Wordproblemsolvingmethods4th5th6thGradesSingaporeMath-1
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Introduction
.......................................................................................................................................
2
1. Remainder Concept (Branching)
.................................................................................................
2
Examples (5th Grade)
...................................................................................................................
4
2. Equal Fractions Concept
..............................................................................................................
6
3. Model Drawing Concept
.............................................................................................................
13
4. The Repeated Variable Concept
................................................................................................
16
5. Constant
Quantity Concept
........................................................................................................
17
a-
Constant Quantity after a Transfer Out
................................................................................
17
b-
Constant Quantity after a Transfer In
...................................................................................
18
6.
Constant Total Concept
...........................................................................................................
23
(A) One-Way Transfer
.................................................................................................................
23
(B) Two-Way Transfer
.................................................................................................................
24
7. Constant Difference Concept
.....................................................................................................
27
a-
Equal Amount Transferred into 2 Variables
.........................................................................
27
b-
Equal Amount Transferred out of 2 Variables
......................................................................
27
c-
Age Difference
......................................................................................................................
28
8. Everything Changed Concept (Units and Parts)
........................................................................
32
9. Part-Whole Concept
...................................................................................................................
33
10.
Excess and Shortage Concept
............................................................................................
34
11.
Gap and Difference Concept
................................................................................................
39
12. The Excess Value Concept
.....................................................................................................
44
13. Grouping Concept
....................................................................................................................
46
14. Number x Value Concept
.........................................................................................................
47
15. Guess and Check / Assumption Concept
................................................................................
48
Solution: https://www.youtube.com/watch?v=Jj_OlRwRsi4
......................................................
50
16. Working Backwards Concept
...................................................................................................
54
17. Simultaneous Equations Concept
............................................................................................
57
18. Double If Concept
....................................................................................................................
60
19. Average Concept
......................................................................................................................
62
20. Percent, Fractions, Ratios
........................................................................................................
63
21. Speed Concept and non-routine questions
.............................................................................
64
Полезные ссылки:
.........................................................................................................................
66
1
Introduction
In this chapter, we will continue using our problem solving plan to solve real-world
problems. These are very powerful strategies in problem solving and probably the most
commonly used in everyday life. Let’s review our problem-solving plan.
Step 1
Understand the problem.
Read the problem carefully. Once the problem is read, list all the components and data that are
involved. This is where you will be assigning your variables
Step 2
Devise a plan – Translate
Come up with a way to solve the problem. Set up an equation, draw a diagram, make a chart or
construct a table as a start to solving your problem.
Step 3
Carry out the plan – Solve
This is where you solve the equation you came up with in Step 2.
Step 4
Look – Check and Interpret
Check to see if you used all your information and that the answer makes sense. Let’s now apply
this plan to a few problems.
1.
Remainder Concept (Branching)
The first type of common PSLE Math questions is on remainder concept. This is usually tested in
Fractions and Percentage questions for PSLE Math. When you see the keyword “remainder”, I
recommend you to use this method called “Branching”. You will learn more in the example
below.
Examples:
1.
John spent 3/5 of his money on books and 1/3 of the remainder on a wallet.
If John had $16 left, how much did he have at first?
2
Video solution:
https://jimmymaths.com/topic/example-1-copy-11/
Written solution:
Fraction of Money Left = 2/3 × 2/5 = 4/15
4 units = $16
15 units = $16 ÷ 4 × 15 = $60 (Answer)
2.
Mrs Tan made some muffins. She sold 2/5 of them in the morning and 4/9 of the
remainder in the evening. She sold 40 more muffins in the morning than evening. How
many muffins did she make altogether?
Before you play the video, try the question first.
Solution:
https://jimmymaths.com/topic/example-2-copy-10/
3.
Solution:
https://jimmymaths.com/topic/example-2-branching-method-copy-2/
4.
4
th
-5
th
Grade
Solution:
https://www.youtube.com/watch?v=QlFiwA2k4eM
5.
5
th
-6
th
Grade
3
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Solution:
https://www.youtube.com/watch?v=c6xcA8OIxIk
Examples (5th Grade)
6.
Anthony bought an 8-ffot board. He cut off 3/4 of the board to build a shelf, and gave the
rest to his brother for an art project. How many inches long was the piece Anthony gave to
his brother?
Solution:
https://youtu.be/kGQKoTlzeiI
7.
There are 48 students going on a field trip. One-fourth are girls. How many boys are going
on the trip?
8.
Three angles are labeled below with arcs. The smallest angle is 3/8 as large as the 160°
angle. Find the value of angle a.
9.
Abbie spent 5/8 of her money and saved the rest. If she spent $45, how much money did
she have at first?
10.
Mrs. Harrison used 16 ounces of dark chocolate while baking. She used 2/5 of the
chocolate to make some frosting and used the rest to make brownies. How much more
chocolate did Mrs. Harrison use in the brownies than in the frosting?
11.
A straight angle is split into two smaller angles as shown. The smaller angle's measure is
1/6 that of a straight angle. What is the value of angle a?
12.
Annabel and Eric made 17 ounces of pizza dough. They used 5/8 of the dough to make a
pizza and used the rest to make calzones. What is the difference between the amount of
dough they used to make pizza, and the amount of dough the used to make calzonee?
13.
The New York Rangers hockey team won 3/4 of their games last season. If they lost 21
games, how many games did they paly in the entire season?
Solutions for problems 7-13:
https://youtu.be/Pn17xXgeEm0
14.
Linda has 75 fliers to post around town. Last week, she posted 1/3 of them. This week,
she posted 1/5 of the remaining fliers. How many posters has she still not posted?
Solution
:
https://youtu.be/w3unjT5U4Zg
15.
A Grand Prix race consists of 180 km. After 30 minutes of racing, driver A completed 3/4 of
the course while driver B completed 5/6 of the course.
a) Which driver had covered the greater distance?
b) How many kilometres does each driver still have to complete?
Solution:
https://youtu.be/_0YQlA9XlKc
16.
1. Janice is preparing a recipe that calls for 3/4 of a cup of oil per serving. If Janice needs
to prepare 2 and 2/3 servings, how many cups of oil will she need?
2. Jamaal is running at a rate of 2 and 2/3 miles per hour. At that rate how far will Jamaal
have run in 1 and 1/8 hours?
Solution:
https://youtu.be/0a3Mjvb7_VE
4
17.
1
3
of
t
h
e
e
ggs
at
a
s
t
or
e
a
r
e
la
rg
e
.
3
4
of
t
h
e
r
e
m
ai
nd
e
r
a
r
e
m
e
d
i
um. Th
e
r
e
s
t
a
r
e
sm
all
e
ggs. Th
e
r
e
a
r
e
20 mor
e
m
e
d
i
um
e
ggs
t
h
a
n
la
rg
e
e
ggs. F
i
nd
t
h
e
t
o
tal
numb
e
r of
e
ggs
at
t
h
e
s
tall
.
Solution:
18.
8
15
of
t
h
e
t
r
ee
s
i
n
a
p
la
n
tati
on
a
r
e
a
pp
le
t
r
ee
s.
3
7
of
t
h
e
r
e
m
ai
n
i
ng
t
r
ee
s
a
r
e
p
ea
r
t
r
ee
s.
3
5
of
t
h
e
r
e
s
t
a
r
e
dur
ia
n
t
r
ee
s. Th
e
r
e
m
ai
n
i
ng 40
t
r
ee
s
a
r
e
or
a
ng
e
t
r
ee
s. F
i
nd
t
h
e
t
o
tal
numb
e
r of
a
pp
le
t
r
ee
s
i
n
t
h
e
p
la
n
tati
on.
Solution:
19.
5
12
of
t
h
e
sh
i
r
t
s
i
n
a
c
upbo
a
rd
a
r
e
r
e
d.
3
5
of
t
h
e
r
e
m
ai
n
i
ng sh
i
r
t
s
a
r
e
b
l
u
e
.
4
7
of
t
h
e
r
e
s
t
a
r
e
gr
ee
n
a
nd
t
h
e
r
e
s
t
a
r
e
y
ell
ow. Th
e
r
e
a
r
e
60 mor
e
b
l
u
e
sh
i
r
t
s
t
h
a
n y
ell
ow
sh
i
r
t
s. How m
a
ny r
e
d sh
i
r
t
s
a
r
e
t
h
e
r
e?
Solution:
5
20.
Solution:
https://www.youtube.com/watch?v=PU6gnyni3gU
21.
Solution:
https://www.youtube.com/watch?v=95i1UlOALGI
2.
Equal Fractions Concept
The Equal Concept is derived from the
Comparison Concept
. It compares two or more fractions,
decimals or percentages, etc that represent equal quantities. In this concept, we first draw a
model to represent the first variable given and mark out the part of it that will be equal in quantity
to a given part in the second variable represented by a second model.
To illustrate this concept, consider the following question,
Example 1:
1/4 of A is equal to 1/3 of B. A is greater than B by 40. What is the value of A and B?
Step 1: Draw a long bar to represent the whole of A. Divide the bar into 4 equal boxes and label
1 box as the equal part.
6
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Step 2: Next, draw a box below the model of A to represent the part of B that is equal to 1/4 of A,
i.e., 1/3 of B.
Step 3: Since the first box of B drawn represents 1/3 of B, we will need to draw another 2 boxes
to its right to represent the remaining 2/3 of B.
Step 4: Since A has 4 units and B has 3 units, the extra 1 unit of A must be equal to 40(given in
question).
7
1 unit ----------> 40
3 units ----------> 3 X 40 = 120
4 units ----------> 4 X 40 = 160
Therefore, A is 160 and B is 120.
Example 2:
There are 836 students in a school. 7/10 of the boys and 7/8 of the girls take bus to school. The
number of boys who do not take bus is
twice
the number of girls who do not take bus. How
many girls do not take bus?
Solution:
Step 1:
Find the fractions of boys and girls who do not take bus
Boys –> 1 – 7/10 = 3/10
Girls –> 1 – 7/8 = 1/8
Step 2:
Compare the boys and girls who do not take bus
3/10 of boys = 2 × 1/8 of girls (The number of boys who do not take bus is
twice
the number of
girls who do not take bus)
3/10 of boys = 1/4 of girls
Step 3:
Make the numerators the same
1/4 = 3/12
Step 4:
Compare the denominators
Boys : Girls
= 10 : 12 = 5 : 6
Step 5:
Find the total number of units and equate it to the total boys and girls
5u + 6u = 11u
11u = 836
Step 6:
Find 1 unit
1u = 76
Step 7:
Find the total number of girls
6u = 456 (Total girls)
8
Step 8:
Find the girls who do not take bus
1/8 × 456 = 57 girls (Answer)
Solved Problems for a practice
1.
1
4
of K
a
r
e
n’s m
a
ss
i
s
e
qu
al
t
o
3
4
of Bobby’s m
a
ss. Th
ei
r
t
o
tal
m
a
ss
i
s 80kg. F
i
nd
Bobby’s m
a
ss.
Solution:
2.
1
3
of
t
h
e
popu
lati
on of Town A
i
s
e
qu
al
t
o
2
5
of
t
h
e
popu
lati
on of Town B. Th
e
2
t
owns
h
a
v
e
a
t
o
tal
popu
lati
on of 22 000. F
i
nd
t
h
e
popu
lati
on of Town A.
Solution:
3.
