This blog is managed by Song Hock Chye, author of Improve Your Thinking Skills in Maths (P1-P3 series), which is published and distributed by EPH.

## Sunday, February 21, 2010

### 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

## Friday, February 19, 2010

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q17

John had some red and blue marbles in a box. The sum of 1/4 of the red marbles and 2/5 of the blue marbles in the box is 64. The sum of 3/4 of the red marbles and 4/5 of the blue marble in the box is 144.
a) How many red marbles are there in the box?
b) How many blue marbles are there in the box?

Solution

Red (all marbles) --> R R R R
Blue (all marbles) --> B B B B B

1/4 or red marbles and 2/5 or blue marbles -- > 64
R + B B --> 64

3/4 of red marble and 4/5 of blue marbles --> 144
R R R + B B B B --> 144

R + B B --> 64 (multiply all by 2)
R R + B B B B --> 128

We now have
R R R + B B B B --> 144
R R + B B B B --> 128

R --> 144 - 128 = 16

a)
Number of Red Marbles
R R R R --> 4 x 16 = 64

Answer: 64 red marbles

b)
R + B B --> 64
16 + B B --> 64
B B --> 64 - 16 = 48
B --> 48 divided by 2 = 24

Number of Blue Marbles
B B B B B --> 5 x 24 = 120

Answer: 120 blue marbles

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q16

The ratio of the number of boys to the number of girls in School A is 4:1. the ratio of the number of boys to the number of girls in School B is 2:3. School A had twice as many pupils as School B.

a) What is the ratio of boys in School A to the number of girls in School B?

b) After 30 girls left School A to join School B, the ratio of the number of boys to the number of girls in School B is now 5:8. How many girls are there in School B now?

Solution * School A is multiplied by 2 to give a total of 10 units. This is because School B has 5 units. School A has twice the number of pupils as School B.

(a)
Boys from School A --> 8 units
Girls from School B --> 3 units

Ratio of number of boys in School A to number of girls in School B
--> 8 : 3

Answer: 8 : 3

b)
School B * Before is multiplied by 5 and After is multiplied by 2 to make the boys have a common unit of 10 for both Before and After, because there was no transfer of boys.

Girls increased by 1 unit after the transfer
1 unit--> 30

Number of girls in School B after transfer
16 units --> 16 x 30 = 480

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q15

Three men, A, B and C, worked together to paint a wall. If the painting was done by one man, the time taken to complete the wall for A, B and C would have been 6 hours, 8 hours and 12 hours respectively. A and B had painted for 3 hours after which A rested. B and C then continued with the painting. What would be the total number of hours taken to complete the wall? (Give your answer as a mixed number.)

Solution

First 3 hours
A painted for 3 h --> 3/6 = 1/2 (of the wall)*
B painted for 3 h --> 3/8 = 3/8 (of the wall)**
1/2 + 3/8 = 7/8 (of the wall was painted in first 3 hours)

* A takes 6h to paint the whole wall, therefore, in 3h, 3/6
** B takes 8h to paint the whole wall, therefore, in 3h, 3/8

A rests, B and C continue to paint remaining 1/8 of wall
Ratio of hours taken to paint whole wall
B : C
8 : 12
2 : 3

B takes less time than C, therefore B would have painted more of the wall
Ratio of amount of wall painted
B : C
3 : 2 --> Total units is 5

Time taken for B to paint whole wall
5/5 of wall --> 8h
Therefore, 3/5 of wall --> (3/5) x 8h = (24/5)h

Since only 1/8 of wall remains
1/8 of wall left --> (1/8) x (24/5)h = 3/5 h

Total time taken
--> 3h + 3/5 h = 3 and 3/5 hours

Answer: 3 and 3/5 hours

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q14

Mr Lim paid \$134.40 for some jackfruits and pomeloes. The cost of the pomelo was 0.8 that of a jackfruit. A pomelo cost \$5.60. If all the pomeloes cost \$22.40 more than the jackfruits, how many fruits did he buy?

