Tutor MathsGuru: Ask me for your burning Maths questions!

[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.

[starlight1968sg]

Hi mathsguru and others,

I need some help:
(1) P, T and V have 960 beads altogether. P gave some of her beads to T and the number of T’s beads was doubled. Then T gave some of her beads to V and the number of V’s beads was doubled. If the 3 girls had the same number of beads in the end, how many beads did P have at first.
(ans: 560)
How to solve it?

(2) Look at the pattern below:
Position Number
1st 1
2nd 2
3rd 3
4th 2
5th 3
6th 4
7th 3
8th 4
9th 5
10th 4
11th 5
12th 6
(a) At what position does the number 12 first appear?
(b) What is the sum of the first 32 numbers?

(ans: (a) 30th, (b) 218)

For (a), do we simply write from position 12th till position 30th ie when we first see the number 12?

For (b), the calculations is too lengthy. Is there a method or a pattern one can recognise to simplify the calculation?

MTIA.

[Dharma]

starlight1968sg:

Hi mathsguru and others,

I need some help:
(1) P, T and V have 960 beads altogether. P gave some of her beads to T and the number of T's beads was doubled. Then T gave some of her beads to V and the number of V's beads was doubled. If the 3 girls had the same number of beads in the end, how many beads did P have at first.
(ans: 560)
How to solve it?


MTIA.

Since P, T and V had the same no. of beads at last ; each will have 960/3 = 320 beads at last

When T gave some of her beads to V and the number of V's beads was doubled
No. of beads T and V before T gave V some beads :
T : 320 + 160 = 480
V : 320 – 160 = 160

When P gave some of her beads to T and the number of T's beads was doubled.
No. of beads P and T before P gave T some beads :
T : 480 – 240 = 240
P : 320 + 240 = 560