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Wednesday, June 28 2017 @ 07:22 AM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 Help on 4 maths questions (updated 13 July)
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By: Kuro (offline)  Wednesday, July 09 2014 @ 11:30 PM CDT (Read 1197 times)  
Kuro

Q1 and Q2. Pls refer to attachment.

Q3. Ms Lee bought DVDs at $4 each and sold them at $9 each. Due to limited stock, customers were only allowed to buy either 1 or 2 DVDs. Those who bought 2 DVDs were given one free DVD.
Ms Lee made a gain of $1250 while giving away 135 free DVDs. How many customers bought only one DVD??

Q4. On the first day of a camp, 600 pupils were divided equally into Teams A and B. The ratio of the number of boys to that of girls in Team A was 1:5. 40% of the pupils in Team B were boys.
On the second day, some of the pupils transferred teams such that there was an equal number of boys and girls in Team A. In Team B, the number of boys was 1/14 of the girls. How many boys transferred from Team B to Team A?

No answer keys given.

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By: echeewh (offline)  Thursday, July 10 2014 @ 02:51 AM CDT  
echeewh

Dear ALL,

For my worked solutions , pls refer to a latter thread as Questions are identical. Thanks.

Cheers,
Edward

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By: Kuro (offline)  Saturday, July 12 2014 @ 10:24 AM CDT  
Kuro

My apology. questions amended.

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By: echeewh (offline)  Sunday, July 13 2014 @ 06:06 AM CDT  
echeewh

Hey <Kuro>

Following are my worked solutions to Q3 , 4 ( will do Q1, 2 in due course soon ) :

Q3.

For each DVD sold:
Amount Gained(Earned) = 9 - 4 = 5

For every 2 DVDs sold, 1 DVD given free:
Amount Gained(Earned) = (2 × 5) - 4 = 10 - 4 = 6

Since 135 DVDs were given away free {referring to customers having bought 2 DVDs each},
Amount Gained(Earned) = 135 × 6 = 810

Given total amount gained(earned) is 1250,
Amount Gained(Earned) (from customers having bought 1 DVD each) = 1250 - 810 = 440

Hence, number of customers having bought 1 DVD each
= 440 ÷ 5 = 88

======

Q4.

<Before>

Team A = 300

B(A) : G(A)
1 : 5

B(A) = (1/6) × 300 = 50
G(A) = 300 - 50 = 250

Team B = 300

B(B ) : G(B )
40 : 60
2 : 3

B(B ) = (2/5) × 300 = 120
G(B ) = 300 - 120 = 180

<After>

B(A) : G(A)
1 : 1

B(B ) : G(B )
1 : 14

Applying <Unchanged Total> concept { Number of Boys (A+B ) <Before> = Number of Boys (A+B ) <After>; Likewise the Girls(G) },

in <After> model, let B(A), G(A) be p(arts) and B(B ), G(B ) be u(nits).

We now have ...

1p + 1u = 50 + 120 = 170 -- (1)
1p + 14u = 250 + 180 = 430 -- (2)

(2)-(1):
14u - 1u --> 430 - 170
13u --> 260
1u --> 20

B(B ) <After> = 1u --> 20

Number of B transferred from Team B:
120 - 20 = 100

=======

Trust these help.

Do let me know again if these are different from your <Answerkey> or if there are further clarifications.

Cheers,
Edward

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By: echeewh (offline)  Monday, July 14 2014 @ 12:08 AM CDT  
echeewh

Hey <Kuro>

Following are my worked solutions to Q1 , 2 (refer to questions in attached file):

Q1.

From graph, 8 lit water were drained in 8 min with A only turned on ...

Rate of flow (Tap A) = 1 lit / min

From graph, in the next 4 min , 20 lit water were drained with both Taps A, B turned on. Given Tap A rate , 4 lit of these were drained by Tap A in this 4 min.

Rate of flow (Tap B ) = 16 ÷ 4 = 4 lit / min

If only Tap B was turned on for 7 1/2 min from start, we have ...

Vol of water drained = 4 × (15/2) = 30 lit

Thus, vol of water left in tank = 40 - 30 = 10 lit

Ht of water in tank = (10 × 1000) ÷ (40 × 60) ~ 4.2 cm (correct to 1 dec place)

=======

Q2.

Given length of each square is a whole number and the sum of areas of the 3 squares is 120 cm^2, we have ...

By <Trial & Error> method,

lets have A , B be 1 and 2 cm respectively (the smallest lengths). Thus, C is 13 cm.

This is not valid as Area C is 169 { exceeds given value of 120 }

Infers length of C cannot be 12, 11 ...and so we have length of C be 10.

Area C = 100; Remaining Areas (A + B ) = 20

Think of smaller lengths for A, B, and squares of numbers that will make 20 (A + B ), we have ...

20 = 4 + 16 , where both 4, 16 are squares of numbers.

Hence, Area B = 16 cm^2

========

Trust these help.

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

Cheers,
Edward

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