Arithmetic (UPSC) Part- V

Total Questions: 50

31. Let p be a two-digit number and q be the number consisting of same digits written in reverse order. If p × q = 2430, then what is the difference between p and q? [2022-II]

Correct Answer: (d) 9
Solution:This question is a tricky question.
To get unit digit zero after multiplying two digits number and number obtained by reversing its digits number must have its digit 5 and factor of 2
Such that 25 × 52 = 1300
45 × 54 = 2430
65 × 56 = 3640
and, 85 × 58 = 4930
Hence, p = 45, q = 54
difference, q – p = 54 – 45 = 9

32. Consider the following statements in respect of two natural numbers p and q such that p is a prime number and q is a composite number: [2022-II]

  1. p × q can be an odd number.
  2. q / p can be a prime number.
  3. p + q can be a prime number.

 

Correct Answer: (d) 1, 2 and 3
Solution:

p = prime number i.e. 2, 3, 5, ......
q = composite number i.e. 4, 6, 8, 9, 10 .....
(I) p × q = 3 × 9 = 27 (odd number)

(II) q/p = 6/2 = 3 (prime number)

(III) p + q = 2 + 9 = 11 (prime number)
Hence, all statements are correct.

33. There are two containers X and Y. X contains 100 ml of milk and Y contains 100 ml of water. 20 ml of milk from X is transferred to Y. After mixing well, 20 ml of the mixture in Y is transferred back to X. If m denotes the proportion of milk in X and n denotes the proportion of water in Y, then which one of the following is correct? [2022-II]

Correct Answer: (a) m = n
Solution:
Container XContainer Y
Initial ÷ Milk = 100 mlWater = 100 ml
First transfer ÷ Milk = 80 mlWater = 100 ml
Second transfer ÷ MilkMilk = 20 ml
= 80 + 20 × (20 / (100 + 20))

∴ m = 250/3 ml

Water = 100 – 20 × (100 / (100 + 20))

∴ n = 250/3

Hence m = n

34. The sum of three consecutive integers is equal to their product. How many such possibilities are there? [2022-II]

Correct Answer: (c) Only three
Solution:For Three consecutive numbers
Sum = Product
∴ (K – 1) + K + (K + 1) = (K – 1) K. (K + 1)
= 3K = K (K² – 1)
K (K² – 1 – 3) = 0
∴ K (K² – 4) = 0
i.e. K = 0, –2 or +2
For, K = –2 three numbers are (–2 –1), (–2) and (–2 +1) = (–3, –2, –1).
For, K = 0, three numbers are (0 – 1), (0) and (0 + 1) = (–1, 0, 1)
For K = 2, three numbers are (2 – 1), (2) and (2 + 1) = (1, 2, 3)

35. What is the number of numbers of the form 0·XY, where X and Y are distinct non-zero digits? [2022-II]

Correct Answer: (a) 72
Solution:

As X and Y are distinct digits
.. Number of numbers = 9 × 8 = 72

36. The average weight of A, B, C is 40 Kg, the average weight of B, D, E is 42 kg and the weight of F is equal to that of B. What is the average weight of A, B, C, D, E and F? [2022-II]

Correct Answer: (c) 41 kg
Solution:Total weight of A, B, C = 40 × 3 = 120 kg.
Total weight of B, D, E = 42 × 3 = 126 kg.
As weight of B = weight of F
∴ Average weight of A to F
= Sum of weight of (A + B + C + D + E + F)/6

(A + B + C) + (B + D + E)/6 = 120 + 126/6 = 41 kg.

37. If 15 × 14 × 13 × ⋯ × 3 × 2 × 1 = 3^m × n where m and n are positive integers, then what is the maximum value of m? [2022-II]

Correct Answer: (b) 6
Solution:15 × 14 × 13 × 12 × ....... × 2 × 1 = 3m × n.
(3 × 5) × (14) × 13 × (4 × 3) × (11) × (10) × (3 × 3) × 8 × 7 ×
(2 × 3) × 5 × 4 × (3) × 2 × 1 = 36 × n.
∴ m = 6

38. What is the remainder when 85 × 87 × 89 × 91 × 95 × 96 is divided by 100? [2023-II]

Correct Answer: (a) 0
Solution:Let P = 85 × 87 × 91 × 95 × 96
= 5 × 17 × 3 × 29 × 7 × 13 × 5 × 19 × 8 × 3 × 4
= 25 × 32 × 52 × 7 × 13 × 17 × 19 × 29
∴ Number of Zeroes at the end of the product
= Highest power of 2 or 5 = 2
Now P ÷ 100 = Remainder = 0

39. What is the unit digit in the expansion of (57242)^9×7×5×3×1? [2023-II]

Correct Answer: (a) 2
Solution:

Here, Periodicity of 2 = 4

9 × 7 × 5 × 3 × 1 = 945 = 236, Remainder = 1
4
∴ Unit digit of 2¹ = 2

40. If ABC and DEF are both 3-digit numbers such that A, B, C, D, E and F are distinct non-zero digits such that ABC + DEF = 1111, then what is the value of A + B + C + D + E + F? [2023-II]

Correct Answer: (d) 31
Solution:A B C
D E F
1 1 1 1
Here, C + F = 11
Possible pair (C, F) = (2, 9), (3, 8), (4, 7), (5, 6)
Similarly B + E = 10
Possible pair (B, E) = (1, 9), (2, 8), (3, 7), (4, 6)
Similarly A + D = 10
Possible pair (A, D) = (1, 9), (2, 8), (3, 7), (4, 6)
∴ A + B + C + D + E + F = (A + D) + (B + E) + (C + F) = 10 + 10 + 11 = 31