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Quantitative Aptitude — Number Systems
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Quantitative Aptitude — Number Systems

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34 mins
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45
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Quantitative Aptitude — Number Systems — Questions with Answers Open after you finish the quiz — all 45 questions, with answers and explanations.
Q1. Which of the following is an irrational number?
  1. √9
  2. 0.333...
  3. -5
  4. π
Show answer
Answer: D. π
An irrational number cannot be expressed as a simple fraction. √9 = 3 (rational), 0.333... = 1/3 (rational), -5 (rational). Pi (π) is a non-terminating, non-repeating decimal, hence it is irrational.
Q2. Which statement is true about rational numbers?
  1. They are closed under division by zero
  2. Their decimal expansion is always terminating
  3. They include all square roots
  4. They can always be written as p/q where p, q are integers and q≠0
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Answer: D. They can always be written as p/q where p, q are integers and q≠0
Rational numbers are defined as numbers that can be expressed in the form p/q, where p and q are integers and q is not equal to zero. They are not closed under division by zero. Their decimal expansion can be terminating or non-terminating repeating. They do not include all square roots (e.g., √2 is irrational).
Q3. Which of the following statements is true regarding the representation of √2 on a number line?
  1. It is a point that cannot be precisely located
  2. It is a rational number
  3. It is located exactly at 1.5
  4. It is a point that can be precisely located
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Answer: D. It is a point that can be precisely located
Every real number, including irrational numbers like √2, corresponds to a unique point on the number line. While its decimal expansion is non-terminating and non-repeating, it can still be precisely located using geometric construction (e.g., using a right-angled triangle with sides 1 and 1).
Q4. How many rational numbers exist between any two distinct rational numbers?
  1. One
  2. None
  3. Finitely many
  4. Infinitely many
Show answer
Answer: D. Infinitely many
The property of rational numbers states that between any two distinct rational numbers, there exist infinitely many rational numbers. This is known as the density property of rational numbers.
Q5. Which of the following statements correctly defines an irrational number like √3?
  1. It is a number that cannot be expressed in the form p/q, where p and q are integers and q ≠ 0
  2. It is a number whose square root is not an integer
  3. It is a number whose decimal expansion is terminating
  4. It is a number that is always negative
Show answer
Answer: A. It is a number that cannot be expressed in the form p/q, where p and q are integers and q ≠ 0
An irrational number is formally defined as a number that cannot be expressed as a simple fraction p/q, where p and q are integers and q is not zero. While its square root is not an integer and its decimal expansion is non-terminating non-repeating, the p/q form is the fundamental definition.
Q6. Which of the following numbers is rational?
  1. √7
  2. π
  3. √16
  4. 0.1010010001...
Show answer
Answer: C. √16
√7 is irrational. π is irrational. 0.1010010001... is a non-terminating, non-repeating decimal, hence irrational. √16 = 4, which can be written as 4/1, making it a rational number.
Q7. The product of a non-zero rational number and an irrational number is always:
  1. A rational number
  2. An irrational number
  3. An integer
  4. A natural number
Show answer
Answer: B. An irrational number
If 'r' is a non-zero rational number and 'i' is an irrational number, then their product 'r * i' is always an irrational number. For example, 2 * √3 = 2√3, which is irrational.
Q8. Which of the following fractions will have a terminating decimal expansion?
  1. 1/3
  2. 2/7
  3. 3/8
  4. 5/6
Show answer
Answer: C. 3/8
A fraction p/q (in simplest form) has a terminating decimal expansion if and only if the prime factorization of the denominator q contains only powers of 2 and/or 5. For 3/8, the denominator 8 = 2^3, which fits the condition.
Q9. The decimal expansion of 1/11 is:
  1. 0.09
  2. 0.0909...
  3. 0.0999...
  4. 0.9090...
Show answer
Answer: B. 0.0909...
Dividing 1 by 11 gives 0.090909... which is a non-terminating repeating decimal, often written as 0.09 with a bar over 09.
Q10. The decimal expansion of 1/7 is:
  1. 0.142857 (bar over 142857)
  2. 0.142857 (bar over 57)
  3. 0.142857 (bar over 7)
  4. 0.142857 (no bar)
Show answer
Answer: A. 0.142857 (bar over 142857)
When 1 is divided by 7, the decimal expansion is 0.142857142857... where the block of digits '142857' repeats. This is represented by placing a bar over the repeating block.
