[144] In the previous question, what is the sum of the number of biscuits hidden by the last 2 men?

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Aptitude

TNPSC G4

10th CBSE Maths

In the previous question, what is the sum of the number of biscuits hidden by the last 2 men?

142

144

180

178

Explanation:

Suppose N=5x+1
A took (x+1) biscuit.

Now 4x is of the form 5y+1 then x must be in the form 5z+4
⇒4(5z+4)=5y+1
⇒y=4z+3 and x=5z+4
The ratio of number of biscuits that A and B took is

[(5z+4)+1]:[(4z+3)+1]=5:4
So, we can say that any two successive persons A, B, C and D take coins in the ratio of 5:4

Let the number of biscuits that A, B, C and D took be a, b, c and d respectively.

a:b=b:c=c:d=5:4
a:b:c:d=125:100:80:64

⇒a=125k
⇒x=125k−1 and N=5x+1=625k−4
⇒N<100, then k=1
⇒N=621
⇒621=(5×124)+3
4×124=(5×99)+1
4×99=(5×79)+1
4×79=(5×63)+1
The number of biscuits hidden by 3rd and the 4th men is 79+1=80 and 63+1=64
i.e. A total of 144.

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