**Finance Word Problems**

- Jack Potts invested $10,000. Part of it he put in the bank at 5 percent interest. The remainder he put in bonds which pay a 9 percent yearly return. Find the absolute value of the difference of the two investments in each vehicle if his yearly income from the two investments was $660?

Solution

Let

x = amount in dollars invested at 5 percent.

10,000 – x = amount in dollars invested at 9 percent.

Convert above information into income.

ie I = P x R x T

Interest = Principal x Rate x Time

If you put $1,000 in the bank at 5 percent, your interest would be 1000(0.05); remember 5 percent = 0.05. Setting up our equation,

0.05x = interest earned from bank investment

0.09(10,000 – x) = interest earned from bond investment

The interest earned from the bank plus the interest raened from the bond equals the total interest earned.

0.05x + 0.09(10,000 – x) = 660

0.05x + 900 – 0.09x = 600

Multiply both sides by 100

5x + 90,000 – 9x = 66,000

-4x = -24,000

x = 6,000

10,000 – x = 4,000

So,

6,000 was invested in the bank

4,000 was invested in the bond.

- Mr. Gold invested $50,000, part at 6 percent and part at 8 percent. The annual interest on the 6 percent investment was $480 more than that from the 8 percent investment. How much was invested at each rate?

Solution

Let x = amount in dollars invested at 6 percent.

50,000 – x = amount in dollars invested at 8 percent.

**Then**

0.06x = income on first investment

0.08(50,000 – x) = income on second investment

The interest on the 6% investment was $480 more than the interest on the 8% investment.

**Equation**

The first income equals the second income plus 480.

0.06x = 0.08(50,000 – x) + 480

0.06x = 4,000 – 0.08x + 480

0.06x = 4480 – 0.08x

6x = 44,8000 – 8x

14x = 448,000

**Answers**

x = 32,000

50,000 – x = 18,000

**So,**

$32,000 was invested at 6% and $18,000 was invested at 8%

- A store advertised dresses on sale at 20 percent off. The sale price was $76. What was the original price of the dress?

Solution

Let x = original price in dollars

If the sale price is 20% off the original price, the sale price is 80% of the original price.

**Equation**

0.08x = 76

80x = 7600

x = 95

**So,**

The original price was $95.

- Tickets to the school play sold at $4 each for adults and $1.50 each for children. If there were four times as many adult tickets sold as children’s tickets, and the total receipts were $3500, how many children’s tickets were sold?

Solution

Let x = number of children’s ticket at $ 1.50 each.

4x = number of adult tickets at $ 4 each.

**Then**

1.50x = total amount of money received from children’s tickets in dollars

4(4x) = total amount of money received from adult tickets in dollars

**Equation**

1.50x + 4(4x) = 3500

1.50x + 16x = 3500

15x + 160x = 35,000

175x = 35,000

x = 200

**So,**

200 tickets were sold at $1.50 and 800 tickets were sold at $4.00

- Art Tillery owns a jewelry store. He marks up all merchandise 50 percent of cost. If he sells a diamond ring for $1500, what did he pay the wholesaler for it?

Solution

Let x = amount in dollars he paid wholesaler.

0.50x = his markup (50%)

Cost plus markup equals selling price.

Equation: x + 0.50x = 1500

Multiply by 10 to clear decimals.

10x + 5x = 15,000

15x + 15,000

x = 1000

**So,**

The wholesaler paid $1,000 for the diamond ring.

- What amount of money invested at 8 ¼ percent yields a $2475 return per year?

Solution

Let x = amount of money invested in dollars

0.0825x = interest of 8 1/4 percent per year.

**Equation**

0.0825 = 2475

x = 30,000

**So,**

$30,000 was invested.

- Orney and Sly Company often price miscellaneous at a price they know will sell fast and determine what they can pay for it by this selling price. The Company bought men’s t-shirts to sell at $10 each. If they allow 40 percent of the
*selling price*for expenses and profit, what will they be willing to pay for the shirts?

