In a race of 1600 m, if the ratio of the speeds of two participants A and C is 4:3, and C has a start of 240 m, then A beats C by:

200 m

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160 m

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180 m

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220 m

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250 m

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Solution

The correct option is B
160 m

As C has a start of 240m, C has to cover 1360m while A has to cover 1600m.

Now, when C covers 3m, A covers 4m.

Hence, when A covers 1600m, C covers $\frac{1600 \times 3}{4}$ = 1200m

Therefore, A beats C by (1600-1200-240) = 160m.

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In a race of 1600 m, if the ratio of the speeds of two participants A and C is 4:3, and C has a start of 240 m, then A beats C by:

The correct option is B 160 m As C has a start of 240m, C has to cover 1360m while A has to cover 1600m. Now, when C covers 3m, A covers 4m. Hence, when A covers 1600m, C covers 1600×34 = 1200m Therefore, A beats C by (1600-1200-240) = 160m.

The correct option is B
160 m

As C has a start of 240m, C has to cover 1360m while A has to cover 1600m.

Now, when C covers 3m, A covers 4m.

Hence, when A covers 1600m, C covers $\frac{1600 \times 3}{4}$ = 1200m

Therefore, A beats C by (1600-1200-240) = 160m.