Question

Find two numbers given that their sum is 48 and the smaller number is equal to one fifth of the larger number.

Answer 1

$\huge\underline\mathfrak\red{Question}$

Find two numbers given that their sum is 48 and the smaller number is equal to one-fifth of the larger number.

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$\huge\underline\mathfrak\blue{Answer:}$

$\huge\underline\green{A=40}$

$\huge\underline\green{B=8}$

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$\huge\underline\mathfrak\pink{Solution :}$

$\sf\large\underline{let :}$ A be the larger number

$\sf\large\underline{let :}$ B be the Smaller number

$\rm{Given:}$

we are given that,

A+B=48

B=(1/5)A......(or,Better, A =5B)

Now, Substitute the value of A in first equation

so, We get,

(5B) + B = 48

therefore,

6B=48

B=48/6

B = 8

$\huge\underline\green{B=8}$

And, if B=8 we know that A+B=48,

Then A=40.......(A=48–8)

$\huge\underline\green{A=40}$

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hope it helps, :))

Answer 2

Let ,

The two numbers be " x " and " y " , x > y

According to the question ,

The sum of two numbers is 48

x + y = 48 --- (i)

The smaller number is equal to one fifth of the larger number

y = (1/5) × x

5y = x ----- (ii)

Put the value of x = 5y in eq (i) , we get

5y + y = 48

6y = 48

y = 48/6

y = 8

Put the value of y = 8 in eq (i) , we get

x + 8 = 48

x = 40

$\therefore \bold{ \underline{The \: two \: numbers \: are \: 40 \: and \: 8}}$