Unmemorable (Non Memory Based) | MCQ Quizzes | Category (T/R/A)
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2026-03-20 08:34:08
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Category UID: 18
Label UID: 42
Category Name: Unmemorable
Category Full Name: Non Memory Based
Category Link/Slug: unmemorable
Total Quizzes: 57
Total Views: 1403
Last Refreshed: 2026-03-20 08:34:08
Category Description: This category is for the informations which we can't memorise.
Q1. Three fourth of a number is more than two third of the number by 5. Then the number is
Q1. Three fourth of a number is more than two third of the number by 5. Then the number is
Answer: (B) 60
Answer: (B) 60
Answer: (B) 60
60
60
60
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Q2. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
Q2. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
Answer: (C) 18
Answer: (C) 18
Answer: (C) 18
18
18
18
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Q3. How many prime numbers exist, which are less than 40?
Q3. How many prime numbers exist, which are less than 40?
Answer: (C) 12
Answer: (C) 12
Answer: (C) 12
12
12
12
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Q4. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
Q4. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
Answer: (D) 3.5 m
Answer: (D) 3.5 m
Answer: (D) 3.5 m
3.5 m
3.5 m
3.5 m
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Q5. In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is
Q5. In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is
Answer: (A) 55 years
Answer: (A) 55 years
Answer: (A) 55 years
25 years + (3*10) years = 55 years
25 years + (3*10) years = 55 years
25 years + (3*10) years = 55 years
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Q6. Find the odd number - 15, 17, 6, 12.
Q6. Find the odd number - 15, 17, 6, 12.
Answer: (D) 17
Answer: (D) 17
Answer: (D) 17
17 which is a Prime number.
17 which is a Prime number.
17 which is a Prime number.
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Q7. The ratio of the number of boys and girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?
Q7. The ratio of the number of boys and girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?
Answer: (B) 120
Answer: (B) 120
Answer: (B) 120
120
120
120
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Q8. The least number by which 2450 must be multiplied to make it a perfect square, is
Q8. The least number by which 2450 must be multiplied to make it a perfect square, is
Answer: (A) 2
Answer: (A) 2
Answer: (A) 2
2
2
2
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Q9. A number when divided by 6 is diminished by 40. The number is
Q9. A number when divided by 6 is diminished by 40. The number is
Answer: (B) 48
Answer: (B) 48
Answer: (B) 48
48
48
48
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Q10. ‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is :
Q10. ‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is :
Answer: (C) 1.2 km
Answer: (C) 1.2 km
Answer: (C) 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B
(10*60)s * 1m/s = 600m
Let A covers a distance of X after starting of B,
Then X + 600m = 2X
=> 2X - X = 600m
=> X = 600m
Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
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Q11. The value of is :
Q11. The value of is :
Answer: (C) 2.73
Answer: (C) 2.73
Answer: (C) 2.73
=
=
= 0.2 + 1.2 + 1.3 + 0.03
= 2.73
= = = 0.2 + 1.2 + 1.3 + 0.03 = 2.73
= = = 0.2 + 1.2 + 1.3 + 0.03 = 2.73
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Q12. Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is
Q12. Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is
Answer: (C) 39
Answer: (C) 39
Answer: (C) 39
39
39
39
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Q13. The number when divided by 2, leaves remainder 1; when divided by 3, leaves remainder 2 and when divided by 4, leaves remainder 3, is
Q13. The number when divided by 2, leaves remainder 1; when divided by 3, leaves remainder 2 and when divided by 4, leaves remainder 3, is
Answer: (D) 11
Answer: (D) 11
Answer: (D) 11
11
11
11
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Q14. If the price of sugar increases by 20%, by what percentage should a household reduce its consumption of sugar so that the budget remains the same?
Q14. If the price of sugar increases by 20%, by what percentage should a household reduce its consumption of sugar so that the budget remains the same?
Answer: (D)
Answer: (D)
Answer: (D)
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Q15. A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is :
Q15. A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is :
Answer: (D) 45 years
Answer: (D) 45 years
Answer: (D) 45 years
The son becomes 15 years old after 3 years.
Hence son's present age = 15 years - 3 years = 12 years
Man is four times as old as his son.
Hence Man's present age = 12 years * 4 = 48 years
Man is 3 years older than his wife.
Hence present age of his wife = 48 years - 3 years = 45 years.
The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.
The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.
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Q16. What is the value of x in the equation 2x + 5 = 11?
Q16. What is the value of x in the equation 2x + 5 = 11?
Answer: (B) 3
Answer: (B) 3
Answer: (B) 3
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
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Q17. The product of two consecutive odd numbers is 19043. Which is the smaller number?
Q17. The product of two consecutive odd numbers is 19043. Which is the smaller number?
Answer: (C) 137
Answer: (C) 137
Answer: (C) 137
137
137
137
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Q18. 45% of a number is same as 30% of another number. The ratio of the first number to the second number is :
Q18. 45% of a number is same as 30% of another number. The ratio of the first number to the second number is :
Answer: (A) 2 : 3
Answer: (A) 2 : 3
Answer: (A) 2 : 3
=
=
= X:Y = 2:3
= = = X:Y = 2:3
= = = X:Y = 2:3
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Q19. Find the value of
Q19. Find the value of
Answer: (D)
Answer: (D)
Answer: (D)
=
=
=
=
=
= = = = =
= = = = =
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Q20. Shyam stored Rs 35 in the form of 1 rupee coin and 50 paise coins in the ratio 2 : 3. The number of 50 paise coins are
Q20. Shyam stored Rs 35 in the form of 1 rupee coin and 50 paise coins in the ratio 2 : 3. The number of 50 paise coins are
Answer: (C) 30
Answer: (C) 30
Answer: (C) 30
1*20 + 0.5*30 = 20+15 = 35
1*20 + 0.5*30 = 20+15 = 35
1*20 + 0.5*30 = 20+15 = 35
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