SOLUTION: Two brothers A and B have their annual incomes in the ratio of 8:5 while their annual spending is in the ratio 5:3.
If they save $1,200 and $1,000 respectively, find their an
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Question 1103973: Two brothers A and B have their annual incomes in the ratio of 8:5 while their annual spending is in the ratio 5:3.
If they save $1,200 and $1,000 respectively, find their annual incomes.
Found 2 solutions by Fombitz, Theo:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
I don't understand the problem setup.
You say their spending is in the ratio of 5:3 but that the values are 1200 and 1000.
Can you check your problem setup and repost the question?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
you have:
ratio of income of A to income of B = 8/5.
ratio of spending of A to spending of B = 5/3.
A saves 1200.
B saves 1000.
all figures are annual figures.
let x = income of A.
let y = income of B.
ratio of income of A to income of B becomes x/y = 8/5
solve for y to get y = 5x/8.
you have:
income of A = x
income of B = 5x/8.
the variable of y disappears from the income ratio and therefore can be used again in the spending ratio.
let y = spending of A.
let z = spending of B.
ratio of spending of A to spending of B becomes y/z = 5/3
solve for z to get z = 3y/5
you have:
spending of A = y
spending of B = 3y/5
the variable of z disappears from the spending ratio and therefore can be used again in the saving ratio.
however, we won't need it anymore because we have the figures for the saving of A and the saving of B.
the saving of A = 1200; the spending of A = y; the income of A = x.
since income minus spending = saving, you get:
x - y = 1200
the saving of B = 1000; the spending of B = 3y/5; the income of B = 5x/8.
since income minus spending = saving, you get:
5x/8 - 3y/5 = 1000
you have two equations that need to be solved simultaneously.
they are:
x - y = 1200 (equation 1)
5x/8 - 3y/5 = 1000 (equation 2)
multiply both sides of equation 2 by 40 to get:
25x - 24y = 40,000 (equation 3)
your equations to solve simultaneously are now:
x - y = 1200 (equation 1)
25x - 24y = 40,000 (equation 3)
multiply both sides of equation 1 by 25 to get:
25x - 25y = 30,000 (equation 4)
your equations to solve simultaneosly are now:
25x - 24y = 40,000 (equation 3)
25x - 25y = 30,000 (equation 4)
subtract equation 4 from equation 3 to get:
y = 10,000
go back to equation 1 and solve for x.
start with x - y = 1200 (equation 1)
replace y with 10,000 to get:
x - 10,000 = 1200
solve for x to get:
x = 11,200
you now have:
x = 11,200 and y = 10,000
11,200 is the income of A.
10,000 is the spending of A.
1200 is the saving of A.
income minus spending = savings, therefore:
11,200 - 10,000 = 1200, confirming the income and spending of A is correct.
from equation 2, you find that:
income of B = 5x/8
spending of B = 3y/8
since x = 11,200, then 5x/8 = 7000.
since y = 10,000, then 3y/5 = 6000.
you have:
income of B = 7000
spending of B = 6000
saving of B = income of B minus spending of B, therefore:
7000 - 6000 = 1000.
this confirms income and spending for B are correct.
you wind up with:
income of A = 11,200
spending of A = 10,000
saving of A = 1,200
income of B = 7,000
spending of B = 6,000
saving of B = 1,000
income ratio of A/B = 11,200 / 7,000 which simplifies to 8/5.
spending ratio of A/B = 10,000 / 6,000 which simplifies to 5/3.
requirements of the problem are satisfied.
your solution is that income of A = $11,200 and income of B = $7,000.
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