ and

$B$ will finish

$\dfrac {1}{80}+\dfrac {1}{48}=\dfrac {1}{30}$ of the work.

To complete the entire work,

$A$ and

$B$ will take

$30$ days.

Taps

$A$ and

$B$ can fill in a tank in

$12$ and

$15$ min respectively. If both are opened and

$A$ is closed after

$3$ min how long will it take for

$B$ to fill in the tank ?

0%
$8$ min $15$ s
0%
$8$ min $5$ s
0%
$7$ min $45$ s
0%
$8$ min $45$ s

Explanation

Tank filled by tab A in 1 minute $=\dfrac{1}{12}$

Tank filled by tab B in 1 minute $=\dfrac{1}{15}$

Tank filled by tab A and B ib 1 minute $=\dfrac{1}{12}+\dfrac{1}{15}=\dfrac{9}{60}$

Pipe A and B is opened for 3 minute

Then part filled by A and B in 3 minute $=\dfrac{9}{50}\times 3=\dfrac{9}{20}$

Remaining part $=1-\dfrac{9}{20}=\dfrac{11}{20}$

Time taken by B to fill this remaining part $=\dfrac{\cfrac{11}{20}}{\cfrac{1}{15}}$

$=\dfrac{11\times 15}{20}=\dfrac{33}{4}=8\cfrac{1}{4}$

$=8 \ min \ 15 \ s$

$A, B$ and

$C$ can do a job in

$11,20$ and

$55$ days respectively. How soon can the work be done if

$A$ is assisted by

$B$ and

$C$ on alternate days?

0%
$7$ days
0%
$9$ days
0%
$8$ days
0%
$10$ days

Explanation

$(A+B)'s$ 1 days work

$=(\dfrac{1}{11}+\dfrac{1}{20})$

$=\dfrac{31}{220}$ .

$(A+C)'s$ 1 days work

$=(\dfrac{1}{11}+\dfrac{1}{55})$

$=\dfrac{6}{55}$ .

Work done in 2 days

$=(\dfrac{31}{220}+\dfrac{6}{55})$

$=\dfrac{55}{220}$

$=\dfrac{1}{4}$

Now,

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

$2$ days.

$\therefore$ Whole work will be done in

$(2\times 4)=8$ days.

$A$ and

$B$ can do a piece of work in

$45$ days and

$40$ days respectively. They begin together but

$A$ leaves after some days and

$B$ completes the rest in

$23$ days. For how many days did

$A$ work ?

0%
$6$ days
0%
$9$ days
0%
$8$ days
0%
$12$ days

Explanation

A's one day's work $= \cfrac{1}{45}$

B's one day's work $= \cfrac{1}{40}$

$\therefore (A+B)$ 's 1 day's work $= \cfrac{1}{45} + \cfrac{1}{40} = \cfrac{8+9}{360} = \cfrac{17}{360}$

Work done by B in 23 days $= \cfrac{1}{40} \times 23 = \cfrac{23}{30}$
Remaining work $= 1 - \cfrac{23}{40} = \cfrac{40-23}{40} = \cfrac{17}{40}$
$(A+B)$ 's 1 day work $= \cfrac{17}{360}$
$\dfrac{17}{40}$ work done by $(A+B)$ in $1 \times \dfrac{360}{17} \times \dfrac{17}{40} = 9 \ Days$

Tap A can fill the tank in two hours while taps B and C can empty it in six and eight hours respectively. All the taps remained open initially. After two hours tap C was closed and after one more hour tap B was also closed. In how much time now would the remaining tank get filled ?

0%
12 min
0%
15 min
0%
30 min
0%
20 min

Explanation

Let us name the

$3$

$Taps$ be

$T_A,T_B,T_C$

where

$T_B,T_C$ are emptying pipes and

$T_A$ is filling Pipe.

