A time to relax a little

Now that school holidays are around, it is time to relax a little. Just a little, not too much, or you will have problems re-adjusting back to school life when the next semester starts.

ACS Primary 2010 SA1 Math Paper 2 Q18

Keane bought some marbles and gave half of them to Leon. Leon then bought some stamps and gave half of the to Keane. Keane used 5 stamps and Leon gave away 11 marbles. The ratio of the number of stamps to the number of marbles Keane had left then became 1:7 and the ratio of the number of stamps to the number of marbles Leon had left became 1:5. How many stamps did Leon buy?

Solution

M ----- Marbles
S ----- Stamps

Step 1
Keane bought marbles and gave half to Leon,
Keane ----- M
Leon ----- M
Both have same number of marbles.

Step 2
Leon bought stamps and gave half to Keane,
Keane ----- M , S
Leon ----- M , S
Now both have same number of marbles and stamps

Step 3
Keane used 5 stamps, Leon gave away 11 marbles,
Keane ----- M , [S - 5]
Leon ----- [M - 11] , S
Now Keane has 5 stamps less, and Leon 11 marbles less

Step 4
Keane's no. of stamps to no. of marbles now is 1:7,
M ----- (7 units)
[S - 5] ----- (1 unit)
Now Keane has 7 times as many marbles as stamps.

Step 5
Leon's no. of stamps to no. of marbles now is 1:5,
[M - 11] ----- (5 parts)
S ----- (1 part)
Now Leon has 5 times as many stamps as marbles.

Step 6
1 unit of Keane's stamps + 5 ----- 1 part of Leon's stamps
1 unit + 5 ----- 1 part
Keane gave away 5 stamps, so to have an equal no. of stamps as Leon now, we add back the 5 stamps Keane gave away.

Step 7
7 units of Keane's marbles ----- 5 parts of Leon's marbles + 11
7 units ----- 5 parts + 11
Leon gave away 11 marbles, so to have an equal no. of marbles as Keane, we add back the 11 marbles Leon gave away.

Step 8
Comparing marbles with stamps
(Stamps) 1 unit + 5 ----- 1 part (refer to Step 6)
(Marbles) 7 units ----- 5 parts + 11 (refer to Step 7)

Step 9
Multiply stamps by 5
(Stamps) 5 units + 25 ----- 5 parts
(Marbles) 7 units --> 5 parts + 11

(Stamps) 5 units ----- 5 parts - 25
(Marbles) 7 units ----- 5 parts + 11

Step 10
(marbles) - (stamps)
7 units - 5 units --> 5 parts + 11 - 5 parts - (-25)
2 units --> 11 + 25
2 units --> 36
1 unit --> 18

Step 11
1 unit ----- S - 5 (refer to Step 4)
18 ----- S - 5
S ----- 18 + 5 = 23
This is the no. of stamps Leon had after giving half away.

Step 12
Leon bought,
2 x 23 = 46

Answer: 46 stamps

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ACS Primary 2010 SA1 Math Paper 2 Q17

The shaded figure below is formed by semicircles, quarter circles and straight lines of 15 cm each. For each of the following, use the calculator value of to find
a) the perimeter of the shaded figure, correct to 2 decimal places.
b) the area of the shaded figure, correct to 2 decimal places.

Solution

(a)
Circumference of 2 semicircles (1 circle)
--> 2r
= 2 x 7.5cm
= 47.12 cm

Circumference of 2 quarter circles (1 semi circle)
-->()2r
= ()(2)() x 15cm
= 47.12 cm

Length of vertical line on the left side of figure
--> 2 x 15cm = 30cm


Perimeter of shaded figure
--> 47.12cm + 47.12cm + 30cm
= 124.25cm

Answer: 124.25 cm


(b)
Area of 2 quarter circles (1 semi circle)
-->()
= () x 15cm x 15cm
= 353.43


Area of 2 semi circles (1 circle)
-->
= x 7.5cm x 7.5 cm
= 176.71

Area of whole figure
--> 30cm x 30cm
= 900

Shaded area
--> 900 - 353.43 - 176.71
= 369.86

Answer: 369.86

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ACS Primary 2010 SA1 Math Paper 2 Q16

Mr Wu had some badges and decided to give them to his two sons, Sean and Matthew. Mr Wu gave 1/3 of the badges and 8 more badges to Sean. He then gave 3/4 of the remainder to Matthew but took back 2 badges. Mr Wu was left with 26 badges. How many badges did Mr Wu have at first?

