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
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Last Refreshed: 2025-07-31 04:35:38
Category Description: This category is for the informations which we can't memorise.
Q1. The sum of first 6 prime numbers is
Q1. The sum of first 6 prime numbers is
Answer: (C) 41
Answer: (C) 41
Answer: (C) 41
41
41
41
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Q2. 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
Q2. 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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Q3. 98 toffees were distributed to some boys in a group. Each boy in the group got twice as many of the toffees as the number of boys. The number of boys in the group was
Q3. 98 toffees were distributed to some boys in a group. Each boy in the group got twice as many of the toffees as the number of boys. The number of boys in the group was
Answer: (B) 7
Answer: (B) 7
Answer: (B) 7
7
=> x * 2x = 98
=> x2 =
=> x2 = 49
=> x = 7
7 => x * 2x = 98 => x2 = => x2 = 49 => x = 7
7 => x * 2x = 98 => x2 = => x2 = 49 => x = 7
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Q4. 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?
Q4. 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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Q5. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
Q5. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
Answer: (D) 28%
Answer: (D) 28%
Answer: (D) 28%
Net change (-40 + 20 + )%
= (-20 + )%
= (-20 - 8)%
= -28%
= 28%
Net change (-40 + 20 + )% = (-20 + )% = (-20 - 8)% = -28% = 28%
Net change (-40 + 20 + )% = (-20 + )% = (-20 - 8)% = -28% = 28%
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Q6. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
Q6. 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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Q7. 4/5 of a number is 64. Then half of the number is
Q7. 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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Q8. A lady deposits money in her savings bank in such a way that every next day her deposit amount is ₹ 12 more than her previous day deposit. If she starts her deposit with ₹ 12 on the first day, the total amount deposited by Liza at the end of 30 days will be :
Q8. A lady deposits money in her savings bank in such a way that every next day her deposit amount is ₹ 12 more than her previous day deposit. If she starts her deposit with ₹ 12 on the first day, the total amount deposited by Liza at the end of 30 days will be :
Answer: (B) 5,580
Answer: (B) 5,580
Answer: (B) 5,580
The formula for the sum of the first n terms of an arithmetic progression is
= 15*(12+360)
= 15 * 372
= 5580
The formula for the sum of the first n terms of an arithmetic progression is = 15*(12+360) = 15 * 372 = 5580
The formula for the sum of the first n terms of an arithmetic progression is = 15*(12+360) = 15 * 372 = 5580
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Q9. Select the number pair in which the two numbers are related in the same way as 35 : 6.
Q9. 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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Q10. 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
Q10. 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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Q11. A number is as much greater than 31 as is less 55. Then the number is
Q11. 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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Q12. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
Q12. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
Answer: (B) r
Answer: (B) r
Answer: (B) r
LCM will be r.
px= q and qy = r, hence r = pxy.
LCM will be r. px= q and qy = r, hence r = pxy.
LCM will be r. px= q and qy = r, hence r = pxy.
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Q13. Find the value of the following expression
Q13. 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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Q14. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
Q14. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
Answer: (A) 13
Answer: (A) 13
Answer: (A) 13
The smallest number amongst them is 13
=> X + (X+2) + (X+4) + (X+6) = 64
=> 4X + 12 = 64
=> 4X = 64 - 12
=> 4X = 52
=> X = 13
The smallest number amongst them is 13 => X + (X+2) + (X+4) + (X+6) = 64 => 4X + 12 = 64 => 4X = 64 - 12 => 4X = 52 => X = 13
The smallest number amongst them is 13 => X + (X+2) + (X+4) + (X+6) = 64 => 4X + 12 = 64 => 4X = 64 - 12 => 4X = 52 => X = 13
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Q15. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
Q15. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
Answer: (C) 20
Answer: (C) 20
Answer: (C) 20
=> 2x * 3x = 96
=> 6x2 = 96
=> x2 = 96/6
=> x2 = 16
=> x = 4
2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
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Q16. A number when divided by 6 is diminished by 40. The number is
Q16. 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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Q17. The simple interest earned by 4,000 in 18 months at 12% per annum is
Q17. 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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Q18. A 100 metre long train moving in a uniform speed of 20 m/sec crosses a bridge of length 1 km. The time taken by the train to cross the bridge is
Q18. A 100 metre long train moving in a uniform speed of 20 m/sec crosses a bridge of length 1 km. The time taken by the train to cross the bridge is
Answer: (C) 55 seconds
Answer: (C) 55 seconds
Answer: (C) 55 seconds
(1000+100)m / (20m/s) = 55 seconds
(1000+100)m / (20m/s) = 55 seconds
(1000+100)m / (20m/s) = 55 seconds
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Q19. When the numerator of a fraction increases by 3, the fraction increases by its three-fourth. The numerator of the fraction is
Q19. When the numerator of a fraction increases by 3, the fraction increases by its three-fourth. The numerator of the fraction is
Answer: (B) 4
Answer: (B) 4
Answer: (B) 4
4
4
4
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Q20. 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 :
Q20. 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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