Arithmetic (UPSC) Part- V

Total Questions: 50

41. D is a 3-digit number such that the ratio of the number to the sum of its digits is least. What is the difference between the digit at the hundred’s place and the digit at the unit’s place of D? [2023-II]

(a) 0
(b) 7
(c) 8
(d) 9

Correct Answer: (c) 8

Solution:A fraction is minimum when numerator is minimum and denominator is maximum
Ratio (R) = 100x + 10y + z/(x + y + z) For a three digit number

D = (100x + 10y + z)
Here, 1 ≤ x ≤ 9 and 0 ≤ (y, z) ≤ 9.
For minimum numerator and maximum denominator,

x - 1, y = z = 9

R = 100 × 1 + 10 × 9 + 9 = 199
1 + 9 + 9 = 19
R = 10.47

∴ Difference = |x - z| = |1 - 9| = 8

42. Three of the five positive integers p, q, r, s, t are even and two of them are odd (not necessarily in order). Consider the following: [2023-II]

p + q + r – s – t is definitely even.
2p + q + 2r – 2s + t is definitely odd.

Which of the above statements is/are correct?

(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2

Correct Answer: (a) 1 only

Solution:We know that
(i) Even ± Even = Even
(ii) Odd ± Odd = Even
(iii) Even ± Odd = Odd
(iv) Odd ± Even = Odd

S1: case I: P, Q, R are even and S, T are odd.
P + Q + R - S - t = (P + Q + R) - (S + t)
= Even - Even = Even.

Case II: P, Q, S are even and r, t are odd.
(P+Q-S) + (R - t) = Even + Even = Even
Case III: P, S, t are even and Q, R are odd
(P- (S + T)) + (Q + R)
= Even - Even + Even = Even
Hence, P + Q + R - S - t is definitely even. Hence, S1 is correct.

S2 : Let P, r, s are even and Q, t are odd
2P + E + 2r - 2S + T
= 2(P + r - s) + (q + t)
= Even + Even = Even
Hence, S2 is not correct.

43. Consider the following in respect of prime number p and composite number c. [2023-II]

p + c/ p - c can be even
2p + c can be odd
pc can be odd

Which of the statements given above are correct?

(a) 1 and 2 only
(b) 2 and 3 only
(c) 1 and 3 only
(d) 1, 2 and 3

Correct Answer: (d) 1, 2 and 3

Solution:Let Prime number, P = 11
Composite number, C = 9

P + C/P - C = 11 + 9/11 - 9 = 20/2 = 10 (Even)
2P + C = 2 × 11 + 9 = 22 + 9 = 31 (Odd)

P.C = 11 × 9 = 99 (Odd)

44. A 3-digit number ABC, on multiplication with D gives 37DD where A, B, C and D are different non-zero digits. What is the value of A + B + C? [2023-II]

(a) 18
(b) 16
(c) 15
(d) Cannot be determined due to insufficient data

Correct Answer: (a) 18

Solution:A B C

D
--------------

3 7 D D
C × D = D
For C = 6, D = 4, ⇒ 6 × 4 = 24
For B = 3, D = 4 ⇒ 3 × 4 = 12
For A = 9, D = 4 ⇒ 9 × 4 = 36

Now, 9 3 6
4
--------
3 7 4 4

∴ A + B + C = 9 + 3 + 6 = 18

45. For any choices of values of X, Y and Z, the 6-digit number of the form XYZXYZ is divisible by: [2023-II]

(a) 7 and 11 only
(b) 11 and 13 only
(c) 7 and 13 only
(d) 7, 11 and 13

Correct Answer: (d) 7, 11 and 13

Solution:XYZXYZ = XYZ × 1001
L.C.M. of 7, 11 and 13 = 7 × 11 × 13 = 1001.
∴ XYZ × 1001 = XYZ × 7 × 11 × 13
Hence, Number 'XYZXYZ' is divisible by 7, 11 and 13.

46. Let x be a positive integer such that 7x + 96 is divisible by x. How many values of x are possible? [2023-II]

(a) 10
(b) 11
(c) 12
(d) Infinitely many

Correct Answer: (c) 12

Solution:Let Z = (7x + 96), Here z is divisible by x.

Tricky Way:
In such type of questions, we should find the factor of constant term: Number of factors of constant term gives the value of X, factors of 96.
= 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96.
∴ Value of X = 12.

47. If p, q, r and s are distinct single digit positive numbers, then what is the greatest value of (p + q)(r + s)? [2023-II]

(a) 230
(b) 225
(c) 224
(d) 221

Correct Answer: (b) 225

Solution:G.M. ≤ A.M.
(p + q) (r + s) ≤ ( (p + q) + (r + s)/2 )²

For largest value, (p, q, r, s) = (9, 8, 7, 6)
∴ Largest value = ( (9 + 8 + 7 + 6)/2 )²
= 225

48. A number N is formed by writing 9 for 99 times. What is the remainder if N is divided by 13? [2023-II]

(a) 11
(b) 9
(c) 7
(d) 1

Correct Answer: (a) 11

Solution:Rule of divisibility by 13:
From the right most group of 3 digits apply the subtraction and addition operations alternatively and find the result. If the result is either 'O' or multiple of 13, then whole number is divisible by 13.
Here N = 99999999 ... 999999
↓
99th times
= 999,999,999, ... 999, 999
33th group
= New, (999 + 999 + ... 17 times) - (999 + 999 ... 16 times)
= 999
when 999 + 13, remainder R = 11.

49. Each digit of a 9-digit number is 1. It is multiplied by itself. What is the sum of the digits of the resulting number? [2023-II]

(a) 64
(b) 80
(c) 81
(d) 100

Correct Answer: (c) 81

Solution:11 × 11 = 121 → Sum of digits
111 × 111 = 12321 → 1 + 2 + 3 + 2 + 1
= 3 + 2 (1 + 2) = 9
1111 × 1111 = 1234321 → 1 + 2 + 3 + 4 + 3 + 2 + 10 = 4 + 2 (3 + 2 + 1) = 16

......................................................

111 - 9 times × 111 - 9 times = 12345678987654321
= 9 + 2 (1 + 2 + 3 + ...8) = 81

Tricky way:
Sum of digits of product of 111 ... n times × 111 ... n time = (n)².
Here, n = 9 → Sum of digits = (9)² = 81

50. What is the sum of all digits which appear in all the integers from 10 to 100? [2023-II]

(a) 855
(b) 856
(c) 910
(d) 911

Correct Answer: (b) 856

Solution:Number's are 10, 11, 12, 13, ... 98, 99, 100.
Sum of its digits:
(1 + 0) + (1 + 1) + (1 + 2) + (1 + 3) + ... (1 + 9)
+ (2 + 0) + (2 + 1) + (2 + 2) + ... (2 + 9) ...
+ (9 + 0) + (9 + 1) + ... (9 + 9) + 1 + 0 + 0
= 1 + 1 + 1 ... 10 times + 2 + 2 + 2 ... 10 times + 9 + 9 + 9 ...
10 times + (1 + 2 + 3 + ... + 9) + (1 + 2 + 3 + ... + 9) ... 9 times
= 10 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
+ 9(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) + 1
= 19(1 + 2 + 3 + ... + 9) + 1
= 19 × 9 × 10/2 + 1
= 856.

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