Water Emptied by

$T_B$ in

$6$

$hours=Volume$

$of$

$Tank$

$or$

$V$

Water Emptied by

$T_B$ in

$1$

$Hour=\dfrac{V}{6}$

Similarly, Water Emptied by

$T_C$ in

$8$

$Hour=V$

Water Emptied by

$T_C$ in

$1$

$Hour=\dfrac{V}{8}$

Similarly, Water Supplied by

$T_A$ in

$2$

$Hour=V$

Water Supplied by

$T_A$ in

$1$

$Hour=\dfrac{V}{2}$

After 2 Hours,

$\Rightarrow 2\dfrac{V}{2}-2\dfrac{V}{8}-2\dfrac{V}{6}=\dfrac{5V}{12}$

After 3 Hours,

$\Rightarrow \dfrac{5V}{12}+\dfrac{V}{2}-\dfrac{V}{6}=\dfrac{3V}{4}$

Now, only

$T_A$ is opened

Remaining Volume

$=V-\dfrac{3V}{4}$

$=\dfrac{V}{4}$

Now Time taken to Fill Remaining Tank

$\Rightarrow \dfrac{\dfrac{V}{4}}{\dfrac{V}{2}}=\dfrac{1}{2}hrs=30 min$

Therefore Answer is

$(C)$

A, B and C can do a piece of work in 36, 54 and 72 days respectively. They started the work but A left 8 days before the completion of the work while B left 12 days before completion. The number of days for which C worked is

0%
4
0%
8
0%
12
0%
24

Explanation

A does the work in 36 days.

$\therefore$ In 1 day A does

$\frac { 1 }{ 36 }$ work.

B does the work in 54 days.

$\therefore$ In 1 day B does

$\frac { 1 }{ 54 }$ work.

C does the work in 72 days.

$\therefore$ In 1 day C does

$\frac { 1 }{ 72 }$ work.

Let the work is finished in x days.

So A works for x-8 days.

$\therefore$ A does

$\frac { x-8 }{ 36 }$ work

B works for x-12 days.

$\therefore$ B does

$\frac { x-12 }{ 54 }$ work.

C works for x days.

$\therefore$ C does

$\frac { x }{ 72 }$ work.

The sum of their works = the whole work = 1 work.

$\therefore$

$\frac { x-8 }{ 36 } +\frac { x-12 }{ 54 } +\frac { x }{ 72 } =1\\ \Longrightarrow \frac { 6x-48+4x-48+3x }{ 216 } =1\\ \Longrightarrow \frac { 13x-96 }{ 216 } =1\\ \Longrightarrow 13x=312\\ \Longrightarrow x=24$

So the work is finished in 24 days.

$A$ does half as much work as

$B$ and

$C$ does half as much work as

$A$ and

$B$ together in the same time. If

$C$ alone can do the work in

$40$ days, all of them together will finish the work in

0%
$13$ days
0%
$15$ days
0%
$20$ days
0%
$13.33$ days

Explanation

$Let \, \,the \, \,total \, \,work \, \,be \, \,x \, \,units.$

$\therefore \, \,speed \, \,of \, \,C \, \,is=\dfrac{x}{40} \, \, work \, \, per \, \, day.$

$\therefore \, \, Speed \, \,of \, \,B=\dfrac{x}{30} \, \,works \, \,per \, \,day.$

$\therefore \, \, Speed \, \,of \, \,A=\dfrac{x}{60} \, \,works \, \,per \, \,day.$

$\therefore \, \, Speed \, \,of \, \,A+B+C=\dfrac{3x}{40} \, \,works \, \,per \, \,day.$

$If \, \, all \, \,of \, \,them \, \,work \, \,together \, \,then \, \,they \, \,will \, \,finish \, \, x \, \,works \, \,in \, \,\dfrac{x}{\dfrac{3x}{40}} \, \,= \, \,13.33 \, \,days.$

Ronald and Elan are working on an assignment Ronald takes

$6$ hours to type

$32$ pages on a computer while Elan takes

$5$ hours to type

$40$ pages. How much time will they take working together on two different computers to type an assignment of

$110$ pages ?

0%
$7$ hours $30$ minutes
0%
$8$ hours
0%
$8$ hours $15$ minutes
0%
$8$ hours $25$ minutes

Explanation

Ronald work at a rate of 32 pages per 6 hrs $=\dfrac{32}{6}=\dfrac{16}{3} pages/hr$

Elan work at a rate of 40 pages per 5 hr $=\dfrac{40}{4}=8 Pages/hr$

If they work together then they work at the rate of $=\dfrac{16}{3}+8=\dfrac{40}{3} pages/hr$

Let t time is required them to type 110 pages then

$\dfrac{40}{3}t=110$

$40t=330$

$t=\dfrac{330}{40}=\dfrac{33}{4}=8\dfrac{1}{4}=8 hr 15 min.$

$A$ does

$\displaystyle \frac{4}{5}$ th of work in

$20$ days. He then calls in

$B$ and they together finish the remaining work in

$3$ days. How long

$B$ alone would take to do the whole work ?

