Q&A - P5 Math

[Applelyn1]

Hi I need help on this


Jewel bought some strawberries and wanted to put them into boxes. if she put 4 strawberries into each box, there would be 3 strawberries left. If she put 6 strawberries into each box, she would have 5 strawberries left.

A) if the number of strawberries was between 100 and 150, how many strawberries did Jewel buy?

B) Jewel decided to put all the strawberries equally into bass of 16. find th minimum number of bags that she would need.

[MathIzzzFun]

Applelyn1:

Hi I need help on this


Jewel bought some strawberries and wanted to put them into boxes. if she put 4 strawberries into each box, there would be 3 strawberries left. If she put 6 strawberries into each box, she would have 5 strawberries left.

A) if the number of strawberries was between 100 and 150, how many strawberries did Jewel buy?

B) Jewel decided to put all the strawberries equally into bass of 16. find th minimum number of bags that she would need.

Hi
More than 1 possible answers -
If packed into boxes of 4, 3 strawberries left --> if we add 1 more, the remainder can be packed into another box of 4

If packed into boxes of 6, 5 strawberries left --> if we add 1 more, the remainder can be packed into another box of 6.

So, if we add 1 more strawberry, the total number of strawberries can be packed into boxes of 4 or boxes of 6 exactly with no remainder --> total number is a multiple of both 4 & 6 --> multiple of 12
Multiples of 12 in the range of 100 to 150 --> 108, 120, 132, 144
Possible number of strawberries --> 107, 119, 131, 143



cheers.

[Applelyn1]

MathIzzzFun:

Applelyn1:

Hi I need help on this


Jewel bought some strawberries and wanted to put them into boxes. if she put 4 strawberries into each box, there would be 3 strawberries left. If she put 6 strawberries into each box, she would have 5 strawberries left.

A) if the number of strawberries was between 100 and 150, how many strawberries did Jewel buy?

B) Jewel decided to put all the strawberries equally into bass of 16. find th minimum number of bags that she would need.

Hi
More than 1 possible answers -
If packed into boxes of 4, 3 strawberries left --> if we add 1 more, the remainder can be packed into another box of 4

If packed into boxes of 6, 5 strawberries left --> if we add 1 more, the remainder can be packed into another box of 6.

So, if we add 1 more strawberry, the total number of strawberries can be packed into boxes of 4 or boxes of 6 exactly with no remainder --> total number is a multiple of both 4 & 6 --> multiple of 12
Multiples of 12 in the range of 100 to 150 --> 108, 120, 132, 144
Possible number of strawberries --> 107, 119, 131, 143



cheers.

Thanks, MathIzzzFun.

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[mummyv]

Hello


It's time from SA2 prep & Would appreciate some help solving the following questions 1b and 2

1. Ms Tan bought 4 times as many toy guns as teddy bears. She spent $1750 altogether. A toy gun cost $10 less than a teddy bear. Total cost of toy guns was $490 more than total cost of teddy bears.
a. How much did Ms Tan spend on teddy bears (Ans$630)
b. How much did one teddy bear cost (Ans $18)

2. There are 9 stack of chairs. There are 12 chairs in each stack. Mohan places each chair equally spaced apart to for a perimeter of a rectangle. If the number of chairs places along the length is thrice the number of chairs placed along the breadth, how many chairs are placed along one of the lenghts of the rectangle. (Ans 42)

Thanks in advance all the maths gurus

[MathIzzzFun]

mummyv:

Hello


It's time from SA2 prep & Would appreciate some help solving the following questions 1b and 2

1. Ms Tan bought 4 times as many toy guns as teddy bears. She spent $1750 altogether. A toy gun cost $10 less than a teddy bear. Total cost of toy guns was $490 more than total cost of teddy bears.
a. How much did Ms Tan spend on teddy bears (Ans$630)
b. How much did one teddy bear cost (Ans $18)

