## Q&A - PSLE Math

### Re: PSLE Math

vvv

A: 1692365

Subject: Math

Level: P6

Tags: Ratios

Source: ???

^^^

1) Mr Ho had a total of 102 pink and white roses in the ratio of 12:5.

After he has given away twice as many pink roses as white roses, the number of pink and white roses left was in the ratio 5:2.

How many pink roses did he give away?

At first, 72 pink roses, 30 white roses.

After giving away 2u pink roses and 1u white roses,

he has 5p pink roses and 2p white roses.

5p -> 72 - 2u

10p -> 144 - 4u

2p -> 30 - 1u

10p -> 150 - 5u

144 - 4u = 150 - 5u

1u = 150 - 144 = 6

2u = 2 x 6 = 12

He gave away 12 pink roses.

A: 1692365

Subject: Math

Level: P6

Tags: Ratios

Source: ???

^^^

1) Mr Ho had a total of 102 pink and white roses in the ratio of 12:5.

After he has given away twice as many pink roses as white roses, the number of pink and white roses left was in the ratio 5:2.

How many pink roses did he give away?

At first, 72 pink roses, 30 white roses.

After giving away 2u pink roses and 1u white roses,

he has 5p pink roses and 2p white roses.

5p -> 72 - 2u

10p -> 144 - 4u

2p -> 30 - 1u

10p -> 150 - 5u

144 - 4u = 150 - 5u

1u = 150 - 144 = 6

2u = 2 x 6 = 12

He gave away 12 pink roses.

### Re: PSLE Math

vvv

A: 1692365

Subject: Math

Level: P6

Tags: Ratios

Source: ???

^^^

2) Last year, the ratio of boys to girls in a school’s computer club was 4 : 7 . This year, 71 new members joined the club.

As a result, the ratio of boys to girls in the club becomes 4 : 3 . There are 72 boys in the club this year. How many of the new members are girls?

This year,

B:G -> 4:3 -> 72:54

Total -> 72 + 54 = 126

126 - 71 = 55 members last year

B:G -> 4:7 -> 20:35

54 - 35 = 19 new girl members

A: 1692365

Subject: Math

Level: P6

Tags: Ratios

Source: ???

^^^

2) Last year, the ratio of boys to girls in a school’s computer club was 4 : 7 . This year, 71 new members joined the club.

As a result, the ratio of boys to girls in the club becomes 4 : 3 . There are 72 boys in the club this year. How many of the new members are girls?

This year,

B:G -> 4:3 -> 72:54

Total -> 72 + 54 = 126

126 - 71 = 55 members last year

B:G -> 4:7 -> 20:35

54 - 35 = 19 new girl members

- questionable_i
- BrownBelt
**Posts:**535**Joined:**Sat Oct 10,**Total Likes:**1

### Re: Q&A - PSLE Math

vvv

Q: 1692726

Subject: Math

Level: P6

Tags: Percentage, Ratios, Geometry

Source: ???

^^^

Can anyone help me with this question?

1. Each side of 10-cm square is increased by 100%. Find the ratio of the original area to the new area of the square.

Q: 1692726

Subject: Math

Level: P6

Tags: Percentage, Ratios, Geometry

Source: ???

^^^

Can anyone help me with this question?

1. Each side of 10-cm square is increased by 100%. Find the ratio of the original area to the new area of the square.

- alfretztay
- KiasuGrandMaster
**Posts:**1199**Joined:**Sun Sep 12,**Total Likes:**5

### Re: Q&A - PSLE Math

vvv

A: 1692726

Subject: Math

Level: P6

Tags: Percentage, Ratios, Geometry

Source: ???

^^^

I would solve the above question as follows :

original area ------- 10cm x 10cm = 100 sq cm

Each side of 10-cm square is increased by 100% ------- 20cm

new area ------- 20cm x 20cm = 400 sq cm

original area : new area

100 : 400

1 : 4

Ans : 1 : 4.

A: 1692726

Subject: Math

Level: P6

Tags: Percentage, Ratios, Geometry

Source: ???

^^^

I would solve the above question as follows :

original area ------- 10cm x 10cm = 100 sq cm

Each side of 10-cm square is increased by 100% ------- 20cm

new area ------- 20cm x 20cm = 400 sq cm

original area : new area

100 : 400

1 : 4

Ans : 1 : 4.

### Re: Q&A - PSLE Math

vvv

Q: 1692859

Subject: Math

Level: P6

Tags: Percentage

Source: ???

^^^

Please can someone help. Thank you in advance!

Q: 1692859

Subject: Math

Level: P6

Tags: Percentage

Source: ???

^^^

Please can someone help. Thank you in advance!