0%
$23$ days
0%
$37$ days
0%
$\displaystyle 37\frac{1}{2}$ days
0%
$40$ days

Explanation

$Let \, \,the \, \,total \, \,work \, \,be \, \,x.$

$\therefore \, \, The \, \, speed \, \, at \, \, which \, \, amount \, \,of \, \,work \, \,A \, \,does \, \,in \, \,20 \, \,days \, \,is=\dfrac{x}{25}$

$Let \, \,the \, \,speed \, \,of \, \,B \, \,be \, \,B \, \,work \, \,per \, \,day.$

$Since, \, \, the \, \, work \, \,done \, \,by \, \,A \, \,is \, \,\dfrac{4}{5}x$

$\therefore \, \,Remaining \, \,work \, \,is \, \,x-\dfrac{4}{5}x=\dfrac{x}{5}$

$If \, \,A \, \,and \, \,B \, \,work \, \,together \, \,then \, \,we \, \,have,$

$\dfrac{\dfrac{x}{5}}{\dfrac{x}{25}+B}=3$

$\therefore \, \, speed \, \, of \, \, B=\dfrac{2x}{75} \, \,work \, \,per \, \,day.$

$If \, \,B \, \, alone \, \,works \, \, then \, \, the \, \,work \, \, will \, \,be \, \, completed \, \,in \, \,\dfrac{x}{\dfrac{2x}{75}}=\, \, 37 \dfrac{1}{2} \, \, days.$

$A$ can do a certain work in the same time in which

$B$ and

$C$ together can do it. If

$A$ and

$B$ together could do it in

$10$ days and

$C$ alone in

$50$ days, then

$B$ alone could do it in

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

Explanation

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

$=\dfrac{1}{10}$

C's 1 day's work

$=\dfrac{1}{50}$

(A+B+C)'s 1 day's work

$=\dfrac{1}{10}+\dfrac{1}{50}=\dfrac{6}{50}=\dfrac{3}{25}.....(I)$

According to the question

A's 1 day's work=(B+C) 1day's work.........(II)

From I and II we get

$2\times$ (A's one day's work)

$=\dfrac{3}{25}$

A's one day's work

$=\dfrac{3}{50}$

$\therefore$ B's 1 day's work

$=\dfrac{1}{10}-\dfrac{3}{50}=\dfrac{2}{50}=\dfrac{1}{25}$

Hence B alone could do the work in 25 days.

A man can do a piece of work in

$5$ days, but with the help of his son, he can do it in

$3$ days. In what time can the son do it alone?

0%
$7.5$ days
0%
$8.5$ days
0%
$9.5$ days
0%
$11.5$ days

Explanation

$Son's \, \,1 \, \,day's \, \,work \, \,= \, \,(\dfrac{1}{3}-\dfrac{1}{5}) \, \,= \, \,\dfrac{2}{15}.$

$\therefore \, \,The \, \,son \, \,alone \, \,can \, \,do \, \,the \, \,work \, \,in \, \,\dfrac{15}{2} \, \,= \, \,7.5 \, \,days$

$A$ and

$B$ together can do a piece of work in

$30$ days.

$A$ having worked for

$16$ days,

$B$ finishes the remaining work alone in

$44$ days. In how many days shall

$B$ finish the whole work alone?

0%
$30$ days
0%
$40$ days
0%
$60$ days
0%
$70$ days

Explanation

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

$Let \, \, the \, \, speed \, \, of \, \, A \, \, in \, \, doing \, \, entire \, \, work \, \, be \, \, x \, \, work/day$

$Let \, \, the \, \, speed \, \, of \, \, B \, \, in \, \, doing \, \, entire \, \, work \, \, be \, \, y \, \, work/day$

$\therefore \, \, \dfrac{W}{x+y}=30$

$\therefore$

$\, \, W=30x+30y............(1)$

$Work \, \, done \, \,by \, \,A \, \,in \, \,16 \, \,days \, \,=16x$

$Remaining \, \,work \, \,done \, \,by \, \,B \, \,in \, \,44 \, \,days=44y$

$\therefore \, \, 16x+44y=W.........(2)$

$From \, \,(1) \, \,and \, \,(2) \, \, ,W=60y$

$Hence, \, \,if \, \,B \, \,works \, \,alone \, \,then \, \,the \, \,work \, \,will \, \,be \, \,completed \, \,in \, \,\dfrac{W}{y} \, \,days \, \,= \, \,\dfrac{60y}{y} \, \,days \, \,=60 \, \,days.$

