5. Operations on Rational Numbers
Step-by-step solutions to Practice Sets 22 to 25. Maharashtra State Board (Balbharati), Grade Seven Mathematics.
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Practice Set 22
(i) 536 + 642 = 536 + 17 = 5 × 7 + 1 × 36252 = 35 + 36252 = 71252
(ii) 123 + 245 = 53 + 145 = 5 × 5 + 14 × 315 = 25 + 4215 = 6715 = 4715
(iii) 1117 + 1319 = 11 × 19 + 13 × 1717 × 19 = 209 + 221323 = 430323 = 1107323
(iv) 2311 + 1377 = 2511 + 8077 = 25 × 777 + 8077 = 175 + 8077 = 25577 = 32477
Use the LCM of the denominators (for example, 252 in (i) and 77 in (iv)) to keep the numbers small. Change mixed numbers into improper fractions first.
(i) 711 − 37 = 7 × 7 − 3 × 1177 = 49 − 3377 = 1677
(ii) 1336 − 240 = 1336 − 120 = 13 × 5 − 1 × 9180 = 65 − 9180 = 56180 = 1445
(iii) 123 − 356 = 53 − 236 = 106 − 236 = 10 − 236 = −136 = −216
(iv) 412 − 313 = 92 − 103 = 27 − 206 = 76 = 116
(i) 311 × 25 = 3 × 211 × 5 = 655
(ii) 125 × 415 = 12 × 45 × 15 = 4875 = 1625
(iii) (−8)9 × 34 = (−8) × 39 × 4 = −2436 = −23
(iv) 06 × 34 = 0 × 36 × 4 = 024 = 0
In (iv), the numerator is 0, so the product is 0. Zero multiplied by any number is zero.
The multiplicative inverse of ab is ba, because their product is 1.
- (i) 25 — 52
- (ii) −38 — −83
- (iii) −1739 — −3917
- (iv) 7 — 17
7 = 71 - (v) −713 — −322
−713 = −223
The inverse of a negative number is also negative, since (−) × (−) = (+) = 1.
To divide, multiply the first number by the multiplicative inverse of the second.
(i) 4012 ÷ 104 = 4012 × 410 = 40 × 412 × 10 = 160120 = 43 = 113
(ii) −1011 ÷ −1110 = −1011 × −1011 = (−10) × (−10)11 × 11 = 100121
(iii) −78 ÷ −36 = −78 × −63 = (−7) × (−6)8 × 3 = 4224 = 74 = 134
(iv) 23 ÷ (−4) = 23 × −14 = −212 = −16
(v) 215 ÷ 536 = 115 ÷ 336 = 115 × 633 = 11 × 65 × 33 = 66165 = 25
(vi) −513 ÷ 726 = −513 × 267 = (−5) × 2613 × 7 = −13091 = −107 = −137
(vii) 911 ÷ (−8) = 911 × −18 = −988
(viii) 5 ÷ 25 = 5 × 52 = 252 = 1212
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Practice Set 23
Method: write both numbers with the same (large enough) denominator, then pick numerators that lie between the two numerators. Many correct answers are possible; one set is given for each.
(i) 27, 67
27 < 37 < 47 < 57 < 67
Three rational numbers: 37, 47, 57
(ii) 45, 23
23 = 2030 and 45 = 2430
2030 < 2130 < 2230 < 2330 < 2430
Three rational numbers: 710, 1115, 2330
(iii) −23, 45
−23 = −1015 and 45 = 1215
−1015 < −415 < 115 < 615 < 1215
Three rational numbers: −415, 115, 25
(iv) 79, −59
−59 < −29 < 19 < 49 < 79
Three rational numbers: −29, 19, 49
(v) −34, 54
−34 < −14 < 14 < 34 < 54
Three rational numbers: −14, 14, 34
(vi) 78, −53
−53 = −4024 and 78 = 2124
−4024 < −2524 < −1024 < 624 < 2124
Three rational numbers: −2524, −512, 14
(vii) 57, 117
57 < 77 < 87 < 97 < 117
Three rational numbers: 1, 87, 97
(viii) 0, −34
−34 = −68 and 0 = 08
−68 < −48 < −38 < −28 < 08
Three rational numbers: −12, −38, −14
Practice Set 24
(i) 134
Denominator 4 = 2 × 2, so the decimal terminates: 134 = 13 × 254 × 25 = 325100
134 = 3.25 (terminating)
(ii) −78
Find 7 ÷ 8 and put the negative sign: 78 = 7 × 1258 × 125 = 8751000
−78 = −0.875 (terminating)
(iii) 735
735 = 385 = 38 × 25 × 2 = 7610
735 = 7.6 (terminating)
(iv) 512
5 ÷ 12: 50 ÷ 12 gives 4 remainder 2; 20 ÷ 12 gives 1 remainder 8; 80 ÷ 12 gives 6 remainder 8; remainder 8 repeats, so 6 repeats.
512 = 0.416666… = 0.416̇ (recurring)
(v) 227
22 ÷ 7 = 3 remainder 1; the remainders then go 3, 2, 6, 4, 5, 1 and repeat, so the six digits 142857 repeat.
227 = 3.142857142… = 3.142857 (recurring)
(vi) 43
4 ÷ 3 = 1 remainder 1; every further step gives 3 remainder 1, so 3 repeats.
43 = 1.3333… = 1.3̇ (recurring)
(vii) 79
70 ÷ 9 = 7 remainder 7; the remainder 7 keeps repeating, so 7 repeats.
79 = 0.7777… = 0.7̇ (recurring)
A dot over a single digit (0.416̇) or a bar over a group of digits (3.142857) shows the part that repeats. A fraction in lowest terms has a terminating decimal only when its denominator has no prime factors other than 2 and 5.
Practice Set 25
50 × 5 ÷ 2 + 24
= 250 ÷ 2 + 24 (multiplication first, left to right)
= 125 + 24
= 149
(13 × 4) ÷ 2 − 26
= 52 ÷ 2 − 26 (bracket first)
= 26 − 26
= 0
140 ÷ [(−11) × (−3) − (−42) ÷ 14 − 1]
= 140 ÷ [33 − (−3) − 1] (× and ÷ inside the bracket)
= 140 ÷ [33 + 3 − 1]
= 140 ÷ 35
= 4
The textbook prints an extra closing bracket after “− 1”. It does not change the working: the square bracket contains (−11) × (−3) − (−42) ÷ 14 − 1.
{(220 − 140) + [10 × 9 + (−2 × 5)]} − 100
= {80 + [90 + (−10)]} − 100 (innermost brackets first)
= {80 + 80} − 100
= 160 − 100
= 60
35 + 38 ÷ 64
= 35 + 38 × 46 (division first)
= 35 + 1248
= 35 + 14
= 12 + 520
= 1720