General Mathematics (General Mathematics) | MCQ Quizzes | Category (S/R/M)
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41 quizzes
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2026-06-19 05:29:01
General Mathematics aims to develop learners' understanding of concepts and techniques drawn from number and algebra, trigonometry and world geometry, sequences, finance, networks and decision mathematics and statistics, in order to solve applied pro
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Category UID: 14
Label UID: 38
Category Name: General Mathematics
Category Full Name: General Mathematics
Category Link/Slug: general-mathematics
Total Quizzes: 41
Total Views: 2191
Last Refreshed: 2026-06-19 05:29:01
Category Description: General Mathematics aims to develop learners' understanding of concepts and techniques drawn from number and algebra, trigonometry and world geometry, sequences, finance, networks and decision mathematics and statistics, in order to solve applied problems.
Q1. 4/5 of a number is 64. Then half of the number is
Q1. 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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Q2. A number is as much greater than 22 as is less than 72. Then the number is
Q2. A number is as much greater than 22 as is less than 72. Then the number is
Answer: (D) 47
Answer: (D) 47
Answer: (D) 47
47
47
47
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Q3. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
Q3. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
Answer: (C) 2
Answer: (C) 2
Answer: (C) 2
In ascending order = -2, -1, 1, 2, 3, 5, 6
median = th term
= 4th term
= 2
In ascending order = -2, -1, 1, 2, 3, 5, 6 median = th term = 4th term = 2
In ascending order = -2, -1, 1, 2, 3, 5, 6 median = th term = 4th term = 2
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Q4. A certain amount invested in a firm would become double at the end of one month, but it deducts an amount of ₹120 on every doubling. A person invests an amount of ₹105 and continues for 3 months without investing any additional amount. At the end of 3 months, his net income is
Q4. A certain amount invested in a firm would become double at the end of one month, but it deducts an amount of ₹120 on every doubling. A person invests an amount of ₹105 and continues for 3 months without investing any additional amount. At the end of 3 months, his net income is
Answer: (B) 0
Answer: (B) 0
Answer: (B) 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90
2nd Month -> (90 * 2) - 120 = 180 - 120 = 60
3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90 2nd Month -> (90 * 2) - 120 = 180 - 120 = 60 3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90 2nd Month -> (90 * 2) - 120 = 180 - 120 = 60 3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
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Q5. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is
Q5. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is
Answer: (B) 44 m
Answer: (B) 44 m
Answer: (B) 44 m
2 * * 7
= 2 * 22
= 44
2 * * 7 = 2 * 22 = 44
2 * * 7 = 2 * 22 = 44
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Q6. A ladder of length 13 m is leaning against a vertical wall with the upper end at the height of 5 m. The horizontal distance between the foot of the wall and the lower end of the ladder is
Q6. A ladder of length 13 m is leaning against a vertical wall with the upper end at the height of 5 m. The horizontal distance between the foot of the wall and the lower end of the ladder is
Answer: (D) 12m
Answer: (D) 12m
Answer: (D) 12m
52 + X2 = 132
=> 25 + X2 = 169
=> X2 = 196 - 25
=> X2 = 144
=> X = 12
52 + X2 = 132 => 25 + X2 = 169 => X2 = 196 - 25 => X2 = 144 => X = 12
52 + X2 = 132 => 25 + X2 = 169 => X2 = 196 - 25 => X2 = 144 => X = 12
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Q7. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
Q7. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
Answer: (C) 23
Answer: (C) 23
Answer: (C) 23
X + (X+2) + (X+4) + (X+6) + (X+8) = 95
=> 5X + 20 = 95
=> 5X = 95-20
=> 5X = 75
=> X = 15
Hence (X+8) = 15+8 = 23
X + (X+2) + (X+4) + (X+6) + (X+8) = 95 => 5X + 20 = 95 => 5X = 95-20 => 5X = 75 => X = 15 Hence (X+8) = 15+8 = 23
X + (X+2) + (X+4) + (X+6) + (X+8) = 95 => 5X + 20 = 95 => 5X = 95-20 => 5X = 75 => X = 15 Hence (X+8) = 15+8 = 23
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Q8. The smallest among is
Q8. The smallest among is
Answer: (C)
Answer: (C)
Answer: (C)
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Q9. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are
Q9. 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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Q10. Three fourth of a number is more than two third of the number by 5. Then the number is
Q10. 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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Q11. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
Q11. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
Answer: (A) 24
Answer: (A) 24
Answer: (A) 24
6*4 = 24
8*3 = 24
12*2 = 24
6*4 = 24 8*3 = 24 12*2 = 24
6*4 = 24 8*3 = 24 12*2 = 24
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Q12. 100% of 100 when added to 200% of 200 would result
Q12. 100% of 100 when added to 200% of 200 would result
Answer: (C) 500
Answer: (C) 500
Answer: (C) 500
100 * 100% + 200 * 200%
=
=
= 100 + 400
= 500
100 * 100% + 200 * 200% = = = 100 + 400 = 500
100 * 100% + 200 * 200% = = = 100 + 400 = 500
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Q13. Find the least number by which 1250 must be multiplied to make it a perfect square.
