GMAT Distance And Speed Questions WIth Solutions

Work done by A and B in one day= 1/12
Work done by B and C in one day = 1/15
Work done by C and A in one day = 1/20
Adding all these, 2 (A + B + C)’s work of one day = 1/12+1/15+1/20=1/5 (4 + B + C)’s one day work = 1/10
∴ A’s one day work = (A + B + C)’s one day work - (B + C)’s one day work ⇒ A’s one day work =1/10-1/15=1/30
Hence, A can do this work in 30 days.

Time taken by 6 men to finish the whole work = 8 days

Let the number of men required be x.

Less days, more men (Inverse Proportion)

Days Men

8 6

2 x

⇒ x =(8 * 6)/2 = 24 men

The additional men required = 24-6 men= 18 men

Suppose pipe A alone takes x hours to fill the tank.

Then, pipes B and C will take x/2 and x/4 hours respectively to fill the tank.

1/x+2/x+4/x=1/5⇔7/x=1/5 ⇔x=35hrs

2 pipes open for 2 hours will fill $$ \frac {2}{3}*\frac{2}{3} *3 $$ buckets = 4/3 buckets.

In one day, work done by them together will be = 1/5 + 1/7 + 1/9 = 143/315
So days taken to complete the unit work = 315/143 days

Work done by a man in a day = $$\frac{1}{800}$$ of the total work
Work done by x men in 8 days = $$\frac{8x}{800}$$ that should be equal to unit amount of work
Hence, x = 100
We need 60 more men to finish the work in 8 days.

If all the pipes are working together, then part of the tank filled in one hr will be = $$\frac{1}{10} + \frac{1}{16} - \frac{1}{32} = \frac{21}{160}$$

Hence, it will take $$\frac{160}{21}$$ hr to fill the complete tank

So answer will be B

6 women complete a piece of work in 3 days.

Hence, one woman can complete $$\frac{1}{18}$$ piece of work in one day.

3 women and 18 children take 2 days to complete the work.

3 women in 2 days complete $$3*2*\frac{1}{18}=\frac{1}{3}$$

Hence, in two days 18 children complete $$1-\frac{1}{3}=\frac{2}{3}$$ of the work.

So, amount of work done by 1 child in a day is $$\frac{2}{3}*\frac{1}{36}=\frac{1}{54}$$

So, 9 children complete $$9*\frac{1}{54}=\frac{1}{6}$$ part of the work in a day.

Hence, the number of days in which 9 children alone complete the work is $$6$$

Let the speed of work done by A, B and C be a, b and c.
Now, b=.5c, a=.5b=.25c
c does work in 7 days. It's speed will be W/7. A's speed will be .25W/7 and B's speed will be .5W/7
Their combined speed is 1.75W/7
Time taken by them to complete the work = $$\frac{W}{(1.75W/7)}$$ = $$\frac{7}{1.75}$$= 4
Hence, together they will complete the work in 4 days.
Therefore, option C is correct.

Let the speed of women be x and that of children be y. Let the piece of work be W.
2x + 10y = W/8
8x = W/6
x = W/48
Hence, we get y= W/120
Speed of 1 child = W/120
Speed of 10 children = W/12
Now time needed for them to complete = W/(W/12) =12days
Therefore, the correct option is B.

6 women take 10 days to complete a piece of work.
So, the amount of work done by 6 women in 1 day is $$\frac{1}{10}$$

10 children take 15 days to complete a piece of work.
So, the amount of work done by 10 children in 1 day is $$\frac{1}{15}$$

So, the amount of work done by 6 women and 10 children in 1 day is $$\frac{1}{10}+\frac{1}{15} = \frac{1}{6}$$

Hence, the number of days taken by 6 women and 10 children to complete the piece of work is 6 days.

Let the amount work done by a woman in a day be W.
So, the amount of work done by a man in a day is W/2.
And the amount of work done by a child in a day is W/4
So, the amount of work done by 3 men, 4 women and six children in a day is 3W/2 + 4W + 6W/4 = 7W.

Hence, 7 women can complete this work in 7 days. So, the correct answer is option (b)

Work done by 18 women in 1 day = $$\frac{1}{12}$$
Work done by 1 woman in 1 day = $$\frac{1}{12\times18}$$
Work done by 12 men in 1 day = $$\frac{1}{9}$$
Work done by 1 man in 1 day = $$\frac{1}{9\times12}$$
Work done by 8 men and 8 women in 1 day = $$8\times(\frac{1}{12\times18} + \frac{1}{9\times12})$$ = $$\frac{1}{9}$$
$$\therefore$$ They will take 9 days to finish the work.

