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 By: Yian Cheng (offline)  Saturday, December 10 2011 @ 12:00 AM CST (Read 2097 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 twenty-cent and fifty-cent coins in the ratio of 4:5.
When 16 fifty-cent coins were taken out and replaced by some twenty-cent coins, the ratio then became 8:7.
The total value of the twenty-cent coins added was the same as the total value of the fifty-cent 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,

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 twenty-cent coins added was the same as the total value of the fifty-cent 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 twenty-cent 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 +(40-16) = No. of coins in the end

(x7)4u +40 = 8p
(x8)5u -16 =7p
28u +280 =56p
40u - 128 = 56p
40-28 = 280 +128
12u = 408
1u = 34
34x4+40 = 176 (20cent coins)
34x5-16 = 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

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