time and distance Model Questions & Answers, Practice Test for ssc steno grade c d 2024

Answer: (b)

Let distance between A and B = x km

According to question

$x/{12} - x/{18} = 2$⇒6x = 2 × 18 × 12

⇒x = ${2 × 18 × 12}/6$ = 72 km

required distance = 72 km

Answer: (c)

Time taken by the 1st athlete

= ${4000}/{16}$ = 250 unit

Distance covered by 2nd Athlete

= 15 × 250 = 3750 m

Distance covered by 3rd Athlete

= 11 × 250 = 2750 m

Distance between 2nd and 3rd Athlete

= 3750 – 2750 = 1000 m

So, option (a) is correct.

Answer: (b)

In a race of 100 m

when A covers total 100 m.

B will cover = 100 – 4 = 96 m

C will cover = 100 – 2 = 98 m

When C covers 98 m, B covers = 96 m

When C covers 100 m, B will cover

= ${96}/{98} × 100 = {4800}/{49}$ m

So C will beat B by = 100 - ${4800}/{49}$

⇒${4900 - 4800}/{49} = {100}/{49}$ = 2.04 m

So, option (b) is correct

Answer: (b)

Let Initial speed = v and

total distance = x km.

Time taken by car = $x/v$ = 8

x = 8v ... (i)

Now, speed is increased by 4 km/hr

⇒$x/{v + 4} = 7 + {30}/{60}⇒x/{v + 4} = 7 + 1/2 = {15}/2$

⇒2x = 15v + 60

Put value of x from equation (i)

⇒16v = 15v + 60; v = 60

Distance covered = 8 × 60 = 480 km

Answer: (b)

Train and man running opposite to each other.

∴ Relative speed = 60 + 6 = 66 km/h

= ${66 × 5}/{18}$ m/s

∴ Required time = ${\text"Distance"}/{\text"Speed"}$

= ${110}/{{66 × 5}/{18}} = {110 × 18}/{66 × 5}$ = 6s