Solution 2 units --> \$134.40 - \$22.40 = \$112
1 unit --> \$112 divided by 2 = \$56

Cost of Pomeloes
--> \$56 + \$22.40 = \$78.40

Number of Pomeloes bought
--> \$78.40 divided \$5.60 (per pomelo)
= 14 pomeloes

Cost of of 1 pomelo is 0.8 of 1 jackfruit
Pomelo --> 8 units
Jackfruit--> 10 units

(Pomelo) 8 units --> \$5.60
1 unit --> \$5.60 divided by 8 = \$0.70
(Jackfruit) 10 units --> 10 x \$0.70 = \$7

Number of Jackfruits bought
--> Total cost - cost of pomeloes
\$134.40 - \$78.40 = \$56
\$56 divided by \$7 (per jackfruit) = 8 (jackfruits)

Total number of fruits
14 (pomeloes) + 8 (jackfruits) = 22 fruits

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q13

The ratio of Jane's allowance to Olivia's allowance was 4:3. After Jane and Olivia were given \$15 and \$8 respectively, the ratio of Jane's allowance to Olivia's allowance became 3:2. How much allowance did Jane have at first?

Solution

Jane
4 units (at first) + \$15 (given) --> 3 parts

Olivia
3 units (at first) + \$8 --> 2 parts
2 parts --> 3 units + \$8
1 part --> 3/2 units + \$4
3 parts --> 3 x 3/2 units + 3 x \$4
= 9/2 units + \$12

(Jane) 3 parts --> 4 units + \$15
(Olivia) 3 parts --> 9/2 units + \$12

(Olivia) 9/2 units + \$12 --> (Jane) 4 units + \$15
9/2 units + \$12 --> 8/2 units + \$15
9/2 units - 8/2 units --> \$15 - \$12
1/2 unit --> \$3
1 unit --> \$6

Jane at first
4 units --> 4 x \$6 = \$24

## Wednesday, February 17, 2010

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q12

Rahim's age is 2/9 of his grandfather's. His grandfather will be 100 years old in 19 years' time. In how many years' time will Rahim's age be 1/4 of his grandfather?

Solution

Now
Rahim --> 2 units
Grandfather --> 9 units

19 years' time (Grandfather)
9 units + 19 --> 100
9 units --> 100 - 19 = 81
1 unit --> 81 divided by 9 = 9

Now
Rahim --> 2 x 9 years = 18 years
Grandfather --> 9 x 9 years = 81 years

Difference between Grandfather and Rahim
--> 81 years - 18 years = 63 years

Future (when Rahim 1/4 of Grandfather's age)
Rahim --> 1 unit
Grandfahter --> 4 units

But Rahim is 63 years younger
4 units - 1 unit --> 63 years
3 units --> 63 years
1 unit --> 63 years divided by 3 = 21 years

Rahim now --> 18 years
Rahim in future --> 21 years

21 years - 18 years = 3 years

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q11

There were some marbles in a box. Sofie took out 2/5 of the marbles and put in 6 more. Then John took out 1/6 of the remaining marbles and put in 5 more. there were 25 marbles left. How many marbles were in the box at first?

Solution 6 units - 1 unit + 6 - 1 + 5 --> 25 (left in box)
5 units + 10 --> 25
5 units --> 25 - 10 = 15
1 unit --> 15 divided by 5 = 3

Marbles at first
--> 10 units x 3
= 30

(At first, there were 10 units and not 5 units because the 5 units have been cut into halves, giving a total of 10 smaller units)

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q10

In the figure shown below, SUVX is a square. STU is an equilateral triangle and TXW is a straight line.
a) Find the value of Angle STX.
b) Find the value of Angle WVX. Solution a)
Line TZ passes through V, while line TY is passes through the centre of Line SU.
Angle STX is 1/4 of Angle STU.
Angle STU is 60 degrees (Triangle STU is equilateral)

Angle STX --> (1/4) x 60 degrees = 15 degrees

b)
Angle SXT = 15 degrees (Triangle STX is isosceles)
Angle TXV --> (90 - 15) degrees = 75 degrees
Angle WXV --> (180 - 75) degrees = 105 degrees

Angle WVX
--> (180 - 105) degrees divided by 2 = 37.5 degrees
(Triangle WVX is isosceles)

### 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

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

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q8

Wilson and Yi Lin had \$71 altogether. Yi Lin and Patrick had \$105 altogether. Wilsom hand 3/5 of the money that Patrick had. How much money did Yi Lin have?

