Q1-3. To complete the table, pick a column where you have been given both the top and the bottom numbers. Work out the relationship between the top and the bottom number. Apply the same rule to all columns. We can see from columns 1 and 2 that: [Bottom number 12 = Top number] OR [Top number 12 = Bottom number] 60 144 72 [132] 120 5 12 [6] 11 [10] Q4. [0.38 11 = 4.10] Principle to use here is Multiplication of Decimal Numbers using Long Multiplication. Click on the links to learn more. There are 2 decimal places altogether your answer should also have 2 decimal places. See working below) 0. 3 8 1 1 0 3 8 0 3 8 0 0 4. 1 8 [Ignore '0' in front of the '4' - unnecessary] [Final answer = 4.18] Q5. Use Long Division techniques to complete this question. See how below. 0. 5 7 9 5. 1 3 4 5 6 3 6 3 0 [5.13 9 = 0.57] 1 Access more @
Q6. (To subtract, align the numbers so that numbers with the same place value are in line directly below each other. See below) (Use borrowing in subtraction) 1 0 3 0 7 5 2 2 7 8 [1030 752 = 278] Q7-10. Conversion graphs are used to convert one set of variables to another e.g. work out the distance travelled in a specific time interval. Read the following instructions carefully to learn how to answer questions on conversion graphs easily. How to use 'distance, speed and time' formula triangle: Cover the value you want to work out, and read off the other values. To work out (distance) 'd', cover 'd' and say: To work out (speed) 's', cover 's' and say: To work out (time) 't' (time), cover 't' and say: d = s x t s = d t t = d s Using the formula triangle above where d = distance, s = speed and t = time, we can see that: [d = s x t] [s = d t] and [t = d s] 2 Access more @
Q7. Tracing you finger from the 2 km mark on the distance (Y)axis, move your finger to the RIGHT, until it touches the line graph. Then move your finger down from that point until it touches the time (X) axis. If you do this correctly, it should give you the answer. [Time to complete 2km = 10 minutes] Q8. Tracing you finger from the 1 km mark on the distance (Y)axis, move your finger to the RIGHT, until it touches the line graph. Then move your finger down from that point until it touches the time (X) axis. If you do this correctly, it should give you the answer. [Time to complete 1 km = 5 minutes] Q9. Start from the 25 min point on the X-axis, then move your finger up till it touches the line graph. Then move your finger to the LEFT, until it touches the Y-axis. Read off the number on the Y-axis. [Distance travelled in 25 minutes = 5 km] Q10. [Speed = distance time] We can see from looking at the line graph that, it takes 20 minutes to cover a distance of 4 km. Use direct proportion to work out the rest of the question. 20 minutes = 4 km ( 3) ( 3) [Times both sides by '3' to work out distance travelled in one hour] 60 minutes = 12 km (1 hour) = 12 km [60 minutes = 1 hour] [Speed in km/h = 12 km/h] 3 Access more @
Q11. We are going to call the unknown number 'w' w 25 = 45 40 w = (45 40) 25 [using inverse] w = 1800 25 [w = 72] Q12. 3 + a = 15 a = 15 3 [using inverse] [a = 12] Q13. x + 4 = 7 x = 7 4 [using inverse] [x = 3] Q14. 5 + a = 7 a = 7 5 [using inverse] [a = 2] Q15. y 7 = 9 y = 9 + 7 [using inverse] [y = 16] Q16. b 4 = 8 b = 8 + 4 [using inverse] [b = 12] Q17. c 1 = 5 c = 5 + 1 [using inverse] [c = 6] 4 Access more @
Q18-21. Use Long Division techniques to complete the questions. Then circle the division sum with the largest answer (quotient). 1 2 8 1 3 1 1 3 2 1 2 9 7 8 9 6 6 7 8 6 9 1 1 8 8 4 5 1 6 7 6 9 4 1 9 1 8 2 8 1 1 1 4 1 8 2 7 8 5 6 0 6 1 8 3 6 5 6 6 1 8 3 6 0 0 0 0 [ 1 2 8 ] [ 1 3 1 ] [ 1 3 2 ] [ 1 2 9 ] 1 3 2 Q22. [The division: 9 1 1 8 8 has the largest quotient] Q23. [Split 19 minutes into 15 mins + 4 mins. Then add them separately] 3:45 pm + 15mins = 4:00pm 4:00 pm + 4 mins = 4:04pm [Time getting home = 4:04 pm] Q24. Monday 10:29pm - Monday 12:00 midnight = 1 hour 31 mins Monday 12:00 midnight - Tuesday 2:05 am = 2 hours 5 mins Adding the 2 'times' together = [3 hours 36 mins] (60 minutes = 1 hour 3 hours 36 minutes = 180 + 36) [Total minutes = 216 minutes] Q25. From the pie chart, ¼ of the circle represents = 72 walkers. (¼ 4 = 1 whole) 72 4 = 288 [Number of pupils in School A = 288] 5 Access more @
