Tutor MathsGuru: Ask me for your burning Maths questions!
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Vanilla Cake:
Let Aw = width of A, Ah = height of A, and so on.Pls :imdrowning: and TIA for your time cum effort.
![](\"http://i52.tinypic.com/2v0dee0.jpg\")
http://i52.tinypic.com/2v0dee0.jpg\">
Answer given: 11 and 2/3 cm2
Ch = 2/Cw = 2/(7/Ah) = 2/7 Ah = 2/7 (21/Bw) = 6/Bw
Aw + Bw = Ah + Ch
7/Ah + Bw = 7/Aw + Ch
7/(21/Bw) + Bw = 7/(2/Ch) + Ch
1/3 Bw + Bw = 7/2 Ch + Ch
4/3 Bw = 9/2 Ch
4/3 Bw = 9/2 (6/Bw) = 27/Bw
Bw*Bw = 27 (3/4) = 81/4
Bw = 9/2
Ch = 6/(9/2) = 4/3
Perimeter of D = 2(9/2 + 4/3) = 2 ((27 + 8)/6) = 35/3 = 11 2/3 cm -
Vanilla Cake:
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=27&t=6373&p=195304Pls :imdrowning: and TIA for your time cum effort.
![](\"http://i54.tinypic.com/96aj5k.jpg\")
Answer given: 150 degree
Please look there. Someone asked this 1 yr ago.
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Vanilla Cake:
S1 = 1Pls :imdrowning: and TIA for your time cum effort.
![](\"http://i51.tinypic.com/2q03s3t.jpg\")
Answer given : 415
S2 = 1 + 11 = 12
S3 = 1 + 11 + 111 = 123
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.
.
S9 = 123456789
S10 = 123456790.
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.
S20 = 12345679012345679010
.
.
.
S30 = 123456790123456790123456790120
.
.
.
S100 = 123456790....1234567901234567890
(10 sets of 123456790)
Sum of digits
= 10(1+ 2+3+4+5+6+7+9+0) + (1+2+3+...+ 8+9 + 0)
= 370 + 45
= 415 -
Vanilla Cake:
Hi there, for the above question on Speed, we need to take into consideration the time traveled by Mary as well. There is a difference of 300m which takes 4.5 mins to travel. Assuming they start at 9am at the same time, John traveled 4km/h faster than Mary which gives him a head start of 300m. Thus, Mary and John would have traveled for 9mins while John traveled 300m more. The answer should be 9:09am instead.
Hi Firebird,firebird:
Dear maths guru
Please help me on the following maths:
Pls refer to the below workings done by Vanilla Cake, this morning. Due to time constraint, she didn't complete the speed question(she was in a rush to go to school). Sorry, pls wait for others like Dharma etc to help.If you don't mind, could you pls advise the source of these questions? The ones on ratio and speed are interesting and I will let DD2 to practise.
Thks.
VC's mumfirebird:
John's printer-> 150%1) John had $180 more than Raymond at first. Both bought a different printer with some of their money. In the end Rayamond was left with $20 more than John. If John's printer cost 50% more than Raymond's printer, how much does John's printer cost?
Raymond's printer -> 100%
$(180+20)= $200
50% -> $200
150% -> $600
Cost of John's printer = $600firebird:
0.3/4 = 0.075 h = 4.5 min2) At 9am John was cycling from Point A to Point B. At the same time, Mary was cycling from point B to point A using the same route as John. Cycliing at a speed of 4km/h faster than Mary, they would pass each other 300 meter away from the mid point.
a) What time did they pass each other?
b) If John took another 3 minutes to reach Point B, what time would Mary reach Poin A?
Add 4.5 min to 9 am -> 9:04:30
(a) They passed each other at 9:04:30
(b) Not enough time to think and solve, pls wait for others to solve.firebird:
Before3) Daniel had 40% more stickers than Brandon. Daniel and Brandon each gave 20% of their stickers to Calvin. As a result, Calvin's stickers increased by 80%. If Daniel has 20 more stickers than Calvin in the end, how many stickers did Brandon have at first?
