Question 1197028: An amount of Rs 10000 is put in three investments at the rate of 10% 12% 15% per annum. the combined income is Rs 1310 and combined income of first and secpnd investment is 190 less of the income from third . find each,s investment
Found 3 solutions by ikleyn, MathTherapy, greenestamps:
Answer by ikleyn(52836)   (Show Source):
You can put this solution on YOUR website! .
x + y + z = 10000 (1)
0.1x + 0.12y + 0.15z = 1310 (2)
-0.1x - 0.12y + 0.15z = 190 (3)
Add equations (2) and (3). You will get
0.3z = 1310 + 190 = 1500, hence z = 1500/0.3 = 5000.
Substitute z= 5000 into equations (1) and (2). You will get
x + y = 5000 (4)
0.1x + 0.12y = 560 (5)
Multiply equation (5) by 10; keep equation (4) as is. You will get then
x + y = 5000 (4')
x + 1.2y = 5600 (5')
Subtract (4') from (5'). You will get
0.2y = 600; hence y = 600/0.2 = 3000.
Now use equation (1) to find x. Substitute y= 3000 and z= 5000 into (1)
x + 3000 + 5000 = 10000.
It gives x = 10000 - 3000 - 5000 = 2000.
ANSWER. $2000 invested at 10%; $3000 invested at 12% and $5000 invested at 15%.
CHECK. Please check it on your own, substituting the values into equations (2) and (3).
Solved.
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website!
An amount of Rs 10000 is put in three investments at the rate of 10% 12% 15% per annum. the combined income is Rs 1310 and combined income of first and secpnd investment is 190 less of the income from third . find each,s investment
Let amount invested in first (10%), second (12%) and third (15%) funds be F, S and T, respectively
Then: F + S + T = 10,000 -- eq (i)
.1F + .12S + .15T = 1,310 --- eq (ii)
.1F + .12S = .15T - 190
.1F + .12S - .15T = - 190 --- eq (iii)
.1F + .12S + .15T = 1,310 -- eq (ii)
.1F + .12S - .15T = - 190 -- eq (iii)
.3T = 1,500 -- Subtracting eq (iii) from eq (ii)
Amount invested in the third (15%) fund, or 
.1F + .12S + .15T = 1,310 --- eq (ii)
.1F + .12S - .15T = - 190 --- eq (iii)
.2F + .24S = 1,120 ---- Adding eqs (iii) & (ii)
.2(F + 1.2S) = .2(5,600)
F + 1.2S = 5,600
F = 5,600 - 1.2S --- eq (iv)
5,600 - 1.2S + S + 5,000 = 10,000 --- Substituting 5,000 for T and 5,600 - 1.2S for F in eq (i)
- .2S + 10,600 = 10,000
- .2S = 10,000 - 10,600
- .2S = - 600
Amount invested in the second (12%) fund, or 
F = 5,600 - 1.2(3,000) ----- Substituting 3,000 for S in eq (iv)
Amount invested in the first (10%) fund, or F = 5,600 - 3,600 = $2,000
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
You have two responses which both show solving the problem using a system of three equations. That is certainly a valid method; however, there are much easier paths to the solution.
The total income is Rs 1310; and the combined income from the first and second investments is Rs 190 less than the income from the third. Do some simple arithmetic to find that the combined income from the first and second is Rs 560 and the income from the third is Rs 750.
Since the rate of return from the third investment was 15%, the income of Rs 750 means Rs 5000 was invested there.
That leaves Rs 5000 to be invested part at 10% and part at 12% to yield income of Rs 560.
There are standard formal algebraic methods for solving a problem like that; but there is also the following informal method which is faster and easier.
The rate of return for an income of Rs 560 from an investment of Rs 5000 is 11.2%. Look at the three percentages 10, 11.2, and 12 on a number line and observe/calculate that 11.2 is 3/5 of the way from 10 to 12. That means 3/5 of the remaining Rs 5000, or Rs 3000, was at the higher rate, leaving Rs 2000 at the lower rate.
ANSWER: Rs 5000 at 15%, Rs 3000 at 12%, and Rs 2000 at 10%
CHECK: .15(5000)+.12(3000)+.10(2000) = 750+360+200 = 1310
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