Solution

Cost plus the markup equals the selling price.

Let x = cost in dollars they will pay for each shirt

0.40(10) = markup (40% of selling price)

**Equation**

x + 0.40(10) = 10

x + 4 = 10

x = 6

**So,**

The Market would be willing to pay $6 each for the shirts.

- Rhoda Davidson invested $50,000. Part of it she put in a gold mine stock from which she hoped to receive a 20 percent return per year. The rest he invested in a bank stock which was paying 6 percent per year. If she received $400 more the first year from the bank stock than from the mining stock, how much did she invest in each stock?

Solution

Let x = amount of dollars invested at 20 percent.

50,000 – x = amount of dollars invested at 6 percent (total minus x)

0.20x = interest on mining stock (smaller income)

0.06(50,000 – x) = interest on bank stock (larger income)

**Equation**

Income from bank stock is interest on the mining stock plus $400.

0.06(50,000 – x) = 0.20x + 400

3000 – 0.06x = 0.20x + 400

300,000 – 6x = 20x + 40,000

-26x = -260,000

**Answer**

x = 10,000

50,000 – x = 40,000

**Check**

0.06(40,000) = 0.20(10,000) + 400

2400 = 2000 + 400 = 2400

- Jim Shortz wished to invest a sum of money so that the interest each year would pay his son’s college expenses. If the money was invested at 8 percent and the college expenses were $10,000 per year, how much should Jim invest?

Solution

Let x = amount in dollars Jim should invest.

0.08 = interest per year (assume simple interest)

**Equation**

0.08 = 10,000

8x = 1,000,000

**Answer**

x = 125,000

**Check**

0.08(125,000) = 10,000

10,000 = 10,000

- The Worn & Torn Store had a sale on T-shirts at which all shirts were sold at 15 percent off the original price. Casey bought a shirt for $7.65.What was the original selling price of the shirt? (Assume no taxes.)

Solution

Let x = original selling price in dollars

0.15x = discount

The original price minus the discount equals the scale price.

**Equation**

x – 0.15x = 7.65

0.85x = 7.65

85x = 765

**Answer**

x = 9

**Check**

9 – 0.15(9) = 7.65

9 – 1.35 = 7.65

7.65 = 7.65

- Stan Back inherited two different stocks whose yearly income was $2 100. The total appraised value of the stocks was $40,000; one was paying 4 percent and one 6 percent per year. What was the value of each stock?

Solution

Let x = value in dollars of stock paying 4%

40000 – = value in dollars of stock paying 6%

0.04x = interest on stock paying 4%

0.06(40000 – x) = interest on stock paying 6%

**Equation**

0.04x + 0.06(40000 – x) = 2100

0.04x = 2400 – 0.06x = 2100

Multiply through by 100

4x + 240000 – 6x = 210000

-2x = -30000

**Answers**

x = 15000

40000 – x = 25000

**Check**

0.04(15,000) + 0.06(25,000) = 2100

600 + 1500 = 2100

2100 = 2100

- A men’s store bought 500 suits, some at $125 each and the rest at $200 each. If the total cost of the suits was $77,500, how many suits were purchased at each price?

Solution

Let x = number of suits purchased at $125 each.

500 – x = number of suits purchased at $200 each.

The total value equals the price times the number at $125 plus the price times the number at $200.

**Equation**

125x + 200(500 – x) = 77,500

125x + 100,000 – 200x = 77,500

-75x = 22,500

**Answer**

x = 300 (suits at $125 each)

500 – x = 200 (suits at $200 each)

**Check**

125(300) + 200(200) = 77,500

37,500 + 40,000 = 77,500

77,500 = 77,500

- When Mary Thon sold her house recently, she received $210,000 for it. This was 40 percent more than she paid for it 10 years ago. What was the original purchase price?

