CUET-General Aptitude-03/04/2025

CUET-General Aptitude-03/04/2025

Dear Students, we welcome you in our post CUET-General Aptitude-03/04/2025.

Our today’s post is on percentage as below:

Q 3562) 40% of 1620 + 30% of 960 = ?% of 5200

(A) 12%    (B) 16%    (C) 18%    (D) 24%

Ans: (C) 18%

Solution:

40/100×1620 + 30/100×960 = ?/100×5200

648+288 = ?x52

936=?x52

?=936/52=18%

Q 3563) If 90% of a number is 621. What will be 50% of 20% of that number?

(A) 54    (B) 69    (C) 79    (D) 84

Ans: (B) 69

Solution:

90% of a number is 621

1% of that number will be 621/90

100% of that number will be 621/90×100=690

20% of 690= 20/100×690=138

50% of 138= 50/100×138= 69

Q 3564) If 80% of a number is 768. What will be 40% of 5% of that number?

(A) 17.6    (B) 18.4    (C) 19.2    (D) 20

Ans: (C) 19.2

Solution:

80% of a number is 768

1% of that number will be 768/80

100% of that number will be 768/80×100=960

5% of 960=5/100×960=48

40% of 48=40/100×48=19.2

Q 3565) A got 78% marks in an examination and B got 64% marks in the same examination. If the sum of the marks obtained by A and B is 923, then find the marks obtained by B in the examination.

(A) 400    (B) 416    (C) 432    (D) 448

Ans: (B) 416

Solution:

78%+64%=923

142%=923

1%=923/142

100%=923/142×100=650

Now,

100%=650

∴ 1%=650/100

∴ 64%=650/100×64=5=416 marks

Q 3566) A got 92% marks in an examination and B got 78% marks in the same examination. If the difference of the marks obtained by A and B is 105, then find the marks obtained by A in the examination.

(A) 600    (B) 615    (C) 645    (D) 690

Ans: (D) 690

Solution:

92%-78%=14%

14%=105

1%=105/14

100%=105/14×100=750

Now,

100%=750

∴ 1%=750/100

∴ 92%=750/100×92=690 marks

Q 3567) If 60% of 80 = y% of 96, then y is….  (CUET-UG 2022)

(A) 40    (B) 45    (C) 50    (C) 55

Ans: (C) 50

Solution:

60% of 80 = y% of 96

48=y/100×96

Multiply both side by 100

48×100=y/100x96x100

4800=96y

y=4800/96=50

Q 3568) In a class, 80% of the students passed in an examination. Out of these successful candidates, 27% of the students passed in the first division and 43% students passed in second division. A total of 90 students failed in the examination. How many students from the class passed in the third division?

(A) 108    (B) 110    (C) 112    (D) 125

Ans: (A) 108

Solution:

20% of the students (i.e. 90) faild in the examination.

1st Step:

20% of a number is 90

1% of that number will be 90/20

100% of that number will be 90/20×100=450

Total number of students are 450.

2nd Step:

80% of 450 students= 80/100×450= 360 students passed.

3rd Step:

Out of 360 students 27% and 43% =70% passed in first and second division and the balance 30% = 30/100×360=108 students passed in third division.

 

20% of 690= 20/100×690=138

50% of 138= 50/100×138= 69

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