Q&A - P5 Math

[Michaelia0816]

11/2=5/1/2

Use guess and check!
Sweets not odd number I guess so

[mummyv]

tianzhu:

mummyv:

Hi, pls help w this qn

John had 993 tables and chairs at first. After he sold 2/5 of the tables and 5/8 of the chairs, he had 459 tables and chairs left. How many tables did he sell? :slapshead: thanks for the help

Hi

Good Morning.

This question touches on simultaneous concepts.

If a student is familiar with SE, he/she may use it.

If not, he/she may solve the two equations pictorially,use the “Alphabet Method” or represent the variables with \"Units and Parts\"

Tables sold ------ TT
Tables left ------ TTT

Chairs sold ------ CCCCC
Chairs left ------ CCC

TTT + CCC ------- 459, hence T+C ------- 153

TT + CCCCC ----- 534 (993 – 459), therefore C -------- 76

T ------- 153 – 76 ------ 77

Tables sold ------- 2*77 ------- 154.

Best wishes

Thanks tianzhu

[Michaelia0816]

Good method of solving it, Tianzhu! Best explainer!

[Michaelia0816]

i suppose someone had answered it, is it

[clblinym]

Pls help with the following question with thanks. It is from Henry Park, P5 2011, SA1, Paper2, Qn15.



http://i42.tinypic.com/302c193.png\">

and the solution is as follows:
http://i40.tinypic.com/30rusty.png\">

From the solution, I do not understand why the area of AOB + COD = that of AOC + BOD. Can somebody help to explain? TIA. Amy

[tianzhu]

clblinym:

From the solution, I do not understand why the area of AOB + COD = that of AOC + BOD. Can somebody help to explain? TIA. Amy

Hi

The bases of AOB + COD are the same as the length of the rectangle (AB=CD).The sum of the heights of these two triangles is equal to the breadth of the rectangle (AC and BD).Hence, their combined areas is equal to half the area of the rectangle.

As an illustration, let the height of triangle AOB be h1 and the height of triangle COD be h2.

Area of triangle AOB ------ ½*AB*h1
Area of triangle COD ------ ½ *CD*h2

Both triangles have similar bases as AB=CD.
AB=CD=length of rectangle ABCD

Their sum of areas is equal to ½ *length of rectangle*(h1+h2), where (hi+h2) is equal to the breadth (AC and BD) of the rectangle.

Therefore, their areas is equal to ½ *CD*BD which is equal to half the area of the rectangle.

The same logic is applicable to the triangles AOC and BOD.

AOB + COD = half the area of rectangle

AOC + BOD = half the area of rectangle

Hence,the area of AOB + COD = that of AOC + BOD.

Best wishes

[SAHMwith2boys]

Hakim and Suresh had some money in the ratio of 13 : 8. When they each received $133, Hakim will have 4 times as much money as Suresh. How much did Hakim had at first?


Answer provided was $273.

Appreciate any help in giving the worked out solution. Thanks.

[tianzhu]

SAHMwith2boys:

Hakim and Suresh had some money in the ratio of 13 : 8. When they each received $133, Hakim will have 4 times as much money as Suresh. How much did Hakim had at first?


Answer provided was $273.

Appreciate any help in giving the worked out solution. Thanks.

Hi

Good Morning.

I think \"received\" should be read as \"gave away\".

Hakim and Suresh had some money in the ratio of 13 : 8. When they each gave away $133, Hakim will have 4 times as much money as Suresh. How much did Hakim had at first?

The answer for the revised question is 273.

Best wishes

[clblinym]

tianzhu:

clblinym:

From the solution, I do not understand why the area of AOB + COD = that of AOC + BOD. Can somebody help to explain? TIA. Amy

Hi

The bases of AOB + COD are the same as the length of the rectangle (AB=CD).The sum of the heights of these two triangles is equal to the breadth of the rectangle (AC and BD).Hence, their combined areas is equal to half the area of the rectangle.

As an illustration, let the height of triangle AOB be h1 and the height of triangle COD be h2.

Area of triangle AOB ------ ½*AB*h1
Area of triangle COD ------ ½ *CD*h2

Both triangles have similar bases as AB=CD.
AB=CD=length of rectangle ABCD

Their sum of areas is equal to ½ *length of rectangle*(h1+h2), where (hi+h2) is equal to the breadth (AC and BD) of the rectangle.

Therefore, their areas is equal to ½ *CD*BD which is equal to half the area of the rectangle.

The same logic is applicable to the triangles AOC and BOD.

AOB + COD = half the area of rectangle

AOC + BOD = half the area of rectangle

Hence,the area of AOB + COD = that of AOC + BOD.

Best wishes

Tianzhu,

Now I and my son understand. I really appreciate your prompt and detailed explanation. Thank you very much.

[SAHMwith2boys]

tianzhu:

SAHMwith2boys:

Hakim and Suresh had some money in the ratio of 13 : 8. When they each received $133, Hakim will have 4 times as much money as Suresh. How much did Hakim had at first?


Answer provided was $273.

Appreciate any help in giving the worked out solution. Thanks.

Hi

Good Morning.

I think \"received\" should be read as \"gave away\".

Hakim and Suresh had some money in the ratio of 13 : 8. When they each gave away $133, Hakim will have 4 times as much money as Suresh. How much did Hakim had at first?

The answer for the revised question is 273.

Best wishes

Oh I see, thanks for the help.