Question

Five years ago, A was twice as old as B. In 4 years time, the sum of their ages will be 78. What are their present ages

Answer 1

Step-by-step explanation:

Let the present age of A be x

Let the present age of B be y

A=2B

x=2y

x-2y=0---(i)

After 4 years,

(x+4)+(y+4)=78

x+y+8=78

x+y=78-8

x+y=70 -----(ii)

Subtract (i) and (ii)

x-2y=0

(x+y. =70)2

2x+2y=140

3x=140

x=140/3

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Answer 2

SOLUTION :-

Let,

Present age of A = x

Present age of B = y

5 years ago,

Age of A = x - 5

Age of B = y - 5

After 4 years,

Age of A = x + 4

Age of B = y + 4

According to the first condition,

$\longrightarrow \sf x-5 = 2(y-5)\\\\\longrightarrow x- 5 = 2y- 10 \\\\\longrightarrow x-2y = -5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ . .........(1)$

According to the second condition,

$\longrightarrow \sf (x+4)+(y+4)= 78 \\\\\longrightarrow x+y+8 = 78 \\\\\longrightarrow x+y = 70 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ . .........(2)$

Eq (2) - Eq (1),

$\longrightarrow\sf 3y = 75 \\\\\longrightarrow y = 25$

Substitute the value of x in eq (2),

$\longrightarrow\sf x+ 25 = 70\\\\\longrightarrow x = 45$

PRESENT AGE OF A = 45 years

PRESENT AGE OF B = 25 years