140%
B: 100%
u
After 112%
B: 80% u+28%+20% = u+48%
80% increased for C-> 180/100xu = 1.8u
1.8u=u+48%
0.8u=48%
u=60% 60%+48% = 108%
Difference between D and C -> D-C=112%-108% = 4%
4% -> 20
100% -> 20/4x100=500
At first, Brandon had 500 stickers.firebird:
This is a trick question and you have to know the common difference is 3u as the number of male and female participants was the same.4) At a conference made up of speakers and participants, there were 20% more men than women. The ratio of male speakers to female speakers was 8:5. There was an equal number of male and female participants.
a) Find the ratio of male speakers to male participants at the conference.
b) Halfway, 40 male participants left the conference and another 60 female participants joined the conference. In the end, there were 3/4 as many male participants remaining behind. How many speakers were there at the conference?
Total
M : W
120 : 100
6 : 5
18: 15
Speakers
M : W
8 : 5
Total speakers = 13u (8u+5u)
Participants
M : W
10 : 10
8: 10
(a) Ratio of male speakers to male participants = 8 : 10 = 4 : 5
I use cross multiplication to solve this, you may wish to use alternative methods to suit you.
(10u-40)/(10u+60)=3/4
30u+180=40u-160
10u=340
u=34
13u=442
Number of speakers at the conference = 442
Check
Speakers
M: 272
W: 170
Participants
M: 340
F: 340
Total
M: 612
W: 510
Phew, done except for speed part (b).Pls check all my calculations, thanks.
How do you solve part b? -
Vanilla Cake:
(1 + 1/2) x (1 + 1/(2)^2)Pls :imdrowning: and TIA for your time cum effort.
![](\"http://i51.tinypic.com/34qoejc.jpg\")
Answer given: 2010
= (1 + 1/((2)^2)^0 ) x (1 + 1/((2)^2)^1 )
= 3/2 x 5/4 = 15/8
= 2 - 1/8
= 2(1 - 1/16)
= 2(1 - 1/((2)^2)^2 )
(1 + 1/2) x (1 + 1/(2)^2) x ( (1 + 1/(2)^4)
=(1 + 1/((2)^2)^0 ) x (1 + 1/((2)^2)^1 ) x ( (1 + 1/((2)^2)^2 )
= 3/2 x5/4 x 17/16 = 255/128
= 2 - 1/128
= 2(1 - 1/256)
= 2(1 - 1/((2)^2)^3 )
(1 + 1/2) x (1 + 1/(2)^2) x ( (1 + 1/(2)^4) x .... x ( (1 + 1/((2)^2)^2009)
= 2(1 - 1/((2)^2)^(2009 +1) )
= 2(1 - 1/((2)^2)^(2010) )
n = 2010 -
Hi, I have one qn. Kindly help.
Mary took x hours to travel from Town A to Town B while Peter took y hours to travel from Town B to Town A. If both of them started their journey at 10am at what time will they meet each other on the way??
Thanks! -
Tania:
Speed of Mary -> 1/x journey per hrHi, I have one qn. Kindly help.
Mary took x hours to travel from Town A to Town B while Peter took y hours to travel from Town B to Town A. If both of them started their journey at 10am at what time will they meet each other on the way??
Thanks!
Speed of Peter -> 1/y journey per hr
Combined Speed -> 1/x + 1/y = (x+y)/xy journey per hour
Hence Time taken -> 1/combinedspeed = xy/(x+y)
Hmm..its kinda weird to express a timing in algebra... -
Hi, I wld like help with this P3 question:
Ali had 3 times as many marbles as Bala. After Ali lost 4 marbles and gave Bala some marbles, the 2 boys have the same number of marbles. Bala has 22 marbles now. How many marbles did Bala have at first? -
wdiong:
Total marbles -> 22x2=44 ( bala has 22 now. they have the same number of marbles now so multiply by 2)Hi, I wld like help with this P3 question:
Ali had 3 times as many marbles as Bala. After Ali lost 4 marbles and gave Bala some marbles, the 2 boys have the same number of marbles. Bala has 22 marbles now. How many marbles did Bala have at first?
at first, ali 3u, bala 1u
4u -> 44
1u -> 44/4=11
hope you get it now..
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hi vanilla cake, may i ask the questions which u hv posted on s1 and n question are they pri sch maths?? Please advice… Tq
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