Q1. a) To work out how many children like Gospel add all the numbers that fall within the Gospel circle: [Gospel = 18 + 9 + 7 + 6 = 40] b) To work out how many children like Country add all the numbers that fall within the Country circle: [Country = 29 + 9 + 7 + 5 = 50] c) To work out how many children like Jazz add all the numbers that fall within the Jazz circle: [Jazz = 17 + 6 + 7 + 5 = 35] d) To work out how many children like both Gospel and Country, add all the numbers that fall in the intersection. [Gospel & Country = 9 + 7 = 16] e) To work out how many children like both Gospel and Jazz, add all the numbers that fall in the intersection. [Gospel & Jazz = 6 + 7 = 13] Q2. a) [5 squared = 5² = 5 5 = 25] b) [20² = 20 20 = 400] c) Call the unknown number 'w'. The means: w w = 121 w² = 121 w = 121 [w = 11] 1
Q3. a) Use long division techniques to work out the answers to these problems and then arrange them in increasing order. Q4. 4 5 4 9 4 7 4 3 7 3 1 5 8 3 9 2 7 3 2 9 9 3 8 7 Question says to rearrange answers in increasing order: Answer = [43 45 47 49] Work out the length of the missing sides first as shown in blue. Then work out the area of the top section and the bottom section separately and then add them together to get area of the whole shape. [Area of a Rectangle = length width] a) Area = top section = 6 2 = 12 cm² Area = bottom = 2 6 = 12 cm² Area of shape = 12 cm² + 12 cm² [Area of shape = 24cm²] Perimeter is adding all the sides together. Starting from the top: b) [Perimeter = 6 + 2 + 2 + 6 + 2 + 6 + 2 + 2 = 28 cm] 2
Q5. Prime numbers are numbers that are only divisible by '1' and themselves. [The Prime Numbers = 13 17] Q6. To turn mixed fractions into decimal number, keep the whole number parts the same. Then use long division to turn the fractions parts into decimals. 9 1 13 3 17 5 25 3 8 5 8 20 [9.125 13.6 17.625 25.15 ] Q8. 1.46 US Dollars 1.42 Euros 118 Turkish Lira 2.13 Canadian Dollars a) To change 3.00 into US Dollars: [times both sides by '3'] 1.00 = 1.46 US Dollars 1.00 3 = 1.46 3 [ 3.00 = 4.38 US Dollars] b) To change 1.50 into Turkish Lira: [times both sides by '1.5'] 1.00 = 118 Turkish Lira 1.00 1.5 = 118 1.5 [ 1.50 = 177 Turkish Lira] c) To change 13.00 into Euros: [times both sides by '13'] 1.00 = 1.42 Euros 1.00 13 = 1.42 13 [ 13.00 = 18.46 Euros] 3
d) To change 50.00 into Canadian Dollars: [times both sides by '50'] 1.00 = 2.13 Canadian Dollars 1.00 50 = 2.13 50 [ 50.00 = 106.50 Canadian Dollars] Q8 Use algebra to work out the length and width of this rectangle. Let's call the width w and the length 3w (length is 3 times as long as width). w 3w [Perimeter of a rectangle = Adding up of all the sides] OR [Perimeter of a rectangle = (length 2) + (width 2)] Perimeter of the rectangle = (3w 2) + (w 2) 0.8 km (as given in question) = 6w + 2 w = 8w 8w = 0.8 [using algebra] w = 0.8 8 [using inverse] [w = 0.1] a) [Length of the rectangle (3w) = 0.3 km = 300 m] (3w = 0.1 3 = 0.3) b) [Width of the rectangle (w) = 0.1 km = 100 m] (To change from kilometres to meters, times by 1000) 4
[Area of a rectangle = length x width] c) (From questions 'a' and 'b', length = 0.3 km and width = 0.1 km) Area of the Rectangle = 0.3 0.1 [Area of the Rectangle = [0.03 km²] Q9. The skill you need to complete this question is inverse proportion. Click here to find out more: Step 1: No. of babies No. of days taken Step 2: 2 babies (twins) = [18 days] Step 3: 1 baby = [2 x 18 days] Step 4: 1 baby = [36 days] (Mrs Bruce adopts a baby so there are now 3 babies) Steps 5/6: 3 babies = [36 days 3] [Answer: 3 babies = 12 days] Q10. 565 656 556 655 566 665 The largest number = 665 The smallest number = + 556 Adding together (sum) = 1221 [Total of largest and smallest number = 1221] Q11. To find the number halfway between 2 given numbers, add the 2 given numbers together and divide the total by 2. 37 + 111 = 148 148 2 = 74 [Middle number = 74] 5
