Number systems: Bank Exam Practice Questions (SBI, IBPS, RRB, PO & Clerk)

Ans: E

Explanation: To solve this, we multiply 412 by 13.
412 * 13 = 5356

Ans: B

Explanation:
Let the five consecutive even numbers be 4, 6, 8, 10, and 12. Their sum is 4+6+8+10+12 = 40. The sum of the odd numbers must be 96 – 40 = 56.
Let the four consecutive odd numbers be x, x+2, x+4, and x+6. Their sum is 4x + 12 = 56.
Solving for x, we get 4x = 44, so x = 11.
The odd numbers are 11, 13, 15, and 17. The smallest odd number is 11, and the largest even number is 12.
The difference between the largest even number and the smallest odd number is 12 – 11 = 1.

Correct Option: B

Ans: C

Explanation: Let the two-digit number be represented as 10a + b, where ‘a’ is the tens digit and ‘b’ is the units digit. When the digits are reversed, the number becomes 10b + a. The problem states that the reversed number is 63 larger than the original number, so:
10b + a = 10a + b + 63
9b – 9a = 63
b – a = 7 —(1)
The sum of the digits is 11, so:
a + b = 11 —(2)
Now we have a system of two equations. We can solve by adding equations (1) and (2):
(b – a) + (a + b) = 7 + 11
2b = 18
b = 9
Substitute b = 9 into a + b = 11:
a + 9 = 11
a = 2
So, the original number x = 10a + b = 10(2) + 9 = 29. Let us check whether its digits reversed i.e. 92, and the condition is satisfied, 92 – 29 = 63. Also the sum of digits 2 + 9 = 11.
Correct Option: C

Ans: A

Explanation: First, we need to find the values of X and Y. We are given that X + Y = 15 and X * Y = 56. We can solve this system of equations. By inspection or by solving the quadratic equation formed (x^2 – 15x + 56 = 0), we find that X and Y are 7 and 8.

Now, we can find the HCF (Highest Common Factor) and LCM (Least Common Multiple) of 7 and 8.
HCF(7, 8) = 1 (since 7 is a prime number and 8 = 2*2*2)
LCM(7, 8) = 7 * 8 = 56 (since 7 and 8 have no common factors other than 1)

Quantity A: HCF of X and Y = 1
Quantity B: LCM of X and Y = 56

Since 1 < 56, Quantity A < Quantity B.

Ans: D

Explanation: Let the two consecutive positive whole numbers be x and x+1. We are given that x(x+1) = 8742. We can approximate x by taking the square root of 8742, which is approximately 93.5. So, we can test 93 * 94. 93 * 94 = 8742. The two consecutive numbers are 93 and 94. The larger number is 94. The sum of the digits of 94 is 9 + 4 = 13.

Correct Option: D

Ans: C

Explanation: We need to find the smallest positive number that, when added to or subtracted from 700, results in a perfect square.
Let’s consider subtracting a number from 700. The largest perfect square less than 700 is 26^2 = 676. The difference is 700 – 676 = 24.
Now consider adding a number to 700. The smallest perfect square greater than 700 is 27^2 = 729. The difference is 729 – 700 = 29.
Comparing the two differences, 24 and 29, we find that 24 is the smaller one.
Therefore, the smallest positive number is 24.

Correct Option: C

Ans: C

Explanation: A number is divisible by 9 if the sum of its digits is divisible by 9. The sum of the digits of 51234 is 5 + 1 + 2 + 3 + 4 = 15. The next multiple of 9 after 15 is 18. The difference between 18 and 15 is 3. However, if we add 6, the sum becomes 21, which divided by 9 gives remainder of 3. We want a number divisible by 9 so need the remainder to be 0 when divided by 9, or a multiple of 9. Adding 3 to 15 gives 18, which is divisible by 9. However, 3 is not an option. Consider 15 + x. We want 15 + x to be divisible by 9. 15+6=21 (not divisible by 9); 15+3=18; if we added 6, we’d have 15+6=21, meaning the sum of the digits in the result would be 2+1=3, which means the remainder is 3 when divided by 9. Adding the next number, we need the sum of the digits to be divisible by 9. 5+1+2+3+4 = 15. The next number divisible by 9 is 18. 18-15 = 3. 51234+3 = 51237, which becomes 5+1+2+3+7=18, which is divisible by 9. The question is asking what needs to be added to 51234. Adding 3 gives us a number divisible by 9. The available answers seem to be erroneous, perhaps the closest is adding 6, however this gives remainder 3 after division by 9. We need a number to be added that changes the remainder. Checking the remainder when the given options is added to 51234: A) 51234+6 = 51240. 5+1+2+4+0 = 12 which is not divisible by 9 B) 51234+4 = 51238. 5+1+2+3+8 = 19 which is not divisible by 9 C) 51234+3 = 51237. 5+1+2+3+7 = 18. This is the correct answer. The options may be wrong. D) 51234+5 = 51239. 5+1+2+3+9 = 20 not divisible by 9 E) 51234+7 = 51241. 5+1+2+4+1 = 13 not divisible by 9

Correct Option: C

Ans: B

Explanation: We need to find the smallest number to add to 4425 to result in a perfect square. First, let’s estimate the square root of 4425. Since 60^2 = 3600 and 70^2 = 4900, the square root of 4425 is between 60 and 70. Let’s try 66: 66^2 = 4356 (too small) and 67^2 = 4489. Since 4489 – 4425 = 64, we need to add 64 to 4425 to get a perfect square (4489). Let’s verify our result: 67 * 67 = 4489.
Correct Option: B