Test: Number System- 1 - SSC CGL MCQ

100a + 10b + c
c = 2b → b = c/2
c = 1.5a → a = c/1.5
c/1.5 + c/2 + c = 13
6.5c = 39
c = 6, b = 3, a = 4 ⇒ 436

Step 1: Define the digits
Let the digit in the ten's place be x and the digit in the unit's place be y. According to the problem, the digit in the unit's place is equal to the square of the digit in the ten's place. Therefore, we can write:
y=x²

Step 2: Write the two-digit number
The two-digit number can be expressed as:
Number=10x+y
Substituting y from step 1, we have:
Number=10x+x²

Step 3: Write the number obtained by interchanging the digits
When we interchange the digits, the new number becomes:
New Number=10y+x
Substituting y from step 1, we have:
New Number=10x²+x

Step 4: Set up the equation for the difference
According to the problem, the difference between the original number and the new number is 54:
(10x+x²)−(10x²+x)=54

Step 5: Simplify the equation
Simplifying the left side:
10x+x²−10x²−x=54
This simplifies to:
−9x²+9x=54
Dividing the entire equation by -9 gives:
x²−x−6=0

Step 6: Factor the quadratic equation
Now we will factor the quadratic equation:
(x−3)(x+2)=0
This gives us two possible solutions:
x=3 or x=−2
Since x must be a positive digit, we take:
x=3

Step 7: Find the unit's place digit
Now, substituting x back to find y:
y=x²=3²=9

Step 8: Determine the original number
The original two-digit number is:
Number=10x+y=10(3)+9=30+9=39

Step 9: Calculate 40% of the original number
To find 40% of the original number:
40% of 39 = 40/100 ×39 = 0.4 × 39 = 15.6

Thus, the original number is 63.

original number – 10x + y
(10x + y) – (10y + x) = 27
9(x – y) = 27
x – y = 3
All the given options not follow the condition.

Given:

The ratio of the two numbers is 11 : 4.

The H.C.F is 16

Concept used:

(1) For the two numbers in the ratio y : z.

The value of first number = H.C.F × y

The value of first number = H.C.F × z

Calculation:

The value of first number = 16 × 11 = 176

The value of second number = 16 × 4 = 64

The required sum = 176 + 64 = 240

∴ The required answer is 240.

To solve this problem, we need to find a number that satisfies the following conditions:

1. When divided by 2, 3, 4, 5, or 6, the remainder is 1.
2. The number is divisible by 7.
3. The number lies between 250 and 350.

Let's start by finding the least common multiple (LCM) of 2, 3, 4, 5, and 6, which is the smallest number divisible by all of these numbers.

LCM(2, 3, 4, 5, 6) = 60

We need to find a number of the form 7k, where k is an integer, that leaves a remainder of 1 when divided by 60. The numbers in this sequence can be expressed as 60n + 1, where n is an integer.

Now, let's find the first few numbers of the form 60n + 1 that are divisible by 7 and lie between 250 and 350:

• For n = 4: 60(4) + 1 = 241 (not divisible by 7)
• For n = 5: 60(5) + 1 = 301 (divisible by 7)

So, the number we're looking for is 301.

Now, let's find the sum of its digits: 3 + 0 + 1 = 4

Therefore, the sum of the digits of the number is 4.

odd numbers – x-2, x, x+2 ; even numbers – y-2, y, y+2
3x + 3y = 231
x + y = 77
(y – 2) – (x – 2) = 11
y – x = 11
x = 33, y = 44
sum of the largest even number and odd number = 46 + 35 = 81

odd numbers — x-8, x-6, x-4, x-2, x, x+2, x+4, x+6
x-8 + x-6 + x-4 + x-2 + x + x+2 + x+4 + x+6 = 656
8x – 8 =656
x = 83
Even numbers — y-2, y, y+2, y+4
4y + 4 = 87 * 4
y = 86
sum of the largest even number and odd number = 89 + 90 = 179

a + b = 6
(10a + b) – (10b +a) = 36, a – b = 4
We get a = 5 and b = 1
So number is 51

let the number be 100a + 10b + c
(100a + 10b +c) – (100a + 10c +b) = 36
b – c = 4

let the fraction is (p-4)/p
now, (p -4 + 15)/(p-4) = 6
we get p = 7
so fraction = 3/7

a + (20/100)*b = (150/100)*b
a:b = 13:10

a – 24 = 2a/3
we get a = 72
so one-eighth of the number = 72/8 = 9

15x – (7x) = 64, we get x = 8
difference between first and third number = 5x – 4x = x = 8

a/3 – a/5 = 24
a = 180, so square = 32400

(25/100)*a = 2*(65/100)*b
b:a = 5:26