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 Forum Index >  Test Paper Related >  Primary 5 Matters P5 Maths 2011 SA2 RGPS Paper 2 Q6, Q10, Q12, Q16, Q17 & Q18
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 By: geniuskids (offline)  Sunday, October 14 2012 @ 09:35 AM CDT (Read 3176 times)
geniuskids

Hi, can someone pls help solved these questions!

Q6 : A pen costs \$1.10 less than a file. Karin can either buy 48 files or 92 pens with all her money.
How much does a pen cost?

Q10 : Sue had \$80 more than her brother. After Sue spent 25% of her money and her brother spent 75% of his money, they still had \$420 left altogether.

(a) How much did Sue left in the end?

(b) Express the amount of money her brother had left as a fraction of the amount of money Sue had in the end. (Give your answer in the simplest form)

Q12 : Kindly refer to the test paper as there are diagrams.

Q16 : The mass of sand in Sack B was 1/4 the mass of sand in Sack A. After 52.5kg of the sand in Sack A and 10kg 150g of sand in Sack B was used, the mass of sand in Sack A was 1/2 the mass of sand in Sack B. What was the total mass in Sack A and Sack B at first?

Q17 :The ratio of the number of \$2 notes and \$5 notes Jamie had was 15:13. She exchanged 60 pieces of \$2 notes for \$5 notes.
She then had an equal number of \$2 and \$5 notes.
What was the total value of \$2 and \$5 notes Jamie had?

Q18: The average number of sit-ups performed by 15 girls and some boys was 50.
The average number of sit-ups performed by the girls was 80% of the average number of sit-ups performed by all the pupils.
The average number of sit-ups performed by the boys was 30% more than the average number of sit-ups performed by the girls.
(a) What was the average number of sit ups performed by the girls?
(b) What percentage of teh pupils were boys?

Help needed Urgently!
Many thanks!

Active Member

Registered: 11/12/11
Posts: 169

 By: echeewh (offline)  Monday, October 15 2012 @ 01:18 AM CDT
echeewh

Hey there,

Im afraid there just too many questions to do at one time. Appreciate that next time you split your 6 questions into 2 or 3 postings in order that others can / may also help respond to them.

Anyway, following are the worked solutions for 3 of these. (Q6, 10,16, 17, 18). Will complete the remaining question shortly after.

=======================

Q6

F - P = 110
F = 110 + P

48F = 48 × (110 + P) = 5280 + 48P
5280 + 48P = 92P
92P - 48P = 5280
44P = 5280
P = 5280 ÷ 44 = 120
Hence cost of 1P = \$1.20

=================

Q10

<Before>
S |-----------|<-80->|
B |-----------|

<process>
S: -25% (1/4)
B: -75% (3/4)
Split the equal portions of S, B into 4 identical units (u).

<After>
S: 3u (of equal portion) + [(3/4) × 80] = 3u + 60
B: 1u

(a)

Hence
3u + 60 + 1u --> 420
4u --> 420 - 60 = 360
1u --> 90

S (left): 3u + 60 --> (3 × 90) + 60 = 270 + 60 = \$330

(b)

B (left): 1u --> \$90

Fraction = 90 / 330 = 3/11

==========================

Q16

A : B
4 : 1

-52.5 -10.15

1 : 2

Apply <Cross-Multiply> method, we have ...

(4u - 52.5) / 1 = (1u - 10.15) / 2
2 × (4u - 52.5) = (1u - 10.15)
8u - 105 = 1u - 10.15
8u - 1u --> 105 - 10.15
7u --> 94.85
1u --> 13.55

Total mass (A+B at first) = 5u --> 5 × 13.55 = 67.75 kg

================================

Q17

<before>
\$2 : \$5
15 : 13

<Process>
60 pieces of \$2 = \$120
120 ÷ 5 = 24 (pieces of \$5)

<before>
\$2 : \$5
15 : 13

-60 +24

<after>
1 : 1

Apply <Cross Multiply> method, we have ...

(15u - 60) / 1 = (13u + 24) / 1
15u - 13u --> 60 + 24
2u --> 84
1u --> 42

At first,
\$2 Qty: 15u --> 15 × 42 = 630
Value : 630 × 2 = \$1260

\$5 Qty: 13u --> 13 × 42 = 546
Value : 546 × 5 = \$2730

Hence, Total value = \$1260 + \$2730 = \$3990

===========================

Q18

(a)
Avg situps (G) = (80/100) × 50 = 40

(b)
Avg situps (B ) = 1.3 x 40 = 52

Let number of boys = 1u
Avg situps (all pupils) = [(40 × 15) + (52 × 1u)] / (15 + 1u) = 50

(600 + 52u) = 50 × (15 + 1u)
600 + 52u = 750 + 50u
52u - 50u --> 750 - 600
2u --> 150
1u --> 75

% pupils (B ) = (75 / 90) × 100 = 83 1/3 %

=========================

Trust these help.

