RGS Primary 2009 PSLE Math Prelim Paper 2 Q18

At first, 25% of Kumar's money was the same as 33 and 1/3 % of Lily's money. Lily's father gave her $80 later, while Kumar spent $325. In the end, Lily had 2 and 1/2 times as much money as Kumar.

a) How much money did Kumar have at first?

b) How much money did Lily have in the end?

Solution

(Kumar) 4 units --> 2 parts + 325 (x5)**

(Lily) 3 units + 80 --> 5 parts (x2)**

**Kumar (x5) and Lily (x2) to make both of them 10 equal parts each.

(K) 20 units --> 10 parts + 1625

(L) 6 units + 160 --> 10 parts

(K) 20 units --> 10 parts + 1625

(L) 6 units --> 10 parts - 160

(K) - (L)

20 units - 6 units --> 10 parts - 10 parts + 1625 - (-160)

14 units --> 1625 + 160

14 units --> 1785
1 unit --> 1785 divided by 14 = 127.5

(a)

Kumar at first

4 units --> 4 x 127.5 = 510

Answer: $510

(b)
Lily in the end
--> 3 units + 80
(3 x 127.5) + 80
= 382.5 + 80
= 462.5

Answer: $462.50

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q17

At 9.30 am, Train A which was 200 m long, pulled out of Nanas Station and travelled towards Dadas Station at a uniform speed of 80 km/h. Half an hour later, Train B which was 150 m long, left Dadas Station and travelled towards Nanas Station at a uniform speed of 90 km/h.

a) How far has Train A travelled when Train B left Dadas Station?

b) The two trains met each other in a tunnel. Both trains took 15 minutes to completely travel through the tunnel. Calculate the length of the tunnel.

Solution

a)

Distance = speed x time

= 80 km/h x (1/2) hour

= 40 km

Answer: 40 km

b)

(Train A) distance

= 80 km x (1/4) hr

= 20 km

(Train B) distance

= 90 km x (1/4) hr

= 22.5 km

Length of Tunnel

--> 20km + 22.5km - 0.2km (Train A's length) - 0.15km (Train B's length)

= 42.15 km

Answer: 42.15 km

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q16

There were some sweets in Boxes X, Y and Z. Box X contained 20% of the total number of sweets in Boxes X, Y and Z. The ratio of the number of sweets in Box Y to the total number of sweets in Boxes X and Z is 2:1. If there are 24 more sweets in Box Y than Box Z, find the total number of sweets in Boxes X, Y and Z.

Solution

Box X --> 20% of total sweets
= 20/100
= 1/5 --> (3/15) of total sweets

Box Y --> 2 units
Box X + Z --> 1 unit
(Total 3 units)

Box Y --> (2 units)/(3 units)
= 2/3 --> (10/15) of total sweets

Z --> 2 units
Y - Z --> 24 sweets
10 units - 2 units --> 24 sweets
8 units --> 24 sweets
1 unit --> 24 sweets divided by 8 = 3 sweets
15 units --> 15 x 3 sweets = 45 sweets

Answer: 45 sweets

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 Q14

John shifted the decimal point of a number twice to the left to obtain a new number. The difference between the new number and the original number was 136.62.

a) How many times the new number is the original number?

b) What is the sum of the 2 numbers?

Solution

a)

Answer: 100 times

b)

Original number --> 1 unit

New number --> 100 units

100 units - 1 unit --> 136.62
99 units --> 136.62

1 unit --> 136.62 divided by 99 = 1.38

Sum of the two numbers

--> 100 units + 1 unit = 101 units

101 units --> 101 x 1.38 = 139.38

Answer: 139.38

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q13

The figure below is made up of a big circle, square and a small circle. The area of the square is 400 square cm. Find the area of the shaded region. (Correct your answer to 2 decimal places)

Solution

Area of square --> 400 square cm
Length of square --> 20 cm (square root of 400 sq cm)

Area of 1/4 square

--> 400 square cm divided by 4

= 100 square cm

Area of one right-angle triangle -->(1/2)(b)(h)

100 square cm = (1/2)(r)(r)

(r)(r) = (100 square cm) x 2 = 200 square cm

Area of circle = (3.14)(r)(r)

=(3.14)(200 square cm)

= 628 square cm

Shaded area -->(628 - 400) square cm

= 228 square cm
= 228.00 square cm (correct to 2 decimal places)

Answer: 228.00 square cm

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q12

Tap A, Tap B, Tap C and an empty rectangle tank are shown below.

Lily turned on Tap A with water flowing at a rate of 5 litres per minute. After 2 minutes, she placed a rock of volume 1250 cubic cm in the tank and turned on Tap B and Tap C as well. Tap C drains the tank at the rate of 2 litres per minute. After 5 more minutes, Lily turned off all the taps and noted that the height of the water level was 30 cm. Find the rate of the flow of water from Tap B.

