Tutor MathsGuru: Ask me for your burning Maths questions!

[CoffeeCat]

tisha:

Is it possible to solve these without algebra? These are P5 Questions from past year exam papers for SA1.

I believe algebra is only taught in P6, so just curious how come these questions can be asked for SA1 of P5 :?

Ken and Patrick had $625 at first. After buying some books, the amount of money Ken has left is 4 times the amount of money he spent and the amount of money Patrick has left is 5 times the amount of money he spent. Find the amount of money Patrick had at first if they have a total of $510 left.


There are 85 plates of chicken rice for 80 people. Each adult eats 2 plates of chicken rice and every 3 children share 1 plate of chicken rice. Find the number of adults and children respectively.

For qns 2. One common non-algebra way to solve such algebraical questions is to use this heuristic called \"make an assumption\".
First assume there are 80 adults. Then these 80 adults will eat 160 plates of rice.
Now we must try to find the number of children.
Notice that if we interchange 3 adults for 3 children, there will be a decrease in (2*3 -1) = 5 plates of rice from the total.
(if you are not convinced you can try 77 adults, 75 adults to see the pattern)
For every group of interchange of 3 adults for 3 children, we are 5 plates nearer the true total of 85. therefore, the number of children is 3*(160-85)*5 =45.

[CoffeeCat]

Daddy:

Hi Maths Monster,


I have question to ask you. Thanks

1. A worker managed to complete 2/5 of his work on the first day and 3/4 of the remainder on second day. How much work was completed at the end of second day?


Daddy.
Thanks alls.

3/4 of the remainder of job = 3/4 * (1- 2/5) = 9/20 of work
amt of work completed at end of second day = 2/5 + 9/20 = 8/20 + 9/20 = 17/20

[small]

Hi tisha,


Hope this clear enough.


Question:
Ken and Patrick had $625 at first. After buying some books, the amount of money Ken has left is 4 times the amount of money he spent and the amount of money Patrick has left is 5 times the amount of money he spent. Find the amount of money Patrick had at first if they have a total of $510 left

Total of $510 left -----> 4 units of Ken + 5 units of Patirck = 510 ---------------> (1)

The amount they spent -----> 1 unit of Ken + 1 unit of patrick = (625-510) = 115
4 units of Ken + 4 units of patrick = 115 x 4 = 460
4 units of Ken + 4 units of patrick = 460 ---------------> (2)

By substitute (2) in (1):
460 + 1 unit of Patrick = 510
1 unit of Patrick = 50

Patrick had 6 units at first.
6 x 50 = 300

Patrick had $300 at first.

[CoffeeCat]

Herbie:

Thanks VC and Dharma Can help with the qn below?


There were some chairs in a school hall. 1/2 of the chairs were placed in short rows while 2/5 of the chairs were placed in long rows. The rest of th chairs were stacked at the back of the hall.

There were 15 short rows and long rows. If each short row contained 7 chairs less tha each kong rows, how many chairs were there in the wall altogether?

Tx

Is there an answer key for this?
At first based on this \"There were 15 short rows and long rows.\", I take it that there are a total of 15 rows altogether.
Unless I am mistaken, this interpretation though interesting, only secondary mo technique can yield the answer of 630. Otherwise guess-and-check will be much easier. Unless this is what the school intended to test upon.
So perhaps it would make sense if it was 15 short and 15 long rows.

[Herbie]

There were some chairs in a school hall. 1/2 of the chairs were placed in short rows while 2/5 of the chairs were placed in long rows. The rest of th chairs were stacked at the back of the hall.


There were 15 short rows and 8 long rows. If each short row contained 7 chairs less tha each kong rows, how many chairs were there in the wall altogether.

Sorry, there is a typo error. There shld be 8 long rows.

[Dharma]

Herbie:

There were some chairs in a school hall. 1/2 of the chairs were placed in short rows while 2/5 of the chairs were placed in long rows. The rest of th chairs were stacked at the back of the hall.


There were 15 short rows and 8 long rows. If each short row contained 7 chairs less tha each kong rows, how many chairs were there in the wall altogether.

Sorry, there is a typo error. There shld be 8 long rows.

Short rows : 15 rows x (1u) chairs per row = 15u chairs
Long rows : 8 rows x (1u + 7) chairs per row = (8u + 56) chairs

No. of chairs in short rows = ½ of total chairs (50%)
No. of chairs in long rows = 2/5 of total chairs (40%)
No. of chairs stacked = 10%

(8u + 56)/ 15u = 4/5
5(8u + 56) = 4(15u)
40u + 280 = 60u
20u = 280
1u = 14

Total no. of chairs = 30u = 30 x 14 = 420

[YLH88]

[quote]YLH88 wrote:

1) In the figure below, which is not drawn to scale, 1/5 of the circle is shaded. The ratio of the area of the square to the sum of area of the rectangle and the circle is 1 : 2. 60% of the square is shaded. 1/3 of the rectangle is shaded. What is the ratio of the area of the circle to the square to the rectangle ?

