Math 41 - Practice for Test #2
1.
Meg is younger than her brother Bob.
a. If you know Meg’s age and Bob’s age, how would you calculate the difference in their
ages?
b. If you know Meg’s age and the difference between their ages, how would you
calculate Bob’s age?
c. In which of the problems above is an additive comparison used? Explain why this is
so.
2.
There are five more children in Ms. Scarlett’s class than in Mr. Gray’s class. Ten children
move from Ms. Scarlett’s class to to Mr. Gray’s class. Now there are twice as many
children in Mr. Gray’s class as in Ms. Scarlett’s class. How many children were in Ms.
Scarlett’s class before the move?
a. List the quantities involved in this problem.
b. Sketch a diagram to show the relevant sums and differences.
c. Solve the problem.
3.
Write a word problem that illustrates the “missing addend” and uses the numbers 27 and
21. Note: You need to provide a “story” for the problem.
4.
Judith had to solve the following problem: “When Jack started his diet, he weighed 182
pounds. Today he weighs 158 pounds. How much weight did he lose?”
Judith replied: “184 minus 160 is 24 so he lost 24 pounds.” Explain Judith’s reasoning.
5.
For the subtraction problem 20 – 13, write word problems to illustrate the
a. Missing addend model for subtraction
b. Comparison model for subtraction
c. Take away model for subtraction
6.
A student performed the subtraction 523 – 297 as shown below. Explain the student’s
reasoning.
523
−297
300
−70
230
−4
226
7.
A student performed the subtraction 523 – 297 as shown below. Explain the student’s
reasoning.
523
−297
374
8.
Use the Australian method to perform the subtraction 523 – 297.
9.
Use the Australian method to perform the addition 147 + 39.
1
10. Give the rest of the family of facts for c – 2 = b.
11. Use the lattice method to add
2576 + 3985
12. Use the “scratch” method to add 569 + 275 + 1967 + 344 + 627 + 3296 + 245.
13. Use the “scratch” method to add the base five numbers
234five + 322five + 21five + 434five
14. A pizza shop offers 2 different kinds of crust, 4 different kinds of cheese and 12 different
toppings.If you order a pizza with one type of cheese and one topping, how many different
orders are possible? (Remember there is a choice on the crust!)
15. Explain the multiplication 3 × 8 using two different models for multiplication.
16. Decide which type of division, partitive (sharing) or measurement (repeated subtraction), is
involved in the following problem. Explain how you decide.
Dang earns $7.00 per hour for working in the college bookstore. Her paycheck
for the past two weeks totalled $175.00. How many hours did she work?
Using the same numbers, write a short story problem illustrating the other type of
division
17. Use the repeated subtraction model to calculate 27 ÷ 6 .
18. Use the scaffolding method to perform the division 4495 ÷ 37 .
19. In each pair of numbers, use number sense to choose the larger of the two numbers
without calculating the values of the expressons. Explain your reasoning.
a. 267 x 15 or 265 x 14
b. 576 – 124 or 576 – 122
c.
7
8
+
5
6
or 2
d. 576 ÷ 2 or 576 ÷ 3
20. Use estimation to tell whether the following calculator answers are reasonable. Explain
why or why not.
a. 657 + 542 + 707 = 543364
b. 26 x 47 = 1222
2
21. Compute 267 x 42 using
a. the lattice method
b. the Russian Peasant Algorithm
c. the method showing all partial products
22. Give the rest of the family of facts for 3 x a = c.
23. Use an area model to represent the multiplication 3 × 2
1
2
ANSWERS – Be sure to try them on your own before looking at the answers! Think!!
1.
a. To find the difference in their ages, subtract Meg’s age from Bob’s age, that is,
difference in ages = Bob’s age – Meg’s age.
b. To find Bob’s age is we know Meg’s age and the difference between them,add the
difference to Meg’s age, that is, Meg’s age + age difference = Bob’s age.
c. An additive comparison is used in b because it looks at how much must be added to
Meg’s age to find Bob’s age.
2.
a. Quantites include:
# children originally in Ms. Scarlett’s class
# children originally in Mr. Gray’s class
difference in the numbers of children in classes before the move
#children in Ms. Scarlett’s class after the move
# children in Mr. Gray’s class after the move
ratio of the numbers of children in the two classes after the move
b.
Original class size
Ms Scarlett has 5
more students than
Mr. Gray.
Class size after the move Comparing sizes after the move
Ten students move from Mr. Gray’s class has twice as many
Ms. Scarlett’s class to students as Ms. Scarlett’s.
Mr. Gray’s class so she
has 5 fewer students
than originally.
Mr. Gray’s class
Mr. Gray’s class
?
+ 10
Mr. Gray’s class
+ 10
?
Ms. Scarlett’s class
?
+ 5 - 10
2 x Ms. Scarlett’s class
-5
?
?
Ms. Scarlett’s class
?
+5
-5
?
c.
?
represents the original size of Mr. Gray’s class and we know that NOW
Mr. Gray’s class =
?
+ 10 =
2 x Ms. Scarlett’s class
?
10 =
?
-10
?
- 10
so that
Which means the original size of Mr. Gray’s class was 20 students.
3
?
= 20
3.
Answers vary.
4.