3
6
of
t
h
e
numb
e
r of
a
pp
le
s
at
a
fru
it
s
tall
i
s
e
qu
al
t
o
1
3
of
t
h
e
numb
e
r of or
a
ng
e
s.
Th
e
r
e
a
r
e
630
a
pp
le
s
a
nd or
a
ng
e
s
alt
og
et
h
e
r. F
i
nd
t
h
e
d
i
ff
e
r
e
n
ce
i
n
t
h
e
numb
e
r of
a
pp
le
s
a
nd or
a
ng
e
s
at
t
h
e
s
tall.
Solution:
9
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4.
4
7
of
t
h
e
numb
e
r of f
e
m
ale
d
a
n
ce
rs
i
s
e
qu
al
t
o
1
5
of
t
h
e
numb
e
r of m
ale
d
a
n
ce
rs.
1080 p
e
op
le
a
r
e
d
a
n
ci
ng. How m
a
ny mor
e
m
ale
s
t
h
a
n f
e
m
ale
s
a
r
e
d
a
n
ci
ng
?
Solution:
5.
A box
c
on
tai
n
e
d
t
w
e
n
t
y-
ce
n
t
c
o
i
ns
a
nd f
i
f
t
y-
ce
n
t
c
o
i
ns.
1
4
of
t
h
e
numb
e
r of
t
w
e
n
t
y-
ce
n
t
c
o
i
ns
i
s
e
qu
al
t
o
4
5
of
t
h
e
numb
e
r of f
i
f
t
y-
ce
n
t
c
o
i
ns. Th
e
r
e
a
r
e
420
c
o
i
ns
alt
og
et
h
e
r
i
n
t
h
e
box. F
i
nd
t
h
e
t
o
tal
v
al
u
e
of
all
t
h
e
c
o
i
ns
i
n
t
h
e
box.
Solution:
6.
P
e
ggy sp
e
n
t
$32 on
a
dr
e
ss. Sh
e
sp
e
n
t
1
3
of h
e
r r
e
m
ai
n
i
ng mon
e
y on
a
sh
i
r
t
.
1
2
of
h
e
r mon
e
y w
a
s
le
f
t
. How mu
c
h mon
e
y h
a
d sh
e
at
f
i
rs
t?
Solution:
10
7.
C
a
ro
l
sp
e
n
t
$25 of h
e
r s
a
v
i
ngs on
a
b
l
ous
e
. Sh
e
sp
e
n
t
1
5
of h
e
r r
e
m
ai
n
i
ng s
a
v
i
ngs on
a
p
e
n
2
5
of h
e
r
t
o
tal
s
a
v
i
ngs w
a
s
le
f
t
. How mu
c
h s
a
v
i
ngs h
a
d sh
e
at
f
i
rs
t?
Solution:
8.
R
a
ndy sp
e
n
t
$480 on
a
r
a
d
i
o. H
e
sp
e
n
t
1
7
of h
i
s r
e
m
ai
n
i
ng mon
e
y on
a
calc
u
lat
or. H
e
h
a
d
2
7
of h
i
s mon
e
y
le
f
t
. How mu
c
h mon
e
y h
a
d h
e
at
f
i
rs
t?
Solution:
9.
Judy sp
e
n
t
$280 on
a
G
a
m
e
boy. Sh
e
sp
e
n
t
1
9
of h
e
r r
e
m
ai
n
i
ng mon
e
y on
a
ca
m
e
r
a
.
Sh
e
h
a
d
4
5
of h
e
r mon
e
y
le
f
t
. How mu
c
h mon
e
y h
a
d sh
e
at
f
i
rs
t?
11
Solution:
10.
B
ill
y sp
e
n
t
$5700 on
a
p
la
sm
a
TV s
et
. H
e
sp
e
n
t
1
9
of h
i
s r
e
m
ai
n
i
ng mon
e
y on
a
r
a
d
i
o.
H
e
h
a
d
2
7
of h
i
s mon
e
y
le
f
t
. How mu
c
h mon
e
y h
a
d h
e
at
f
i
rs
t?
Solution:
11.
12.
13.
Solution for 11, 12, 13:
https://www.youtube.com/watch?v=DlZSGfht8sM
12
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14.
Solution:
https://www.youtube.com/watch?v=ebL_EjaRv6s
15.
Solution:
https://www.youtube.com/watch?v=Vkul2qA7H9I
13
3.
Model Drawing Concept
Model drawing
is used extensively in Singapore Math. I recommend you to use this method
when you see the word “more than” or “less than” in Fractions questions. Watch the video below
to learn how to cut models to compare them easily.
Example 1
“Mrs Tan made some muffins. She sold 2/5 of them in the morning and 4/9 of the remainder in
the evening. She sold 40 more muffins in the morning than evening. How many muffins did she
make altogether?”
Firstly, I draw a model and cut into 5 parts. I used the pink region to represent the muffins that
are sold in the morning and the blue region to represent the remainder.
Next, I
cut the remainder
into 9 units because the question said that she sold 4/9 of the
remainder in the evening. To do that, I will need to cut each blue unit into 3 units to get 9 units in
total, So 4 units out of these 9 units are sold in the evening.
Now, to make things
consistent
, I also need to cut the pink unit into 3 units each.
14
From here, we can clearly see that 6 units are sold in the morning while 4 units are sold in the
evening. And the difference between both of them is 2 units.
Since the question says, “She sold 40 more muffins in the morning than evening,” we can say
that
2 units = 40
1 unit = 20
From the model, we can see that there are 15 units in total after cutting the model. So,
Total Number of Muffins = 15 × 20 = 300
Another Method
Of course, there are other methods to solve this question. Some children might find it
cumbersome to draw models and prefer to use statements instead.
Remainder = 1 – 2/5 = 3/5
4/9 of the remainder = 4/9 × 3/5 = 4/15 (This will represent the fraction of Total muffins that were
sold in the evening)
Difference between Morning and Evening = Fraction sold in the morning – Fraction sold in the
evening
= 2/5 – 4/15 = 2/15
2/15 of Total Muffins = 40
(
Warning:
Many students write “2/15 = 40”. This statement is wrong because 2/15 is not equal to
40. Be careful of your presentation!)
Total Muffins = 40 ÷ 2 × 15 = 300
Using statements can be a faster way to solve the question. However, for children who are
visually-inclined learners, parents might want to use the model method to teach them instead.
Another questions:
1.
Grade 3
15
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Solution:
https://www.youtube.com/watch?v=nhdmgru84ps
2.
Grade 4
Solution:
https://www.youtube.com/watch?v=lAWtVR41sGA
3.
Primary 3
Solution
:
https://www.youtube.com/watch?v=UlxjqEVsBeQ
4.
5.
6.
Solution for
4, 5 and 6th
:
https://www.youtube.com/watch?v=gYNlEVdkZPo
16
4.
The Repeated Variable Concept
The Repeated Variable Concept is derived from the
Comparison Concept
and the
Part-Whole Concept
. This concept is applicable when the problem
deals with one or more variables that is repeated, i.e., it is related to more than 1 variable in the
question. For example, A is 1/2 of B and B is 2/3 of C. Thus B is the repeated variable.
To illustrate this concept, consider the following question.
Example 1:
Jane has 3 times as much money as Cindy. Lea has twice as much money as Jane. If Lea has
$30 more than Cindy, how much money does Jane have?
Solution:
Take note that Jane is the repeated variable,i.e., she
is related to Cindy and then to Lea.
Step 1: Draw 1 box to represent Cindy's money and
3 boxes to represent Jane's money.
Step 2: Since Lea has twice as much as Jane, we
draw twice as many number of units as Jane to
represent Lea's money,i.e., 2 x 3 units = 6 units.
Step 3: From the question, we know that Lea has
$30 more than Cindy. Since Lea has 5 units more
than Cindy, then 5 units must be equal to $30.
5 units -----------> $30
1 unit ------------> $30 / 5 units = $6
17
3 units -----------> 3 units X $6 = $18
Therefore, Jane has $18.
5.
Constant
Quantity Concept
The Constant Quantity Concept is derived from a combination of
the
Part-Whole Concept
, the
Comparison Concept
and/or
the
Change Concept
. The Constant Quantity Concept is applicable
when the problems deal with quantities being transferred in or
transferred out of one of the two variables concerned. This leaves
the other variable with its quantity unchanged, hence we refer to it
as the "Constant Quantity Concept". The unique feature in this
concept lies in the fact that after the transfer in or transfer out of
quantities, the value of the second variable remains unchanged.
To illustrate this concept, consider the following problems:
a-
Constant Quantity after a Transfer Out
Example 1:
Pamela had thrice as many hats as Christine. After Pamela gave away 10 hats, she had half as
many hats as Christine. How many hats did Christine and Pamela each had at first?
Solution:
Step 1: Draw 1 box to represent the number of hats Christine had and 3 boxes to represent the
number of hats Pamela had at first.
Step 2: Since Christine did not receive or give away any hats, the model (1 box) representing her
hats remains constant.
We also know that after Pamela gave away 10 hats, she had only half the number of hats that
Christine had, i.e., Christine would have twice the number of hats that Pamela had.
Thus, Christine would have 2 units and Pamela 1 unit. We can then divide the 1 unit that
Christine had into 2 equal smaller units and mark out 1 equivalent small unit for Pamela.
18
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Step 3: Next, we also know that Pamela had given away 10 hats. The parts to the right of the 1
small unit in Pamela's model bar must be equal to 10 hats.
Step 4: After we had put all the information into the model, we need to check if the unknown units
of the model are equal. If not, can we sub-divide them to make them equal.
In the question, we can sub-divide the part Pamela gave away into 5 equal parts as follow:
Hence, from the model,
5 units ----------> 10 hats
1 unit -----------> 10 hats / 5 = 2 hats
2 units ----------> 2 units X 2 hats = 4 hats
6 units ----------> 6 units X 2 hats = 12 hats
Therefore, Pamela had 12 hats and Christine had 4 hats at first.
b-
Constant Quantity after a Transfer In
19
Example 2:
Sean had thrice as many marbles as Irwyn. After Irwyn bought another 10 marbles, he had twice
as many marbles as Sean. How many marbles did they each have at first?
Solution:
Step 1: Draw 1 box to represent the number of marbles Irwyn had and 3 boxes to represent the
number of marbles Sean had at first.
Step 2: Since Sean did not buy or give away any marbles, the model (3 boxes) representing his
marbles remains constant.
We also know that after Irwyn bought another 10 marbles, he had twice as many marbles as
Sean. Since Sean had 3 units, Irwyn would have twice as many units as Sean, i.e., 6 units. So,
we add another 5 boxes to Irwyn's model bar to make it 6 units.
20
Step 3: Next, since Irwyn bought another 10 marbles, these 10 marbles must be represented by
the extra 5 units added to his model bar.
Thus,
5 units ----------> 10 marbles
1 unit -----------> 10 marbles / 5 = 2 marbles
3 units ----------> 3 units X 2 marbles = 6 marbles
Therefore, Sean had 6 marbles and Irwyn had 2 marbles at first.
Example 3:
Ali and Billy have money in the ratio of 5 : 6. After Billy spent $16, the ratio became 3 : 2. How
much money does Billy have in the end?
