Question

The cake weighing 1.98kg is cut into two parts so that the first part is heavier than the second by 20 %, find the weight of each part.

Answer 1

let the heavier part of cake be X and left part of the cake Be (1.98-X)
according to question
X =[( 1.98-X)×120] ÷100
on solving the equation we get X = 1.08 kg therefore left part of the cake weighs 1.98- 1.08 kg = 0.90 kg

Answer 2

Answer:

Weight of first part is 1.08kg

weight of second part is 0.9 kg.

Given:

Weight of cake = 1.98 kg

First part is heavier than second by 20%.

To find:

weight of each part.

Step-by-step explanation:

Hint:

the percentage x of y is given by $\frac{x}{100} * y$.

let the weight of second part be x kg.

It is given that,

First part is heavier than second by 20%.

so,

the weight of first part = x + 20% of x

⇒ $x + \frac{20}{100} *x$

⇒ $x + \frac{x}{5}$

⇒ $\frac{6x}{5}$

it is also given that,

Weight of cake is 1.98 kg

∴ $x + \frac{6x}{5} = 1.98\\\frac{11x}{5} = 1.98\\11x = 1.98 * 5\\11x = 9.90\\x = \frac{9.90}{11}\\ x = 0.9$

so,

weight of second piece is x = 0.9 kg.

weight of second piece:

$\frac{6x}{5} = \frac{6*0.9}{5} = \frac{5.4}{5} = 1.08$ kg.

#SPJ2