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Answers
Step-by-step explanation:
Given :-
Sunita is as twice as old as Ashima. If 6 years are subtracted fr Ashima' s age and 4 years are added to Sunita' s age then Sunita will be four times of Ashima.
To find :-
How old were they two years ago?
Solution:-
Let the present age of Ashima be X years
Then the present age of Sunita = 2X years
Since Sunita is as twice as old as Ashima
If 6 years are subtracted from Ashima' s age then her age will be (X-6) years
If 4 years are added to Sunita's age then her age will be (2X+4) years
According to the given problem
If 6 years are subtracted fr Ashima' s age and 4 years are added to Sunita' s age then Sunita will be four times of Ashima.
=> 2X+4 = 4×(X-6)
=> 2X+4 = 4X-24
=> 4+24 = 4X-2X
=> 28 = 2X
=> 2X = 28
=> X = 28/2
=> X = 14 years
And 2X = 2×14 = 28 years
Ashima's present age = 14 years
Two years ago her age = 14-2 = 12 years
Sunita's present age = 28 years
Two years ago her age = 28-2 = 26 years
Answer:-
Two years ago the ages of Ashima and Sumita are 12 years and 26 years respectively.
Check:-
Two years ago the ages of Ashima and Sumita are 12 years and 26 years respectively.
Ashima' s present age = 12+2=14 years
Sunita's present age = 26+2 = 28 years
Sunita's present age = Twice of Ashima's present age
If 6 years subtracted from 14 years = 14-6 = 8 years
If 4 years are added to 28 = 28+4 = 32 years
=> 4×8 years
Sunita's age= 4 times the age of Ashima
Verified the given relations in the given problem