time and work Model Questions & Answers, Practice Test for ssc cgl tier 1 2023

Question :21

A thief running at 8 km/hr is chased by a policeman whose speed is 10 km/hr. If the thief is 100 m ahead of the policeman, then the time required for the policeman to catch the thief will be

Answer: (c)

If we consider the difference of speeds, policeman is 2km/hr leading speed and he can catch the thief at 100m ahead by $({100 m}/{2 kmph}) = {100}/{1000 × 2}$ × 60 = 3 min.

Question :22

24 men working 8 hours a day can finish a work in 10 days. Working at the rate of 10 hours a day, the number of men required to finish the same work in 6 days is :

Answer: (c)

$m_1 × d_1 × t_1 × w_2 = m_2 × d_2 × t_2 × w_1$

24 × 10 × 8 × 1 = $m_2$ × 6 × 10 × 1

⇒$m_2 = {24 × 10 × 8}/{6 × 10}$ = 32men

Question :23

10 men can complete a piece of work in 15 days and 15 women can complete the same work in 12 days. If all the 10 men and 15 women work together, in how many days will the work get completed ?

Answer: (b)

10 men's 1 day's work = $1/{15}$

15 women's 1 day's work = $1/{12}$

(10 men + 15 women)'s 1 day's work

= $(1/{15} + 1/{12}) = 9/{60} = 3/{20}$

∴ 10 men and 15 women will complete the work in

${20}/3 = 6 2/3$ days.

Question :24

Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A , B and C discharge chemical solutions P, Q and R respectively. What is the proportion of solution R in the liquid in the tank after 3 minutes ?

Answer: (c)

Part filled by (A + B + C) in 3 minutes

= 3$(1/{30} + 1/{20} + 1/{10}) = (3 × {11}/{60}) = {11}/{20}$

Part filled by C in 3 minutes = $3/{10}$

∴ Required ratio = $(3/{10} × {20}/{11}) = 6/{11}$

Question :25

The speed of a car increases by 2 km/h after every one hour. If the distance travelled in the first one hour was 35 kms, what was the total distance travelled in 12 hours ?

Answer: (b)

Total distance travelled in 12 hours

= (35 + 37 + 39 + ..... upto 12 terms)

This is an A.P. with first term a = 35,

number of terms n = 12, common difference d = 2.

∴ Required distance

= ${12}/2$[2 × 35 + (12 - 1) × 2] = 6(70 + 22) = 552km.

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