CUET UG 2025 — General Test Geometry

2025mcqmedium

If the radius of a sphere is increased by 50%, find the percent increase in surface area.

Official previous-year question

Held on 23 May 2025 · Verified 13 Jul 2026.

Options

A
125%
B
250%
C
237.5%
D
220%

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Solution

Original radius = $r$

Radius increased by 50% means new radius = $r + 0.5r = 1.5r$

Surface area of a sphere = $4\pi r^2$

Original surface area:

$A_1 = 4\pi r^2$

New surface area with radius $1.5r$ :

$A_2 = 4\pi(1.5r)^2$

$A_2 = 4\pi \times 2.25 \times r^2$

$A_2 = 9\pi r^2$

Increase in surface area:

$\text{Increase} = 9\pi r^2 - 4\pi r^2$

$\text{Increase} = 5\pi r^2$

Percent increase:

$\text{Percent Increase} = \dfrac{\text{Increase}}{\text{Original}} \times 100$

$= \dfrac{5\pi r^2}{4\pi r^2} \times 100$

$= \dfrac{5}{4} \times 100$

$= 1.25 \times 100$

$= 125$

Therefore, the percent increase in surface area is 125%.

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Solution

Original radius = $r$

Radius increased by 50% means new radius = $r + 0.5r = 1.5r$

Surface area of a sphere = $4\pi r^2$

Original surface area:

$A_1 = 4\pi r^2$

New surface area with radius $1.5r$ :

$A_2 = 4\pi(1.5r)^2$

$A_2 = 4\pi \times 2.25 \times r^2$

$A_2 = 9\pi r^2$

Increase in surface area:

$\text{Increase} = 9\pi r^2 - 4\pi r^2$

$\text{Increase} = 5\pi r^2$

Percent increase:

$\text{Percent Increase} = \dfrac{\text{Increase}}{\text{Original}} \times 100$

$= \dfrac{5\pi r^2}{4\pi r^2} \times 100$

$= \dfrac{5}{4} \times 100$

$= 1.25 \times 100$

$= 125$

Therefore, the percent increase in surface area is 125%.

CUET UG 2025 — General Test Geometry

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CUET UG 2025 — General Test Geometry

Options

Did you get this right?

Solution

Your note

CUET UG Geometry in other years

More CUET UG Geometry questions

Other General Test topics for CUET UG

General Knowledge

Logical Reasoning

Arithmetic

Quantitative Reasoning

Current Affairs

Keep practicing Geometry