Q11. For a rational number p/q to have a terminating decimal expansion, the prime factorization of q must be of the form:
  1. 2^m * 3^n
  2. 2^m * 5^n
  3. 3^m * 5^n
  4. Any prime factors
Show answer
Answer: B. 2^m * 5^n
A rational number p/q (where p and q are coprime) has a terminating decimal expansion if and only if the prime factorization of the denominator q is of the form 2^m * 5^n, where m and n are non-negative integers.
Q12. The number 0.123456789101112... (where digits are consecutive natural numbers) is:
  1. A terminating decimal
  2. A non-terminating repeating decimal
  3. A non-terminating non-repeating decimal
  4. An integer
Show answer
Answer: C. A non-terminating non-repeating decimal
The decimal expansion 0.123456789101112... neither terminates nor repeats a fixed block of digits. Therefore, it is a non-terminating non-repeating decimal, which is characteristic of an irrational number.
Q13. The fractional form of 0.4 (bar over 4) is:
  1. 4/9
  2. 4/10
  3. 1/4
  4. 2/5
Show answer
Answer: A. 4/9
Let x = 0.444... Multiply by 10: 10x = 4.444... Subtract x from 10x: 9x = 4, so x = 4/9.
Q14. Every real number is either:
  1. Rational or integer
  2. Irrational or natural
  3. Rational or irrational
  4. Integer or natural
Show answer
Answer: C. Rational or irrational
The set of real numbers is composed of the union of rational numbers and irrational numbers. There are no other types of real numbers.
Q15. Based on the bar chart showing powers of 2, what is the value of 2^4?
  1. 8
  2. 16
  3. 32
  4. 4
Show answer
Answer: B. 16
The bar chart shows the values: 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32. From the chart, the value corresponding to 2^4 is 16.
Q16. Simplify: 3^2 * 3^5
  1. 3^7
  2. 3^10
  3. 9^7
  4. 6^7
Show answer
Answer: A. 3^7
According to the law of exponents, a^m * a^n = a^(m+n). So, 3^2 * 3^5 = 3^(2+5) = 3^7.
Q17. Simplify: (√5 + √2)(√5 - √2)
  1. √3
  2. 7
  3. 3
  4. √7
Show answer
Answer: C. 3
This expression is in the form (a+b)(a-b) = a^2 - b^2. Here, a = √5 and b = √2. So, (√5)^2 - (√2)^2 = 5 - 2 = 3.
Q18. Rationalize the denominator of 1/(√2 + 1)
  1. √2 + 1
  2. 1 - √2
  3. 1/√2
  4. √2 - 1
Show answer
Answer: D. √2 - 1
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is (√2 - 1). So, [1/(√2 + 1)] * [(√2 - 1)/(√2 - 1)] = (√2 - 1) / ((√2)^2 - 1^2) = (√2 - 1) / (2 - 1) = √2 - 1.
Q19. What is the value of √12 / √3?
  1. 2
  2. √4
  3. 3
  4. √36
Show answer
Answer: A. 2
Using the property √(a)/√(b) = √(a/b), we have √12 / √3 = √(12/3) = √4. The square root of 4 is 2.
Q20. Expand and simplify (√5 + √2)^2
  1. 7 + √10
  2. 7 + 2√10
  3. 5 + 2√10
  4. 7 + 2√5
Show answer
Answer: B. 7 + 2√10
Using the identity (a+b)^2 = a^2 + 2ab + b^2, where a = √5 and b = √2. So, (√5)^2 + 2(√5)(√2) + (√2)^2 = 5 + 2√(5*2) + 2 = 5 + 2√10 + 2 = 7 + 2√10.
Q21. Find the least common multiple (LCM) of 24, 36, and 40.
  1. 120
  2. 240
  3. 360
  4. 720
Show answer
Answer: C. 360
To find the LCM of 24, 36, and 40: Prime factorization: 24 = 2^3 × 3 36 = 2^2 × 3^2 40 = 2^3 × 5 LCM is the product of the highest powers of all prime factors involved: 2^3 × 3^2 × 5 = 8 × 9 × 5 = 360.
Q22. What is the highest common factor (HCF) of 108, 288, and 360?
  1. 36
  2. 18
  3. 12
  4. 72
Show answer
Answer: A. 36
To find the HCF of 108, 288, and 360: Prime factorization: 108 = 2^2 × 3^3 288 = 2^5 × 3^2 360 = 2^3 × 3^2 × 5 HCF is the product of the lowest powers of common prime factors: 2^2 × 3^2 = 4 × 9 = 36.