Vanilla Cake wrote :
Area of square : Area of rectangle+Area of circle
1 : 2
10 : 20 [/quote]
Hi Vanilla Cake,

May I know why you change the ratio for area of square : area of Rectangle + Area of circle from 1 : 2 to 10 : 20 ?

[kitty2]

Please help to solve these questions,thanks.


1)Audrey and Belle have some money each.If Audrey spends $18 and Belle spends $24 each day.Audrey will have $25 left.when belle has spent all her money.If Audrey spends $13 and belle spends $30 each day,Audrey will still have $139 left.when Belle has spent all her money.How much money do they have altogether?

2)Figure No of cakes

1 1
2 4
3 10


a)How many cubes will ther be in Fig 6?

b)Which figure will have 120 cubes?

[Vanilla Cake]

CoffeeCat:

Herbie:

Thanks VC and Dharma Can help with the qn below?


There were some chairs in a school hall. 1/2 of the chairs were placed in short rows while 2/5 of the chairs were placed in long rows. The rest of th chairs were stacked at the back of the hall.

There were 15 short rows and long rows. If each short row contained 7 chairs less tha each kong rows, how many chairs were there in the wall altogether?

Tx

Is there an answer key for this?
At first based on this \"There were 15 short rows and long rows.\", I take it that there are a total of 15 rows altogether.
Unless I am mistaken, this interpretation though interesting, only secondary mo technique can yield the answer of 630. Otherwise guess-and-check will be much easier. Unless this is what the school intended to test upon.
So perhaps it would make sense if it was 15 short and 15 long rows.

Hi CoffeeCat
Refer to Herbie's post dated 05 May 2010 08:10, this question was wrongly typed but it is still possible to solve using basic algebra at P6 level to solve the wrongly typed question by Herbie. Correct me if I am wrong and pls check the logic flow of my workings.

There were some chairs in a school hall. 1/2 of the chairs were placed in short rows while 2/5 of the chairs were placed in long rows. The rest of th chairs were stacked at the back of the hall.
There were 15 short rows and long rows. If each short row contained 7 chairs less tha each kong rows, how many chairs were there in the wall altogether?

Look for the unknown which is the total number of chairs in the hall and assume that it is C.

Number of Chairs
Short rows: 1/2 = 50% -> 0.5C
Long rows: 2/5 = 40% -> 0.4C

Assume the number of short rows to be S and the number of long rows to be 15-S

Difference between the chairs in the short rows and long rows = 0.5C-0.4C = 0.1C
Difference between number of chairs in a short row and a long row = 7

7S = 0.1C
35S = 0.5C
28S = 0.4C

Form an equation based on info given by the question:
0.5C/S + 7 = 0.4C/(15-S)
(0.5C+7S)/S = 0.4C/(15-S)
(35S+7S)/S = 28S(15-S)
42S/S = 28S/(15-S)
28S² = 42S(15-S)
28S² = 630S-42S²
70S² = 630S
70S = 630
S = 9

Sub S = 9 into 0.5C/S + 7 = 0.4C/(15-S)

0.5C/9 + 7 = 0.4C(15-9)
0.5C/9 + 7 = 0.4C/6
C + 126 = 1.2C
0.2C = 126
C = 630

There were 630 chairs in the hall altogether.

Submitted by VC's mum.

[Vanilla Cake]

kitty2:

1)Audrey and Belle have some money each.If Audrey spends $18 and Belle spends $24 each day.Audrey will have $25 left.when belle has spent all her money.If Audrey spends $13 and belle spends $30 each day,Audrey will still have $139 left.when Belle has spent all her money.How much money do they have altogether?


2)Figure No of cakes

1 1
2 4
3 10

a)How many cubes will ther be in Fig 6?
b)Which figure will have 120 cubes?

Hi kitty2,
Pls refer to the links below while waiting for Mathsguru's solutions:

http://psle2010a.blogspot.com/2010/03/14.html
http://psle2010a.blogspot.com/2010/03/mathsp62009sa1p2ny.html
http://www.onsponge.com/forum/35-thinkingmathonsponge/2528-p6-maths.html
http://www.onsponge.com/forum/35-thinkingmathonsponge/1917-p6-maths-2009-nanyang-sa1-paper-2.html

Submitted by VC's mum.