Preparing for the OSSC CGL exam requires strong proficiency in arithmetic and numerical ability. Focusing on the top 30 multiple-choice questions (MCQs) in this area can significantly enhance your performance. These MCQs cover essential topics such as percentages, ratios, averages, simple and compound interest, profit and loss, time and work, and data interpretation. Practicing these questions helps in understanding key concepts and improves speed and accuracy. Comprehensive preparation with these critical MCQs ensures a solid foundation in arithmetic and numerical ability, crucial for excelling in the OSSC CGL exam
Top 30 Arithmetic and Numerical Ability MCQs for OSSC CGL
- Question: The HCF of two numbers is 8 and their LCM is 240. If one of the numbers is 32, find the other number.
[A] 60
[B] 72
[C] 80
[D] 96
Answer: [A] 60]
Solution: Let the other number be x. We know that
HCF×LCM=Product of the numbers.
So, 8×240=32×x. Solving for x, we get
x=(8×240)/32 =60 - Question: The LCM of two numbers is 720 HCF is 4 and their ratio is 4:5. Find the sum of the numbers.
[A] 180
[B] 240
[C] 108
[D] 400
Answer: C [108]
Solution: Let the numbers be 4x and 5x.
We know that
HCF×LCM=Product of the numbers.
So, 720 × 4 =4x×5x
. Since the ratio is given, we know the HCF is 4. Solving for
x, we get x=12
x=9. Hence, the sum of the numbers is
4x+5x=9x=12x 9=108 - Question: The HCF of two numbers is 12, and their LCM is 144. If one of the numbers is 36, find the other number.
[A] 48
[B] 72
[C] 96
[D] 108
Answer: A [48]
Solution: Let the other number be
HCF×LCM=Product of the numbers. So,
12×144=36×x. Solving for x, we get
x= 12×144/36 =48. - Question: The LCM of two numbers is 180. If their HCF is 15 and one number is 45, find the difference of 2 numbers.
[A] 25
[B] 30
[C] 15
[D] 40
Answer: C [15]
Solution: Let the other number be x. We know that
HCF×LCM=Product of the numbers. So,
15×180=45×x. Solving for
x, we getx= 15×180/ 45 =60.
Hence, the other number is
x=60−45=15. - Question: The HCF of two numbers is 6, and their LCM is 210. If one of the numbers is 18, find the difference of 2 numbers.
[A] 52
[B] 40
[C] 45
[D] 50
Answer: A [52]
Solution: Let the other number be x. We know that
HCF×LCM=Product of the numbers. So,
6×210=18×x. Solving for x, we get
x= 6×210/18 =70. Hence, the difference of number is
x=70−18=52. - A prime number N. in the range 10 to 50, remains unchanged when its digits are reversed. The square of such a number is
(1) 1936
(2) 1089
(3) 484
(4) 121
Answer:(4) 121
Sol: A prime number N from 10 to 50 which on interchanging the digits remains same is 11, so its square is 121. - A four-digit number formed by repeating a two digit number is always exactly divisible by
(1) 7
(2) 19
(3) 83
(4) 101
Answer:(4) 101
Sol: All the four digit numbers which are formed by repeating a two digit number are always divisible by 101. - If a, b, c,…. x26 natural numbers, then the value of (x – a)(x – b)(x – c) … (x – y)(x – z) is
(1) 0
(2) 1
(3) 13
(4) 26
Answer:(1) 0
Sol: 14. If a, b, c, …, x, y, z are 26 natural numbers, then
(x-a)(x-b)…. (x – x)(x – y)(x – z) =(x-a)(x-b)… 0(x – y)(x – z) = 0 - If a six-digit number 93p 25q is divisible by 88, then the values of p and q are respectively
(1) 2 and 8
(2) 8 and 2
(3) 8 and 6
(4) 6 and 8
Answer:(3) 8 and 6
Sol: 15. If 93p 25q is divisible by 88 then it must be divisible by 8 and 11 both. We test the divisibility by 8 first, for that the last three digits must be divisible by 8 i.e., 25q is divisible by this gives a = 6 When the number is divisible by 11, then (9 + p + 5) – (3 + 2 + q) must be zero or multiple of 11, this gives us 14 + p – 11 =3+p Rightarrow p=8 - Consider the following statements
1. The sum of two prime numbers is a prime number.
2. The product of two prime numbers is a prime number. Which of these statements is/are correct?
(1) Neither 1 nor 2
(2) 1 alone
(3) 2 alone
(4) Both 1 and 2
(5) None of these
Answer:(1) Neither 1 nor 2
Sol: Neither the sum of two prime numbers is always a prime number nor their product is prime. - If a and b are such numbers that a > 0 and b <0, then which one of the following is always correct?
(1) a-b>0
(4) a – b < 0
(2) a + b > 0
(3) a + b < 0
(5) None of these
Answer: (1) a-b>0
Sol: Ifa > 0 and b < 0 then a – b > 0 is always true as if suppose b = – p < 0 then a – b = a – (- p) = a + p > 0 which is always true. - The sum of the numerator and denominator of a certain fraction is 8. If 2 is added to both the numerator and the denominator the value of the fraction is increased by 4/35, then the fraction is
(1 ) 1/7
(2) 3/5
(3) 5/3
(4) cannot be determined
Answer:(4) cannot be determined
Sol: cannot be determined - In a number of three digits, the digits in the unit’s place and the hundred’s place are equal and the sum of all the digits is 8. The number of /such numbers is
(1) 3
(2) 4
(3) 5
(4) 6
Answer:(1) 3
Sol: The numbers of three-digits having unit’s place and the hundred’s place equal can be 611, 422 and 233. So, there are only three such numbers. - In (337)^337, the unit digit is occupied by
(1) 1
(2) 3
(3) 7
(4) 9
Answer:(4) 9
Sol: To find the unit digit of 337^337
, we can use the concept of cyclicity of unit digits.