Use direct proportion to complete this question. Q12. No. of cans Cost 9 cans = 6.30 Divide 'both sides by '9' to work out the cost of 1 can. (9 9 = 1) ( 6.30 9 = 0.70) 1 can = 0.70 Times both sides by '11' to work out the cost of 11 items. 1 11 = 0.70 11 [11 cans = 7.70] Q13. Percentage means, a 'number over 100'. For example, 16% = 16 5% = 5 12% = 12 35% = 35 100 100 100 100 a) Work out 60 % of 30 (total number of pupils in Mrs Nicholson's class) to find number of girls. 60% of 30 = 60% 30 ['of' in Maths means ' '] (Use multiplication of fractions; simplify fraction by cancelling out the zeros) 60 30 = 6 3 = 18 100 [Number of girls = 18] b) Work out 40 % of 30 (total number of pupils in class to find number of boys. (Note: If girls make 60%, then boys will be 40%: '%' adds up to 100) 40% of 30 = 40% 30 ['of' in Maths means ' '] 6
(Use multiplication of fractions; simplify fraction by cancelling out the zeros) 40 30 = 4 3 = 12 100 [Number of boys = 12] An alternative way to get number of boys is to simply subtract number of girls worked out (18) from total number of pupils (30). [30 pupils 18 girls = 12 boys] Q14. The first thing to note here is that a dice has numbers 1, 2, 3, 4, 5 & 6. Each number occurs just once. This means each of the 6 numbers has a one in 6 chance of coming up when the dice is rolled. a) The probability of Jason rolling a '5' or a '6' = 2/6 = [⅓] [1 2 3 4 5 6] ['5' and '6' are two numbers out of the six possible numbers] b) The probability of Jason rolling a '3' = [1/6] [1 2 3 4 5 6] [There is only one number '3' out of the six possible numbers] c) The probability of tossing a number between 0 and 7 = [6/6 = 1] [0 1 2 3 4 5 6 7] [There are six possible numbers bigger than '0' and less than '7'] 7
Q15. Follow the steps below to work out the mixed fraction problems. a) 7 7 + 5 13 8 16 Add the whole numbers first and place this on the side to use later: [7 + 5 = 12] Now add the fractions by working out the LCM: LCM of 8 and 16 = 16 7 + 13 8 16 14 + 13 = 27 = 1 11 [change improper fraction] 16 16 16 16 [to mixed fraction] Bring back the whole number and add to the 1 11 worked out. 16 [12 + 1 11 = 13 11] 16 16 b) 7 1 3 11 5 15 Subtract the whole numbers first and place this on the side to use later: [7 3 = 4] Now subtract the fractions by working out the LCM: LCM of '5 and 15' = 15 1 11 3 11 [You can't do (3 11)] 5 15 15 15 So borrow '1' from the '4' on the side and add this '1' to 3 and then minus 11. Write down this '1' as 15 because 15 is the same as '1'. 15 15 15 15 [ 15 + 3 11 ] [18 11 = 7 ] 15 15 15 15 15 15 8
[Final answer = 3 7 ] [You have only 3 on the side now because] 15 [you borrowed '1' from the 4] Alternatively, you could change the 2 mixed fractions into improper fractions, work out the LCM and then subtract to get the answer. Q16. This question is testing your skill of working out Area and Perimeter of regular and irregular shapes. See working below: [Flag is a rectangle] [Area of a rectangle = length x width] [length of flag = 20 cm: (8 cm + 4 cm + 8 cm) [width of flag = 16 cm: (6 cm + 4 cm + 6 cm) Area of flag = 20 cm 16 cm a) [Area of flag = 320 cm²] b) To work out the area of the cross, work out the total area of the four white corners and subtract from area of whole flag. Length of each white corner = 8 cm Width of each white corner = 6 cm Area of each white corner = 8 cm 6 cm = 48 cm² 9
Realise that there are '4' white corners in total in the flag! *[Area of the '4' white corners = (48 cm² 4 = 192 cm²)] Area of the cross = Area of flag Area of the white corners [Area of the cross = 320 192 = 128 cm²] c) [Area of the '4' white corners = 192 cm²] See * above d) Perimeter of the flag is the total perimeter of the rectangle: [Perimeter of a rectangle = Adding up of all the sides] OR [Perimeter of a rectangle = (length x 2) + (width x 2)] Perimeter of flag (rectangle) = (20 2) + (16 2) Perimeter of flag = 40 cm + 32 cm [Perimeter of flag = 72 cm] e) Perimeter of the cross is adding the length of all the sides on the cross. Looking carefully, you can see 4 outer widths, 4 inside lengths and 4 inside widths. Add up all these sides together to get perimeter of the cross. 1. [outer width of cross = 4 cm: (4 4 cm)] 2. [inside length of cross = 8 cm: (4 8 cm)] 3. [inside width of cross = 6 cm: (4 6 cm)] Perimeter of cross = [(4 4) + (4 8) + (4 6)] Perimeter of cross = 16 cm + 32 cm + 24 cm [Perimeter of cross = 72 cm] 10
