Question 68

A number is first decreased by 30% and then increased by 30%. If the number so obtained is 72 less than the original number, then what is the value of the original number?

Solution

Let the original number = $$100x$$

If number is first decreased by 30%, then number = $$100x-(\frac{30}{100}\times100x)$$

= $$100x-30x=70x$$

Now, the number is again increased by 30%, number = $$70x+(\frac{30}{100}\times70x)$$

= $$70x+21x=91x$$

According to ques,

=> $$100x-91x=72$$

=> $$9x=72$$

=> $$x=\frac{72}{9}=8$$

$$\therefore$$ Original number = $$100\times8=800$$

=> Ans - (B)

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