Q23. The product of two numbers is 2160. If their HCF is 12, what is their LCM?
  1. 120
  2. 180
  3. 240
  4. 360
Show answer
Answer: B. 180
For any two positive integers, the product of the numbers is equal to the product of their HCF and LCM. Product of numbers = HCF × LCM 2160 = 12 × LCM LCM = 2160 / 12 = 180.
Q24. Find the smallest number which when divided by 15, 20, 24, and 32 leaves a remainder of 4 in each case.
  1. 480
  2. 484
  3. 476
  4. 496
Show answer
Answer: B. 484
First, find the LCM of 15, 20, 24, and 32. 15 = 3 × 5 20 = 2^2 × 5 24 = 2^3 × 3 32 = 2^5 LCM(15, 20, 24, 32) = 2^5 × 3 × 5 = 32 × 15 = 480. The smallest number which leaves a remainder of 4 in each case is LCM + 4 = 480 + 4 = 484.
Q25. Find the largest number that divides 62, 132, and 237 leaving the same remainder in each case.
  1. 15
  2. 25
  3. 35
  4. 45
Show answer
Answer: C. 35
If a number divides a, b, and c leaving the same remainder, then it must divide the differences (b-a), (c-b), and (c-a). Differences: 132 - 62 = 70 237 - 132 = 105 237 - 62 = 175 Now, find the HCF of 70, 105, and 175. 70 = 2 × 5 × 7 105 = 3 × 5 × 7 175 = 5^2 × 7 HCF(70, 105, 175) = 5 × 7 = 35.
Q26. Which of the following numbers is divisible by 9?
  1. 23578
  2. 34567
  3. 45678
  4. 56781
Show answer
Answer: D. 56781
A number is divisible by 9 if the sum of its digits is divisible by 9. 1) 23578: 2+3+5+7+8 = 25 (not divisible by 9) 2) 34567: 3+4+5+6+7 = 25 (not divisible by 9) 3) 45678: 4+5+6+7+8 = 30 (not divisible by 9) 4) 56781: 5+6+7+8+1 = 27 (divisible by 9) Therefore, 56781 is divisible by 9.
Q27. Which of the following numbers is divisible by 8?
  1. 123456
  2. 234567
  3. 345678
  4. 456789
Show answer
Answer: A. 123456
A number is divisible by 8 if its last three digits are divisible by 8. 1) 123456: Last three digits are 456. 456 ÷ 8 = 57. So, 123456 is divisible by 8. 2) 234567: Last three digits are 567 (not divisible by 8). 3) 345678: Last three digits are 678 (not divisible by 8). 4) 456789: Last three digits are 789 (not divisible by 8).
Q28. Which of the following numbers is divisible by 6?
  1. 12345
  2. 23456
  3. 34567
  4. 45678
Show answer
Answer: D. 45678
A number is divisible by 6 if it is divisible by both 2 and 3. For divisibility by 2, the number must be even (end in 0, 2, 4, 6, 8). For divisibility by 3, the sum of its digits must be divisible by 3. 1) 12345: Not even (not divisible by 2). 2) 23456: Even. Sum of digits = 2+3+4+5+6 = 20 (not divisible by 3). 3) 34567: Not even (not divisible by 2). 4) 45678: Even. Sum of digits = 4+5+6+7+8 = 30 (divisible by 3). Therefore, 45678 is divisible by 6.
Q29. Find the HCF of 2/3, 4/5, and 6/7.
  1. 2/105
  2. 12/105
  3. 2/35
  4. 12/35
Show answer
Answer: A. 2/105
The HCF of fractions is given by HCF(Numerators) / LCM(Denominators). HCF(2, 4, 6): 2 = 2 4 = 2^2 6 = 2 × 3 HCF(2, 4, 6) = 2. LCM(3, 5, 7): Since 3, 5, and 7 are prime numbers, their LCM is their product: 3 × 5 × 7 = 105. So, HCF(2/3, 4/5, 6/7) = 2/105.
Q30. Find the LCM of 1/3, 5/6, and 2/9.
  1. 5/18
  2. 10/3
  3. 2/3
  4. 10/9
Show answer
Answer: B. 10/3
The LCM of fractions is given by LCM(Numerators) / HCF(Denominators). LCM(1, 5, 2): 1 = 1 5 = 5 2 = 2 LCM(1, 5, 2) = 1 × 5 × 2 = 10. HCF(3, 6, 9): 3 = 3 6 = 2 × 3 9 = 3^2 HCF(3, 6, 9) = 3. So, LCM(1/3, 5/6, 2/9) = 10/3.