12 men can complete the work in $$(3\times8 = 24)$$ days
Work done by 12 men in 1 day = $$\frac{1}{24}$$
Work done by 1 man in 1 day = $$\frac{1}{24\times12}$$
Work done by 16 men in 1 day = $$16 \times(\frac{1}{24\times12}) = \frac{1}{18}$$
Hence, 16 men will take 18 days to finish the work.

Let the work be 120 units.
A+B = 120/30 = 4
B+C = 120/24 = 5
C+A = 120/20 = 6
A+B+C = (4+5+6)/2 = 7.5
Time taken = 120/7.5 = 16 days

Let the volume of the tank be 30. So inlet pipe fills 3 units per minute while outlet pipe empties 2 units per minute. So, in the 4 minutes after 11, the inlet pipe will be able to fill 12 units water in the tank. Now that out let pipe is also opened, the effective rate at which water goes into the tank is 1 unit per minute. So the remaining 18 units will take 18 more minutes. Thus, the tank will be first full after 22 minutes. Hence the required time is 11:22 am.

Let the total work to be done is 120 units.
=> 8 men and 12 women can do 120/10 = 12 units work per day
Or 8m+12w=12
Since a man can complete the work in 120 days, he must do 1 unit work per day.
=> 12w=12-8=4 or w=1/3
So, 10 men and 6 women will do (10+6/3) 12 units work/ day.
Thus they will take 120/12 = 10 days to complete the work.

Work done by Arjun in a day = $$\frac{1}{12}$$
Work done by Arnab in a day = $$\frac{1}{15}$$
Work done in 2 days by both of them combined = $$\frac{1}{12}+\frac{1}{15}=\frac{3}{20}$$
After 12 days the work done = $$\frac{3}{20}*6=\frac{18}{20}=\frac{9}{10}$$
The work remaining to be done = $$\frac{1}{10}$$
On the 13th day the work done by Arjun = $$\frac{1}{12}$$
After the 13th day the amount of work remaining = $$\frac{1}{10}-\frac{1}{12}=\frac{1}{60}$$
This can be completed on the next day by Arnab.
Thus, the work will be completed on the 14th day.

5 men working together can finish the work in 10 days. Let each man does x units of work in 1 day. So total work to be done = 50x
Now 10 women finish the work in 7.5 days. Hence, if y is the amount of work done by 1 women in 1 day then total work = 10*7.5 = 75y
50x = 75y
=> x/y = 3/2
So, let y = 2x/3
When 3 men and 3 women work, then the amount of work done by them in 1 day = 3x + 2x = 5x
Hence it will take 10 days to complete 50 units of work.

Work done by 28 men in 1 day = $$\frac{1}{15}$$
Work done by 1 man in 1 day = $$\frac{1}{28\times15}$$
Work done by 30 men in 1 day = $$\frac{30}{28\times15} = \frac{1}{14}$$
Work done by 1 woman in 1 day = $$\frac{1}{15\times24}$$
Work done by 18 women in 1 day = $$\frac{18}{15\times24} = \frac{1}{20}$$
Required ratio = $$\frac{20}{14} = \frac{10}{7}$$

$$\frac{M1D1}{W1}$$ = $$\frac{M2D2}{W2}$$

W1=W2 = Q

$$\frac{18Mx30}{Q}$$ = $$\frac{16Wx36}{Q}$$

M = $$\frac{32}{30}$$W ....(1)

Let the days required by 18 women to complete the remaining work = y days

so $$\frac{(15M\times9)+(18W \times y)}{Q}$$ = $$\frac{16W\times36}{Q}$$ ......(2)

using equation 1 and 2

$$\frac{(16W\times9)+(18W \times y)}{Q}$$ = $$\frac{16W\times36}{Q}$$

144W + 18Wy = 576W

18Wy = 432 W

y = 24 days

Time taken by 24 children is twice the time taken by 16 women to finish the same project.
Hence, 24 children's work can be completed by 8 women.
Work done by 1 woman in 1 day = $$\frac{1}{28\times 8}$$
Work done by 28 women and 24 children in 1 day = Work done by (28+8 = 36) women in 1 day
= $$\frac{36}{28 \times 8}$$
= $$\frac{9}{56}$$
$$\therefore$$ Number of days required by them = $$\frac{56}{9} = 6 \frac{2}{9}$$

