1,000+ TNPSC Previous Year Model Question Papers with Answers - 1

Question: 1

The sum of two numbers is 528 and their H.C.F. is 33. The numbers of pairs of numbers satisfying the above conditions is

(A) 2

(B) 4

(C) 6

(D) 7

Ans: B

Let the required numbers be 33a and 33b.

Then, 33a + 33b = 528 ⇔ a + b = 16.

Now, co-primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7,9).

∴ Required numbers are (33 × 1, 33 × 15), (33 × 3, 33 × 13),
(33 × 5, 33 × 11), (33 × 7, 33 × 9)

The number of such pairs is 4.

Question: 2

The H.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the numbers is 77, find the other.

(A) 66

(B) 99

(C) 119

(D) 909

Ans: B

Required number = $${L.C.M x H.C.F.}/{Given number}$$ = $${693 × 11} / {77}$$ = 99.

Question: 3

The least number, which when increased by 1 is exactly divisible by 12, 18, 24, 32, 40 is

(A) 1439

(B) 1440

(C) 1449

(D) 1459

Ans: A

LCM of 12, 18, 24, 32, 40 is 1440.

∴ Required number is 1440 - 1 = 1439.

Question: 4

The G.C.D. of two whole numbers is 5 and their LCM is 60. If one of the number is 20, then the other number would be

(A) 13

(B) 15

(C) 23

(D) 25

Ans: B

1st number x 2nd number = GCD x LCM

∴ 2nd number = $${5 × 60} / {20}$$ = 15.

Question: 5

The product of two 2 digit numbers is 2028 and their GCM is 13. The numbers are

(A) 13, 156

(B) 26, 78

(C) 36, 68

(D) 39, 52

Ans: D

Let the numbers be 13x and 13y.

Then 13x × 13y = 2028 or xy = 12.

Co-primes with product 12 are (1, 12) and (3, 4).

So, the numbers with HCF 13 and product 2028

are (13 x 1, 13 x 12) and (13 x 3, 13 x 4)

∴ Required 2 digit numbers are 39 and 52.

Related Questions