Mathematics 7 Solved Excercise PDF

Mathematics-7 2 1. Represent the following numbers in rational form. Solution: (i) 16 (ii) (– 13) (iii) (– 525) (iv) 0.04 16 –13 –525...

Mathematics-7 2 1. Represent the following numbers in rational form. Solution: (i) 16 (ii) (– 13) (iii) (– 525) (iv) 0.04 16 –13 –525 4 = ____ = ____ = ______ = ______ 1 1 1 100 (v) 1.24 (vi) (3.5) (vii) 0.42 (viii) 0 124 35 42 0 = ______ = ____ = ______ = ____ 100 10 100 1 2. Identify he positive and negative rational numbers: Solution: (i) 3 (ii) 6 (iii) –5 –4 12 13 = negative rational numbers = positive rational numbers = negative rational numbers (iv) – 13 (v) 2 (vi) 7 – 16 5 – 11 = positive rational numbers = positive rational numbers = negative rational numbers (vii) –1 (viii) 12 (ix) – 100 – 11 – 19 5 = positive rational numbers = negative rational numbers = negative rational numbers (x) 1 (xi) 17 (xii) –2 –6 –6 –7 = negative rational numbers = negative rational numbers = positive rational numbers 3. Compare the following rational numbers: (i) 3 and – 10 6 8 Solution: as negative number is always less then positive numbers. Hence, 3 > – 10 6 8 (ii) –6 and – 2 14 7 Solution: Let us make the denomenator of both fraction same. –6 , – 2 x 2 14 7 2 3 Mathematics-7  –6 , – 2 14 7 Compare digits of Numerator Hence –6 > – 2 14 7 Solution: as negative number is always less then positive numbers. Hence, > (iv) 7 and – 1 8 15 Solution: as negative number is always less then positive numbers. Hence, 7 > –1 8 15 (v) –2 and 7 3 –4 Solution: Let us make the denominator same. –2 x 4 ; 7 x 4 3 4 –4 4 By comparing Numerator –8 > –21 ; –2 > 7 3 –4 (vi) –1 and 2 8 –5 Solution: Let us make the denominator same. –1 x 5 ; 2 x 8 8 5 –5 8 –5 ; –16 40 40 By comparing Numerator –5 > –16 –1 > 2 8 –5 4. Write equivalent fractions for the following fractions: (i) –2 5 Solution: –2 x 2 = –4 5 2 10 Mathematics-7 4  –2 x 3 = –6 5 3 15 –2 x 4 = –8 5 4 20 –2 x 5 = –10 5 5 25 = –4 , –6 , –8 , –10 Ans 10 15 20 25 Solution: 4 x 2 8 = 3 2 6 4 x 3 12 = 3 3 9 4 x 4 16 = 3 4 12 4 x 5 20 = 3 5 15 8 , 12 , 16 , 20 Ans = 6 9 12 15 2 7 Solution: 2 x 2 4 = 7 2 14 2 x 3 6 = 7 3 21 2 x 4 8 = 7 4 28 2 x 5 10 = 7 5 35 4 , 6 , 8 , 10 Ans = 14 21 28 35 6 –7 Solution: 6 x 2 12 = –7 2 –14 5 Mathematics-7  6 x 3 = 18 –7 3 –21 6 x 4 = 24 –7 4 –28 6 x 5 = 30 –7 5 –35 = 12 , 18 , 24 , 30 Ans –14 –21 –28 –35 –4 8 Solution: –4 x 2 –8 = 8 2 16 –4 x 3 –12 = 8 3 24 –4 x 4 –16 = 8 4 32 –4 x 5 –20 = 8 5 40 –8 , –12 , –16 , –20 = Ans 16 24 32 40 5 25 Solution: 5 x 2 = 10 25 2 50 5 x 3 = 15 25 3 75 5 x 4 = 20 25 4 100 5 x 5 = 25 25 5 125 10 , 15 , 20 , 25 Ans = 50 75 100 125 5. Represent the following rational numbers on number line: (i) –5 3 Solution: –5 –1.66 / 3 –3 –2 –1 0 1 2 3 Mathematics-7 6  (ii) –2 3 Solution: –2 3 –3 –2 –1 0 1 2 3 (iii) –4 3 Solution: –4 3 –3 –2 –1 0 1 2 3 (iii) 1 2 Solution: 1 2 –2 –1 0 1 2 (iii) 3 2 Solution: 3 2 –2 –1 0 1 2 (ii) 5 2 Solution: 5 2 –4 –3 –2 –1 0 1 2 3 4 6. Put >, < or = between each of the following set of rational numbers. (i) 25 25 4 8 Solution: Let us make the denominator same. 25 x 2 50 4 2 8 100 50 8 8 As 100 > 25 (ii) –3 3 2 2 Solution: As negative number is always less than positive number. –3 3 2 2 7 Mathematics-7  (iii) –4 –8 5 10 Solution: Let us make the denominator same. –4 x 2 , –8 5 2 10 100 , 50 8 8 Hence –8 = –8 –4 = –8 5 10 (iv) 1 –2 2 4 Solution: As negative number is always less than positive number. 1 Hence > –2 2 4 (v) 7 1 3 3 Solution: As denominators are same. 7 1 So > 3 3 (vi) 2 –2 3 –3 Solution: As denominators are same. 2 –2 So = 3 –3 1. Convert: (i) 11 3 7 , 6 , 5 2 , 7 4 into improper fractions. 8 13 11 15 (i) 11 37 (iii) 2 5 11 = 80 Ans = 57 Ans 7 11 (ii) 6 8 13 (iv) 4 7 15 = 110 Ans = 120 Ans 13 15 Mathematics-7 8 1. Convert: (ii) 28 , 65 , 54 , 125 into mixed fractions. 