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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

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 : 1617 15 18

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Solution

The total ratio of the sweets distributed among A, B, C, and D is 6:8:5:4.

Step 1: Add up all the parts of the ratio. 6 (for A) + 8 (for B) + 5 (for C) + 4 (for D) = 23 parts in total.

Step 2: We know that B gets 10 sweets more than D. The ratio tells us that B gets 8 parts and D gets 4 parts. So, the difference in their parts (8 - 4 = 4 parts) is equal to 10 sweets.

Step 3: Now, we can find out how many sweets each part of the ratio represents. 10 sweets / 4 parts = 2.5 sweets per part.

Step 4: To find out A's share, we multiply the number of parts A gets (6) by the number of sweets each part represents (2.5). 6 parts * 2.5 sweets/part = 15 sweets.

So, A's share is 15 sweets.

This problem has been solved

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