$A$ can do a piece of work in

$9$ days and

$B$ in

$18$ days. They began the work together but

$3$ days before the completion of work

$A$ leaves. The time taken to complete the work is:

0%
$7$ days
0%
$5$ days
0%
$8$ days
0%
$11$ days

Explanation

Work done by

$A$ in 1 day

$=\displaystyle \frac{1}{9}$

Work done by

$B$ in 1 day

$=\displaystyle \frac{1}{18}$

Work done by

$B$ in last 3 days

$=\displaystyle \frac{1}{18}\times 3=\frac{1}{6}$

Remaining work

$=1-\displaystyle \frac{1}{6}=\frac{5}{6}$

Work done by A and B in 1 day

$=\displaystyle \frac{1}{9}+\frac{1}{18}=\frac{3}{18}=\frac{1}{6}$

Hence no. of days in which A and B worked together is

$=\displaystyle \dfrac{\cfrac{5}{6}}{\cfrac{1}{6}}=5 \ days$

Then total time to complete the work

$=3+5=8 \ days$ .

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

Part of tanker fill by the pipe A in 1 minute

$=\displaystyle \frac{1}{60}$

Part of tanker fill by the pipe B in 1 minute

$=\displaystyle \frac{1}{40}$

Let the time taken to fill the tank is

$2x$ minute.

$\therefore$ Part of tanker filled

$=\displaystyle \frac{1}{40}\times x+\left(\frac{1}{40}+\frac{1}{60} \right)x=1$

$\Rightarrow\displaystyle \frac{x}{40}+\frac{(2+3)}{120}x =1$

$\Rightarrow \displaystyle \frac{x}{40}+\frac{x}{24}=1$

$\Rightarrow\displaystyle \frac{3x+5x}{120}=1$

$\Rightarrow x=\displaystyle \frac{120}{8}=15$ min

$\therefore$ Time taken to fill the tank

$=2x=2\times 15=30$ min

A can do a piece of work in 24 days If B is 60% more efficient then the number of days required by B to do the twice as large as the earlier work is

0%
24
0%
36
0%
15
0%
30

Explanation

A's 1 day's work

$= \frac{1}{24}$
B's 1 day's work

$= \frac{1}{24}+60 Percent of \frac{1}{24} = \frac{1}{24} \times \frac{160}{100} = \frac{1}{15}$

$\therefore B Can do same work in 15 days$

$\therefore B can do double work in 2 \times 15 = 30 days$

$A$ can do a work in

$15$ days and

$B$ in

$20$ days. If they work on it together for

$4$ days, then the fraction of the work that is left is:

0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{10}$
0%
$\displaystyle \frac{7}{15}$
0%
$\displaystyle \frac{8}{15}$

Explanation

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

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

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

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

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

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

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

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

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

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

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

Two workers A and B working together completed a job in

$5$ days. If A worked twice as efficiently as he actually did and B worked

$\displaystyle \frac{1}{3}$ as efficiently as he actually did the work would have been completed in

$3$ days. A alone could complete the work in

0%
$\displaystyle 5\frac{1}{4}$ days
0%
$\displaystyle 6\frac{1}{4}$ days
0%
$\displaystyle 7\frac{1}{2}$ days
0%
None of these

Explanation

Let the total work be

$W$ .
Let the speed of

$A$ be

$x$ work/day.
Let the speed of

$B$ be

$y$ work/day.
Two workers

$A$ and

$B$ working together completed a job in

$5$ days.

$\therefore \, \, \dfrac{W}{x+y}=5$

$\therefore\,\,W=5x+5y$ ..........(1)
If

$A$ worked twice as efficiently as he actually did and

$B$ worked

$\dfrac{1}{3}$ as efficiently as he actually did the work would have been completed in

$3$ day.

$\therefore\,\,\dfrac{W}{2x+\dfrac{1}{3}y}=3$

$\therefore\,\,5W=30x+5y$ .........(2)
Solving (1) and (2), we get

$\dfrac{W}{x}=\dfrac{25}{4}$

$\therefore$

$A$ alone could complete the work in

$\dfrac{25}{4}=6\dfrac{1}{4}$ days.