Solution

Tap A (water filled into tank)

--> 5 litres x 7 (min) = 35 litres

Tap B (water drained from tank)

-->2 litres x 5 (min) = 10 litres

Volume of water in tank in the end

--> (50 x 30 x 30) cubic cm - 1250* cubic cm

= (45 000 - 1250) cubic cm

= 43 750 cubic cm

* volume of stone

43 750 cubic cm is the total volume of water filled through Tap A and Tap B less the volume of water drained from Tap C.
Tap A + Tap B - Tap C --> 43 750 ml

Tap B --> 43 750 ml - Tap A + Tap C
= 43 750 ml - 35 000 ml + 10 000 ml

= 18 750 ml (after 5 min)

1 min --> 18 750 ml divided by 5

= 3750 ml/min

= 3.75 litres/min

Answer: 3.75 litres/min

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q11

Mrs Kee baked some cookies and packed all the cookies in 12 small boxes and 5 big boxes. There were equal number of cookies in each small box and equal number of cookies in each big box. Each big box contained 14 more cookies than each small box. 18/29 of the cookies baked were packed in small boxes. How many cookies were there in each small box?

Solution

12 small boxes --> 18/29 of cookies
1 small box --> 18/29 divided by 12
= 3/58 of cookies

5 big boxes --> 11/29 of cookies (29/29 - 18/29)
1 big box --> 11/29 divided by 5
= 11/145 of cookies

There were 14 more cookies in each big box than in each small box
11/145 - 3/58 --> 14 cookies
7/290 --> 14 cookies
1/290 --> 14 divided by 7 = 2

(Small box) 3/58 --> 15/290
15/290 --> 15 x 2 = 30

Answer: 30 cookies

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q10

Some secondary one boys were asked to name their favourite sport. Their choices were represented on the pie chart below.

There was an equal number of boys who liked athletics and swimming. 80 boys chose football as their favourite sport.
a) What fraction of the boys liked swimming?
b) Find the total number of secondary one pupils who took part in the survey.

Solution

100% --> 360 degrees
1% --> 360 degrees divided by 100 = 3.6 degrees
(Rugby) 5% --> 5 x 3.6 degrees = 18 degrees


a)
Athletics --> (90 - 18) degrees = 72 degrees
Athletics --> same number as Swimming
Swimming --> 72 degrees

Fraction of boys who liked swimming
--> 72/360 = 1/5

Answer: 1/5 of the boys


b)
Football --> (100 - 25* - 20** - 15)% = 40%
40% --> 80
10 % --> 80 divided by 4 = 20
100% --> 10 x 20 = 200

*25% --> Athletics + Rugby (90 degress is 25% of 360 degrees)
**20% --> Swimming is 1/5 (20%)

Answer: 200 boys

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q9

Tricia has some pink, red and yellow ribbons. 1/3 of them are pink ribbons. Four fewer than 1/3 of the remainder are red ribbons and the remaining 24 are yellow ribbons. How many pink ribbons does Tricia have?

Solution

4 units --> 24 - 4 = 20
1 unit --> 20 divided by 4 = 5
3 units --> 3 x 5 = 15

Answer: 15 pink ribbons

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q8

In the figure below, O is the centre of the circle where OCD is an equilateral triangle. Given that Angle OAB = 20 degrees and Angle AOD = 127 degrees, find Angle BOC.

Solution

Angle DOC --> 60 degrees (Triangle OCD is equilateral)
Angle AOB --> (180 - 20 - 20) degrees
= 140 degrees (Triangle OAB is isosceles)

Angle BOC --> (360 - 140 - 127 - 60) degrees
= 33 degrees

Answer: 33 degrees

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q7

Last year the total age of Mr Tan and his wife was p years old. His wife is 1 year younger than him. What is his wife's age 2 years from now? Express your answer in terms of p.

Solution

Last year

2 units + 1 --> p
2 units --> p - 1
1 unit --> (p -1)/2


Two years from now
--> (p - 1)/2 + 3
= (p - 1)/2 + 6/2
= (p -5)/2

Answer: [(p - 5)/2] years old

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

RGS Primary 2009 PSLE Math Prelim Paper 2 Q4

Find the number that is exactly between 4/5 and 1 and 1/3.

Solution

4/5 + 1 and 1/3
= 12/15 + 1 and 5/15
= 2 and 2/15

2 and 2/15 divided
= 1 and 1/15

Answer: 1 and 1/15

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q3

When a number is divided by 8, the quotient is 121 with no remainder. What is the remainder when the same number is divided by 9?

Solution

? divided by 8 --> 121
? --> 121 x 8
? --> 968

968 divided by 9
= 107 R5

Answer: 5

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q2

A dress cost $139.95. During a sale, a 20% discount was given. Calculate the sale price of the dress.

Solution

Discount --> 20%
Cost after discount --> 80%

Sale price after discount
--> (80/100) x $139.95 = $111.96

Answer: $111.96

Printer Friendly Version

RGS Primary 2009 PSLE Math Prelim Paper 2 Q1

Find the value of each of the following expressions when y = 6.
a) 3y - 4
b) y + 2y/3

Solution

a)
3(6) - 4
= 18 - 4
= 14

Answer: 14


b)
(6) + 2(6)/3
= 6 + 4
= 10

Answer: 10

Printer Friendly Version