time and distance Model Questions & Answers, Practice Test for ssc cgl tier 1 2024

Answer: (d)

Let time taken by passenger train = t

time taken by express train = t + 3

when distance = 540

ATQ

${540}/t - {540}/{t + 3}$ = 15 ∵ S = $D/T$

540 $[{t + 3 - t}/{t^2 + 3t}]$ = 15

108 = $t^2$ + 3t

$t^2$ + 3t – 108 = 0

$t^2$ + 12t – 9t – 108

(t – 9) (t + 12)

t = 9 hr

Express train takes 9 hr

9pm + 9 hr = 6 am

Answer: (c)

Let total distance = x km.

Average speed = ${\text"total distance"}/{\text"total time"}$

$x/{t_1 + t_2 + t_3} = x/{{x_1}/{v_1} + {x_2}/{v_2} + {x_3}/{v_3}}$

$x_1 = x/3, x_2 = x/3, x_3 = x/3 = x/{x/3(1/{10} + 1/{20} + 1/{60})}$

= $3/{({6 + 3 + 1}/{60})} = {3 × 60}/{10}$ = 18km/h

Answer: (b)

Let the length of the journey be x km.

Suppose speed of the train be y km/h.

∴ Time taken to cover x km = $x/y$ hours

∴ $x/{y+6}=x/y-4,x/{y-6}=x/y+6$

Solving these equations, we get

y = 30, x = 720.

∴ Length of the journey = 720 km.

Answer: (d)

Let usual speed = v

increased speed = v + 5

Now, ${300}/v - {300}/{v + 5}$ = 2

${300(v + 5) –300v}/{v(v + 5)}$ = 2

⇒$2v^2$ + 10v – 1500 = 0

⇒(v + 30)(2v – 50) = 0

∴ v = –30, 25

∴ v = 25 km/hr {since V is always + ve value}

Answer: (c)

Relative velocity = (50 – 32) km/h = 18 km/h

Elapse time = 15s = 15 × $1/{60} × 1/{60}h = {15}/{3600}h$

Time = ${\text"Distance"}/{\text"Speed"}⇒{15}/{3600} = x/{18}$

⇒x × 3600 = 18 × 15

∴x = ${18 × 15}/{3600}km = {18 × 15}/{3600}$ × 1000m = 75 m.