Test: Number System- 2 - SSC CGL MCQ

Let the digits be x and y respectively.
Therefore, x + y = 5 ----------- (1)
Original Number : 10x + y
Reversed Number : 10y + x
10x + y - (10y + x) = 27
10x + y - 10y - x = 27
9x - 9y = 27
9 (x - y) = 27
x - y = 3 ------------ (2)
Adding equation (1) and (2)
2x = 8
x = 4
Substituting value for x in equation (1), we get
4 + y = 5
y = 1
The numbers are 41 and 14.

Let the two-digit number be 10a + b
(10a + b)/(a+b) = 3/1, 7a = 2b
And also given b = 5 + a
7a = 2(5+a)
7a =10 + 2a
5a = 10  
a = 2
b = 5 + a
b = 5 + 2
b = 7
so number 10a + b = 10x2 + 7 = 27
Solve both equations to get the number

LCM of 4,6 and 8 is 24
Divide 1000 by 24, we get quotient = 41 and 16 as remainder
so 41 numbers are there which are divisible by 4,6 and 8 together.

[(129)²]²⁹ = (1)²⁹ = 1 (9² always result in 1 on unit place)

Le the number be 5a + 4
square of the number = 25a² + 16 + 40a
so remainder = 1 (16 divided by 5 leaves a remainder 1)

Let the number be 527a + 21
when divided by 17, 527a is divisible by 17 and leaves remainder as 4 when 21 is divided by 17

(a +4)/b = 1/3 and a/(b+3) = 1/6 solve both the equations, u will get a = 5 and b = 27

Let the number be x.

According to the given condition:

- (1/4)x + 16 = (3/4)x

Solve for x:

• (1/4)x + 16 = (3/4)x
• 16 = (3/4)x - (1/4)x
• 16 = (2/4)x
• 16 = (1/2)x
• x = 32

Therefore, the number is 32.

Answer: B: 32

Let total marks = M
(3/8)*M = 300 + 36 = 336
M = 112*8 = 896

Let us say that the two numbers are 'a' and 'b' and the divisor is 'd'

We are given that

Rem [a/d] = 11 and Rem [b/d] = 21

We are also given that the Remainder [(a + b)/d] = 4

=> Rem[(11 + 21)/d] = 4

=> Rem[32/d] = 4

=> 32 - 4 = 28 is divisible by 'd' or 'd' is a factor of 28

=> 'd' could be 1, 2, 4, 7, 14, or 28

We also know that 'd' is greater than 21 because 'b' when divided by 'd' leaves a remainder of 21.

=> The value of 'd' is 28

561 = 3*11*17
So the number must be divided by 3, 11 and 17
Only B option is divided by all.

Check the options in the number 10x47y
all numbers will be divisible by 5 because in end it is 5 and 0
for number to be divisible by 11, (y+4+0) – (7+x+1) should be divisible by 11
from option A, y = 5, x = 1 gives (y+4+0) – (7+x+1) as 0 which is divisible by 11

Let the number is 10x+y
So x+y = 6
And (10x+y)/(10y+x) = 4/7
Solve, 2x = y and from above we have x+y = 6
Solve both equations, x = 2, y = 4

13x + 13 which is divisible by 5, or 13(x+1) should be divisible by 5.
The smallest value of x = 4 to be put here to make it divisible by 5.
So the number is 13(4+1)

Smaller no = x, then larger = x+2577
Now x+2577 = 26x + 2
Solve, x = 103
So larger no is = 103+2577

Let number is 10x+y
Then x-y = 5 or y-x = 5
Now given that, 10x+y = 2(10y+x) + 18 Solve, 8x – 19y = 18
Now solve: 8x – 19y = 18 and x-y = 5. In this y = 2, x = 7
And also solve; 8x – 19y = 18 and y-x = 5. In this y come to be negative which is not possible so discard this
So number is 10*7 + 2

Let the number be N.

From the given information:

• When N is divided by 462, the remainder is 25. This means:

  N = 462k + 25

  where k is an integer.

Now, we need to find the remainder when N is divided by 14. Substituting N = 462k + 25:

• First, divide 462 by 14:

  462 ÷ 14 = 33 (remainder 0)

  So, 462k is divisible by 14, leaving us with:

  N = 462k + 25

Now, divide 25 by 14:

• 25 ÷ 14 = 1 (remainder 11).

Therefore, the remainder when N is divided by 14 is 11.

Answer: The remainder is 11.

Let she attend x correct answers out of 40, then incorrect = 40-x
So x*3 – (40-x)*1 = 96
Solve, x = 34

Let fraction = x/y
Then (x+1)/y = 1/4
And x/(y-1) = 1/5
Solve both equations, x = 3, y = 16