1. The sum of all prime numbers between 58 and 68 is

(a) 179 (b) 178(c) 187 (d) 183

**S.S.C. Online CHSL (T-I) 16 Jan., 2017 (II-Shift)**

**S.S.C. Online CHSL (T-I) 7, 20, 22 Jan., 2017 (I-Shift)**

**Ans.(c)**

**Solution: Prime number between 58 and 68 are 59, 61, 67.**

**Sum of prime number 59 +61 + 67=187 Hence option (c)will be correct answer.**

2. Between 50and 150 how many numbers are divisible by 7?

(a)15 (b) 16 (c) 17 (d) 14

**S.S.C. Online CHSL (T-I) 23,27 Jan., 2017 (IlI-Shift)**

**Ans.(d)**

** Solution: The number which is divisible by 7 between 50 and 150, first number =56 and last number-147**

**Thus, 56,63,……147 is in A.P. series **

**Last number (1)= a + (n – 1) d**

**147 =56 + (n 1) 7****147-56 =(n-1) 7****91 = (n-1) 7****(n-1) =91/7 = 13****n=13+1=14**

**Thus, option (d) is correct**

**answer.**

3. Which of the following numbers is not a prime number?

(a) 197 (b) 313 (c) 439 (d) 391

** S.S.C. OnlineCHSL(T-I) 23 Jan., 2017 (IlII-Shift)**

** Ans.(d)**

** Solution: By options number 391 is divisible by 17 and other numbers 197,313 and 439 are prime number**

**Hence, option (d) is becorrect answer.**

4. Number of composite numbers lying between 67 and 101 is

(a) 27 (b) 24 (c) 26 (d) 23

**S.S.C. C.G.L. (T-I) Online 8 Sept., 2016 (lI-Shift)**

** Ans.(a)**

**Solution : Composite Numbers lying 67 and 101 are**

**68,69,70,72,74,75,76,77,78,80,81,82,84,85,86,87,88,,90,91,92,93,94,95,96,98,99,100**

**Therefore, total composite number between 67 and 101=27**

5.Which of the following numbers is not a Prime Number?

(a) 731 (b) 227 (c) 347 (d) 461

**S.S.C Online CHSL (T-1) 1 Feb.,2017 (I-Shift) **

**Ans.(a)**

**Solution :By Options number 731 is divisible by 17 and other numbers 227,347 and 461 are prime number.**

6.What smallest number should be added to 2401 so that the sum is completely divisible by 14?

(a) 8 (b) 7 (c) 4 (d) 5

**S.S.C. Online CHSL (T-I) 7, 9 Feb., 2017 (III-Shift) S.S.C. Online CHSL (T-I) 3 Feb., 2017 (II-Shift) **

**Ans.(b)**

** Solution : If 2401 is divide by 14, then we get 7 as remainder Thus, the number 2401 is completely divisible by 14 when we added 7 on it. = 14 – 7⇒ 7 **

**Hence option (b)will be correct answer. **

7. Which of the following numbers is completely divisible by 99?

(a) 57717 (b) 57627 (c) 55162 (d) 56982

**S.S.C. Online CHSL (T-I) 15 Jan., 2017 (III-Shift) S.S.C. Online CHSL (T-I) 8 Feb., 2017 (II-Shift)**

**Ans.(a)**

**Solution: Divisor=99 Factor of divisor=9 × 11 It means that the number is divisible by both 9 and 11. We know that, Divisibility by 9 ⇒ If sum of the digits of the number is divisible by 9, then number is also divisible by 9. **

**Divisibility by 11 ⇒ If difference between sum of its digits at even place and at odd place is either 0 or divisible by 11. Then numberis divisibley 11. **

**Now number, 57717 = 5 + 7+7+ 1 + 7 = 27 It is divisible by 9 , 27/9 = 3 Again,**

** number 57717 = (7+ 1)–(5+ 7+ 7) = 19 – 8 = 11 Which is also divisibleby 11.**

** Hence required numberis 57717 Thus,option (a) will becorrect answer. **

8. The largest 5 digit number exactly divisibleby 89 is:

(a) 99947 (b) 99940 (c) 99938 (d) 99939

**S.S.C. Online CHSL (T-I) 16 Jan., 2017 (II-Shift) S.S.C. Online CHSL (T-I) 19 Jan., 2017 (I-Shift) **

**Ans.(a)**

**Solution : Five digit greatest number=99999 When 99999 is divideby 89,then we get ‘52’ as remainder. **

** ∴ Required number =99999–52 ⇒ 99947**

** Hence option (a)will be correct answer.**

9. The least number to be added to 13851 to get a number which is divisibleby 87 is :

(a) 18 (b) 43 (c) 54 (d) 69

** S.S.C. C.G.L. (T-I) Online 10 Sept., 2016 (III -Shift)**

** Ans.(d)**

**Solution: On dividing 13851 by 87, we get 18 as remainder**

** ∴ Required number to be added = (87–18) = 69**

**Hence, Option (D) is correct answer.**

10. How many 3-digit numbers are there,each one of which is divisibleby 6?

(a) 149 (c) 151 (b) 150 (d) 166

** S.S.C. Online CPO, 7 June, 2016 (II-Shift)**

** Ans.(b) **

**Solution : The desired numbers are 102, 108, 114, …… 996. **

**This is an A.P. series in which a = 102 and d= 6. **

**Let the number of terms in A.P.be n**

**. Then, a +(n–1). d=996 **

**⇒ 102 +(n – 1) ×6 = 996 **

**⇒ 6(n – 1)=894**

** ⇒ (n – 1) = 149**

**⇒ n = 150.**

** Hence option (b) will be correct answer.**