Arithmetic

Mean, Median & Mode

Medium

There are three persons $\mathrm{A}, \mathrm{B}$ and $C$ in a room. If a person $D$ joins the room, the average weight of the persons in the room reduces by $x$ kg . Instead of $D$ , if person $E$ joins the room, the average weight of the persons in the room increases by $2 \mathrm{x} \mathrm{kg}$ . If the weight of $E$ is $12 \mathrm{~kg}$ more than that of $D$ , then the value of $x$ is

$2$

$1$

$0.5$

$1.5$

✅ Correct Option: 2

We have three people (A, B, C) in a room. When person D joins, the average weight decreases by x kg. When person E joins instead, the average weight increases by 2x kg. We also know E weighs 12 kg more than D.

Let's solve this systematically using the relationship between totals and averages.

Let the average weight of A, B, and C = a kg

Since average = total ÷ number of people:

A + B + C = 3a

This is our foundation - we'll use this total in all our calculations.

When D joins, there are 4 people and the average becomes (a - x) kg.

Using the average formula:

$\frac{A + B + C + D}{4} = a - x$

A + B + C + D = 4(a - x)

= 4a - 4x

Since A + B + C = 3a, we can substitute:

3a + D = 4a - 4x

D = 4a - 4x - 3a

= a - 4x

When E joins instead, there are 4 people and the average becomes (a + 2x) kg.

Using the average formula:

$\frac{A + B + C + E}{4} = a + 2x$

A + B + C + E = 4(a + 2x)

= 4a + 8x

Since A + B + C = 3a, we can substitute:

3a + E = 4a + 8x

E = 4a + 8x - 3a

= a + 8x

We're told that E weighs 12 kg more than D.

So: E - D = 12

(a + 8x) - (a - 4x) = 12

a + 8x - a + 4x = 12

12x = 12

x = 1

The key insight is that when we add someone to a group, the change in average tells us exactly how that person's weight compares to the original average:

D causes average to drop by x, so D weighs 4x less than the original average
E causes average to rise by 2x, so E weighs 8x more than the original average
The difference (12x) equals the given weight difference (12 kg)

Therefore, x = 1 kg

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