420+ Solved Time and Work Problems with Solutions and Explanation

Q:

If a soldier fires 7 shots from a gun in 12 minutes then find the total number of shots fired by the man in 3/2 hrs.

A) 45	B) 44
C) 46	D) 47

Answer & Explanation Answer: C) 46

Explanation:

Here given soldier shots 7 shots in 12 min

=> 1 shot doen't take any time in 12 min

=> Only 6 shots take 12 min

12 min ------ 6 shots

90 min ------ ? shots

=> 90 x (6/12) = 45

Therefore, total shots fired in 90 minutes = 45 + 1 = 46 shots.

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11 8131

Q:

If 6 engines consume 24 metric tonnes of coal, when each is working 8 hours day, how much coal will be required for 9 engines, each running 13hours a day, it being given that 2 engines of former type consume as much as 3 engines of latter type ?

A) 45 metric tonnes	B) 47 metric tonnes
C) 55 metric tonnes	D) 34 metric tonnes

Answer & Explanation Answer: A) 45 metric tonnes

Explanation:

2 engines of former type for one hour consumes (2 x 24)/(6 x8) = 1 metric ton

i.e. 3 engines of latter type consumes 1 ton for one hour

Hence 9 engines consumes 3 tons for one hour

For 15 hours it is 15 x 3 = 45 metric tonnes.

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3 8027

Q:

P and Q were assigned to do a work for an amount of 1200. P alone can do it in 15 days while Q can do it in 12 days. With the help of R they finish the work in 6 days. Find the share of R ?

A) 120	B) 240
C) 360	D) 180

Answer & Explanation Answer: A) 120

Explanation:

1/15 + 1/12 + 1/R = 1/6, we got R = 60 (it means R will take 60 days to complete the work alone)

so ratio of work done by P:Q:R = 1/15 : 1/12 : 1/60 = 5 : 4 : 1

so R share = (1/10) x 1200 = 120.

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5 7967

Q:

A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained?

A) 1/5	B) 1/6
C) 1/7	D) 1/8

Answer & Explanation Answer: B) 1/6

Explanation:

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 = 1/6.

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7 7916

Q:

A and B together finish a wor in 20 days.They worked together for 15 days and then B left. Afer another 10 days,A finished the remaining work. In how many days A alone can finish the job?

A) 30	B) 40
C) 50	D) 60

Answer & Explanation Answer: B) 40

Explanation:

(A+B)'s 15 days work = $(\frac{1}{20} \times 15) = \frac{3}{4}$

Remaining work = 1/4

Now, 1/4 work is done by A in 10 days.

Whole work will be done by A in (10 x 4) = 40 days.

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9 7898

Q:

Three men, four women and six children can complete a work in 9 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 this work in 9 days?

A) 7	B) 8
C) 9	D) 6

Answer & Explanation Answer: A) 7

Explanation:

Given (3 Men + 4 Women + 6 Children) -----> 9 days

But W = 2M and C = M/2

Now, convert Men and Children into Women by

$3 \frac{W}{2} + 4 W + 6 \frac{W}{4} = > \frac{3}{2} + 4 + \frac{3}{2} = 7 Women$

Therefore, 7 women alone can complete this work in 9 days.

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12 7645

Q:

Mr. stenley employed a certain number of typist for his project. 8 days later 20% of the typist left the job and it was found that it took as much time to complete the rest work from then as the entire work needed with all the employed typists. The average speed of a typist is 20 pages/hour. Minimum how many typist could be employed?

A) 10	B) 5
C) 15	D) 4

Answer & Explanation Answer: B) 5

Explanation:

Since 20% i.e 1/5 typists left the job. So, there can be any value which is multiple of 5 i.e, whose 20% is always an integer. Hence, 5 is the least possible value.

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6 7502

Q:

16 women can do a piece of work in 10 days while 15 men can do the same work in 12 days. All men and all women work alternately starting with all men, then find the time taken by them to complete the whole work?

A) 10 5/2 days	B) 11 days
C) 11 2/5 days	D) 12 1/2 days

Answer & Explanation Answer: A) 10 5/2 days

Explanation:

Let efficiency of every man and every woman be 'm' unit/day and 'w' unit/day respectively

15×12×m = 10×16×w

⇒ m/w = 8/9

Total work = 15 × 12 × 9 = 1620 units

In 2 days, total work done = 15 x 9 + 16 x 8 = 263 units

So, in 10 days work done will be = 263 × 5 = 1315 units

Remaining work will be done in = (1620-1315)/(15×8) = 5/2 days

Total days = 10 5/2 days.

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