Question

Ken and ivan had some sweets. After Ken gave away
5/7 of his sweets, he had 2/3 as many sweets as ivan.
ivan then gave away 120 sweets and she had 1/4 as
many sweets as Ken. How many sweets did each of
them have at first?​

Answer 1

$Let \: number \: of \: sweets \: Ken \: have = x$

$number \: of \: sweets \: Ivan \: have = y$

/* After Ken gave away

After Ken gave away5/7 of his sweets, he had 2/3 as many sweets as ivan. */

$i) x - \frac{5x}{7} = \frac{2y}{3} \\\implies \frac{7x-5x}{7} = \frac{2y}{3} \\\implies \frac{2x}{7} = \frac{2y}{3}\\\implies \frac{x}{7} = \frac{y}{3}\\\implies x = \frac{7y}{3} \: --(1)$

/*ivan then gave away 120 sweets and she had 1/4 as

ivan then gave away 120 sweets and she had 1/4 asmany sweets as Ken */

$ii) \frac{x}{4} = y - 120$

$\implies x = 4(y-120)$

$\implies x = 4y- 480 \--(2)$

$Put \: x = \frac{7y}{3} \: in \: equation \: (2) , we \:get$

$\implies \frac{7y}{3} = 4y - 480$

$\implies 7y = 3(4y-480)$

$\implies 7y = 12y- 1440$

$\implies 1440 = 12y - 7y$

$\implies 1440 = 5y$

$\implies 5y = 1440$

$\implies y = \frac{1440}{5}$

$\implies y = 288$

/* Now, put y = 288 in equation (1), we get */

$x = \frac{7}{3} \times 288$

$\implies x = 7 \times 96$

$\implies x = 672$

Therefore.,

$\red{ number \: of \: sweets \: Ken \: have} \green { = 672 }$

$\red{number \: of \: sweets \: Ivan \: have}\\\green { = 288}$

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