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 Need help in these few questions.
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By: drstevewu (offline)  Thursday, February 05 2015 @ 11:46 PM CST (Read 1381 times)  
drstevewu

Dear all, kindly help to solve the following questions. Thank you very much.

1. Tom spent a total of $36.40 on some stickers, pens and pencils. 1/4 of them were stickers and cost $0.50 each. The number of pens was 7 fewer than half the number of stickers and the rest of the items were pencils. A pen cost $1.20 and a pencil cost $0.40. How many pens did Tom buy?

2. Sam had a number of 20 cents and 50 cents coins. The number of 50 cents coins were three times as many as the number of 20 cents. He spent half of the total number of coins. The number of 50 cents coins spent was five times as many as the number of 20 cents spent. The total value of the coins left was $14.40. Find the total number of coins he had at first.

3. Jim had a number of 20 cents and 50 cents coins. The number of 50 cents coins was three times as many as the number of 20 cents coins. After spending 5/6 of the 50 cents coins and some 20 cents coins, he had 1/5 of the coins left. The total value of the coins left was $12.40. Find the total number of coins he spent altogether.

Thank you very much.


Yours sincerely,
Steve Wu

Forum Junior
Junior


Registered: 04/20/10
Posts: 27

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By: echeewh (offline)  Friday, February 06 2015 @ 04:50 AM CST  
echeewh

Hey Steve,

Following are my worked solutions:

Q1.

S=Stickers; P=Pens; C=Pencils
* - for alignment purpose

S *** |----|
P+C |----|----|----|

Given number of pens(P) was 7 fewer than half the number of stickers(S), we split the above model for S into 2 units(u). Hence,

S: 2u
P+C: 6u

But,
(1/2)S: 1u

P: 1u - 7

C: (P+C) - P = 6u - (1u - 7)
= 5u + 7

Given total cost of all these items was $36.40 and the unit cost of each item was known, apply <Number x Value> method and we have ...

(2u × 50) + [(1u - 7) × 120] + [(5u + 7) × 40] = 3640
100u + 120u - 840 + 200u + 280 = 3640
420u --> 3640 + 840 - 280 = 4200
1u --> 10

P = 1u - 7 --> 10 - 7 = 3

======

Q2.

<Before>
Total number of coins = 4p
50cents = 3p
20cents = 1p

<Spent>
Half the total coins = 2p -- (1)
50cents = 5u
20cents = 1u
Total coins (u) = 5u + 1u = 6u -- (2)

But (1)=(2) {half the total number of coins at first = total coins spent}

2p = 6u
1p = 3u

Hence,
<Before>
Total number of coins = 4p --> 4 × 3u = 12u
50cents = 3p --> 3 × 3u = 9u
20cents = 1p --> 3u

<After>
50cents = 9u - 5u = 4u
20cents = 3u - 1u = 2u

Given total value of the coins left was $14.40, apply <Number x Value> method and we have ...

(4u × 50) + (2u × 20) = 1440
200u + 40u --> 1440
240u --> 1440
1u --> 6

Total number of coins at first = 12u
12u --> 12 × 6 = 72

======

Q3.

<Before>
Total number of coins = 4p
50cents = 3p
20cents = 1p

<After>
(1/5)total coins left.

i.e. total coins at first = (5/5)

common multiple of 4, 5 is 20.

Hence,
<Before>
Total number of coins = 4p --> 4 × 5 = 20u
1p --> 5u
50cents = 3p --> 3 × 5 = 15u
20cents = 1p --> 5u

<Spent>
(5/6)50cents: (5/6) × 15u = (25/2)u

<After>
(1/5)total coins left = (1/5) × 20u = 4u
50cents: 15u - (25/2)u = (5/2)u
20cents: 4u - (5/2)u = (3/2)u

Given total value of the coins left was $12.40, apply <Number x Value> method and we have ...

[(5/2)u × 50] + [(3/2)u × 20) = 1240
125u + 30u --> 1240
155u --> 1240
1u --> 8

Total number of coins spent = 20u - 4u = 16u
16u --> 16 × 8 = 128

=======

Trust these help.

Do let me know again if these are different from the <Anskey> or if there's further clarification.

