Mathematics Paper 1 (L.C.M & H.C.F) - CTET & State TET MCQ

Given:

3 consecutive numbers are in the ratio 1 : 2 : 3

Calculation:

Let the 3 consecutive numbers be y, (y + 1), (y + 2)

y : (y + 1) : (y + 2) = 1 : 2 : 3

On comparing both the sides, we get the value of,

y = 1

So, the three consecutives numbers are 1, 2, 3

1 + 2 + 3 = 6

∴ The sum of these numbers is 6.

Given:

HCF is 1 and LCM is 20

One number is 4

Formula Used:

HCF × LCM = Multiplication of two numbers

Calculation:

Let, the other number is A

According to the formula,

⇒ 1 × 20 = 4 × A

⇒ A = 5

∴ The other number is 5

Calculation:

Let the least number = a

When we doubled the number it becomes 2a

Now, 2a is exactly divisible by 12, 14, 16 and 18

⇒ 2a = 1008 (LCM of 12, 14, 16 and 18 = 1008)

⇒ a = 504

∴ The correct answer is 504.

Formula used:

HCF of different fractions = HCF of numerators / LCM of denominators

Calculations:

HCF = HCF of (3,4,7)/LCM of (4,5,60) 

⇒ 1/60

Hence HCF of  and $\frac{7}{60}$  is 1/60.

Given:
Ratio of the numbers = 5 : 7 : 9
LCM of the numbers = 34650

Calculation:-
Let the numbers be 5x, 7x, and 9x.
LCM of 5x, 7x and 9x = 315x
According to the question,
⇒ 315x = 34650
⇒ x = 34650/315 = 110
Now, HCF of 5x, 7x and 9x = x = 110
∴ The correct answer is 110

Given:

LCM = 144, HCF = 8, and One of the numbers is 16.

Formula used:

First Number × Second Number = LCM × HCF

Calculation:

Let the second number be x.

First Number × Second Number = LCM × HCF

⇒ 16 × x = 8 × 144

⇒ x = (8 × 144)/16

⇒ x = 72

∴ The second number is 72.

Calculation:

16 = 2 × 2 × 2 × 2 

10 = 2 × 5

12 = 2 × 2 × 3

27 = 3 × 3 × 3

LCM = 2 × 2 × 2 × 3 × 3 × 3 × 5 = 2160

Before leaving remainder 9, the number required is 2169.

2169/13 = 166.84

So, this is not a divisible by 13.

2160 × 2 = 4320 + 9 = 4329

⇒ 4329 ÷ 13 = 333

Correct number is 4329.

Sum of the digit = 4 + 3 + 2 + 9

We know that:

Formula:

L.C.M of 

⇒ L.C.M of  = 

⇒ L.C.M of  =  

Given:

HCF = 11 & LCM = 609

Concept used:

Product of numbers = HCF × LCM

Calculation:

Let, 2nd number be m,

⇒ m × 77 = 11 × 609

⇒ m = 609/7

⇒ m = 87

∴ The 2nd number is 87.

Concept:

Least number divisible by a group of numbers is given by their LCM.

Calculation:

14 = 2 × 7

15 = 5 × 3

18 = 2 × 3²

LCM(14, 15, 18) = 2 × 3² × 5 × 7

LCM(14, 15, 18) = 630

∴ Least number divisible by 14, 15 and 18 is 630