Answers to problems sums in NY p6 prelim paper
I hope some of you attempted the Math Challenge I issued yesterday! Lilian posted Brian's workings for some of the problem sums in the Nanyang paper. Since she has asked for my workings, here they are, at least the ones that I got right. The sums where I erred, please refer to her post for the solutions. Do note however that my aim is to try and explain how the answers are derived, so they are text-heavy. In an actual exam, the kids just need to show the numbers.
The hardest questions were in Section C, from Q42 onwards, so here they are.
Q42 (a): The figure shows a rectangular field ABCD. Mike walked from A to B to C to D and he covered a distance of 57m. Sandy walked from B to C to D to A and she covered a distance of 48m. What is the perimeter of the field?
A: I used algebra, B = breadth, L = length. Based on Sandy's distance, 2B + L = 48 (equation 1). Based on Mike's distance, 2L + B = 57 (equation 2).
Using equation 1, L = 48 - 2B. Substitute this into equation 2, you'll get:
2 (48 - 2B) + B = 57
96 - 4B + B = 57
4B - B = 96 - 57
3B = 39
B = 39 ÷ 3 = 13
Mike's distance is just short of one breadth from the perimeter of the field, so 57 + 13 = 70
Answer: a) The perimeter of the field is 70m.
(b): Some construction work was undertaken on the same field and a semi-circular part was removed from it. What is the perimeter of the field after the construction? (Take ╥ = 22/7)
Note: I got this one wrong due to a careless mistake, but here's the correct answer.
A: First, find the perimeter of the semi-circle, which is ½ (╥d)
½ (22/7 x 10.5) = ½ (321/7) = 16.5
The two remaining bits of the breadth are B - 10.5 = 13 - 10.5 = 2.5
The perimeter of the field is Mike's distance + perimeter of semi-circle + remaining bits of B
57 + 16.5 + 2.5 = 76
Answer: b) The perimeter of the field after construction is 76m.
Q43: got this wrong, refer to Lilian's post.
Q44: The tickets for a show are priced at $10 and $5. The number of ten-dollar tickets available is 1½ times the number of five-dollar tickets. 5 out of 6 ten-dollar tickets and all the five-dollar tickets were sold. The ticket sales amounted to $5600. How much more would have been collected if all the tickets were sold?
A: This is the only sum in the entire paper where I used a model (bah, so much for being a model mum!) First, note that there are two different variables here, the number of tickets and the value of the tickets.
There are 1½ the number of ten-dollar tickets vs five-dollar tickets available, ie for every five-dollar ticket ($5), there is 1½ ten-dollar ticket ($15). In terms of value, the ten-dollar tickets are three times that of the five-dollar tickets. I drew a model for the value of the tickets available (right pic).
Next, we know 1/6 of the ten-dollar tickets were not sold, so I cut the ten-dollar ticket portion into six portions. Since I halve each of the 3 portions for the ten-dollar ticket, I have to do the same for the five-dollar tickets. (If you're not clear about the rules for math models, refer to my earlier post.) The shaded part is the part that is unsold.
Since ticket sales amounted to $5,600, $5,600 is represented by the 7 unshaded parts.
1 unknown part = 5600 ÷ 7 = 800 (which is equivalent to the shaded part)
Answer: $800 more would have been collected.
Q45 A rectangular tank measuring 60cm by 35cm by 40cm is half-filled with water. If Tap A is turned on, it will take 6 min to fill the remaining half of the tank to its brim. Tap B drains water from the tank at a rate of 12 litres per min. How long will it take for the tank to be filled to 1/8 of its capacity if both taps are turned on at the same time?
A: First, find how much water is in the tank, which is total volume of tank divided by 2.
(60 x 35 x 40) ÷ 2 = 84,000 ÷ 2 = 42,000 cm³ or 42 litres
The rate of Tap A is 42 ÷ 6 = 7 litres/min
The rate of Tap B is -12 litres/min (take note of the minus, since Tap B drains water, not adds water).
This means that every minute both taps are on, the tank loses 5 litres (7-12)
We need to find out how much time it takes for the tank to reach 1/8 of its capacity (which is 42 ÷ 4 = 10.5). The amount of water it needs to lose is 42 - 10.5 = 31.5.
- 5 litres takes 1 min
- 31.5 litres takes 31.5/5 = 6.3 mins
Answer: It takes 6.3 mins for the tank to be filled to 1/8 of its capacity.
Q46: got this wrong, refer to Lilian's post.
Q47. At a school carnival, there were 520 more girls than boys. 1/8 of the girls and 20% of the boys left the carnival. In the end, there were 488 more girls than boys. (a) Did more girls or boys leave the carnival? How many more?
A: If we put aside the 520 more girls at first, we have an equal number of boys and girls. So we convert 1/8 and 20% (1/5) into a common denominator to be able to compare the number of boys and girls who left the carnival, which is:
Girls - 5/40 + 1/8 x 520 = 5/40 + 65
Boys - 8/40
So the number of boys and girls still at the carnival is:
Girls - 35/40 + 520 - 65 = 35/40 + 455
Boys - 32/40
This means there were 3/40 + 455 more girls than boys still at the carnival, which is equivalent to 488.
3/40 + 455 = 488
3/40 = 488 - 455 = 33
1/40 = 33/3 = 11
Since 1/40 = 11, the number of girls who left the carnival is 5 x 11 + 65 = 120
The number of boys who left the carnival is 8 x 11 = 88
120 - 88 = 32
Answer: a) 32 more girls left the carnival than boys. (Note: This is the roundabout way! Lilian's is way more direct but I couldn't work it out at that time).
(b) How many children were there at the carnival in the end?
