Time AND Work Questions FOR Bank Exams with Detailed Solutions

Given:

Priya  alone can complete a piece of work in 18 days

Monika alone can complete a piece of work in 30 days

Formula Used: 

Total work = Efficiency x Time

Explanation:

(Priya and Monika)’s 1 day work alternatively
=1/18+1/30=8/90
(Priya and Monika)’s 22 days work
=(8×11)/90=88/90
Remaining work=1 –88/90=1/45
∴1/45 work done by Priya = 2/5 days
Total time = 22 (2/5) days.

Exam Hall Method: 

Information:
6 women or 8 Boys can do a price of work in 12 days

Formula Used:
Efficiency of boys ×time taken by boys=efficiency of woman ×time taken by women
Explanation:
Let the work done by each boy and a woman in a day is B and W units respectively.
6 W = 8 B
W/B=4/3
Efficiency of women: Boys = 4: 3
Total work if each women does 4 units of work each day or boy will be 3 units each day =>
= 4 × 6 × 12 or = 3 × 8 × 12
= 288 units = 288 units
Let ‘D’ Days are required to finish the entire work when 3women and 5 boys will work on it,
D × [3W + 5B] = 288
D [3 × 4 + 5 × 3] = 288
D [27] = 288
d = 288/27
= 10 2/3 days.

Given
10 men and 15 women time = 6 days
1 man time = 100 days

Formula Used
Total Work = (Number of People) × Efficiency × Time

Solution
Let the efficiency of a man be M and a woman be W.
Total Work = (10M + 15W) × 6
Total Work = 1M × 100
Equating both:
(10M + 15W) × 6 = 100M
60M + 90W = 100M
90W = 40M
9W = 4M
M = $$\frac{9}{4}W$$​​
Total work in terms of W = $$100 × (\frac{9}{4}W)$$​ = 225W
Time for 1 woman = Total Work / Efficiency of 1 woman
Time =$$\frac{225W}{W}$$​= 225 days

Final Answer
So the correct answer is (d)

Given
A's time = 11 days
B's time = 20 days
C's time = 55 days

Formula Used
Efficiency = Total Work / Time

Solution
Total work = {LCM}(11, 20, 55) = 220 units
Efficiency of A = $$\frac{220}{11}$$​= 20 units/day
Efficiency of B = $$\frac{220}{20}$$​ = 11 units/day
Efficiency of C = $$\frac{220}{55}$$​= 4 units/day
Day 1: A + B work together
Efficiency = 20 + 11 = 31 units
Day 2: A + C work together
Efficiency = 20 + 4 = 24 units
Total work in 2 days = 31 + 24 = 55 units
Number of 2-day cycles required = $$\frac{220}{55}$$​ = 4 cycles
Total time = 4 × 2 = 8 days

Final Answer
So the correct answer is (b)

Given
12 men + 16 boys time = 5 days
13 men + 24 boys time = 4 days

Formula Used
Total Work = (M × D)

Solution
Let M be the efficiency of a man and B be the efficiency of a boy.
Total Work = (12M + 16B) × 5 = (13M + 24B) × 4
60M + 80B = 52M + 96B
60M - 52M = 96B - 80B
8M = 16B
​$$\frac{M}{B} = \frac{16}{8} = \frac{2}{1}$$​​
The ratio is 2 : 1

Final Answer
So the correct answer is (a)

Exam Hall Approach

Given
Time taken by 6 women = 10 days
Time taken by 10 children = 15 days

Formula Used
​$$\frac{1}{Total\, Time} = \frac{1}{T_1} + \frac{1}{T_2}$$​​

Solution
Work done by 6 women in 1 day = $$\frac{1}{10}$$​​
Work done by 10 children in 1 day = $$\frac{1}{15}$$​

Work done by (6 women + 10 children) in 1 day = $$\frac{1}{10} + \frac{1}{15}$$​​
LCM of 10 and 15 is 30.
Combined 1 day work = $$\frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$$​​
Total time taken = 6 days.

Final Answer
So the correct answer is (c)

​$$\text{Given:}\\A = 10 \text{ hours}\\B = 15 \text{ hours}\\C = 30 \text{ hours (emptying)}\\[6pt]\text{Concept Used:}\\\text{Net work rate of filling and emptying pipes}\\[6pt]\text{Formula Used:}\\\text{Net rate} = \text{Filling rate} - \text{Emptying rate}\\[6pt]\text{Solution:}\\\text{Rate of A} = \frac{1}{10}\\\text{Rate of B} = \frac{1}{15}\\\text{Rate of C} = \frac{1}{30}\\[6pt]\text{Net rate}= \frac{1}{10} + \frac{1}{15} - \frac{1}{30}\\= \frac{3+2-1}{30}\\= \frac{4}{30}\\= \frac{2}{15}\\[6pt]\text{Time} = \frac{1}{2/15} = \frac{15}{2}\\[6pt]\text{Final Answer:}\\\boxed {7.5}$$​​
Exam Hall Approach

​$$\textbf{Given:}\\(16M + 12W) \text{ complete work in } 45 \text{ days}\\(40M + 30W) \text{ must complete } 75\% \text{ work}\\[6pt]\textbf{Concept Used:}\\\text{Work and time}\\[6pt]\textbf{Formula Used:}\\\text{Work} = \text{Efficiency} \times \text{Time}\\[6pt]\textbf{Solution:}\\(16M + 12W) \times 45 = 1\\\frac{40M + 30W}{16M + 12W}= \frac{10(4M + 3W)}{4(4M + 3W)}= \frac{10}{4}= 2.5\\\text{Time for full work} = \frac{45}{2.5} = 18\\\text{Time for } 75\% = 18 \times \frac{3}{4} = 13.5\\[6pt]\textbf{Final Answer:}\\\boxed{13.5 \text{ days}}$$

Exam Hall Method:

​$$\text{Given:}\\A = 4,\;B = 6,\;C = 12\\[6pt]\text{Concept Used:}\\\text{Combined work using rates of pipes}\\[6pt]\text{Formula Used:}\\\text{Rate} = \frac{1}{\text{Time}}\\[6pt]\text{Solution:}\\\text{Rate of A} = \frac{1}{4}\\\text{Rate of B} = \frac{1}{6}\\\text{Rate of C} = \frac{1}{12}\\[6pt]\text{Combined rate}\\= \frac{1}{4} + \frac{1}{6} + \frac{1}{12}\\= \frac{3+2+1}{12}\\= \frac{6}{12}\\= \frac{1}{2}\\[6pt]\text{Time} = \frac{1}{1/2} = 2\\[6pt]\text{Final Answer:}\\\boxed{2 \text{ hours}}$$​​

Exam Hall Approach