ACS Primary 2010 SA1 Math Paper 2 Q15

Both Joanne and Joseph had an equal amount of money at first. Every month, Joanne spent $850 and Joseph spent $912. After a few months, Joanne was left with $1550 while Joseph had 4/5 as much as Joanne. How much money did Joseph have at first?

Solution

Every month,
Joanne spends $850
Joseph spends $912
Difference every month
--> $912 - $850 = $62


After a few months, Joseph had 4/5 as much as Joanne,

5 units ----- $1550
1 unit ------ $1550 divided by 5 = $310
($310 is the difference between Joanne and Joseph after a certain number of months)


Number of months to achieve the $310 difference,
$310 divided by $62 = 5 (mths)
It also takes 5 months for Joseph to have 4/5 as much as Joanne.


Amount Joseph has at first,
Joseph ------ 4 units + (5 mths x $912 spent per mth)
----- (4 x $310) + (5 x $912)
= $1240 + $4560
= $5800

Answer: $5800


Printer Friendly Version

ACS Primary 2010 SA1 Math Paper 2 Q6

At first, Matthew has twice as many soccer cards as Ivan. Each of them then bought the same number of cards. As a result, both of the now have 160 cards in total. If Matthew now has 30 more cards than Ivan, find the number of cards each of them bought.

Solution

bought + bought ----- 160 - 30 - 30 - 30 = 70
bought ----- 70 divided by 2 = 35


Answer: 35 soccer cards

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q15

The equilateral triangles below are formed using 2 cm sticks.

a) How many sticks are needed to form Pattern 5?
b) In which pattern will each side of the triangle measure 30 cm?
c) Calculate the number of shaded triangles in Pattern 100.

Solution

b)
Pattern 1 --> 1 x 2cm = 2 cm
Pattern 2 --> 2 x 2cm = 4 cm
Pattern 3 --> 3 x 2cm = 6 cm
:
:
:
Pattern 15 --> 15 x 2cm = 30 cm

Answer: Pattern 15

From above
1 + 2 + 3 ..... + 100

1 + 99 = 100
2 + 98 = 100
3 + 97 = 100
:
:
49 + 51 = 100
(only 50 is not paired)

49 groups of 100 --> 4900
Add the unpaired 50 --> 4900 + 50 = 4950

Answer: 4950 shaded triangles

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q6

Alice, Beth and Claire had 600 stamps altogether. After Beth had given 30 stamps to Alice, Beth had twice as many stamps as Claire and Alice had to 20 stamps more than Claire. How many stamps did Claire have?

Solution

4 units --> 600 - 20 = 580
1 unit --> 580 divided by 4 = 145

Answer: 145

Printer Friendly Version

ACS Primary 2009 PSLE Math Prelim Paper 2 Q18

There were some black and white marbles in a bag. If 20 black marbles are removed from the bag, the total number of marbles will be 7 times the number of black marbles left. If 50 white marbles are removed from the bag, the total number of marbles left will be 5 times the number of black marbles left. How many marbles are there in the bag?

Solution

20 black marbles (removed)
Black --> 1 unit + (20)
White --> 6 units

50 white marbles (removed)
Black --> 1 part
White --> 4 parts + (50)

(Black) 1 unit + 20 --> 1 part ...... (x4)*
(White) 6 units --> 4 parts + 50

(Black) 4 units + 80 --> 4 parts
(White) 6 units --> 4 parts + 50

*(x4) to make an equal number of 4 parts for both Black and White Marbles


(White Marbles) - (Black Marbles)
6 units - 4 units - 80 --> 4 parts + 50 - 4 parts
2 units - 80 --> 50
2 units --> 50 + 80
2 units --> 130
1 unit --> 130 divided by 2 = 65

Total number of marbles
Black --> 1 unit + 20
White --> 6 units
--> 7 units + 20
= (7 x 65) + 20
= 475

Answer: 475 marbles

Printer Friendly Version

ACS Primary 2009 PSLE Math Prelim Paper 2 Q11

Container A has 150 more marbles than Container B. If 30 marbles are being transferred from Container B to Container A, there will be thrice as many marbles in Container A as Container B. How many marbles are there in Container A in the beginning?

Solution

2 units --> 30 + 150 + 30 = 210
1 unit --> 210 divided by 2 = 105

Container A in the beginning
--> 3 units - 30
= (3 x 105) - 30
= 315 - 30
= 285

Answer: 285 marbles

Printer Friendly Version

CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q1

The table below shows the difference between the digits in some 2-digit numbers.

List all the 2-digit numbers, from 40 to 100, in which the digits have a difference of 3. Arrange them in ascending order.