Solution Yi Lin + Patrick
? + 5 units --> \$105

Yi Lin + Wilson
? + 3 units --> \$71

5 units - 3 units --> \$105 - \$71
2 units --> \$34
1 unit --> \$34 divided by 2 = \$17

Yi Lin + Wilson
? + 3 units --> \$71
? + (3 x \$17) --> \$71
? + \$51 --> \$71
? --> \$71 - \$51 = \$20

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q6

(a) Arif is 2x years old. His father is 4 times as old as he. His mother is 3 years younger than his father. What is their total age in terms of x?

(b) If x is = 4, find their total age.

Solution

(a)
Arif --> 2x
Father --> 4 x 2x = 8x
Mother --> 8x - 3

Total --> 2x + 8x + 8x - 3 = 18x - 3

Answer: (18x - 3) years

(b)
18(4) - 3
= 72 - 3
= 69

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q5

A container with Bottle A placed in it has a mass of 4.27 kg. An identical container with Bottle B placed in it has a mass of 6.58 kg. The mass of Bottle A is 1/3 that of Bottle B. What is the mass of the Bottle A? Give your answer correct to 2 decimal places.

Solution

Bottle A --> 1 unit
Bottle B --> 3 unit

Container + Bottle A (1 unit) --> 4.27 kg
Container + Bottle B (3 units) --> 6.58 kg

3 units - 1 unit --> 6.58 kg - 4.27 kg
2 units --> 2.31 kg
1 unit --> 2.31 kg divided by 2 units = 1.155 kg
1.155 kg --> 1.16 kg correct to 2 decimal places

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q4

Box A contains only one-dollar coins. Box B contains only fifty-cent coins and Box C contains only twenty-cent coins. Box A has 5 times as many coins as Box C. Box B contains 12 coins fewer than Box A. Box C contains half the number of coins in Box B. How much money is there in Box B?

Solution 2 units --> 5 units - 12
3 units --> 12
1 unit --> 12 divided by 3 = 4

(Box B)
5 units - 12
-->(5 x 4) - 12
= 20 - 12
= 8 (coins)

8 coins x \$0.50 = \$4

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q3

The ratio of the number of Chloe's stickers to the number of Faith's stickers is 3:5. The ratio of the number of Faith's stickers to Melissa's stickers is 6:7. If Melissa has 204 stickers more than Chloe, how many stickers do they have altogether?

Solution Melissa - Chloe --> 204
35 units - 18 units --> 204
17 units --> 204
1 unit --> 204 divided by 17 = 12

All stickers
Chloe + Faith + Melissa
18 + 30 + 35 = 83
83 units --> 83 x 12 = 996

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q1

LMNO is a square. PQN and PLQ are isosceles triangles. Angle QNM is 23 degrees. Find Angle NPQ. Solution

Angle PNO --> 23 deg (mirror image of Angle MNQ)
Angle OPN --> 90 deg - 23 deg = 67 deg
Angle LPQ --> 45 deg (Triangle LQP is isosceles and Angle LPQ is a rt angle)
Angle NPQ --> (180 - 45 - 67) deg = 68 deg

## Wednesday, February 10, 2010

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q18

Ai Tong School organized a 2-day camp. On the first day, the number of boys was 600 more than the girls. On the second day, the number of boys decreased by 10% but the number of girls increased by 10%. If there were 2540 children on the second day, how many children were there on the first day?