Solution:
Step 1:
Make the ratio for Ali the same
Before: A : B = 5 : 6 =
15
: 18
After: A : B = 3: 2 =
15
: 10
Step 2:
Find the difference between Billy’s starting amount and ending amount
18u – 10u = 8u
Step 3:
Find 1 unit
8u = $16
1u = $2
21
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Step 4:
Find the amount for Billy in the end
10u =
$20
(Ans)
Solved problems for practice
1.
Mr T
a
n h
a
d
a
t
o
tal
of 126 r
e
d
a
nd b
l
u
e
m
a
rb
le
s
i
n
t
h
e
r
ati
o of 1
:
2 r
e
sp
ecti
v
el
y. Som
e
r
e
d m
a
rb
le
s w
e
r
e
g
i
v
e
n
a
w
a
y
a
nd
t
h
e
numb
e
r of r
e
d m
a
rb
le
s r
e
m
ai
n
i
ng w
a
s
2
9
of
t
h
e
r
e
m
ai
n
i
ng m
a
rb
le
s. How m
a
ny m
a
rb
le
s w
e
r
e
g
i
v
e
n
a
w
a
y
?
2.
Th
e
numb
e
r of
a
pp
le
s
i
n
a
fru
it
s
tall
w
a
s
t
w
ice
t
h
e
numb
e
r of m
a
ngo
e
s. Af
te
r 156
a
pp
le
s w
e
r
e
a
dd
e
d,
t
h
e
numb
e
r of
a
pp
le
s b
eca
m
e
4
5
of
t
h
e
t
o
tal
numb
e
r of fru
it
s.
How m
a
ny fru
it
s w
e
r
e
t
h
e
r
e
i
n
t
h
e
s
tall
at
f
i
rs
t?
3.
Th
e
numb
e
r of r
e
d, b
l
u
e
a
nd y
ell
ow sh
i
r
t
s
i
n
a
shop w
a
s
i
n
t
h
e
r
ati
o 12
:
7
:
3. Wh
e
n 56 r
e
d
sh
i
r
t
s w
e
r
e
so
l
d,
1
3
of
t
h
e
r
e
m
ai
n
i
ng sh
i
r
t
s w
e
r
e
r
e
d. How m
a
ny y
ell
ow sh
i
r
t
s w
e
r
e
t
h
e
r
e
i
n
t
h
e
shop
at
f
i
rs
t?
4.
S
all
y h
a
d som
e
r
e
d, b
l
u
e
a
nd gr
ee
n m
a
rb
le
s. Th
e
r
ati
o of r
e
d m
a
rb
le
s
t
o b
l
u
e
m
a
rb
le
s
w
a
s 5
:
2. Th
e
r
ati
o of b
l
u
e
m
a
rb
le
s
t
o gr
ee
n m
a
rb
le
s w
a
s 3
:
5. Af
te
r S
all
y r
e
mov
e
d 126
r
e
d m
a
rb
le
s,
1
3
of
t
h
e
r
e
m
ai
n
i
ng m
a
rb
le
s w
e
r
e
r
e
d. How m
a
ny m
a
rb
le
s w
e
r
e
le
f
t
i
n
t
h
e
box
?
5.
Th
e
r
e
w
e
r
e
som
e
s
t
ud
e
n
t
s
i
n
a
s
c
hoo
l
ca
n
tee
n
at
f
i
rs
t.
5
8
of
t
h
e
s
t
ud
e
n
t
s
le
f
t
t
h
e
ca
n
tee
n
a
f
te
r 1 hour.
1
6
of
t
h
e
r
e
m
ai
n
i
ng s
t
ud
e
n
t
s
le
f
t
t
h
e
ca
n
tee
n
a
f
te
r
a
no
t
h
e
r hour.
Af
te
r
a
no
t
h
e
r 225 s
t
ud
e
n
t
s
e
n
te
r
e
d
t
h
e
ca
n
tee
n,
t
h
e
r
ati
o of
t
h
e
numb
e
r of s
t
ud
e
n
t
s
i
n
t
h
e
e
nd
t
o
t
h
e
numb
e
r of s
t
ud
e
n
t
s
at
f
i
rs
t
w
a
s 7
:
8. How m
a
ny s
t
ud
e
n
t
s w
e
r
e
t
h
e
r
e
i
n
t
h
e
ca
n
tee
n
at
f
i
rs
t?
Solution:
1.
R : B = 1 : 2 = 7 : 14
R : B = 2 : 7 = 4 : 14
7 + 14 = 21
21u = 126
3u = 18 maarbles (Ans)
2.
A : M = 2 : 1
A : M = 4 : 1
2u = 156
3u = 234 fruits (Ans)
3.
R : B : Y = 12 : 7 : 3
22
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R : B + Y = 1 : 2 = 5 : 10
12 – 5 = 7
7u = 56
3u = 24 (Ans)
4.
R : B = 5 : 2 = 15 : 6
B : G = 3 : 5 = 6 : 10
R : B : G = 15 : 6 : 10
After:
R : B + G = 1 : 2 = 8 : 16
15 – 8 = 7
7u = 126
24u = 432
5.
1-
5
8
=
3
8
5
8
х
3
8
=
5
16
7:8 = 14 : 16
14 – 5 = 9
9u = 225 1u = 25
16u = 400 (Ans)
Question with video solution
6.
23
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S:
https://www.youtube.com/watch?v=cHFB5TotBK0
7.
S:
https://www.youtube.com/watch?v=bllBofF71P4
8.
S:
https://www.youtube.com/watch?v=9MdkVf5Pr4Y
9.
S:
https://www.youtube.com/watch?v=X9-Ff7Li3d0
10.
11.
12.
24
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Solutions to 10, 11, 12 :
https://www.youtube.com/watch?v=nFpSabvIpEM
6.
Constant Total Concept
The Constant Total Concept is derived from a combination of the
Comparison Concept and theChange Concept . This concept is
applicable when the problems deal with two variables transferring
quantities to each other leaving the total unchanged, hence the
name "Constant Total Concept". The unique feature in this
concept lies in the fact that in the problem, it can be a one-way or
two-way transfer. In a one-way transfer, one variable transfer
some quantities to the other variable leaving the total unchanged.
In a two-way transfer, the two variables each gives some quantities to the other party, leaving
both variables' total unchanged.
To illustrate the Constant Total Concept, take a look at the following problems.
(A) One-Way Transfer
Example 1:
Class A had 1/3 the number of pupils in Class B. After 14 pupils were transferred from Class B to
Class A, Class A had 4/5 the number of pupils in Class B. How many pupils were there in Class A
at first?
Solution:
Step 1:
Draw the "Before" model for both classes, 1 box to represent Class A and 3 boxes to
represent Class B, since Class A is 1/3 of Class B.
Step 2 :
Draw the "After" model for both classes, 4 boxes to represent Class A and 5 boxes to
represent Class B, since Class A is 4/5 of Class B.
From the model, we know that before the transfer, the total number of units is 4 units and after
the transfer, the total number of units is 9 units. As this is purely an internal transfer of quantities
not involving any external parties, the 4 units in the "Before" model must be equivalent to the 9
units in the "After" model.
We need to find a common multiple for the two totals to make the comparison meaningful. A
common multiple of 4 and 9 would be 36.
25
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Step 4:
Thus, we divide both the "Before" and "After" models into 36 units each.
From the model, we can see that 7 units was transferred from Class B to Class A. Since 14
pupils were transferred from Class B to Class A,
7 units -------> 14 pupils
1 unit --------> 14 pupils / 7 units = 2 pupils
9 units -------> 9 units X 2 pupils = 18 pupils
Therefore, Class A had 18 pupils at first.
(B) Two-Way Transfer
Example 2:
Roger had twice as many marbles as Mark at first. Mark gave 1/2 of his marbles to Roger and
Roger gave 3/5 of his marbles back to Mark. In the end, Mark had 8 more marbles than Roger.
How many marbles did each of them have at first?
Solution:
Step 1:
Draw 2 boxes to represent the number of marbles that Roger had and 1 box to represent
the number of marbles that Mark had.
Step 2:
Since Mark gave 1/2 of his marbles to Roger, we divide his bar into 2 boxes and transfer
1 box to Roger.
Step 3:
For ease of comparison, we divide all the other boxes that Roger had into 2 smaller
boxes as well. Now we have all the unknown boxes as equal units.
Step 4:
Since Roger then gave 3/5 of his marbles back to Mark, we transfer 3 units out of his 5
units back to Mark.
Step 5:
In the end, Mark had 8 more marbles than Roger (given in the question).
From the model,
2 units -------> 8 marbles
1 unit --------> 8 marbles / 2 units = 4 marbles
4 units -------> 4 units X 4 marbles = 16 marbles
Therefore, Roger had 16 marbles and Mark had 8 marbles at first.
Example 3:
Ali and Billy have money in the ratio of 5 : 4. After Ali gave Billy $20, they have an equal amount
of money. How much money does Billy have in the end?
Solution:
Step 1:
Make the total for Ali and Billy to be the same
Before:
A : B : Total = 5 : 4 : 9 = 10 : 8 :
18
After: A : B : Total = 1 : 1 : 2 = 9 : 9 :
18
Step 2:
Find the difference between Ali’s starting amount and ending amount
26
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10u – 9u = 1u
Step 3:
Find 1 unit
1 unit = $20
Step 4:
Find Billy’s amount in the end
9 units = $180 (Ans)
Solved Questions for practice
1.
Th
e
r
ati
o of Ann’s mon
e
y
t
o B
e
rn
ice
’s mon
e
y w
a
s 6
:
5
at
f
i
rs
t
. Af
te
r Ann g
a
v
e
som
e
of h
e
r
mon
e
y
t
o B
e
rn
ice
,
t
h
e
r
ati
o b
eca
m
e
4
:
7. G
i
v
e
n
t
h
at
Ann h
a
d $99
le
f
t
, how mu
c
h d
i
d sh
e
g
i
v
e
t
o B
e
rn
ice?
2.
Th
e
r
ati
o of Jun
e
’s mon
e
y
t
o Bob’s mon
e
y
i
s 4
:
5. If Jun
e
g
i
v
e
s Bob $40, Bob w
ill
h
a
v
e
t
w
ice
a
s mu
c
h mon
e
y
a
s Jun
e
. How mu
c
h mon
e
y h
a
v
e
t
h
e
y
alt
og
et
h
e
r
?
3.
A
lice
, B
ett
y
a
nd C
i
ndy h
a
d p
e
n
cil
s
i
n
t
h
e
r
ati
o 1
:
2
:
4. Af
te
r C
i
ndy g
a
v
e
96 of h
e
r p
e
n
cil
s
t
o A
lice
a
nd som
e
t
o B
ett
y,
all
of
t
h
e
m h
a
d
t
h
e
s
a
m
e
numb
e
r of p
e
n
cil
s. How m
a
ny p
e
n
cil
s
d
i
d C
i
ndy h
a
v
e
at
f
i
rs
t?
4.
A
li
a
nd B
e
n h
a
v
e
som
e
ca
rds
i
n
t
h
e
r
ati
o 3
:
1. Af
te
r A
li
l
os
t
som
e
ca
rds
t
o B
e
n, A
li
h
a
d
4
7
a
s m
a
ny
ca
rds
a
s B
e
n. G
i
v
e
n
t
h
at
B
e
n h
a
d 72 mor
e
ca
rds
t
h
a
n A
li
i
n
t
h
e
e
nd, how m
a
ny
ca
rds d
i
d A
li
l
os
e?
