IPMAT Indore 2023 - QA (SA) Q1 Explanation

Explanation:

(312) Let the speed of car in gear1, gear2, gear 3, and gear 4, be s1, s2, s3 and s4.

Given that, s4: s1 = 5:1

So let s4 = 5x and s1 = x

Also, t2: t3 = 2 ∶ 1(where t2 & t1, be the time taken by car to travel that certain distance)

∴ We can say, s2: s3 = 1/2 ∶ 1/1 = 1 ∶ 2

Let s2 = y and s3 = 2y

According to the question, t1 +t2 +t3 +t4 = 585 min.

25/x + 25/y + 25/y + 25/x = 525 min {time = distance/speed}

= 6/5x + 3/2y = 585/25 .............(1)

We have to find the total time, when Vinita travels 4 km in gear 1, 4 km in gear 2, 32 km in gear 3 and 60km in gear 4.

i.e. the value of 4/x + 4/y + 32/2y + 60/5x = total time required

4/x + 4/y +16/y + +12/x = ?

or 16/x +20/y =? = ?

Multiply eqn. (1) by 10, we get 12/x + 15/y = 5850/25 = 234 min. ---(2)

Multiply eqn. (2) by 4/3, we get 16/x + 20/y = 234 × 4/x = 78 × 4 = 312min. Ans.