SSC CGL 20201)What is the smallest interger that is a multiple of 5, 8 and 15 ?
120
LCM of 5, 8, 15 is 120
SSC CGL 20202)When 2 is subtracted from each of the given n numbers, then the sum of the numbers so obtained is 102. When 5 is subtracted from each of them, then the sum of the numbers so obtained is 12. What is the average of the given n numbers?
5.4
Let for 'n' numbers the average be 'x'.
So, the total sum of 'n' numbers would be 'nx'.
If 2 is subtracted from each 'n' numbers, then the resulted value to be subtracted becomes = 2n;
Thus, value of the total sum now = (nx - 2n);
Given that, this value equals to 102.
So, nx - 2n = 102 ...(1);
Again when 5 is subtracted from each 'n' numbers, then the resulted value to be subtracted becomes = 5n;
Thus, value of the total sum now = (nx - 5n);
Given that, this value equals to 12.
So, nx - 5n = 12 ...(2);
Subtracting (2) from (1), we get:
nx - 2n - (nx - 5n) = 102 - 12; ⇒ -2n + 5n = 90; ⇒ 3n = 90 ;⇒ n = 90/3 = 30;
There are 30 numbers.
Putting n = 30, in eqn.(1), we get:
(30)x - 2(30) = 102; ⇒ 30x - 60 = 102; ⇒ 30x = 162; ⇒ x = 162/30 = 5.4
SSC CGL 20203)If integer n is divided by 5, the remainder is 2. What will be the remainder if 7n is divided by 5?
4
Let n = 5k + 2 where k = quotient; 7n = 7(5k + 2) = 35k + 14 = \(5\times7k+10+4\) = 5(7k + 2) + 4; so remainder = 4
SSC CGL 20204)The greatest digit which may replace * in the number 1190*6 to make the number divisible by 9 is :
1
1190*6, is divisble by p. so 1+1+9+0+*+6=(17+*) is a multiple of 9. ⇒17+* =18; ⇒ *= 18 - 17 = 1
SSC CGL 20205)What is the remainder when we divide \(5^{70}+7^{70}\) by 74 ?
0
\(5^{70}+7^{70}={(5^2)}^{35}+{(7^2)}^{35}=(25)^{35}+(49)^{35}\); When n is odd then \((x^n+a^n)\) is divisible by (x + a). Here, n = 35 (odd). So \((25)^{35}+(49)^{35}\) is divisible by (25 + 49) = 74;
Remainder = 0
SSC CGL 20206)If a positive integer n is divided by 7 the remainder is 2. Which of the following numbers gives a remainder of 0 when divided by 7?
n + 5
Dividing n by 7, remainder = 2; n + (7 - 2) = n + 5 is exactly divisible by 7.
Look: \(16\div7\), Remainder = 2; \(21\div7\), Remainder = 0
SSC CGL 20207)If the given number 925x85 is divisible by 11, then the smallest value of x is:
4
925x85 is divisible by 11. Then, Sum of digits at even places - sum of digits at odd places = 11. ⇒(9 + 5 + 8)-(2 + x + 5) = 11;⇒ x = 4
SSC CGL 20208)The sum of the squares of 3 natural numbers is 1029, and they are in the proportion 1 : 2 : 4. The difference between greatest number and smallest number is:
21
Let the smallest number be x.
Numbers are x, 2x, 4x.
The sum of the squares of 3 natural numbers = 1029;
\(x^2 + (2x)^2 + (4x)^2 = 1029\);
\(x^2 + 4x^2 + 16x^2 = 1029\);
\(x^2 = 1029/21\);
\(x^2 = 49\);
x = 7;
Smallest number = 7;
Greatest number = 4x = 4 \times 7 = 28;
The difference between greatest number and smallest number = 28 - 7 = 21
SSC CGL 20209)What is the smallest integer that is divisible by 3, 7 and 18?
126
LCM of of 3, 7 and 18 = 126;
\(\therefore\)126 is the smallest integer that is divisible by 3, 7 and 18.
SSC CGL 202010)The largest number which should replace * in the number 2365*4 to make the number divisible by 4 is:
8
2365*4 is divisible by 4. For divisibility by 4, *4 should be divisible by 4. Possible value of * = 2, 4, 6, 8 Maximum value of * = 8