5.
M
a
ry w
a
s so
l
v
i
ng qu
e
s
ti
ons
i
n
a
M
at
h qu
iz
. Sh
e
so
l
v
e
d
1
4
of
t
h
e
qu
e
s
ti
ons
i
n
t
h
e
f
i
rs
t
hour. Sh
e
so
l
v
e
d
a
no
t
h
e
r 28 qu
e
s
ti
ons
i
n
t
h
e
s
ec
ond hour. Th
e
t
o
tal
numb
e
r of qu
e
s
ti
ons sh
e
so
l
v
e
d
i
n
t
h
e
2 hours
w
a
s
2
5
mor
e
t
h
a
n
t
h
e
numb
e
r of qu
e
s
ti
ons
t
h
at
w
e
r
e
unso
l
v
e
d.
How m
a
ny qu
e
s
ti
ons w
e
r
e
unso
l
v
e
d
?
Solutions:
1.
A : B : Total = 6 : 5 : 11
A : B : Total = 4 : 7 : 11
4u = $99
6 – 4 = 2
2u = $49.50 (Ans)
2.
Before
After
Before
J: B : Total
4 : 5 : 9u
After
J : B : Total
1×3 : 2×3 : 3×3
3 : 6 : 9
Difference →4u – 3u = 1u
Transfer from Bob to June →1 unit →40
27
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Amount of money both Bob and June had→9 units →9 × 40
= 360 (Ans)
3.
A : B : C : Total = 1 : 2 : 4 : 7 = 3 : 6 : 12 : 21
A B : C : Total = 1 : 1 : 1 : 3 = 7 : 7 : 7 : 21
7 – 3 = 4 4u = 96 1u = 24
12u = 288 (Ans)
4.
A : B : Total = 3 : 1 : 4 = 33 : 11 : 44 A : B : Total = 4 : 7 : 11 = 16 : 28 : 44
28 – 16 = 12
12u = 72
1u = 6
33 – 16 = 17
17u = 102 (Ans)
5.
First hour
Solved : Unsolved : Total = 1 : 3 : 4 = 3 : 9 : 12
First 2 hours
Solved : Unsolved : Total = 7 : 5 : 12
7 – 3 = 4
4u = 28
1u = 7
5u = 35 (Ans)
6.
Figo had $541 more than Gordon at first. After Figo bought a computer game, he had $98
less than Gordon. What is the cost of the Computer game?
Solution:
https://www.youtube.com/watch?v=0ZPt1qn5bew
28
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7.
Constant Difference Concept
The Constant Difference Concept is derived from the
Comparison Concept. This concept is applicable when the problems deal with an equal quantity
being transferred in or transferred out of the two variables concerned. This leaves the two
variables with an equal increase or decrease in value. The unique feature in this concept lies in
the fact that after the transfer in or transfer out of quantities, the difference between the two
variables remains unchanged, hence the name "Constant Difference" Concept.
To illustrate this concept, take a look at the following problems.
https://www.youtube.com/watch?v=pDaR9VGlftE
https://www.youtube.com/watch?v=nCdjuzF0ols
a-
Equal Amount Transferred into 2 Variables
Example 1:
Ken had 14 pens and Ben had 2 pens. When they
received an equal number of pens from their teacher, the ratio of
Ken's pens to Ben's pens became 3:1. How many pens did each
of them receive from their teacher?
Solution: For this question, we will work backwards. It is always
easier to start drawing models where a multiples-relationship
exist, i.e., a variable is a multiple of another variable. In this
example, Ken is 3 times of Ben after the transfer in.
Step 1:
Since the ratio of Ken's pens to Ben's pens after the transfer in is 3:1, we draw 3 boxes
to represent the number of units that Ken had and 1 box to represent the number of units that
Ben had.
Step 2:
Since an equal amount was transferred in, we mark out an equal amount from both
Ken's and Ben's model bars to show this amount.
Step 3:
Next, we label the models with the number of pens they each had at first.
Step 4:
After all information have been put into the model, we can then mark out all the known
parts and try to make all the unknown parts equal.
Step 5:
From Ken's bar, we can see that,
2 units + 2 pens + 2 pens + 2 pens ----------> 14 pens
2 units + 6 pens ----------> 14 pens
2 units ----------> 14 pens - 6 pens = 8 pens
29
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1 unit ----------> 8 pens / 2 units = 4 pens
Thus, they reach receive 4 pens from their teacher.
________________________________________
b-
Equal Amount Transferred out of 2 Variables
Example 2:
Chloe had 18 stickers and Jane had 6 stickers. When they both gave away an
equal number of stickers, Chloe had 4 times as many stickers as Jane. How many stickers did
they each gave away?
Solution: For this question, we will again work backwards. It is always easier to start drawing
models where a multiples-relationship exist, i.e., a variable is a multiple of another variable. In
this example, Chloe is 4 times of Jane after the transfer out.
Step 1:
Since the ratio of Chloe's pens to Jane's pens after the transfer out is 4:1, we draw 4
boxes to represent the number of units that Chloe had and 1 box to represent the number of
units that Jane had.
Step 2:
Since an equal amount was transferred out, we add an equal amount back to represent
the part that each of them gave away.
Step 3:
Next, we label the models with the number of stickers they each had at first.
Step 4:
Rearrange the units in Chloe's bar.
From the model, we can see that after we shift the last unit of Chloe's bar forward, the remaining
3 unknown but equal units add up to 12.
3 units ----------> 12 stickers
1 unit -----------> 12 stickers / 3 = 4 stickers
Step 5: Putting the value of 4 stickers per unit back into the model, we get:
Looking at Jane's model,
6 stickers - 4 stickers = 2 stickers
Therefore, they each gave away 2 stickers.
________________________________________
c-
Age Difference
Example 3:
Matthew is 29 years old and his son is 5 yrs old now. In how many years will
Matthew be thrice as old as his son?
Solution: In such questions, do bear in mind that the age difference between any 2 persons will
always remain a constant.
Step 1:
It is easier to draw the "After" model first and work backwards to the "Before" model.
First, we draw 3 boxes to represent Matthew's age and 1 box to represent the son's age after
their age increases.
Step 2:
Since an equal amount was added to their ages, we mark out an equal box on both their
model bars to represent the number of years they have grown from their present age.
Step 3:
Next, we label the models with their present age.
30
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Step 4:
We then divide the model to reflect the known and unknown units.
From the model,
2 units + 5 years + 5 years + 5 years ----------> 29 years
2 units + 15 years ----------> 29 years
2 units ----------> 29 years - 15 years
2 units ----------> 14 years
1 unit ----------> 14 years / 2 = 7 years
Therefore, Matthew will be thrice as old as his son in 7 years' time.
For questions relating to age, the age difference between 2 people will always remain the same.
Example 4
The ages of Ali and Billy are in the ratio of 4 : 7. In 3 years’ time, their ages will be in the ratio of 3
: 5. How old is Billy now?
Solution:
Step 1:
Make the difference for Ali and Billy the same
Before: A : B : Difference = 4 : 7 : 3 = 8 : 14 :
6
After: A : B : Difference = 3 : 5 : 2 = 9 : 15 :
6
Step 2:
Find the difference between Ali’s starting age and final age
9u – 8u = 1u
Step 3:
Find 1 unit
1 unit = 3 years
Step 4:
Find Billy’s age now
14 units =
42 years old
(Ans)
Examples:
1.
M
a
ry h
a
d $75
a
nd Sus
a
n h
a
d $48. Af
te
r
eac
h sp
e
n
t
t
h
e
s
a
m
e
a
moun
t
of mon
e
y,
t
h
e
r
ati
o
of M
a
ry’s mon
e
y
t
o Sus
a
n’s mon
e
y b
eca
m
e
7
:
4. How mu
c
h d
i
d
t
h
e
y h
a
v
e
le
f
t
alt
og
et
h
e
r
?
2.
Bob
i
s
t
w
ice
a
s o
l
d
a
s Ch
a
r
lie
. In 10 y
ea
rs’
ti
m
e
,
t
h
e
r
ati
o of Bob’s
a
g
e
t
o Ch
a
r
lie
’s
a
g
e
w
ill
b
e
8
:
5. How o
l
d
i
s Ch
a
r
lie
now
?
3.
Th
e
r
ati
o of D
a
n’s mon
e
y
t
o Fr
e
ddy’s mon
e
y w
a
s 5
:
4. Af
te
r
eac
h of
t
h
e
m sp
e
n
t
$60,
t
h
e
r
ati
o b
eca
m
e
13
:
10. How mu
c
h d
i
d D
a
n h
a
v
e
at
f
i
rs
t?
4.
A
l
fr
e
d’s
a
g
e
i
s
1
4
of h
i
s f
at
h
e
r’s
a
g
e
wh
e
n h
i
s f
at
h
e
r
i
s 48 y
ea
rs o
l
d. In how m
a
ny
y
ea
rs’
ti
m
e
w
ill
A
l
fr
e
d’s
a
g
e
b
e
1
3
of h
i
s f
at
h
e
r’s
a
g
e?
5.
W
i
ns
t
on
i
s 6
ti
m
e
s
a
s o
l
d
a
s X
e
rx
e
s. In 8 y
ea
rs
ti
m
e
, h
e
w
ill
b
e
t
w
ice
a
s o
l
d
a
s
X
e
rx
e
s. How o
l
d
i
s W
i
ns
t
on now
?
Solutions:
31
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1.
75 – 48 = 27
M : S = 7 : 4
7 – 4 = 3
3u = 27
1u = 9
11u = $99 (Ans)
2.
B : C : Difference = 2 : 1 : 1 = 6 : 3 : 3
B : C : Difference = 8 : 5 : 3
8 – 6 = 2
2u = 10
1u = 5
3u = 15 years old (Ans)
3.
D : F : Difference = 5 : 4 : 1 = 15 : 12 : 3
D : F : Difference = 13 10 : 3
15 – 13 = 2
2u = $60
15u = $450 (Ans)
4.
A : F : Difference = 1 : 4 : 3 = 2 : 8 : 6
A : F : Difference = 1 : 3 : 2 = 3 : 9 : 6
8u = 48
1u = 6 (Ans)
5.
W : X : Difference = 6 : 1 : 5
W : X : Difference = 2 : 1 : 1 = 10 : 5 : 5
10 – 6 = 4 4u = 8
1u = 2
6u = 12 years old (Ans)
More Examples:
6.
There were 48 more boys than girls at a party. Another 10 boys and 10 girls joined the party.
After that, there were 4 times as many boys as girls. How many boys were there at the party
initially?
Solution:
https://www.youtube.com/watch?v=Wdbe1xirq3s
7
Solution:
https://www.youtube.com/watch?v=UxxMcF_e2eM
8.
32
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Solution:
https://www.youtube.com/watch?v=BoOWxiNQePk
9.
Solution:
https://www.youtube.com/watch?v=HM5AhiOoI4o
10.
Solution:
https://www.youtube.com/watch?v=VWluIvdCynI
33
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8.
Everything Changed Concept (Units and Parts)
This is a more challenging type of PSLE Math questions. Both
sides of the ratio changed by different amounts. I recommend
“Units and Parts” to solve this type of questions.
Example 1
The ratio of Ali’s money to Billy’s money was 2 : 1. After Ali
saved another $60 and Billy spent $150, the ratio became 4 : 1.
