Question 41

Four children A, B, C and D are having some chocolates each.
A gives B as many as he already has, he gives C twice of what C already has and he gives D thrice of what D already has.
Now, D gives (1/8)th of his own chocolates to B.
Then A gives 10% chocolates he now owns to C and 20% to B.
Finally, all of them have 35 chocolates each. What is the original number of chocolates each had in the beginning?

Solution

Let A, B, C, D have a, b, c, d chocolates respectively initially.

D gets thrice as many as he already has from A which makes a total of 4d chocolates with D.

He gives one-eigth of that to B which leaves D with 3.5d Chocolates.

Now 3.5 d = 35 => d = 10

Now, A has given b, 2c, 3d chocolates to B, C, D respectively at first which left him with (a-b-2c-3d) chocolates.

Then he gave 10% of the remaining chocolates to C and 20% to B.

He finally has 0.7(a-b-2c-3d) chocolates.

0.7(a-b-2c-3d) = 35

=>(a-b-2c-3d) = 50 = k (say)

C got 2c chocolates at first and then again 10% of 'k' from A.

C finally has 3c + 0.1k = 35

=> c = 10

B got b chocolates at first and then again 20% of 'k' from A and 0.5d from D.
B finally has 2b + 0.5d + 0.2k = 35

=> b = 10

For A, 0.7(a-b-2c-3d) = 35

=> a = 110


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