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[CJM]

Mr Su bought 56 apples and pears altogether. He paid 3 times as much for the apples as for the pears. But the price of each pear was twice that for each apple.

How many apples and how many pears did he buy ?

[iFruit]

CJM:

Mr Su bought 56 apples and pears altogether. He paid 3 times as much for the apples as for the pears. But the price of each pear was twice that for each apple.

How many apples and how many pears did he buy ?

Let Apples bought by Su = a
Pears = 56-a
ratio of apples:pears = a:56-a

Ratio Apple's price: Pear's price = 1:2

Ratio of money paid for Apples:pears = a x1 : (56-a) x2 = 3:1

so a/(56-a)2 = 3:1---> a = (56-a)6----> 7a = 6x56--> a =6x8 = 48

So Apples = 48; Pears = 8

[Dharma]

CJM:

Mr Su bought 56 apples and pears altogether. He paid 3 times as much for the apples as for the pears. But the price of each pear was twice that for each apple.

How many apples and how many pears did he buy ?

Cost of apples = 1u x 1 = 1u
Cost if pears = 1p x 2 = 2p

1u = 6p

1u + 1p = 56
7p = 56
1p = 8 ( No. of pears)
1u = 48 ( No. of apples

[Muffins]

tianzhu:

Hi muffins


2 5/6 is correct.

It helps to draw a diagram for this question.

Beginning from zero hours, he took the cough mixture 7 times on the first day or first 24 hours, followed by 6 times on the second day and 5 times on the third day.

Best wishes

oh. thanks for all the help Tianzhu!

[yongjianrong]

tamarind:

Thanks in advance. My friend sent me this question :


I am a three digit number. The digit in the tens place is 6 less than the digit in the hundreds place. The tens place is 2 more than the digit in the ones place. If the digit in the ones place is 5, then I am the number ______.

If the digit in the ones place is 5, then the digit in the tens place will be 7.

The digits in tens place is 6 less than digit in hundreds place. 7+6=13???

Then it'll be 4 digit? Weird.

[capricorn28]

Hi,


Please help with these questions.

1) A cuboid has a square base of side 16cm. The ratio of its breadth to its height is 2:5 respectively. Find its volume.

2)The diagram below, not drawn to scale, shows 2 containers. 3960 ml of water was poured into them. The heights of water in both containers are the same.
What is the height of water in each container?

http://postimage.org/image/6t0gg5qc/

3) The figure below, not drawn to scale, shows a rectangular garden measuring 112m by 87m. Three triangular plots of land A, B and C each has different plants growing in them. Four tracks (unshaded), each of constant width and of different measurements, lined the garden as shown below.
(a) Find the area of triangular plot B.
b) Find the total area covered by the four tracks.


http://postimage.org/image/56dvxdk4/

[iFruit]

capricorn28:

Hi,


Please help with these questions.


3) The figure below, not drawn to scale, shows a rectangular garden measuring 112m by 87m. Three triangular plots of land A, B and C each has different plants growing in them. Four tracks (unshaded), each of constant width and of different measurements, lined the garden as shown below.
(a) Find the area of triangular plot B.
b) Find the total area covered by the four tracks.


http://postimage.org/image/56dvxdk4/

a) The base of the traingle B = 112 - 4.2 - 2 = 105.8
Height of triangle B = 87 - 3.5 -2.8 = 80.7

Area = 1/2 bxh = 1/2x 105.8 x 80.7


b) Sum of the bases ( the lines parallel to length of rectangle) of plot A & Plot C = 112 - 4.2 - 2 = 105.8
Height of triangles A & C = 87 - 3.5 -2.8

Area of A+c = 1/2 x105.8 x 80.7

Area covered by tracks = Area of rectangle - Area of (A+B+C)

112x87 - (1/2x 105.8 x 80.7) - 1/2 x105.8 x 80.7

= 112x87 - 105.8x80.7

[Vanilla Cake]

capricorn28:

Hi,

Please help with these questions.

Hi capricorn28,
Thanks for sharing these P5 questions and these are the solutions worked out by my P5 younger sister. Pls advise if her answers are correct or not.

capricorn28:

1) A cuboid has a square base of side 16cm. The ratio of its breadth to its height is 2:5 respectively. Find its volume.

Height = 16/2x5=40 cm
Volume = 16x16x40=10240 cm³

capricorn28:

2)The diagram below, not drawn to scale, shows 2 containers. 3960 ml of water was poured into them. The heights of water in both containers are the same.What is the height of water in each container?
http://postimage.org/image/6t0gg5qc/

Total base area of both containers = (20x6)+(24x10)=120+240 = 360 cm²
Height of water in each container = 3960÷360 = 11 cm

capricorn28:

3) The figure below, not drawn to scale, shows a rectangular garden measuring 112m by 87m. Three triangular plots of land A, B and C each has different plants growing in them. Four tracks (unshaded), each of constant width and of different measurements, lined the garden as shown below.
(a) Find the area of triangular plot B.
b) Find the total area covered by the four tracks.
http://postimage.org/image/56dvxdk4/

Base of B=112-4.2-2=105.8 m
Height of B = 87-3.5-2.8=80.7 m
Area of B = ½x105.8x80.7 = 4269.03 m²
(a) Area of triangular plot B = 4269.03 m²

Area of shaded area = 105.8x80.7=8538.06 m²
Area of tracks = 112x87-8538.06 = 1205.94 m²
(b) Total area covered by the four tracks = 1205.94 m²

[wkong]

Hi All,


Need helps for the following questions:

1. Last year, the enrolment of school X was 90% that of school Y. This year, the enrolment of school X decreased by 220 while that of school Y increased by 200. In the end, the enrolment of school X is 70% that of school Y this year, what was the enrolment of each school last year?

2. Daniel’s savings was 75% as much as Annabel. After Annabel saved another $40 and Daniel saved another $20, Daniel’s savings became 5/8 as much as Annabel’s savings. Find Daniel’s savings at first.

3. A farmer had 5/8 as many chickens as ducks at first. After selling 50 chickens and 8 ducks, he had 25% as many chickens and ducks. How many ducks did the farmer have at first?

4. In May, Mrs Tan’s salary was 90% of Mrs Ismal’s salary. In June, Mrs Tan’s salary increased by $400 while Mrs Ismal’s salary increased by $600. If Mrs Tan’s salary became 80% of Mrs Ismal’s salary, what was Mrs Tan’s salary in May?

Thanks.

[Vanilla Cake]

wkong:

1. Last year, the enrolment of school X was 90% that of school Y. This year, the enrolment of school X decreased by 220 while that of school Y increased by 200. In the end, the enrolment of school X is 70% that of school Y this year, what was the enrolment of each school last year?

Last year
School X : School Y
90% : 100%
9u : 10u

This year
School X : School Y
9u-220 : 10u+200
70% : 100%
7 : 10

10(9u-220)=7(10u+200)
90u-2200=70u+1400
20u=3600
u=180

Last year
School X -> 9u= 9x180=1620
School Y -> 10u=10x180=1800

Last year, the enrolment of school X and school Y was 1620 and 1800 respectively.

wkong:

2. Daniel's savings was 75% as much as Annabel. After Annabel saved another $40 and Daniel saved another $20, Daniel's savings became 5/8 as much as Annabel's savings. Find Daniel's savings at first.

Before
D : A
75% : 100%
3 : 4

Change
D : A
+20 : +40

After
D : A
5 : 8

D : A
3u : 4u
+20 : +40
5p : 8p

D : A
24u : 20u
+160 : +200
40p : 40p

24u+160=20u+200
4u=40
u=10

At first, Daniel's savings = 3u = 3x10 = $30