Ten children are standing in a line. Each child has some chocolates with him. If the first child attempted to double the number of chocolates with each of the others he would fall short by two chocolates. If the second child took two chocolates from each of the remaining, he would have three chocolates less than what the first child initially had. Find the total number of chocolates with the third to the tenth child.

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answer is C.

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Detailed Solution

Let the first child have x number of chocolates, the first child has y number of chocolates and the sum of the number of chocolates with the remaining children is z.
Since, the first child will try to double up all other chocolates i.e., he will give y number of chocolates to the second and z
number of chocolates to the remaining children.
And the first child will fall short of 2 chocolates.
Now, we will form the equation with the above condition.
⇒ x – y – z = − 2 ………. (1)
The second condition is that when the second child took two chocolates from the remaining children, he got 3 chocolates less than the number of chocolates of the first child.
The number of chocolates taken by the second child is equal to 2 × 9 = 18.
Therefore, the equation formed will be
⇒x−y−18=3 ………… (2)
Subtracting equation (2) from equation (1), we get
- y – z = − 2
- x ∓ y ∓ 18 = − 3
− z + 18 = − 5.
Now we got the equation as− z + 18 = − 5.
On adding and subtracting the terms, we get
⇒ z = 23
Therefore, the total number of chocolates with the third to the tenth child is equal to 23.

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