The number of children still at the carnival is:
35/40 + 455 (girls) + 32/40 (boys)
= 67/40 + 455
= 67 x 11 + 455 = 1192
Answer: b) There were 1,192 children at the carnival in the end.
Q48: Couldn't do this, refer to Lilian's post.
The hardest questions were in Section C, from Q42 onwards, so here they are.
Q42 (a): The figure shows a rectangular field ABCD. Mike walked from A to B to C to D and he covered a distance of 57m. Sandy walked from B to C to D to A and she covered a distance of 48m. What is the perimeter of the field?A: I used algebra, B = breadth, L = length. Based on Sandy's distance, 2B + L = 48 (equation 1). Based on Mike's distance, 2L + B = 57 (equation 2).
Using equation 1, L = 48 - 2B. Substitute this into equation 2, you'll get:
2 (48 - 2B) + B = 57
96 - 4B + B = 57
4B - B = 96 - 57
3B = 39
B = 39 ÷ 3 = 13
Mike's distance is just short of one breadth from the perimeter of the field, so 57 + 13 = 70
Answer: a) The perimeter of the field is 70m.
(b): Some construction work was undertaken on the same field and a semi-circular part was removed from it. What is the perimeter of the field after the construction? (Take ╥ = 22/7) Note: I got this one wrong due to a careless mistake, but here's the correct answer.
A: First, find the perimeter of the semi-circle, which is ½ (╥d)
½ (22/7 x 10.5) = ½ (321/7) = 16.5
The two remaining bits of the breadth are B - 10.5 = 13 - 10.5 = 2.5
The perimeter of the field is Mike's distance + perimeter of semi-circle + remaining bits of B
57 + 16.5 + 2.5 = 76
Answer: b) The perimeter of the field after construction is 76m.
Q43: got this wrong, refer to Lilian's post.
Q44: The tickets for a show are priced at $10 and $5. The number of ten-dollar tickets available is 1½ times the number of five-dollar tickets. 5 out of 6 ten-dollar tickets and all the five-dollar tickets were sold. The ticket sales amounted to $5600. How much more would have been collected if all the tickets were sold?
A: This is the only sum in the entire paper where I used a model (bah, so much for being a model mum!) First, note that there are two different variables here, the number of tickets and the value of the tickets.
There are 1½ the number of ten-dollar tickets vs five-dollar tickets available, ie for every five-dollar ticket ($5), there is 1½ ten-dollar ticket ($15). In terms of value, the ten-dollar tickets are three times that of the five-dollar tickets. I drew a model for the value of the tickets available (right pic).
Next, we know 1/6 of the ten-dollar tickets were not sold, so I cut the ten-dollar ticket portion into six portions. Since I halve each of the 3 portions for the ten-dollar ticket, I have to do the same for the five-dollar tickets. (If you're not clear about the rules for math models, refer to my earlier post.) The shaded part is the part that is unsold.Since ticket sales amounted to $5,600, $5,600 is represented by the 7 unshaded parts.
1 unknown part = 5600 ÷ 7 = 800 (which is equivalent to the shaded part)
Answer: $800 more would have been collected.
Q45 A rectangular tank measuring 60cm by 35cm by 40cm is half-filled with water. If Tap A is turned on, it will take 6 min to fill the remaining half of the tank to its brim. Tap B drains water from the tank at a rate of 12 litres per min. How long will it take for the tank to be filled to 1/8 of its capacity if both taps are turned on at the same time?
A: First, find how much water is in the tank, which is total volume of tank divided by 2.
(60 x 35 x 40) ÷ 2 = 84,000 ÷ 2 = 42,000 cm³ or 42 litres
The rate of Tap A is 42 ÷ 6 = 7 litres/min
The rate of Tap B is -12 litres/min (take note of the minus, since Tap B drains water, not adds water).
This means that every minute both taps are on, the tank loses 5 litres (7-12)
We need to find out how much time it takes for the tank to reach 1/8 of its capacity (which is 42 ÷ 4 = 10.5). The amount of water it needs to lose is 42 - 10.5 = 31.5.
- 5 litres takes 1 min
- 31.5 litres takes 31.5/5 = 6.3 mins
Answer: It takes 6.3 mins for the tank to be filled to 1/8 of its capacity.
Q46: got this wrong, refer to Lilian's post.
Q47. At a school carnival, there were 520 more girls than boys. 1/8 of the girls and 20% of the boys left the carnival. In the end, there were 488 more girls than boys. (a) Did more girls or boys leave the carnival? How many more?
A: If we put aside the 520 more girls at first, we have an equal number of boys and girls. So we convert 1/8 and 20% (1/5) into a common denominator to be able to compare the number of boys and girls who left the carnival, which is:
Girls - 5/40 + 1/8 x 520 = 5/40 + 65
Boys - 8/40
So the number of boys and girls still at the carnival is:
Girls - 35/40 + 520 - 65 = 35/40 + 455
Boys - 32/40
This means there were 3/40 + 455 more girls than boys still at the carnival, which is equivalent to 488.
3/40 + 455 = 488
3/40 = 488 - 455 = 33
1/40 = 33/3 = 11
Since 1/40 = 11, the number of girls who left the carnival is 5 x 11 + 65 = 120
The number of boys who left the carnival is 8 x 11 = 88
120 - 88 = 32
Answer: a) 32 more girls left the carnival than boys. (Note: This is the roundabout way! Lilian's is way more direct but I couldn't work it out at that time).
(b) How many children were there at the carnival in the end?
The number of children still at the carnival is:
35/40 + 455 (girls) + 32/40 (boys)
= 67/40 + 455
= 67 x 11 + 455 = 1192
Answer: b) There were 1,192 children at the carnival in the end.
Q48: Couldn't do this, refer to Lilian's post.
0 Response to "Answers to problems sums in NY p6 prelim paper"
Post a Comment