time and work Model Questions & Answers, Practice Test for ibps rrb clerk asst multipurpose prelims 2024

Answer: (a)

Part of the capacity of the cistern emptied by the leak in one hour

= $(1/6 - 1/7) = 1/{42}$ of the cistern.

The whole cistern will be emptied in 42 hours.

Answer: (b)

6M + 8B = 10 days ...(i)

26M + 48B = 2 days ... (ii)

15M + 20 B = ?

By to formula⇒20B = 10 days

$M_1 D_1 = M_2 D_2$⇒(6M + 8B) × 10 = (26M + 48B) × 2

⇒60M + 80B = 52M + 96B⇒8M = 16B

M = 2B

Now in eq. (i), put M = 2B

6 × 2B + 8B = 10 days

12B + 8B = 10 days

Again

15M + 20B = 15 × 2B + 20B = 30B + 20B = 50B

From formula, $M_1 D_1 = M_2 D_2$

⇒20 × 10 = 50 × $D_2$

$D_2 = {20 × 10}/{50} = D_2$ = 4 days.

Answer: (a)

[(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.

Answer: (b)

Let h be the length of water column discharged in 1 hour or 1 minute.

Volume discharged by the 4 pipes = Volume discharged by the single pipe.

4 × π × $(1.5)^2 × h = π × (r)^2$ × h

∴ $r^2$ = 9

∴ r = 3, Diameter = 6 cm.

Answer: (d)

Given that

1 Goats = $3/2$ sheeps.

Now, 2 goats + 9 sheeps

= 2 × $3/2$ sheeps + 9 sheeps = 12 sheeps

Here $M_1D_1 = M_2 D_2$

⇒6 × 50 = 12 × $D_2$

$D_2$ = 25 days