PSLE Maths question. How to solve?

[Hiphiphoray]

1/4 of the fishes in an aquarium are goldfish.

There are 4 more guppies than goldfish in the aquarium.

The remaining 16 fishes are craps.

How many fishes are there in the aquarium?

 

[Samshio]

1/4 of the fishes in an aquarium are goldfish.

There are 4 more guppies than goldfish in the aquarium.

The remaining 16 fishes are craps.

How many fishes are there in the aquarium?

 

 

 

40

 

using models:

 

[  gold: 1/4  ] [  1/4 ]  [ 1/4 ]  [ 1/4 ]

 

guppies = gold + 4 = [ 1/4 ] + 4

 

remaining = carp = [1/4] [1/4] - 4 = 16

 

thus [1/4] = 10

 

4/4 = 40 fishes

[Porker]

1/4 of the fishes in an aquarium are goldfish.

There are 4 more guppies than goldfish in the aquarium.

The remaining 16 fishes are craps.

How many fishes are there in the aquarium?

 

Cannot be solved because never heard of a fish called crap

[Coltplussport]

1/4 of the fishes in an aquarium are goldfish.

There are 4 more guppies than goldfish in the aquarium.

The remaining 16 fishes are craps.

How many fishes are there in the aquarium?

 

I try try.

 

Let y be number of fishes.

 

1/2 y - 4 = 16

 

Y= 40

[Hiphiphoray]

Draw 4 units of boxes.

No. of Goldfish = 1 unit

No. of Guppies = 1 unit +4

Since 2/4 units already identified, the other 2/4 units = Craps +4

Hence 2 units = 16 + 4 = 20 units.

 

4 units = total number of fishes = 20 x 2 = 40.

 

 

 

1/4 of the fishes in an aquarium are goldfish.

There are 4 more guppies than goldfish in the aquarium.

The remaining 16 fishes are craps.

How many fishes are there in the aquarium?

 

 

Edited July 11, 2016 by Hiphiphoray

[Ender]

1/4 of the fishes in an aquarium are goldfish.

There are 4 more guppies than goldfish in the aquarium.

The remaining 16 fishes are craps.

How many fishes are there in the aquarium?

 

Stucked at the "The remaining 16 fishes are craps".

[Hiphiphoray]

For those whom are bored....lol

 

There are children in a funfair.

1/4 left after a while which consists of 40% of girls and 10% boys.

Of the remaining there are 18 boys more than girls.

How many girls were there at first?

[1fast1]

For those whom are bored....lol

 

There are children in a funfair.

1/4 left after a while which consists of 40% of girls and 10% boys.

Of the remaining there are 18 boys more than girls.

How many girls were there at first?

60 girls (and 60 boys also) at first.

[Hiphiphoray]

Steady.....a tiny winy more challenging one.

 

i had $10 & $2 in the ratio of 3/8

 

After exchanging two $10 with $2. The ratio became 1:3.

 

How much did i have?

[1fast1]

Steady.....a tiny winy more challenging one.

 

i had $10 & $2 in the ratio of 3/8

 

After exchanging two $10 with $2. The ratio became 1:3.

 

How much did i have?

 

24 tens and 64 twos = $368 at the start.

 

These are all very easy lah. But if you can't use algebra, have to rack brains for the stupid model.

Edited July 15, 2016 by Turboflat4

[SuPerBoRed]

24 tens and 64 twos = $368 at the start.

 

These are all very easy lah. But if you can't use algebra, have to rack brains for the stupid model.

 

x/y = 3/8

(x-2)/(y+2) = 1/3

 

still algebra ah.. 

 

ps... i still cannot formulate the first qn tho.. 

Edited July 15, 2016 by SuPerBoRed

[1fast1]

x/y = 3/8

(x-2)/(y+2) = 1/3

 

still algebra ah.. 

 

ps... i still cannot formulate the first qn tho.. 

 

x = B + G

1/4 x = 2/5 B + 1/10 B

 

Prove B = G. Rest easy.

[Hiphiphoray]

is the stupid model I cant figure out.........

 

Btw...Is the answer suppose to be $736 instead?

 

24 tens and 64 twos = $368 at the start.

 

These are all very easy lah. But if you can't use algebra, have to rack brains for the stupid model.

 

[Vid]

60 girls (and 60 boys also) at first.

Fail.

You are supposed to draw rectangles and show workings.

[Hiphiphoray]

Answering by Model as per PSLE marking scheme....like this tiok bor ????? 

 

[  1/4  ]     [  1/4  ][  1/4  ][  1/4  ]

   left                    remains

 

On the 1/4 left portion:

Girls: 40% = 2/5 or 2 units

Boys: 10% = 1/10 or 1 part

 

On the 3/4 remain portion:

Girls: 60% = 3/5 or 3 units

Boys: 90% - 9/10 or 9 parts

 

Putting this information back into the model:

 

 

        [  1/4  ]              [  1/4  ][  1/4  ][  1/4  ]

2 units + 1 parts           3 units + 9 parts  (which also means 1/4 = 1 unit + 3 parts)

 

Since now all are in 1/4 form, we can compare apple to apple: "Left 1/4" to "1/4 remaining"

 

[   1u   ][   1u   ][1p]

[   1u   ][1p][1p][1p]

 

Model tells us: 1u = 2p

 

Therefore of the remaining portion:

 

Girls = 3 units = 3 x 2p = 6p

Boys = 9p.

 

Since 18 boys more than girls:

 

Girls: [  6p  ]

Boys:[  6p  ][3p]

 

Therefore 3p = 18.   P = 6

 

Total No. of boys: 10p = 10 x 6 = 60

 

Total No. of girls: 5 units = 5 x 2p = 5 x 2 x 6 = 60

 

For those whom are bored....lol

There are children in a funfair.
1/4 left after a while which consists of 40% of girls and 10% boys.
Of the remaining there are 18 boys more than girls.
How many girls were there at first?

 

[Ender]

Steady.....a tiny winy more challenging one.

 

i had $10 & $2 in the ratio of 3/8

 

After exchanging two $10 with $2. The ratio became 1:3.

 

How much did i have?

u and p method.

 

3u - 2 = 1p

8u + 10 = 3p

 

9u - 6 = 3p

 

9u - 6 = 8u+10

u = 16

 

$10 => 3u => 48 pcs=> $480

$2 => 8u => 128pcs => $256

 

Total $736

[Ender]

For those whom are bored....lol

 

There are children in a funfair.

1/4 left after a while which consists of 40% of girls and 10% boys.

Of the remaining there are 18 boys more than girls.

How many girls were there at first?

The u and p method again. But I use G and B instead.

 

Let girls be 10G and Boys be 10B at the start.

 

10G +10B = 4(4G + 1B)= 16G + 4B

6B = 6G

B=G

9B - 6B = 18, since B = G

B = 6 =G

 

At first, number of girls is 10G = 60

[1fast1]

u and p method.

 

3u - 2 = 1p

8u + 10 = 3p

 

9u - 6 = 3p

 

9u - 6 = 8u+10

u = 16

 

$10 => 3u => 48 pcs=> $480

$2 => 8u => 128pcs => $256

 

Total $736

Not sure how you're getting this.

 

My answer was 24 tens and 64 twos.

 

24:64 = 3:8

 

Swapping 2 tens for 2 twos would cause the number of tens to decrease by 2 and the number of twos to increase by 2.

 

The ratio would become 22:66 = 1:3

 

As far as I can tell I'm satisfying the conditions.