How much money did Ali have at first?
Solution:
Step 1:
Write down the starting ratio and apply the changes.
A : B = 2u : 1u
2u + 60 : 1u – 150
Step 2:
Compare the final units with the final ratio.
A : B = 2u + 60 : 1u – 150 = 4:1
Step 3:
Cross multiply the final units with the final ratio
1 × (2u + 60) = 4 × (1u – 150)
2u + 60 = 4u – 600
Step 4:
Solve for 1 unit
4u – 2u = 600 + 60
2u =
$660 (Ans)
Solved problems for practice
1.
S:
https://www.youtube.com/watch?v=yrp0sLTPIbU
34
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9.
Part-Whole Concept
This is another common type of PSLE Math questions which you need to use one part to find
one whole.
Example 1:
Kelly spent 1/3 of her money on 5 pens and 11 erasers. The cost of each pen is 3 times the cost
of each eraser. She bought some more pens with 3/4 of her remaining money. How many pens
did she buy altogether?
Solution:
Step 1:
Write down the ratio of the cost of pen: eraser
P : E = 3u : 1u
Step 2:
Find the fraction spent on the extra pens
1 – 1/3 = 2/3 (Remainder)
3/4 × 2/3 = 1/2 (Fraction spent on extra pens)
Step 3:
Find the total cost of 5 pens and 11 erasers
5 × 3u + 11 × u = 26u (Total cost of 5 pens and 11 erasers)
Step 4:
Find the total amount of money in terms of units
26u × 3 = 78u (Total amount of money)
Step 5:
Find the total cost of the extra pens
1/2 × 78u = 39u (Total cost of extra pens)
Step 6:
Find the number of extra pens
39u ÷ 3u = 13
35
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Step 7:
Find the total number of pens
13 + 5 = 18 pens (Answer)
1.
Excess and Shortage Concept
Short video introduction in this topic:
https://www.youtube.com/watch?v=W4PXNMQqwbg
Example 1:
Tom packed 5 balls into each bag and found that he had 8
balls left over. If he packed 7 balls into each bag, he
would need another 4 more balls.
a) How many bags did he have?
b) How many balls did he have altogether?
Solution:
Step 1:
Find the difference in the number of balls in
each bag
7 – 5 = 2
Step 2:
Find the number of bags by distributing the 8 extra balls
8 ÷ 2 = 4
Step 3:
Find the number of bags by distributing the shortage of 4 balls
4 ÷ 2 = 2
Step 4:
Find the total number of bags
4 + 2 = 6 bags (Ans for a)
Step 5:
Find the total number of balls
6 × 5 + 8 = 38
Or
6 × 7 – 4 = 38 balls (Ans for b)
36
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1.
Tom packed 5 balls into each bag and found that he had 7 balls left over. If he packed 6 balls
into each bag, he would need another 5 more balls. How many balls did he have altogether?
Solution:
https://jimmymaths.com/topic/excess-and-shortage-example-1-copy-2/
2.
Tom packed 5 balls into each bag and found that he had 8 balls left over. If he packed 7 balls
into each bag, he would need another 4 more balls. How many bags did he have? How many
balls did he have altogether?
Solution:
https://jimmymaths.com/topic/excess-and-shortage-example-2-copy-2/
3.
When a stack of books were packed into bags of 4, there would be 3 books left over. When
the same number of books were packed into bags of 6, there would still be 3 books left over.
What could be the least number of books in the stack?
Solution:
https://jimmymaths.com/topic/excess-and-shortage-example-3-copy-2/
4.
Mary had some money to buy some pens. If she bought 12 pens, she would need $8 more. If
she bought 9 pens, she would be left with $4. How much money did Mary have?
Solution:
https://jimmymaths.com/topic/excess-and-shortage-example-4-copy-2/
37
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5.
Mrs. Chan has some stickers. If she gives each student in her class 3 stickers, she will have
25 stickers left. If she gives each student in her class 5 stickers, she will be short of 21
stickers. How many stickers does Mrs. Chan have?
Solution:
https://www.youtube.com/watch?v=2btgl6BqmpI
6.
Mrs. Tan has some lollipops. If she gives 6 lollipops to each of her students, she will have an
excess of 4 lollipops. If she gives 8 lollipops to each of her students, she will need another 12
lollipops. How many students does Mrs. Tan give the lollipops to? How many lollipops does
Mrs. Tan have?
Solution:
https://www.youtube.com/watch?v=UwvWEu2j7tM
Self-practice problems
7.
Ms L
ee
bough
t
som
e
cl
o
t
h
t
o m
a
k
e
cl
owns of
t
h
e
s
a
m
e
s
ize
. If sh
e
m
a
d
e
45
cl
owns, sh
e
wou
l
d h
a
v
e
21m of
cl
o
t
h
le
f
t
. If sh
e
m
a
d
e
63
cl
owns, sh
e
wou
l
d n
ee
d
a
no
t
h
e
r 33m. How
mu
c
h
cl
o
t
h d
i
d sh
e
buy
?
Solution:
8.
Mr. S
a
mu
el
w
a
n
t
s
t
o d
i
s
t
r
i
bu
te
a
box of
c
o
l
our p
e
n
cil
s
t
o h
i
s K2 pup
il
s. If h
e
g
i
v
e
s 4
c
o
l
our
p
e
n
cil
s
t
o
eac
h pup
il
, h
e
w
ill
h
a
v
e
8
c
o
l
our p
e
n
cil
s
le
f
t
. If h
e
g
i
v
e
s 6
c
o
l
our p
e
n
cil
s
t
o
eac
h
pup
il
, h
e
w
ill
b
e
shor
t
of 68 p
e
n
cil
s. How m
a
ny pup
il
s
a
r
e
t
h
e
r
e
i
n
t
h
e
cla
ss
?
Solution:
9.
A s
c
hoo
l
i
s org
a
n
i
s
i
ng
a
Pr
i
m
a
ry 6 In
te
ns
i
v
e
R
e
v
i
s
i
on C
la
ss. A
ce
r
tai
n numb
e
r of
cla
ssrooms
h
a
d b
ee
n
all
o
cate
d for
t
h
e
p
a
r
tici
p
ati
ng s
t
ud
e
n
t
s. If 35 s
t
ud
e
n
t
s w
e
r
e
t
o b
e
s
eate
d
i
n 1
cla
ssroom, 15 s
t
ud
e
n
t
s wou
l
d b
e
le
f
t
w
it
hou
t
a
s
eat
. If 40 s
t
ud
e
n
t
s w
e
r
e
t
o b
e
s
eate
d
i
n 1
cla
ssroom,
t
h
e
r
e
wou
l
d b
e
1
e
mp
t
y
cla
ssroom
le
f
t
.
a.
How m
a
ny
cla
ssrooms w
e
r
e
all
o
cate
d for
t
h
e
cla
ss
?
b.
How m
a
ny s
t
ud
e
n
t
s
e
nro
lle
d for
t
h
e
cla
ss
?
Solution:
10.
John bough
t
1 bo
ttle
of v
ita
m
i
n p
ill
s for h
i
s f
a
m
il
y. If 6 v
ita
m
i
n p
ill
s
a
r
e
ta
k
e
n
e
v
e
ry d
a
y, 25
v
ita
m
i
n p
ill
s w
ill
b
e
le
f
t
a
f
te
r som
e
d
a
ys. If 9 v
ita
m
i
n p
ill
s
a
r
e
ta
k
e
n
eac
h d
a
y,
t
h
e
f
a
m
il
y w
ill
n
ee
d
a
no
t
h
e
r 8 v
ita
m
i
n p
ill
s for
t
h
e
s
a
m
e
numb
e
r of d
a
ys.
a.
How m
a
ny d
a
ys
ca
n
t
h
e
bo
ttle
of v
ita
m
i
n p
ill
s
la
s
t?
38
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b.
How many vitamin pills are there in 1 bottle?
Solution:
11.
Ph
il
a
nd h
i
s fr
ie
nds sh
a
r
e
d
a
box of s
tic
k
e
rs
a
mong
t
h
e
m. If
eac
h on
e
of
t
h
e
m g
et
s 10
s
tic
k
e
rs,
t
h
e
r
e
w
ill
b
e
8 s
tic
k
e
rs
e
x
t
r
a
. If 12 s
tic
k
e
rs w
e
r
e
d
i
s
t
r
i
bu
te
d
t
o
eac
h on
e
,
t
h
e
la
s
t
p
e
rson w
ill
r
ecei
v
e
2 s
tic
k
e
rs. F
i
nd
t
h
e
numb
e
r of s
tic
k
e
rs
i
n
t
h
e
box.
Solution:
12.
J
a
n
e
b
a
k
e
d som
e
c
up
ca
k
e
s for h
e
r Pr
i
m
a
ry 6 F
a
r
e
w
ell
P
a
r
t
y. Sh
e
t
r
ie
d pu
tti
ng 12
c
up
ca
k
e
s
on
a
t
r
a
y
a
nd found
t
h
at
t
h
e
la
s
t
t
r
a
y h
a
d on
l
y 1
c
up
ca
k
e
. If sh
e
pu
t
8
c
up
ca
k
e
s on
eac
h
t
r
a
y,
sh
e
wou
l
d h
a
v
e
33
c
up
ca
k
e
s
le
f
t
. How m
a
ny
c
up
ca
k
e
s d
i
d sh
e
b
a
k
e?
Solution:
13.
Am
i
n
a
h
c
y
cle
s from h
e
r hous
e
t
o
t
h
e
L
i
br
a
ry
t
o work d
ail
y. If sh
e
c
y
cle
s
at
t
h
e
sp
ee
d of 2
m
/
s, sh
e
w
ill
b
e
late
for work by 4 m
i
nu
te
s. If sh
e
i
n
c
r
ea
s
e
d h
e
r sp
ee
d
t
o 3 m
/
s, sh
e
w
ill
a
rr
i
v
e
at
work 3 m
i
nu
te
s b
e
for
e
h
e
r sh
i
f
t
s
ta
r
t
s. How f
a
r
i
s
t
h
e
li
br
a
ry from h
e
r hous
e?
Solution:
39
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14.
Mr T
a
n h
a
d som
e
r
ice
. H
e
pour
e
d 36 kg of r
ice
i
n
t
o 3
la
rg
e
ja
rs
a
nd 5 sm
all
ja
rs. W
it
h
t
h
e
r
e
m
ai
n
i
ng r
ice
, h
e
i
s
a
b
le
t
o f
ill
a
sm
all
ja
r
a
nd h
a
v
e
1.5 kg of r
ice
le
f
t
. How
e
v
e
r, h
e
wou
l
d b
e
shor
t
of 2.5 kg of r
ice
i
f h
e
w
a
n
t
s
t
o f
ill
a
la
rg
e
ja
r.
a.
F
i
nd
t
h
e
d
i
ff
e
r
e
n
ce
i
n
t
h
e
m
a
ss of r
ice
wh
ic
h
a
la
rg
e
ja
r
a
nd sm
all
ja
r
ca
n
c
on
tai
n.
b.
Wh
at
i
s
t
h
e
m
a
ss of r
ice
t
h
at
Mr T
a
n h
a
v
e
alt
og
et
h
e
r
?
Solution:
15.
