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

To find the last digit of the product 41×42×43×44×45×46×47×48×49, we only need to consider the last digits of each number:

• 41 → last digit is 1
• 42 → last digit is 2
• 43 → last digit is 3
• 44 → last digit is 4
• 45 → last digit is 5
• 46 → last digit is 6
• 47 → last digit is 7
• 48 → last digit is 8
• 49 → last digit is 9

Now we multiply these last digits together and focus on the last digit of the result:

1×2×3×4×5×6×7×8×9

Since 45 is in the product and its last digit is 5, and any number multiplied by 5 results in a number ending in 0 (provided there is another even number in the multiplication to make the product divisible by 10), we notice there are multiple even numbers (42, 44, 46, 48).

Thus, the product will definitely end in 0.

Therefore, the last digit of the number obtained by multiplying these numbers is: 0

Place value of 5 in 65231 = 5 x 1000
= 5000
Face value = 5
required product = 5000 x 5 = 25000

When a number is divisible by 11, then the sum of numbers at even place - sum of numbers at odd place = 0 or divisible by 11.
( y + y + 1 ) - ( 1+ 5 + 0 ) = 11
⇒ 2y - 5 = 11
∴ y = 8

By checking options
91-19=72
Or
Let the number be xy
So 10x + y = 10 ..... (i)
When the number is reversed the new number is yx
So (10y + 10 ) - ( 10x +y ) = 72 ....(ii)
From Eqs. (i) & (ii)
x = 1 and y =9
∴number = 19

To determine which pair of numbers is relatively prime, we need to check if their highest common factor (HCF) or greatest common divisor (GCD) is 1.

• For option A (68, 85):
  • Prime factors of 68 = 2 × 2 × 17
  • Prime factors of 85 = 5 × 17
  • Common factor: 17
  • HCF = 17 (not relatively prime)
• For option B (65, 91):
  • Prime factors of 65 = 5 × 13
  • Prime factors of 91 = 7 × 13
  • Common factor: 13
  • HCF = 13 (not relatively prime)
• For option C (92, 85):
  • Prime factors of 92 = 2 × 2 × 23
  • Prime factors of 85 = 5 × 17
  • No common factors other than 1
  • HCF = 1 (relatively prime)
• For option D (102, 153):
  • Prime factors of 102 = 2 × 3 × 17
  • Prime factors of 153 = 3 × 3 × 17
  • Common factors: 3 and 17
  • HCF = 51 (not relatively prime)

Thus, the correct answer is option C, as 92 and 85 are coprime.

The number given is 43836.

• Identify the digit 8, which is in the hundreds place.
• Face value of 8 = 8.
• Place value of 8 (hundreds place) = 800.
• Sum = 800 + 8 = 808.

dividend = divisor x quotient + remainder
= ( 13 x 30) + 12
= 390 + 12
= 402

First whole number is 0 and when any number is multiplied with zero, then result is zero.

Let number be the x.
According to the equation , x - x/7 = 180
⇒ 6x/7 = 180
∴ x = 210

Prime number less than 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
∴ The number of all prime numbers less than 40 = 12.

Let four consecutive even numbers are x, x+2, x+4, x+6.
According to the question
⇒ x + x+2 + x+4 + x+6 = 284
⇒ 4x + 12 = 284
⇒ 4x = 272
∴ x = 272 / 4 = 68

prime numbers between 0 and 50 are 2 3 5 7 11 13 17 19 23 29 31 37 41 43 and 47
∴ required number of prime number is 15

Let be the ten's digit be x and unit's digit be y.
So the two digit number = 10x + y ( where x > y )
According to question
x + y = 14 ....(i)
x - y = 2 ....(ii)

Solving Eqs. (i) and (ii), we get
x = 8 and y = 6
∴ Required product = 8 x 6 = 48

There are infinitely many rational numbers between 1 and 1000.

Explanation: Rational numbers are numbers that can be expressed as the ratio of two integers (i.e., p/q​, where p and q are integers, and q≠0). Since between any two numbers (whether integers or rational numbers), we can always find another rational number, the set of rational numbers between 1 and 1000 is infinite.

For example, between 1 and 2, you can have 1.5, and between 1 and 1.5, you can have 1.25, and so on. This process can continue indefinitely. Therefore, the number of rational numbers between 1 and 1000 is infinite.