Q&A - PSLE Math

[anneshirleygilbert]

I need some help with this question. Can somebody please help?


Joe had a piece of wire. He cut it into lengths of 55cm and 60cm. On each of the 55cm wire, he stringed 1 large bead after every 4 small ones as shown below. There were 5 large beads on every 55cm wire.

ssssBssssBssssBssssBssssB (Total length of wire is 55cm)

On the 60cm wire, he stringed 1 large bead after every small one and found that he had an equal number of small and large beads. The lengthof each bead is a whole number. The total number of beads on all the 55cm wires was equal to the total number of beads on all the 60cm wires. What was the least possible length of the wire?

TIA

[lesskiasu]

Hi, can anyone help on to this question? The question is Catholic High Prelim 2011. http://i45.tinypic.com/2ryfq6o.jpg\">

[MathIzzzFun]

Shimmer:

2)Machine P and Q are use to print a book in 5h. When printing a book , P broke down. As a result, it took 7 h to finsh a book. It took Q 8h 20 min to finish printing a book alone.
A. How long P was used before it broke down?
B. How does p take to print a book?

using units approach :
1 book --> 500 units
P+Q takes 5h --> 500/5 = 100 units/h
Q takes 8h 20min ie 25/3 h --> 500/25 x 3 = 60 units/h
So, P --> 100units-60units = 40 units/h
a)
in 7h, Q--> 7 x 60 units = 420 units
P --> 500 units - 420 units = 80 units
80 /40 (*P prints at 40 units/h) = 2h --> P was used for 2h
b)
time taken for P to print a book --> 500/40 = 12.5h

cheers.

[MathIzzzFun]

anneshirleygilbert:

I need some help with this question. Can somebody please help?


Joe had a piece of wire. He cut it into lengths of 55cm and 60cm. On each of the 55cm wire, he stringed 1 large bead after every 4 small ones as shown below. There were 5 large beads on every 55cm wire.

ssssBssssBssssBssssBssssB (Total length of wire is 55cm)

On the 60cm wire, he stringed 1 large bead after every small one and found that he had an equal number of small and large beads. The lengthof each bead is a whole number. The total number of beads on all the 55cm wires was equal to the total number of beads on all the 60cm wires. What was the least possible length of the wire?

TIA

using G&C... key consideration is that the length of each bead is a whole number

For the 55 cm wire, there are 5 sections of 4 small beads + 1 big bead
so, 4 small bead + 1 big bead --> 11 cm
possible length combinations:
1) 1 small bead --> 1cm, 1 big bead --> 7 cm
ie 1 small + 1 big --> 8 cm
2) 1 small bead --> 2 cm, 1 big bead 3 cm
ie 1 small + 1 big --> 5 cm

1) is not possible because 60 cm / 8 cm is not a whole number

so, 1 small bead --> 2cm , 1 big bead --> 3 cm

total number of beads on one 55 cm wire --> 5 x 5 = 25 beads
total number of beads on 60 cm wire --> 60/5 x 2 = 24 beads
lowest common multiple of 24 & 25 = 600
Shortest length of wire = 24 x 55 cm + 25 x 60 cm = 2.82m

cheers,

[MathIzzzFun]

Winx5015:

Hi,can anyone help me solve the following questions.


2)The distance between point A and point B of an island is 50km.At 11am, Afiq left point A and cycled towards point B.Benny left point A half an hour later and cycled towards point B.Benny's cycling speed was 4km/h faster than Afiq's.At 1pm,Benny caught up with Afiq.When Benny was 10km away from point B,he stopped to wait for Afiq.
a)What was Benny's speed?
b))How far was Afiq from Benny when Benny stopped to wait for him?

Thanks:)

Time taken for Benny to catch up with Afiq --> from 1130 to 1300 --> 1.5h
Time taken by Afiq, when Benny caught up --> from 1100 to 1300 --> 2h
Time ratio of Benny : Afiq --> 1.5 : 2 = 3 :4
Speed ratio of Benny : Afiq --> 4u : 3u
1u --> 4km/h
a) Benny's speed--> 16 km/h

When Benny stopped to wait for Afiq,
distance he cycled --> 50 km-10km = 40km

time taken --> 40/16 = 2.5h
2.5h-1.5h = 1h --> Benny and Afiq cycled for 1h after they \"met\"
b) 1h x 4km/h = 4km --> Afiq was 4km behind Benny when Benny stopped.

cheers.

[zanen]

Hi I’m Zan En’s sister and would like to ask this qns on her behalf. Mrs Leong had white and orange ping pong balls in the ratio 7:5. She then replaced several white ping pong balls with orange ones keeping the total number of ping pong balls the same. The ratio of the number of white ping pong balls to orange ping pong balls became 3:5.

(a) If the total number of ping pong balls is not more than 30, how many white ping pong balls did Mrs Leong have at first? (b) How many new orange ping pong balls were used to replace the white ones?

[tianzhu]

zanen:

Hi I'm Zan En's sister and would like to ask this qns on her behalf. Mrs Leong had white and orange ping pong balls in the ratio 7:5. She then replaced several white ping pong balls with orange ones keeping the total number of ping pong balls the same. The ratio of the number of white ping pong balls to orange ping pong balls became 3:5.

(a) If the total number of ping pong balls is not more than 30, how many white ping pong balls did Mrs Leong have at first? (b) How many new orange ping pong balls were used to replace the white ones?

Hi

White:Orange ------ 7:5(12)------- 14:10 (24)

White:Orange ------ 3:5(8) ------- 9:15(24)

Number of white balls@first ------- 14

Number of orangeballs@replacing white balls ------- 15-10 ------ 5

Best wishes

[tianzhu]

Shimmer:

One more question pls!


Dean,Ellen and Andy shared a box of sweets in the ratio of 3:4:5 and there was 4 left .They put the 4 sweets into a second box which contained 1/8 fewer sweets than the first box, and shared the sweets in the same ratio as before,and this time there were no remainder.
A)Find the total number of sweets they shared in the first round.
B)Find the total number of sweets each child received in both rounds of sharing.

Hi

Where is this question from?

What are the answers in the WS?

I think there is a possibility of multiple answers.

One set of possible answers

a) 60 sweets

b) 30,40,50

Best wishes

[tianzhu]

Winx5015:

1)There were a total of 800 paper clips in 3 containers, A, B and C.Alex transferred 3/8 of the paper clips from container A to container B.He then transferred 40 paper clips from container B to container C.As a result, the number of paper clips in container C is 1/5 the number of paper clips in container A There were 264 paper clips in container B at first. How many paper clips were there in container A at first?

Hi

It’s useful to draw a simple diagram to show the flow.

After the transfer

A ------- 5 units

B ------- 224 + 3 units (40 transferred to C, hence 264-40 ----- 224, and 3 units received from A)

C ------- 1 unit (1/5 the number of paper clips in container A)

9 units + 224 ------- 800

1unit ------- 64

A ------- 8*64 -------- 512

Best wishes

[Shimmer]

tianzhu:

Shimmer:

One more question pls!


Dean,Ellen and Andy shared a box of sweets in the ratio of 3:4:5 and there was 4 left .They put the 4 sweets into a second box which contained 1/8 fewer sweets than the first box, and shared the sweets in the same ratio as before,and this time there were no remainder.
A)Find the total number of sweets they shared in the first round.
B)Find the total number of sweets each child received in both rounds of sharing.

Hi

Where is this question from?

What are the answers in the WS?

I think there is a possibility of multiple answers.

One set of possible answers

a) 60 sweets

b) 120 sweets

Best wishes[/quote

This question was given by my TUTION teacher.
Y