Q17. Follow the steps below to work out this questions. Number of street lights Number of gaps 2 1 3 2 4 3 5 4 6 5 Looking at the pattern above, you can see that the number of gaps is one less than the number of street lights. Using this same deduction, we can see that if: Number of street lights Number of gaps 25 = 24 Size of 1 gap = 85 metres 24 gaps = 85 m 24 = 2040 m [Total gap between all 25 poles = 2040 metres = 2.04 km] (To change meters to km, divide by 1000) Q18. [5% = 5 = 1 = ( 20)] 100 20 Original salary = 16,000.00 Percentage increase = 5 % Work out the percentage increase first. Then add this to the original salary. Increase in salary = 5 % of 16 000 = 5 of 16000 100 5 16 000 = 5 160 = 800 100 Cross out the zeros in the fraction. Remember 'of' in Maths means ' ' [Increase in salary = 800.00] Add this increase to the original wage. [New salary = ( 16 000.00 + 800.00) = 16 800.00] 11
Q19. Number of games = 6 games [Average/Mean score of the games = 12 points] Note: If you know the Mean, and you know how many games were played, multiply them together to get the Total number of points. Find out more here. Multiply mean score by 6 (number of games) to get the total number of points. Total number of points in 6 games = 12 6 = [72] If the Mean score went up to '13' in the next game, this means number of games played has also gone up by one to '7' games. This means: Total number of points in 7 games = 13 7 = [91] Score in seventh game = Total of 7 games Total of 6 games. [Score in seventh game = 91 72 = 19 points] Q20. If there are 6 girls to every 7 boys, this means the ratio of girls to boys is: Girls Boys 6 7 (ratios) 1. Add up the ratios [6 + 7 = 13] 2. Total number of of children in the school = 351 3. Divide total number of children by total of ratio 4. [351 13 = 27] 5. Multiply each of the ratios by '27' 6. The number of boys and girls should total '351' Girls Boys 6 7 (6 x 27 = 162) (7 x 27 = 189) a) [Girls = 162] b) [Boys = 189] 12
Q21. Daniel Jesse Stewart 9w 3w w [Call Stewart's goals 'w'. This means Jesse scored '3w' goals (3 times as many as Stewart and Daniel scored '9w' (3 times as many as Jesse)] 9w + 3w + w = 78 goals [using algebra] 13w = 78 goals w = 78 13 [w = 6 goals] If w = 6, then Daniel Jesse Stewart 9w 3w w [9 6 = 54] [3 6 = 18] [1 6 = 6] [Daniel scored 54 goals] [Jesse scored 18 goals] [Stewart scored 6 goals] Check: [If you add up all the goals the boys scored, they should total 78] Q22. Congregation of church = 364 [18 more women than men] [call number of men 'w'] [If No. of men = w, then No. of women will be (w + 18)] No. of men. + No. of women = 364 w + (w + 18)= 364 [using algebra] 2w = 364-18 2w = 346 w = 346 2 [w = 173] No. of men. + No. of women = 364 w + (w + 18) = 364 173 + 173 + 18 = 364 [173 men + 191 women = 364] 13
Q23. Total number of farm animals = 17 222 (represent the cattle by 'x', the pigs by 'y' and the sheep by 'z') cattle pigs sheep x y z (1) x + y + z = 17 222 [Total of cattle, pigs & sheep] Question says: (2) x + z = 9142 [cattle + sheep = 9142] (3) y + z = 13 201 [pigs + sheep = 13 201] Substitute 'y + z' = in equation (1) by 13 201 in equation (3) (1) x + (y + z) = 17 222 x + (13 201) = 17 222 x = 17 222 13 201 [using inverse] [x = 4 021] = [4 021 cattle] Substitute 4 021 for 'x' in equation (2) (2) x + z = 9142 4021 + z = 9142 z = 9142 4021 [using inverse] [z = 5121] = [5121 sheep] Substitute 5121 for 'y' in equation (3) (3) y + z = 13 201 y + 5121 = 13 201 y = 13 201 5121 [using inverse] [y = 8080] = [8080 pigs] The order in which you work out the number of animals does not matter, as long as you get the correct answers for all 3 types of farm animals. This means you can work out the number of pigs first, then cattle and then sheep. 14