- alfretztay
- KiasuGrandMaster
**Posts:**1199**Joined:**Sun Sep 12,**Total Likes:**5

### Re: Q&A - PSLE Math

vvv

A: 1692859

Subject: Math

Level: P6

Tags: Percentage

Source: ???

^^^

I would solve the above question as follows :

3600/3 = 1200 (marbles in Box B at first)

1200 x 2 = 2400 (marbles in Box A at first)

Before :

In Box A, there were 25% as many red marble as the total number of blue and yellow marbles -- 1u (red), 4u (blue + yellow), totalling 5u

Box A --

2400/5 = 480 (red)

480 x 4 or 2400 - 480 = 1920 (blue + yellow)

Box B --

In Box B, 60% of the number of marbles were blue and yellow.

60% x 1200 = 720 (blue + yellow)

40% x 1200 or 1200 - 720 = 480 (red)

After :

Box A --

25% (red)

75% (blue + yellow)

Box B --

40% (red)

60% (blue + yellow)

25%A + 40%B ------- 480 + 480 = 960

75%A + 120%B ------- 960 x 3 = 2880

75%A + 60%B ------- 1920 + 720 = 2640

60%B -------- 2880 - 2640 = 240

100%B ------- (240/60) x 100 = 400

3600 - 400 = 3200

Ans : 3200 marbles.

A similar question with the solution provided @ onsponge.

http://www.onsponge.com/forum/35-thinki ... -equations

There was a total of 2400 red, blue and green marbles in Box A and Box B at first. The number of marbles in Box A was twice the number of marbles in Box B. In Box A, the number of red marbles is 25% of the total number of blue and green marbles. The number of blue and green marbles in Box B is 60% of the total number of marbles in Box B. After transferring some marbles between each other, 25% of the marbles in Box A and 75% of marbles in Box B were red. How many marbles were there in Box A after the change?

I would present my solution to the above problem as follows :

2400/3 = 800 (marbles in Box B at first)

800 x 2 = 1600 (marbles in Box A at first)

Before :

Box A --

In Box A, the number of red marbles is 25% of the total number of blue and green marbles--- 1u (red) and 4u (blue + yellow)

1600/5 = 320 (red)

320 x 4 = 1280 (blue + yellow)

Box B --

The number of blue and green marbles in Box B is 60% of the total number of marbles in Box B ------- 6p (blue + yellow) and 4p (red).

800/10 = 80

80 x 6 = 480 (blue + yellow)

80 x 4 = 320 or 800 - 480 = 320 (red)

After 25% of the marbles in Box A and 75% of marbles in Box B were red :

Box A --

25% (red)

75% (blue + yellow)

Box B --

75% (red)

25% (blue + yellow)

25%A + 75%B ------- 320 + 320 = 640

75%A + 225%B ------- 640 x 3 = 1920

75%A + 25%B ------- 1280 + 480 = 1760

200%B -------- 1920 - 1760 = 160

100%B ------- 160/2 = 80

2400 - 80 = 2320

Ans : 2320 marbles.

A: 1692859

Subject: Math

Level: P6

Tags: Percentage

Source: ???

^^^

I would solve the above question as follows :

3600/3 = 1200 (marbles in Box B at first)

1200 x 2 = 2400 (marbles in Box A at first)

Before :

In Box A, there were 25% as many red marble as the total number of blue and yellow marbles -- 1u (red), 4u (blue + yellow), totalling 5u

Box A --

2400/5 = 480 (red)

480 x 4 or 2400 - 480 = 1920 (blue + yellow)

Box B --

In Box B, 60% of the number of marbles were blue and yellow.

60% x 1200 = 720 (blue + yellow)

40% x 1200 or 1200 - 720 = 480 (red)

After :

Box A --

25% (red)

75% (blue + yellow)

Box B --

40% (red)

60% (blue + yellow)

25%A + 40%B ------- 480 + 480 = 960

75%A + 120%B ------- 960 x 3 = 2880

75%A + 60%B ------- 1920 + 720 = 2640

60%B -------- 2880 - 2640 = 240

100%B ------- (240/60) x 100 = 400

3600 - 400 = 3200

Ans : 3200 marbles.

A similar question with the solution provided @ onsponge.

http://www.onsponge.com/forum/35-thinki ... -equations

There was a total of 2400 red, blue and green marbles in Box A and Box B at first. The number of marbles in Box A was twice the number of marbles in Box B. In Box A, the number of red marbles is 25% of the total number of blue and green marbles. The number of blue and green marbles in Box B is 60% of the total number of marbles in Box B. After transferring some marbles between each other, 25% of the marbles in Box A and 75% of marbles in Box B were red. How many marbles were there in Box A after the change?