9 17 11 8 (i) 28 (ii) 65 9 17 Solution: Solution: 3 3 = 9 28 = 17 65 27 _____ 51 _____ 1 14 = 3 1 Ans = 3 14 Ans 9 17 (iii) 54 (iv) 125 11 8 Solution: Solution: 4 15 = 11 54 = 8 125 44 _____ 8 _____ 10 45 40 _____ = 4 10 Ans 5 11 5 8 Ans = 15 2. Arrange the following integers in ascending and descending order. (i) – 3, – 5, 0, – 4, 2 (ii) – 5, 1, 3, 5, 0 Solution: Solution: In ascending form = –5, –4, –3, 0, 2 In ascending form = –5, 0, 1, 3, 5 In descending form = 2, 0, –3, –4, –5 In descending form = 5, 3, 1, 0, –5 (iii) – 401, – 101, – 512, 418, 317 Solution: In ascending form = –512, –401, –101, 317, 418 In descending form = 418, 317, –101, –401, –512 (iv) 0, – 1, 5, – 5, 7 (v) 12, – 17, 13, –14, 15 Solution: Solution: In ascending form = –5, –1, 0, 5, 7 In ascending form = –17, –14, 12, 13, 15 In descending form = 7, 5, 0, –1, –5 In descending form = 15, 13, 12, –14, –17 (vi) 10, – 10, 11, – 11, 13 (vii) 5, – 5, – 3, 4, 2, 1 Solution: Solution: In ascending form = –11, –10, 10, 11, 13 In ascending form = –5, –3, 1, 2, 4, 5 In descending form = 13, 11, 10, –10, –11 In descending form = 5, 4, 2, 1, –3, –5 9 Mathematics-7  (viii) 211, – 103, – 213, – 15, + 18 Solution: In ascending form = –213, –103, –15, 18, 211 In descending form = 211, 18, –15, –103, 213 (i) 37.02, 37.021, 3.702, 73.02, 37.002 Solution: In ascending form = 3.702, 37.002, 37.02, 37.021, 73.02 In descending form = 73.02, 37.021, 37.02, 37.002, 3.702 (ii) 49.534, 49.354, 49.453, 49.05, 54.543, 45.0543 Solution: In ascending form = 45.0543, 49.354, 49.453, 49.534, 54.543 In descending form = 54.543, 49.534, 49.453, 49.354, 49.05, 45.0543 (iii) 0.93, 1.87, 1.9, 1.78, 0.39, 5.6 Solution: In ascending form = 0.39, 0.93, 1.78, 1.87, 0.39, 5.6 In descending form = 5.6, 1.9, 1.87, 1.78, 0.93, 0.39 4. Write or = to complete the sentence. (i) 0.3 0.03 (ii) 0.7 0.5 (iii) 2.03 2.30 0.3 > 0.03 Ans 0.7 > 0.5 Ans 2.03 > 2.30 Ans (iv) 0.8 0.88 (v) 5 0.5 (vi) 1.37 1.49 0.8 > 0.88 Ans 5 > 0.5 Ans 1.37 > 1.49 Ans 5. Place the numbers in order from least to greatest. (i) 3564, 2198, 2975, 3529, 2340 Solution: 2198, 2340, 2975, 3529, 3564 (ii) 23562, 21728, 20953, 32529, 23642 Solution: 20953, 21728, 23562, 23642, 32629 (iii) 398945, 431565, 426190, 312320, 322112 Solution: 322320, 322112, 398945, 426190, 431565 6. Which rational number is smaller in each of the following pairs. (i) 1 ,–1 5 5 Solution: As negative number is always less or smaller than positive number. Mathematics-7 10  – 1 is smaller number. 5 (ii) –5,–4 6 3 Solution: Let us make the denominator same. – 5 , –8 6 6 –5 > –8 – 4 is smaller number. 3 (iii) –3 2 , –3 4 7 7 Solution: –19 , –17 7 7 Compare the numerator –3 2 is smaller number. 7 (iv) 0 , –3 4 Solution: –3 is smaller number. 4 (v) 0 , –9 –5 Solution: 0 is smaller number. (v) 7 , –21 –5 –23 Solution: 7 , –21 –5 –23 Positive number is always greater. 7 is smaller number. –5 7. Arrange the following rational numbers in ascending order. (i) 9 , 5 , – 7 ,– 3 – 24 – 12 16 4 Solution: Let us make the denominator same. 11 Mathematics-7  9 x 2 , 5 x 4 , –7 x 3 , –3 x 12 –24 2 –12 4 16 3 4 12 –18 , –20 , –21 , –36 48 48 48 48 In ascending order –3 , –7 , –5 , –9 Ans 4 16 12 24 (ii) 2 , –7 , 19 , –11 –5 10 –30 15 Solution: Let us make the denominator same. 2 x 6 , 7 x 3 , 19 , –11 x 2 –5 6 –10 3 –30 15 2 12 , 21 , 19 , 22 –30 –30 –30 –30 Compare the numerator. –22 , –21 , –19 , –12 30 30 30 30 –11 , –7 , 19 , 2 Hence Ans 15 10 30 –5 (iii) – 11 , 17 , – 3 , – 7 20 – 30 10 15 Solution: Let us make the denominator same. –11 x 3 , 17 x 2 , –3 x 6 , –7 x 4 20 3 –30 2 10 6 15 4 33 , 34 , –18 , –28 –60 –60 60 60 In ascending order. –34 , –33 , –18 , –28 60 60 60 60 –17 , –11 , –7 , –3 Hence Ans 30 20 15 10 8. Arrange the following rational numbers in descending order. (i) 5 2 1 6 , -1 3 ,-0.1, 1.9, – 1 5 Solution: 1 5 , 1 2 , –0.1 , 1.9 , –1 6 3 5 Mathematics-7 12  5 x 11 , –5 x 10 , –1 x 3 , 19 x 3 , –1 x 6 5 6 3 10 10 3 10 3 5 6 55 , –50 , –3 , 57 , –6 30 30 30 30 30 In ascending order. 