ExamVeda
Login
Home

Number System - Aptitude MCQ Questions and Solutions with Explanations

Number System MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Number System with easy and logical explanations. Number System MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-1 section-6

Each section contains 100 MCQs on Number System .
1
The sum of the perfect square between 120 and 300 is :
Discuss
Answer & Solution
Answer: Option D
Solution:
Perfect squares number between 120 and 300
= 121, 144, 169, 196, 225, 256 and 289
Sum of perfect square between 120 and 300
= 121 + 144 + 169 + 196 + 225 + 256 + 289
= 1400
2
What is the remainder when 461 is divided by 51 ?
Discuss
Answer & Solution
Answer: Option D
Solution:
461 = 4 × 460 = 4 × (44)15 = 4 × (256)15
Now, (xn - an) is divisible by (x - a) for all values of n.
∴ (25615 - 1) is divisible by (256 - 1) i.e., 255 and hence by 51
⇒ On dividing (256)15 by 51, we get 1 as remainder
⇒ On dividing 460 by 51, we get 1 as remainder
⇒ On dividing 461 by 51, remainder obtained = (4 × 1) = 4
3
If n is a natural number and n = $${p_1}^{{x_1}}$$   $${p_2}^{{x_2}}$$   $${p_3}^{{x_3}}$$ where p1, p2, p3 are distinct prime factors, then the number of prime factors for n is :
Discuss
Answer & Solution
Answer: Option B
Solution:
Given n = $${p_1}^{{x_1}}{p_2}^{{x_2}}{p_3}^{{x_3}}$$
Where p1, p2, p3 are distinct prime factors
Number of prime factors form :
= (x1 × x2 × x3)
= x1 x2 x3
Hence option (B) is correct
4
The number of prime factors in the expression 610 × 717 × 1127 is equal to :
Discuss
Answer & Solution
Answer: Option B
Solution:
610 × 717 × 1127
= (2 × 3)10 × 717 × 1127
= 210 × 310 × 717 × 1127
Number of prime factors in the given expression
= (10 + 10 + 17 + 27)
= 64
5
A number when divided by 195 leaves a remainder 47. If the same number is divided by 15, the remainder will be :
Discuss
Answer & Solution
Answer: Option B
Solution:
Let the number be x and the quotient be q
Then, x = 195q + 47
= (15 × 13q) + (15 × 3) + 2
= 15 (13q + 3) + 2
So, the given number when divided by 15 gives 2 as remainder.
6
A young girl counted in the following way on the fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8, thumb 9 and then back to the index finger for 10, middle finger for 11, and so on. She counted upto 1994. She ended on her
Discuss
Answer & Solution
Answer: Option B
Solution:
Number of thumbs = 1, 9, 17, 25, .....
This is an AP in which a = 1 and d = (9 - 1) = 8
∴ Tn = a + (n - 1) d
        = 1 + (n - 1) 8
        = (8n - 7)
∵ 8n - 7 = 1994
⇒ 8n = 2001
⇒ n = 250
T250 = 1 + (250 - 1) 8
        = 1 + 249 × 8
        = 1993
So, 1993 lies on thumb and 1994 on index finger.
7
The numbers 1, 2, 3, 4, ......, 1000 are multiplied together. The number of zeros at the end (on the right) of the product must be :
Discuss
Answer & Solution
Answer: Option D
Solution:
Let N = 1 × 2 × 3 × 4 × ..... × 1000 = 1000!
Clearly, the highest power of 2 in N very high as compared to that of 5.
So, the number of zeros in N will be equal to the highest power of 5 in N.
∴ Required number of zeros
= $$\left[ {\frac{{1000}}{5}} \right] + \left[ {\frac{{1000}}{{{5^2}}}} \right] + \left[ {\frac{{1000}}{{{5^3}}}} \right]$$      $$ + \left[ {\frac{{1000}}{{{5^4}}}} \right]$$
= 200 + 40 + 8 + 1
= 249
8
The number 89715938* is divisible by 4. The unknown non-zero digit marked as * will be :
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the missing digit be x.
Then, (80 + x) must be divisible by 4.
Hence, x = 4
9
If p is a prime number greater than 3, then (p2 - 1) is always divisible by :
Discuss
Answer & Solution
Answer: Option C
Solution:
⇔ p = 5
⇒ (p2 - 1) = (25 - 1)
⇒ (p2 - 1) = 24, which is divisible by 24
⇔ p = 7
⇒ (p2 - 1) = (49 - 1)
⇒ (p2 - 1) = 48, which is divisible by 24
⇔ p = 11
⇒ (p2 - 1) = (121 - 1)
⇒ (p2 - 1) = 120, which is divisible by 24
Hence, (p2 - 1) is always divisible by 24
10
The unit's digit of 132003 is :
Discuss
Answer & Solution
Answer: Option C
Solution:
34 gives unit digit 1
So, (34)500 gives unit digit 1
And, 33 gives unit digit 7
∴ (13)2003 gives unit digit = (1 × 7) = 7