Q13. Find the least number by which 1250 must be multiplied to make it a perfect square.
Answer: (D) 2
Answer: (D) 2
Answer: (D) 2
1250 = 2*5*5*5*5
1250*2 = 2*2*5*5*5*5 = 2500
= 2*5*5 = 50
1250 = 2*5*5*5*5 1250*2 = 2*2*5*5*5*5 = 2500 = 2*5*5 = 50
1250 = 2*5*5*5*5 1250*2 = 2*2*5*5*5*5 = 2500 = 2*5*5 = 50
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Q14. 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
Q14. 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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Q15. Which of the following is a prime number?
Q15. Which of the following is a prime number?
81,33,71,93
81,33,71,93
81,33,71,93
Answer: (C) 71
Answer: (C) 71
Answer: (C) 71
71
71
71
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Q16. If the average age of A, B and C is 22 years and the average age of B and C is 25 years, then find the age of A after 9 years from now.
Q16. If the average age of A, B and C is 22 years and the average age of B and C is 25 years, then find the age of A after 9 years from now.
Answer: (A) 25 years
Answer: (A) 25 years
Answer: (A) 25 years
25 years
=> A+B+C = 22*3
=> A+(B+C) = 66
=> A+(25*2) = 66
=> A = 66-50
=> A = 16
After 9 years A = 16+9 = 25 years
25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years
25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years
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Q17. Consider the four numbers given numbers. If the digits of each number are arranged in ascending order, which new number would be the smallest?
Q17. Consider the four numbers given numbers. If the digits of each number are arranged in ascending order, which new number would be the smallest?
I. 376
II. 629
III. 921
IV. 397
I. 376 II. 629 III. 921 IV. 397
I. 376 II. 629 III. 921 IV. 397
Answer: (A) III
Answer: (A) III
Answer: (A) III
376 -> 367
629 -> 269
921 -> 129 *
397 -> 379
376 -> 367 629 -> 269 921 -> 129 * 397 -> 379
376 -> 367 629 -> 269 921 -> 129 * 397 -> 379
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Q18. If is equal to
Q18. If is equal to
Answer: (A) 2
Answer: (A) 2
Answer: (A) 2
Hence, X:Y:Z = 3:4:7
=
=
= 2
Hence, X:Y:Z = 3:4:7 = = = 2
Hence, X:Y:Z = 3:4:7 = = = 2
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Q19. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q19. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Answer: (C) 1:9
Answer: (C) 1:9
Answer: (C) 1:9
Area = πr2
π12 : π32
= π1 : π9
= 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
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Q20. The difference of two numbers is 5 and the difference of their square is 135. Then the sum of the numbers is
Q20. The difference of two numbers is 5 and the difference of their square is 135. Then the sum of the numbers is
Answer: (D) 27
Answer: (D) 27
Answer: (D) 27
27
27
27
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