Number System- 1 - SSC CGL Quant Aptitude Free MCQ Test with solutions

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

Let the tens digit be t and the units digit u.
We have
• u = t²
• The reversed number minus the original is 54 (since units > tens):
 (10u + t) – (10t + u) = 54
⇒ 9u – 9t = 54
⇒ u – t = 6

Substitute u = t²:
 t² – t = 6 ⇒ t² – t – 6 = 0
⇒ (t – 3)(t + 2) = 0
⇒ t = 3

Then u = t² = 9, so the number is 39.
40% of 39 = 0.4 × 39 = 15.6

Answer: 15.6 (option a)

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

Let the first of the eight consecutive odd numbers be a. Then the numbers are a, a+2, a+4, a+6, a+8, a+10, a+12, a+14.

Their sum is 8a + (2+4+6+8+10+12+14) = 8a + 56. This equals 656, so

8a + 56 = 656 → 8a = 600 → a = 75.

The largest odd number = a + 14 = 75 + 14 = 89.

For the four consecutive even numbers let the first be n. Then the numbers are n, n+2, n+4, n+6.

Their average is n + 3. Given this average is 87, so n + 3 = 87 → n = 84.

The largest even number = n + 6 = 84 + 6 = 90.

Therefore the sum of the largest even and largest odd numbers = 90 + 89 = 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

Step 1
Let the number be N. According to the problem, N = 143k + 31, where k is some integer.

Step 2
We need to find the remainder when N is divided by 13.

Step 3
Substitute the expression for N: N = 143k + 31.

Step 4
Take this modulo 13: N ≡ 143k + 31 (mod 13).

Step 5
Since 143 ≡ 0 (mod 13), we have N ≡ 31 (mod 13). The remainder of 31 when divided by 13 is 5.

Final Answer:
The remainder when the same number is divided by 13 is 5.

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

Let the first number be a and the second number be b. According to the problem:

a + (20/100)b = (150/100)b.

Simplifying this equation:

• a + 0.2b = 1.5b.
• Subtract 0.2b from both sides:
• a = 1.5b - 0.2b.
• a = 1.3b.

Thus, the ratio of a to b is:

a/b = 1.3 = 13/10.

Therefore, the ratio between the first and second number is 13:10.

Step-by-Step Solution:

Step 1: Represent the numbers
Let the divisor be d, and the two numbers be a and b.

• When a is divided by d, remainder = 3.
• When b is divided by d, remainder = 4.

Step 2: Sum of the numbers
The sum of a and b gives:
a + b = (some multiple of d) + 7
When the sum is divided by d, the remainder is given as 2.
7 ≡ 2 (mod d)

Step 3: Solve for the divisor
Simplify the equation:
7 - 2 = 5
This means d must divide 5 evenly.
The only possible value of d is 5.

The divisor is 5.

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

Let the number be a. According to the problem:

(a/3) - (a/5) = 24.

To solve this, we need to find a common denominator:

• Common denominator: 15
• Rewrite the equation: (5a/15) - (3a/15) = 24

This simplifies to:

(2a/15) = 24.

Now, solving for a gives:

• 2a = 24 * 15
• 2a = 360
• So, a = 180.

The square of the number is:

a² = 180² = 32400.

Let the first number be a and the second number be b. According to the problem:

0.25a = 2 × (0.65b)

Simplifying this equation:

• 0.25a = 1.30b

Dividing both sides by 0.05 to simplify further:

• 5a = 26b

Thus, the ratio b:a is:

• b/a = 5/26

Therefore, the ratio of the second number to the first number is 5:26.