A packet of sweets is distributed among A, B, C, D in the proportion of 6:8:5:4. If B gets 10 sweets more than D then what is A’s share?Options :16171518
Question
A packet of sweets is distributed among A, B, C, D in the proportion of 6:8:5:4. If B gets 10 sweets more than D then what is A’s share?Options :16171518
Solution
To solve this problem, we first need to understand the proportions in which the sweets are distributed.
The proportions are A:B:C:D = 6:8:5:4.
The total proportion is 6+8+5+4 = 23 parts.
We know that B gets 10 sweets more than D. So, the difference in their proportions (8-4 = 4 parts) is equal to 10 sweets.
Therefore, 1 part is equal to 10/4 = 2.5 sweets.
A's share is 6 parts, so A gets 6 * 2.5 = 15 sweets.
So, the answer is 15.
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