Cheers,
Edward

Forum Active Member
Active Member

Registered: 04/21/11
Posts: 623

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By: drstevewu (offline)  Friday, February 06 2015 @ 10:22 AM CST  
drstevewu

Hi Edward,

Thank you very much for your detailed workings. By the way, i have two more questions that i am looking for a better working solutions. Hope you can help to solve them:

1. Mr Wong numbered forty-two of his pupils by writing 1,2,3,...40, 41 and 42 on their name tags. She arranged all the forty-two pupils to stand in a row, standing from number 1 to number 42.
Next, for every second pupil, he asked each of the pupil to sit. Then for every third pupil, he asked each of the standing pupil to sit and each sitting pupil to stand.
She followed the same pattern (which is to ask each standing pupil to sit and each sitting pupil to stand) for every fourth pupil, and every fifth pupil and so on till the 42nd pupil.
How many pupil were standing after the whole process was completed?

2. Ann had two strings. One string is thrice as long as the other. She cut one string in equal parts of length 60 cm and to each part she tied 2 Big balloons. After that, she cut the other string into equal
parts of length 90 cm and to each part she tied 5 Small balloons. When she finished tying, she counted that there were 48 fewer big balloons than small balloons. How many balloons were there
altogether?


Thank you for your help.


Yours sincerely,
Steve Wu

Forum Junior
Junior

Registered: 04/20/10
Posts: 27

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By: echeewh (offline)  Friday, February 06 2015 @ 11:41 PM CST  
echeewh

Hey Steve,

Your next 2 questions are answered as follows:

Q1.

<Similar question appeared in NY 2013 Prelim>

Apply the knowledge of Factors and/or Multiples to solve this.

Start by listing it out and analyze the pattern.

For multiple of 2 (every 2 pupils):
Seated: ALL Even
Stand: ALL Odd

For multiple of 3 (every 3 pupils):
Even Stand: 6, 12, 18, 24, 30, 36, 42
Odd Seated: 3, 9, 15, 21, 27, 33, 39

For multiple of 4 (every 4 pupils):
Even Stand: 6, 18, 30, 42, 4, 8, 16, 20, 28, 32, 40
Odd Seated: 3, 9, 15, 21, 27, 33, 39

For multiple of 5 (every 5 pupils):
Even Stand: 6, 18, 42, 4, 8, 16, 28, 32, 10
Odd Seated: 3, 9, 21, 27, 33, 39, 5, 25, 35

For multiple of 6 (every 6 pupils):
Even Stand: 4, 8, 16, 28, 32, 10, 12, 24, 30, 36
Odd Seated: 3, 9, 21, 27, 33, 39, 5, 25, 35

Based on the pattern,

1 is always standing;

Look at 3. It was first standing (after 2), seated (after 3) and remains seated. Its factors are 1, 3.

Same with 5, 6.

This infers that a number that has an even number of factors will be seated.

Conversely speaking, a number that has an odd number of factors will be standing.

Take a look at 4. Factors of 4 are 1, 2, 4. So pupil 4 will be standing in the end.

Same with 9.

From 1 to 42, identify those numbers that have an odd number of factors.

These are ....

1, 4, 9, 16, 25, 36

Incidentally, all these are also square of numbers.

Hence 6 pupils were standing in the end.

======

Q2.

<Qty Balloons>
S - B = 48

Given that each part of length 90cm(L90) had 5 small(S) balloons and each part of length 60cm(L60) had 2 big(B ) balloons, this infers that L90 part comes from the longer string (LS).

<length>
L90 : L60
90 : 60
3 : 2

Given Long String (LS) : Short String (SS) is 3 : 1, and that L90 part comes from the LS, multiply the ratio component of L90 by 2x, and this gives the 3 : 1 comparison. Thus we have ...

2 parts of L90 and 1 part of L60 that gives the 3 : 1 (LS : SS) comparison (in terms of its total length)

L90 : L60
2 : 1

In terms of length, it becomes ...
180 : 60
3 : 1

In terms of number of balloons, we have ...
L90(S) : L60(B )
10 : 2

This gives a difference of 8 balloons {8 fewer B than S}.

Given that this difference is 48, which is a multiple of 6, we have ...

L90(S) : L60(B )
60 : 12 {48 fewer B than S}

therefore, there were 72 balloons altogether {60 + 12}

=======

Trust these help.

Do let me know again if these are different from your <Anskey> or if there's further clarification.

Cheers,
Edward

Forum Active Member
Active Member

Registered: 04/21/11
Posts: 623

Profile Email    
   


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