Let the extra men required be M

using the formula

$$\frac{M1D1}{W1}$$ = $$\frac{M2D2}{W2}$$

$$\frac{56 \times14 }{W}$$ = $$\frac{8(M+14)}{W}$$

M +56 = 98

M = 42 men

Pipe A 1 hour work = 1/8

Pipe B 1 hour work = 1/16

Pipe C 1 hour work = - 1/32

Here negative sign shows that it is doing negative work

Now when all pipes are opened simultaneously then 1 hour combined work = 1/8 + 1/16 - 1/32

= 5/32

So time taken by all pipes to fill the tank = 32/5 hours = 6.4 hours

Total quantity = 160 units
Speed pipe 1 = 16 $$\frac{units}{hours}$$

Speed pipe 2 = 10 $$\frac{units}{hours}$$

Speed pipe 3 = 5 $$\frac{units}{hours}$$

Time reqd = Total Units / [Speed (pipe 1 + pipe 2) - Speed Pipe 3]

Time = 160/21 = 7 $$\frac{13}{21}$$ hours

Total Man-days = 56*14
Number of days = 8

Number of workers = 56*14/8 = 7*14
Total number of workers = 98
Additional workers = 98-56 = 42

12 women take 5 days to do one item. Therefore each woman contributed 1/60th of the item per day.

3 women working for 10 days = 30/60ths of the item = 1/2.
9 children therefore make 1/2 an item in 10 days.
36 children will make 2 items in 10 days, so they can make 1 item in 5 days.

It is given that 6 woman can complete the work in 10 days. Let the work completed by each woman be 'w'.
Work = 10*6w = 60w.

10 children and 6 woman complete the work in 6 days. Let the amount of work completed by each child be 'c'.
6*(10c+6w) = 60w
60c+36w = 60w
60c = 24w

Multiplying both sides by 2.5, we get,
150c = 60w
Therefore, 10 children will take 15 days to complete the work.
Option D is the right answer.

Given that , 3 men(M) or 8 boys(B) can do the work in 17 days i.e the capacity of 3 men and 8 boys is same.
Hence, 3M = 8B , which gives , M = $$\frac{8}{3}$$*B

We need to find , the time taken by 2M + 6B . Replacing M with $$\frac{8}{3}$$*B

we get 2*$$\frac{8}{3}$$ *B + 6B = $$\frac{34}{3}$$*B.
But as 8 boys can do the job in 17days , $$\frac{34}{3}$$ boys can do the same job in (18*7)$$\frac{34}{3}$$

= 12 days.

It is clear from the question,
16 men do 5/9th of work in 30 days.
Let 'n' no. of more men are required to complete the remaining work.
Hence, (n+16) men do 4/9th of work in 8 days.
We know that,
$$\frac{Amount of work}{No. of men\times{No. of days}}=Constant$$.
Hence,
$$\frac{5/9}{16\times30}=\frac{4/9}{(n+16)\times8}$$.
$$n=32$$.
Hence, Option C is correct.

Rate of doing work by A and B is $$\frac{1}{A} $$ and $$\frac{1}{B}$$
$$\frac{1}{A} $$ = $$\frac{3}{B}$$
In 4 days A would have completed $$\frac{4}{A}$$ or $$\frac{12}{B}$$ amount of work
In 15 days B would have completed $$\frac{15}{B}$$ amount of work
Since total work done was 75%.
$$\frac{12}{B} + \frac{15}{B} = \frac{3}{4}$$
$$\frac{27}{B} = \frac{3}{4}$$
$$\frac{1}{B} = \frac{1}{36}$$
B alone would take 36 days to complete the work.
Option D is the correct answer.

Let total work be 1 unit

Let efficiency of A = $$3x$$ units/day

=> efficiency of B = $$x$$ units/day

Acc to ques :

=> $$\frac{1}{x} - \frac{1}{3x} = 10$$

=> $$\frac{2}{3x} = 10$$

=> $$x = \frac{1}{15}$$

$$\therefore$$ Time taken by B = 15 days

(Shortcut)

Since, A takes 10 days less than B, => B definitely takes more than 10 days

=> Options (B), (C) & (D) are eliminated.

