MONEY AND INVESTMENT WORD PROBLEMS

Problem 1 :

Michaela has 5-cent and 10-cent stamps with a total value of $5.75. If she has 5 more 10-cent stamps than 5-cents stamps, how many of each stamp does she have?

Solution:

Number of 5 cent coins = x

Number of 10 cent coins = x + 5

Converting 5.75 dollars as 575 cents.

5x + 10(x + 5) = 575

5x + 10x + 50 = 575

15x = 575 - 50

15x = 525

x = 525/15

x = 35 stamps

10-cent stamp = (x + 5)

= 35 + 5

= 40 stamps

So, there are 35 5-cent stamps and 40 10-cent stamps.

Problem 2 :

The school tuck-shop has milk in 600 mL and 1 liter cartons. If there are 54 cartons and 40 L of milk in total, how many 600 mL cartons are there?

Solution:

Let x be the number of 600 ml cartons and y be the number of 1 liter cartons.

x + y = 54 ---> (1)

The tuck-shop has 40 L of milk in total.

From equation (1),

y = 54 - x

Substitute y = 54 - x in equation (2)

0.6x + 54 - x = 40

0.4x = 14

x = 35 cartons

Problem 3 :

Aaron has a collection of American coins. He has three times as many 10 cent coins as 25 cent coins, and he has some 5 cent coins as well. If he has 88 coins with total value $11.40, how many of each type does he have?

Solution:

Let x represent the number of 25-cent coins

then 3x represents the number of 10-cent coins

and 88 - 4x represents the number of 5 cent coins

0.25(x) + 0.1(3x) + 0.05(88 - 4x) = 11.40

0.25x + 0.3x + 4.4 - 0.2x = 11.40

0.25x + 0.3x - 0.2x + 4.4 = 11.40

0.55x - 0.2x + 4.4 = 11.40

0.35x + 4.4 = 11.40

0.35x = 11.40 - 4.4

0.35x = 7

x = 20

25 cent coins = 20

10 cent coins = 3x = 3(20)

= 60 

5 cent coins = 88 - 4x

= 88 - (4 × 20)

= 88 - 80

= 8

Therefore, Aaron has 8 five-cent, 60 ten-cent and 20 25-cent coins.

Problem 4 :

Tickets at a football match cost $8, $15 or $20 each. The number of $15 tickets sold was double the number of $8 tickets sold. 6000 more $20 tickets were sold than $15 tickets. If the total gate receipts were $783000, how many of each type of tickets was sold?

Solution:

Let x represent the $8 tickets

then y represents the $15 tickets

and z represents the $20 tickets.                     

y = 2x

z = y + 6000

8x + 15y + 20z = 783000

8x + 15(2x) + 20(2x + 6000) = 783000

8x + 30x + 20x + 120000 = 783000

78x + 120000 = 783000

78x = 783000 - 120000

78x = 663000

x = 8500

y = 2x = 2(8500)

y = 17000

z = y + 6000

= 17000 + 6000

z = 23000

Problem 5 :

Kelly blends coffee. She mixes brand A costing $6 per kilogram with brand B costing $8 per kilogram. How many kilograms of each brand does she has to mix to make 50 kg of coffee costing her $7.20 per kg?

Solution:

Let x be the cost of brand A, per kilogram

Let y be the cost of brand B, per kilogram.

x + y = 50 ---> (1)

The total cost of the mix = 6x + 8y

So, the total cost per kg of the mix will be

Solve equation (1) and (2),

2y = 60

y = 30

By applying y = 30 in equation (1),

x + 30 = 50

x = 50 - 30

x = 20

So, 20 kg of A and 30 kg of B.

Problem 6 :

Su Li has 13 kg of almonds costing $5 per kg. How many kg of cashews costing $12 per kg should be added to get a mixture of the two nut types which would cost $7.45 per kg?

Solution:

5 × 13 + 12x = 7.45 (13 + x)

12x + 65 = 96.85 + 7.45x

12x - 7.45x = 96.85 - 65

4.55x = 31.85

x = 31.85/4.55

x = 7 kg

So, 7 kg of cashew cost $12.