Tutor MathsGuru: Ask me for your burning Maths questions!

[clblinym]

Dharma:

clblinym:

Hi, Can anyone help in solving the following 4 questions.


They are from SMOPS 2004. Thanks in advance!!
http://postimage.org/image/11lki3290/

Thank you very much.


The answers are
Q19 (360 m)
Q28 (24 cm2)
Q29 (64665912)
Q30 (8, 11, 81) (12,4,84) (4, 18, 78 )


Regards,
Amy

Assume that telephone number is ABCDEFGH
ABC + DEFGH = 66558
ABCDE + FGH = 65577
A = 6
D = 6

60000 + 1000E + 100(6 + F) + 10(B + G) + (C + H) = 66,558
60000 + 1000B + 100(C + F) + 10(6 + G) + (E + H) = 65,577

E is either 6 or 5
B is either 5 or 4
C + H => 8 or 18
E + H => 7 or 17

6 + G = 7. So, G = 1
We know that
C + H = 8 and
E + H = 7

B + G = 5
B = 4
C + F = 15
6 + F = 15
C = 6
F = 9
1+ E = 6
E = 5
C + H = 8
H = 8 – 6 = 2
Tel number = ABCDEFGH = 64665912

Hi, Dharma
Thank you for your detailed explanation. Appreciate your help.

Regards,
Amy

[Dharma]

clblinym:

Hi, Can anyone help in solving the following 4 questions.


http://postimage.org/image/11lm5meqs/

Is there any easier or shorter way to solve Q30? Thank you very much.


The answers are
Q19 (360 m)
Q28 (24 cm2)
Q29 (64665912)
Q30 (8, 11, 81) (12,4,84) (4, 18, 78 )


Regards,
Amy

A = Chocolate
B = Cheese
C = Durian

A + B + C = 100
5A + 3B + 1/3C = 100
15A + 9B + C = 300
14A + 8B = 200
7A + 4B = 100

7A + 4B = A + B + C
C = 6A + 3B = 3(2A + B)

3A + 3B + 3C = 300
5A + 3B + 1/3C = 100
8/3C – 2A = 200
2A = 8/3C – 200
A = 4/3C – 100

4/3C > 100
C > 75
Use C = 78
3(2A + B) = 78
2A + B = 26
A + B = 22
A = 26 – 22 = 4
A = 4, B = 18, C = 78
If C = 81,
2A + B = 27
A + B = 19
A = 8
B = 11
A = 8, B = 11 , C = 81
If C = 84,
2A + B = 28
A + B = 16
A = 12
B = 4
A = 12, B = 4, C = 84
If C = 87
2A + B = 29
A + B = 13
A = 16 (Not possible - reject)

If C = 90
2A + B = 30
A + B = 10
A = 20 (Not possible - reject)

[clblinym]

Dharma:

clblinym:

Hi, Can anyone help in solving the following 4 questions.


http://postimage.org/image/11lm5meqs/

Is there any easier or shorter way to solve Q30? Thank you very much.


The answers are
Q19 (360 m)
Q28 (24 cm2)
Q29 (64665912)
Q30 (8, 11, 81) (12,4,84) (4, 18, 78 )


Regards,
Amy

A = Chocolate
B = Cheese
C = Durian

A + B + C = 100
5A + 3B + 1/3C = 100
15A + 9B + C = 300
14A + 8B = 200
7A + 4B = 100

7A + 4B = A + B + C
C = 6A + 3B = 3(2A + B)

3A + 3B + 3C = 300
5A + 3B + 1/3C = 100
8/3C – 2A = 200
2A = 8/3C – 200
A = 4/3C – 100

4/3C > 100
C > 75
Use C = 78
3(2A + B) = 78
2A + B = 26
A + B = 22
A = 26 – 22 = 4
A = 4, B = 18, C = 78
If C = 81,
2A + B = 27
A + B = 19
A = 8
B = 11
A = 8, B = 11 , C = 81
If C = 84,
2A + B = 28
A + B = 16
A = 12
B = 4
A = 12, B = 4, C = 84
If C = 87
2A + B = 29
A + B = 13
A = 16 (Not possible - reject)

If C = 90
2A + B = 30
A + B = 10
A = 20 (Not possible - reject)

Hi Dharma,

Thank you so much for your detailed explanation and patience.

Good night!

Amy

[clblinym]

Hi


Can someone check whether my working for this question is correct? Thanks in advance.

Question:

Mrs Wong is preparing gifts for the coming party.
She buys 540 notepads, 720 pens and 900 pencils.
The total cost of each type of stationery is equal.
She divides the notepads equally into red boxes, pens equally in yellow gift boxes and pencils equally in blue gift boxes.
She wishes that the cost of each gift box is equal and as low as possible.
Find the number of notepads she needs to put in each red gift box.

My working:

A  Notepad
B  Pen
C  Pencil

Ratio of number
A: B : C
3 : 4: 5

The LCM of 3,4 and 5 is 60.

3 x 4 = 12
4 x 5 = 20
3 x 5 = 15

Ratio of cost per item
A: B: C
20:15:12  Cost
3 : 4: 5  Number put in each box

For the cost of each gift box to be equal, the cost multiplied by the number should result in same answer for all 3 cases.

20x3=60
15x4=60
12x5=60

Answer: 3

Thank you and have a nice day.
Amy

[Dharma]

clblinym:

Hi


Can someone check whether my working for this question is correct? Thanks in advance.