2. There are 9 stack of chairs. There are 12 chairs in each stack. Mohan places each chair equally spaced apart to for a perimeter of a rectangle. If the number of chairs places along the length is thrice the number of chairs placed along the breadth, how many chairs are placed along one of the lenghts of the rectangle. (Ans 42)

Thanks in advance all the maths gurus

Hi
Q1.
Amount spent on teddy bears --> $(1750-490)/2 = $630
Amount spent on toy guns --> $1750 - $630 = $1120
Ms Tan bought 4 times as many toy guns as teddy bears
--> 4 groups of toy guns, 1 group of teddy bear
1 group of toy guns --> $ 1120/4 = $280
$630-$280= $350 --> 1 group of teddy bears cost $350 more than 1 group of toy guns
Number of teddy bears/toy guns in 1 group --> $350/$10 = 35
Cost of 1 teddy bear = $630/35= $18

Q2.
You may want to draw a rectangle with 9 beads on each length and 3 beads on each breadth (total 20 beads), when the number of beads on 1 length and 1 breadth are added, we add 2 more beads to 1/2 the total number ie 9+3=12 = 10 + 2.

Total number of chairs --> 12 x 9 = 108
Length + breadth --> 4 units --> (108/2) + 2
1 unit --> 56 / 4 = 14
Length --> 3 units --> 3 x 14 = 42 chairs.

cheers.

[Zack7]

mummyv:

Hello


It's time from SA2 prep & Would appreciate some help solving the following questions 1b and 2

1. Ms Tan bought 4 times as many toy guns as teddy bears. She spent $1750 altogether. A toy gun cost $10 less than a teddy bear. Total cost of toy guns was $490 more than total cost of teddy bears.
a. How much did Ms Tan spend on teddy bears (Ans$630)
b. How much did one teddy bear cost (Ans $18)

2. There are 9 stack of chairs. There are 12 chairs in each stack. Mohan places each chair equally spaced apart to for a perimeter of a rectangle. If the number of chairs places along the length is thrice the number of chairs placed along the breadth, how many chairs are placed along one of the lenghts of the rectangle. (Ans 42)

Thanks in advance all the maths gurus

let x be number of teddy bear, no of guns = 4x
let y be cost of each teddy bear, cost of gun = y-10

so

4x(y-10) + xy = 1750 --- eq1
4x(y-10) - xy = 490 --- eq2

eq1 - eq2

2xy = 1260
xy = 630 (cost of bears)

cost of guns = 1750 - 630 = 1120 = 4x(y-10)

so sub x = 630/y into the above

so

1120 = 4(630/y) (y-10)
y = 18 bucks... thats pretty expensive.



question 2

this is a perimeter question

let breathe be x
length = 3x

perimeter = 8x

9 x 12 = 108

but notice that based on the way we formalise the answer, i.e we let breathe be x, length be 3x,

if the perimeter is 8x, then we would have double counted the 4 corners. i.e, our 4 corner chairs can act as both the length and the breadth.

so the total chairs is 9 x 12 + 4 = 112.

so 8x = 112, x = 14, length = 3x = 42.



* i have to say these questions are quite tricky, harder than the questions i see at the p6 section

[MathIzzzFun]

imacsg:

Can someone help?

Mark's current age is 1/4 of Ronald's. In 24 years' time, Ronald's age will be 14 years less than twice Mark's age. What is Ronald's age now?

Thanks,

Hi
Now:
Mark's age --> 1 unit
Ronald's age --> 4 units

24 years' time:
Ronald's age --> 4 units + 24
Mark's age --> 1 unit + 24
4 units + 24 = 2 units + 48 - 14
1 unit --> 5

Ronald's age now --> 20

cheers.

[mathnoobs]

Zack7:

* i have to say these questions are quite tricky, harder than the questions i see at the p6 section

yes, pity our poor children.... so little time for play now....it's just mug, mug, mug...