HOW TO SOLVE THE MOST DIFFICULT WORD PROBLEMS
NOTE TO TEACHERS
In solving story problems (oral or written) children should be encouraged to:
A. Think of the mathematical sentence, number sentence or equation.
- What bits of information are given in the problem?
- What is the relationship between information and the question?
- What mathematical sentence can we write to show this?
B. Perform the computation C. State the answer in a complete sentence
In solving story problems (oral or written) children should be encouraged to:
A. Think of the mathematical sentence, number sentence or equation.
- What bits of information are given in the problem?
- What is the relationship between information and the question?
- What mathematical sentence can we write to show this?
B. Perform the computation C. State the answer in a complete sentence
EXAMPLE 1
Joe won 12 marbles. He had 10 marbles before. How many marbles does he have?
answers:
A. 12 + 10 = ____
B 12
+10
_____
22
C.Joe has 22 marbles
Joe won 12 marbles. He had 10 marbles before. How many marbles does he have?
answers:
A. 12 + 10 = ____
B 12
+10
_____
22
C.Joe has 22 marbles
EXAMPLE 2: Age Problems
Mary's father is four times as old as Mary. Five years ago he was seven times as old. How old is each now?
First fact: Mary's father is four times as old as Mary
Let: x = Mary's age now (smaller number)
4x = Father's age now
Second fact: Five years ago...(STOP!) The problem tells you about their ages at another time. Five years ago your age would be five less than your age now. So...
x-5 = Mary's age 5 years ago
4x - 5 = father's age 5 years ago
Now read on: Five years ago he was seven times as old (as she was)
4x - 5 = 7 (x - 5)
4x - 5 = 7x - 35
-3x = -30
x = 10
4x = 40
ANSWERS: Mary is ten, her father is forty
Mary's father is four times as old as Mary. Five years ago he was seven times as old. How old is each now?
First fact: Mary's father is four times as old as Mary
Let: x = Mary's age now (smaller number)
4x = Father's age now
Second fact: Five years ago...(STOP!) The problem tells you about their ages at another time. Five years ago your age would be five less than your age now. So...
x-5 = Mary's age 5 years ago
4x - 5 = father's age 5 years ago
Now read on: Five years ago he was seven times as old (as she was)
4x - 5 = 7 (x - 5)
4x - 5 = 7x - 35
-3x = -30
x = 10
4x = 40
ANSWERS: Mary is ten, her father is forty
EXAMPLE 3 Work Problem:
Bill can mow the lawn in two hours, and his younger brother Tom can mow it in three hours. If both brothers mow the lawn together, how many hours will it take them?
Solution:
Person Hours to mow lawn Part done in one hour
Bill 2 1/2
Tom 3 1/3
Together x 1/x
Equation: 1/2 + 1/3 = 1/x
answer:
L.C.D = 6x
3x + 2x = 6
5x = 6
x = 1 1/5 hour
or one hour and twelve minutes (1/5 of an hour) to finish the job together
Bill can mow the lawn in two hours, and his younger brother Tom can mow it in three hours. If both brothers mow the lawn together, how many hours will it take them?
Solution:
Person Hours to mow lawn Part done in one hour
Bill 2 1/2
Tom 3 1/3
Together x 1/x
Equation: 1/2 + 1/3 = 1/x
answer:
L.C.D = 6x
3x + 2x = 6
5x = 6
x = 1 1/5 hour
or one hour and twelve minutes (1/5 of an hour) to finish the job together
EXAMPLE 4 Lever Problem:
A lever is a stiff board or bar which is balanced upon a support called a fulcrum
Formula: w(a) x D1 = D2 x w(b)
A D1 distance C D2 distance B
weight from Fulcrum weight
Example: A forty pound weight placed 2 feet from the fulcrum of a lever balances an unknown weight placed 4 feet from the fulcrum.
Find the unknown weight
w(a) w(b)
40 lbs D1 D2 x lbs.
2 ft. 4 ft.
answer:
Solution: 40 x 2 = 4 x X
80 = 4x
x = 20 lbs.
A lever is a stiff board or bar which is balanced upon a support called a fulcrum
Formula: w(a) x D1 = D2 x w(b)
A D1 distance C D2 distance B
weight from Fulcrum weight
Example: A forty pound weight placed 2 feet from the fulcrum of a lever balances an unknown weight placed 4 feet from the fulcrum.
Find the unknown weight
w(a) w(b)
40 lbs D1 D2 x lbs.
2 ft. 4 ft.
answer:
Solution: 40 x 2 = 4 x X
80 = 4x
x = 20 lbs.
