Question

If Rati travels at his usual speed for ‘x’ hours, he covers the same distance as he covered when he travels at a speed 30 km/hr less than usual, but for (x + 3) hours. Which of the following can be determined?

(A) If he travels at 20 km/hr less speed than usual, he can cover a distance of 240 km in (x - 1) hours. What is the value of ‘x’?

(B) If travels at his usual speed for (t + 2) hours, and at 50% of his usual speed for (t - 2) hours, his average speed for the journey is 68 km/hr. What is his usual speed?

(C) His usual speed (in km/hr) is a multiple of 5 between 40 and 50 (both inclusive). If ‘x’ is a whole number, what is his usual speed?

(D) If he travels at his usual speed for ‘t’ hours, he covers a distance of 400 km. What is the value of ‘x’?

1. Only A, B and D
2. All A, B, C and D
3. Only B and D
4. Only A and D
5. Only B and C

Answer 1

Correct Answer - Option 1 : Only A, B and D

Let the usual speed of Rati be ‘x’ km/hr

So, x × t = (x – 30) × (t + 3)

xt = xt + 3x – 30t – 90

x – 10t = 3      ----(i)

(A) : Now, (x – 20) (t – 1) = 240

xt – x – 20t + 20 = 240

xt – x – 20t = 220

(10t + 30) × t – (10t + 30) – 20t = 220

10t² + 30t – 10t – 30 – 20t = 220

10t² = 250

T = 5

So, the value of ‘t’ is determined.

(B) : So, total distance travelled = x(t + 2) + (x/2) × (t – 2) = (3xt/2 + x) km

Total time taken = (t + 2) + (t – 2) = 2t

So, (3xt/2 + x)/2t = 68

3t(10t + 30)/2 + (10t + 30) = 136t

15t² + 45t + 10t + 30 = 136t

15t² – 81t + 30 = 0

5t² – 27t + 10 = 0

(5t – 2) (t – 5) = 0

So, t = 2/5 or 5 but t = 2/5 is not possible

For t = 5, x = 80

So, the value of ‘x’ can be determine.

(C) : Multiples of 5 between 40 and 50 are 40, 45 and 50

For x = 40, t = 1

For x = 45, t = 1.5

For x = 50, t = 2

Since, ‘t’ is a whole number, t = 1 or 2

So, the value of ‘x’ cannot be determined.

(D) : So, xt = 400

T = 400/x

Putting in (i)

x – 400/x = 30

x² – 30x – 400 = 0

(x – 40) (x + 10) = 0

x = 40

So, the value of ‘x’ is determined.