Let’s first find the cyclicity of the unit digit of 337:
337^1 =337→unit digit=7
337^2 =113369 →unit digit=9
337^3 =38272753→unit digit=3
337^4 =28880041761→unit digit=1
Now, notice the pattern: the unit digit of powers of 337 follows the cycle: 7, 9, 3, 1.
Since the exponent 337 is one more than a multiple of 4, the unit digit of 337^337 will be the next number in the cycle after 7, which is 9.
Therefore, the unit digit of 337^337=9 - 36 * 45 + sqrt ? =1700
(1) 640
(2) 360
(3) 80
(4) 6400
Answer: (4) 6400 - 12 ^ 2 + 16 =
(1) 180
(2) 160
(3) 240
(4) 156
Answer: (2) 160 - 2 whole 3/5 + 4 whole 1/5 – 3 whole 2/5 =
(1) 3 whole 1/5
(2) 3 whole 4/5
(3) 4 whole 2/5
(4) None of these
Answer: (4) None of these - 3785+?-4297= 5523
(1) 6125
(2) 6035
(3) 6025
(4) 6135
Answer: (2) 6035 - 11.02 * 8 + 1.5 * 0.6 =
(1) 89.06
(2) 89.60
(3) 89.96
(4) 89.26
Answer: (1) 89.06 - 3/5 of 4/7 of 7/9 of 1380 = ?
(1) 378
(2) 376
(3) 368
(4) 386
Answer: (3) 368 - 75% of 580 + 58% of 750 =
(1) 840
(2) 860
(3) 680
(4) 870
Answer: (4) 870 - Among the following fractions, the largest fraction is
(1) 10/11
(2) 0.91
(3) (7 * 3)/6
(4) (8 * 3)/9
Answer:(2) 0.91
Sol. 10/11 = 0 (7 * 3)/8 = 0 * 9125 , , (8 * 3)/9 = 0 * 9222 - The rational numbers lying between 1/3 and 3/4 are
(1) 97/300 ,299 /500
(2) 99/300 , 301/400
(3) 95/300, 301/400
(4) 117/300, 287/400
(5) None of these
Answer:(4) 117/300, 287/400
Sol. 1/3 = 0 * 33334 117/300 = 0 .3900 and 287/400 = 0 . 7175, 117/300, 287/400 lies between 1/3 and 3/4 - Arrangement of – 3/7, 2/3 and – 1/3 in the ascending order is
(1) – 3/7, – 1/3, 2/3
(2) 2/3, – 1/3, – 3/7
(3) – 1/3, – 3/7, 2/3
(4) – 3/7, 2/3, – 1/3
Answer:(1) – 3/7, – 1/3, 2/3
Sol. – 3/7 = 0.4285, 2/3 = 0.6667 and – 1/3 = 0 . 3334 So, the ascending order is given by 0.4285 < – 0.3334 < 0.6667 – 3/7 < – 1/3 < 2/3 - (6.5×6-5-45.5+3.5×3.5) is equal to
(1) 10
(2) 9
(3) 7
(4) 6
Answer:(2) 9
Sol.6. 5 * 6 . 5 – 45 .5 + 3.5 * 3.5 =(6 .5) ^ 2 – 2 * 6 .5 * 3.5 + (3.5) ^ 2 = (6.5 – 3.5) ^ 2 = (3) ^ 2 = 9 - Given that 0.111…= 1/9 , 0.444 is equal to
(1) 1/90
(2) 2/45
(3) 1/99
(4) 4/9
Answer:(4) 4/9
Sol.0*1111…= 1 /9 , so 0*444…= 4 * 0 .111 … = 4 * 1/9 = 4/9 - The smallest possible three place decimal is
(1)0.012
(2)0.123
(3)0.111
(4) 0.001
Answer:(4) 0.001
Sol.The smallest possible three place decimal is 0-001. - (16 ^ 0.16 * 2 ^ 0.36) is equal to
(1) 2
(2) 16
(3) 32
(4) 64
Answer:(1) 2
Sol. 16 ^ (0 * 16) * 2 ^ 36 = 2 ^ (4 * 0 * 16) * 2 ^ 36 = 2 ^ (64 + 36) = 2 ^ 100 = 2 - On simplification of{ ( 2. 644 )^ 2 -(2.356)^2} /0.288 we get
(1) 1
(2) 4
(3) 5
(4) 6
Answer:(3) 5
Sol. { ( 2. 644 )^ 2 -(2.356)^2} /0.288= (2. 644 +2.356)(2. 644 -2.356)/0.288=5*0.288/0.288=5 - The value of ( 243 )^ 0.16 x (243)^0.04 is equal to
(1)0.16
(2) 3
(3) 1/3
(4)0.04
Answer:(2) 3
Sol.( 243 )^ 0.16 x (243)^0.04=( 243 )^0.2=(3)^5*0.2=3