Q31. The greatest number of three digits which is exactly divisible by 3, 4, 5 and 8 is?
  1. 120
  2. 960
  3. 900
  4. 980
Show answer
Answer: B. 960
First, find the LCM of 3, 4, 5, and 8. LCM(3, 4, 5, 8) = 120. The greatest 3-digit number is 999. To find the greatest 3-digit multiple of 120, divide 999 by 120: 999 = 120 × 8 + 39. So, the greatest 3-digit multiple is 120 × 8 = 960.
Q32. If (√2 + 1)³ = a + b√2, then the correct relation between a and b is?
  1. a - b = 2
  2. a + b = 1
  3. a = b + 1
  4. a² + b = 0
Show answer
Answer: A. a - b = 2
Expand (√2+1)³ using the formula (x+y)³ = x³ + y³ + 3x²y + 3xy²: (√2)³ + 1³ + 3(√2)²(1) + 3(√2)(1)² = 2√2 + 1 + 3(2)(1) + 3√2(1) = 2√2 + 1 + 6 + 3√2 = 7 + 5√2. Comparing this with a + b√2, we get a=7 and b=5. Then, a-b = 7-5 = 2.
Q33. In an examination, 90% of students passed and 260 students failed. The total number of students is?
  1. 2600
  2. 1500
  3. 3700
  4. 500
Show answer
Answer: A. 2600
If 90% of students passed, then the percentage of students who failed is 100% - 90% = 10%. We are given that 260 students failed. So, 10% of the total number of students is 260. Total number of students = (260 / 10) × 100 = 26 × 100 = 2600.
Q34. The HCF of two numbers is 4 and their product is 48. Find the numbers.
  1. 4, 8
  2. 4, 12
  3. 12, 16
  4. 4, 24
Show answer
Answer: B. 4, 12
We know that for any two numbers, Product of numbers = HCF × LCM. Given HCF = 4 and Product = 48. So, LCM = Product / HCF = 48 / 4 = 12. Let the two numbers be 4x and 4y, where x and y are coprime integers. Their product is (4x)(4y) = 16xy = 48, which means xy = 3. Since x and y must be coprime, the only pair (x,y) is (1,3). Thus, the numbers are 4×1 = 4 and 4×3 = 12.
Q35. The least number that should be added to 1549 so that the sum is exactly divisible by 2, 3, 5 and 7 is?
  1. 210
  2. 131
  3. 1339
  4. 79
Show answer
Answer: B. 131
For a number to be exactly divisible by 2, 3, 5, and 7, it must be divisible by their LCM. LCM(2, 3, 5, 7) = 210. Now, divide 1549 by 210: 1549 = 210 × 7 + 79. The remainder is 79. To make 1549 divisible by 210, we need to add (210 - 79) = 131 to it. The sum will be 1549 + 131 = 1680, which is 210 × 8.
Q36. There are 20 people in a party. If every person shakes hand with every other person, then what will be the total number of handshakes?
  1. 145
  2. 190
  3. 180
  4. 155
Show answer
Answer: B. 190
This is a combination problem. Each handshake involves 2 people, and the order doesn't matter. So, the total number of handshakes is given by the combination formula nC₂ = n(n-1)/2, where n is the number of people. Here, n=20. So, ²⁰C₂ = (20 × 19) / (2 × 1) = 380 / 2 = 190.
Q37. The largest number of four digits that is divisible by 12, 15 and 18 is?
  1. 9000
  2. 9900
  3. 9450
  4. 9750
Show answer
Answer: B. 9900
First, find the LCM of 12, 15, and 18. LCM(12, 15, 18) = 180. The largest 4-digit number is 9999. To find the largest 4-digit number divisible by 180, divide 9999 by 180: 9999 = 180 × 55 + 99. The remainder is 99. Subtract the remainder from 9999 to get the desired number: 9999 - 99 = 9900.
Q38. Find the ninth term of an arithmetic progression with the first term as 5 and the common difference as 4.
  1. 35
  2. 37
  3. 39
  4. 41
Show answer
Answer: B. 37
The formula for the n-th term of an arithmetic progression (AP) is Tₙ = a + (n-1)d, where 'a' is the first term, 'n' is the term number, and 'd' is the common difference. Given a=5, d=4, and n=9. So, T₉ = 5 + (9-1)×4 = 5 + 8×4 = 5 + 32 = 37.
