PSLE Math Problem Sums that Most P6 Pupils Stumble Over (Part 7)

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Baldwin Hancock
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1 1. Mandy has 23 one-dollar coins. Nicole has 148 one-dollar coins. Every day, Nicole gives Mandy 12 one-dollar coins each time while Mandy gives Nicole 6 one-dollar coins. After how many days would Mandy have 25% more money than Nicole? Mandy: Nicole Total At first Change (+6) : (-6) for every day... In the end 5 units : 4 units 9 units 25% ; Mandy has 1 unit more than Nicole 9 units unit 19 4 units 76 (Nicole) = 72 one-dollar coins given to Mandy Since the number of Mandy s one-dollar coins increased by 6 every day, 72 6 = 12 days Ans: After 12 days, Mandy will have 25% more money than Nicole.

2 2. The ratio of the number of $2 notes to the number of $5 notes to the number of $10 notes Roy had was 1 : 6 : 4. After spending of the $2 notes and $210 using only $5 notes, the value of the $10 notes Roy had became of the total amount of money he had left. Find the total amount of money Roy had at first. Method 1: Number of $5 notes spent = $210 $5 = 42 $2 : $5 : $10 Ratio of the number of notes 1 : 6 : 4 x3 (At first) 3 units : 18 units : 12 units Change -1 unit -42 Ratio of the number of notes 2 units : : 12 units (In the end) Ratio of the value of money 4 units : : 120 units (In the end) 2 $2 notes = $4 12 $10 notes = $120 Ratio of value of $2 : $10 = 4 : 120 Since the value of $10 notes in the end was of the total amount of money left, ; the total amount of money left 144 units = 20 units (value of $5 notes) Number of $5 notes 20 5 = 4 units 18 4 = 14 units 14 units 42 1 unit 3 3 units 9 ($2 notes) ; Value $18 18 units 54 ($5 notes) ; Value $ units 36 ($10 notes) ; Value $360 Total amount of money at first $18 + $270 + $360 = $648 Ans: Roy had $648 at first.

3 Method 2: $2 : $5 : $10 Ratio of the number of notes 1 : 6 : 4 Grouping $2 : $30 : $40 Ratio of the value of money 1 unit : 15 units : 20 units x3 (At first) 3 units : 45 units : 60 units Change -1 unit -$210 Ratio of the value of money (In the end) 2 units : 10 units : 60 units Since the value of $10 notes in the end was of the total amount of money left, ; the total amount of money left 72 units = 10 units (value of $5 notes) 35 units $210 1 unit $6 Total number of units (value of money) at first = 108 units 108 units $648 Ans: Roy had $648 at first.

4 3. In the final year examination, Janet sat for a total of 5 papers based on 5 different subjects. Her best subject was Mathematics, followed by Mother Tongue, Science, Social Studies and English. Janet obtained the sum of the scores of 2 subjects. The following are all the possible sums: 157, 163, 166, 170, 171, 173, 178, 179, 184, 187 What score did Janet obtain for her Science paper? Given 5 different subjects as A, B, C, D and E, the 10 possible sums include A+B, A+C, A+D, A+E, B+C, B+D, B+E, C+D, C+E, and D+E. Since each subject is counted 4 times among the 10 possible sums which means 4A + 4B + 4C + 4D + 4E = total of 10 sums Total of 10 sums = 1728 total score of all 5 papers = 432 Combined score of English and Social Studies papers 157 Combined score of Mother Tongue and Mathematics papers 187 Score of Janet s Science paper = 88 Ans: Janet obtained a score of 88 for her Science paper.

5 4. Boys and girls attended a birthday party. If 36 boys left the party, of the children remaining would be girls. If 36 girls left the party, 48% of the children remaining would be boys. How many children were there at the birthday party? In both cases, the total number of children remaining does not change. Case 1: If 18 boys left the party Boys : Girls Total?? Change -36 Case 1 3 : 7 10 units 30 units : 70 units 100 units x10 Case 2: If 18 girls left the party Boys : Girls Total?? Change -36 Case 2 48 units : 52 units 100 units Comparing the number of boys in both cases, = 18 units 18 units 36 1 unit 2 Total number of children 70 units + 48 units = 118 units 118 units 236 Ans: There were 236 children at the birthday party.