Solution

Let x = original purchase price in dollars

0.40x= increase in value in dollars

**Equation**

x + 0.40x = 210,000

1.40x = 210,000

14x = 2,100,000

**Answer**

x = 150,000 (original purchase price)

**Check**

150,000 = 0.4(150,000) = 210,000

150,000 + 60,000 = 210,000

210,000 = 210,000

- Minnie Sota inherited $20,000 which she invested in stocks and bonds. The stocks returned 6 percent and the bonds 8 percent. If the return on the bonds was $80 less than the return on the stocks, how much did she invest in each?

Solution

Let x = amount in dollars invested in stock at 6%

20000 – x = amount in dollars invested in bonds at 8%

0.06x = interest on stocks

0.08(20000 – x) = interest on bonds

**Equation**

Interest on stock equals interest on bonds plus $80

0.06x = 0.08(20000 – x) + 80

0.06x = 1600 – 0.08 + 80

Multiply through by 100

6x = 160000 – 8x + 8000

14x = 168000

**Answers**

x = 12000 (stocks)

20000 – x = 8000 (bonds)

**Check**

0.06(12,000) = 0.08(8000) + 80

720 = 640 + 80 = 720

- The total of two investments is $25,000. One amount is invested at 7 percent and one at 9 percent. The annual interest from the 7 percent investment is $470 more than from the 9 percent investment. How much is invested at each rate?

Solution

Let x = amount in dollars invested at 7 percent.

25,000 -x = amount in dollars invested at 9 percent.

**Equation**

The amount at 7 percent equals the amount at 9 percent plus $470.

0.07x = 0.09 (25,000 – x) + 470

0.07x = 2250 – 0.09x + 470

7x = 225,000 – 9x + 47,000

16x = 272,000

x = 17,000 (amount invested at 7 percent)

25,000 – x = 8000 (amount invested at 9 percent)

**So,**

$17,000 was invested at 7 percent and $8,000 was invested at 9 percent.

**Check**

0.07(17,000) = 0.09(8000) + 470

1190 = 720 = 470

1190 = 1190

- A taxpayer’s state and federal income taxes plus an inheritance tax totaled $14,270. His California state income tax was $5780 less than his federal tax. His inheritance tax was $2750. How much did he pay in state and federal taxes?

Solution

Let x = amount in dollars of state tax.

x + 5780 = amount in dollars of federal tax.

**Equation**

x + x + 5780 + 2750 = 14,270

2x + 8530 = 14,270

2x = 5740

x = 2870 (state tax)

x + 5780 = 8650 (federal tax)

**So,**

He paid $2,870 in state tax and $8,650 in federal tax.

**Check**

2870 + 2870 + 5780 + 2750 = 14,270

14,270 = 14,270

- Ann Athlete had saved $6000 which she wished to invest. She put part of it in a term Certificate of Deposit (CD) at 8 percent and part in a regular savings account at 5 1/2 percent. How much was invested in each account if her total yearly income amounted to $425?

Solution

Let x = amount in dollars invested at 8 percent.

6000 – x = amount in dollars invested at 51/2 percent.

0.08x = interest on 8 percent investment.

0.055(6000 – x) = interest on 5 ½ percent investment.

**Equation**

Total interest (income) was $425.

0.08x + 0.055(6000 – x) = 420

0.08x + 330 – 0.055x = 425

Multiply by 1000 to clear decimals.

80x + 330,000 – 55x = 425,000

25x = 95,000

x = 3800

6000 – x = 2200

**So,**

$3,800 was invested at 8% and $2,200 was invested at 5 1/2%

**Check**

0.08(3800) + 0.055(2200) = 425

304 + 121 = 425

425 = 425

- Mel Ting had $10,000 invested at 5 percent. How many dollars more would he have to invest at 8 percent so that his total interest per year would equal 7 percent of the two investments?

Solution

Let x = amount in dollars at 8 percent.

10,000 = total amount invested at 5 percent.

x + 10,000 = total amount invested at 7 percent.

0.08x = interest on amount at 8 percent.

0.05(10,000) = interest on amount at 5 percent.

0.07 (x + 10,000) = interest on entire investment at 7 percent.