Answer: 41, 47, 52, 58, 63, 69, 74, 86, 96

Printer Friendly Version

Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q18

Ace Drama Company sold some tickets for a children's performance. It sold the same number of $8 and $5 tickets in Week 1 and collected a total of $1664. In Week 2, it sold 96 $8 and $5 tickets. If the company collected $632 more from the sale of $8 tickets than the $5 tickets in the two weeks, how many $8 tickets were sold altogether?

Solution

Week 1
$8 tickets --> 8 units
$5 tickets --> 5 units
Total 13 units --> $1664
1 unit --> $1664 divided by 13 = $128

Value of tickets sold in Week 1
$8 tickets --> 8 x $128 = $1024
$5 tickets --> 5 x $128 = $640

Difference between $8 tickets and $5 tickets in Week 1
$1024 - $640 = $384

Difference between $8 tickets and $5 tickets in Week 2
$632 - $384 = $248

If 96 tickets in Week 2 were equally sold between $5 tickets and $8 tickets, there will be 48 tickets each.
$5 tickets --> 48 x $5 = $240
$8 tickets --> 48 x $8 = $384

Difference between $8 tickets and $5 tickets in Week 2 if equal number of $5 tickets and $8 sold, would be
$384 - $240 = $144

But the difference is $248 and not $144. Instead, we have,
$248 - $144 = $104 (more)

For every $8 sold instead of $5, there would be an increase of ($8 + $5) = $13
$104 (more) divided by $13 = 8 (tickets more)

48 tickets + 8 tickets (more) = 56 $8 tickets sold on 2nd day.
1st Week --> $1024 divided by $8 = 128 (tickets)

Total number of $8 tickets
--> 128 + 56 = 184

Answer: 184 $8-tickets

Printer Friendly Version

Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q9

Nigel had a total of 227 durians and pears. He sold half of the durians and bought another 40 pears. As a result, he had an equal number of durians and pears.
a) How many durians were there at first?
b) How many pears were there at first?

Solution

(a)

3 units --> 267
1 unit --> 267 divided by 3 = 89

(Durians) 2 units --> 2 x 89 = 178

Answer: 178 durians


(b)
(Pears) 1 unit - 40
--> 89 - 40 = 49

Answer: 49 pears

Printer Friendly Version

Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q14

For every 200 books Johnson sells, he earns $8. He will receive an addition of $20 for every 3000 books sold. How many books must he sell to earn $700?

Solution

1 group of 200 books --> $8

3000 books divided by 200 books --> 15 groups of 200 books

15 groups of 200 books

--> 15 x $8 + $20 commission

= $120 + $20

= $140

To earn $700,

$700 divided by $140 --> 5 groups of $140

1 group of $140

--> 15 groups of 200 books

5 groups of $140 --> 5 x 15 groups of 200 books

=75 groups of 200 books

75 groups of 200 books

--> 75 x 200 books

= 15 000 books

Answer: 15 000 books

Printer Friendly Version

Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q10

The diagrams below show tiling patterns. Each tile is a square of side 1 cm.

What is the perimeter of Pattern 10?

Solution

(Add 6 cm for every new pattern)

Pattern 1 --> 10 cm

Pattern 2 --> 16 cm

Pattern 3 --> 22 cm

Pattern 4 --> 28 cm

Pattern 5 --> 34 cm

Pattern 6 --> 40 cm

Pattern 7 --> 46 cm

Pattern 8 --> 52 cm

Pattern 9 --> 58 cm

Pattern 10 --> 64 cm

Answer: 64 cm

Printer Friendly Version

Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q9

A snail fell into a well that is 300 cm deep.

In the first hour, it climbed 80 cm up the well.

In the second hour, it climbed 70 cm up the well.

Each hour, it managed to climb 10 cm less than the hour before.

How many hours did it take to climb out of the well?

Solution

1st h --> 80 cm

2nd h --> 70 cm

3rd h --> 60 cm

4th h --> 50 cm

5th h --> 40 cm

(80 + 70 + 60 + 50 + 40) cm = 300 cm

--> 5 hours

Answer: 5 h

Printer Friendly Version

Anglo Chinese School 2007 PSLE Math Prelim Q43

Derrick had 2/3 as many stickers as Benedict. After Derrick bought another 8 stickers and Benedict lost 5 stickers, Derrick now has 4/5 as many stickers as Benedict. Find the number of stickers Derrick and Benedict had at first.

Solution



(Derrick after)
4 parts ----- 2 units + 8
2 parts ----- 1 unit + 4
2 parts ----- 1 part + 13 + 4
2 parts – 1 part ----- 17
1 part ----- 17

1 unit – 13 ----- 17
1 unit ----- 17 + 13 = 30
5 units ----- 30 x 5 = 150

Answer: 150 stickers

Anglo Chinese School 2007 PSLE Math Prelim Q40

Fara started collecting stamps in January. In each month from February to May, she collected 30 stamps more than the month before. She saved a total of 750 stamps from January to May. How many stamps did she collect in January?

