Tutor MathsGuru: Ask me for your burning Maths questions!

[trytry]

wkong:

Hi Maths Guru,


i have another Maths question that need helps:

Belinda has a collection of some stamps. If she gave 36 stamps to Samantha, she would have 4 times as many stamps as Samantha. If she gave 23 stamps to Samantha, she would have 5 times as many stamps as Samantha. How many stamps did Belinda have at first?

Thanks.

1st condition
B:[...][...][...]
S:[...]

2nd conditon
B:[...][...][...][...][...]
S:[...]

Total remain the same for both conditions
therefore
1st condition
B:[.6u.][.6u.][.6u][6u]..........24u
S:[.6u.].................................6u
.....................................total 30u

2nd conditon
B:[.5u][5u.][.5u][5u.][5u.].......25u
S:[5u..]....................................5u
.......................................total 30u

6u-5u=36-23
.....1u=13

Belinda has =24u+36=24x13 + 36
...............................=348

[CoffeeCat]

Jolyn:

Hi Maths Guru


I have the following of 2 questions which needs your help:

1. Zul has 18 sweet. He has thrice as many sweets as Shania. How Many more sweets does Zul have than Shania?

2. A fruit seller sold some fruits on Tuesday morning. He sold 3 times as many oranges as apples. He sold 684 fruits that day.

a) How many oranges did he sell?
b) How many more oranges than apples did he sell?


Thank you very much.....

1) Zul has 18 sweets. Shania has 18/3 = 6 sweets. Difference is 18-6=12 sweets. Alternatively using a model approach, draw 1 unit for shania and 3 units for Zul.

2)Similarly draw 1 unit to represent the number of apples and 3 units to represent oranges (because 3 times as many oranges as apples).
Together the 3+1=4 units represent 684.
1 unit = 684 / 4 = 171
a) no . of oranges = 171 X 3 = 513
b) difference = 171 X 2 = 342

[YLH88]

Hi Mathsguru,


With ref to the following question posted by Firebird, how did you derive the shaded are is 420 cm2 ?

Question 11) In the figure below, not drawn to scale, PQRS is a rectangle with a length of 45 cm and a width of 22 cm. The area of te TUVW is 75 cm2. Find the ratio of the shaded area to the unshaded area.

http://www.postimage.org/image.php?v=Pq46L9A


Thank you very much!

p/s now my ds is practicing the school test paper and come across this question which I recalled you have helped solved the question, but hor, I don't really understand the solution

[adhdadhd]

trytry:

wkong:

Hi Maths Guru,


i have another Maths question that need helps:

Belinda has a collection of some stamps. If she gave 36 stamps to Samantha, she would have 4 times as many stamps as Samantha. If she gave 23 stamps to Samantha, she would have 5 times as many stamps as Samantha. How many stamps did Belinda have at first?

Thanks.

1st condition
B:[...][...][...]
S:[...]

2nd conditon
B:[...][...][...][...][...]
S:[...]

Total remain the same for both conditions
therefore
1st condition
B:[.6u.][.6u.][.6u][6u]..........24u
S:[.6u.].................................6u
.....................................total 30u

2nd conditon
B:[.5u][5u.][.5u][5u.][5u.].......25u
S:[5u..]....................................5u
.......................................total 30u

6u-5u=36-23
.....1u=13

Belinda has =24u+36=24x13 + 36
...............................=348

GR8 approach, btw, how to decide on using 5u and 6u?

[adhdadhd]

YLH88:

Hi Mathsguru,


With ref to the following question posted by Firebird, how did you derive the shaded are is 420 cm2 ?

Question 11) In the figure below, not drawn to scale, PQRS is a rectangle with a length of 45 cm and a width of 22 cm. The area of te TUVW is 75 cm2. Find the ratio of the shaded area to the unshaded area.

http://www.postimage.org/image.php?v=Pq46L9A


Thank you very much!

p/s now my ds is practicing the school test paper and come across this question which I recalled you have helped solved the question, but hor, I don't really understand the solution

Let me try...
to work out unshaded area first:-
referring to the given diagram..
Area of ▲SVR = 1/2 x 45 x 22 = 495 cm²
Area of ▲STR= 495/2 cm² = 247.5 cm²
Area of VSTR = ▲VSR - ▲STR (495 - 495/2 ) cm² = 247.5 cm²
Area of ▲PTQ= Area of ▲STR
Area of ▲PWV+▲VUG=▲PTQ - TUVW = 495/2 - 75 = 172.5 cm²
Total unShaded Area = (247.5 + 172.5) = 420 cm²
So total Shaded = Area of Rect - unShaded Area= (990-420)cm²= 570cm²

[Dharma]

YLH88:

Hi Mathsguru,


With ref to the following question posted by Firebird, how did you derive the shaded are is 420 cm2 ?

