Advertisements
Advertisements
Question
Anil and Sunita have incomes in the ratio 3 : 5. If they spend in the ratio 1 : 3, each saves T 5000. Find the income of each.
Advertisements
Solution
Let Anil's income = Rs. x and Sunita's income = Rs. y
According to given information, we have
`x/y = (3)/(5)`
⇒ 5x = 3y
⇒ 5x - 3y = 0 ....(i)
And,
`(x - 5000)/(y - 5000) = (1)/(3)` ....[Expense = Income - Saving]
⇒ 3x - 15000 = y - 5000
⇒ 3x - y = 10000 ....(ii)
Multiplying eqn. (ii) by 3, we get
9x - 3y = 30000 ....(iii)
Subtracting eqn. (i) from (iii), we get
4x = 30000
⇒ x = 7500
⇒ 5(7500) - 3y = 0
⇒ 37500 - 3y = 0
⇒ 3y = 37500
⇒ y = 12500
Hence, Anil's income is Rs.7500 and Sunita's income is Rs.12,500.
APPEARS IN
RELATED QUESTIONS
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
13x+ 11y = 70
11x + 13y = 74
If 10y = 7x - 4 and 12x + 18y = 1; find the values of 4x + 6y and 8y - x.
Solve the following simultaneous equation :
8v - 3u = 5uv
6v - 5u = -2uv
Solve the following simultaneous equations :
3(2u + v) = 7uv
3(u + 3v) = 11uv
Solve the following simultaneous equations:
13a - 11b = 70
11a - 13b = 74
Solve the following pairs of equations:
`(2)/(x + 1) - (1)/(y - 1) = (1)/(2)`
`(1)/(x + 1) + (2)/(y - 1) = (5)/(2)`
Solve the following pairs of equations:
`(2)/(3x + 2y) + (3)/(3x - 2y) = (17)/(5)`
`(5)/(3x + 2y) + (1)/(3x - 2y)` = 2
Solve the following pairs of equations:
`(xy)/(x + y) = (6)/(5)`
`(xy)/(y - x)` = 6
Where x + y ≠ 0 and y - x ≠ 0
If the following three equations hold simultaneously for x and y, find the value of 'm'.
2x + 3y + 6 = 0
4x - 3y - 8 = 0
x + my - 1 = 0
Sunil and Kafeel both have some oranges. If Sunil gives 2 oranges to Kafeel, then Kafeel will have thrice as many as Sunil. And if Kafeel gives 2 oranges to Sunil, then they will have the same numbers of oranges. How many oranges does each have?
