Tutor MathsGuru: Ask me for your burning Maths questions!
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tisha:
Oh yea opps what a typo. Sorry for the late reply. First we assume that there 80 adults and calculate the total. Then do the same for 77 adults+ 3 children , and 74 adults + 6 children. You will notice the pattern. The total will get smaller and nearer the required total.Thank you very much coffeecat for the non algebric solution.
But I'm not so clear on this part
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
I suppose it is a typo and should be 3*(160-85)/5, but I still don't understand it. Sorry to bother you so much.
160-85 is the no. of dishes which must be lost during our interchanges.
And since we lost 5 dishes for every 1 interchange, we can find the number of interchanges by (160-85)/5. Then remember for every interchange 3 children are added to the group. So the number of children will be 3*(160-85)/5. -
YLH88:
Hi YLH88,Hi VC's mum,
No worries, there is no need to apologies, really. In fact, I'm very very grateful for all the help the she, mathsguru and all other parents have given in this forum. Greatly appreciated!
Sorry, i'm still abit blur in this question. Is it possible to multiple by 5 instead of 10 ? ie 5 x 3/5 = 3 ?
Thank you.
I am VC. There is no difference if you prefer to use 5 instead of 10.
Pls see my revised workings as per your request:
Area of square : Area of rectangle+Area of circle
1 : 2
5 : 10
Shaded area of square = 60% = 3/5 = 3/5x5units = 3 units
1/3R+1/5C = 3 units
(R=Rectangle and C = Circle)
Multiply by 3
R+3/5C = 9 units - (1)
R+C = 10 units - (2)
(2)-(1)
2/5C = 1 unit
C = 5/2x1 = 2.5 units
R=10units-2.5units = 7.5 units
C : S : R
2.5 : 5 : 7.5
5 : 10 : 15
1 : 2 : 3
Ratio of the area of the circle to the square to the rectangle = 1:2:3
Hope this helps you to understand and pls feel free to tell me which part that still makes you blur.Thanks. -
Vanilla Cake:
Hi CoffeeCat
Is there an answer key for this?CoffeeCat:
[quote=\"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
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.
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.[/quote]Interestingly u got the answer. But I don't understand how you can get this.
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 -
Dear maths guru
Please help me with the following question:
1) The distance between village A and Village B is 360km. John drove from Village A to Village B and completed the journey in 4 hours. Peter drove from Village B to Village A and completed the journey in 6 hours. If both of them started the journey at 9.30am, at what time would they pass each other?
2) Chive, Parsley and Thyme bought a present for their mother. Chive paid 1/3 of the total amount paid by Parsely and Thyme. Parsley paid 1/3 of the total amount paid by Chive and Thyme. If Thyme paid $60 more than Parsely, how much did the present cost?
3) The ratio of the number of $5 notes to the number of $10 notes that Libby had was 5:3. She bought a dress for $65 and paid it with 7 pieces of $5 notes and some $10 notes. The ratio of the number of $5 notes to the number of $10 notes Libby had left was 3:2.
How much did Libby have at first?
Thank you
firebird -
CoffeeCat:
Interestingly u got the answer. But I don't understand how you can get this.
Hi CoffeeCatVanilla Cake:
[quote=\"CoffeeCat\"]
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.
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.
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[/quote]Hi CoffeeCat,
All the equations are formed based on info from the question itself. Since Herbie advised that the question was wrongly typed, let's not waste time to pursue this question further.
Thanks for your understanding.
Submitted by VC's mum -
Dear Firebird,
I am VC's mum and pls allow me to post the solutions to your questions while waiting for Mathsguru's solutions.
Thanks.firebird:
Assume that both John and Peter had travelled on constant speed.1) The distance between village A and Village B is 360km. John drove from Village A to Village B and completed the journey in 4 hours. Peter drove from Village B to Village A and completed the journey in 6 hours. If both of them started the journey at 9.30am, at what time would they pass each other?