D
a
rw
i
n
i
s s
a
v
i
ng up
t
o buy
a
phon
e
. If h
e
i
n
c
r
ea
s
e
s h
i
s s
a
v
i
ngs by 20%, h
e
wou
l
d s
till
n
ee
d
a
no
t
h
e
r $25. If h
e
i
n
c
r
ea
s
e
s h
i
s s
a
v
i
ngs by 45%, h
e
wou
l
d h
a
v
e
$30 mor
e
t
h
a
n h
e
n
ee
ds.
How mu
c
h mor
e
mon
e
y do
e
s h
e
n
ee
d
t
o buy
t
h
e
phon
e?
Solution:
40
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16.
P
ete
r h
a
d som
e
sug
a
r. H
e
f
ille
d up 3
la
rg
e
p
ac
k
et
s
a
nd 5 sm
all
p
ac
k
et
s w
it
h 7.2 kg of
sug
a
r. H
e
c
ou
l
d no
t
f
ill
up
a
no
t
h
e
r
la
rg
e
p
ac
k
et
a
s h
e
w
a
s shor
t
of 0.5 kg. Ins
tea
d, h
e
f
ille
d
up
a
no
t
h
e
r sm
all
p
ac
k
et
a
nd h
a
d 0.3 kg of sug
a
r
le
f
t
.
a.
How m
a
ny mor
e
kg of sug
a
r
ca
n
eac
h
la
rg
e
p
ac
k
et
c
on
tai
n
t
h
a
n
eac
h sm
all
p
ac
k
et?
b.
How mu
c
h sug
a
r d
i
d P
ete
r h
a
d
i
n
t
o
tal?
Solution:
2.
Gap and Difference Concept
Brief theory
:
https://practicle.sg/gap-and-difference/
This type of PSLE Math questions requires you to find the
difference and use it to solve the question.
Example 1:
Betty bought 12 erasers from a bookstore. If the price of
1 eraser was 5 cents less than the original, she could
have bought 4 more erasers. What was the original price
of each eraser?
Step 1:
Find the difference in cost for 12 erasers
12 × 0.05 = 0.6
Step 2:
Find the new price of each eraser
0.6 ÷ 4 = 0.15
Step 3:
Find the original price of each eraser
0.15 + 0.05 = $0.20 (Answer)
41
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Example 2:
Solution:
https://www.youtube.com/watch?v=W4PXNMQqwbg
Example 3:
Solution:
https://www.youtube.com/watch?v=JvpI6nQdwUI
Solved problems for practice
1.
Ch
e
ry
l
bough
t
15 p
e
n
cil
s from
a
books
t
or
e
. If
t
h
e
pr
ice
of 1 p
e
n
cil
w
a
s 10
ce
n
t
s
le
ss
t
h
a
n
t
h
e
or
i
g
i
n
al
, sh
e
c
ou
l
d h
a
v
e
bough
t
5 mor
e
p
e
n
cil
s. Wh
at
w
a
s
t
h
e
or
i
g
i
n
al
pr
ice
of
eac
h
p
e
n
cil?
Solution:
$0.10 × 15 = $1.50
$1.50 ÷ 5 = $0.30
$0.30 + $0.10 = $0.40 (Ans)
2.
In a school hall, chairs were arranged in rows such that there were exactly 9 chairs in each
row. For a concert, Mr Ong brought 6 more chairs into the school hall and rearranged all the
chairs. There are now exactly 7 chairs in each row and 12 more rows than before. How
many chairs are there in the school hall for the concert?
Solution 1:
12 more rows of 7 chairs each row ---> 12 x 7 = 84 chairs
Substracting the 6 chairs that were added ---> 84 - 6 = 78 chairs
Difference of each row with 9 chairs and 7 chairs ---> 2 chairs per row
No. of rows with 9 chairs in school hall at first ---> 78/2 = 39 rows
No. of chairs for the concert ---> (39rows x 9) + 6 more chairs = 357 chairs
Solution 2:
42
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3.
In
a
ca
n
tee
n,
ta
b
le
s
ca
n b
e
a
rr
a
ng
e
d
i
n
e
x
actl
y 11
ta
b
le
s p
e
r row. If
t
h
e
s
a
m
e
numb
e
r of
ta
b
le
s
i
s
a
rr
a
ng
e
d
i
n rows of 8
ta
b
le
s
eac
h,
t
h
e
r
e
w
ill
b
e
6 mor
e
rows
a
nd 3
ta
b
le
s
le
f
t
ov
e
r.
How m
a
ny
ta
b
le
s
a
r
e
t
h
e
r
e?
Solution:
8 × 6 = 48
48 + 3 = 51
51 ÷ 3 = 17
17 × 11 = 187 (Ans)
4.
S
a
r
a
h h
a
d
t
h
e
e
x
act
a
moun
t
of mon
e
y
t
o buy 30 p
e
n
cil
ca
s
e
s. E
ac
h p
e
n
cil
ca
s
e
c
os
t
s
$9.30 mor
e
t
h
a
n
a
no
te
book. If sh
e
us
e
d
t
h
e
s
a
m
e
a
moun
t
of mon
e
y
t
o buy no
te
books
i
ns
tea
d of p
e
n
cil
ca
s
e
s, sh
e
c
ou
l
d buy 66 no
te
books. How mu
c
h mon
e
y d
i
d sh
e
h
a
v
e?
Solution:
$9.30 × 30 = $279
$279 ÷ 36 = $7.75
$7.75 × 66 = $511.50 (Ans)
5.
C
i
ndy
a
nd D
ai
sy h
a
d
t
h
e
s
a
m
e
a
moun
t
of mon
e
y
at
f
i
rs
t
. C
i
ndy sp
e
n
t
$20
eac
h d
a
y
a
nd
D
ai
sy sp
e
n
t
$22
eac
h d
a
y. Wh
e
n C
i
ndy h
a
d $500
le
f
t
, D
ai
sy h
a
d $480
le
f
t
. How mu
c
h d
i
d
D
ai
sy h
a
v
e
at
f
i
rs
t?
Solution:
$22 – $20 = $2
43
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$500 – $480 = $20
$20 ÷ $2 = 10 (days)
$20 × 10 + $500 = $700
6.
Th
e
r
e
a
r
e
som
e
c
h
ai
rs
i
n
a
h
all
.
3
7
of
t
h
e
c
h
ai
rs
a
r
e
a
rr
a
ng
e
d
i
n 18 shor
t
rows
a
nd
1
2
of
t
h
e
c
h
ai
rs
a
r
e
a
rr
a
ng
e
d
i
n 7
l
ong rows. Th
e
r
e
s
t
of
t
h
e
c
h
ai
rs
a
r
e
s
tac
k
e
d
i
n
t
h
e
c
orn
e
r.
Th
e
r
e
a
r
e
16 mor
e
c
h
ai
rs
i
n
eac
h
l
ong row
t
h
a
n
eac
h shor
t
row. How m
a
ny
c
h
ai
rs
a
r
e
a
rr
a
ng
e
d
i
n
t
h
e
h
all?
Solution:
7.
Th
e
ta
b
le
s
i
n
a
cla
ssroom w
e
r
e
or
i
g
i
n
all
y
a
rr
a
ng
e
d
i
n rows of 15. Dur
i
ng
a
M
at
h
cla
ss, 6
ta
b
le
s w
e
r
e
r
e
mov
e
d from
t
h
e
cla
ssrooms
a
nd
t
h
e
r
e
m
ai
n
i
ng
ta
b
le
s w
e
r
e
r
ea
rr
a
ng
e
d
i
n
t
o
rows of 12. As
a
r
e
su
lt
,
t
h
e
r
e
w
e
r
e
now 9 mor
e
rows
t
h
a
n b
e
for
e
. How m
a
ny
ta
b
le
s w
e
r
e
t
h
e
r
e
for
t
h
e
M
at
h
cla
ss
?
Solution:
44
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8.
A f
a
rm
e
r
a
rr
a
ng
e
d som
e
p
la
n
t
s
i
n rows of 10. Wh
e
n h
e
a
dd
e
d 4 mor
e
p
la
n
t
s
a
nd
t
h
e
p
la
n
t
s
w
e
r
e
r
ea
rr
a
ng
e
d
i
n
t
o rows of 8,
t
h
e
r
e
w
e
r
e
2 mor
e
rows
t
h
a
n b
e
for
e
. How m
a
ny mor
e
p
la
n
t
s mus
t
h
e
a
dd
t
o h
a
v
e
a
t
o
tal
of 12 rows of 8 p
la
n
t
s
?
1
st
way to solve:
8 × 2 = 16
16 – 4 = 12
12 ÷ 2 = 6
6 × 10 = 60
12 × 8 = 96
96 – 60 = 36 (Ans)
2
nd
way to solve:
Let there was x rows initially, and y rows after rearrangement.
10x+4=8y
y=x+2
10x+4=8(x+2)=8x+6
y=x+2
x= 6
y = 8
So, initially there were 6*10 = 60 plants. If we want to plant 12*8 = 96 plants we have to
add 36 plants more.
Ans: 36
9.
28
c
h
il
dr
e
n sh
a
r
e
d
a
b
a
g of
ca
nd
ie
s
e
qu
all
y. Wh
e
n 4
c
h
il
dr
e
n g
a
v
e
up
3
4
of
t
h
ei
r sh
a
r
e
,
t
h
e
r
e
m
ai
n
i
ng
c
h
il
dr
e
n r
ecei
v
e
d 2 mor
e
ca
nd
ie
s
eac
h. How m
a
ny
ca
nd
ie
s w
e
r
e
t
h
e
r
e
i
n
t
h
e
45
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b
a
g
at
f
i
rs
t?
Solution:
1
st
step: 24*2 = 48 candies were rearranged among 24 kids
2
nd
step: 48 candies is
3
4
of all candies that 4 kids have, so each 4 kid have 48:3*4=64
candies.
3
rd
step: we can find how many candies has each kid initially, so 64: 4 = 16 candies
4
th
step:
Now we can find total amount of candies that were shared among 28 kids, so
28*16 = 448 candies.
Ans: 448 candies
3.
The Excess Value Concept
The Excess Value Concept is derived from a combination of the
46
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Part-Whole Concept
, the
Comparison Concept
and the
Change Concept
. A classic example of
a question involving the Excess Value Concept is the "Chicken & Rabbit Legs" question. In these
types of questions, both variables have a set of common values attached to them while one of
the variables has an extra set of values attached to it. The idea is to isolate this extra set of
excess value so that it could be subtracted from the total value of the two variables, thereby
allowing us to divide the remaining unknown units equally to solve for its value.
To illustrate this concept, consider the following problem.
Example 1:
A farmer has 36 animals on his farm. They are either chickens or rabbits. Altogether, the
chickens and rabbits have 100 legs. How many chickens are there?
Solution:
Step 1:
First, let's assume all the animals on the
farm are chickens. Each animal would have 2
legs. Draw a box to represent 1 animal with 2
legs.
Step 2:
Draw an arrow to the right to show that
this box is repeated 36 times (label "36" at the
end of the arrow).
Step 3:
Since all the animals have only 2 legs.
36 animals X 2 legs = 72 legs
Thus, they have a total of 72 legs (if they are all chickens).
Step 4:
Since there are 100 legs in reality, the
extra legs must belong to the rabbits(rabbits
have 4 legs instead of 2). So, let's draw
another long box to the right to represent the
extra legs.
100 legs - 72 legs = 28 legs
Hence, there are 28 "extra legs".