Solution 2nd Day
Boys --> 0.9 unit + 540
Girls --> 1.1 unit
Total --> 2540

0.9 unit + 540 + 1.1 unit --> 2540
2 units + 540 --> 2540
2 units --> 2540 - 540 = 2000
1 unit --> 2000 divided by 2 = 1000

1st Day
2 units + 600
(2 x 1000) + 600
= 2000 + 600
= 2600

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q17

75% of the children in the stadium were girls. After 52 girls and 4 boys left, the remaining children formed groups of 8. In each group, there were 3 boys. How many children were there in the stadium at first?

Solution Note - "Before" for girls is 3 parts, while "Before" for boys is 1 part. This is to represent 75% girls and 25% boys, before 52 girls and 4 boys left.

From the table above
4 units --> 40
1 unit --> 40 divided by 4 = 10

Total boys (after)
3 units --> 3 x 10 = 30

Total girls (after)
5 units --> 5 x 10 = 50

Total children (after)
30 + 50 = 80

Total children at first
80 + 52 + 4 = 136

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q16

The number of pupils in Team A to the number of pupils in Team B is in the ratio of 7:6. If 45 pupils are transferred from Team A to Team B, the ratio will become 2:3. How many pupils are there altogether?

Solution Note - "Before" is multiplied by 5, while "After" is multiplied by 13 so as to have an equal number of total units of 65 each. The total number of units for "Before" and "After" are the same because there was no increase or decrease in the total number of pupils before or after the transfer.

Team A had a decrease of

35 units – 26 units --> 45 (pupils)
9 units --> 45
1 unit --> 45 divided by 9 = 5

Total number of pupils
Team A + Team B
26 units + 39 units --> 65 units
65 units --> 5 x 65 = 325

## Tuesday, February 09, 2010

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q15

At a party, only 4/9 of the invited guests came. The ratio of the number of women to the number of men present was 3 : 4. If 80 more men turned up for the party, the number of men would be twice the number of women. How many guests were invited?

Solution

Before 80 more men turned up
Women : Men
3 : 4

After 80 more men turned up
Women : Men
3 : 6

There was an increase of 2 units for men
2 units --> 80 (men)
1 unit --> 80 divided by 2 = 40

Total present at the party before 80 more men turned up
7 units --> 280

4/9 of invited guests --> 280
1/9 --> 280 divided by 4 = 70

All invited guests
9/9 --> 9 x 70 = 630

### 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

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q13

The length of a rectangle is thrice as long as its width. Its perimeter is 16p cm.
a) Find the length of the rectangle in terms of p.
b) Find the area of the rectangle if p = 4.

Solution Perimeter
--> 3 units + 1 unit + 3 units + 1 unit
= 8 units

8 units --> 16p cm
1 unit --> 16p cm divided 8 = 2p cm

a)
Length
3 units --> 3 x 2p cm = 6p cm Answer: 192 square cm

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q12

20 similar pails of water can fill 5/12 of a container. Another 8 similar pails and 105 similar bowls of water are needed to fill the container to its brim. How many such bowls of water are needed to fill the empty container completely?

Solution

20 pails --> 5/12 of container (x2)
8 pails + 105 bowls --> remaining 7/12 of container (x5)

40 pails --> 10/12 of container (after x2)
40 pails + 525 bowls --> 35/12 containers (after (x5)

From the above we can see that 525 bowls can fill up:-
525 bowls --> 35/12 - 10/12 = 25/12 containers

25/12 --> 525 bowls
1/12 --> 525 divided 25 = 21 bowls
(Full container) 12/12 --> 12 x 21 = 252

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q11

James spends 20% of his monthly income on transport, 30% of it on food and 10% of the remainder on clothes. He saves the rest of his income. If his monthly savings is \$900, find his monthly income.

