time and work Model Questions & Answers, Practice Test for ssc cgl tier 1 2023
ssc cgl tier 1 2023 SYLLABUS WISE SUBJECTS MCQs
Number System
Ratio Proportion & Partnership
Averages
Percentages
Profit & Loss
Time & Work
Time & Distance
Simple Interest & Compound Interest
Mensuration: Area & Volumes
Algebraic Expressions
Linear Equations
Squareroots & Cuberoots
Two taps can fill a tank in 12 and 18 minutes respectively. Both are kept open for 2 minutes and the first is turned off. In how many minutes more will the tank be filled ?
Answer: (d)
Part filled by first tap in one min = $1/{12}$th
Part filled by second tap in one min = $1/{18}$th
Now, $2[1/{12} + 1/{18}]$ unfilled part = 1
⇒ unfilled part = ${13}/{18}$th
∵$1/{18}$th part of tank is filled by second tap in 1min.
∴${13}/{18}$th part of tank is filled by second tap in 1 min.
= 18 × ${13}/{18}$ min = 13 min.
A contractor undertakes to do a piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the additional men, how many days behind schedule would it be finished ?
Answer: (c)
[(100 × 35) + (200 × 5)] men can finish the work in 1 day.
∴ 4500 men can finish the work in 1 day. 100 men can finish it in ${4500}/{100}$ = 45 days.
This is 5 days behind schedule.
A and B can do a job in 16 days and 12 days respectively. 4 days before finishing the job, A joins B. B has started the work alone. Find how many days B has worked alone ?
Answer: (b)
A's one day's work = $1/{16}$ th work
B's one day's work $1/{12}$ th work
Let B has worked alone = x days. Then,
A's amount of work + B's amount of work = 1
⇒ $4(1/{16}) + (x + 4)(1/{12}) = 1$
⇒ $1/4 + {x + 4}/{12} = 1⇒x = 3/4 × 12 - 4$
⇒ x = 5 days
A train 100 metres long takes 3$3/5$ seconds to cross a man walking at the rate of 6 km/h in a direction opposite to that of train. Find the speed of the train.
Answer: (a)
Let speed of train be S km/h.
Speed of train relative to man = [S - (-6)] km/h
= (S + 6) × $5/{18}$ m/s
Now (S + 6) × $5/{18} = {100}/{18/5}$
⇒S = 94 m/s
A and B can do a piece of work in 40 days. After working for 10 days they are assisted by 'C' and work is finished in 20 days more. If 'C' does as much work as B does in 3 days, in how many days A alone can do the work.
Answer: (c)
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