Pages printed by P, Q and R in one minute

= $$\frac{1}{8} + \frac{1}{18} + \frac{1}{24}$$

= $$\frac{9 + 4 + 3}{72} = \frac{16}{72}$$

= $$\frac{2}{9}$$

Number of pages to be printed = 368

$$\therefore$$ Time taken = $$\frac{9}{2} \times 368$$

= $$1656$$ minutes

= $$\frac{1656}{60} = 27.6$$ hours

A's 1 day's work = $$\frac{1}{27}$$

=> A's 12 days work = $$\frac{1}{27} \times 12 = \frac{4}{9}$$

Remaining work = $$1 - \frac{4}{9} = \frac{5}{9}$$

Similarly, B's 11 days work = $$\frac{1}{33} \times 11 = \frac{1}{3}$$

Remaining work = $$\frac{5}{9} - \frac{1}{3} = \frac{2}{9}$$

Now, days taken by C to complete the work = 45 days

=> Days taken to complete $$\frac{2}{9}$$ th of the work = $$\frac{2}{9} \times 45 = 10$$ days

$$\therefore$$ Days in which the whole work is completed = 12 + 11 + 10 = 33 days

(A + C)'s 1 day's work = $$\frac{1}{18} + \frac{1}{27}$$

= $$\frac{3 + 2}{54} = \frac{5}{54}$$

=> (A + C)'s 6 day's work = $$\frac{5}{54} \times 6$$

= $$\frac{5}{9}$$

=> Remaining work = $$1 - \frac{5}{9} = \frac{4}{9}$$

Acc to ques,

=> $$\frac{16}{x} = \frac{4}{9}$$

=> $$x = \frac{16 \times 9}{4}$$

=> $$x = 4 \times 9 = 36$$ days

Let B's efficiency = $$x$$ units/day

=> A's efficiency = $$4x$$ units/day

and C's efficiency = $$3x$$ units/day

Time taken by C to complete the work alone = 16 days

=> Total work done = $$3x \times 16 = 48x$$ units

Sum of efficiencies of A, B and C = $$x + 4x + 3x = 8x$$ units/day

$$\therefore$$ Time taken by all to complete the work = $$\frac{48x}{8x}$$

= 6 days

1 minute work of tap A = $$\frac{1}{60}$$

1 minute work of tap B = $$\frac{1}{30}$$

=> (A + B)'s 1 minute work = $$\frac{1}{60} + \frac{1}{30}$$

= $$\frac{1 + 2}{60} = \frac{3}{60}$$

= $$\frac{1}{20}$$

$$\therefore$$ Time taken by (A + B) to empty the tank = 20 min

Let the time taken to complete the work = $$x$$ days

=> $$(4 M + 3 W) \times 6 = (5 M + 7 W) \times 4$$

=> $$24 M + 18 W = 20 M + 28 W$$

=> $$4 M = 10 W$$ ---------------Eqn(1)

Acc to ques,

=> $$(4 M + 3 W) \times 6 = (1 M + 1 W) \times x$$

Using eqn(1), we get :

