Q1. The difference of the two numbers is 425. On dividing the larger number by the smaller, we get 2 as the quotient and 125 as the remainder. What is the larger number?
(a) 750
(b) 725
(c) 775
(d) 800
Solution:
Let the two numbers be x and y.
x – y = 425 —–(1)
x – 2y = 125 —–(2)
From (1) and (2), we get y = 300 and x = 725.
Hence, the larger number = 725
Hence, option (b) is the required answer.
Q2. On dividing a number by 42, we get 10 as the remainder. On dividing the same number by 6, what will be the remainder?
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
Let us take the number be x and the quotient be y.
So, x = 42y + 10 = 6* (7y +1) + 4
Therefore, the number divided by 6, we get 4 as the remainder.
Hence, (c) is the required answer.
Q3. Eight distinct prime numbers are chosen, each of whose value is less than 100. The sum of these numbers is a three-digit odd number A. If the value of A is maximum, then the sum of the least and the highest of these numbers is equal to _
(a) 96
(b) 97
(c) 98
(d) 99
Solution:
Since, the sum is an odd number, the least of these prime numbers must be 2.
Now, for A to be the maximum, the highest among these prime numbers must be 97.
Therefore, the required sum is 99.
Hence, (d) is the required answer.
Q4. How many integers from 5000 to 6999 have at least one of its digits repeated?
(a) 992
(b) 993
(c) 994
(d) 995
Solution:
Total numbers with none of its digits repeated = 2 × 9 × 8 × 7 = 1008.
So, numbers having at least one of its digits repeated = 2000 – 1008 = 992.
Hence, (a) is the required answer.
Q5. If x and y are two natural numbers such that x≤y and mn=25^1240. How many pairs (x, y) exist?
(a) 1240
(b) 1241
(c) 1242
(d) 1243
Solution:
Here, 25^1240=(〖5^2)〗^1240=3^2480
Total factors of 5^2480 is 2481.
a×b=5^2480
Out of 2481 factors, for 1240 factors x < y and in one case x = y.
Therefore, the total possible such cases = 1241
Hence, (b) is the required answer.
Q6. If n is a natural number between 1 and 28 (both included), how many natural numbers exist between 1 and 28 such that n! is divisible by (n+ 1)?
(a) 17
(b) 18
(c) 19
(d) 20
Solution:
According to the corollary of Wilson’s Theorem, n! is divisible by (n+1) where (n+1) is a composite number except 4. There are 17 such numbers 5, 7, 8, 9, 11, 13, 14, 15, 17, 19, 20, 21, 23, 24, 25, 26, and 27.
So, the total number of such numbers = 17
Hence, (a) is the required answer.
Q7. Some numbers can be expressed as the sum of three of their factors. For example, 12 can be expressed as the sum of 2, 4, and 6. How many other such numbers are there which are less than 120?
(a) 15
(b) 17
(c) 18
(d) 19
Solution:
Let N be one such number. The factors of N in decreasing order can be listed as N/2, N/3, N/4, N/5, …
Case 1: N is not divisible by 2.
In this case, the three largest factors of N would be N/3, N/5, and N/7.
N/3 + N/5 + N/7 = (35N + 21N + 15N)/105 = 71N/105
Since their sum is less than N. This implies N/2 must be a factor of N.
Case 2: N is not divisible by 3.
In this case, the next two largest factors of N after N/2 would be N/4 and N/5.
N/2 + N/4 + N/5 = (10N + 5N + 4N)/20 = 19N/20
Since their sum is less than N, this implies N/3 must be a factor of N.
Now, N/2 and N/3 must be added to N/6 to get N.
N/2 + N/3 + N/6 = N
Hence, the only possibility of 3 factors of N, whose sum is N are N/2, N/3, and N/6.
So, N must be divisible by 6.
The total number of multiples of 6 below 120 is 19, including the number 12.
So, there are 18 such numbers other than 12.
Hence, (c) is the correct answer.
Q8. What is the sum of all the numbers up to 30, which have exactly 5 or 3 or 2 factors?
(a) 184
(b) 185
(c) 181
(d) 183
Solution:
N = p4 (P is prime) has exactly five factors
N = p2 (P is prime) has exactly three factors
and also, all the prime numbers have exactly two factors.
Therefore, we have to find the sum of all prime numbers, squares of prime numbers, and fourth power of prime numbers up to 30.
= (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 ) + (22 + 32 + 52) + 24
= 129 + 38 + 16 = 183
Hence, (d) is the required answer.
Q9. How many natural numbers are there between 1 to 95, which have exactly 2 factors?
(a) 22
(b) 24
(c) 26
(d) 28
Solution:
Each prime number has exactly 2 factors.
So, the total number of prime numbers between 1 to 95 is 24.
Hence, (b) is the required answer.
Q10. Let ‘n’ be a factor of 360. How many positive integral solutions does the equation abc = n have?
(a) 760
(b) 780
(c) 800
(d) 820
Solution:
