CSS Primary Standard “Mathematics” 101 Q3. Find the sum of the following integer on number line. (i) –15, –5 Ans: (–15) + (–5) = –20 (ii) –60, –10 Ans: (–60) + (–10) = –70 (iii) –12, –2 Ans: (–12) + (–2) = –14 Q4. Use the number line to find the sum. (i) Ans: 4 + (–5) = –1
CSS Primary Standard “Mathematics” 102 (ii) Ans: –7 + 2 = –5 Q5. Add following. (i) 923, –713 Ans: 923 + (–713) = 210 (ii) –133, +115 Ans: (–133) + (+115) = –18 (iii) –78434, –443 Ans: (–78434) + (–443) = –78877 (iv) –8434, –3474 Ans: (–8434) + (–3474) = –11908 (v) 4444, –474 Ans: 4444 + (–474) = 3970 (vi) –8434, –3474 Ans: (–133) + (+115) = –11908 Q6. Find the sum of the following by three step method. (i) +745, –392 Ans:
CSS Primary Standard “Mathematics” 103 Step - 1: Take the absolute values of the integers |745| = 745, |–392| = 392 Step - 2: Subtract the smaller absolute value from the larger absolute value. 745 – 392 = 353 Step - 3: Number 353 take the positive sign as 743 is larger number and it is positive. Hence (745) + (–392) = 353 (ii) 1496 – 1135 Ans: Step - 1: Take the absolute values of the integers |1496| = 1496, |–1135| = 1135 Step - 2: Subtract the smaller absolute value from the larger absolute value. 1496 – 1135 = 361 Step - 3: Number 361 will be positive sign as 1496 is a larger number and it is positive. Hence (1496) + (–1135) = 361 (iii) –87632 – 92523 Ans: Step - 1: Take the absolute value of the integers. |–87632| = 87632, |–92528| = 92528 Step - 2: Subtract the smaller absolute value from the larger absolute value. 87632 + 92528 = 180160 Step - 3: Number 180160 take the negative sign as both integers have negative signs. Hence (–87632) + (–92528) = –180160 (iv) –8434 – 3474 Ans: Step - 1: Take the absolute value of the integers. |–8434| = 8434, |–3474| = 3474 Step - 2:
CSS Primary Standard “Mathematics” 104 Add the smaller absolute value and larger absolute value. 8434 + 3474 = 11908 Step - 3: Number 11908 will take negative sign as both integers have negative signs. Hence (–8434) + (–3474) = –11908 Q7: If Ahmad borrows Rs. 300 from his mother. If he borrows another Rs. 900 from her. How much he own her altogether? Show his loan on numbers line. Ans: Rs. 300 + Rs 900 = Rs 1200 Exercise 4c Exercise 4c Q1. Subtract the following. (i) 64 from 89 Ans: 89 – 64 = 25 (ii) 1834 from 3397 Ans: 3397 – 1834 = 1563 (iii) –8643 from –9837 Ans: –9837 – (–8643) = –9837 + 8643 = –1194 (iv) –10937 from –35935 Ans: (–35935) – (–10937) = –35935 + 10937 = –24998 (v) –89325 from 93654 Ans: 93654 – (–89325) = 93654 + 89325 = 182979 (vi) –96384 from 118937
CSS Primary Standard “Mathematics” 105 Ans: 118937 – (–96384) = 118937 + 96384 = 215321 (vii) –65937 from –89563 Ans: –89563 – 65937 = –89563 –65937 = –155500 (viii) 80371 from –32596 Ans: –32596 – 80371 = –32596 – 80371 = –112967 Week 2 Exercise 4d Q1. Fill the missing integers. (i) Ans: 3 × –6 = –18 (ii) Ans: –15 × ( –2) = 30 (iii) Ans: (–5) × 15 = –75 (iv) Ans: 12 × 9 = 108 (v) Ans: 6 × 4 = 24 (vi) Ans: 7 × –13 = –91 (vii) Ans: –9 × (–1) = 9 (viii) Ans: (–2) × 5 = –10
CSS Primary Standard “Mathematics” 106 Q2. Simply. (i) (–7) × (–8) × 1 Ans: Step - 1: |–7| = 7, |–8| = 8 Step - 2: 7 × 8 × 1 = 56 Step - 3: 56, we take sign due to rule no. 1 as both integer have same signs. Hence (–7) × (–8) × 1 = 56 (ii) (–8) × 9 × 7 Ans: Step - 1: |–8| = 8, |9| = 9, |7| = 7 Step - 2: 8 × 9 × 7 = 504 Step - 3: 504 take the negative signs as the sign of two integers are unlike Hence (–8) × 9 × 7 = –504 (iii) 3 × (–8) × 6 × 1 Ans: Step - 1: |3| = 3, |–8| = 8, |6| = 6, |1| = 1 Step - 2: 3 × 8 × 6 × 1 = 144 Step - 3: 144 will take negative sign as the signs of two integers are unlike. 3 × (–8) × 6 × 1 = – 144 (iv) (–16) × (–7) × (–9) Ans: Step - 1: |–16| = 16, |–7| = 7, |–9| = 9 Step - 2: 16 × 7 × 9 = 1008 Step - 3: 1008 will take the negative sign. (–16) × (–7) × (–9) = –1008
CSS Primary Standard “Mathematics” 107 (v) 4 × 0 × 7 Ans: Step - 1: 4 × 0 × 7 = 0 (because one of three terms is zero). Step - 2: 0 has no sign. (vi) 2 × 3 × (–5) × (–16) × 2 Ans: Step - 1: |4| = 2, |3| = 3, |–5| = 5, |–16| = 16, |2| = 2 (because one of three terms is zero). Step - 2: 2 × 3 × 5 × 16 × 2 = 960 Step - 3: 960 will take positive sign. Hence 2 × 3 × (–5) × (–16) × 2 = 960 (vii) (–16) × (–7) × (–9) Ans: Step - 1: |–16| = 16, |–7| = 7, |–9| = 9 (because one of three terms is zero). Step - 2: 16 × 7 × 9 = 1008 Step - 3: 1008 will take negative sign. Hence, (–16) × (–7) × (–9) = –1008 (vii) (–16) × (–7) × (–9) Ans: Step - 1: |–16| = 16, |–7| = 7, |–9| = 9 Step - 2: 16 × 7 × 9 = 1008 Step - 3: 1008 will take negative sign. Hence, (–16) × (–7) × (–9) = –1008 (viii) 15 × (–3) × 14 × (–12) Ans: Step - 1:
CSS Primary Standard “Mathematics” 108 |15| = 15, |–3| = 3, |14| = 14, |–12| = 12 Step - 2: 15 × 3 × 14 × 12 = 7560 Step - 3: 7560 will take the positive sign Hence 15 × (–3) × 14 × (–12) = 7560 (ix) (–14) × (–5) × 3 × 8 × 10 Ans: Step - 1: |–14| = 14, |–5| = 5, |3| = 3, |8| = 8, |10| = 10 Step - 2: 14 × 5 × 3 × 8 × 10 = 16800 Step - 3: 16800 take positive sign. Hence (–14) × (–5) × 3 × 8 × 10 = 16800 (x) 7 × (–11) × (20) ×10 Ans: Step - 1: |7| = 7, |–11| = 11, |20| = 20, |10| = 10 Step - 2: 7 × 11 × 20 × 10 = 15400 Step - 3: 15400 will take the negative sign as. Hence, 7 × (–11) × 20 × 10 = –15400 Q3. Simply. (i) 15 × 15 Ans: Step - 1: 15 × 15 = 225 Step - 2: 225 will take positive sign. Hence 15 × 15 = 225 (ii) (14) × (–13) Ans: Step - 1: |14| = 14, |–13| = 13 Step - 2:
CSS Primary Standard “Mathematics” 109 14 × 13 = 182 Step - 3: 182 will take negative sign because both intege5rsw have unlike signs. Hence (14) × (–13) = –182 (iii) –150 × –14 Ans: Step - 1: |–150| = 150, |–14| = 14 Step - 2: 150 × 14 = 2100 Step - 3: 2100 will take the positive sign as both integers have like signs. Hence –150 × –14 = 2100 (iv) 130 × 130 Ans: Step - 1: 130 × 130 = 16900 Step - 2: 16900 has positive sign as both integers have like sign. Hence 130 × 130 = 16900 (v) (–70) × (25) Ans: Step - 1: |–70| = 70, |25| = 25 Step - 2: 70 × 25 = 1750 Step - 3: 1750 will take negative sign as both integers have unlike signs. Hence (–70) × 25 = –1750 (vi) –141 × –89 Ans: Step - 1: |–141| = 141, |89| = 89 Step - 2: 141 × 89 = 12549 Step - 3: 12549 has positive sign because both integers have like signs.
CSS Primary Standard “Mathematics” 110 Hence –141 × (–89) = 12549 (vii) –170 × 800 Ans: Step - 1: |–170| = 170, |800| = 800 Step - 2: 170 × 800 = 136000 Step - 3: 136000 has negative sign as both integers have unlike signs. Hence, –170 × 800 = –136000 (viii) 7539 × –189 Ans: Step - 1: |7539| = 7539, |–189| = 189 Step - 2: 7539 × 189 = 1424871 Step - 3: 1424871 has negative sign, as both integers have unlike signs. Hence 7539 × –189 = –14248171 Exercise 4e Q.1 Solve it. (i) 55 ÷ 10 = (ii) 400 ÷ 2 = Ans : Ans : 11 55 ÷ 10 = 400 2 200 2 (iii) 700 ÷ 14 = (iv) 75 + ( 15) = Ans : Ans : 700 ÷ 14 = 50 75 15 90 (v) 10000 ÷ 5 = (vi) 175 ÷ 7 = Ans : Ans : 10000÷ 5 = 2000 175 7 25 (vii) 128 ÷ 8 = (viii) 128 ÷ 49 = Ans : Ans :
CSS Primary Standard “Mathematics” 111 128 128 ÷ 8 = 16 128 49 49 28 364 ÷ 65 = 209556 5821 36 5 (ix) 364 ÷ 65 = (x) 209556 ÷ 5821 = Ans : Ans : Q.2 Fill the box with suitable integer. (i) –1800 ÷ = –18 (ii) –1400 ÷ 2 = (iii) ÷ 258 = 236 (iv) ÷ 3 = 888 (v) ÷ –1 = 420 (vi) 12500 ÷ = –25 (vii) –2063635 ÷ = 58961 Q.3 Find the quotient of the following. (i) –9504 ÷ 4 Ans: 2376 4 9504 8 15 12 30 28 24 24 0 Hence –9504 ÷ 4 = –2376 (ii) 96768 ÷ –216 Ans:
CSS Primary Standard “Mathematics” 112 448 216 96768 864 1036 864 1728 1728 0 Hence 96768 ÷ –216 = –448 (iii) –96768 ÷ –216 Ans: 964 253 243,892 2277 1619 1518 1012 1012 0 Hence –243,892 ÷ –253 = 964 (iv) 335,750 ÷ 850 Ans: 395 850 335,750 2550 8075 7650 4250 4250 0 Hence 335,750 ÷ 850 = 395 Q.4 440 ÷ 0 is this possible? Ans:
CSS Primary Standard “Mathematics” 113 No it is not possible because division by 0 does not work as any number divided by 0 is infinity. Q.5 Explain if any number divide by 1 remain the same. Ans: Since 1 is multiplicative identity, therefore any number when multiply by 1 remain the same. Review Exercise 4 Q 1: Choose the correct answer and fill the circle: i. On the left side of zero, on number line, integers are: negative zero whole number positive ii. Natural numbers are also called __________ numbers: whole number counting number integers real numbers iii. Negative numbers added to __________ numbers become integers: natural number counting numbers whole numbers real numbers iv. Absolute value is the distance of a number from____________: number line total sum zero 1 v. Integers were denoted by: Z W I F vi. To add two positive numbers, add their absolute values, the result is _______: positive zero negative none of these vii. |–12| =_________: 12 –12 ± 12 –|12| viii. (–1) + (–3) = _________ + 4 –4 ± 4 3 ix. (60) × (–50) =_________: 300 3000 –3000 –300
CSS Primary Standard “Mathematics” 114 viii. (–60) + (–50) = _________ –185 150 3000 –3000 Q2. Fill the missing integers. (i) (–120) + _________ = 13 (ii) _________ + (–67894) = –20 Ans: Ans: (iii) 190 – _________ = 185 (iv) _________ – (–150) = –50 Ans: Ans: (v) 18 – _________ = 14 (vi) _________ (–340) = –6 Ans: Ans: (vii) _________ –220 = +340 (viii) 9 × _________ = 18 Ans: Ans: (ix) 4 – _________ = –9 (x) (–1) ×_________ = –10 Ans: Ans: (xi) 3 + _________ = –16 (xii) _________ + (–4) = –4 Ans: Ans: (xiii) 4 – _________ = –9 (xiv) _________ – 17 = –1 Ans: Ans: Q3. Represent on number line to describe each situation. (i) Karachi king loss the math by 50 runs. Ans: (ii) Peshawar Zalmi won the match by 5 runs. Ans: (iii) Anoral ran 1 meter to right. Ans:
CSS Primary Standard “Mathematics” 115 (iv) Alishba ran 2 meter toward right. Ans: Q4. Draw the following numbers on number line. 0, ±10, ±30, ±50, ±60 Ans: Q5. Add –50 and 20 on the number line. Ans: – 50 + 20 = –30 Q6. Display the sum of –9 and 8 on a number line. Ans: –50 + 20 = –30 Q7. Given integers –18, 10, 19, –20, 400, –1, 0, 35, –15. Ans: Step 1: Arrange integer in ascending order. –20, –18, –15, –1, 0, 10, 19, 35 400 Step 2: Find absolute values of integers. |–18| = 18, |10| = 10, |19| = 19, |–20| = 20, |400| = 400, |–1| = 1, |0| = 0, |35| = 35, |– 15| = 15. 18, 10, 19, 20, 400, 1, 0, 35, 15 Step 3:
CSS Primary Standard “Mathematics” 116 Arrange the absolute values of integers in ascending order. 0, 1, 10, 15, 18, 19, 20, 35, 400 Q8. Add the following . (i) +78945, –38493 Ans: 78945 + (–38493) = 78945 – 38493 = 40452 (ii) –937546, –105935 Ans: –937546 + (–105935) = –937546 – 105935 = –1043481 (iii) 345968, +698324 Ans: 345968 + 698324 = 1044292 (iv) –923540, +564239 Ans: –923540 + 564239 = –359301 Q9. Find the following products. (i) 9378 × 325 Ans: 9378 325 46890 187560 2813400 3047850 (ii) 10056 × 136 Ans: 10056 136 60336 301680 1005600 1367616 Hence 10056 × 136 = 1367616 (iii) 1235 × 123 Ans: 1235 123 3705
CSS Primary Standard “Mathematics” 117 24700 133500 151905 Hence 1235 × 123 = 151905 Q9 Find the quotient in the following. (i) 98547 ÷ 307 Ans: 321 307 98547 921 644 614 307 307 0 Hence 98547 ÷ 307 = 321 (ii) 338520 ÷ 546 Ans: 620 546 338520 3276 10920 10920 0 Hence 338520 ÷ 546 = 620 (iii) 1032400 ÷ 145 Ans: 7120 145 1032400 1015 174 145 2900 2900 0
CSS Primary Standard “Mathematics” 118 Hence 1032400 ÷ 145 = 7120 Week 3 Unit:5 Simplification Teacher Objectives: ☻ To introduce perfect square. ☻ To explain the method to check whether a number is a perfect square or not. ☻ To introduce and explain the following properties of perfect square of a number. The square of an even number is even. The square of an odd number is odd. The square of a proper fraction is loss then itself. The square of a decimal less than 1 is smaller item decimal. ☻ To introduce the square root of natural number and its notation. ☻ Explain the method to find square root, by division method and factorization method of, natural number, fraction and decimal, which is perfect square. ☻ Explain the solution of real life problems involving square roots. Learning outcome: ☻ Know that the following four kinds of brackets –––– vinculum, ( ) parentheses or curved brackets or round brackets, { } braces or curly brackets, { } square brackets or box brackets, are used to group two or more numbers together with operations. ☻ Know the order of preference as, ––––, ( ), { } and [ ], to remove (simplify) them from an expression. ☻ Recognize BODMAS rule to follow the order in which the operations, to simplify mathematical expressions, are performed. ☻ Simplify mathematical expressions involving fractions and decimals grouped with brackets using BODMAS rule. ☻ Solve real life problems involving fractions and decimals. Teacher materials. ☻ CSS Primary Standard Mathematics Book 6. ☻ Writing Board. ☻ Marker. ☻ Eraser.
CSS Primary Standard “Mathematics” 119 Classroom Activity: The following table shows the numbers from 10 to 20, complete the squares. Numbers Sequences 10 11 12 13 14 15 16 17 18 19 20 102 = 10 × 10 = 100 112 = 11 × 11 = 121 122 = 12 × 12 = 144 132 = 13 × 13 = 169 142 = 14 × 14 = 196 152 = 15 × 15 = 225 162 = 16 × 16 = 256 172 = 17 × 17 = 289 182 = 18 × 18 = 324 192 = 19 × 19 = 361 202 = 20 × 20 = 400 If we take decimal number which is less then 1 then its square will be less than that decimal number. Take the square of the following number. Number Squares 0.1 0.2 0.6 0.25 0.5 0.01 0.04 0.36 0.0625 0.25 What will be the number of zeroes in the square of the following numbers? (i) 20 (ii) 200 (iii) 2000 (iv) 20000 (i) (20)2 = 400 (ii) (200)2 = 40000 (iii) (2000)2 = 4000000 (iv) (20000)2 = 400000000 From the above result, it is clear that number of zeroes will become double after taking the square of number. Exercise 5a Q1. Simplify the following. (i) 157 – {170 × 10 (144 + 850 – 70)} Ans:
CSS Primary Standard “Mathematics” 120 = 157 – {170 ÷ 10 (994 – 707)} = 157 – {170 ÷ 10 (287)} = 157 – {170 ÷ 2870} = 157 – {170 ÷ 2870} 170 157 2870 17 157 287 45059 17 287 45042 287 (ii) 234740 + {3750 + (570550 ÷ 5)} Ans: = 234740 + {3750 + 114110} = 234740 + 117860 = 352600 (iii) 100 – {50 – (10 + 2)} Ans: = 100 – {50 – 12} = 100 – 38 = 62 (iv) 410000 – {2345 + 11745 (1050 ÷ 10)} Ans: = 410000 – {2345 + 1745(105)} = 410000 – {2345 + 183225} = 410000 – 185570 = 224430 (v) 147484 + [18760 × 3 –{1474 – (875 ÷ 5)}] Ans: = 147484 + [18760 × 3 – {1474 – 175}] = 147484 + [18760 × 3 – 1299} = 147484 + [56280 – 1299] = 147484 + 54981 = 202465 (vi) 53474 + [843 × 5 –{1500 ÷ (3000 ÷ 100)}] Ans: = 53474 + [843 × 5 –1500 ÷ 30}]
CSS Primary Standard “Mathematics” 121 = 53474 + [843 × 5 – 50} = 53474 + [4215 – 50] = 53474 + 4165 = 57639 (vii) 3764564 + [14450 ÷ 100 {3987 – (3987 – 274 + (880 ÷ 10)}] Ans: = 3764564 + [14450 ÷ 100 ÷{3713 + (880 ÷ 10)}] = 3764564 + [14450 ÷ 100 + {3713 + 88}] = 3764564 + [14450 ÷ 100 + 3801] 14450 3764564 3801 100 14450 3801 3764564 10 39455 3764564 10 3764564 39455 37685095 10 10 Classroom Activity: Find the square root of the following numbers. 81, 121, 100, 64, 169, 225, 196. 2 81 9 9 9 9 2 121 11 11 11 11 2 100 10 10 10 10 2 64 8 8 8 8 2 169 13 13 13 13 2 225 15 15 15 15 2 196 14 14 14 14 2 81 9 9 9 9 2 81 9 9 9 9 2 81 9 9 9 9 Find the square root of 36 and square of 6. 36 6 6 2 = 36 Square root is the inverse operation of square.
CSS Primary Standard “Mathematics” 122 Exercise 5b Q1. Simplify the following. 4 3 4 16 100 (i) + × ÷ 14 20 9 15 130 4 3 4 16 100 + × ÷ 14 20 9 15 130 Ans : 2 3 4 16 10 + × ÷ 7 20 9 15 13 2 3 4 16 10 + × ÷ 7 20 9 15 13 2 1 10 + 7 16 13 416 91 1120 1456 613 1456 4 3 4 16 100 (ii) + × ÷ 14 20 9 15 130 10 13 9 25 13 + 8 8 10 40 Ans : 114 13 9 5 + 8 8 10 8 114 13 9 5 + 8 8 10 8 114 13 36 + 25 8 8 40 114 13 61 8 8 40 114 65 61 8 40 114 4 8 40
CSS Primary Standard “Mathematics” 123 114 1 8 10 57 1 4 10 285 2 20 283 20 1 11 100 1 14 (iii) 12 ÷ 15 2 7 4 3 181 11 100 1 14 15 2 7 4 3 Ans : 181 11 100 7 14 15 2 28 3 181 11 393 14 × 15 2 28 3 181 11 131 15 2 2 181 11 2 15 2 131 181 11 15 131 114 4 8 40 114 1 8 10 23711 - 165 1965 23546 1965 1931 11 1965 14 16 14 3 (iv) 8 150 23 23 500
CSS Primary Standard “Mathematics” 124 1214 16 14 300 150 23 15 500 Ans : 1214 16 14 150 23 25 1214 16 322 150 515 1214 78 150 575 1214 575 150 78 1214 575 150 78 13961 234 155 or 59 234 (v) 150 180 15 12 13 +16 150 180 15 12 29 Ans : 150 180 15 17 150 180 15 17 150 180 32 150 148 2 1 1 18 50 3 (vi) 4 180 1 ÷ 8 8 7 9 40 4 33 1 18 50 7 180 8 8 7 9 40 4 Ans : 33 1 5 7 1 180 2 8 7 4 4 8 33 1 35 180 2 8 7 128
CSS Primary Standard “Mathematics” 125 33 1 70 180 8 7 128 33 1 35 180 8 7 64 33 64 245 180 8 448 33 390 180 8 448 33 80640 390 8 448 33 80331 8 448 1848 80331 448 82179 448 195 or183 448 7 1 3 1 2 7 (vii) 11+ + 180 15 10 3 8 14 3 2 7 541 123 3 28 7 11 10 3 8 42 2 Ans : 7 541 123 31 7 11 10 3 8 42 4 7 541 123 31 147 11 10 3 8 42 7 541 123 116 11 10 3 8 42 7 541 123 58 11 10 3 8 21 7 541 2583 464 11 10 3 168
CSS Primary Standard “Mathematics” 126 7 541 2119 11 10 3 168 7 1146379 11 10 504 27720 1764 5731895 2520 5761379 2520 13 1 14 1 16 12 13 (viii) 15 + 1 5 6 15 13 12 17 10 7 103 1 14 1 16 12 13 1 1 6 15 13 12 17 10 7 5 Ans : 103 1 14 13 16 78 6 15 13 8 17 175 103 1 14 13 1248 6 15 13 12 2975 103 1 14 38675 14976 6 15 13 35700 103 1 14 23699 6 15 13 35700 103 1 1823 6 15 2550 103 170 1823 6 2550 103 1653 6 2550 43775 1653 2550 42122 21061 661 16 2550 1275 1275 70 2 4 1 1 3 1 (ix) 15 + + 4 8 1 2 100 5 12 5 5 25 5
CSS Primary Standard “Mathematics” 127 70 22 100 1 6 53 15 5 100 5 12 5 5 25 Ans : 7 22 25 1 6 53 15 10 5 3 5 5 5 7 22 25 1 6 53 15 10 5 3 5 7 22 25 60 15 10 5 3 5 7 22 15 100 10 5 7 22 500 15 10 5 7 522 15 10 5 15 7 1044 10 1201 10 Exercise 5c Q1. Simplify the following. (i) 130.75 × [4.22 + {7.14 × (1.96 ÷ 1.4 × 3.05)}] Ans: =130.75 × [4.22 + {7.14 × (1.4 × 3.05)}] =130.75 × [4.22 + {7.14 × 4.27}] =130.75 × [4.22 + 30.488] =130.75 × 34.7078 =4538.045 (ii) 24 180.3 + 123.14 + 8.6-(19.6 ÷ 32.2) 18 144 Ans: 2616 =180.3+ 123.14 + 8.6 0.609 144 109 =180.3+ 123.14 + 7.991 6
CSS Primary Standard “Mathematics” 128 = 180.3+131.131 18.1667 = 293.26 (iii) 16.29 2.15 14.4 ÷ 3.44 (18.8×0.8 0.3) Ans: = 16.29 2.15 14.4 ÷ 3.44 1.44 0.3 = 16.29 - 2.15 + 14.4 ÷ 3.44 -1.14 =16.29- 2.15+ 14.4 ÷ 2.3 = 16.29 2.15 6.261 = 16.29 6.261 2.15 = 20.401 (iv) 14 21 3.21 + 349.682 + 2.59 840.321 5889 +1.20 66 Ans: 1400 = 3.21 349.682 + 2.59 840.321 5889 1.203 66 = 21.21 3.21 349.682 + 2.59 840.321 5890.20 = 21.21 3.21 349.682 +842.911 5890.203 = 21.21 3.21 349.682 + 4964916.901 = 21.21 3.21 4965266.583 = 496534.003 (v) 850.17 14.2× 110.20 + 14.689 1000.1 981.2 Ans: = 850.17 14.2 110.20 + 14.689 1981.3 = 850.17 14.2 110.20 +14.689 1981.3 = 850.17 14.2 124.889 1981.3 = 850.17 14.2 1856.411 = 850.17 2636.036 = 25510.866 (vi) 334.70 170.0 ÷ 11.1 + 18.0 14.1 110.2 110.1 75 Ans: = 4.463 170.0 11.1+ 18.0 14.1 220.3 = 4.463 170.0 11.1+ 3.9 220.3
CSS Primary Standard “Mathematics” 129 = 4.463 170.0 11.1+859.17 = 4.463 170.0 870.27 = 4.463 0.1961 = 4.649 (vii) 2 130.5 62.52 20.5 2.92 × 5.6 × 3 14 Ans: 44 =130.5 62.52 20.5 2.92 5.6 14 = 130.5 62.52 20.5 16.35 3.143 =130.5 62.52 36.85 3.143 = 130.5 25.67 3.143 = 130.5 80.68 = 211.18 Week 4 Exercise 5d Q1. Three cousings shared a sum of money amounting Rs. 10,000, they received on passing the annual exam. 1st received 1 6 of it, the second got 1 4 of this money, remaining is given to the third one. Find the amount of each cousin. Ans: Total Money = 10,000 Share of 1st cousin = 1 10000 6 = 5000 3 Share of 2nd cousin = 7 10000 12 17500 3 Q.2 Amna gets Rs. 20,000 per week. She saves 5 8 of it. Find (i) How much she saves. (ii) How much she spends. (iii) Her monthly saving. Ans:
CSS Primary Standard “Mathematics” 130 Total amount of Amna = Rs. 20,000 She saves = 5 20000 8 = 12500 Amna’s spends amount = 20,000 – 12500 = Rs. 7500 Amna’s monthly saving = 4 × 12500 = Rs. 50000 Q.3 The lawn of a house is feet long and feet wide. What is the total area of the lawn? Is the total area of the lawn? Ans: Length of lawn = 1 12 2 25feet 5 Width of lawn = 1 11 6 67 feet 6 25 67 × 5 6 1675 2 feet 12 2 139.583feet Q.4 Ans: Total distance from Lahore to Gujrawala = 70 km Distance traveled by train = 9 70 10 63km Distance traveled by bus 1 ×70 = 5.83 km 12 Distance traveled by Rickshaw = (70 – 63 – 5.83) km = 1.167 km Q.5 Ibrahim has to pay zakat of Rs. 20,000. He decide to pay the Zakat 40 NGO, hospital and widow such as 1 5 of Zakat to NGO and 0.667 to hospital. What amount of Zakat has been left for widow. Ans: Ibrahim has Zakat = Rs. 20,000 Amount of Zakat to NGO = 1 20,000 5 Rs. 4,000
CSS Primary Standard “Mathematics” 131 Amount of Zakat to hospital 0.6667 20,000 Distance traveled by Rickshaw = Rs. 13334 Amount of Zakat to widow = 20,000 – (4,000 + 13334) = Rs. 2666 Q.6 Sidara has Rs. 1300 blance in her mobile. She gave 3 5 to samia. Samia gave 1 6 to her own share to her friend Mariam. How much blance Mariam got. Ans: Sidara has blance = Rs. 1300 She gave to samia = 3 1300 5 Rs. 780 Samia gave to mariam 1 780 6 = Rs. 130 Review Exercise 5 Review Exercise 5 Q.1 Choose the correct answer and fill the circle. (i) 14 1 + is 7 16 3.124 2.1667 2.186 4.423 ii. 5 8 of 40 is: 26 29 25 30 iii. 0.5 ÷ 0.2 is: 3.5 2.5 4.1 1.3 iv. 13.2 – 8.32 is equal to: 11.2 6.88 4.88 none of these v. 1 6 of 10,000 is _______: 500 678.11 589.17 1666.667
CSS Primary Standard “Mathematics” 132 vi. 116.5 – 8.91 is: 107.59 113.14 110.89 89.794 vii. 1300 10 is _________: 133.3 130 1300 none of these viii. BODMAS stand for: brackets observation division multiplication addition subtraction brackets operation division multiplication addition subtraction all of them none of them ix. [ ] is called _______ brackets: vinculum square brackets braces parenthese x. 15.1 + (14.2 – 13.2) =: 18.2 16.1 15.1 14.2 Q.2 Simplify the followings. (i) 1300 – 24 [150 – {150 ÷ 30 × 40 – (300 + 530 – 630)}] Ans: = 1300 – 24 [150 – {150 ÷ 30 × 40 – (300 + 530 – 630)}] = 1300 – 24 [150 – {150 ÷ 30 × 40 – (300 – 100)}] = 1300 – 24 [150 – {150 ÷ 30 × 40 – 200}] = 1300 – 24 [150 – {5 × 40 – 200}] = 1300 – 24 [150 – {200 – 200}] = 1300 – 24 [150 – 0] = 1300 – 24 (150) = 1300 – 3600 = 2300 1 50 250 210 (ii) 135 120 150 250 1 50 250 210 × 135 120 150 250 Ans : 1 5 25 21 × 135 12 15 25
CSS Primary Standard “Mathematics” 133 1 5 5 21 × 135 12 3 25 1 5 125 63 × 135 12 75 1 5 62 × 135 12 75 1 31 × 135 90 31 12150 31 12150 = 0.0025 1 2 1 3 480 (iii) 41 12 11 10 1.44 ÷ 22 890 15 16 80 903 10682 166 163 80 + × + 1.44× 22 890 15 16 480 Ans : 903 10682 166 163 6 + × + 22 890 15 16 25 903 10682 13529 6 + + 22 890 120 25 903 10682 + 112.74 + 0.24 22 890 903 10682 + 112.98 22 890 903 + 12.00 112.98 22 903 100.98 22 41.045 100.98 = – 59.935 11 1 1 340 (iv) 44 6 12 3.778 22 30 6 90
CSS Primary Standard “Mathematics” 134 979 181 73 340 3.778 22 30 6 90 Ans : 979 181 73 3.778 3.778 22 30 6 979 181 73 0 22 30 6 979 181 73 22 30 6 979 13213 22 180 1175957 22 3266.55 (v) 80.792 + [139.49 – {140.78 × 150.0 + (150.4 – 4.10 + 79.4}] Ans: = 80.792 + [139.49 – {140.78 × 150.0 + (150.4 – 83.5)}] = 80.792 + [139.49 – {140.78 × 150.0 + 66.9}] = 80.792 + [139.49 – {21117 + 66.9}] = 80.792 + [139.49 – 21183.9] = 80.792 – 21044.41 = – 20963.618 (vi) 1890 – {1432.10 × 100 + 71.09 ÷ (174 × 1.25)} Ans: =1890 – {1432.10 × 100 + 71.09 ÷ 217.5} =1890 – {1432.10 × 100 + 0.327} =1890 – {143210 + 0.327} =1890 – 143210.327 = –141320.33 (vii) 1311.01 + [1410.1 × 22.00 + {11.87 – (180.89 + 17.89)}] Ans: =1311.01 + [1410.1 × 22.00 + {11.87 – 198.78}] =1311.01 + [1410.1 × 22.00 – 186.91] =1311.01 + [31022.2 – 186.91] =1311.01 + [30835.29] = 32146.3
CSS Primary Standard “Mathematics” 135 (viii) 1768.142 ÷ [1182.1 + {18.732 – 4.893 + (43.25 ÷ 23.25)}] Ans: =1768.142 ÷ [1182.1 + {18.732 – 4.893 + (43.25 ÷ 23.25)}] =1768.142 ÷ [1182.1 + {18.732 – 4.893 + 1.860}] =1768.142 ÷ [1182.1 + {20.592 – 4.893}] =1768.142 ÷ [1182.1 + 15.699] =1768.142 ÷ 1197.799 =1.476 Q.3 A telephone company has 18000 employees 0.5 are in HR, 0.4 are in marketing and 0.06 in quality control. The rest are in IT. Tell how many employees are in IT. Ans: Total employees = 18000 Employees in HR = 0.5 of total = 0.5 × 18000 = 9000 No. of employees in marketing = 0.4 of total = 0.4 × 18000 = 7200 No. of Employees in quality control = 0.06 of total = 0.06 × 18000 = 1080 No. of employess in IT = 18000 – (9000 + 7200 + 1080) = 18000 – 17280 = 720 Q.4 Alisha bought 2 purses in Rs. 3500 with a discount of 171. She also bought one suit of Rs. 2700 with discount of 131. Tell what was the amount that she paid? Ans: Q.5 Sum of two fractions is 3 15 4 . If one fraction is 6 4 8 find the other fraction.? Ans:
CSS Primary Standard “Mathematics” 136 Sum of two fraction = 3 15 4 = 63 4 One fraction = 6 4 8 Other fraction = 38 8 Other fraction = 63 38 40 8 = 126 38 8 88 8 11 Q.6 Sum of two fractions is 3 15 4 . If one fraction is 6 4 8 find the other fraction? Ans: Price of Ticket = Rs. 2100 Distance between Lahore and Peshawar = 550 km Rate of 550 killometers = Rs. 100 Rate of 1 killometer = 2100 550 = 42 11 = Rs. 3.82 Q.7 Alishaba ate 1 1 2 pizzas, Anoral ate 1 1 2 pizzas and Ibrahim ate 1 2 4 pizzas. How much pizzas did they eat altogether? Ans: Alishaba eat pizza = 1 1 2 = 3 2 Anoral eat pizza = 1 1 2 = 3 2 Ibrahim eat pizza = 1 2 4 = 9 4 They eat pizza altogether = 3 3 9 224
CSS Primary Standard “Mathematics” 137 = 21 4 = 1 5 4 Q.8 Ali has a cable wire of length 55.55m. He want to distribute his wire into five houses equally. Find the legthe of wire distributed to each house. If the price of each house. If the price of each part of wire is Rs. 15.25 find the cost of total wire? Ans: Length of wire = 55.55 m The length of wire distributed to each house = 55.55 m 5 = 11.11 m The cost of distributed wire = 11.11 × 15.25 = 169.4275 The cost of total wire = 169.4275 × 5 = 847.1375 Unit:6 Ratio Teacher Objectives: ☻ To introduce continued ratio and revise the direct and inverse proportion. ☻ To explain the unitary and proportion method to solve the real life problems. ☻ To explain the solution of real life problems related to time and work using proportion. ☻ To explain relation between time and distance. ☻ To explain the conversion of units of speed kilometer per hour into meter per second and vice versa. ☻ To explain the method to solve variation related problems involving time and distance. Learning Outcome: ☻ Define ratio as a relation which one quantity bears to another quantity of the same kind with regard to their magnitudes. ☻ Know that of the two quantities forming a ratio, the first one is called antecedent and the second one consequent. ☻ Know that a ratio has no units. ☻ Calculate ratio of two numbers.
CSS Primary Standard “Mathematics” 138 ☻ Reduce given ratio into lowest (equivalent) form. ☻ Describe the relationship between ratio and fraction. ☻ Know that an equality of two ratios constitutes a proportion, e.g., a : b :: c : d, where a, d are known as extremes and b, c are called the means. ☻ Find proportion (direct and inverse). ☻ Solve real life problems involving direct and inverse proportion. Teacher materials: CSS Primary Standard Mathematics Book 6. Writing Board. Marker. Eraser .Exercise 6a Q.1 Simplify the following. (i) : 150 70 160 100 Ans: = 150 70 : 160 100 = 15 7 : 16 10 = 15 7 × 80 : 80 16 10 = 75 : 56 (ii) : 13 39 40 80 Ans: = 13 39 × 80 : 80 40 80 = 26 : 39 = 2 : 3 (iii) 0.450 : 0.750 Ans: = 450 750 : 1000 1000 = 450 750 ×1000 : 1000 1000 1000
CSS Primary Standard “Mathematics” 139 = 450 : 750 = 225 : 375 = 45 : 75 = 9 : 15 = 3: 15 (iv) 82 : 112 : 142 Ans: 41: 56: 71 (v) 1.4 : 2 5 : 1 3 Ans: = 14 2 1 : : 10 5 3 = 14 2 1 × 30 : 30 : 30 10 5 3 = 42: 12: 10 = 21: 12: 10 (vi) 120 : 240 : 280 Ans: =120 : 240 : 280 = 60: 120 : 140 = 30: 60: 70 = 15: 30: 35 = 3: 6 : 7 Q.2 Find the ratio between the following and make them as the lowest form. (i) 250 cm, 1m Ans: 1m = 100 cm 250 cm , 100 cm 250 : 100 25 : 10 5 : 2 (ii) 800g to 1kg Ans: 1kg = 1000g So required ratio is 800 : 1000 4 : 5 (iii) 364maths books to 445 English books 364:455.
CSS Primary Standard “Mathematics” 140 Ans: (iv) 2 kilometers, 8 meters and 672 meters. Ans: As 2kilometers = 2000 meters so the required ratio is. 2000 : 8 : 672 1000 : 4 : 336 500 : 2 : 168 250 : 1 : 74 (v) 4 hours, 8 minutes and 3 hours. x Ans: As 1 hour = 60 minutes So 3 hours = 3 × 60 = 180 minutes And 4 hours = 4 × 60 = 240 minutes Now required ratio is. 240 : 8 : 180 120 : 4 : 90 60 : 2 : 45 Q.3 The ratio of monthly income to the savings in a family is 5:4. If the saving is Rs. 9000 find the income and the express. Ans: Income to saving = 5 : 4 Saving = 9000 Income : Saving = 5 : 4 Income 5 Saving 4 Income = 5 9000 4 Income = Rs. 11250 Expenses = Icome – saving = 11250 – 9000 = Rs. 2250 Q.4 A sum of Eidi is divided between Ibrahim and Hadi in the ratio 4:7 of Hadi in the ratio 4:7. If Hadi’s share is Rs. 6160, find the total money and share of Ibrahim. Ans: Sum of Eidi divided between Ibrahim Hadi = 4 : 7 Hadi’s Share = Rs. 6160 Ibrahim’s share : Hadi’s share = 4 : 7 Ibrahim's share 4 Hadi's share 7
CSS Primary Standard “Mathematics” 141 Ibrahim’s share = 4 6160 7 Ibrahim’s share = 3520 Total money = 6160 + 3520 = Rs. 9680 Q.5 Mosa has notes of Rs. 100, Rs. 50 and Rs. 10. The ratio of these notes is 2 : 3 : 5 respectively and total amount is Rs. 20,00,000. Find the numbers of notes of each kind. Ans: Total amount = 20000 Ratio of notes of Rs 100, Rs 50, Rs 10 = 2 : 3 : 5 Sum of ratio = 2 + 3 + 5 = 10 Notes of Rs 100 = 2 20000 10 = 4000 Number of notes of Rs 100 = 400 Notes of Rs. 50 = 3 20000 10 = 60000 Number of notes of Rs 50 = 1200 Notes of Rs 10 = 5 20000 10 = 10000 Number of the notes of Rs 10 = 1000 Q.6 A certain sum of money is divided among Amna, Fatima and Umer in the ratio 2 : 3 : 4. If Amna share is Rs 25000 find the share of Fatima and Umar? Ans: Ratio among Amna, Fatima and Umar = 2 : 3 : 4 Amna share = Rs 25000 Sum of share = 2 + 3 + 4 = 9 Amna share = Rs 25000 2 9 of total 25000 Total = 9 25000 2 = 112500 Fatima’s share = 3 112500 9 = Rs. 37500
CSS Primary Standard “Mathematics” 142 Umer’s share = 4 112500 9 = 50000 Q.7 The ratio of make and female. Students in a wiresity is 3 : 4 If these are 16000 female students find the number of male students? Ans: Ratio of male and female = 3 : 4 Sum of Ratio = 3 + 4 = 7 No. of female students = 16000 4 7 of female students = 16000 4 7 × Total = 16000 Total = 16000 × 7 4 = 28000 No. of male students = 28000 × 3 7 = 12000 Q.8 In a library the ratio of English books to Math books is 3 : 4. If there are 12000 books of English what will be the number of math books? Ans: Ratio of English to Math books = 3 : 4 Sum of Ratio = 3 + 4 = 7 No. of English books = 12000 3 7 of total = 12000 3 7 × total = 12000 Total = 7 12000 28000 3 Week 5 Exercise 6b Q.1 Which of the following proportion are trues?
CSS Primary Standard “Mathematics” 143 (i) True (ii) False (iii) False (iv) True (v) False (vi) True Q.2 Solve the formula following. (i) 10 : 2 :: x : 4 Ans: 10 : 2 :: x : 4 10 × 4 = 2 × x 40 = 2x x = 40 2 x = 20 (ii) 15 : 6 :: 5 : x Ans: 15 : 6 :: 5 : x 15x = 6 × 5 = 30 x = 30 15 x = 2 (iii) 8 : x :: 2 : 6 Ans: 8 : x :: 2 : 6 2x = 2 × 6 = 48 2x = 48 x = 48 2 x = 24 (iv) x : 16 :: 12 : 2 Ans: x : 16 :: 12 : 2
CSS Primary Standard “Mathematics” 144 2x = 10 × 12 2x = 120 x = 120 2 x = 60 Q.3 If 3 men earns Rs. 480 in a day find how much will 7 men earn in day? Ans: Men : Rs 3 : 480 7 : x 7 3 : x 480 x = 7 3 × 480 = 7 × 160 = 1120 Q.4 6 typists working 5 hours a day can type the manuscript of a book in 16 days. How many days will 4 typists take to do the same job, each working 5 hours a day? Ans: Typists : Days 6 : 16 4 : x 6 4 : 16 x x = 6 16 4 x = 24 Q.5 6 oxen can graze a field in 28 days. How long would 9 oxen take to graze the same field? Ans: Oxen : Days 6 : 28 9 : x 6 9 = 28 x x = 28 9 6
CSS Primary Standard “Mathematics” 145 = 18.7 days Q.6 The cost of 16 packets of besan, each wright 900 grams is Rs 84. What will be the cost of 27 packets of besen, each weight 900 grams? Ans: Packets : Rs 16 : 84 27 : x 27 16 = 84 x x = 84 16 27 = 141.75 days Q.7 To improve her vocabulary, Sumera is asked to learn 39 words a day from dictionary. If she learn one word in 13 minute. How long will she take to complete her task? Ans: Words : Minut 1 : 13 39 : x 39 1 = 13 x x = 39 × 13 x = 507 words Q.8 12 men, working 8 hours a day, complete a piece of work in 10 days. Find the number of men required to complete the same work in 8 days, working 8 hours a day. Ans: Men : Days 12 : 10 x : 8 12 x = 8 10 x = 12 8 10 x = 15 Q.9 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 5 hours a day, complete the work?
CSS Primary Standard “Mathematics” 146 Ans: Days : Persons 12 : 39 x : 30 12 x = 30 39 x = 12 30 39 = 15.6 Q. 10 Food for 150 students in a boarding school is sufficient for 20 days. 100 more students join them, for how many days the food will be sufficient for students. Ans: Students : Days 150 : 20 250 : x 250 150 = 20 x x = 20 25 15 x = 12 Review Exercise 6 Q.1 Choose the correct answer. (i) If 14kg of pulses cost Rs. 441, what is cost of 22 kg of pulses? Rs. 627 Rs. 649 Rs. 671 Rs. 693 ii. If 36 men can do a price of work in 25 days, in how many days will 15 men do it? 50 56 60 72 iii. If 20 men can but a wall 56 meter is long in 6-days. What length: 49 meters 16 meters 52 meters 42 meters iv. If a car covers 102km in 6.8 litres of petrol, now much distance will it cover in 242 liters of petrol? 363km 330km 375km 396km
CSS Primary Standard “Mathematics” 147 v. On a penticular day, 200 US dollars are worth Rs.9666. On that day, how many dollars could be bought for Rs. 5074.65? 115 US dollar 117 US dollar 172 US dollar 105 US dollar vi. If 5 men of 7 women earn Rs. 525 per day, how much would 7 men would earn per day? Rs. 1331 Rs. 735 Rs. 710 Rs. 1420 vii. The 16 bags of washing power, each weighing 1.5kg is Rs. 672. Find the cost of 18 bags of same kind and weight. Rs. 82 Rs. 1128 Rs. 756 Rs. 1000 viii. If 3 persons can weave 168 shows in 14 days, how many shawls will be woven by 8 person in the same number of day? 153 484 727 448 ix. If the cost f transporting 160kg of goods for 125km is Rs.60, what will be the cost of transporting 200kg of goods for the same distance. Rs. 85 Rs. 50 Rs. 75 Rs. 175 x. If 18 dolls cost Rs 630, how many dolls can be bought for Rs 455? 9 11 13 15 xi. If a man earn Rs 805 per week in how many days he will earn Rs. 1840? 7 days 16 days 19 days 23 days Q.2 If 4A = 5B = 6C, find the ratio of A:B:C. Ans: Q.3 Divid Rs. 43000 into 3 parts such that ratio of A,B, and C is 15:12:10? Ans: Total Money = Rs. 43000 Ratio of A, B, C = 15 : 12 : 10 Sum of ratio = 15 + 12 + 10 = 37
CSS Primary Standard “Mathematics” 148 Share of A = 43000 37 15 Share of B = 43000 37 12 Share of C = 43000 37 10 Q.4 A certain sum of money is divided among A,B,C in the ratio 2 : 3 : 4. If A’s share is Rs. 20000 find the share of B and C. Ans: A : B : C = 2 : 3 : 4 Sum of ratio = 2 + 3 + 4 = 9 A’s share = 20,000 9 2 of total money = 20, 000 Total money = 20,000 × 2 9 Total money = 90, 000 B’s share = 9 2 of total money = 90000 9 3 = 30000 C’s Share = 9 4 of total money = 9 4 Q.5 Divid Rs. 15000 among A,B,C in the ratio 3 1 , 4 1 , 5 1 ,. Ans: A : B : C = 3 1 : 4 1 : 5 1 = 60 3 1 : 60 4 1 : 60 5 1 = 20 : 15 : 12 Sum of ratio = 20 + 15 + 12 = 47 A’s share = 150000 47 20 Q.6 The ratio of number of male and female teachers in a school is 5:6. If there are 16 female teachers, find the number of male teachers.
CSS Primary Standard “Mathematics” 149 Ans: Ratio of male and female = 5:6 No. of male : No. of female = 5:6 5 = No.of female 6 No. of male No. of male = ×No.of female 6 5 = 16 6 5 Q.7 In Rs. 166.50 is the cost of 9kg of rice, how much rice can be purchased for Rs. 259. Ans: Rs : kg 166.50 : 9 259 : x 9 x = 9 166.50 259 = 14kg Q.8 In a house of 10 persons food is sufficient for 30 days, 3 persons left the house. For how many days the food would be sufficient? Ans: Days : Persons 30 : 10 x : 7 30 x = 10 7 x = 30 7 10 = 42.86 =43 days Q.9 9 hens can lay 9 eggs in 10 days. How many eggs can 4 hens lay in 40 days. Ans: Hens : Eggs 9 : 9 4 : x 9 4 = 9 x x = 9 9 4 = 4, x = 4 eggs
CSS Primary Standard “Mathematics” 150 Unit: 7 Financial Arithmetic Teachers Objectives: ☻ To explain property tax and general sale and tax. ☻ To explain the solution the tax-related problems. ☻ To introduce profit and makeup. ☻ To explain the method to find rate of profit and makeup per annum. ☻ To explain solution of real life problems involving profit makeup. ☻ To introduce zakat and usher. ☻ To explain solution of problems related to zakat and usher. Learning Outcomes: ☻ Recognize percentage as a fraction with denominator of 100. ☻ Convert a percentage to a fraction by expressing it as a fraction with denominator 100 and then simplify. ☻ Convert a fraction to a percentage by multiplying it with 100%. ☻ Convert a percentage to a decimal by expressing it as a fraction with denominator. 100 and then as a decimal. ☻ Convert a decimal to a percentage by expressing it as a fraction with denominator 100 then as a percentage. ☻ Solve real life problems involving percentage. ☻ Define: selling price and cost price profit, loss and discount profit and loss percentage. ☻ Solve real life problem involving profit, loss and discount. Teacher materials. CSS Primary Standard Mathematics Book 6. Writing Board. Marker. Eraser. Classroom Activity: Find out the property tax Ali’s father pays for his house of Rs. 9000000 at the rate 0.8%? Solution: Property value = Rs. 9000000 Tax rate = 0.8%