Question

Richie read 4/9th of a book on one day and 3/5th of the remaining the next day . If 100 pages of the book were left unread . How many pages did the book contain ??


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Answer 1

Answer:

☆Let the number of pages the book contained be x.

♧Part read in 1st day=4x/9

♤Remainder=x-4x/9=9x/9 - 4x/9=5x/9

♡Part read in next day=5x/9 × 3x/5=x/3

◇Remainder=x-x/3=3x/3-x/3=2x/3

ACQ,

》》2x/3=100

》》x=100×3/2=50×3=150 pages

Answer 2

Let the total no. of pages in the book be $x.$

(4/9)th of a book is read on first day.

So no. of pages read on this day $=\dfrac{4x}{9}.$

And no. of pages remaining $=x-\dfrac{4x}{9}=\dfrac{5x}{9}.$

(3/5)th part of the remaining is read on next day.

No. of pages read on this day $=\dfrac{3}{5}\times\dfrac{5x}{9}=\dfrac{x}{3}.$

And no. of pages remaining unread $=x-\dfrac{x}{3}=\dfrac{2x}{3}.$

These remaining pages are those 100 pages of the book which were left unread.

$\longrightarrow\dfrac{2x}{3}=100$

$\longrightarrow x=100\times\dfrac{3}{2}$

$\longrightarrow\underline{\underline{x=150}}$

Hence the book contains a total of 150 pages.