Q&A - PSLE Math
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Angel25:
I am a rusty old woman attempting here:Henry Park Prelim 2011 Q15.
Mr Lee packed some oranges into big boxes and apples into small boxed. Each big box contained 50 oranges and each small box contained 30 apples. After packing, there were 12 more big boxes than small boxes. Given that there were 1240 fewer apples than oranges, how many oranges were there ?
Ratio of packing: Or: Ap = 5 : 3
And Or - Ap = 1240
After packing there are additional 12 x 50 = 600 oranges
Using ratio diff of 5-3 = 2, this equates to the diff of 1240-600 = 640.
2u = 640
1u = 320
Oranges initially = 5 x u +600 = 2220
I did a counter-chcek theerafter to make sure. I need some expert help to exhibit my answers more mathmathically more correctly...these are just logical workings in my brains.... -
Angel25:
HiPei Chuan Public School Prelim 2011 P2 Q18
At first, Ella had $52 more than Felix. After Ella had spent 25% of her money and Felix had spent 3/5 of his, Ella had $123 more than Felix. How much money did Ella have at first?
You can solve this using MD too.. I am using the units method here as I do not time to draw the model...
Felix = 20u
Ella = 20u + 52
Ella spent 25%, had 3/4 left --> 15u + 39
Felix spent 3/5, had 2/5 left --> 8u
So, 15u + 39 = 8u + 123, 7u = 84, 1u = 12
At first, Ella had 20u + 52 = $292
cheers. -
Angel25:
HiCHIJ Prelims 2011 P2 Q12
At 8.30am, Myke and Jerome set off at the same time from Town X to Town Y. At 11.00am, Myke completed his journey but Jerome had covered only 5/8 of the journey. Jerome's speed was 54km/h slower than Myke's.
a) Find the distance between Town X and Town Y.
b) At what time would Jerome complete his journey?
8.30am --> 11.00am = 2.5h, Myke reached Town Y in 2.5h.
2.5h x 54km = 135 km --> Jerome was still 135 km from Town Y at 11.00am
so 3/8 of the distance = 135 km, distance between Town X & Y = 135/3 x 8 = 360 km
Jerome took 2.5h to cover 5/8 of the distance, for remaining 3/8 of the distance, time taken = 3/5 x 2.5 = 1.5h = 1h 30 min
Jerome will reach Town Y at 11.00 am + 1h 30min = 12.30 pm
cheers. -
Hi
HP 2011 SA2 Q11
Outer fence ----- 40cm
Inner fence ------ 180cm
Use a common length of 360cm as comparison
Inner fence ------- 2180 ------- 18 metal spokes
Outer fence ------- 940 --------27 wooden spokes
Difference ------- 9 wooden spokes
198/9 -------22
22(18+27) ------- 990
Best wishes -
Hi
CHIJ 2011 Prelims P2 Q15
You may solve it using MD.
Donovan ----- 1 unit + 48
Ethan ------1 unit
Freddy ------ 2 units
Gilbert ------- 432
1 unit + 48 -------20%
4 units + 192 ------ 80%
4 units + 192 -------- 3 units + 432
1 unit ------- 240
Total number of trading cards -------- 4 units + 48 + 432 --------1440
487 – (240+48) -------199
199/288* 100% --------- 69.1%
Best wishes -
Angel25:
http://i53.tinypic.com/sfey6o.jpg\"> http://i53.tinypic.com/sfey6o.jpg\"> Henry Park 2011 Prelim P2 Q11
Hi
every 360cm stretch,
Number of metal spokes = 9 x 2 = 18
Number of wooden spokes = 3 x 9 = 27
27 – 18 = 9 --> 9 more wooden spokes than metal spokes every 360cm.
198 ÷ 9 = 22 of 360 cm-stretch
Number of metal spokes = 22 x 18 = 396
Number of wooden spokes = (22 x 27) x 4 = 2376
2376 + 396 = 2772
There a total of 2772 wooden and metal spokes.
cheers. -
Angel25:
http://i52.tinypic.com/10ruv0z.jpg\"> Henry Park 2011 Prelim P2 Q10
Hi
Rearrange the figure.
Area of square (12*12) ------ area of circle (r6) -------- A
Area of eye -------- area of semi circle (r6) – square (6*6) -------- B
The answer is (A+B)
I’ll leave it to students to work out the calculations
Best wishes -
Hi - Could someone h http://i56.tinypic.com/9tcx1t.jpg\"> elp me with the sum below.
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Hi,
Need help!! Thanks
A group of boys tool their pets to a park. Each boy either has a dog or cat. Between the boys and theirs pets, there were a total of 204 leags.
a) How many boys were there?
b) If there were 15 cats, how many more dogs than cats were there?
Thanks all the masters -
Daddy:
Hi,
Need help!! Thanks
A group of boys tool their pets to a park. Each boy either has a dog or cat. Between the boys and theirs pets, there were a total of 204 leags.
a) How many boys were there?
b) If there were 15 cats, how many more dogs than cats were there?
Thanks all the masters
Every group of Boy + Cat/Dog will have 6 legs.
So total no of such groups - 204/6 = 34.
Therefore a) 34 boys
b) If there were 15 cats - there are 19 dogs. So 4 more dogs than cats.
Am I missing something here?