John's speed = 360/4 = 90 km/h
Peter's speed = 360/6 = 60 km/h
Time lapsed for them to meet each other = 360/(90+60) = 2.4 h = 2 h 24 min
Add 2 h 24 min to 9.30 am = 11.54 am
At 11.54 am, they would pass each other.firebird:
C : P+T2) Chive, Parsley and Thyme bought a present for their mother. Chive paid 1/3 of the total amount paid by Parsely and Thyme. Parsley paid 1/3 of the total amount paid by Chive and Thyme. If Thyme paid $60 more than Parsely, how much did the present cost?
1 : 3
P : C+T
1 : 3
C=1u
P=1u
T=2u
Thyme paid $60 more than Parsely: T-P = (2-1)u = 1u
1u = $60
4u = $240
Cost of the present was $240firebird:
$5 : $103) The ratio of the number of $5 notes to the number of $10 notes that Libby had was 5:3. She bought a dress for $65 and paid it with 7 pieces of $5 notes and some $10 notes. The ratio of the number of $5 notes to the number of $10 notes Libby had left was 3:2.
How much did Libby have at first?
5u : 3u
-7 : -3
3p : 2p
$5 : $10
10u : 9u
-14 : -9
6p : 6p
10u-14 = 9u-9
u = 5
Numbers
$5: 5u = 5x 5 = 25
$10: 3u = 3x5 = 15
Value
$5: 25x$5 = $125
$10: 15x$10 = $150
$(125+150) = $275
At first, Libby had $275
Submitted by VC's mum -
firebird:
Q1)Dear maths guru
Please help me with the following question:
1) The distance between village A and Village B is 360km. John drove from Village A to Village B and completed the journey in 4 hours. Peter drove from Village B to Village A and completed the journey in 6 hours. If both of them started the journey at 9.30am, at what time would they pass each other?
2) Chive, Parsley and Thyme bought a present for their mother. Chive paid 1/3 of the total amount paid by Parsely and Thyme. Parsley paid 1/3 of the total amount paid by Chive and Thyme. If Thyme paid $60 more than Parsely, how much did the present cost?
3) The ratio of the number of $5 notes to the number of $10 notes that Libby had was 5:3. She bought a dress for $65 and paid it with 7 pieces of $5 notes and some $10 notes. The ratio of the number of $5 notes to the number of $10 notes Libby had left was 3:2.
How much did Libby have at first?
Thank you
firebird
John ‘s speed = 360km/4hrs = 90 km/h
Peter’s speed = 360km/6hrs = 60 km/h
Time to meet (hrs) = 360km/ (90 + 60)km/h = 2.4 hrs
Meeting time is 2.4 hrs after 9.30am = 11.54 am
Q2)
Clive paid : ¼ of cost
Parsley paid : ¼ of cost
So, Thyme paid : 1 – ¼ - ¼ = ½ the cost
(½ - ¼ = ¼ ) of cost = $60
Cost of present = $ 60 x 4 = $240
Q3)
$65 = (7 x $5) + (N x $10)
Number of $10 notes, N used to buy dress = ($65 - $35) / $10 = 3
$5 notes : 5u – 7 = 3p
$10 notes : 3u – 3 = 2p
10u – 14 = 9u – 9
1u = 5
Amt Libby had at first = ($5 x 5 x 5) + ($10 x 3 x 5) = $125 + $150 = $275 -
YLH88:
Square : Circle + Rectangle = 1 : 2 (Area ratio)Hi Mathsguru,
Need your help with the following questions :
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 ?
(ans : 5 : 10 : 15)
http://www.postimage.org/image.php?v=TsM_an9
Thank you very much!
Total area
Circle : 5u
Rectangle : 3p
60% of area of Square = 1u + 1p
100% of area of Square = 5u/3 + 5p/3
5u + 3p = 2(5u/3 + 5p/3)
15u + 9p = 10u + 10p
1p = 5u
Circle : 5u
Rectangle : 3p (3 x 5u = 15u)
Square : (5u + 15u)/2 = 10u
Area of circle : Area of square : Area of rectangle = 5 : 10 : 15 = 1 : 2 : 3 -
Dear VC’s mum and Dharma
Many thanks for the solutions.
With best regards
Firebird -
Hi VC, Dharma,
Thank you so much for the solution !! Am trying to 'disgest' and understand the method
May need to trouble you again if still blur .. :?
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