Solution (Saves)
90% of 50% --> \$900
(90/100) x (50/100) x 100% --> \$900
45% --> \$900
1%--> \$900 divided by 45 = \$20

(Monthly Income)
100% --> 100 x \$20 = \$2000

### 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

## Friday, February 05, 2010

### 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

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q8

The rectangle ACEG is divided into 4 parts. BCEF is a square. Each part has a different area. Find the area X. Area of X
--> 4 cm x 3 cm = 12 square cm

Answer: 12 square cm

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q7

The table below shows the rates for water consumption.

a) Find the amount paid for 50 cubic metres of water used.
b) If 7% GST is imposed on the total amount, how much is the GST correct to the nearest 10-cent?

Water Consumption Rates
First 20 cubic metres ----- \$1.33 per cubic metre
Next 20 cubic metres ----- \$1.46 per cubic metre
Additional amount above 40 cubic metres ----- \$1.73 cubic meter

Solution

a)
1st 20 cubic metres --> 20 x \$1.33 = \$26.60
2nd 20 cubic metres --> 20 x \$1.26 = \$29.20
Next 10 cubic metres --> 10 x \$1.73 = \$17.30

Total Amount
--> \$26.60 + \$29.60 + \$17.10 = \$73.30

b)
7 % x \$73.10 = \$5.117
Approximately --> \$5.10 (to nearest 10-cent)

## Thursday, February 04, 2010

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q6

Zen had a piece of rectangular paper. He cut away 1/9 of the breadth and 2/5 of the length. Then he measured the remaining piece of rectangular paper. He found that the breadth to be 16 cm and the length to be 15 cm. Find the area of the original piece of paper.

Solution 8/9 --> 16 cm
1/9 --> 16 cm divided by 8 = 2 cm
9/9 --> 9 x 2 cm = 18 cm

Length
3/5 --> 15 cm
1/5 --> 15 cm divided by 3 = 5 cm
5/5 --> 5 x 5 cm = 25 cm

Original area --> 25 cm x 18 cm = 450 square cm

Answer: 450 square cm

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q5

The figure below is made up of 2 squares, A and B. The ratio of Square A to Square B is 1:4. The shaded part is 1/5 of Square A. What fraction of the whole figure is shaded? Solution

Shaded part --> 1 unit

Area of Square A --> 5 units (shaded area is 1/5 of Area of Square A)
Note that Square A has 4 units of unshaded and 1 unit of shaded area.

Area of Square B --> 4 x 5 units = 20 units (Area of Square B is 4x that of Square A)
Note that Square B has 19 units of unshaded and 1 unit of shaded area.

Total Area
--> 1 unit (shaded) + 4 units (unshaded Square A) + 19 units (unshaded Square B)
= 24 units

Fraction of shaded area to area of whole figure
--> 1/24

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q4

In a right-angled triangle, the two sides which form the right angle are 16 cm by 12 cm respectively. How many such triangles are needed to form the smallest square? Horizontal --> 4 x 12 cm = 48 cm
Vertical --> 3 x 16 cm = 48 cm
The above figure (not drawn to scale) is therefore a square.

The smallest 1 unit rectangle has 2 triangles.
There are 3 rows and 4 columns of the smallest 1 unit rectangle.

3 x 4 = 12 smallest unit of rectangles
2 triangles x 12 = 24 triangles

## Wednesday, February 03, 2010

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q3

The ratio of Paul's age to Sue's age is 4:3. In 8 years' time the sum of their ages will be 86. What is Paul's age now?

Solution 7 units + 8 + 8 --> 86
7 units --> 86 - 8 - 8 = 70
7 units --> 70
1 unit --> 70 divided by 7 = 10

Paul's age now
4 units --> 4 x 10 = 40

Answer: 40 years old

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q2

Each skirt costs \$m and 3 blouses cost \$62.50. Find the cost of 2 skirts and 6 blouses.

Solution

Skirt --> \$m
2 skirts --> 2 x \$m = \$2m

3 blouses --> \$62.50
6 blouses --> 2 x \$62.50 = \$125

Total --> \$2m + \$125m
= \$(2m + 125) (Answer)

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q1

Express the perimeter of the figure below in terms of t. Solution Perimeter
--> (6 + t + 5 + 7 +11 + 7 + t) cm
= (2t + 36) cm (Answer)