$A$ alone can do a piece of work in

$6$ days and

$B$ alone in

$8$ days.

$A$ and

$B$ undertook to do it for

$Rs. 3200.$ With the help of

$C,$ they completed the work in

$3$ days. How much is to be paid to

$C$ ?

0%
$300$
0%
$400$
0%
$600$
0%
$800$

Explanation

C's 1 day's work

$=$

$\dfrac{1}{3}-(\dfrac{1}{6}+\dfrac{1}{8})=\dfrac{1}{3}-\dfrac{7}{24}=\dfrac{1}{24}$

A's wages : B's wages : C's wages

$=\dfrac{1}{6}:\dfrac{1}{8}:\dfrac{1}{24}=4:3:1$

$\therefore$ C's share (for 3 days)

$=Rs.\, (3 \times \dfrac{1}{24} \times 3200)=Rs.\, 400$

Ravi and Kumar are working on an assignment. Ravi takes $6$ hours to type $32$ pages on a computer, while Kumar takes $5$ hours to type $40$ pages. How much time will they take, working together on two different computers to type an assignment of $110$ pages?

0%
$20$ hours
0%
$5$ hours $40$ minutes
0%
$8$ hours $15$ minutes
0%
$23$ hours

Explanation

Number of pages typed by Ravi in 1 hour

$=\dfrac{32}{6}=\dfrac{16}{3}$

Number of pages typed by Kumar in 1 hour

$=\dfrac{40}{5}=8$

Number of pages typed by both in 1 hour

$=(\dfrac{16}{3}+8)=\dfrac{40}{3}.$

$\therefore$ Time taken by both to type 110 pages

$=(110 \times\, \dfrac{3}{40})\,$ hours.

$=8\dfrac{1}{4}$ hours (or) 8 hours 15 minutes..

A cistern can be filled by a tap in

$4$ hours while it can be emptied by another tap in

$9$ hours. If both the taps are opened simultaneously then after how much time will the cistern get filled ?

0%
$4.5$ hrs
0%
$5$ hrs
0%
$6.5$ hrs
0%
$7.2$ hrs

Explanation

A cistern can be filled by a tap in

$4$ hours while it can be emptied by another tap in

$9$ hours.

Let tap

$A$ fill the cistern and tap

$B$ empty the cistern.

Work done by tap

$A$

$=\dfrac{1}{4}$

Work done by tap

$B$

$=\dfrac{1}{9}$

Work done to fill the tap when both are opened

$=\dfrac{1}{4}-\dfrac{1}{9}$

$=\dfrac{9-4}{36}$

$=\dfrac{5}{36}$

Thus, time taken to completely fill the cistern

$=\dfrac{36}{5}$

$=7.2$ hrs

A man a woman and a boy can complete a job in

$3, 4$ and

$12$ days respectively. How many boys must assist 1 man and 1 woman to complete the job in

$\displaystyle \frac{1}{4}$ of a day?

0%
$1$
0%
$4$
0%
$19$
0%
$41$

Explanation

(

$1$ man

$+$

$1$ woman)'s

$1$ day's work

$=\left (\dfrac{1}{3}+\dfrac{1}{4}\right)=\dfrac{7}{12}.$
Work done by

$1$ man and

$1$ woman in

$\dfrac{1}{4}$ day

$=\left (\dfrac{7}{12} \times \dfrac{1}{4}\right)=\dfrac{7}{48}$
Remaining work

$=\left (1-\dfrac{7}{48}\right)=\dfrac{41}{48}$
Work done by

$1$ boy in

$\dfrac{1}{4}$ day

$=\left (\dfrac{1}{12} \times \dfrac{1}{4}\right)$

$=\dfrac{1}{48}$

$\therefore$ Number of boys required

$=\left (\dfrac{41}{48} \times 48\right)=41$ .

$A$ works twice as fast as $B.$ If $B$ can complete a work in $12$ days independently, the number of days in which $A$ and $B$ can together finish the work in

0%
$4$ days
0%
$10$ days
0%
$8$ days
0%
$13$ days

Explanation

Part of work done by

$B$ in

$1$ day

$=\dfrac1{12}$

Since,

$A$ works twice as

$B$

So, Part of work done by

$A$ in

$1$ day

$=\dfrac2{12}=\dfrac16$

$\therefore$ Part of work done by

$A+B$ in

$1$ day

$=\dfrac16+\dfrac1{12}=\dfrac14$

So,

$A$ and

$B$ together can finish the work in

$4$ days.