Judith reasoned that the difference between 182 and 158 is the same as the diffrence
between 184 (which is 2 larger than 182) and 160 (which is two larger than 158) so that
182 – 158 = 184 – 160 = 24. Using 184 – 160 to calculate the difference made the
problem easier to do.
5.
a. Problem for Missing addend model should involve looking at how many more you need
to get the desired total.
b. Problem for Comparison model should involve looking at how many more or how much
bigger one group is than another.
c. Problem for Take Away Model should involve removing some from a larger group or set.
6.
523
−297
300
−70
230
−4
226
In each place value, the student subtracts the lower number (subtrahend) from
the minuend, recognizing that in the tens and the ones the answer will be
negative.
500 – 200 leaves 300
Now she takes 70 from 300 because 20 – 90 = -70 (too few to subtract)
leaving 230
Again, looking at the ones she figues 3 – 7 = -4
This leaves 226
Another way to look at it:
523 − 297 = ( 500 − 200 ) + (20 − 90 ) + (3 − 7 )
=
=
7.
523
− 29 7
300
226
−
70
−
4
In each place value, the student found the difference between the larger
and smaller numbers. However, she did not pay attention to which
number was the subtrahend and which was the minuend. She actually
calculated 597 – 223!
374
-70
8.
226
-10
296
-10
306
-100
316
-100
416
-4
516
-3
520
523
So 523 – 297 = 226. Note: you can make your “jumps” is various ways. You could
even do 523 – 100 – 100 – 100 + 3
4
+10
+10
+10
+9
9.
147
157
167
177
186
So 147 + 39 = 147 + 30 + 9 = 186
10. Given c – 2 = b, we know c – b = 2,
11.
c=b+2
Lattice addition:
+
12.
c = 2 + b,
2
5
7
6
3
9
8
5
0
1
1
1
5
4
5
1
6
5
6
1
Scratch addition
13. Scratch addition, base five. Recall that here, we “scratch” anytime we get a group of 5.
5
14. 2 different kinds of crust could be used with any of 4 cheese to make 2 x 4 = 8 types of
pizza with only crust and cheese. Since any one of the 12 toppings can be put on any one
of these 8 pizzas, the total number of pizza orders possible is 2 x 4 x 12 = 96.
15. Use any two of the models for multiplication (repeated addition, rectangular array or
fundamental counting principal) to explain the multiplication 3 x 8.
a.
Repeated addition: 3 x 8 means 3 groups of 8 so 3 x 8 = 8 + 8 + 8 = 24
b.
Rectangular array (area) model: area of rectange with dimensions 3 by 8 is 24
c.
Fundamental Counting Principal: For the “fast lunch” at the school cafeteria a child can
choose any one of 3 sandwiches with any one At the cafeteria you can choose any one
of 8 drinks for a total of 24 possible lunches.
16. To determine the number of hours Dang worked to earn $175.00 when she earns $7.00 per
hour, we need to decide how many groups of $7.00 are in $175.00. This is the
measurement or repeated subtraction model. Dang works 175 ÷ 7 = 25 hours.
A problem using the partitive or sharing model of division would involve sharing 175 among
7 groups or individuals and asking how much or how many eacg group or individual would
get.
17. Performing division by repeated subtraction:
27
−6
21
−6
15
−6
9
There are 4 groups of 6 in 27
with 3 left over.
So, 27 ÷ 6 = 4, R. 3
Using the division algorithm,
we’d write 27 = 6 x 4 + 3
−6
3
6
18.
Performing division using the scaffolding method:
37 4495
3700
100 groups of 37
795
370
10 groups of 37
425
370
10 groups of 37
55
37
18
1 group of 37
121 groups of 37 with 18 left over
19. a. 267 x 15 > 265 X 14 because 267 > 265 and 15 > 14 so product of two larger is larger
b. 576 – 124 < 576 – 122 because subtracting a smaller number from 576 leaves more
c.
7
8
+
5
6
< 2 because
7
8
< 1 and
5
6
<1
so
7
+
8
5
6
<1+1
d. 576 ÷ 2 > 576 ÷ 3 because dividing 576 into 2 parts rather than 3makes larger parts
20. a. 657 + 542 + 707 ≈ 700 + 500 + 700 = 1900 so 543364 is too big. Also since each of the
addends is less than 1000, the sum must be less than 3000.
b. 26 x 47 ≈ 25 x 50 = 1250 so the answwwer seems reasonable.
21. a. Lattice multiplication
2
1
1
0
6
2
8
0
4
1
7
×
2
8
4
1
4
2
4
2
1
4
7
2
Our result says 267 x 42 = 11,214
21. b. Russian peasant algoritm
42
267
21
534
10
1068
5
2136
2
4274
1
8544
11214
22. c.
267
×42
14
120
400
280
2400
8000
11214
= 2x7
= 2 x 60
= 2 x 200
= 40 x 7
= 40 x 60
= 40 x 200
add partial products to get the entire product
23. Given 3 x a = c, we know also that a x 3 = c, c ÷ a = 3, c ÷ 3 = a
23. Area model for 3 × 2
2
+
1
2
1
2
Area = 3 × 2
1
2
= 3×2 + 3×
2
3
1
1
2
1
2
=
6
+
=
6
+
=
8
7
1
2
1
2
3
2
1
1
2