CUET UG 2025 — General Test Arithmetic

The class has 30 students who scored 61 or more marks and 39 students who scored below 61 marks. The average marks of the class is 48.

Total students $= 30 + 39 = 69$=30+39=69

Total marks of entire class $= 48 \times 69 = 3,312$=48×69=3,312

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Let the average marks of students who scored 61 or more be $A_1$A1​ and the average marks of students who scored below 61 be $A_2$A2​.

Total marks $=$= Marks of high scorers $+$+ Marks of low scorers

$$3,312 = 30 \times A_1 + 39 \times A_2$$3,312=30×A1​+39×A2​

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To maximize $A_2$A2​, the total marks of the high-scoring group must be minimized.

Since students in the high-scoring group scored 61 or more, the minimum possible value of $A_1 = 61$A1​=61.

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$$3,312 = 30 \times 61 + 39 \times A_2$$3,312=30×61+39×A2​

$$3,312 = 1,830 + 39 \times A_2$$3,312=1,830+39×A2​

$$1,482 = 39 \times A_2$$1,482=39×A2​

$$A_2 = \dfrac{1,482}{39}$$A2​=391,482​

$$A_2 = 38$$A2​=38

Therefore, the largest possible average marks of students who scored below 61 is $38$38 marks.