## Q&A - PSLE Math

### Re: Q&A - PSLE Math

Hi,

Please help on this question. Thank you.

Patrick, Devi and Jiamin each have the same number of beads. How many beads must Devi give to Patrick and Jiamin so that Patrick has 20 beads more than Devi and Jiamin has 4 fewer beads than Patrick?

Please help on this question. Thank you.

Patrick, Devi and Jiamin each have the same number of beads. How many beads must Devi give to Patrick and Jiamin so that Patrick has 20 beads more than Devi and Jiamin has 4 fewer beads than Patrick?

### Re: Q&A - PSLE Math

taringal wrote:Hi,

Please help on this question. Thank you.

Patrick, Devi and Jiamin each have the same number of beads. How many beads must Devi give to Patrick and Jiamin so that Patrick has 20 beads more than Devi and Jiamin has 4 fewer beads than Patrick?

### Re: Q&A - PSLE Math

Please help the following qtn, thks!

Find the smallest positive integer, n, such that 42n is a multiple of 180

Find the smallest positive integer, n, such that 42n is a multiple of 180

### Re: Q&A - PSLE Math

Reduce 180 and 42 to multiples of prime numbers.S-H wrote:Please help the following qtn, thks!

Find the smallest positive integer, n, such that 42n is a multiple of 180

180 = 2 * 2 * 3 * 3 * 5

42 = 2 * 3 * 7

Some numbers must be multipled to 42 in order to become a multiple of 180.

These numbers must include at least one 2, one 3 and one 5, since these are missing from the prime numbers of 42.

The smallest n = 2*3*5 = 30.

Ans: 30

### Re: Q&A - PSLE Math

Understand, thank you very much!Jmaths.SG wrote:Reduce 180 and 42 to multiples of prime numbers.S-H wrote:Please help the following qtn, thks!

Find the smallest positive integer, n, such that 42n is a multiple of 180

180 = 2 * 2 * 3 * 3 * 5

42 = 2 * 3 * 7

Some numbers must be multipled to 42 in order to become a multiple of 180.

These numbers must include at least one 2, one 3 and one 5, since these are missing from the prime numbers of 42.

The smallest n = 2*3*5 = 30.

Ans: 30

### Re: Q&A - PSLE Math

Can someone pls help, thanks!

(a) Solve the following inequality

4x-3

------- less than. 5x+2

7

(b) write down the smallest integer that satisfies this inequality

(a) Solve the following inequality

4x-3

------- less than. 5x+2

7

(b) write down the smallest integer that satisfies this inequality

### Re: Q&A - PSLE Math

Hi, hope the below helps...taringal wrote:Hi,

Please help on this question. Thank you.

Patrick, Devi and Jiamin each have the same number of beads. How many beads must Devi give to Patrick and Jiamin so that Patrick has 20 beads more than Devi and Jiamin has 4 fewer beads than Patrick?

### Re: Q&A - PSLE Math

Hi, can someone pls help, thks!

(a) solve the following inequality

4x-3

---------- < 5x + 2

7

(b) write down the smallest integer that satisfies this inequality

(a) solve the following inequality

4x-3

---------- < 5x + 2

7

(b) write down the smallest integer that satisfies this inequality

- MathWithModels
- YellowBelt
**Posts:**11**Joined:**Sun Dec 28,

### Re: Q&A - PSLE Math

Question:

We work backwards for Anne. She had $129 at the end.

This was after giving half her money to Bernard. So, she gave $129 to Bernard.

Before giving $129 to Bernard, she had $129 + $129 = $258.

Anne tripled her money after getting 1/4 of Catherine's money.

So Anne had $258/3 = $86 at first.

Anne tripled her money after getting 1/4 of Catherine's money.

As Anne had $86 at first, she received 2 x $86 = $172 from Catherine.

This was 1/4 of Catherine's money

So Catherine had $172 x 4 = $688 at first

Bernard received $129 from Anne, and had 1 1/2 times as much as before.

So the $129 was half what he had at first.

So Bernard had $129 x 2 = $258 at first.

*Catherine had some money at first. She gave 1/4 of her money to*

Anne. As a result, Anne had thrice as much as before. Then, Anne

gave 1/2 of her money to Bernard. Bernard then had 1 1/2 times as

much money as before. Anne still had $129 left. How much money

did each child have at first?

Answer:Anne. As a result, Anne had thrice as much as before. Then, Anne

gave 1/2 of her money to Bernard. Bernard then had 1 1/2 times as

much money as before. Anne still had $129 left. How much money

did each child have at first?

**Anne**We work backwards for Anne. She had $129 at the end.

This was after giving half her money to Bernard. So, she gave $129 to Bernard.

Before giving $129 to Bernard, she had $129 + $129 = $258.

Anne tripled her money after getting 1/4 of Catherine's money.

So Anne had $258/3 = $86 at first.

*Anne had $86 at first***Catherine**Anne tripled her money after getting 1/4 of Catherine's money.

As Anne had $86 at first, she received 2 x $86 = $172 from Catherine.

This was 1/4 of Catherine's money

So Catherine had $172 x 4 = $688 at first

*Catherine had $172 x 4 = $688 at first***Bernard**Bernard received $129 from Anne, and had 1 1/2 times as much as before.

So the $129 was half what he had at first.

So Bernard had $129 x 2 = $258 at first.

*Bernard had $129 x 2 = $258 at first*