Unmemorable (Non Memory Based) | MCQ Quizzes | Category (S/R/A)
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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: 1584
Last Refreshed: 2026-05-04 20:41:22
Category Description: This category is for the informations which we can't memorise.
Q1. The least number by which 2450 must be multiplied to make it a perfect square, is
Q1. 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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Q2. Select the number pair in which the two numbers are related in the same way as 35 : 6.
Q2. Select the number pair in which the two numbers are related in the same way as 35 : 6.
Answer: (D) 120 : 11
Answer: (D) 120 : 11
Answer: (D) 120 : 11
62-1 : 6 = 35 : 6
112-1 : 11 = 120 : 11
62-1 : 6 = 35 : 6 112-1 : 11 = 120 : 11
62-1 : 6 = 35 : 6 112-1 : 11 = 120 : 11
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Q3. 4/5 of a number is 64. Then half of the number is
Q3. 4/5 of a number is 64. Then half of the number is
Answer: (D) 40
Answer: (D) 40
Answer: (D) 40
40
40
40
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Q4. The product of two consecutive odd numbers is 19043. Which is the smaller number?
Q4. 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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Q5. Find the value of the following expression
Q5. Find the value of the following expression
48÷12+4×25÷5
48÷12+4×25÷5
48÷12+4×25÷5
Answer: (B) 24
Answer: (B) 24
Answer: (B) 24
24
24
24
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Q6. The conversion of in the fractional form is
Q6. The conversion of in the fractional form is
Answer: (C)
Answer: (C)
Answer: (C)
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Q7. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are
Q7. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are
Answer: (D) 7 and 8
Answer: (D) 7 and 8
Answer: (D) 7 and 8
7 and 8
7 and 8
7 and 8
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Q8. 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
Q8. 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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Q9. If '÷' means ‘addition’, ‘+’ means ‘subtraction’, ‘–’ means ‘multiplication’ and ‘×’ means ‘division’, then the value of 18 ÷ 12 × 4 – 5 is
Q9. If '÷' means ‘addition’, ‘+’ means ‘subtraction’, ‘–’ means ‘multiplication’ and ‘×’ means ‘division’, then the value of 18 ÷ 12 × 4 – 5 is
Answer: (D) 33
Answer: (D) 33
Answer: (D) 33
33
33
33
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Q10. 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
Q10. 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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Q11. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
Q11. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
Answer: (C) 5
Answer: (C) 5
Answer: (C) 5
5
5
5
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Q12. 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?
Q12. 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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Q13. The simple interest earned by 4,000 in 18 months at 12% per annum is
Q13. The simple interest earned by 4,000 in 18 months at 12% per annum is
Answer: (B) 720
Answer: (B) 720
Answer: (B) 720
720
720
720
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Q14. 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
Q14. 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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Q15. 5 : 27 :: 9 : ________
Q15. 5 : 27 :: 9 : ________
Fill the blank.
Fill the blank.
Fill the blank.
Answer: (A) 83
Answer: (A) 83
Answer: (A) 83
5 : (52 + 2) = 5 : 27
Hence
9 : (92 + 2) = 9 : 83
5 : (52 + 2) = 5 : 27 Hence 9 : (92 + 2) = 9 : 83
5 : (52 + 2) = 5 : 27 Hence 9 : (92 + 2) = 9 : 83
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Q16. The LCM of two numbers is 40 and their HCF is 4. If the difference between the two numbers is 12, then the sum of the numbers is
Q16. The LCM of two numbers is 40 and their HCF is 4. If the difference between the two numbers is 12, then the sum of the numbers is
Answer: (C) 28
Answer: (C) 28
Answer: (C) 28
X * (X-12) = 40 * 4
X = 20
X + (X-12) = 20 + 20 - 12 = 28
X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28
X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28
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Q17. A hostel has 120 students and food supplies are for 45 days. If 30 more students joined the hostel, then how many days the hostel will run with the existing food?
Q17. A hostel has 120 students and food supplies are for 45 days. If 30 more students joined the hostel, then how many days the hostel will run with the existing food?
Answer: (C) 36 days
Answer: (C) 36 days
Answer: (C) 36 days
36 days
36 days
36 days
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Q18. The remainder when –76 is divided by 3, is
Q18. The remainder when –76 is divided by 3, is
Answer: (C) 2
Answer: (C) 2
Answer: (C) 2
2
Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
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Q19. A number is as much greater than 31 as is less 55. Then the number is
Q19. A number is as much greater than 31 as is less 55. Then the number is
Answer: (C) 43
Answer: (C) 43
Answer: (C) 43
43
43
43
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Q20. ‘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 :
Q20. ‘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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