Q&A - P5 Math

[shawnlim88]

hi all,

how to solve this Rosyth 2011 SA2 Paper 2, Qn16 using models.

Q16: Don and Larry had 315 stickers altogether. Don gave 1/4 of his stickers to Larry. After that, Larry gave 1/2 of all that he had to Don. In the end, Don had twice as many as Larry
a) How many stickers did Don have at first? ( 140 )
b) How many stickers did Larry have at last? ( 175 )

Suz855:

D[ ][ ] 210
L[ ]105

315/3=105 (b)

Since L give 1/2 away, half is left,
L: 105*2=210
105

Since D give 1/4, 3/4--> 105
1/4--> 35
Thus D 105+35=140

It is a working backward problem, modeling not v useful in solving, cheers

hi Suz855,
thanks for the reply. now i understand what you mean by working backwards.

how to solve this Rosyth 2011 SA2 Paper 2, Qn16 using models.

Q16: Don and Larry had 315 stickers altogether. Don gave 1/4 of his stickers to Larry. After that, Larry gave 1/2 of all that he had to Don. In the end, Don had twice as many as Larry

D has twice as Larry
2u
L: 1u total 315 so 1u=315/3 105
210
L: 105

Larry gave 1/2 of all that he had to Don.
so 210/2 =105
L: 105 + 105 = 210

Don gave 1/4 of his stickers to Larry
3/4 = 105, so 4/4(1u) = 105*4/3 = 140 ( nett is 35)
L : 210 - 35 = 175

thanks.
now i have to find a way to explain to my DS.

cheers

[shawnlim88]

papilion:

Hello, can someone help me to solve.

This can be done using guess and check, but is there any other method, like algebraic expression? thanks ! :imcool:

Mr Lim wants to buy some boxes of chocolates which are sold in boxes of 10 and 24. Each box of 10 pieces is sold for $5.35 and each box of 24 pieces is sold for $12.50.
Mr Lim and his class of 37 pupils will be given 2 pieces of chocolates each.

a) How many boxes of each type of chocolates should Mr Lim buy so that the number of pieces of chocolates left over is the least?

Ans: (a) ____ box(es) of 10 and ____ box(es) of 24

b) How much will Mr Lim pay for the chocolates? :stupid:

hi papilion
guess and check is the best
total choc needed = 37*2 = 74 chocs
10 box | 24 box | total | extra
5 (50) | 1 (24) | 74 | 0

a) 5 boxes of 10 and 1 box of 24
b) 10*5.35 + 12.50 = $66

hope it helps
shawnlim88

[Suz855]

[quote=\"papilion\"]Hello, can someone help me to solve.

This can be done using guess and check, but is there any other method, like algebraic expression? thanks ! :imcool:

Mr Lim wants to buy some boxes of chocolates which are sold in boxes of 10 and 24. Each box of 10 pieces is sold for $5.35 and each box of 24 pieces is sold for $12.50.
Mr Lim and his class of 37 pupils will be given 2 pieces of chocolates each.

a) How many boxes of each type of chocolates should Mr Lim buy so that the number of pieces of chocolates left over is the least?

Ans: (a) ____ box(es) of 10 and ____ box(es) of 24

b) How much will Mr Lim pay for the chocolates?

Use logic deduction,
Since it is cheaper to buy choc in box of 24 than 10 but at the same time if we buy the max number of boxes of 24 then box of 10 we will get more choc then require thus, best still workout both to compare
38x2=76

2box 2x12.50=25 (48).
3x5.35=16.05 (30)

A) 3 boxes of 10 n 2 boxes of 24
B) $41.05

Not a better option, if we consider buying all in boxes of 24 first
76/24=3r4
3x12.5=37.5
37.5+5.35=42.85

[papilion]

@Suz855 and @shawnlim88

:thankyou:

thanks to both of you for the alternative answers,

cause during exam, kid can guess and check, but it is time consuming and likely to make careless mistakes, if not properly tabulated. :xedfingers:

[mama_10]

Hi,


Need help on below question :-
Ryan, Sean and Tristan collected a total of 492 bookmarks. Ryan had the least number of bookmarks. At first, the ratio of the number of bookmarks Sean had to the ratio of bookmarks Tristan had was 3:2. After Ryan and Sean each gave away 50% of their bookmarks, Sean had 74 bookmarks more than Ryan. If the three boys had 326 bookmarks left, howmany bookmarks did Ryan have in the beginning?

Thank you.

[mama_10]

Need help on this question too.

Mrs Ho had 2 boxes of oranges.At first, there are 228 more oranges in Box A than in Box B. Mrs Ho then transfers 600 oranges from Box A toBox B. Now, Box B contained 5 times as many oranges as Box A. How many oranges are there ineach box at first?

Thank you.

[dazzlego]

mama_10:

Hi,


Need help on below question :-
Ryan, Sean and Tristan collected a total of 492 bookmarks. Ryan had the least number of bookmarks. At first, the ratio of the number of bookmarks Sean had to the ratio of bookmarks Tristan had was 3:2. After Ryan and Sean each gave away 50% of their bookmarks, Sean had 74 bookmarks more than Ryan. If the three boys had 326 bookmarks left, howmany bookmarks did Ryan have in the beginning?

Thank you.

Hi mama_10,

At first,
R: 2p
S: 6u
T: 4u
2p + 10u -> 492
In the end,
R: 1p
S: 3u
T: 4u
1p + 7u -> 326
2p + 14u -> 652

4u -> 652 - 492 = 160 (T)
1u -> 40
S (in the end): 3u -> 3 x 40 = 120
R (in the end): 120 - 74 = 46
R (at first): 46 x 2 = 92

Hope it helps
Cheers :celebrate:

[mama_10]

:thankyou:

[mama_10]

Hi Dazzlego, thank you very much.

I worked out the solution too but not sure if its correct

R S T
?
3 : 2 Total 492

-50% -50% ? Total 326
<74
-------------------------------------------
166

x x+74 166

x+(x+74) = 166
2x=92
x=46
Ryan @ beginning = 46x2=92.

[dazzlego]

mama_10:

Need help on this question too.

Mrs Ho had 2 boxes of oranges.At first, there are 228 more oranges in Box A than in Box B. Mrs Ho then transfers 600 oranges from Box A toBox B. Now, Box B contained 5 times as many oranges as Box A. How many oranges are there ineach box at first?

Thank you.

Easier way to solve this question is to draw the model.
Here is my handdrawing model. Hope it still can be seen and understood

http://i48.tinypic.com/14age4y.jpg\">

4u -> 372 + 600 = 972
1u -> 243
At first,
A: 243 + 600 = 843
B: 243 + 372 = 615



Another way is to assume
At first,
A: 1u + 228
B: 1u

So in the end,
A: 1u - 372
B: 1u + 600

5 x (1u - 372) = 1u + 600
5u - 1860 = 1u + 600
4u = 2460
1u = 615

B: 615
A: 615 + 228 = 843