57 , 55 , –3 , –6 , –50 30 30 30 30 30 5 , –1 , –1 2 Hence 1.9 , 1 –0.1 , Ans 6 5 3 (ii) –5 , –7 , –13 , 23 6 12 18 – 24 Solution: Let us make the denominator same. –5 x 12 , –7 x 6 , –13 x 4 , –23 x 2 6 12 12 6 18 4 24 2 –60 , –42 , –52 , –69 72 72 72 72 In descending –42 > –52 > –60 > –69 –7 , –13 , –5 , –23 Hence Ans 12 8 6 24 (iii) – 10 , – 19 , –23 , –37 11 22 33 44 Solution: Let us make the denominator same. –10 x 12 , –19 x 6 , –23 x 4 , –37 x 3 11 12 22 6 33 4 44 3 –120 , –114 , –92 , –111 132 132 132 132 In descending order. –92 > –111 > –114 > –120 –23 , –37 , –19 , –10 Hence Ans 33 44 22 11 Solution: Solution: 215 » 220 367714 » 367710 13 Mathematics-7 Solution: Solution: last deleted number. Indicated digit is a last deleted 81252149 » 80000000 number. digit 0 < 5 8900000 891865 ; hundreds 2316878 ; milions Solution: Solution: As last deleted number. As last deleted number. digit 6 < 5 digit 6 < 5 891865 » 391900 891865 » 391900 8906 ; hundreds 23133143 ; hundreds Solution: Solution: As last deleted number. As last deleted number. digit 0 < 5 digit 1 < 5 8906 » 8900 23133143 » 23000000 Solution: As last deleted number. digit 1 < 5 52132 » 52000 4.109 Solution: Solution: As last deleted number. As last deleted number. digit 7 < 5 digit 1 < 5 31.7 » 32 4.109 » 4 Solution: Solution: As last deleted number. As last deleted number. digit 3 < 5 digit 8 < 5 53.3 » 53 22.89 » 23 Solution: Solution: As last deleted number. As last deleted number. Mathematics-7 14  digit 1 < 5 digit 8 < 5 3.156 » 3 19.82 » 20 Solution: Solution: As last deleted number. As last deleted number. digit 5 = 5 digit 5 = 5 5.5 » 6 0.506 » 1 Solution: As last deleted number. digit 35 hence 5>5 54.378 » 54.380 0.002770 0.07897 (2 decimal places) 0.0007897 (3 decimal places) Solution: Solution: As last deleted number. 724 are three significant figures. digit 7>5 Hence 6>5 0.07897 » 0.07900 0.0007246 » 0.0007250 3.726 + 5.892 (2 decimal places) 14.348 – 11.104 (2 decimal places) Solution: Solution: As last deleted number. 14.348 » 14 11.104 » 11 digit 7>5 8>5 14 + 11 = 3 3.726 » 4 5.892 » 6 4 + 6 = 10 3.281 x 6.573 (2 decimal places) 9.281 3.173 (2 decimal places) Solution: Solution: 3.281 » 3 6.573 » 7 9.281 » 9 3.173 » 3 3 x 7 = 21 9 3=3 3.281 x 6.573 9.281 3.173 15 Mathematics-7 21.56013 » 22 2.921 » 3 Estimeted velue is wrong. Estimeted velue is wrong. Review Exercise Solution: 1.7, –0.75, 1.5, –0.666, –0.6 Compare the digits 1.7, 1.5, –0.6, –0.666, –0.75 In descending form 1.7, 1 , –0.6, –0.666, –3 1 2 4 Solution: 1.83, –0.666, 1.9, –0.5 and 1.3 –0.5, –0.666, 1.3, 1.83, 1.9 Ascending order –0.5, 1 , 1.3, 11 , 1.9 2 4 Mathematics-7 16  Solution: Solution: Negative number is always less –5 –15 than positive number. 4 8 –1 Let us make denominator same. is smaller number. 11 –5 x 2 –15 4 2 8 –10 –15 8 8 Hence –10 > –13 Solution: Negative number is always less is smaller number. than positive number. –1 is smaller number. 100 Solution: 0.03 is smaller number. Solution: Solution: Negative number is always less Negative number is always less than positive number. than positive number. is smaller number. –0.1 is smaller number. Solution: Solution: Solution: As last digit 7 > 5 As last digit 1 > 5 As last digit 5 = 5 97.832883 » 100 468814 » 468800 29955945 » 30,000,000 Solution: Solution: Solution: As last digit 4 < 5 As last digit 0 < 5 As last digit 5 = 5 614436381 » 614,000,000 9051 » 9000 391865 » 391870 < = > < < > 17 Mathematics-7 Solution: –3 –2 –1 0 1 2 Solution: –1 0 1 2 3 4 5 6 7 8 Solution: –3 –2 –1 0 1 2 Solution: –3 –2 –1 0 1 2 Solution: –3 –2 –1 0 1 2 Solution: –2 –1 0 1 2 Mathematics-7 18  Solution: Solution: Solution: As last deleted number. As last deleted number. As last deleted number. digit 5 = 5 digit 9 > 5 digit 7 > 5 5.285 » 5.290 9.789 » 9.790 0.007 » 0.010 Solution: Solution: Solution: As last deleted number. As last deleted number. As last deleted number. digit 6 < 5 digit 8 > 5 digit 7 > 5 11.256 » 11.260 4.218 » 4.220 9.241 » 9.240 Solution: Solution: Solution: As last deleted number. As last deleted number. As last deleted number. digit 4 < 5 digit 3 < 5 digit 6 > 5 80.0344 » 80.0340 33.3333 » 33.3330 5.2456 » 5.2460 Solution: Solution: Solution: As last deleted number. As last deleted number. As last deleted number. digit 5 = 5 digit 8 > 5 digit 1 < 5 27.4235 » 27.4240 9.2378 » 9.2380 5.6161 » 5.6160 19 Mathematics-7  2 Laws of Operations After Completing this unit students will be able to: verify associative, commutative and distributive properties of rational numbers solve real-world word problems involving operations on rational numbers recognise the order of operations and use it to solve mathematical expressions involving whole numbers, decimals, fractions and integers 20 Mathematics-7  Exercise 2.1 1. Verify commutative property of addition and multiplication on the following pairs of numbers. –7 and —— (i) —— 5 –7 and —— (ii) 1—— 4 –20 –14 8 5 Solution: Solution: w.r.t addition –15 4 —— and — a + b = b +a 8 5 –7 5 –5 –7 w.r.t addition —— + —— = —— + —— a + b = b +a –20 –14 14 –20 –7 x 7 5 x 10 –5 –7 –15 x 5 4 x 8 4 –15 —— – —— = —— + —— —— + — = — + —— –20 x 7 –14 x 10 14 –20 8 x5 5 x8 5 8 –75 + 32 32 – 75 –1 –1 ———— = ———— —— = —— 40 40 140 140 –43 = –43 Hence L.H.S = R.H.S —— —— 40 40 w.r.t multiplication Hence L.H.S = R.H.S ab = ba –7 5 –5 –7 w.r.t multiplication —— —— = —— —— –20 –14 14 20 ab = ba 1 1 1 1 –7 5 –5 –7 1 –3 –3 1 —— 1—— = —— —— — — = — — –20 –14 14 20 2 4 4 4 4 2 2 4 1 1 –3 –3 — = — — = — 8 8 8 8 Hence L.H.S = R.H.S Hence L.H.S = R.H.S –1 and —— +3 1 2 (iii) —— (iv) 1—— and —— –2 –4 2 3 Solution: Solution: –1 –3 — and — 3 2 –2 4 — and — 2 3 w.r.t addition w.r.t addition a + b = b +a a + b = b +a 4 –3 –3 4 — + — = — + — 3 2 3 2 5 4 4 5 — + — = — + — 2 3 2 3 2–3 –3 + 2 ——— = ——— 9+4 4+9 4 4 ——— = ——— 6 6 –1 = –1 —— —— 13 13 4 4 —— = —— 6 6 Hence L.H.S = R.H.S Hence L.H.S = R.H.S Mathematics-7 21  w.r.t multiplication w.r.t multiplication ab = ba ab = ba –15 4 4 –15 3 2 2 3 — — = — — — — = — — 8 5 5 8 2 3 3 2 3 1 1 3 1 1 –15 4 4 –15 3 2 2 3 — — = — — — — = — — 8 2 51 52 8 1 2 1 31 3 2 –3 –3 — = — 1 = 1 2 2 Hence L.H.S = R.H.S Hence L.H.S = R.H.S 3 2 and —— 1 14 (v) 2—— (vi) 17—— and —— 3 16 2 21 Solution: Solution: 8 3 35 14 — and — — and — 3 16 2 21 w.r.t addition w.r.t addition a + b = b +a a + b = b +a 8 3 3 8 35 14 14 35 — + — = — + — — + — = — + — 3 16 16 3 2 21 21 2 128 + 9 ——— 9 + 128 = ——— 735 + 28 ——— 28 + 735 = ——— 48 48 42 42 137 = 137 763 = 763 —— —— —— —— 48 48 42 18 Hence L.H.S = R.H.S Hence L.H.S = R.H.S w.r.t multiplication w.r.t multiplication ab = ba ab = ba 8 3 3 8 35 14 14 35 — — = — — — — = — — 3 16 16 3 2 21 21 2 1 1 1 1 5 7 7 5 8 3 3 8 35 14 14 35 — — = — — — — = — — 31 162 16 3 21 213 21 2 2 1 3 1 1 1 35 35 — = — — = — 2 2 3 2 Hence L.H.S = R.H.S Hence L.H.S = R.H.S 2. Verify commutative property of addition and multiplication on the following pairs of numbers. (i) –2 and —— 2 (ii) –3 and –5 3 Solution: Solution: 3 –3 and –5 –2 and — 16 Commutative property w.r.t addition Commutative property w.r.t addition a + b = b +a a + b = b +a –3 –5 = –5 –3 22 Mathematics-7  –6 + 2 ——— 2–6 = ——— –8 = –8 3 3 Hence L.H.S = R.H.S 4 = 4 —— —— w.r.t multiplication 3 3 ab = ba Hence L.H.S = R.H.S (–3)(–5) = (–5)(–3) w.r.t multiplication 40 = 40 ab = ba 2 2 Hence L.H.S = R.H.S –2 — = — –2 3 3 –4 4 — = — 3 3 Hence L.H.S = R.H.S –7 1 and — 2 (iii) 21 and —— (iv) — 3 9 7 Solution: Solution: –7 1 2 21 and — — and — 3 7 9 Commutative property w.r.t addition Commutative property w.r.t addition a + b = b +a a + b = b +a –7 –7 1 2 2 2 21 + — = — + 21 — + — = — + — 3 3 9 7 7 7 63 – 7 ——— –7 + 63 = ——— 7 + 18 18 + 7 3 3 ——— = ——— 63 63 56 = 56 25 25 —— —— —— = —— 3 3 63 63 Hence L.H.S = R.H.S Hence L.H.S = R.H.S w.r.t multiplication w.r.t multiplication ab = ba ab = ba 7 –7 –7 7 21 — = — 21 1 2 2 1 31 13 — — = — — 9 7 7 9 –49 = –49 2 2 — = — 63 63 Hence L.H.S = R.H.S Hence L.H.S = R.H.S 9 1 and — 5 1 and — (v) 2— (vi) 1— 3 7 2 3 Solution: Solution: 7 9 3 5 — and — — and — 3 7 2 3 Commutative property w.r.t addition Commutative property w.r.t addition a + b = b +a a + b = b +a 7 9 9 7 3 5 5 3 — + — = — + — — + — = — + — 3 7 7 3 2 3 3 2 Mathematics-7 23  49 + 27 ——— 27 + 49 = ——— 9 + 10 ——— 10 + 9 = ——— 21 21 6 6 76 = 76 76 = 76 —— —— —— —— 21 21 21 21 Hence L.H.S = R.H.S Hence L.H.S = R.H.S w.r.t multiplication w.r.t multiplication ab = ba ab = ba 7 9 9 7 3 5 5 3 — — = — — — — = — — 3 7 7 3 2 3 3 2 1 3 3 1 1 1 7 9 9 7 3 5 5 3 — — = — — — — = — — 3 1 71 71 31 2 31 31 2 3 = 3 5 5 — = — 2 2 Hence L.H.S = R.H.S Hence L.H.S = R.H.S Exercise 2.2 1. Verify associative property of additional and multiplication on the following set of rational numbers. (i) 1 ,— — 3 and — 7 (ii) 1 ,— — 3 and — 4 2 4 8 7 14 7 Solution: Solution: Associative w.r.t addition Associative w.r.t addition as: (a + b) + c = a + (b + c) as: (a + b) + c = a + (b + c) 1 3 7 1 3 7 1 3 4 1 3 4 — + — +— = — + — +— — + — +— = — + — +— 2 4 8 2 4 8 7 14 7 7 14 7 2+3 7 1 6+7 2+3 4 1 3+8 ——- + — = — + ——- ——- + — = — + ——- 4 8 2 8 14 7 7 14 5 7 1 13 5 4 1 11 —+— = —+— —+— = —+— 4 8 2 8 14 7 7 14 10 +7 4 + 13 5+8 2 + 11 ——- = ——- ——- = ——- 8 8 14 14 17 17 13 13 —=— —=— 8 8 14 14 Hence L.H.S = R.H.S Hence L.H.S = R.H.S Associative w.r.t addition Associative w.r.t addition as: (a x b) x c = a x (b x c) as: (a x b) x c = a x (b x c) 1 3 7 1 3 7 1 3 4 1 3 4 —x — x— =— x — x— —x — x— =— x — x— 2 4 8 2 4 8 7 14 7 7 14 7 3 7 1 21 3 4 1 7 —x — = —x — —x — = —x — 8 8 2 32 98 7 7 98 24 Mathematics-7  21 21 6 6 —=— —— = —— 64 64 343 343 Hence L.H.S = R.H.S Hence L.H.S = R.H.S (iii) 2 ,— — 3 and — 7 (iv) 1 ,— — 3 and —4 3 6 9 5 10 15 Solution: Solution: Associative w.r.t multiplication Associative w.r.t addition 1 1 1 as: (a + b) + c = a + (b + c) 2 3 7 2 3 7 —x — x — = — x — x— 1 3 4 1 3 4 3 6 9 31 63 9 — + — +— = — + — +— 3 1 5 10 15 5 10 15 2 7 1 7 —x — = —x — 2+3 4 1 9+8 3 189 3 9 ——- + — = — + ——- 10 15 5 30 1 7 1 7 — x — = —x — 5 4 1 17 3 9 3 9 — + — = — +— 10 15 5 30 7 7 — =— 15 + 8 6 + 17 27 27 ——- = ——- 30 30 Hence L.H.S = R.H.S 23 23 —=— Associative w.r.t addition 30 30 as: (a + b) + c = a + (b + c) Hence L.H.S = R.H.S Associative w.r.t addition 2 3 7 2 3 7 — + — +— = — + — +— as: (a x b) x c = a x (b x c) 3 6 9 3 6 9 1 2 4+3 7 2 9 + 14 1 3 4 = 1 3 4 —x — x— — x — x— ——- + — = — + ——- 5 10 15 5 10 15 6 9 3 18 5 5 1 2 7 7 2 23 3 4 1 2 —+— = —+— —x — = —x — 6 9 3 18 5025 155 5 25 21 + 14 14 + 21 2 2 ——- = ——- —= — 18 18 75 75 35 35 —=— Hence L.H.S = R.H.S 18 18 Hence L.H.S = R.H.S (iii) —2 ,— 3 and —8 (iv) 1 ,— — 3 and —7 13 13 13 5 10 15 Solution: Solution: Associative w.r.t addition Associative w.r.t addition as: (a + b) + c = a + (b + c) as: (a + b) + c = a + (b + c) 2 3 8 2 3 8 1 3 7 1 3 7 — + — +— = — + — +— — + — +— = — + — +— 13 13 13 13 13 13 5 10 15 5 10 15 2+3 8 2 3+8 2+3 7 1 9 + 14 ——- + — = — + ——- ——- + — = — + ——- 13 13 13 13 10 15 5 30 5 8 2 11 5 7 1 23 — + — = — +— — + — = — +— 13 13 13 13 10 15 5 30 Mathematics-7 25  5+8 2 + 11 15 + 14 6 + 23 ——- = ——- ——- = ——- 13 13 30 30 13 13 29 29 —=— —=— 13 13 30 30 1=1 1=1 Hence L.H.S = R.H.S Hence L.H.S = R.H.S Associative w.r.t multiplication Associative w.r.t multiplication 1 2 3 8 2 3 8 1 3 7 1 3 7 —x— x — = — x — x— —x— x — = — x — x— 13 13 13 13 13 13 5 10 15 5 10 15 1 5 6 8 2 24 3 7 1 7 — x — =— x — — x — =— x — 13 13 13 13 50 15 5 5 50 48 48 7 7 — =— — =— 13 13 250 250 Hence L.H.S = R.H.S Hence L.H.S = R.H.S 2. Verify the following. (i) 3 x –4 — — = –4 3 — x — (ii) –8 — x 11 — = 11 — x –8 — 7 8 8 7 7 9 9 7 Solution: Solution: L.H.S L.H.S 1 3 –4 –8 11 =— x — =— x — 7 82 7 9 3 –1 –88 =— x — = —— 7 2 63 –3 R.H.S =— 14 11 –8 R.H.S =— x — 1 9 7 –4 3 –88 =— x — = —— 82 7 63 –1 3 L.H.S = R.H.S =— x — 2 7 Hence, verified –3 =— 14 L.H.S = R.H.S Hence, verified (i) –12 —— x —6 =— 6 x –12 —— (ii) –8 7 =— — x — 7 x –8 — 5 13 13 5 3 5 5 3 Solution: Solution: L.H.S L.H.S –12 6 –8 7 = —— x — =— x — 5 13 3 5 26 Mathematics-7  –72 –56 = —— = —— 65 15 R.H.S R.H.S 6 –12 7 –8 = —— x — =— x — 13 5 5 3 –72 –56 = —— = —— 65 15 L.H.S = R.H.S L.H.S = R.H.S Hence, verified Hence, verified 3. Verify each of the following. (i) 5 x 12 — — x —7 = — 5 x 12 7 — x — (ii) –12 —— x —4 x — 25 — = –12 —— x —4 x —25 — 7 13 25 7 13 25 5 15 –16 5 15 –16 Solution: Solution: L.H.S L.H.S 1 5 5 12 7 –12 4 25 = — x — x — = —— x — x —— 7 13 25 5 15 –16 3 4 12 1 60 7 –12 1 –5 = — x — = —— x — x — 1391 25 5 3 4 5 12 1 1 1 = — x — –12 –5 13 5 = —— x — 51 121 12 = — = –1 x –1 65 = 1 R.H.S R.H.S 5 12 7 4 = — x — x — –12 4 25 7 13 25 = —— x — x — 5 155 –16 1 12 5 84 –4 4 25 = — x — = —— x — x — 1 7 325 65 5 5 16 1 12 1 1 = — x — –16 25 1 65 = — x— 25 1 –161 12 = — = –1 x –1 65 = 1 L.H.S = R.H.S L.H.S = R.H.S Hence, verified Hence, verified Mathematics-7 27  Exercise 2.3 1. Find the additive inverse of the following: 1 (i) –17 (ii) — 3 Solution: Solution: 1 –1 Additive inverse of –17 is 17 Additive inverse of — is — 3 3 –2 –8 (ii) — (iii) — 3 –7 Solution: Solution: –2 2 –8 –8 Additive inverse of — is — Additive inverse of — is — 3 3 –7 7 19 (i) 0 (ii) — –8 Solution: Solution: 19 19 Additive inverse of 0 is 0 Additive inverse of — is — –8 8 3. Verify and name the property used in the following: 2 x –4 2 39 52 2 39 2 52 (i) — — + –8 2 x –4 — = — 2 x –8 — + — — (ii) — x — 13 4 +— 8 = 13 — x —4 – 13 — x —8 5 7 9 5 7 5 9 Solution: Solution: 1 3 1 2 –36 + (–56) –8 16 2 39 52 2 39 2 52 — x — = — – — — x —+— = — x — – — x — 5 63 35 45 13 4 8 131 4 2 13 84 3 1 2 –92 –72 + 112 2 78 – 52 39 52 — x —— = — — x — = — – — 5 63 315 5 8 86 2 521 1 2 184 184 2 26 3 —— = —— — x —— = — x –1 315 315 131 84 2 1 2 1 L.H.S = R.H.S —— = —— 42 2 Distributive property w.r.t multiplication 1 1 —— = —— over addition 2 2 L.H.S = R.H.S Distributive property w.r.t multiplication over addition 1. Find the additive inverse of the following: (i) –13 (ii) –5 — 7 Solution: Solution: 1 –5 –5 Multiplicative inverse of –13 is — Multiplicative inverse of — is — –13 7 7 28 Mathematics-7  (iii) –3 — (iv) 1 — 5 2 Solution: Solution: –3 –5 1 1 Multiplicative inverse of — is — Multiplicative inverse of — is — 5 3 2 2 (i) 2 Solution: Multiplicative inverse of 2 is –2 4. Solve these using distributive property: (i) 7 x –5 — 7 x — — –— 6 (i) 9 x — 1 – —9 x –1 5 14 5 14 — — 16 3 16 3 Solution: Solution: 1 1 1 3 3 1 7 –5 7 6 9 1 9 1 = — x — –— x — = — x — –— x — 51 142 5 14 2 1 16 3 1 16 31 –1 3 3 3 =— – — =— – — 2 5 16 16 –5 – 6 3–3 = ———— = ———— 10 16 –11 = 0 = — 10 4. Find the rational numbr that should be added and subtracted so that they will make the 9 3 5 sum — + — + — to the nearest whole number. 2 4 7 Solution: Let x be the number that 9 is added in — 3 5 + — + — 2 4 7 so it become 9 x 14 3 x 7 5 x4 = —x 14 + —x 7 + — x 4 2 4 7 126 + 21 + 20 = ———— ———— 28 167 = — 28 6. The product of two rational numbers ias –5 8 —. If one number is — , then find the other. 13 10 Solution: 8 –5 x x — = — 10 13 25 –5 10 –50 –25 x = — x — x = — 52 x =— 13 8 104 52 Mathematics-7 29  Exercise 2.4 1. Solve: (i) 4x6+8¸4 (ii) 10 – 3 x 2 + 4 Solution: Solution: 4x6+8¸4 10 – 3 x 2 + 4 = 24 + 2 = 10 – 6 + 8 = 26 = 4+8 = 12 (iii) 5x4+6–2¸2 Solution: 5x4+6–2¸2 = 20 + 6 – 1 = 25 3. Simplyfy the following: (i) 12 x 18 – [ 64 – { 9 + ( –13 – 4 )}] (ii) 50 – [ 5 + { 28 – ( 20 ¸ 4 + 5 )}] Solution: Solution: 12 x 18 – [ 64 – { 9 + ( –13 – 4 )}] 50 – [ 5 + { 28 – ( 20 ¸ 4 + 5 )}] = 12 x 8 – [64 – {9 + (–9)}] = 50 – [5 + {28 – ( 5 + 5)}] = 12 x 8 – [64 – {9 – 9}] = 50 – [5 + {28 – 10}] = 12 x 8 – [64 – 0] = 50 – [5 + 18] = 12 x 8 – 64 = 50 – 23 = 96 – 64 = 27 = 32 = 32 1 7 12 4 7 (iii) 9 x 9 – [ 45 – { 64 ¸ (8 – 4 + 2)}] (iv) 1 2 ¸ [ 6 + { 5 – ( 