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

Answer: (a)

Let Average speed = x km/hr

Time taken = t hr

Distance = 14 km

⇒x = ${14}/t$ ...(1)

According to question,

Now Average speed = (x – 1) km/hr

Time taken = $(t + {20}/{60})hr = (t + 1/3)hr$

Distance = 14 km

⇒ x - 1 = ${14}/{t + 1/3}$

⇒${14}/t - 1 = {14}/{t + 1/3}$ (∵ x= ${14}/t$)

⇒${14 - t}/t = {42}/{3t + 1}$

⇒(14 – t) (3t + 1) = 42t

⇒42t – $3t^2$ + 14 – t = 42t

⇒ $–3t^2$ + 14 – t = 0

⇒(3t + 7)(t – 2) = 0

⇒t = 2 hr $(t ≠ - 7/3)$

⇒ x = ${14}/t = {14}/2$ Stat

x = 7 km/hr

∴ Option (c) is correct

Answer: (c)

Let average speed of flight = v

Time taken by flight (t) = ${600}/v$ ... (i)

Now, flight speed is reduced by 200 km/hr

= ${600}/{v - 200} = t + {30}/{60}$ ... (ii)

Now, put value of t in eqn (ii)

⇒${600}/{{600}/t - 200} = t + 1/2$

⇒${600t}/{600 - 200t} = t + 1/2$

⇒$600 t – 200t^2$ + 300 – 100t = 600 t

⇒$2t^2$ + t – 3 = 0

$t = {-1 ± √{1 + 24}}/{2 × 2} = {-1 ± 5}/4, {-6}/4, 4/4,$; t = 1 hour

Duration of flight = 1 hour

Answer: (e)

Speed of train A = $280/14$ = 20 meter/second

Length of train B = 20 × 35 – 280 meter

= 700 – 280 meter

= 420 meter

Answer: (b)

Let length of A train = 2l

length of B train = l

effective speed = (48 + 42) × $5/{18}$ m/s

Case I

${31}/{90 × {5/{18}}}$ = 12

l = 100 m

Length of A train = 200 m

Case II

let length of plat for m = P

${200 + P}/{48 × {5/{18}}}$ = 45

200 + P = 45 × ${48 × 5}/{18}$

l = 600 m

P = 400 m

Answer: (c)

Let speed of motorboat be B km/h.

Speed of water = D km/h.

According to question

B + D = $9/2$ = 4.5 ....(i)

B – D = $9/6 = 3/2$ = 1.5 .... (ii)

Now, on solving eqs. (i) and (ii), we get

B = 3 km/h and D = 1.5 km/h

speed of boat = 3km/h.