"> Word Problem 12 | Job Exam Rare Mathematics (JEMrare)

Word Problem 12

Problem 12: A lady gave away 1/2 of her stock of oranges she had and half the extra orange to the first customer. Half of her remaining stock and half the orange extra she gave to the second customer. She is left with 15 oranges. The number of oranges she had at first was;

….  numbers math —

a.67                b.63                c.65                          d.70

Solution:

Here we will have to notice and take care of  the words ‘oranges’ and ‘ half the extra orange’

So we start by given information;

First customer =1/2 of the oranges+1/2 orange

Second customer = 1/2 of the remaining oranges +1/2 orange

Remaining orange=15

Original=?

Let the lady have ‘y’ oranges

First Customer=\frac{y}{2}+\frac{1}{2}

Remaining=y-(\frac{y}{2}+\frac{1}{2})

\Rightarrow Remaining = \frac{y}{2}-\frac{1}{2}

Coming to second customer;

Second Customer = \frac{1}{2}(\frac{y}{2}-\frac{1}{2})+\frac{1}{2}

\Rightarrow Second Customer = \frac{y}{4}+\frac{1}{4}

So, total oranges she gave away can be calculated by taking some of oranges given to both customers;

Total Oranges Given=(\frac{y}{2}+\frac{1}{2})+(\frac{1}{4}+\frac{1}{2})

Total Oranges Given=(\frac{y+1}{2}+\frac{y+1}{4})=\frac{3(y+1)}{4}

Again remaining oranges = total – given away

\Rightarrow remaining = y- \frac{3(y+1)}{4} =\frac{y-3}{4}

But remaining oranges are 15

so,

\Rightarrow 15 =\frac{y-3}{4}

\Rightarrow y = 63


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