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Step 5
: Since the legs belong to the rabbits, each rabbit should get 2 extra legs each.
28 extra legs / 2 legs per rabbit = 14 rabbits
36 animals - 14 rabbits = 22 chickens
Therefore, there are 22 chickens.
Solved Problems for practice
1.
The Farmer has 30 chickens and rabbits altogether. There are only 100 legs. Find the
number of chickens and the number of rabbits the farmer has.
Solution:
https://www.youtube.com/watch?v=v7wJSLwHgkY
2.
In a farm there are thrice as many chickens as cows. If there are 980 legs altogether,
find the number of chickens in the farm.
Solution:
https://www.youtube.com/watch?v=lPuGmCqghJ8
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13.Grouping Concept
This is another common concept which needs you to group items
together, followed by finding the total number of groups.
Example 1:
Mark bought an equal number of shorts and shirts for $100.
A shirt cost $8 and each pair of shorts cost $12. How much did he spend
on the shirts?
Solution:
Step 1: Group 1 shirt and 1 pair of shorts
8 + 12 = 20
Step 2:
Find the number of groups
100 ÷ 20 = 5
Step 3:
Find the amount spent on the shirts
5 × 8 = $40 (Ans)
14.Number x Value Concept
Under this concept, you multiply the number of units by the value
of each unit to find the total value of 1 group. From here, you can
find the total number of groups.
49
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Example 1:
The ratio of the number of 50 cents coins to 1 dollar coin is 3 : 1. The total value of the coins is
$12.50. How many coins are there in total?
Solution:
Step 1:
Write down the ratio of 50 cents : $1
3 : 1
Step 2:
Group three 50 cents coins and one $1 coin into 1 group
3 × 0.5 = $1.50
1 × 1 = $1
Step 3:
Find the total value of 1 group
$1.50 + $1 = $2.50
Step 4:
Find the number of groups
12.50 ÷ 2.5 = 5
Step 5:
Find the total number of coins
5 × 4 = 20 coins (Ans)
Example 2:
As a birthday gift, Zoey gave her niece an electronic piggy bank that displays the total amount of
money in the bank as well as the total number of coins. After depositing some number of nickels
and quarters only, the display read:
Money: $2.00
Number of coins: 16
How many nickels and quarters did Zoey put in the bank?
Solution:
https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:systems-of-
equations/x2f8bb11595b61c86:systems-of-equations-word-problems/v/substitution-method-3
15.Guess and Check / Assumption Concept
The strategy for the method “Guess and Check” is to guess a solution and use the guess
in the problem to see if you get the correct answer. If the answer is too big or too small, then
make another guess that will get you closer to the goal. You continue guessing until you arrive at
the correct solution. The process might sound like a long one, however the guessing process will
often lead you to patterns that you can use to make better guesses along the way.
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Example 1:
Miss Lee bought some pencils for her class of 8 students. Each girl received 5 pencils and each
boy received 2 pencils. She bought a total of 22 pencils. How many boys were there in the
class?
Step 1:
Start with an assumption (You can start with girls or boys).
Suppose there are 8 girls
Step 2:
Find the total number of pencils
8 × 5 = 40
Step 3:
Change your assumption
Suppose there are 7 girls, 1 boy.
Step 4:
Find the total number of pencils
7 × 5 + 1 × 2 = 37
Step 5:
Spot the pattern
40 – 37 = 3 (When the boys increase by 1, the total pencils decrease by 3)
Step 6:
Find the total difference
40 – 22 = 18
Step 7:
Find the number of boys
18 ÷ 3 = 6 boys (Ans)
Check your answer!
Number of girls = 8 – 6 = 2
Total pencils = 2 × 5 + 6 × 2 = 22 (Correct)
Example 2
N
a
d
i
a
t
a
k
e
s
a
r
i
bbon
t
h
a
t
i
s
48
i
n
c
h
e
s
l
on
g
a
nd
c
u
t
s
i
t
i
n
t
w
o
p
i
e
c
e
s
.
O
n
e
p
i
e
c
e
i
s
t
h
r
ee
t
i
m
e
s
a
s
l
on
g
a
s
t
h
e
o
t
h
e
r
.
H
o
w
l
on
g
i
s
e
ac
h
p
i
e
c
e
?
S
o
l
u
t
i
o
n
S
t
e
p
1.
U
nd
e
r
st
a
d
W
e
n
ee
d
t
o
fi
nd
t
w
o
nu
m
b
e
r
s
t
h
a
t
a
dd
t
o
48.
O
n
e
nu
m
b
e
r
i
s
t
h
r
ee
t
i
m
e
s
t
h
e
o
t
h
e
r
nu
m
b
e
r
.
S
t
e
p
2.
S
t
r
a
t
e
g
y
W
e
g
u
e
ss
t
w
o
r
a
ndo
m
nu
m
b
e
r
s
,
on
e
t
h
r
ee
t
i
m
e
s
b
i
gg
e
r
t
h
a
n
t
h
e
o
t
h
e
r
a
nd
fi
nd
t
h
e
s
u
m
.
I
f
t
h
e
s
u
m
i
s
t
oo
s
m
a
ll w
e
g
u
e
ss
l
a
r
g
e
r
nu
m
b
e
r
s
a
nd
i
f
t
h
e
s
u
m
i
s
t
oo
l
a
r
g
e
w
e
g
u
e
ss
51
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s
m
a
ll
e
r
nu
m
b
e
r
s
.
T
h
e
n
, w
e
s
ee
i
f
a
n
y
p
a
tt
e
r
n
s
d
e
v
e
l
op
f
r
o
m
ou
r
g
u
e
ss
e
s
.
S
t
e
p
3.
A
pp
l
y
S
t
r
a
t
e
g
y
/S
o
l
v
e
G
u
e
ss
5
a
nd
15
G
u
e
ss
b
i
gg
e
r
nu
m
b
e
r
s
6
a
nd
18
t
h
e
s
u
m
i
s
5
+
15
=
20
t
h
e
s
u
m
i
s
6
+
18
=
24
w
h
i
c
h
i
s
t
oo
s
m
a
ll
w
h
i
c
h
i
s
t
oo
s
m
a
ll
H
o
w
e
v
e
r
,
y
ou
c
a
n
s
ee
t
h
a
t
t
h
e
a
n
s
w
e
r
i
s
e
x
a
c
t
l
y
h
a
l
f
o
f
48.
M
u
l
t
i
p
l
y
6
a
nd
18
b
y
t
w
o
.
O
u
r
n
e
x
t
g
u
e
ss
i
s
12
a
nd
36
t
h
e
s
u
m
i
s
12
+
36
=
48
T
h
i
s
i
s
c
o
rr
e
c
t
:
A
n
s
w
e
r T
h
e
p
i
e
c
e
s
a
r
e
12
i
n
c
h
e
s
a
nd
36
i
n
c
h
e
s
l
on
g
.
S
t
e
p
4
C
h
e
c
k
12
+
36
=
48
36
=
3
(
12
)
52
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Solved questions:
1.
(Primary 3): A candy cost $7 and a lollipop cost $4. Mum bought a total of 27
candies and lollipops. She paid a total of $141 for all of them. How many
lollipops did Mum buy?
Solution:
https://www.youtube.com/watch?v=Jj_OlRwRsi4
2.
Iron Man made a total of 21 gadgets and masks for $3340. It cost him $190 to
make each gadget and $140 to make each mask. How many masks did he
make?
Solution:
https://www.youtube.com/watch?v=m8jUq-Kcq2k
3.
Andrew took a handful of change out of his pocket and noticed that he was only
holding dimes and quarters in his hand. He counted that he had 22 coins that
amounted to $4. How many quarters and how many dimes does Andrew have?
Solution: Let’s solve this problem using “Guess and Check Method”
Dime =
10c
Quarter =
25 c
Amount of
coins
Amount of
money, $
0,1
0,25
22
1
st
step
22
0
22
$2,20
2
nd
step
11
11
22
$3,85
3
rd
step
12
10
22
$3,70
4
th
step
10
12
22
$4,00
1
st
step: let there are
22
dimes and
none
of quarters, then the amount of money
is calculated as follows:
22
*0,1+
0
*0,25 =
$2,20
2
nd
step: let there are
11
dimes and
11
of quarters, then the amount of money is
calculated as follows:
11
*0,1+
11
*0,25 =
$3,85
3
rd
step: let there are
12
dimes and
10
of quarters, then the amount of money is
calculated as follows:
12
*0,1+
10
*0,25 =
$3,70
4
th
step: let there are
10
dimes and
12
of quarters, then the amount of money is
calculated as follows:
10
*0,1+
12
*0,25 =
$4,00 –
Solved!!!
Ans: there are
10
dimes and
12
of quarters
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4.
Anne wants to put a fence around her rose bed that is one and a half times as
long as it is wide. She uses 50 feet of fencing. What are the dimensions of the
garden?
Solution: Let’s solve this problem using “Guess and Check Method”
Length,
feet
Width,
feet
Perimetr,
feet
L = 1,5*W
W
1 step
1,5
1
5
2 step
3
2
10
3 step
4,5
3
15
4 step
6
4
20
5 step
7,5
5
25
6 step
9
6
30
7 step
10,5
7
35
8 step
12
8
40
9 step
13,5
9
45
10
step
15
10
50
11
step
16,5
11
55
12
step
18
12
60
1
st
step: let width is equal to
1
, then the length is
1,5
times bigger than width, so
its size is
1,5
. Let’s find the Perimeter. Perimeter = 2* (Width + Length) =
2*(1+1,5)=
5
is less than 50, so we have to try another combination.
Let’s try another combinations till we came to our goal of 50 feets:
10
st
step: let width is equal to
10
, then the length is
1,5
times bigger than width,
so its size is
15
. Let’s find the Perimeter. Perimeter = 2* (Width + Length) =
2*(1+1,5)=
50
is equal to 50, so we have a solution!
Ans: width is 10, length is 15
5.
Peter is outside looking at the pigs and chickens in the yard. Nadia is indoors and
cannot see the animals. Peter gives her a puzzle He tells her that he counts 13
heads and 36 feet and asks her how many pigs and how many chickens are in
the yard. Help Nadia find the answer.
Solution: Let’s solve this problem using “Guess and Check Method”
Pigs
chickens
Total
number of
legs
number of
4
2
54
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legs
Number of
animals
13
0
1 step
13
0
52
2 step
12
1
50
3 step
11
2
48
4 step
10
3
46
5 step
9
4
44
6 step
8
5
42
7 step
7
6
40
8 step
6
7
38
9 step
5
8
36
10 step
4
9
34
11 step
3
10
32
12 step
2
11
30
1
st
step: let’s number of pigs is 13 and number of chickens is zero, then the total
number of legs is calculated by this formula: N = 4*13+2*0 = 52 is greater than
36, so we have to try another combination.
Let’s try another combinations, till we came to our goal of 36 legs:
9
th
step: let’s number of pigs is 5 and number of chickens is 8, then the total
number of legs is calculated by this formula: N = 5*13+8*0 = 36. Problem is
solved
Ans: there were 5 pigs and 8 chickens
6.
Andrew invests $8000 in two types of accounts. A savings account that pays
5.25% interest per year and a more risky account that pays 9% interest per year.
At the end of the year he has $450 in interest from the two accounts. Find the
amount of money invested in each account.
