## Q&A - PSLE Math

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

### Re: Q&A - PSLE Math

∠CAD + ∠ACD = 90°, From (180° - 90°)

∠ACB = ∠ACD + 22°

∠BAC = ∠CAD + ∠DAB

∠ACB + ∠BAC = 180° - 41° = 139°

(∠ACD + 22°) + (∠CAD + ∠DAB) = 139°

∠CAD + ∠ACD + 22° + ∠DAB = 139°

90° + 22° + ∠DAB = 139°

∠DAB = 139 - 90° - 22°

= 27° ##

∠ACB = ∠ACD + 22°

∠BAC = ∠CAD + ∠DAB

∠ACB + ∠BAC = 180° - 41° = 139°

(∠ACD + 22°) + (∠CAD + ∠DAB) = 139°

∠CAD + ∠ACD + 22° + ∠DAB = 139°

90° + 22° + ∠DAB = 139°

∠DAB = 139 - 90° - 22°

= 27° ##

- MathIzzzFun
- KiasuGrandMaster
**Posts:**2358**Joined:**Wed Mar 09,**Total Likes:**5

### Re: Q&A - PSLE Math

At point D, angle ADC (90°) + angle ADC (reflex) = 360°questionable_i wrote:Can anyone help me with this question? Thank you

In quadrilateral BCDA,

angle DCB + angle CBA + angle BAD + angle ADC (reflex) = 360°

∴ angle DCB + angle CBA + angle BAD =90°

angle BAD = 90° - 22° - 41° = 27°

cheers.

### Re: Q&A - PSLE Math

Hi , pls help on the below question .

Diana had 200 coins in her coin box. There were only 10-cents,20-cents and 50cents coins. First she

Replaced all her 20-cent coins with 50-cent coins of the same value

And there were 146 coins in the box. Then she replaced all her 10-cent coins with 50-cent coins

Of the same value and there were 66 coins left in the box. How many 50-cent coins were there

In the box at first ?

Thanks very much !

Fly2High

Diana had 200 coins in her coin box. There were only 10-cents,20-cents and 50cents coins. First she

Replaced all her 20-cent coins with 50-cent coins of the same value

And there were 146 coins in the box. Then she replaced all her 10-cent coins with 50-cent coins

Of the same value and there were 66 coins left in the box. How many 50-cent coins were there

In the box at first ?

Thanks very much !

Fly2High

### Re: Q&A - PSLE Math

vvv

A: 1708482

Subject: Math

Level: P6

Tags: Money

Source:

^^^

For every 5 20c coins, Diana exchanged for 2 50c coins -> 3 fewer coins with each exchange

200 - 146 = 54

54 / 3 = 18 exchanges of 5 20c coins

18 x 5 = 90 (20c coins)

For every 5 10c coins, Diana exchanged for 1 50c coin -> 4 fewer coins with each exchange

146 - 66 = 80

80 / 4 = 20 exchanges of 5 10c coins

20 x 5 = 100 (10c coins)

Number of 50c coins -> 200 - 90 - 100 = 10

A: 1708482

Subject: Math

Level: P6

Tags: Money

Source:

^^^

For every 5 20c coins, Diana exchanged for 2 50c coins -> 3 fewer coins with each exchange

200 - 146 = 54

54 / 3 = 18 exchanges of 5 20c coins

18 x 5 = 90 (20c coins)

For every 5 10c coins, Diana exchanged for 1 50c coin -> 4 fewer coins with each exchange

146 - 66 = 80

80 / 4 = 20 exchanges of 5 10c coins

20 x 5 = 100 (10c coins)

Number of 50c coins -> 200 - 90 - 100 = 10

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

### Re: Q&A - PSLE Math

Let number of tin at first be uquestionable_i wrote:Can anyone help me with this question?

20 X u = 16 X (u+5)

20u = 16u + 80

4u = 80

u = 20 ###

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