Q39. Find the value of x if √(1 + x/144) = 13/12
  1. 20
  2. 16
  3. 30
  4. 25
Show answer
Answer: D. 25
To solve for x, first square both sides of the equation: (√(1 + x/144))² = (13/12)². This gives 1 + x/144 = 169/144. Now, isolate x/144: x/144 = 169/144 - 1. Convert 1 to 144/144: x/144 = 169/144 - 144/144 = (169 - 144)/144 = 25/144. Since the denominators are equal, the numerators must be equal: x = 25.
Q40. How many of the first 100 positive integers are divisible by 3 or 4 without a remainder?
  1. 85
  2. 50
  3. 5
  4. 58
Show answer
Answer: B. 50
Use the principle of inclusion-exclusion. Number of integers divisible by 3 = ⌊100/3⌋ = 33. Number of integers divisible by 4 = ⌊100/4⌋ = 25. Number of integers divisible by both 3 and 4 (i.e., by LCM(3,4)=12) = ⌊100/12⌋ = 8. Total integers divisible by 3 or 4 = (Divisible by 3) + (Divisible by 4) - (Divisible by both) = 33 + 25 - 8 = 58 - 8 = 50.
Q41. The sum of two numbers is 20 and their product is 96. What is the difference between the two numbers?
  1. 8
  2. 6
  3. 5
  4. 4
Show answer
Answer: D. 4
Let the two numbers be x and y. We are given x+y=20 and xy=96. We need to find x-y. We know the algebraic identity: (x-y)² = (x+y)² - 4xy. Substitute the given values: (x-y)² = (20)² - 4(96) = 400 - 384 = 16. Taking the square root of both sides, x-y = √16 = 4 (since we are looking for a positive difference).
Q42. The LCM of two numbers is 96 and their HCF is 8. If one of the two numbers is 32, then what is the other number?
  1. 48
  2. 28
  3. 16
  4. 24
Show answer
Answer: D. 24
The product of two numbers is equal to the product of their LCM and HCF. Let the two numbers be N1 and N2. So, N1 × N2 = LCM × HCF. Given N1=32, LCM=96, HCF=8. Substitute these values: 32 × N2 = 96 × 8. N2 = (96 × 8) / 32 = 768 / 32 = 24.
Q43. The sum of the first four prime numbers is?
  1. 17
  2. 19
  3. 16
  4. 18
Show answer
Answer: A. 17
Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves. The first few prime numbers are 2, 3, 5, 7, 11, 13, ... The first four prime numbers are 2, 3, 5, and 7. Their sum = 2 + 3 + 5 + 7 = 17.
Q44. Find the smallest natural number N such that the product 288 × N is a perfect cube.
  1. 12
  2. 8
  3. 9
  4. 6
Show answer
Answer: D. 6
First, find the prime factorization of 288: 288 = 2 × 144 = 2 × 12² = 2 × (2² × 3)² = 2 × 2⁴ × 3² = 2⁵ × 3². For a number to be a perfect cube, the exponent of each prime factor in its prime factorization must be a multiple of 3. In 2⁵ × 3², the exponent of 2 is 5 (which needs 1 more 2 to become 6, a multiple of 3) and the exponent of 3 is 2 (which needs 1 more 3 to become 3, a multiple of 3). So, N must be 2¹ × 3¹ = 2 × 3 = 6.
Q45. Four bells begin to toll together at 8.00 am and then each one at intervals of 5s, 6s, 7s and 8s respectively. How many times will they toll together in a shift of 7 hours?
  1. 21 times
  2. 25 times
  3. 15 times
  4. 31 times
Show answer
Answer: D. 31 times
To find when the bells will toll together, we need to find the LCM of their intervals: LCM(5, 6, 7, 8). Prime factorization: 5=5, 6=2×3, 7=7, 8=2³. LCM = 2³ × 3 × 5 × 7 = 8 × 3 × 5 × 7 = 840 seconds. Convert 840 seconds to minutes: 840/60 = 14 minutes. The bells toll together every 14 minutes. The total shift duration is 7 hours = 7 × 60 = 420 minutes. Number of times they toll together after the initial toll = Total time / Interval = 420 / 14 = 30 times. Since they also tolled together at the beginning (8:00 am), we add 1 to this count. So, total times = 30 + 1 = 31 times.
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