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: 2026-11-05 01:39:29
Category Description: This category is for the informations which we can't memorise.
Q1. Find the median of the data set.
Q1. Find the median of the data set.
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
Answer: (C) 5
Answer: (C) 5
Answer: (C) 5
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
= 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17
Total numbers of values N = 13
Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th
7th Number = 5
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12 = 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17 Total numbers of values N = 13 Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th 7th Number = 5
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12 = 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17 Total numbers of values N = 13 Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th 7th Number = 5
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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. ‘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 :
Q3. ‘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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Q4. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
Q4. 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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Q5. When the numerator of a fraction increases by 3, the fraction increases by its three-fourth. The numerator of the fraction is
Q5. 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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Q6. 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
Q6. 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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Q7. The angles of a quadrilateral are in the ratio of 1 : 3 : 4 : 7. The difference between the largest and the smallest angle is
Q7. The angles of a quadrilateral are in the ratio of 1 : 3 : 4 : 7. The difference between the largest and the smallest angle is
Answer: (C) 144°
Answer: (C) 144°
Answer: (C) 144°
(360°/15) * (7-1) = 144°
(360°/15) * (7-1) = 144°
(360°/15) * (7-1) = 144°
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Q8. Find the value of the following expression
Q8. 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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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. If '÷' means ‘addition’, ‘+’ means ‘subtraction’, ‘–’ means ‘multiplication’ and ‘×’ means ‘division’, then the value of 18 ÷ 12 × 4 – 5 is
Q10. 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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Q11. 45% of a number is same as 30% of another number. The ratio of the first number to the second number is :
Q11. 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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Q12. If 16℅ of 40℅ of a number is 8, then the number is
Q12. If 16℅ of 40℅ of a number is 8, then the number is
Answer: (B) 125
Answer: (B) 125
Answer: (B) 125
125
125
125
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Q13. 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?
Q13. 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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Q14. The missing term in the sequence 7, 11, 19, 31, ______, 67 is
Q14. The missing term in the sequence 7, 11, 19, 31, ______, 67 is
Answer: (B) 47
Answer: (B) 47
Answer: (B) 47
47
47
47
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Q15. The remainder when –76 is divided by 3, is
Q15. 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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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. The 8th term of the sequence 2, 6, 18, 54, ...... is
Q17. The 8th term of the sequence 2, 6, 18, 54, ...... is
Answer: (B) 4374
Answer: (B) 4374
Answer: (B) 4374
4374
4374
4374
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Q18. The sum of first 6 prime numbers is
Q18. The sum of first 6 prime numbers is
Answer: (C) 41
Answer: (C) 41
Answer: (C) 41
41
41
41
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Q19. If 20% of a is b, then b% of 20 is the same as :
Q19. If 20% of a is b, then b% of 20 is the same as :
Answer: (A) 4% of a
Answer: (A) 4% of a
Answer: (A) 4% of a
20% of a is b
Hence b = 20% of a = 20a/100
b% of 20
= 20% of b
= (20/100) * 20a/100
= (20*20*a)/(100*100)
= 4a/100
= (4/100) * a
= 4% of a
20% of a is b Hence b = 20% of a = 20a/100 b% of 20 = 20% of b = (20/100) * 20a/100 = (20*20*a)/(100*100) = 4a/100 = (4/100) * a = 4% of a
20% of a is b Hence b = 20% of a = 20a/100 b% of 20 = 20% of b = (20/100) * 20a/100 = (20*20*a)/(100*100) = 4a/100 = (4/100) * a = 4% of a
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Q20. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
Q20. 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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