Word problems -?

Question

1)You invested $7000 in two accounts paying 6% and 8% annual interest, respectively. If the total interest earned for the year was $520, how much was invested at each rate?

2) Thing did not go quite as planned. You invested $8000, part of it in stock that paid 12% annual interest. However, the rest of the money suffered a 5% loss. If the total annual income from both investment was $620, how much was invested at each rate?

Please help? How do I solve?

Answer 1

Hi,

1)You invested $7000 in two accounts paying 6% and 8% annual interest, respectively. If the total interest earned for the year was $520, how much was invested at each rate?

Let x = money at 6% and y = money at 8%
Then .06x = interest on 6% money and .08y = interest on 8% money

Equations are

x + y = 7000
.06x + .08y = 520

Multiply first equation by -6 and multiply 2nd equation by 100 to eliminate decimals. Then add the equations and solve.

-6(x + y = 7000)
100(.06x + .08y = 520)


-6x - 6y = -42000
6x + 8y = 52000
-------------------------
2y = 10000
y = 5000, so 5000 was invested at 8%
The other 2000 was invested at 6%


2) Thing did not go quite as planned. You invested $8000, part of it in stock that paid 12% annual interest. However, the rest of the money suffered a 5% loss. If the total annual income from both investment was $620, how much was invested at each rate?

Let x = money at 12% and y = money suffering 5% loss
Then .12x = interest on 12% money and -.05y = loss on other account

Equations are

x + y = 8000
.12x - .05y = 620

Multiply first equation by 5 and multiply 2nd equation by 100 to eliminate decimals. Then add the equations and solve.

5x + 5y = 40000
12x - 5y =62000
------------------------
17x = 102000
x = 6000, so $6000 was invested at 12%.
The other $2000 suffered a 5% loss.

I hope that helps!! :-)

Answer 2

Question 1.
x + y = 7000
.06x + .08y = 520

Subtracting .06 of the first row from the second row gives:
x + y = 7000
.02y = 100

Multiplying the second row by 50 gives:
x + y = 7000
y = 5000

Back substituting the second equation into the first gives:
x + 5000 = 7000

So,
x = 2000

Conclusion: 2000 was invested at 6% and 5000 was invetesed at 8%
------
Question 2.
We begin as before:
x + y = 8000
.12x + (-.05y) = 620

Subtracting .12 the first row from the second gives:
x + y = 8000
-.17y = -340

Dividing the second row by -.17 gives:
x + y = 8000
y = 2000

Back substitution yields:
x + 2000 = 8000

So x = 6000

Conclusion: 6000 was invested at 12% interest and 2000 was invested at a 5% loss.

Answer 3

1. 0.06x + 0.08(7000 - x) = 520
...560 - 0.02x = 520
...0.02x = 40
...x = 2000
...2000 @ 6% and 5000 @ 8%

2. 0.12x - 0.05(8000 - x) = 620
...0.17x - 400 = 620
...0.17x = 1020
...x = 6000
...6000 @ 12% and 2000 @ -5%

Answer 4

1>Let amount @ 6%=x
so amount @8%=7000-x

given that 6x/100+8(7000-x)/100=5200
or 6x+56000-8x=52000

or 2x=4000 so x=2000
so other amount=5000

2>Let amount at 12%=x
Interest for the year =12x/100
Balance amount @5% loss=8000-x
given 12x/100-5(8000-x)/100=620

or 17x=62000+40000=102000
or x=6000
so other amount=2000

Answer 5

1) Let x = invested at 6%, $7,000 + x = invested at 8%
0.06x + 0.08($7,000 - x) = $520
0.06x + $560 - 0.08x = $520
- 0.02x = - $40
x = $2,000
$7,000 - $2,000 = $5,000

Answer: $2,000 was invested at 6% while $5,000 was invested at 8%

Proof:
= 0.06($2,000) + 0.08($5,000)
= $120 + $400
= $520

2) Let x = invested money which suffered loss

Equation:
0.12($8,000) + (- 0.05x) = $620
$960 - 0.05x = $620
- 0.05x = - $340
x = $6,800
Loss = (0.05 * $6,800) or $340

Answer: $6,800 was invested at a 5% loss. $8,000 earned $960; $6,800 lost $340

Proof: $960 - $340 = $620

Answer 6

q1. 520=0.08x+0.06(7000-x)
=0.02x+420
x=100/0.02=5000

6%==>2000
8%==>5000

q2. 620=0.12x-0.05(8000-x)
=0.17x-400
x=1020/0.17=6000

12%==>6000
5%==>2000