Question 11) In the figure below, not drawn to scale, PQRS is a rectangle with a length of 45 cm and a width of 22 cm. The area of te TUVW is 75 cm2. Find the ratio of the shaded area to the unshaded area.

http://www.postimage.org/image.php?v=Pq46L9A


Thank you very much!

p/s now my ds is practicing the school test paper and come across this question which I recalled you have helped solved the question, but hor, I don't really understand the solution

Hi YLH88,

Think there's typo error in solution ...unshaded area = 420cm2 (instead of shaded area)

[trytry]

adhdadhd:

trytry:

[quote=\"wkong\"]Hi Maths Guru,


i have another Maths question that need helps:

Belinda has a collection of some stamps. If she gave 36 stamps to Samantha, she would have 4 times as many stamps as Samantha. If she gave 23 stamps to Samantha, she would have 5 times as many stamps as Samantha. How many stamps did Belinda have at first?

Thanks.

1st condition
B:[...][...][...]
S:[...]

2nd conditon
B:[...][...][...][...][...]
S:[...]

Total remain the same for both conditions
therefore
1st condition
B:[.6u.][.6u.][.6u][6u]..........24u
S:[.6u.].................................6u
.....................................total 30u

2nd conditon
B:[.5u][5u.][.5u][5u.][5u.].......25u
S:[5u..]....................................5u
.......................................total 30u

6u-5u=36-23
.....1u=13

Belinda has =24u+36=24x13 + 36
...............................=348

GR8 approach, btw, how to decide on using 5u and 6u?[/quote]1st condition
B:[...][...][...][...].....}
S:[...]......................}................total 5 UNITS

2nd conditon
B:[...][...][...][...][...]....}
S:[...]..........................}...........total 6 PARTS

Since total stamps remain the same, find common multiples of 5 and 6 which is 30. Make 5 UNITS into 30u and 6PARTS into 30u by mutliplying 6 and 5 respectively.

[Dharma]

wkong:

Hi Maths Guru,


i have another Maths question that need helps:

Belinda has a collection of some stamps. If she gave 36 stamps to Samantha, she would have 4 times as many stamps as Samantha. If she gave 23 stamps to Samantha, she would have 5 times as many stamps as Samantha. How many stamps did Belinda have at first?

Thanks.

The no. of stamps Belinda and Samantha had initially
B : 4u + 36
S : 1u – 36

No. stamps each of them had after Belinda gave Samantha 23 stamps
B : 4u + 36 – 23 = 4u + 13
S : 1u – 36 + 23 = 1u – 13

(4u + 13) /(1u – 13) = 5/1
4u + 13 = 5u – 65
1u = 65 + 13 = 78

No. of stamps Belinda had at first = 4 (78 ) + 36 = 348

[tinasen]

I’m breaking my head to give a model method solution for this question. I tried to get a solution from some friends but it still doesn’t help me understand how to solve it.


Joseph had 5 times as much money as Lyn at first. After their mom gave $1002 to each of them , Joseph had twice as much as Lyn How much did Joseph had at first?

[adhdadhd]

tinasen:

I'm breaking my head to give a model method solution for this question. I tried to get a solution from some friends but it still doesn't help me understand how to solve it.


Joseph had 5 times as much money as Lyn at first. After their mom gave $1002 to each of them , Joseph had twice as much as Lyn How much did Joseph had at first?

Before:

Jo [u][u][u][u][u]
Ly [u]

After $1002 each
Jo [1002][u][u][u][u][u]
Ly [1002][u]

Since Jo : Ly = 2 : 1, referring to the after model
4u = 1002 + u
u=1002/3=334

Joseph had $1670.