Question

1. ) Raju brothers is twice the age of Raju. After 10 years his brother will be one and half times of Raju.
Find their present ages.

Please, help me to get the answer question please....

the first who will answer correctly will get a reward...

Answer 1

hey
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raju age be x
brother age be 2x

after 10 years

raju age = x+10 years
brother age = 2x+10 years

a/q

2x+10 = 1 whole 1/2 (x+10)
2x+10 = 3/2(x+10)
2x+10 = 3/2x + 3/2×10
2x+10 = 3/2x +15
2x-3/2x = 15-10
4x-3x/2 = 5
x/2 = 5
x = 10

present ages are

raju = 10 yrs
brother age = 20 years

hope helped
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Answer 2

Heya !

Here's your answer !!

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Let Raju's age be x years

His brother's age = 2x years


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10 years later :

Raju's age = ( x + 10 ) years

His brother's age = ( 2x + 10 ) years


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A.T.Q.

The equation formed :

=> 2x + 10 = 3/2 ( x + 10 )

=> 2x + 10 = 3x/2 + 15

=> 2x - 3x/2 = 15 - 10

=> x/2 = 5

=> x = 10 years


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Ans ) Raju's present age = 10 years

Ans ) His brother's present age =

=> 10 × 2

=> 20 years


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Thanks !!