**Equation**

Total interest per year equals 7 percent of the entire investment.

0.08x + 0.05(10,000) = 0.07(x + 10,000)

0.08x + 500 = 0.07x + 700

Multiply both sides by 100.

8x + 50,000 = 7x + 70,000

x = 20,000

**So,**

$20,000 would have to be invested at 8%

**Check**

0.08(20,000) + 0.05(10,000) = 0.07(30,000)

1600 + 500 = 2100

2100 = 2100

- Ms. Twinkle invested part of her savings at 6% and the rest at 9%. The amount at 9% was twice the amount at 6%. If her total return after one year was $72, find the amount invested at each rate.

Solution

Let x = amount in dollars at 6 percent.

Let 2x = amount in dollars at 9 percent.

**Equation**

0.06x + 0.09(2x) = 72

Multiply both sides by 100.

6x + 9(2x) = 7,200

6x + 18x = 7,200

24x = 7,200

x = 300

and

2x = 600

**So,**

$300 would have to be invested at 6%

$600 would have to be invested at 9%

**Check**

0.06(300) + 0.09(600)

$18 + $54 = $72

- Rockjaw invested part of his savings at 7% and the rest at 13%. The amount at 7% was $200 more than the amount at 13%. If his total return after one year was $84, find the amount invested at each rate.

Solution

Let x = amount in dollars at 13 percent.

Let x +200 = amount in dollars at 7 percent.

**Equation**

0.13x + 0.07(x + 200) = 84

Multiply both sides by 100.

13x + 7(x + 200) = 8,400

13x + 7x + 1400 = 8400

20x + 1400 = 8,400

20x = 7,000

x = 350

**So,**

$350 would have to be invested at 13%

$550 would have to be invested at 7%

**Check**

0.13(350) + 0.07(550)

$45.5 + $38.5 = $84

- Carol invested part of her savings at 10% and the rest at 8%. The amount at 8% was $1500 more than the amount at 10%. If the total annual income is $480, how much was invested at each rate?

Solution

Let x = amount in dollars at 10 percent.

Let x +1500 = amount in dollars at 8 percent.

**Equation**

0.10x + 0.08(x + 1500) = 480

Multiply both sides by 100.

10x + 8(x + 1500) = 48,000

10x + 8x + 12,000 = 48,000

18x + 12,000 = 48,000

18x = 36,000

x = 2,000

**So,**

$2,000 would have to be invested at 10%

$3,500 would have to be invested at 8%

**Check**

0.10(2,000) + 0.08(3,500)

$200 + $280 = $480

- Patty Wack had $900. She invested part of it at 12% and the rest at 9%. If her total annual return was $96, how much did she invest at each rate?

Solution

Let x = amount in dollars at 12 percent.

Let 900 – x = amount in dollars at 9 percent.

**Equation**

0.12x + 0.09(900 – x) = 96

Multiply both sides by 100.

12x + 9(900 – x) = 9,600

12x + 8,100 – 9x = 9,600

3x + 8,100 = 9,600

3x = 1,500

x = 500

**So,**

$500 would have to be invested at 12%

$400 would have to be invested at 9%

**Check**

0.12(500) + 0.09(400)

$60 + $36 = $96

- Dr. Beaker invested $3000, part at 8% and the rest at 7 ½%The total return for one year was $231 How much was invested at each rate?

Solution

Let x = amount in dollars at 8 percent.

Let 3000 – x = amount in dollars at 7.5 percent.

**Equation**

0.08x + 0.075(3000 – x) = 231

Multiply both sides by 1000.

80x + 75(3000 – x) = 231,000

80x + 225,000 – 75x = 231,000

5x + 225,000 = 231,000

5x = 6,000

x = 1,200

**So,**

$1,200 would have to be invested at 8%

$1,800 would have to be invested at 7.5%

**Check**

0.08(1200) + 0.075(1800)

$96 + $135 = $231

24. A scholarship fund raised $7000 in contributions. Part was invested in bonds paying 6% interest, and the rest was invested in bank certificates paying 8 ½%. It the total annual income is $520, find the amount invested at each rate.

