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Option 1 : Quantity A > Quantity B

Quantity A:

Suppose ‘s’km/hr is the speed of the car and ‘D’ km is the distance;

If he increases his speed by 8 km/hr, he takes 2 hours less time;

∴ D/s – D/(s + 8) = 2

⇒ D = /8      ----(1)

If he drives 8 km/hr slower, he takes 4 hours more time;

∴ D/(s – 8) – D/s = 4

⇒ D = /8      ----(2)

From equation (1) and (2);

/8 = /8

⇒ 2s + 16 = 4s – 32

⇒ 2s = 48

⇒ s = 24

∴ D = /8 = 192 km

Quantity B:

Suppose ‘s’ km/hr is the speed of the car and ‘D’ km is the distance;

If he increases his speed by 10 km/hr, he takes 1 hour less time;

∴ D/s – D/(s + 10) = 1

⇒ D = /10        ----(1)

If he further increases his speed by 5 km/hr, he takes further 20 minutes ( = 1/3 hours) lesser time;

∴ D/s – D/(s + 15) = (1 + 1/3)

⇒ D = /15      ----(2)

From equation (1) and (2);

/10 = /15

⇒ 9(s + 10) = 8(s + 15)

⇒ 9s + 90 = 8s + 120

⇒ s = 30

∴ D = /10 = 120 km

∴ Quantity A > Quantity B

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