Solution



5 units + 10 groups of 30 ---- 750

5 units + 300 ---- 750

5 units ---- 750 – 300 = 450

1 unit ---- 450 divided by 5 = 90

Answer: 90 stamps

Singapore Chinese Girls Sch 2008 PSLE Math Prelim Q43

There were 9 chairs in each row. 8 rows of chairs were rearranged equally spaced, to form the perimeter of a square. There were same number of chairs on each side of the square. How many chairs were there on each side of the square?

Solution

Total number of chairs ---- 8 x 9 = 72
No. of chairs less 4 corners ---- 72 - 4 = 48
No. of chairs per row less 4 corners ---- 68 divided by 4 = 17
No. of chairs on 1 side ---- 17 + (2 corner chairs) = 19

Answer: 19 chairs per side.

Rosyth Sch 2006 PSLE Math Prelim Q45

Ali, Billy and Caven had some cards. Ali would have twice as many cards as Billy if Billy gave 28 cards to Ali. Both Billy and Caven would have the same number of cards if Caven gave 84 cards to Billy. Given that Caven had 112 more cards than Ali at the beginning, find the number of cards each of them had at the beginning.

Solution

Ali ----- 2 units – 28 (at first)
Billy ----- 1 unit + 28 (at first)
Caven ----- 1 unit + 28 + 84 + 84 (at first)

At first,
(Caven) – (Ali) ---- 112
(1 unit + 28 + 84 + 84) – (2 units – 28) ----- 112
(1 unit + 196) – (2 units – 28) ----- 112
224 – 1 unit ----- 112
224 – 112 ----- 1 unit
1 unit ---- 112

Ali at first
(2 x 112) – 28 = 196

Billy at first
(112) + 28 = 140

Caven at first
(112) + 28 + 84 + 84 = 308

Answer: Ali had 196, Billy had 140 and Caven had 308

Rosyth Sch 2006 PSLE Math Prelim Q27

A number of coloured balls are in a basket. There are 145 more red balls than white balls. There are 358 more white balls than black balls. How many more red balls than black balls are there?

Solution

Red ----- 1 unit + 358 + 145
White ----- 1 unit + 358
Black ----- 1 unit

Red more than black ----- 358 + 145 = 503

Answer: There are 503 more red than black balls.

Tao Nan School P5 SA2 2006 Math Q47

Ricci saved $200 from her salary and spent the rest. She spent 1/9 of the expenditure on a blouse, $40 on a scarf and the rest on books. The amount spent on the scarf was $20 less than that spent on the blouse. What was her salary?

Solution



1/9 of expenditure (blouse) ----- $20 + $40 (scarf) = $60

9/9 of expenditure ----- 9 x $60 = $540

Salary ----- $540 + $200 (saved) = $740

Answer: Her salary was $740.

Tao Nan School P5 SA2 2006 Math Q46

Cherie had 3 times as much money as Jolene. After Cherie spent $40 and Jolene spent $8, Jolene had 3 times as much money as Cherie. How much money did Cherie have at first?

Solution


Jolene (before) ----- 3 units + $8
*Cherie (before) had 3 times as much as Jolene -----
(3 x 3 units) + (3 x $8) = 9 units + $24

Cherie (before) – Cherie (after) ----- $40 (spent)
(9 units + $24) – (1 unit) ----- $40
8 units + $24 ------ $40
8 units ----- $40 - $24 = $16
1 unit ----- $16 divided by 8 = $2

Cherie (at first)
9 units + $24
= (9 x $2) + $24
= $18 + $24
= $42

Answer: Cherie had $42 at first.

Tao Nan School P5 SA2 2006 Math Q45

A box containing 58 iron balls weighed 2145g . when 46 of the iron balls were removed, the mass of the box and the remaining iron balls weighed 489g. What was the mass of the box?

Solution

Number of iron balls left when 46 were removed
58 – 46 = 12 (iron balls)

Box + 58 iron balls ----- 2145 g
Box + 12 iron balls ----- 489 g

(58 – 12) iron balls ----- 2145 g – 489 g
46 iron balls ----- 1656 g
1 iron ball ----- 1656 g divided by 46 = 36 g

Box + 58 iron balls ----- 2145 g
Box + (58 x 36 g) ----- 2145 g
Box + 2088 g ----- 2145 g
Box ----- 2145 g – 2088 g = 57 g

Answer: The box was 57 g.