Solution

Let x = amount in dollars at 6 percent.

Let 7000 – x = amount in dollars at 8.5 percent.

**Equation**

0.06x + 0.085(7000 – x) = 520

Multiply both sides by 1000.

60x + 85(7000 – x) = 520,000

60x + 595,000 – 85x = 520,000

-25x + 595,000 = 520,000

-25x = -75,000

x = 3,000

**So,**

$3,000 would have to be invested at 6%

$4,000 would have to be invested at 8.5%

**Check**

0.06(3000) + 0.085(4000)

$180 + $340 = $520

- Sam Quirk invested $7000, part at 7% and the rest at 11%. If his total return for one year was $690, how much was invested at each rate?

Solution

Let x = amount in dollars at 7 percent.

Let 7000 – x = amount in dollars at 11 percent.

**Equation**

0.07x + 0.11(7000 – x) = 690

Multiply both sides by 100.

7x + 11(7000 – x) = 69,000

7x + 77,000 – 11x = 69,000

-4x + = -8,000

x = 2,000

**So,**

$2,000 would have to be invested at 7%

$5,000 would have to be invested at 11%

**Check**

0.07(2000) + 0.11(5000)

$140 + $550 = $690

- An investment fund has $3000 more invested at 8% than it does at 10%. If the annual return from the 8% investment is the same as the annual return from the 10% investment, how much is invested at each rate?

Solution

Let x = amount in dollars at 8 percent.

Let x – 3000 = amount in dollars at 10 percent.

**Equation**

0.08x = 0.10(x – 3000)

Multiply both sides by 100.

8x = 10(x – 3000)

8x = 10x – 30,000

-2x + = -30,000

x = 15,000

**So,**

$15,000 would have to be invested at 8%

$12,000 would have to be invested at 10%

**Check**

0.08(15000) = 0.10(12000)

1200 = 1200

- Ms. Smyle has $200 less invested at 9% than she does at 6 ½%. If the annual return from the two investments is the same, how much is invested at each rate?

Solution

Let x = amount in dollars at 9 percent.

Let x + 200 = amount in dollars at 6.5 percent.

**Equation**

0.09x = 0.065(x + 200)

Multiply both sides by 1000.

90x = 65(x + 200)

90x = 65x + 13000

25x + = 13,000

x = 520

**So,**

$520 would have to be invested at 9%

$720 would have to be invested at 6.5%

**Check**

0.09(520) = 0.065(720)

46.8 = 48.6

- Sally Snuggle has $1600 more invested at 5% than she does at 8%. The annual return from the 5% investment is $17 more than the annual return from the 8% investment. How much is invested at each rate?

Solution

Let x = amount in dollars at 5 percent.

Let x – 1600 = amount in dollars at 8 percent.

**Equation**

0.05x -17 = 0.08(x – 1600)

Multiply both sides by 100.

5x – 1700= 8(x – 1600)

5x – 1700 = 8x – 12800

3x = 11100

x = 3700

**So,**

$3700 would have to be invested at 5%

$2100 would have to be invested at 8%

**Check**

0.05(3700) – 17 = 0.08(2100)

185 – 17 = 168

168 = 168

- Merlin invested half of his money at 12%, one fourth at 8%, and the rest at 6%. If the total annual income is $570, how much was invested altogether?

Solution

Let x = amount in dollars at 6 percent.

Let x = amount in dollars at 8 percent.

Let 2x = amount in dollars at 12 percent.

**Equation**

.06x + .08x + .12(2x) = 570

Multiply through by 100

6x + 8x + 12(2x) = 57000

6x + 8x + 24x = 57000

38x = 57000

x = 1500

**So,**

$1500 would have to be invested at 6%

$1500 would have to be invested at 8%

$3000 would have to be invested at 12%

**Check**

0.06(1500) + 0.08(1500) + .12(3000) = 570

90 + 120 + 360 = 570

570 = 570