Q26. From the pie chart, ½ of the circle represents = 117 walkers. (½ 2 = 1 whole) 117 2 = 234 [Number of pupils in School B = 234] Q27. From the pie chart, 1/6 of the circle represents = 47walkers. (1/6 6 = 1 whole) 47 6 = 282 [Number of pupils in School C = 282] Q28. From the pie chart, 1/8 of the circle represents = 33 walkers. (1/8 8 = 1 whole) 33 8 = 264 [Number of pupils in School D = 264] 6 Access more @
Q29 30. Total angle about a STRAIGHT line = 180 2x + x = 180 [using algebra] 3x = 180 Q29. [x = 60 ] x = 180 3 [using inverse] x = 60 Q30. [2x = (60 2) = 120 ] Q31-32. Total angle about a STRAIGHT line = 180 a + a + a = 90 [using algebra] 3a = 90 (right angle) Q31. [a = 30 ] a = 90 3 [using inverse] Angles b + b = 90 2b = 90 Q32. [b = 45 ] b = 90 2 7 Access more @
Q33-43. Starting from the end of the 5 th lesson (12:50pm), subtract 40 minutes first to work out the start of the 5 th lesson (12:10am). The start of the 5 th lesson is the same as the end of the 4 th lesson. Subtract 40 minutes again to get the start of the 4 th lesson (11:30am). This is the same as the end of the third lesson. Subtract 40 minutes again to get the start of the 3 rd lesson (10:50am). This is the same as the end of break period. Then from 10:50 am (when the break ends), subtract 15 minutes to work out the start of the break time (10:35 am). This is the same as end of the 2 nd lesson. From here, subtract 40 minutes each time. See working below. Lesson begins Lesson ends 1 st Lesson 09:15:00 09:55:00 2 nd Lesson 09:55:00 10:35:00 Break 10:35:00 10:50:00 3 rd Lesson 10:50:00 11:30:00 4 th Lesson 11:30:00 12:10:00 5 th Lesson 12:10:00 12:50:00 Q44-45. Boys Girls 5 4 (ratios) 1. Add up the ratios [5 + 4 = 9] 2. Total number of children in school = 630 3. Divide total number of children by total of ratio 4. [630 9 = 70] 5. Multiply each of the ratios by '70' 6. The number of boys and girls should total '630' Boys Girls 5 4 (5 70 = 350) (4 70 = 280) [Boys = 350] [Girls = 280] [Check: Total number of boys and number of girls should give 630] 8 Access more @
Q46 47. Perimeter of Rectangle = 40 cm Length Width [call width 'w' cm] 4w w 4w + w + 4w + w = 40 cm [using algebra] 10w = 40 w = 40 10 [using inverse] [w = 4 cm] Q46. [Length '4w' = (4 4) = 16 cm] Q47. [Width 'w' = 4 cm] Q48. Use direct proportion to complete this question. Click the link to find out more on direct proportion. 1 kg of Washing powder = 1.20 (120p) [1kg = 1000g] Mass of powder (kg) Cost of powder ( ) 1 120p = 1.20 To move from 1.20 to 6.00, you will need to multiply by '5'. So times both sides by '5' to work out the amount of powder to get for the 6.00 of money. (1 5 = 5) (120 5 = 600p) 5kg 600p ( 6.00) [ 6.00 = 5kg of Washing powder] 9 Access more @
Q49. Use direct proportion to complete this question. Click the link to find out more on direct proportion. 1 kg of Washing powder = 1.20 (120p) [1kg = 1000g] Mass of powder (kg) Cost of powder ( ) 1 = 1000g 120p = 1.20 To move from 1.20 to 0.30, you will need to divide by '4'. So divide both sides by '4' to work out the amount of powder to get for the 0.30 of money. (1000 4 = 250g) (120 4 = 30p) 250g 30p ( 0.30) [ 0.30 = 250g of Washing powder] Q50. Use direct proportion to complete this question. Click the link to find out more on direct proportion. 1 kg of Washing powder = 1.20 (120p) [1kg = 1000g] Mass of powder (kg) Cost of powder ( ) 1 = 1000g 120p = 1.20 To move from 1 kg to 3.5kg, you will need to multiply by '3.5'. So times both sides by '3.5' to work out the amount of money to pay for 3.5 kg of of washing powder. (1 3.5 = 3.5) (120 3.5 = 420p) 3.5kg 420p ( 4.20) [ 3.5kg of Washing powder = 4.20] 10 Access more @