Q&A - P5 Math

[Easy-going]

Hi

Please help me in this P5 Nanyang Primary School Maths Question:
Question16) Mr Chen bought some pencil cases and bags for the children at a gathering. The price of each pencil case was $2.40 while the price of each bag was $10.80. For every 4 bags bought, he was given one such pencil case for free. After receiving the free pencil cases, the number of pencil cases was 1/3 of the number of bags bought.
If Mr Chen paid a total of $1584, how much more did he pay for the bags than the pencil cases? :? :? :?:

Thanks in advance :thankyou:

[lapapillion]

dazzlego:

The shaded area in this diagram is actually the number of students.
Therefore, Blue shaded = Red shaded
Blue shaded = 24 x 1 unit + 32 x 5 = 24 units + 160
Red shaded = 26 x 1 unit + 26 x 5 = 26 units + 130
2 units = 30
1 unit = 15
total no. of class now: 15 + 5 = 20
average no. of students now = 26
total no. of students = 20 x 26 = 520

Cheers :celebrate:

Wow, brilliant! I never thought of that!

[shawnlim88]

Easy-going:

Hi

Please help me in this P5 Nanyang Primary School Maths Question:
Question16) Mr Chen bought some pencil cases and bags for the children at a gathering. The price of each pencil case was $2.40 while the price of each bag was $10.80. For every 4 bags bought, he was given one such pencil case for free. After receiving the free pencil cases, the number of pencil cases was 1/3 of the number of bags bought.
If Mr Chen paid a total of $1584, how much more did he pay for the bags than the pencil cases? :? :? :?:

Thanks in advance :thankyou:

every 4 bags 1 free pencil case
no of bags = 3 x No of pencil case
so Mr chen must buy 12bags and 1 pencil case ( +3 free )
ie: 12 bags = 3 x ( 1 -bought + 3 - free)
he must buy 12 bags and 1 pencil case ( 1 set )
cost of 1 set : = 12*10.80 + 2.40 = 132
Either:
A
>>>>>>>>>>>
he can buy 1584 / 132 sets = 12 sets
each set, he pay 12 bags = $129.6
1 pencil case: 2.4
so each set he pays ( 129.6 - 2.4) more $127.2

he bought 12 sets: so 12 * 127.2 = 1526.4
>>>>>>>>>>>>>
or B:
>>>>>>>>>>>>>>>
he can buy 1584 / 132 = 12 sets of ( 12 bags & 1 pencil case)
for 12 sets: he paid: 12*12* 10.80 for bags = 1555.2 for bags
for 12 sets: he paid : 12 * 1 * 2.4 for pencil cases: = 28.8 for pencil cases

so he paid ( 1555.2 - 28.8 ) = 1526.4 more for bags than pencil case.
>>>>>>>>>>>>>>>

Answer: 1526.4

[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 )

thanks.

[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

[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:

[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: