Math, asked by nishitaverma13, 10 months ago

A and B are friends and their ages differ by 2 years. A's father D is twice as old as A and B is twice as old
as his sister C. The age of D and C differ by 40 years. Find the ages of A and B.​

Answers

Answered by raneemmustaf19
2

Answer:

 years and B's age =29  

3

1

​  

 years.

Step-by-step explanation:

Let the ages of A and B be x and y years respectively. Then,

x−y=±2 [Given]

D's age =2x years and, C's age =  

2

y

​  

 years.

Clearly, D is older than C

2x−  

2

y

​  

=40⇒4x−y=80

Thus, we have the following two systems of linear equations

x−y=2 .(i)

and, 4x−y=80 (ii)

or

x−y=−2 (iii)

and, 4x−y=80 ..(iv)

Substracting equation (i) from equation (ii), we get

3x=78⇒x=26Putting x=26 in equation (i), we get y=24

or

Substracting equation (iv) from equation (iii), we get

−3x=−82⇒x=  

3

82

​  

=27  

3

1

​  

 

Putting x=  

3

82

​  

 in equation (iii), we get

y=  

3

82

​  

+2=  

3

88

​  

=29  

3

1

​  

 

Hence, A's age =26 years and B's age =24 years

or

A's age =27  

3

1

​  

 years and B's age =29  

3

1

​  

 years.

Answered by madhutiwari793
2

Answer:

24years and 26 years

Step-by-step explanation:

D-C =40

C =B/2

D=2A

2A-B/2 =40

4A-B = 80

GIVEN THAT

A-B =2

B = A-2

SUBSTITUTING IN ABOVE EQ

4A -A +2 =80

A =26

B =A-2

B =24

PLEASE MARK BRAINIEST

Similar questions