In an election, there were two candidates A and B. The total number of voters in the constituency was 90000 and 70% of the total votes were polled. If 60% of the polled votes were cast in favour of A, how many votes were received by B?

Correct option is B

Given:
Total number of voters (N) = 90000
Percentage of votes polled = 70%
Percentage of votes in favor of candidate A = 60%
Formula Used:
$$\text{Votes} = \frac{\text{Percentage of Votes} \times \text{Total Number of Voters}}{100}$$​​
Solution:
Number of votes polled = $$\frac{70}{100}$$​ × 90000 = 63000
Number of votes in favor of candidate A = $$\frac{60}{100}$$​× 63000 = 37800
Since there were only two candidates, the remaining votes would be in favor of candidate B.
Number of votes received by candidate B = Total number of votes polled - Number of votes in favor of candidate A
Number of votes received by candidate B = 63000 - 37800 = 25200

Alternate Solution:

Total voters = 90,000
Polled votes = $$70\% \times 90000 = \frac{70}{100} \times 90000 = 63000$$​​
B received = $$(100 - 60)\% \times 63000 = \frac{40}{100} \times 63000 = 25200$$​​