5 + 49 ¸ 5 )}] Solution: Solution: 7 9 x 9 – [ 45 – { 64 ¸ (8 – 4 + 2)}] 1 = 1 2 ¸ [ 7 + { 12 – ( 4 + 49 x 5 )}] 6 5 5 71 = 9 x 9 – [45 – {64 ¸ (8 – 6)}] = 3 7 12 4 = 9 x 9 – [45 – {64 ¸2}] 2 ¸ [ 6 + { 5 – ( 5 + 35 )}] = 9 x 9 – [45 – 32] = 3 7 12 4 + 175 )}] = 9 x 9 – 13 2 ¸[ 6 +{ 5 – ( 5 = 81 – 13 3 7 12 179 = 2 ¸ [ 6 + { 5 – ( 5 )}] = 63 30 Mathematics-7  2 1 (v) 3 3 ¸ {1 1 + ( 1 1 + 3 3 ¸ 2 1 )} = 3 ¸ [ 7 + { 12 – ( 12 – 179)}] 3 3 2 2 6 5 5 Solution: 2 1 = 3 ¸ [ 7 – { 167 { [ 3 3 ¸ {1 1 + ( 1 1 + 3 2 – 2 1 )} 2 6 5 3 3 2 = 3 ¸ [ 35 + 1002 [ = 11 ¸ { 4 + ( 4 + 7 – 5 )} 2 30 3 3 3 2 2 3 –967 = ¸ + 21 – 15 2 30 = 11 ¸ { 4 + ( 8 )} 15 3 3 6 = 3 x 30 2 1 967 = 11 ¸ { 4 + ( 14 )} 45 3 3 6 = 967 = 11 ¸ { 8 + 14 { 3 6 = 11 22 3 ¸{ 6 { 1 2 = 11 x 6 31 22 2 (vi) [2.35 + { 12.95 ¸ ( 1.45 – 2.1 x 1.25)}] (vii) 0.8 x [ 4.9 x { 10.2 ¸ ( 0.2 + 0.02 + 0.002)}] Solution: Solution: [2.35 + { 12.95 ¸ ( 1.45 – 2.1 x 1.25)}] 0.8 x [4.9 x {10.2 ¸ ( 0.2 + 0.02 + 0.002)}] = [2.35 + {12.95 ¸ (1.45 – 2.625)}] = 0.8 x [4.9 x {10.2 ¸ ( 0.222 )}] = [2.35 + {12.95 ¸ (–1.175)}] = 0.8 x [4.9 x {10.2 ¸ 0.222 }] = [2.35 + {12.95 ¸ –2.625}] = 0.8 x [4.9 x {45.946}] = [2.35 + {–11.021}] = 0.8 x [4.9 x 45.946] = [2.35 – 11.021] = 0.8 x 225.135 = –8.671 = 180.108 (viii) 15.16 ¸ [ 2.125 + 3.04 – (2.75 x 2.06 + 2.02)] Solution: 15.16 ¸ [ 2.125 + 3.04 – (2.75 x 2.06 + 2.02)] = 15.16 ¸ [ 2.125 + 3.04 – (2.75 x 4.08)] = 15.16 ¸ [ 2.125 + 3.04 – (11.22)}] = 15.16 ¸ [ 2.125 + 3.04 – 11.22] = 15.16 ¸ [ 2.125 + –8.18] = 15.16 ¸ [ –6.055] = 15.16 ¸ –6.055 = –2.503 Mathematics-7 31 3. If 40 is subtracted from 60 and then divided by 5, what is the result? Solution: 60 – 40 = 20 20 = 4 5 = 4 ans 4. If 48 is divided by 4 and then multiplied by 3, what is the result? Solution: 48 = 12 4 12 x 3 = 12 = 36 ans 5. If 15 is subtracted from 30 and then multiplied by 9, find the result. Solution: 30 – 15 = 15 15 x 9 = 135 = 135 ans 2 Review Exercise 1. Tick (ü) the correct option: (i) The additive inverse of – 3: 7 3 3 7 –7 (a) (üb) (c) (d) –7 7 3 3 (ii) The name of the property used in: 1 x –2– 5 1 x– 2 – 1 x 5 = 3 5 6 3 5 3 6 (a) Commutative property of rational numbers over subtraction (b) Commutative property of integers over subtraction (c) Associative property of rational numbers over subtraction (üd)Distributive property of rational numbers with respect to multiplication over subtraction (iii) The additive identity of rational number is: (a) 1 (b) – 1 (üc) 0 (d) ± 1 (iv) The multiplicative identity of rational number: (a) – 1 (b) 0 (üc) 1 (d) +1 7 (v) The multiplicative inverse of : 15 –7 – 15 15 7 (a) (b) (üc) (d) 15 7 7 – 15 32 2. Prove associative property with respect to multiplication over addition and subtraction of the following set of rational numbers: 1 , 3 and – 6 – 5 , 4 and – 3 (i) (ii) 2 5 7 10 5 4 Solution: Solution: Associative w.r.t addition Associative w.r.t addition as: a + (b + c) = (a + b) +c as: a + (b + c) = (a + b) +c 1 3 –6 1 3 –6 –5 4 –3 –5 4 –3 — + —+ — = — + — + — —+ — + — = — + — + — 2 5 7 2 5 7 10 5 4 10 5 4 1 21– 30 5+6 6 –5 16 – 15 –5 + 6 3 — + ——— = — — –— — + ——— = — — –— 2 35 10 7 10 20 10 4 1 –9 11 6 –5 1 3 3 —+— = — –— —— + —— = — —–— 2 35 10 7 10 20 10 4 35 – 18 77 – 60 –10 + 1 6 – 15 ——- =——- ——- ——- =——- 70 70 20 20 17 17 –9 9 —=— —=— — 70 70 20 20 Hence L.H.S = R.H.S Hence L.H.S = R.H.S Associative w.r.t Subtarction Associative w.r.t Subtarction as: a – (b – c) = (a – b) – c as: a – (b – c) = (a – b) – c 1 3 –6 1 3 –6 –5 4 –3 –5 4 –3 — – —– — = — – — – — —– —– — = — – — – — 2 5 7 2 5 7 10 5 4 10 5 4 –5 16 +15 –5 – 8 3 1 3 6 1 3 6 — – ——— = — — +— — – — +— = — – — + — 10 20 10 4 2 5 7 2 5 7 –5 31 13 3 1 21 – 30 5–6 6 —— – —— = — — +— — – ——— = — — +— 10 20 10 4 2 35 10 7 –10 + 31 6 + 15 1 –9 –1 6 ——- ——- =——- —+ — = — +— 20 20 2 35 10 7 –41 –11 1 9 –7 + 60 —— ¹ — — — + — = ——- 20 20 2 35 70 Hence L.H.S ¹ R.H.S 35 + 18 53 ——- = — 70 70 53 53 —=— 70 70 