Question

Weight of A and B are in the ratio of 3:5. If the weight of A is increased by 20 percent and then the total weight becomes 132 kg with an increase of 10 percent. B weight is increased by what percent.

Answer 1

Let the total weight of A and B be x kg
Then,

125%of x =80

X=80*100/125=64kg


A’s initial weight

=(64*3/8)kg=24kg

B’s initial weight
=(64*5/8)kg=40kg


A’s new weight

=120%of24kg=28.8kg

B’s new weight

=(80-28.8kg=51.2kg

Increase in B’s weight

=(51.2-40)kg=11.2kg


Therefore increase%

=(11.2/40*100)%= 28%

Answer 2

B weight is increased by 4% percent.

Step-by-step explanation:

Consider the provided information.

Weight of A and B are in the ratio of 3:5.

Therefore, Weight of A and B are 3x and 5x.

The total weight becomes 132 kg with an increase of 10 percent.

Let x is the original weight.

$x=\frac{132\times100}{110}=120$

The total weight is: 3x+5x=8x

$8x = 120\\x = 15$

Thus, the Initial weight of A and B are 3x=45 and 5x=75 kg respectively.

Weight of A is increase by 20%.

$45\times\frac{120}{100}=54$

New weight of A = 54

So weight of B = 132 – 54 = 78.

$\% increase = \frac{78-75}{75}\times100 = 4 \%$

Hence, B weight is increased by 4% percent.

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