Do let me know if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: echeewh (offline)  Wednesday, October 17 2012 @ 07:27 PM CDT
echeewh

Hey there,

18/10 - Post worked solution for final Q12 ...

Q12.

(Fig1):
Vol of water + Vol of 8 blocks = 10 × 448
= 4480 cm3 -- (1)
Base Area = 448 = 8u (blocks) × B (where B = breadth of container)
B = 448 ÷ 8u --> (56/u)

(Fig2):
Vol of water + Vol of 3 blocks = 22 x B x 3u
= 22 x (56/u) x 3u
= 3696 cm3 -- (2)

Vol of 5 blocks = 4480 - 3696 = 784
Vol of 8 blocks = (784 / 5) x 8 = 1254.4 cm3

in (1):
Vol of water = 4480 - Vol of 8 blocks
= 4480 - 1254.4
= 3225.6 cm3

=================

Trust this helps.

Do let me know again if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: geniuskids (offline)  Friday, October 19 2012 @ 08:04 AM CDT
geniuskids

Quote by: echeewh

Hey there,

Im afraid there just too many questions to do at one time. Appreciate that next time you split your 6 questions into 2 or 3 postings in order that others can / may also help respond to them.

Anyway, following are the worked solutions for 3 of these. (Q6, 10,16, 17, 18). Will complete the remaining question shortly after.

=======================

Q6

F - P = 110
F = 110 + P

48F = 48 × (110 + P) = 5280 + 48P
5280 + 48P = 92P
92P - 48P = 5280
44P = 5280
P = 5280 ÷ 44 = 120
Hence cost of 1P = \$1.20

=================

Q10

<Before>
S |-----------|<-80->|
B |-----------|

<process>
S: -25% (1/4)
B: -75% (3/4)
Split the equal portions of S, B into 4 identical units (u).

<After>
S: 3u (of equal portion) + [(3/4) × 80] = 3u + 60
B: 1u

(a)

Hence
3u + 60 + 1u --> 420
4u --> 420 - 60 = 360
1u --> 90

S (left): 3u + 60 --> (3 × 90) + 60 = 270 + 60 = \$330

(b)

B (left): 1u --> \$90

Fraction = 90 / 330 = 3/11

==========================

Q16

A : B
4 : 1

-52.5 -10.15

1 : 2

Apply <Cross-Multiply> method, we have ...

(4u - 52.5) / 1 = (1u - 10.15) / 2
2 × (4u - 52.5) = (1u - 10.15)
8u - 105 = 1u - 10.15
8u - 1u --> 105 - 10.15
7u --> 94.85
1u --> 13.55

Total mass (A+B at first) = 5u --> 5 × 13.55 = 67.75 kg

================================

Q17

<before>
\$2 : \$5
15 : 13

<Process>
60 pieces of \$2 = \$120
120 ÷ 5 = 24 (pieces of \$5)

<before>
\$2 : \$5
15 : 13

-60 +24

<after>
1 : 1

Apply <Cross Multiply> method, we have ...

(15u - 60) / 1 = (13u + 24) / 1
15u - 13u --> 60 + 24
2u --> 84
1u --> 42

At first,
\$2 Qty: 15u --> 15 × 42 = 630
Value : 630 × 2 = \$1260

\$5 Qty: 13u --> 13 × 42 = 546
Value : 546 × 5 = \$2730

Hence, Total value = \$1260 + \$2730 = \$3990

===========================

Q18

(a)
Avg situps (G) = (80/100) × 50 = 40

(b)
Avg situps (B ) = 1.3 x 40 = 52

Let number of boys = 1u
Avg situps (all pupils) = [(40 × 15) + (52 × 1u)] / (15 + 1u) = 50

(600 + 52u) = 50 × (15 + 1u)
600 + 52u = 750 + 50u
52u - 50u --> 750 - 600
2u --> 150
1u --> 75

% pupils (B ) = (75 / 90) × 100 = 83 1/3 %

Understand up to last portion: do not understand why 75/90 x100 (where do u get 90? Can explain pls.tks

Trust these help.

Do let me know if there's further clarification.

Cheers,
Edward

Active Member

Registered: 11/12/11
Posts: 169

 By: echeewh (offline)  Friday, October 19 2012 @ 07:47 PM CDT
echeewh

Hi there,

Question is (b) What percentage of the pupils were boys?

Boys = 75; Girls =15 (given in question)
Total (B+G) = 75 + 15 = 90
Hence, Fraction (B/Total) = (75/90)

=================

Trust the explanation helps.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: geniuskids (offline)  Friday, October 19 2012 @ 09:02 PM CDT
geniuskids

Thanks Edward. I managed to figure out 90 last night.

Thanks alot

Active Member

Registered: 11/12/11
Posts: 169

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