6 5. Mr Chew had 3 large wooden cubes painted grey on all faces. However, he decided to cut each large wooden cube into 8 identical smaller wooden cubes and paint the unpainted faces of the smaller cubes grey. He realized that he had painted a total area of cm 2 on the unpainted faces of the smaller wooden cubes. What was the total volume, in cubic metres, of the 3 large wooden cubes Mr Chew had at first? Since each smaller wooden cube has 3 unpainted faces, and each large wooden cube can be cut into 8 smaller wooden cubes, Total number of square faces unpainted = 72 Area of 1 square face = 2025 cm 2 Length of the edge of a small wooden cube = 45 cm Length of the edge of a large wooden cube 45 2 = 90 cm = 0.9 m Total volume of 3 large wooden cubes 0.9 m 0.9 m 0.9 m 3 = m 3 Ans: The total volume of the 3 large wooden cubes was m 3.

7 6. Mrs Smith and Mrs Tate went shopping with a total sum of $4580. They went to a boutique and each bought the same dress. Mrs Smith, who had membership, was entitled to a 20% discount on the dress. After the purchase, she had 30% of her money left. Mrs Tate, who was not a member, was entitled to a 5% discount. After the purchase, she had 40% of her money left. (a) Who had more money at first, and how much more? (b) What was the price of the dress they bought? Mrs Smith: 30% of money left ; 70% or spent Paid 80% of dress price of her money 4 20% of dress price 4 = of her money 5 100% of dress price 5 = of her money Mrs Tate: 40% of money left ; 60% or = spent Paid 95% of dress price of her money 19 5% of dress price 19 = of her money % of dress price 20 = of her money Equivalent (Dress price) of Mrs Smith s money of Mrs Tate s money Mrs Smith = = Mrs Tate = 229 units Difference = 37 units 229 units $ unit $20 37 units $740 Ans: (a) Mrs Tate has $740 more than Mrs Smith at first.

8 (continued) Dress price was of Mrs Smith s money at first 96 units $1920 (Mrs Smith s money at first) Dress price $1920 = $1680 Ans: (b) The price of the dress was $1680.

9 7. Ivan and Jeth shared a bag of marbles. Ivan received 65% of the marbles. Later, Ivan bought another 40 marbles while Jeth lost 2 of his marbles. In the end, 30% of Ivan s marbles was equal to 80% of Jeth s marbles. What was the total number of marbles both of them had in the end? 65% = ; 30% = ; 80% = In the end, of Ivan s marbles was equal to of Jeth s marbles. = Ivan s marbles : 40 units ; = Jeth s marbles : 15 units Ivan : Jeth (In the end) 40 : 15 = 8 : 3 Ivan : Jeth At first 13 units : 7 units Change OR 30% of Ivan = 80% of Jeth Ivan Jeth In the end 8 : 3 80% 30% = 8 : 3 3 (13 units + 40) 8 (7 units 2) 39 units units units = unit 8 At first, the total number of marbles (20 units) 160 Total number of marbles in the end = 198 Ans: Both of them had 198 marbles in the end.

10 8. The diagram below, not drawn to scale, shows the floor plan of a walkway in a park. The walkway is 1.12 m wide and consists of circular, identical cemented tiles of negligible thickness. Two identical grass patches, made up of artificial grass mats which cost $16.50 per square metre, are placed alongside the walkway. Given that the grass patches cost a total of $6468, (a) how many circular cemented tiles are there altogether; and (b) what is the total area, in square centimetres, of the walkway covered by all the circular cemented tiles? (Take π = ) m Grass Patch Walkway Grass Patch

11 Diameter of a circular tile 1.12 m 2 = 0.56 m Cost of a grass patch $ = $3234 Area of 1 grass patch $3234 $16.50 = 196 m 2 Breadth of a grass patch (5 circular tiles) 0.56 m 5 = 2.8 m Length of a grass patch 196 m m = 70 m Number of circular tiles forming up the length of a grass patch 70 m 0.56 m = 125 Number of circular tiles forming up the breadth of a grass patch Grass Patch Grass Patch Total number of circular cemented tiles ( ) = 794 Ans: (a) There are 794 circular cemented tiles altogether.

12 (continued) Radius of a circular tile 28 cm Area of a circular tile 28 cm 28 cm = 2464 cm 2 Total area of walkway covered by all the circular cemented tiles 2464 cm = cm 2 Ans: (b) The total area of the walkway covered by all the circular cemented tiles is cm 2.

