A businessman divided his P360,000 into 2 amounts and invested each at 8% and 10%. If the interest earned from each investment was the same, how much was invested at each rate?
2007-07-03 19:44:05 · 6 answers · asked by ? 2 in Education & Reference ➔ Homework Help
So the given is P360,000
Let us use x to represent amount invested at 8% or 0.08 therefore, the amount invested at 10% or 0.01 is (P360,000 - x)
It is stated in the problem that the interest earned from each investment is the same or in other words, equal. Therefore:
0.10(P360,000-x)=0.08(x)
Distributing the 0.10 and 0.08 will give us:
P36,000-0.10x=0.08x
By arranging the equation, it will give us:
0.18x = P36,000
Divide the whole equation by the coefficient of x.
0.18x/0.18 = P36,000/0.18
x = P200,000 ==> this is the amount invested at 8%
Therefore P360000-P200000=P160000 ==>this is the amount invested at 10%
by checking our answers:
P200000*0.08=P16000
P160000*0.10=P16000
we got equal answers and therefore we are sure that our answer is correct.
2007-07-04 06:06:48 · answer #1 · answered by Gerard 2 · 0⤊ 0⤋
Let A and B the amounts out of 3,60,000.
So, A + B = 360000
8A = 10B (we can ignore the 100 in the denominators since it is common)
4A = 5B
A = 1.25B
Substituting in the first equation,
1.25B + B = 360,000
2.25B = 360000
B = 360000 / 2.25
= 160000
A = 360000 - B = 200000
So, 200000 at 8% and 160,000 at 10% are the invested amounts.
2007-07-04 02:58:32 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
P360,000 he has...
Let x represent amount invested at 8%
and (P360,000 - x) represent amount invested at 10%
(0.08)(x) = (0.10)(P360,000 - x) so that the interest earned is the same...
0.08x = P36,000-0.10x
0.18x = P36,000
x = P200,000 invested at 8%
and P360,000 - P200,000 = P160,000 invested at 10%
The interest earned on each amount invested is P16,000... because P200,000 * 8% = P16,000....
and P160,000 * 10% = P16,000
2007-07-04 02:50:57 · answer #3 · answered by blueskies 7 · 0⤊ 0⤋
Let X= first amount
Y=second amount
eq. 1 --------- 8X=10Y
X=10/8Y
X=1.25Y
eq. 2 --------- X+Y=360000
substituting value of X from eq.1 to eq.2 , ergo
1.25Y+Y=360000
2.25Y=360000
Y=360000/2.25
Y=160000 (1st amt) -ans
therefore: from eq.2
X+Y=360000
X+160000=360000
X=200000 (2nd amt) -ans
2007-07-04 03:50:17 · answer #4 · answered by jose Izxc 3 · 0⤊ 0⤋
I thought the interst was 8% and 10%. I'm confused already. Next question please.
2007-07-04 02:48:33 · answer #5 · answered by JeffG 3 · 0⤊ 0⤋
im so glad to be done with my math class...
Do you know any fomulas that can help you .
2007-07-04 02:50:33 · answer #6 · answered by Imdatchick 3 · 0⤊ 1⤋