Hence, A is the correct option.

$24$ men can complete a work in

$16$ days,

$32$ women can complete the same work in

$24$ days.

$16$ men and

$16$ women started working and worked for

$12$ days. How many more men are to be added to complete the remaining work in

$2$ days?

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

Explanation

24 men can complete the work in 16 days

1 man can complete the work in

$=16\times 24=384 Days$

32 women cam complete the work in 24 days

1 woman can complete the work in a

$=32\times 24=768 days$

Ratio between man and woman

$=384:768=1:2$

then 1 man =2 women

The work is done br 16 men and 16 women in 12 days

16 women=8 men

thus the number of men complete the work in 12 days=

$16+8=24 men$

Since 24 men takes 16 days to complete the work

So after 12 days,4 days work will be left for 24 Men

Thus the work left can be completed in 2 days by

$=24\times \dfrac{4}{2}=48 Days$

Additional men required to complete the work in 2 days

$=48-24=24 men$

$12$ men can complete a piece of work in

$4$ days while

$15$ women can complete the same work in

$4$ days

$6$ men start working on the job and after working for

$2$ days all of them stopped working. How many women should be put on the job to complete the remaining work if it is to be completed in

$3$ days ?

0%
$15$
0%
$18$
0%
$22$
0%
None of these

Explanation

$12$ men can complete a piece of work in

$4$ days

$15$ women can complete the same work in

$4$ days
Let the productivity of men be

$m$ and the productivity of women be

$w$

Thus,

$12m \times 4 = 15w \times 4$

$=>4m=5w$

$=>m=\displaystyle \frac{5}{4}w$

Total work done by women

$=60w$

Work done by 6 men in 2 days

$=6m \times 2$

$=6 \times \displaystyle \frac {5}{4} \times 2$

$=15w$

Thus, remaining work

$=60w-15w$

$=45w$

Let F women complete the remaining job in 3 days

Thus,

$Fw \times 3 = 45w$

$=>F=\displaystyle \frac{45}{3}$

$=>F=15$

Thus,

$15$ women are required to complete the job.

$A$ and

$B$ together can do a piece of work in

$12$ days which

$B$ and

$C$ together can do in

$16$ days. After

$A$ has been working at it for

$5$ days and

$B$ for

$7$ days,

$C$ finishes it in

$13$ days. In how many days

$C$ alone will do the work ?

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

Explanation

Work done by (A+B)'s in 1 day

$=\dfrac{1}{12}$

work done by (B+C)'s in 1 day

$=\dfrac{1}{16}$

Let C does a work in x days

Then work done by C in a day

$=\dfrac{1}{x}$

According to the questio

A's 5 day's work+B's 7 day's work+C's 13 day's work=1

(A+B)'s 5 day's+(B+C)'s 2 days+C's 11 day's

$\dfrac{5}{12}+\dfrac{2}{16}+\dfrac{11}{x}=1$

$\dfrac{11}{x}=1-\dfrac{5}{12}-\dfrac{2}{16}$

$\dfrac{11}{x}=\dfrac{22}{48}$

$x=24 days$

$12$ children take

$16$ days to complete a work which can be completed by

$8$ adults in

$12$ days.

$16$ adults started working and after

$3$ days,

$10$ adults left and

$4$ children joined them. How many days will they take to complete the remaining work ?

0%
$3$
0%
$4$
0%
$6$
0%
$8$

Explanation

$12$ children take

$16$ days to complete a work

$8$ adults take

$12$ days to complete the same job.

Let the productivity of children be

$c$ and the productivity of adult be

$m$

Thus,

$12c\times 16=8m \times 12$

$=>c=\displaystyle \frac{1}{2}m$

Work done by

$8$ men in

$12$ days

$8 m \times 12$

$=96m$

Work done by

$16$ men in

$3$ days

$=16 m \times 3$

$=48m$

Work left

$=96m - 48 m =48m$

After

$3$ days

$10$ adults left and

$4$ children joined them

Thus,

$6$ men and

$4$ children have to complete the remaining job.

Let the time taken by them be

$t$ days.

Thus

$(6m+4c)t=48m$

$=>t=\displaystyle \frac{48m}{6m+4\times \frac{1}{2}m}$

$=>t=\displaystyle \frac{48m}{6m+2m}$

$=>t=6$

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 6 - 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 6 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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