Hence L.H.S = R.H.S 3. Verify the following: – 4x 8 8 x–4 – 1x( 5 7 –1x 5 –1 7 (i) = (ii) + )= + x 5 10 10 5 6 3 12 6 3 6 12 Solution: Solution: 2 2 –1 5 7 –1 5 –1 7 –4 8 8 –4 — x (— + —( = — x — + — x — — x — =— x — 6 3 12 6 3 6 12 5 10 5 105 5 Mathematics-7 33  –1 20 + 7 –5 7 –2 8 8 –2 — x ( ———— ( = — – — — x — =— x — 6 12 18 72 5 5 5 5 –16 –16 –1 27 –20 – 7 —— = —— — x — = ———— 25 25 6 12 72 –27 –27 L.H.S = R.H.S —— = —— 72 72 Hence, verified L.H.S = R.H.S Hence, verified 3 x (– 2 – 1 ) = 3 x– 2– 3 x 1 (iii) 5 9 3 5 9 5 3 Solution: 1 3 x (– 2 – 1 ) = 3 x– 2 – 3 x 1 5 9 3 5 9 5 31 3 x ( –2 – 3 ) = –6 – 1 5 9 45 5 1 3 x –5 = –2 – 3 = 1 5 –15 = –5 5 9 15 45153 153 1 = 1 3 3 L.H.S = R.H.S Hence, verified 4. Find the additive and multiplicative inverse of the following numbers: 5 (i) – 15 (ii) 7 Solution: Solution: Additive inverse of –15 is 15 Additive inverse of 5 is –5 Multiplicative inverse of –15 is –1 7 7 Multiplicative inverse of 5 is 7 15 7 5 3 –1 (iii) (iv) – 11 10 Solution: Solution: Additive inverse of –3 is 3 Additive inverse of –1 is 1 11 11 10 10 Multiplicative inverse of –3 is –11 Multiplicative inverse of –1 is –10 11 3 10 8 (v) 9 (ii) 13 Solution: Solution: Additive inverse of 9 is –9 Additive inverse of 8 is –8 Multiplicative inverse of 9 is 1 13 13 Multiplicative inverse of 8 is 13 9 13 8 34 Mathematics-7 3 SQUARE AND SQUARE ROOT After Completing this unit students will be able to: Mathematics-7 35  Exercise 3.1 1. Find the square of the following numbers. (i) 17 (ii) 49 Solution: Solution: As, 17 can be written as As, 49 can be written as 17 = 20 – 3 49 = 50 – 1 So, by taking square So, by taking square 2 2 (17) = (20 – 3) (49)2 = (50 – 1)2 As, (a-b)2 = a2 – 2ab + b2 As, (a-b)2 = a2 – 2ab + b2 2 2 2 2 = (20) -2(20)(3) +(3) = (50) -2(50)(1) +(1) = 400 – 120 + 9 = 2500 – 100 + 1 2 2 (17) = 289 ans (49) = 2401 ans (iii) 110 (iv) 203 Solution: Solution: As, 110 can be written as As, 203 can be written as 110 = 100 + 10 203 = 200 + 3 So, by taking square So, by taking square 2 2 2 2 (110) = (100 +10) (203) = (200 +3) 2 2 2 2 2 2 As, (a+b) = a + 2ab + b As, (a+b) = a + 2ab + b 2 2 2 2 = (100) +2(100)(10) +(10) = (200) +2(200)(3) +(3) = 100,00 + 2000 + 100 = 40,000 + 1200 + 9 (110)2 = 12100 ans (203)2 = 41209 ans (v) 118 (vi) 891 Solution: Solution: As, 118 can be written as As, 891 can be written as 118 = 120 – 2 891 = 900 – 9 So, by taking square So, by taking square 2 2 (118) = (120 – 2) (891)2 = (900 – 9)2 As, (a-b)2 = a2 – 2ab + b2 As, (a-b)2 = a2 – 2ab + b2 = (120)2 -2(120)(2) +(2)2 = (900)2 -2(900)(9) +(9)2 = 14400 – 480 + 4 = 8100000 – 16200 + 81 (118)2 = 13924 ans (891)2 = 793881 ans 36 Mathematics-7 2. Find the square of the following integers: (i) –19 (ii) –103 Solution: Solution: 2 2 (–19) = –19 x –19 (–103) = –103 x –103 = 381 = 10609 (iii) 117 (iv) –27 Solution: Solution: (117)2 = 117 x 117 (–27)2 = –27 x –27 = 13689 = 729 (v) 81 (vi) –700 Solution: Solution: 2 2 (81) = 81 x 81 (–700) = –700 x –700 = 6561 = 490000 3. Find the square of the following proper fraction and compare with proper fraction: 23 1 (i) —— (ii) —— 25 5 Solution: Solution: 23 2 23 x 23 1 2 1 x 1 —— = ———— — = ———— 25 25 x 25 5 5 x 5 529 1 = —— = —— 625 25 2 9 (iiii) —— (iv) —— 7 10 Solution: Solution: 2 2 2 x 2 9 2 9 x 9 — = ———— —— = ———— 7 7 x 7 10 10 x 10 4 81 = —— = —— 49 100 7 12 (v) —— (vi) —— 11 17 Solution: Solution: 7 2 7 x 7 12 2 12 x 12 —— = ———— —— = ———— 11 11 x 11 17 17 x 17 49 144 = —— = —— 121 289 Mathematics-7 37 2. Find the square of the following decimal numbers: (i) 2.2 (ii) 1.3 Solution: Solution: 2 2 (2.2) = 2.2 x 2.2 (1.3) = 1.3 x 1.3 = 4.84 = 1.69 (iii) 0.25 (iv) 1.09 Solution: Solution: (0.25)2 = 0.25 x 0.25 (1.09)2 = 1.09 x 1.09 = 0.0625 = 1.1881 (v) 2.07 (vi) 23.21 Solution: Solution: 2 2 (2.07) = 2.07 x 2.07 (23.21) = 23.21 x 23.21 = 4.2849 = 538.7041 Exercise 3.2 1. Find the squares root of the following numbers: (i) 225