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

Answer: (b)

Radius of the wheel = 2.1 m

Distance covered in 1 revolution = 2πr

= 2 × ${22}/7$ × 2.1

∴ Distance covered in 75 revolutions

= 2πr × 75 = 2 × ${22}/7$ × 2.1 × 75

= 990 m = 0.99 km

Time = $1/{60}$ h

∴ Required speed = ${\text"Distance"}/{\text"Time"} = {0.99}/{1/{60}}$ km/h

59.4 km/h

Answer: (a)

Distance = 64 × 8

= 512 km

∴ Speed = $512/6$

= 85 km/hr (approx.)

Answer: (d)

Let usual speed of train = 4 km/hr

Time taken by train = t hr

Distance travelled = 120 km

So, u = ${120}/t⇒t = {120}/4$ hr ...(1)

According to question

New speed = (u + 10) km/hr.

Distance = 120 km

Time taken = (t – 1) hr

So u + 10 = ${120}/{t - 1}$

(using (1)

u + 10 = ${120}/{{120}/4 - 1} = {120u}/{120 - u}$

⇒(u + 10) (120 – 4) = 1204⇒1204 + 1200 – $u^2$ – 104

= 1204

⇒$u^2$ + 10 u – 1200 = 0

⇒(u + 40) (u – 30) = 0⇒u = – 40 is not possible

⇒u = 30 km/hr

∴ Option (d) is correct.

Answer: (a)

Let the speed of bike = v km/h

∴ Time taken to cover 200 km at a speed of v km/

h = ${200}/v$h

New speed of bike = (v + 5) km/h

∴ Time taken to cover 200 km at a speed of

(v + 5) km/h = ${200}/{v + 5}$

According to question, ${200}/v - {200}/{v + 5}$ = 2

⇒${(v + 5 - v)200}/{v^2 + 5v} = 2 ⇒500 v^2 + 5v$

$⇒v^2 + 5v - 500 = 0⇒v^2 + 25v - 20v - 500 = 0$

⇒v (v + 25) – 20 (v + 25) = 0

⇒(v – 20) (v + 25) = 0 (∵v ≠ -25)

∴ = v 20km / h

So the original speed of a bike is 20 km/h

Answer: (c)

Let speed of passenger train be x km/h and speed of goods train be y km/h

Speed in same direction = x – y km/h

Speed in opposite direction = (x + y) km/h

Let total length of trains be 100 m

According to the question

${100}/{x - y} = ({100}/{x + y})3$

${100}/{x - y} = {300}/{x + y}$

⇒100x + 100y = 300x – 300y

⇒200x = 400y

∴ x : y = 2 : 1