This question was previously asked in

SSC Graduation Level Previous Paper (Held on: 9 Nov 2020 shift 3)

Option 4 : 15

GK Chapter Test 1 - Ancient History

20047

15 Questions
30 Marks
10 Mins

**Given :**

93, 119 and 158 when divided by n, give remainder d in each case

**Concept used :**

HCF of 3 different numbers a, b and c = HCF of (b -a), (c -b) and (c -a)

**Calculations :**

93, 119 and 158 gives remainder d when divided by n

Then, (93 - d), (119 -d) and (158 - d) must be divisible by n

Now

HCF of (93 - d), (119 - d) and (158 - d) = HCF of (119 - d - (93 - d), (158 -d -(119 -d) and (158 - d - (93 -d)

⇒ HCF of 26, 39 and 65

⇒ 13

So value of n will be 13

Remainder we get on dividing 93, 119 and 158 by 13 is 2

So d = 2

S0,

n + d = 13 + 2

⇒ 15

**∴ The value of (n + d) is 15**