=> $$(10 + 3) \times 6 = (\frac{10}{4} + 1) \times x$$

=> $$78 = \frac{7}{2} x$$

=> $$x = \frac{78 \times 2}{7} = \frac{156}{7}$$

=> $$x = 22\frac{2}{7}$$ days

Let work = 1 units

Let time taken by A to complete the work = $$x$$ days

=> Time taken by B = $$4x$$ days

Acc to ques,

=> $$\frac{12}{4x} + \frac{6}{x} = 1$$

=> $$\frac{6}{x} (\frac{1}{2} + 1) = 1$$

=> $$\frac{3}{2} = \frac{x}{6}$$

=> $$x = \frac{3}{2} \times 6$$

=> $$x = 9$$ days

Part of tank filled by A and B in 1 hour

= $$\frac{1}{5} - \frac{1}{8} = \frac{3}{40}$$

=> Time taken to fill the tank completely = $$\frac{40}{3}$$ hrs

$$\therefore$$ Time taken to fill two-fifth part of tank

= $$\frac{2}{5} \times \frac{40}{3}$$

= $$\frac{16}{3} = 5\frac{1}{3}$$ hrs

Time taken by A and B together = 12 days

Time taken by B alone = 30 days

=> A's 1 day's work = $$\frac{1}{12} - \frac{1}{30}$$

= $$\frac{5 - 2}{60} = \frac{3}{60} = \frac{1}{20}$$

$$\therefore$$ Required time = 20 days

Using the formula, $$M_1 D_1 = M_2 D_2$$

=> $$36 \times 14 = M_2 \times 8$$

=> $$M_2 = \frac{36 \times 14}{8} = 63$$

$$\therefore$$ Additional workers = 63 - 36 = 27

Time taken by A alone = 15 days

Efficiency of A : B = 100 : 160

= 15 : 24

=> Time taken by B = 24 days

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

= $$\frac{1}{15} + \frac{1}{24}$$

= $$\frac{8 + 5}{120} = \frac{13}{120}$$

$$\therefore$$ Required time = $$\frac{1}{\frac{13}{120}}$$

= $$\frac{120}{13} = 9\frac{3}{13}$$ days

Priti's second day typing speed = $$\frac{115}{100} \times 60$$ = 69 wpm

Required difference in time

= $$(\frac{5000}{60} - \frac{5000}{69})$$

= $$5000(\frac{69 - 60}{60 \times 69})$$

$$\approx$$ 10 minutes

Let the total work to be done = 30 units

Rate at which A alone finishes the task = $$\frac{30}{10}$$ = 3 units/day

Rate at which A & B together finishes the work = $$\frac{30}{6}$$ = 5 units/day

=> Rate at which B alone finishes the work = 5 - 3 = 2 units/day

$$\therefore$$ Time taken by B to complete the task = $$\frac{30}{2}$$ = 15 days

Time taken by a single person to write a letter = 3 min

=> No. of letters written from 10 a.m. to 12.00 noon (i.e. 120 min) by a single person = $$\frac{120}{3}$$ = 40 letters

$$\therefore$$ No. of persons required to write 1960 letters in 120 min

= $$\frac{1960}{40}$$ = 49

Let total work to be done = 6 units

=> A completes $$\frac{5}{6} \times 6 = 5$$ units in 10 days

=> A's efficiency = $$\frac{5}{10} = \frac{1}{2}$$ units /day

Now, B finishes $$\frac{1}{6}$$th of the task in 3 days

=> B completes $$\frac{1}{6} \times 6 = 1$$ units in 3 days

=> B's efficiency = $$\frac{1}{3}$$ units /day

Now, time taken by A alone to complete the entire task = $$\frac{6}{\frac{1}{2}} = 12$$ days

Time taken by B alone to complete the entire task = $$\frac{6}{\frac{1}{3}} = 18$$ days

$$\therefore$$ Required ratio = $$\frac{12}{18} = 2 : 3$$

Let efficiency of A = $$3x$$ units/day

=> Efficiency of B = $$\frac{4}{3} \times 3x = 4x$$ units/day

Let Work to be done = 8 units

=> Work done by A in 15 days = $$15 \times 3x = \frac{5}{8} \times 8$$

=> $$45x = 5$$

=> $$x = \frac{5}{45} = \frac{1}{9}$$

Thus, B's 1 day work = $$4 \times \frac{1}{9} = \frac{4}{9}$$ units

Work done by B in 10 days = $$\frac{4}{9} \times 10 = \frac{40}{9}$$ units

=> Work left = $$8 - \frac{40}{9} = \frac{32}{9}$$

$$\therefore$$ Fraction of work left = $$\frac{\frac{32}{9}}{8}$$

= $$\frac{4}{9}$$

A alone can make 100 baskets in 6 days and B alone in 12 days

=> Rate of efficiency of A = $$\frac{100}{6}$$ baskets/day

Rate of efficiency of B = $$\frac{100}{12} = \frac{50}{6}$$ baskets/day

=> Baskets made by A and B together in 1 day = $$\frac{100}{6} + \frac{50}{6}$$

= $$\frac{150}{6} = 25$$ baskets/day

$$\therefore$$ Time taken by A and B together to make 100 baskets = $$\frac{100}{25} = 4$$ days

If only A and B were opened they would have filled $$\frac{1}{A}+\frac{1}{B}$$ = $$\frac{1}{12}+\frac{1}{18}$$ = $$\frac{5}{36}$$th of the tank in 1 hour i.e. they would have filled half the tank in $$\frac{36}{5*2}$$ = $$3.6$$ hours.
Time required by A to fill rest of the tank is 6 hours(as it can fill whole tank in 12 hours).
Hence, total time required is = 3.6+6 = 9.6 hours.