Sign in
Please select an account to continue using cracku.in
↓ →
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?
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
Was this solution helpful?
Create a FREE account and get:
Book Free CAT Mentorship
Get personalized CAT strategy from a 99%iler
500+ students mentored
OTP Verification
Enter the 6-digit code sent to your phone
Booking Summary
Enter OTP
Didn't receive the OTP?
Start your IIM journey with the right preparation and crack CAT 2026.
Educational materials for CAT preparation