I would present my solution to the above problem as follows :

2400/3 = 800 (marbles in Box B at first)

800 x 2 = 1600 (marbles in Box A at first)

Before :

Box A --

In Box A, the number of red marbles is 25% of the total number of blue and green marbles--- 1u (red) and 4u (blue + yellow)

1600/5 = 320 (red)

320 x 4 = 1280 (blue + yellow)

Box B --

The number of blue and green marbles in Box B is 60% of the total number of marbles in Box B ------- 6p (blue + yellow) and 4p (red).

800/10 = 80

80 x 6 = 480 (blue + yellow)

80 x 4 = 320 or 800 - 480 = 320 (red)

After 25% of the marbles in Box A and 75% of marbles in Box B were red :

Box A --

25% (red)

75% (blue + yellow)

Box B --

75% (red)

25% (blue + yellow)

25%A + 75%B ------- 320 + 320 = 640

75%A + 225%B ------- 640 x 3 = 1920

75%A + 25%B ------- 1280 + 480 = 1760

200%B -------- 1920 - 1760 = 160

100%B ------- 160/2 = 80

2400 - 80 = 2320

Ans : 2320 marbles.

Last edited by alfretztay on Sun Jun 26, 2016 10:51 am, edited 4 times in total.

### Re: Q&A - PSLE Math

vvv

Q: 1692859

Subject: Math

Level: P6

Tags: Percentage

Source: ???

^^^

Since A has twice of B, we can have 3 units with A having 2 units and B having 1 unit, giving a total 3 units.

B = 3600/3 = 1200 beads

A = 1200 X 2 =2400 beads.

Now we can establish the number of red colors and Blue/Yellow (BY) beads.

In Box A, Red is 25% of Blue/Yellow(BY). We let BY be 4u, and 25% of BY will be 1u.

1u + 4u = 2400

Red beads in A = 1u = 2400/5 = 480

BY beads in A = 1920

In Box B. 60% are BY,

BY beads in Box B = 0.6 X 1200 = 720

Red beads in Box B = 0.4 X 1200 = 480

Total Red beads = 480 + 480 = 960

Total BY Beads = 1920 + 720 = 2640

KEy point, constant total of red and BY after the transfer.

After the transfer

Total Red beads,

960 = 0.25A + 0.4B

2880 = 0.75A + 1.2B ---(1) <-multiply by 3 to the above eqn

Total BY beads

2640 = 0.75A + 0.6B ---(2)

(1) - (2)

240 = 0.6B

B = 400

Final A = 3600 - 400 =3200 ##

Q: 1692859

Subject: Math

Level: P6

Tags: Percentage

Source: ???

^^^

Since A has twice of B, we can have 3 units with A having 2 units and B having 1 unit, giving a total 3 units.

B = 3600/3 = 1200 beads

A = 1200 X 2 =2400 beads.

Now we can establish the number of red colors and Blue/Yellow (BY) beads.

In Box A, Red is 25% of Blue/Yellow(BY). We let BY be 4u, and 25% of BY will be 1u.

1u + 4u = 2400

Red beads in A = 1u = 2400/5 = 480

BY beads in A = 1920

In Box B. 60% are BY,

BY beads in Box B = 0.6 X 1200 = 720

Red beads in Box B = 0.4 X 1200 = 480

Total Red beads = 480 + 480 = 960

Total BY Beads = 1920 + 720 = 2640

KEy point, constant total of red and BY after the transfer.

After the transfer

Total Red beads,

960 = 0.25A + 0.4B

2880 = 0.75A + 1.2B ---(1) <-multiply by 3 to the above eqn

Total BY beads

2640 = 0.75A + 0.6B ---(2)

(1) - (2)

240 = 0.6B

B = 400

Final A = 3600 - 400 =3200 ##

### Re: Q&A - PSLE Math

vvv

Q: 1693098

Subject: Math

Level: P6

Tags: Speed

Source: ???

^^^

Hi

Pls help on the below question.

Xiang Bao and Daniel took part in a go-kart competition. Xiang Bao drove at a speed of

30km/h . Both of them did not change their speeds throughout the competition.

When Daniel covered half the distance,Xiang Bao was 2.5km in front of him.Xiang Bo reached the finishing line

At 2.55pm. What time did Daniel reach the finishing line ?

Thanks

Fly2high

Q: 1693098

Subject: Math

Level: P6

Tags: Speed

Source: ???

^^^

Hi

Pls help on the below question.

Xiang Bao and Daniel took part in a go-kart competition. Xiang Bao drove at a speed of

30km/h . Both of them did not change their speeds throughout the competition.

When Daniel covered half the distance,Xiang Bao was 2.5km in front of him.Xiang Bo reached the finishing line

At 2.55pm. What time did Daniel reach the finishing line ?

Thanks

Fly2high