Question:

Mrs Wong is preparing gifts for the coming party.
She buys 540 notepads, 720 pens and 900 pencils.
The total cost of each type of stationery is equal.
She divides the notepads equally into red boxes, pens equally in yellow gift boxes and pencils equally in blue gift boxes.
She wishes that the cost of each gift box is equal and as low as possible.
Find the number of notepads she needs to put in each red gift box.

My working:

A  Notepad
B  Pen
C  Pencil

Ratio of number
A: B : C
3 : 4: 5

The LCM of 3,4 and 5 is 60.

3 x 4 = 12
4 x 5 = 20
3 x 5 = 15

Ratio of cost per item
A: B: C
20:15:12  Cost
3 : 4: 5  Number put in each box

For the cost of each gift box to be equal, the cost multiplied by the number should result in same answer for all 3 cases.

20x3=60
15x4=60
12x5=60

Answer: 3

Thank you and have a nice day.
Amy

Let unit cost of notepads, pens and pencils be a, b and c
540a = 720b = 900c (Total cost of each type of stationery is the same)

Since the cost of each gift box is equal;
3a = 4b = 5c (Lowest cost for each gift box; number of gifts in each box must be lowest possible) since the unit cost for each type of gift does not change.

Number of note pads = 3

[Daddy]

Hi Brenda10


O : P
9 : 5
-240 -200
12 : 5


O 45u 1200
P 60u 2400
Diff 15u = 1200
1u = 80

How did you get O 45u 1200 & P 60u 2400 ? I not understand? Thanks

[Brenda10]

Daddy:

Hi Brenda10


O : P
9 : 5
-240 -200
12 : 5

O 45u 1200
P 60u 2400
Diff 15u = 1200
1u = 80

How did you get O 45u 1200 & P 60u 2400 ? I not understand? Thanks

O 45u 1200 (9 x 5, 240 x 5)
P 60u 2400 (5 x 12, 200 x 12)

[Vanilla Cake]

clblinym:

Mrs Wong is preparing gifts for the coming party.

She buys 540 notepads, 720 pens and 900 pencils.
The total cost of each type of stationery is equal.
She divides the notepads equally into red boxes, pens equally in yellow gift boxes and pencils equally in blue gift boxes.
She wishes that the cost of each gift box is equal and as low as possible.
Find the number of notepads she needs to put in each red gift box.

My working:

A  Notepad
B  Pen
C  Pencil

Ratio of number
A: B : C
3 : 4: 5

The LCM of 3,4 and 5 is 60.

3 x 4 = 12
4 x 5 = 20
3 x 5 = 15

Ratio of cost per item
A: B: C
20:15:12  Cost
3 : 4: 5  Number put in each box

For the cost of each gift box to be equal, the cost multiplied by the number should result in same answer for all 3 cases.

20x3=60
15x4=60
12x5=60

Answer: 3

Dharma:

Let unit cost of notepads, pens and pencils be a, b and c
540a = 720b = 900c (Total cost of each type of stationery is the same)

Since the cost of each gift box is equal;
3a = 4b = 5c (Lowest cost for each gift box; number of gifts in each box must be lowest possible) since the unit cost for each type of gift does not change.

Number of note pads = 3

Alternative http://www.onsponge.com/forum/35-thinkingmathonsponge/6568-p6-math-olympiad.html from OnSponge forum for this question from SMOPS 2005 1st round - Q30.

[clblinym]

Vanilla Cake:

clblinym:

Mrs Wong is preparing gifts for the coming party.

She buys 540 notepads, 720 pens and 900 pencils.
The total cost of each type of stationery is equal.
She divides the notepads equally into red boxes, pens equally in yellow gift boxes and pencils equally in blue gift boxes.
She wishes that the cost of each gift box is equal and as low as possible.
Find the number of notepads she needs to put in each red gift box.

My working:

A  Notepad
B  Pen
C  Pencil

Ratio of number
A: B : C
3 : 4: 5

The LCM of 3,4 and 5 is 60.

3 x 4 = 12
4 x 5 = 20
3 x 5 = 15

Ratio of cost per item
A: B: C
20:15:12  Cost
3 : 4: 5  Number put in each box

For the cost of each gift box to be equal, the cost multiplied by the number should result in same answer for all 3 cases.

20x3=60
15x4=60
12x5=60

Answer: 3

Dharma:

Let unit cost of notepads, pens and pencils be a, b and c
540a = 720b = 900c (Total cost of each type of stationery is the same)

Since the cost of each gift box is equal;
3a = 4b = 5c (Lowest cost for each gift box; number of gifts in each box must be lowest possible) since the unit cost for each type of gift does not change.

Number of note pads = 3

Alternative http://www.onsponge.com/forum/35-thinkingmathonsponge/6568-p6-math-olympiad.html from OnSponge forum for this question from SMOPS 2005 1st round - Q30.

Hi, Vanilla Cake

Thank you very much.

Regards,
Amy

[clblinym]

Hi, Can anyone help to solve the problems below. Thank you.


1) There are buses travelling to and fro between Station A and Station B. The buses leave the stations at regular interval and a bus will meet another bus coming in the opposite direction every 6 minutes.
Peter starts cycling from A towards B at the same time Jane starts cycling from B towards A.
Peter and Jane will meet a bus coming in the opposite direction every 7 and 8 minutes respectively.
After 56 minutes of cycling on the road, they meet each other.
Find the time taken by a bus to travel from A to B.
Answer : 68

2)


http://postimage.org/image/1ww3enfvo/
Answer : 666


Thank you for your time and effort.
Amy