Question

consider two no sum is 135 hcf is 27 and lcm 162

Answer 1

Answer:

the two number are  54 and 81

Step-by-step explanation:

let the number be x and y  

x + y = 135  

x =135-y  

and we know that  

product of two numbers = HCF×LCM  

x = 27×162  

put the value of x from (1)  

(135-y) y = 27×162  

135y-y² = 4374  

y² -135y + 4374 = 0  

y² - 81y -54y + 4374=0  

y(y -81) - 54(y-81) = 0  

(y-81)(y-54) = 0  

y-81=0, y-54= 0  

y=81 , y= 54  

put these values in (1)  

when y= 81 then , x = 54 and  

when y = 54 ,then x=81  

hence the two number are

54 and 81

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Answer 2

SOLUTION

COMPLETE QUESTION

Consider the two numbers whose sum is 135 and their HCF is 27. If their LCM is 162, then what will be the larger number?

GIVEN

• Consider the two numbers whose sum is 135

• Their HCF is 27.

• Their LCM is 162

TO DETERMINE

The larger number

EVALUATION

Here it is given that the HCF of two given numbers are 27

Let the numbers are 27a and 27b ( a > b )

The sum of the numbers = 135

⇒ 27a + 27b = 135

⇒ a + b = 5 - - - - - (1)

Now it is given that LCM = 162

HCF × LCM = Product of the numbers

⇒ 27 × 162 = 27a × 27b

⇒ 27a × 27b = 27 × 162

⇒ ab = 6 - - - - - - (2)

From Equation 1 and Equation 2 we get

a( 5 - a) = 6

$\displaystyle \sf{ \implies 5a - {a}^{2} = 6}$

$\displaystyle \sf{ \implies {a}^{2} - 5a + 6 = 0}$

$\displaystyle \sf{ \implies {a}^{2} - 3a - 2a + 6 = 0}$

$\displaystyle \sf{ \implies a(a - 3) - 2(a - 3) = 0}$

$\displaystyle \sf{ \implies (a - 3) (a - 2) = 0}$

a - 3 = 0 gives a = 3

a - 2 = 0 gives a = 2

From Equation 1

When a = 3 we have b = 2

When a = 2 we have b = 3

We have assumed a > b

∴ a = 3 , b = 2

Consequently the required numbers are 81 and 54

FINAL ANSWER

Hence the required larger number = 81

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