CUET-General Aptitude-14/05/2025
Dear Students, we welcome you in our post CUET-General Aptitude-14/05/2025.
Today’s post is a Test Paper as below:
Q 3755)
(A) 24 (B) 28 (C) 30 (D) 36
Ans: (C) 30
Solution:
Proceed clockwise
9+3=12
12+(3×2)=18
18+(6×2)=30
30+(12×2)=54
54+(24×2)=102
Q 3756) The smallest natural number by which 135 will be divided to get a perfect cube is —
(A) 2 (B) 3 (C) 4 (D) 5
Ans: (D) 5
Solution: When we divide 135 by 5, we get 27, which is a perfect cube.
Q 3757) ———— is the smallest natural number by which 392 must be multiplied so that the product will be a perfect cube
(A) 5 (B) 6 (C) 7 (D) 8
Ans: (C) 7
Solution: 392 = 2 x 2 x 2 x 7 x 7
The smallest natural number 7 does not appear in group of three.
Q 3758) Each prime factor appears ——- times in its cube.
(A) 2 (B) 3 (C) 4 (D) 5
Ans: (B) 3
Q 3759) ———— is the smallest natural number by which 243 must be multiplied so that the product will be a perfect cube
(A) 2 (B) 3 (C) 4 (D) 5
Ans: (B) 3
Solution: 243 = 3 x 3 x 3 x 3 x 3
The smallest natural number 3 does not appear in group of three.
Q 3760) If the volume of a cube is 64 cm3 then what would be the length of its side ?
(A) 2 cm (B) 3 cm (C) 4 cm (D) 5 cm
Ans: (C) 4 cm
Solution: Volume of cube is 64 cm3
Therefore side length of cube = 4 cm (4 x 4 x 4 =64)
Q 3761) If the volume of a cube is 64 cm3 then the diagonal of cube is ?
(A) 4 cm (B) ∛4 cm (C) √4 cm (D) None
Ans: (B) ∛4 cm
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General Aptitude
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