# Finance Word Problems

Finance Word Problems

1. 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.

1. 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

x = 32,000

50,000 – x = 18,000

So,

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

1. 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.

1. 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

1. 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.

1. 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.

1. 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.

1. 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

x = 10,000

50,000 – x = 40,000

Check

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

2400 = 2000 + 400 = 2400

1. 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)

0.08 = 10,000

8x = 1,000,000

x = 125,000

## Check

0.08(125,000) = 10,000

10,000 = 10,000

1. 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

x = 9

Check

9 – 0.15(9) = 7.65

9 – 1.35 = 7.65

7.65 = 7.65

1. 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

x = 15000

40000 – x = 25000

Check

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

600 + 1500 = 2100

2100 = 2100

1. 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

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

1. 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

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

1. 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

x = 12000 (stocks)

20000 – x = 8000 (bonds)

Check

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

720 = 640 + 80 = 720

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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