EXAMPLE 5 Rate Problem:
A car leaves San Francisco for Los Angeles, traveling at an average of 60 miles per hour. At the same time another car leaves Los Angeles for San Francisco at 50 miles per hour. If it is 440 miles between San Francisco and Los Angeles, how long before the two cars meet, assuming that each maintained its average speed.
Let x = time in hours for the two cars to meet
Car from Los Angeles is equal to 50 times x
50 times x = 50x
Car from San Francisco is equal to 60 times x
60 times x = 60x
Note: The total distance is 440 miles, so add the two distances which is 50x + 60x = 440
Solution:
50x + 60x = 440
110x = 440
x = 4 hours
A car leaves San Francisco for Los Angeles, traveling at an average of 60 miles per hour. At the same time another car leaves Los Angeles for San Francisco at 50 miles per hour. If it is 440 miles between San Francisco and Los Angeles, how long before the two cars meet, assuming that each maintained its average speed.
Let x = time in hours for the two cars to meet
Car from Los Angeles is equal to 50 times x
50 times x = 50x
Car from San Francisco is equal to 60 times x
60 times x = 60x
Note: The total distance is 440 miles, so add the two distances which is 50x + 60x = 440
Solution:
50x + 60x = 440
110x = 440
x = 4 hours
EXAMPLE 6
Extend addition and subtraction to include amounts of money.
Since the elementary child has not had experience in decimal notation it may be advisable at this time to think of money in terms of dollars, dimes, and pennies (which require no decimal point) and then translate the answer into dollars and cents using the dollar sign and point.
$1.13 + $4.39 = ____
$1.13 = 1 dollar, 1 dime, 3 pennies
+$4.39 = 4 dollars, 3 dimes, 9 pennies
_______
$5.52 = 5 dollars, 4 dimes, 12 pennies
Read answer: 5 dollars and 52 cents
Extend addition and subtraction to include amounts of money.
Since the elementary child has not had experience in decimal notation it may be advisable at this time to think of money in terms of dollars, dimes, and pennies (which require no decimal point) and then translate the answer into dollars and cents using the dollar sign and point.
$1.13 + $4.39 = ____
$1.13 = 1 dollar, 1 dime, 3 pennies
+$4.39 = 4 dollars, 3 dimes, 9 pennies
_______
$5.52 = 5 dollars, 4 dimes, 12 pennies
Read answer: 5 dollars and 52 cents
EXAMPLE 7 INVESTMENT PROBLEM:
Mr Johnson invested part of his $ 40,000 investment in high yields bonds paying 9.5% and preferred stock paying 8%. If his annual income from both investments is $3,620.00, how much did he invest in the bonds and in the stock?
Solution: x + y = $40,000
.095x + .08y = 3620.00
(move the decimal over two spaces)
9.5x + 8y = 362,000
multiply (1st equation) by -8 (this is to get the y's to cancel out)
-8x - 8y = -320,000
9.5x + 8y = 362,0000
1.5 X = 42,000
3/2 X = 42,000
X = 28,000 invested in 9.5 % bonds
Substitute: $28,000 for x in (1)
28,000 + y = 40,000
y = $12,0000 at 8 %
Mr Johnson invested part of his $ 40,000 investment in high yields bonds paying 9.5% and preferred stock paying 8%. If his annual income from both investments is $3,620.00, how much did he invest in the bonds and in the stock?
Solution: x + y = $40,000
.095x + .08y = 3620.00
(move the decimal over two spaces)
9.5x + 8y = 362,000
multiply (1st equation) by -8 (this is to get the y's to cancel out)
-8x - 8y = -320,000
9.5x + 8y = 362,0000
1.5 X = 42,000
3/2 X = 42,000
X = 28,000 invested in 9.5 % bonds
Substitute: $28,000 for x in (1)
28,000 + y = 40,000
y = $12,0000 at 8 %
EXAMPLE 8
If the length of the rectangle is 5 feet longer than the width and the area is 50 square feet, what are the dimensions of the rectangle?
Hint: a = length times the width
Let x = width
Let x + 5 = length
x(x + 5) = area
Solution:
x(x + 5) = 50
x(x) + 5(x) - 50 = 0
(x + 10)(x -5) = 0
x+10 = 0
x = -10
x - 5 = 0
x= 5
measurements are 5 by 10 (width is 5 and length is 10)
If the length of the rectangle is 5 feet longer than the width and the area is 50 square feet, what are the dimensions of the rectangle?
Hint: a = length times the width
Let x = width
Let x + 5 = length
x(x + 5) = area
Solution:
x(x + 5) = 50
x(x) + 5(x) - 50 = 0
(x + 10)(x -5) = 0
x+10 = 0
x = -10
x - 5 = 0
x= 5
measurements are 5 by 10 (width is 5 and length is 10)
