LCM of 2 no. is 90 & their HCF is 6 if the sum of 2 numbers is 48 then the difference of those numbers is ??-?

Answer 1

LCM = 90
HCF = 6

so product = HCF *LCM = 6* 9 = 540

sum = 48

ab = 540

a+b = 54
(a-b)^2 = (a+b)^2 - 4ab = 48^2- 4*540 = 16(144-135) or 16* 9

(a-b) = 12

check
so numbers are 30 and 18 and LCM = 90 HCF = 6

Answer 2

LCM = 90
HCF = 6

so product = HCF *LCM = 6* 9 = 540

sum = 48

ab = 540

a+b = 54
(a-b)^2 = (a+b)^2 - 4ab = 48^2- 4*540 = 16(144-135) or 16* 9

(a-b) = 12

check
so numbers are 30 and 18 and LCM = 90 HCF = 6

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Answer 3

Look at the numbers that can divide into 90 --
1,2,3,5,6,9,10,15,18,30, 45,and 90.
Now, which of these two could addup to 48?
Obviously, only 3 and 45 or 18 and 30.
Which have a HCF of 6?
Obviously only 18 and 30.

Answer 4

Hey there Manisha,

The Product of the HCF and the LCM of two numbers is equal to the product of the two nos.

Take one no. as x and the other as 48-x
{Since the sum is 48}

Hence,

HCF x LCM = x(48-x)
90 x 6 = x^2 -48x
540 = x^2 -48x
x^2-48x-540 = 0

Now you can solve the equation to get the two nos. :)

Then since you know the two nos., you find their difference.

Hope that helped...^^

Answer 5

i would suggest you a method and show the solution.
in a question if L.C.M & H.C.F are given then to find the nos. follow this method.

first, divide L.C.M by H.C.F.=90/6 =15
now split it into factors and multiply it to H.C.F
15 can be splitted into 5 and 3.
on multiplying with H.C.F 6 =15 * 5 & 15 * 3
required nos. are 75 and 45.

their differance is 30.

Answer 6

I know the answer but I'm not telling you. Knowing that their highest common factor is 6 should help (you know both are multiples of 6). Finding two of those multiples that equal 48 should help you find the answer to this question.

Answer 7

i dont know how u dint figure out the ans it z so simple if u think gud
ne way the numbers r 18 n 30 obviously