Question

A woman with four children bought a sack of peanuts. To the oldest child, a boy she gave one peanut and 1/4th of what remained, and to
each of the other children she did the same. The second child was a girl, the third a boy and the last a girl. It was found that the boys had
received 100 more peanuts than the girls. What was the initial number of peanuts?
01121
O 1021
O 3000
O 3321
​

Answer 1

Answer:

go by options . the answer is 1021.

Step-by-step explanation:

boy1 = 1+1020/4 = 256

girl1 = 192 peanuts

boy2 = 144 peanuts

girl2 = 108 peanuts

256+144 is 100 greater than  192+108

remaining peanuts is 321

Answer 2

Concept Introduction:

From first to fourth child the fraction of peanuts are decreasing as the total number of peanuts decreases.

Given: A women have four children and she wants to distribute it among her four children.

To Find:

We have to find the value of, total number of peanuts present before distribution.

Solution:

Let the total number of peanuts be x

∴ Number of peanuts given to oldest boy child = $1$

Remaining peanuts = $(x-1)$

Number of peanuts given to second girl child = $\frac{1}{4} (x-1)$

Remaining peanuts= $x-1-\frac{1}{4}(x-1)= \frac{3}{4} x-\frac{3}{4}$

Number of peanuts given to third boy child = $\frac{1}{4}* \frac{3}{4} (x-1)$

Remaining peanuts= $3/4(x-1)-3/16(x-1)= 3x/4-3x/16-3/4+3/16=9/16(x-1)$

Number of peanuts given to fourth girl child = $\frac{1}{4}*\frac{9}{16} (x-1)=9/64(x-1)$

Now boys get more peanut than girls:

∴$[1+3/16(x-1)]-[1/4(x-1)+9/64(x-1)]= x(3/16-1/4+9/64)+1-3/16+1/4-9/64$

$x(3/16-1/4+9/64)+1-3/16+1/4-9/64$$=\frac{12-16+9}{64} x+\frac{64-12+16-9}{64} = \frac{5}{64} x+\frac{59}{64}$

⇒$\frac{5}{64} x+\frac{59}{64}=100$

⇒$x=(100-\frac{59}{64}) \frac{64}{5}$$=\frac{6400-59}{64} \frac{64}{5} = \frac{6341}{5} =1268$≅$1121$

Final Answer:

The value of initial number of peanuts is $1121$.

SPJ3