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NIMCET 2008 Question Paper with solutions

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Question : 65 of 120

Marks: +1, -0

A woman took a certain number of eggs to the market and sold some of them. The next day through her poultry industry, the number left over had been doubled, and she sold the same number as the previous day. On the third day, the new remainder was tripled, and she sold the same number as before. On the fourth day, the remainder was quadrupled, and her sale were the same as before. On the fifth day, what had been left over were quintupled, yet she sold exactly the same as on all the previous occasions and so disposed of her entire stock. What is the smallest number of eggs she could have taken to the market the first day, and how many did she sell daily?

110.50
127.65
100.60
103.60

Solution:

If

y

was the total number of eggs and

x

is the daily sale.

Then,

Left over egg on day

1=y-x

Left over egg on day

2=2(y-x)-x=2 y-3 x

Left over egg on day

3=3(2 y-3 x)-x

=6 y-10 x

Left over egg on day

4=4(6 y-10 x)-x

=24 y-41 x

Left over egg or say total eqgs on day 5

=5(24 y-41 x)=x

\Rightarrow \;\; 120 y-205 x=x

\Rightarrow \;\; 120 y=206 x

\Rightarrow \;\; 60 y=103 x

\Rightarrow y=103

and

x=60

as minimum values.

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