
By: Yian Cheng (offline) Saturday, December 10 2011 @ 12:00 AM CST (Read 2040 times)



Yian Cheng 
Hi,
Help needed . Can anyone help me to solve the following problem sums?
1) The total length of two poles is 5.1m. 30% of the length of the shorter pole and 50% of the length of the longer pole add up to 2.25m.
Find the length of the shorter pole in cm. Ans: 150cm
2) A coin box contained some twentycent and fiftycent coins in the ratio of 4:5.
When 16 fiftycent coins were taken out and replaced by some twentycent coins, the ratio then became 8:7.
The total value of the twentycent coins added was the same as the total value of the fiftycent coins taken out.
Find the sum of money in the coin box. Ans: $112.20
Thank you.

Newbie
Registered: 09/30/11 Posts: 2





By: wenying (offline) Sunday, December 11 2011 @ 10:26 AM CST



wenying 
Hi, Yang Cheng,
My answer about 1) is:
50%=1/2 & 30%=3/10,
1/2 longer pole + 3/10shorter pole = 2.25m=225cm,
whole longer pole 2x 1/2 + 2x(3/10) shorter ploe = 2x225 =450cm,
5.1m=510cm,
510  450 = 60cm,
4/10 shorter pole : 60cm,
60 x 10/4 = 150cm
Have a cheer holiday,
Wen Ying

Newbie
Registered: 12/31/06 Posts: 1





By: pkrdd (offline) Tuesday, December 13 2011 @ 02:34 AM CST



pkrdd 
Hi Cheng,
To solve Q2, you have to take note that the value of the total coins will remain same before and after.
Applying this relation in Algebra, below are my steps
At First: (Total coins are 'u' units)
20C : 50C = 4u : 5u
Total Value = 4u * (20/100) + 5u * (50/100) = 3.3u (A)
At End: (Total coins are 'p' units)
20C : 50C = 8p : 7p
Total value = 8p * (20/100) + 7p * (50/100) = 5.1p  (
Value of coins in step A & B remains same (The total value of the twentycent coins added was the same as the total value of the fiftycent coins taken out.)
3.3u = 5.1p
> 1.1u = 1.7p
simplest form is 11u = 17p (E)
Now coming to each individual coins total (not value)
5u  16 = 7p (16 50C coins taken out or $8 value taken out)
5u = 7p + 16 (C)
4u + 40 = 8p ($8 value added or 40 20C coins added)
u + 10 = 2p (simplest form)
u = 2p  10
5u = 10p  50 (multiply equation by 5)  (D)
Equating (C) and (D)
7p + 16 = 10p 50
> 3p = 66
> p = 22
> u = 34 (Using step (E) )
Total 20C coins = 4u = 126 (value = 126 * 20/100 = $27.2)
Total 50C coins = 5u = 170 (value = 170 * 50/100 = $85.0)
Total value of coins in box = $27.2 + $85.0 = $112.2
(Actually, you can make use any 2 quations in (C, D, E) to get the number coins and calculate from their the value. but I have given all the 3 steps so that similar problems but questioned differently can also be solved applying the above steps.
For ex: You can skip step E and equate C & D or skip step C and equate D & E etc
then directly use 'p value also and calculate which comes to $35.2 & $77 for 176 & 154, 20C & 50C coins respectively)
Hope this helps....

Junior
Registered: 01/11/10 Posts: 16





By: Yian Cheng (offline) Tuesday, December 13 2011 @ 02:41 AM CST



Yian Cheng 
Thank you so much , pkrdd and wenying.
Pkrdd, can you solve this Q2 using models. My girl has not learned algebra.

Newbie
Registered: 09/30/11 Posts: 2





By: pkrdd (offline) Tuesday, December 13 2011 @ 02:48 AM CST



pkrdd 
I cannot post solution in models here. but if your girl do not know Algebra, she can relate the solution in units and parts method which is normally taught from P5.

Junior
Registered: 01/11/10 Posts: 16





By: KQ (offline) Wednesday, December 21 2011 @ 07:17 PM CST



KQ 
20cent 50cent
4u : 5u
+40 16
_____________________
8parts : 7parts
Value of twentycent added  16 x $0.50 = $8 (since the value of the 20 cent added is the same as the value of the 50 cent taken out)
No.of twenty cent added  $8 divide $0.20 = 40
Amount of money at first = Amount of money in the end
No. of coins at first +(4016) = No. of coins in the end
(x7)4u +40 = 8p
(x8)5u 16 =7p
28u +280 =56p
40u  128 = 56p
4028 = 280 +128
12u = 408
1u = 34
34x4+40 = 176 (20cent coins)
34x516 = 154 (50cent coins)
176x $0.20 +154x$0.50 =$35.20+$77
= $112.20[/u
That's how I worked it out...

Newbie
Registered: 11/15/11 Posts: 1