13 9. Brandon and Carell can complete a painting assignment together in 20 minutes. If both of them started the painting assignment for 5 minutes, followed by Brandon who continued painting for 21 minutes, Carell would take another 7 minutes to complete the entire painting assignment. How many minutes would it take each of them to complete the painting assignment alone? 20 mins 1 painting mins painting B + C B C C painting 21 mins 7 mins After the first 5 mins, if C had continued painting with B for 7 mins, his part will have been completed. 1 min painting (B and C together) 7 mins = mins = painting Remaining part of painting to be completed by B 1 = Brandon : painting 21 7 = 13 mins painting 13 3 = = mins painting 8 = 36 mins Carell : 1 min = = = painting painting 45 mins Ans: Brandon : 36 mins ; Carell : 45 mins

14 10. At 6.30 p.m., Jack and Jill were at the park and they started to run towards the seafood centre along the same route. Jill was running at a speed of 180 m/min. After 12 minutes, Jack arrived at a bicycle kiosk to get a bicycle. He cycled back at a speed 200 m/min faster than his running speed to meet Jill. At 6.45 p.m., both of them met at a point mid-way between the park and the seafood centre. Jack picked Jill up and continued cycling to the seafood centre at a speed 3 times as fast as Jill s running speed. (a) At what time did Jack and Jill arrive at the seafood centre? (b) Find the ratio of Jack s running speed to Jill s running speed p.m. 15 min 6.45 p.m. Park Meeting Bicycle Seafood 6.30 p.m. Point Kiosk Centre 6.45 p.m. 12 min Jack Jill 2 min 12 min 2 min (when Jack returned) 15 min Distance between park and meeting point (mid-way) 180 m/min 15 min = 2700 m Distance between meeting point and seafood centre 2700 m Cycling speed = 540 m/min Time taken to cycle from meeting point to seafood centre 6.45 p.m. 5 min 6.50 p.m. = 5 min Ans: (a) Jack and Jill arrived at the seafood centre at 6.50 p.m.

15 (continued) Let Jack s running speed be 1 unit (distance) per min ; and his cycling speed would be 1 unit m (distance) per min For 12 min, Jack ran 1 unit 12 min = 12 units At bicycle kiosk, Jack cycled back (1 unit m/min) 2 min = 2 units m Distance between park and meeting point (mid-way) 2700 m 12 units (2 units m) 2700 m 10 units 2700 m m = 3200 m 1 unit 320 m Jack s running speed 320 m/min Jill s running speed 180 m/min 320 : 180 = 16 : 9 Ans: (b) The ratio of Jack s running speed to Jill s running speed was 16 : 9.

16 11. A school conducts a fund-raising event each year. Compared to last year, the number of boys who participated remains the same this year and there are 25% more girls who participated this year. The total amount of funds raised last year was $78 800, whereas the total amount of funds raised this year is $ Given that the average amount of funds raised by each boy and that of each girl this year increased by 10% and 20% respectively as compared to last year, find (a) the amount of funds raised by the girls this year, and (b) the increase in the number of girls who participated, if the average amount of funds raised by each girl last year was $115. Let the amount of funds raised by girls last year be 100 units This year, assume average funds remains unchanged, due to 25% more girls amount of funds raised by girls should be 100 units = 125 units As average funds raised by each girl increased by 20%, amount of funds raised by girls this year 125 units = 150 units Last Year : This Year Total $ $ Funds raised by girls -100 units : -150 units Funds raised by boys 10 : 11 Funds raised by boys increased by 10% (this year versus last year) without any change in number of boys 110% = 11 ($ units) 10 ($ units) $ units $ units 400 units $ unit $ units $ Ans: (a) The girls raised a total of $ this year.

17 (continued) 100 units $ (raised by girls last year) Number of girls who participated last year $ $115 = 320 Increase in the number of girls (last year to this year) 320 = 80 Ans: (b) The number of girls who participated in fund-raising increased by 80.

18 12. A watermelon, with 93% of its weight being water, was left under the Sun at noon. At 1 p.m., some of its water evaporated such that only 92% of the watermelon s weight is due to water. Given that the amount of water evaporated every hour is the same, what will the percentage of water by weight of the watermelon be at 4 p.m.? 92% = = ; 93% = Watermelon : Water At noon (12 p.m.) 7 units : 93 units 8 8 Weight of the fruit itself without water 56 units 744 units -? 100 units of water evaporated At 1 p.m. 8 units : 92 units 7 7 At 1 p.m. 644 units of water 56 units 644 units Since each hour, amount of water evaporated is 100 units, at 4 p.m. (3 hours later) 644 (3 100) = 344 units Weight of watermelon 56 units units = 400 units Percentage of water 100% = 86% Ans: The percentage of water at 4 p.m. will be 86%.

Cumulative Test. Name. Score. Show all work on this paper. Please use the Student Reference Guide.

Cumulative Test. Name. Score. Show all work on this paper. Please use the Student Reference Guide. Name Score Math Course 1 1A 1. Use the numbers 6, 12, and 18 to make two addition facts and two subtraction facts. 12 + 12 12 + 12 12 12 12 12 2. Use the numbers 5, 15, and 75 to make two multiplication