Solution

Assume all badges --> 1 unit

Mr Wu was left with,

----- 1/4 of Remainder + 2

----- 1/4 x (2/3 unit - 8) + 2

= (1/6 unit - 2) + 2

= 1/6 unit - 2 + 2

= 1/6 unit

1/6 unit --> 26 badges

1 unit ----- 26 divided by 1/6
= 26 x 6
= 156

Answer: 156 badges

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ACS Primary 2010 SA1 Math Paper 2 Q15

Both Joanne and Joseph had an equal amount of money at first. Every month, Joanne spent $850 and Joseph spent $912. After a few months, Joanne was left with $1550 while Joseph had 4/5 as much as Joanne. How much money did Joseph have at first?

Solution

Every month,
Joanne spends $850
Joseph spends $912
Difference every month
--> $912 - $850 = $62


After a few months, Joseph had 4/5 as much as Joanne,

5 units ----- $1550
1 unit ------ $1550 divided by 5 = $310
($310 is the difference between Joanne and Joseph after a certain number of months)


Number of months to achieve the $310 difference,
$310 divided by $62 = 5 (mths)
It also takes 5 months for Joseph to have 4/5 as much as Joanne.


Amount Joseph has at first,
Joseph ------ 4 units + (5 mths x $912 spent per mth)
----- (4 x $310) + (5 x $912)
= $1240 + $4560
= $5800

Answer: $5800


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Another Problem Sum from a reader

Received another question from a reader.
Can anyone help this reader out

Question:
Mrs tan bought 4 times as many toys as teddy bears . She spent $1750 altogether . A toy gun cost $10 less than a teddy bear . The total cost of toy guns was $490 more than the total cost of teddy bears .
(a) How much did Mrs tan spend on the teddy bears?

(B) How much did one teddy bear cost ?

ACS Primary 2010 SA1 Math Paper 2 Q14

Tony and Charles took part in a car race. Tony drove at a speed of 90km/h. Both of them did not change their speed throughout the race. When Charles had covered 1/3 the distance, Tony was 15 km in front of him. Tony reached the finishing line at 9.35 a.m. At what time did Charles recah the finishing line?

Solution

* For every 1/3 of the race Charles covered, Tony was 15 km ahead. When Charles completed the whole race, Tony would have been 3 x 15km = 45km ahead of Charles, if we assume Tony continued to travel beyond the finishing line.

Time = 45 km divided 90 km/h
= 0.5 hour

Half hour after 9.35 am ------ 10.05 am

Answer: 10.05 am

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ACS Primary 2010 SA1 Math Paper 2 Q12

Weiming had 2/3 as many stickers as Shiyang. After Weiming gave 52 stickers to Shiyang, Weiming had 2/5 as many stickers as Shiyang. How many stickers did Weiming have at first?

Solution

*(x7) and (x5) to give a common total of 35 units for both "At first" and "In the end", since there is no change in the total number of stickers.

Weiming gave
14 units - 10 units ---- 52
4 units ----- 52
1 unit ----- 52 divided by 4 = 13

Weiming at first
14 units x 13 = 182

Answer: 182 stickers

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ACS Primary 2010 SA1 Math Paper 2 Q9

The graph below shows the mass of a container when it is empty and when different combinations of objects X, Y and Z are placed in it.

Solution

Empty container --> 80g
Y + Z ----- 140g - 80g = 60g
X + Z ----- 300g - 80g = 220g
X + Y ----- 340 g - 80g = 260g

Total of all objects
----- 60g + 220g + 260g = 540g

Average ----- 540g divided by 6 = 90g


Answer: 90g

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ACS Primary 2010 SA1 Math Paper 2 Q6

At first, Matthew has twice as many soccer cards as Ivan. Each of them then bought the same number of cards. As a result, both of the now have 160 cards in total. If Matthew now has 30 more cards than Ivan, find the number of cards each of them bought.

Solution

bought + bought ----- 160 - 30 - 30 - 30 = 70
bought ----- 70 divided by 2 = 35


Answer: 35 soccer cards

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