Geom question:
Given: ABCD is a square, EG=EF=EH. Find angle GFH.
Since ABCD is a square, AB=BC=CD=DA
As EF=BC=DA and AB=GH, hence EG=EH=AB=GH.
Therefore, triangle EGH is equilateral since EG=GH=EH.
angle GEH=60°
angle GEF=60/2=30°
Triangle EGF is isosceles. Hence, angle FGE= angle GFE = (180-30)/2 = 75°
angle GFE = angle HFE = 75°
Therefore angle GFH= 75X2=150°
Balloon question:
Given: String of 2 big balloons is 90cm String of 5 small balloons is is 1.2m If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether?
105 small balloons would have a length of (105/5) X 1.2 = 25.2m = 2520cm
1 big balloon has a length of 90/2=45cm
1 small balloon has a length of 1.2m/5= 0.24m= 24cm
Difference in length of big and small balloon = 45-24= 21 cm
Number of big balloons = 2520/21 = 120
Number of small balloons = 120+105 = 225
Total number of balloons = 120+225 = 345
13 comments:
What do you of the answer (211) for the balloon question from this blog.
http://prischoolmaths.blogspot.com/2009/10/psle-challenging.html
All the P6 kids are praying that your answer (345) is correct.
Why EG=EH=AB=GH?
GH is a chord while EG and EH are radii
Anon 21.33: I checked the blog stated. The answer given was 345, not 211. Perhaps the author has changed the answer after realising the original answer was wrong.
Anon 21.39: There is no need to look at radius or chord (in any case, there is nothing wrong with radius=chord). As ABCD is a square, EF is same as 1 side of the square, GH also same as 1 side of the square. So GH=EF. Hence Triangle EGH is equilateral.
Thank you Lim Jeck.
The author has amended the answer as additional information has been provided and added to the question.
Eg. balloons were not at 0 and 90 cm for the big balloons and 0 and 120 cm for the small balloons. The two strings are of the same length.
Mei and Lin were in a bicycle race. Mei was travelling at a constant speed of 20km/hr and they both did not change their speed. When Lin completed half the race, Mei was 3.5km ahead. Mei completed the race at 10.45am. What time did Lin complete the race?
Can I trouble you to solve this problem sum? Thanks.
FC
Hi I am FC who left the question on Mei & Lin for you. Below is my solution, can you help me verify if it is correct? Thanks:
At half way time ahead for Mei is:
3.5km/20kmph = 0.175hr
By end of journey, Mei would be 0.175 x 2 = 0.35hr faster than Lin (i.e. Lin would have reached 21min later)
10 hr 45 min + 21 min = 10 hr 66 min = 11 hr 06 min
Ans: Lin would have reached at 11.06am
hi FC,
Yes, your answer is correct.
Regards.
Thanks Jeck.
Cheers!
FC
ps:For the benefit of those who are buffled by why x2 - Just imagine moving Mei back to halfway point and Lin would need 0.175hr to reach the same halfway point. And if both were to start at halfway point afresh at the same speed towards finishing line, Mei would still end up 0.175hr ahead of Lin at the end of 2nd half of the journey, therefore x2.
Actually,
If so many students can get 345 & 68 & some also got 150deg as answers to their PSLE questions, I don't understand why everyone is shouting & complaining that it is 'hard'. If it is really that difficult to many, there won't be so many students getting similar (correct) answers!
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