Solution: Let’s solve this problem using “Guess and Check Method”
Account
#1
Account
#2
Total
amount of
money
Total
interest, $
Interest rate
5,25 %
9,00 %
8000
1 step
$8 000,00
$0,00
$8 000,00
$420,00
2 step
$5 000,00
$3 000,00
$8 000,00
$532,50
3 step
$6 000,00
$2 000,00
$8 000,00
$495,00
55
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4 step
$7 000,00
$1 000,00
$8 000,00
$457,50
5 step
$7 100,00
$900,00
$8 000,00
$453,75
6 step
$7 150,00
$850,00
$8 000,00
$451,88
7 step
$7 200,00
$800,00
$8 000,00
$450,00
Solution:
1
st
step: Let’s Andrew invests $8000 in 1
st
accounts and zero in the 2
nd
, then the total
amount of interest can be calculated by following formula: $ 8000* 5,25/100+ $ 0*9/100
= $420, is less than $450, so we have to try another combination.
Let’s try another combinations, till we came to our goal of $450:
7
th
step: Let’s Andrew invests $7200 in 1
st
accounts and 800 in the 2
nd
, then the total
amount of interest can be calculated by following formula: $ 7200* 5,25/100+ $
800*9/100 = $450, is our goal. Problem is solved!
Ans: He has to invest $7200 in 1
st
account and $800 in the 2
nd
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16.Working Backwards Concept
The “Work Backward” method works well for problems in which a series of
operations is applied to an unknown quantity and you are given the resulting number.
The strategy in these problems is to start with the result and apply the operations in
reverse order until you find the unknown. Let’s see how this method works by solving
the following problem.
Example 1
A bus left an interchange carrying some passengers with it.
At the first stop, 1/4 of the people in it alighted and 5 people boarded it.
At the 2nd stop, 1/2 of the people in it alighted and 20 people boarded the bus.
When it left the 2nd stop, there were 60 passengers in it.
How many passengers were there in the bus when it left the interchange?
Step 1:
Find the number of people before the 2
nd
stop
60 – 20 = 40
40 × 2 = 80
Step 2:
Find the number of people before the 1
st
stop
80 – 5 = 75
75 ÷ 3 × 4 = 100 people (Ans)
Example 2.
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57
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$451.25.
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Examples for self-solving
:
1.
Finish the problem we started in Example 1.
2.
Peter bought several notebooks at Staples for $2.25 each and he bought a few
more notebooks at Rite-Aid for $2 each. He spent the same amount of money in
both places and he bought 17 notebooks in total. How many notebooks did
Peter buy in each store?
Solution:
https://youtu.be/s1QN7sSfBM8
3.
There is a bowl of candy sitting on our kitchen table. This morning Nadia takes
one-sixth of the candy. Later that morning Peter takes one-fourth of the candy
that’s left. This afternoon, Andrew takes one-fifth of what’s left in the bowl and
finally Anne takes one-third of what is left in the bowl. If there are 16 candies left
in the bowl at the end of the day, how much candy was there at the beginning of
the day?
Solution
: Начнем решать эту задачу с конца
Шаг 1: Так как в конце осталось всего 16 конфет после того как Анна взяла одну
треть всех конфет, это значит, что до этого на тарелке оставалось 24 конфеты
Шаг 2: Andrew взял одну пятую всех оставшихся конфет, после чего в тарелке
остается 24 конфеты. Это значит он взял 6 конфет, значит до того как он пришел
за конфетами в тарелке было 30 конфет.
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Шаг 3: Peter взял одну четвертую часть всех оставшихся конфет, после чего в
тарелке остается 30 конфеты. Это значит он взял 10 конфет, значит до того как он
пришел за конфетами в тарелке было 40 конфет.
Шаг 4: Nadia взялf одну шестую часть всех оставшихся конфет, после чего в
тарелке остается 40 конфеты. Это значит она взяла 8 конфет, значит до того как
он пришел за конфетами в тарелке было 48 конфет.
Ответ: изначально в тарелке было 48 конфет.
4.
Nadia can completely mow the lawn by herself in 30 minutes. Peter can
completely mow the lawn by himself in 45 minutes. How long does it take both of
them to mow the lawn together?
5.
Solution:
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Nadia 30 minutes
Peter = 45 minutes
..
Nadia = 1/30 of job in 1 minute
Peter = 1/45 of the job in1 minute.
Together = 1/30 +1/45
LCD = 90
=(3+2)/90
=5/90
=1/18 of the job in 1 minute.
together they will take 18 minutes to complete the job.
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17.Simultaneous Equations Concept
In this type of PSLE Math questions, you need to form 2 equations to solve for 2
unknowns.
Video introduction in simultaneous equations topic:
https://www.youtube.com/watch?v=V_S7M4kjZfE
To models to solve an SE:
1
st
way to solve an equation:
https://youtu.be/wskxzaApiH4
2
nd
way to solve an equation:
https://youtu.be/45gBIpj_DHo
Example 1:
Amy and Billy had a total of $400. Amy spent 1/4 of her sum and Billy spent 2/5 of his.
They then had a total of $255 left. How much did Amy spend?
Step 1:
Let Amy’s money be 4 units, Billy’s money be 5 parts
A –> 4u
B –> 5p
Step 2:
Form a first equation using their total amount of money at first
4u + 5p = 400
(Equation 1)
Step 3:
Find the amount of money Amy and Billy have left
A –> 4u – u = 3u
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B –> 5p – 2p = 3p
Step 4:
Form a second equation using their total amount of money left
3u + 3p = 255
Step 5:
Simplify the second equation to make the number of units the same as the first
equation
3u + 3p = 255
u + p = 85 (Divide every term by 3)
4u + 4p = 340 (Multiply every term by 4)
(Equation 2)
Step 6:
Use the first equation minus the second equation to find 1 part
4u + 5p = 400
– (4u + 4p = 340)
–> 1p = 400 – 340 = 60
Step 7: Find 1 unit
85
– 60 = $25 (Ans)
Example 2:
Ann and Beth together have $57. Ann and Cindy together have $131. If Cindy has 3
times as much money as Beth, how much money does Ann have?
Solution:
https://www.youtube.com/watch?v=fBP328qBMEY
Solved problems:
1.
Solution:
https://www.youtube.com/watch?v=8t3IHlM2sUo
2.
(
P5/P6
) Four classes A, B, C and D participated in a fundraising event for the
school. Classes A and B collected a total of $120. Together, classes B, C and D
collected a total of $200. Total amount of money collected by all 4 classes is 4
times the amount that class B raised. How much money did class B raised?
Solution:
https://www.youtube.com/watch?v=r3bTEBeU1vg
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18.Double If Concept
This is another type of PSLE Math
questions which most students get stuck
in. It involves 2 “ifs” which represent 2
scenarios.
Example 1:
A farmer has some chickens
and ducks. If he sells 2 chickens and 3
ducks every day, there will be 50
chickens left when all the ducks have
been sold. If he sells 3 chickens and 2
ducks every day, there will be 25
chickens left when all the ducks have
been sold.
a) how many ducks are there?
b) how many chickens are there?
Solution:
Step 1:
Write down the selling ratio for both cases
Case 1: Chicken : Duck = 2u : 3u
Case 2: Chicken : Duck = 3u : 2u
Step 2:
Make the ratio of ducks to be the same as all the ducks are sold out in both
cases.
Case 1 (Times 2 to both sides)
Chicken : Duck = 2u : 3u = 4u : 6u
Case 2 (Times 3 to both sides)
Chicken : Duck = 3u : 2u = 9u : 6u
Step 3:
Form an equation using the chickens left
4u + 50 = 9u + 25
Step 4:
Find 1 unit
5u = 50 – 25
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1u = 5
Step 5:
Find the total ducks
5 × 6 = 30 ducks
Step 6:
Find the total chickens
5 × 4 + 50 = 70 chickens
Or
5 × 9 + 25 = 70 chickens
Example 2:
Solution:
https://www.youtube.com/watch?v=x-U3ncJYoOY
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19.Average Concept
The average concept is a simple concept,
“Total Amount = Average × Number of Items”.
But the questions can be pretty challenging. I
share a common type of PSLE Math questions
below which most students are unable to do.
Example 1:
In January, the average number of
students in one class was 24. In February, 5
new classes were added. The average number
of students in all the classes became 26. If
there were 32 students in each of the new
classes, find the total number of students in all
the classes in February.
Solution:
5 × 32 = 160
5 × 26 = 130
160 – 130 = 30
Number of classes in Jan –> 30 ÷ 2 = 15
Number of students in Jan –> 15 × 24 = 360
Number of students in Feb –> 360 + 160 = 520 (Ans)
Or if you are comfortable using Algebra…
Increase in students –> 5 × 32 = 160
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Number of classes in Jan –> u
Number of students in Jan –> 24 × u = 24u
Number of classes in Feb –> u + 5
Number of students in Feb –> 26 × (u + 5) = 26u + 130
24u + 160 = 26u + 130
26u – 24u = 160 – 130
2u = 30
u = 15
Number of students in Feb –> 26 × 15 + 130 = 520 (Ans)
20.Percent, Fractions, Ratios
1.
Solution:
https://www.youtube.com/watch?v=fjwik9F6jl4
2.
Solution:
https://www.youtube.com/watch?v=cw6AAV1hBK8
3.
Ramesh is 40% heavier than Lionel. If Ramesh weight 49 kg, find the weight of
Lionel?
Solution:
https://www.youtube.com/watch?v=Mq_u_BVTNNk
4.
Solution:
https://www.youtube.com/watch?v=j-oiDR9sxdg
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21.Speed Concept and non-routine questions
1.
Introduction to speed topic:
https://www.youtube.com/watch?v=oS20ssiWhC4
2.
Solution:
https://www.youtube.com/watch?v=sWbbsdxssfU
3.
Solution:
https://www.youtube.com/watch?v=Si8N27qRD-k
4.
Solution:
https://www.youtube.com/watch?v=8ynagEOTQCU
5.
Speed catching Up
https://www.youtube.com/watch?v=mLSGsu7ij7Y
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6.
https://www.youtube.com/watch?v=_kbJ6gPQ3Z8
7.
Solution:
https://www.youtube.com/watch?v=4j8GPVTxO34
8.
Solution 1:
https://www.youtube.com/watch?v=PU6gnyni3gU
Solution 2:
https://www.youtube.com/watch?v=HYaEXe78l4I
9.
Solution:
https://www.youtube.com/watch?v=NbGs10DfQ3I
10.
Solution:
https://www.youtube.com/watch?v=0uIHqW1P0ak
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11.
Nadia is at home and Peter is at school which is 6 miles away from home. They
start traveling towards each other at the same time. Nadia is walking at 3.5 miles
per hour and Peter is skateboarding at 6 miles per hour. When will they meet and
how far from home is their meeting place?
Solution:
Let’s find the approach speed: V(ap)=3,5+6 =9,5 km/h
Let’s time needed to meet each other: T=
6
1
:
19
2
=
6
1
:
2
19
=
12
19
Let’s find the distance to home: D =
12
19
×
7
2
=
84
38
km
Полезные ссылки:
Дополнительные ссылки:
Introduction into BAR MODEL Video
https://www.youtube.com/watch?v=TbayTZvS_bc
Очень полезный ютуб канал
https://www.youtube.com/user/edufrontsg/videos
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