SSC CGL 201921)If \(x=(164)^{169}+(333)^{337}\)\(-(727)^{726}\), then what is the unit's digit of x ?

Correct Option: C

8

SSC CGL 201922)Let x be the least number which when divided by 15, 18, 20 and 27, the remainder in each case is 10 and x is a multiple of 31. What least number should be added to x to make it a perfect square?

Correct Option: A

39

SSC CGL 201923)When a two-digit number is multiplied by the sum of its digits, the product is 424. When the number obtained by interchanging its digits is multiplied by the sum of the digits, the result is 280. The sum of the digits of the given number is:

Correct Option: C

8

SSC CGL 201924)If \(\sqrt{10-2\sqrt{21}}\)\(+\sqrt{8+2\sqrt{15}}\)\(=\sqrt a+\sqrt b\), where a and b are positive integers, then the value of \(\sqrt{ab}\) is closest to :

Correct Option: B

5.9

\(\sqrt{10-2\sqrt{21}}=\sqrt{(\sqrt{7})^2+(\sqrt3)^2-2\sqrt{7}\times \sqrt{3}}=\sqrt{(\sqrt{7}-\sqrt{3})^2}\)

Similarly

\(\sqrt{8+2\sqrt{15}}=\sqrt{(\sqrt{5})^2+(\sqrt3)^2+2\sqrt{5}\times \sqrt{3}}=\sqrt{(\sqrt{5}+\sqrt{3})^2}\)

therefore

\(\sqrt a+\sqrt b = \sqrt 5+\sqrt 7\);

a= 5; b=7

there approx value of \({\sqrt{35}} \) = 5.9 (approx)

SSC CGL 202025)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?

Correct Option: B

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 202026)If integer n is divided by 5, the remainder is 2. What will be the remainder if 7n is divided by 5?

Correct Option: D

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 202027)The greatest digit which may replace * in the number 1190*6 to make the number divisible by 9 is :

Correct Option: B

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 202028)What is the remainder when we divide \(5^{70}+7^{70}\) by 74 ?

Correct Option: D

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 202029)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?

Correct Option: A

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 202030)If the given number 925x85 is divisible by 11, then the smallest value of x is:

Correct Option: D

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

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