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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’?

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’? Correct Answer 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.

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