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Sunita as twice as old as ashima.if six years is subtracted from ashima's age and four years added to sunita's age,then Sunita will be four times ashima's age.How old are they two years ago?
Verified Answer
Sunita as twice as old as ashima.if six years is subtracted from ashim...
Ashima's age be x
Sunita's age be 2x
x - 6
2x + 4
then 
4( x - 6) = 2x + 4
4x - 24 = 2x + 4
4x - 2x = 4 + 24
2x = 28
x = 14
2x = 28
Ashima's present age is 14 and Sunit's present age is 28.
Now Ashima 2 years ago = 14 - 2 = 12 years
Sunita "" = 28 - 2 = 26 years
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Most Upvoted Answer
Sunita as twice as old as ashima.if six years is subtracted from ashim...
Given information:
- Sunita is twice as old as Ashima.
- If 6 years are subtracted from Ashima's age and 4 years are added to Sunita's age, then Sunita will be four times Ashima's age.

To find:
- The ages of Sunita and Ashima two years ago.

Step-by-step solution:

Step 1: Assign variables to the ages of Sunita and Ashima.
Let's say Sunita's age is S and Ashima's age is A.

Step 2: Translate the given information into equations.
- Sunita is twice as old as Ashima: S = 2A
- If 6 years are subtracted from Ashima's age and 4 years are added to Sunita's age, then Sunita will be four times Ashima's age: S + 4 = 4(A - 6)

Step 3: Solve the system of equations.
Substitute the value of S from the first equation into the second equation:
2A + 4 = 4(A - 6)

Simplify the equation:
2A + 4 = 4A - 24

Rearrange the equation:
2A - 4A = -24 - 4

Combine like terms:
-2A = -28

Divide by -2:
A = 14

Substitute the value of A into the first equation to find S:
S = 2(14)
S = 28

Step 4: Find their ages two years ago.
Two years ago, Ashima's age would be A - 2 = 14 - 2 = 12.
Two years ago, Sunita's age would be S - 2 = 28 - 2 = 26.

Answer:
Two years ago, Ashima was 12 years old and Sunita was 26 years old.
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Sunita as twice as old as ashima.if six years is subtracted from ashima's age and four years added to sunita's age,then Sunita will be four times ashima's age.How old are they two years ago?
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