Question

Find 2 numbers differing by 10 whose sum is equal to twice their difference
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Answer 1

Answer:

Let the number s be x and y. It is given

x-y =10 equation 1

x+y=2(x-y)=2×10=20 equation 2

Add equation 1 and 2, you will get

2x=30

x=30/2=15

Put this value of x in any of the above equations, you will get

y=5

Ans numbers are 15 and 5

Answer 2

Step-by-step explanation:

Given :-

Two numbers differing by 10 whose sum is equal to twice their difference .

To find :-

Find the two numbers ?

Solution :-

Let the one of the numbers be X

According to the given problem

The other number = X-10

Their sum = X+X-10 = 2X-10

Twice their difference = 2(10) = 20

According to the given problem

Their sum = 2×Their difference

=> 2X-10 = 20

=> 2X = 20+10

=> 2X = 30

=> X = 30/2

=> X = 15

Now

The first number = 15

The other number = X-10 = 15-10 = 5

Alternative Method:-

Let the two numbers be X and Y

Let X>Y

Their difference = X-Y = 10 --------(1)

According to the given problem

Their sum = 2×Their difference

=> X+Y = 2(10)

=> X+Y = 20 -------------(2)

On adding (1) & (2) then

X+Y = 20

X-Y = 10

(+)

________

2X+0 = 30

_________

=> 2X = 30

=> X = 30/2

=> X = 15

On Substituting the value of X in (2) then

=> 15+Y = 20

=> Y = 20-15

=> Y = 5

Therefore, X = 15 and Y = 5

Answer:-

The required two numbers are 15 and 5

Check:-

The two numbers = 15 and 5

Their difference = 15-5 = 10

Twice their difference = 2(10) = 20

Their sum = 15+5 = 20

Their sum is equal to the twice their difference

Verified the given relations in the given problem.