# QUANT QUIZ For SSC EXAM(Numbers System)

As the SSC CGL Form are out so all of you are now very concerned about your prepration.So Here we will help you in this.We will Conduct Topic wise Quizes on daily basis and will provide study notes also.We will cover each and every section of
the syllabus. So keep yourself updated.

Here we start with Maths.In the following quiz We will cover Numbers system part.

1. n! has 23 zeroes . what is the maximum possible value of n ?
A. 99
B. 100
C. 98
D. none of these

2. n! has 13 zeroes. The  highest and least  value of n are ?

A. 59 and 55
B. 59 and 56
C. 59 and 57
D. none of these
3. Find the maximum value of n such that 50! is perfectly divisible by 2520n.
A. 6
B. 8
C. 7
D. none of these
4.The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ?
A. 240
B. 270
C. 295
D. 360
5. If the number 517*324 is completely divisible by 3, then the smallest whole number in the place of * will be:
A. 0
B. 1
C. 2
D. None of these
6.The sum of first 45 natural numbers is:
A. 1035
B. 1280
C. 2070
D. 2140
7. The difference between the local value and the face value of 7 in the numeral 32675149 is
A. 75142
B. 64851
C. 5149
D. 69993
8. On dividing a number by 56, we get 29 as remainder. On dividing the same number by 8, what will be the remainder ?
A. 4
B. 5
C. 6
D. 7
9.What will be remainder when (67^67 + 67) is divided by 68 ?
A. 1
B. 63
C. 66
D. 67
10. How many natural numbers are there between 23 and 100 which are exactly divisible by 6 ?
A. 8
B. 11
C. 12
D. 13

1(d) This can never happen at 99! number of zeroes is 22 and at 100! the number of zeroes is 24.
2(a)

3(B)Explanation:
2520 = 7×3^2×2^3×5
The value of n would be given by the value of the number of 7s in 50!
This value is equal to (50/7) + 50/49)= 7+1 = 8

4(B)Explanation:
Let the smaller number be x. Then larger number = (x + 1365).
so  x + 1365 = 6x + 15
=>5x = 1350
=> x = 270
Smaller number = 270.

5(C)Explanation:
Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = (22 + x), which must be divisible by 3.
x = 2.

6(A)Explanation:
Let Sn =(1 + 2 + 3 + … + 45). This is an A.P. in which a =1, d =1, n = 45.
Sn = n/2 [2a + (n – 1)d] = 45/2 x [2 x 1 + (45 – 1) x 1]
=> (45/2 x 45)= (45 x 23)
= 45 x (20 + 3)
= 45 x 20 + 45 x 3
= 900 + 135
= 1035.
Shorcut Method:
Sn = n(n + 1)/2 = 45(45 + 1)/2 = 1035.

7(D)Explanation:
(Local value of 7) – (Face value of 7) = (70000 – 7) = 69993

8(B)Explanation:
29 as remainder is divided by 8 than remainder is 5

9(C)Explanation:
(x^n + 1) will be divisible by (x + 1) only when n is odd.
so (67^67 + 1) will be divisible by (67 + 1)
(67^67 + 1) + 66, when divided by 68 will give 66 as remainder.

10(D)Explanation:
Required numbers are 24, 30, 36, 42, …, 96
This is an A.P. in which a = 24, d = 6 and l = 96
Let the number of terms in it be n.
Then tn = 96    a + (n – 1)d = 96
=>24 + (n – 1) x 6 = 96
=> (n – 1) x 6 = 72
=> (n – 1) = 12
=> n = 13
Required number of numbers = 13.

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