Number System

Number System

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

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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

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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

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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

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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

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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

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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

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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.