Hints for solving the problems (5.2)
Word Problems
Problem No. 3 – The sum of two numbers is 135. One-fourth the larger number is equal to two times the smaller number. What are the numbers?
Let x = the first number (the larger)
Let y = the second number (the smaller)
x + y = 135 the first equation
(1/4)x = 2y the second equation
(4)(1/4)x = (4)2y Multiplying by 4 to get rid of the fraction.
x = 8y
8y + y = 135 Substituting 8y for x in the first equation.
9y = 135
y = 15 the smaller number
x + 15 = 135 according to the first equation
x = 120 the larger number
120 + 15 = 135 checking the problem
135 = 135 true
(1/4)120 = 2(15)
30 = 30 true
The answers are 15 & 120
Problem No. 11 – Sally has a total of 40 quarters and dimes worth $6.55. How many of each does she have?
Let x = number of quarters
Let y = number of dimes
Then x + y = 40 first equation
and .25x + .10y = 6.55 second equation
25x + 10y = 655 multiply the second equation by 100 to eliminate the decimal.
5x + 2y = 131 divide the equation by 5.
-2(x + y = 40) multiply the first equation by -2
-2x - 2y = -80 add to second equation
5x + 2y = 131 eliminates the y
3x = 51
x = 17 no. of quarters
17 + y = 40
y = 23 no. of dimes
check 17 + 23 = 40 yes
.25(17) + .10(23) = 6.55 yes
The answer is: 17 quarters and 23 dimes.
Problem No. 17 – Zoe has a solution that is 75% sulfuric acid and a solution that is 25% sulfuric acid. How much of each should she use to obtain 400 ml of a solution that is 45% sulfuric acid.
Make a chard with three rows and three columns.
No. of ml %(dec.) Amt. of pure acid
75% x .75 .75x
25% y .25 .25y
45% 400 .45 180
Remember the (%) is always changed to a decimal when working the problem. The third column is derived from the product of the numbers or letters in the first column and the second column in each row. The first equation comes from the first column and the second equation comes from the third column.
x + y = 400
.75x + .25y = 180 multiply this equation by 100
75x + 25y = 18000 divide by 25
3x + y = 720
-1(x + y = 400) Multiply the first equation by -1 and add to the second equation to get rid of the y.
3x + y = 720
-x - y = -400
2x = 320
x = 160 no. of ml of the 75% solution to be added.
160 + y = 400 substituting into the first equation.
y = 240 no. of ml of the 25% solution to be added.
I will let you check it. The answers are 160 ml @ 75% & 240 ml @ 25%.
Problem No. 27 – Lena has 16 pounds of coffee that sells for $6.50 per pound. If she has a second coffee that sells for $8.00 per pound, how many pounds of the second coffee should be added to the first coffee to obtain a blend that will sell for $7.00 per pound? Make a chart.
No. of lbs. Price per pound Total value in cents
$6.50 16 6.50 10400
$8.00 x 8.00 800x
$7.00 y 7.00 700y
In these problems, like the last one, the first equation comes from the first column and the second equation comes from the third column.
16 + x = y first equation
10400 + 800x = 700y second equation
104 + 8x = 7y divide second equation by 100
104 + 8x = 7(16 + x) Substitute first equation into the second equation for y and solve for x.
104 + 8x = 112 + 7x
8x - 7x = 112 - 104
x = 8
The answer is 8 lbs of coffee to be added.