Exercise 1 B
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Question 1
Answer True (T) or False (F)
(i)
Subtraction is commutative for rational numbers. F
(ii) To subtract cd from ab , we add the additive inverse of ab to cd . F
(iii) 0 is its own additive inverse . T
(iv) The additive inverse of −21−30 is −21−30 . T
(v) While subtracting three or more rational numbers , they can be grouped in any order . F
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Question 2
Add the following rational numbers
(i)
49 and 29
Sol :
⇒49+29
⇒69=23
(ii) −1423 and 923
Sol :
⇒−1423+923
⇒−14+923=−523
(iii) −910 and 34
Sol :
L.C.M of 4 and 10 is = 2×2×5=20
24,1022,551,51,1
⇒−9×2+3×520
⇒−18+1520
⇒−320
(iv) −512 and −79
Sol :
⇒−512+(−7)9
L.C.M of 12 and 9 is = 2×2×3×3=36
212,926,933,931,31,1⇒−5×3+(−7)×436
⇒−15−2836
⇒−4336
(v) 710 and −815
Sol :
⇒710+(−815)
L.C.M of 10 and 15 is = 2×3×5=30
210,1535,1555,51,1⇒7×3−8×230
⇒21−1630
⇒530=16
(vi) −7311 and (−8)
Sol :
⇒−7×11+311+−81
⇒−8011−81
⇒−80−8×1111
⇒−80−8811
⇒−16811=−15311
(vii) 413 and 257
Sol :
⇒133+197
L.C.M of 3 and 7 is 21
⇒13×7+19×321
⇒91+5721
⇒14821=7121
(viii) 356 and −5−12
Sol :
⇒236+512
L.C.M of 6 and 12 is =2×2×3=12
26,1223,633,31,1⇒46+512
⇒5112=174=414
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Question 3
Simplify:
(i)
14+−67+314
Sol :
⇒14−67+314
L.C.M of 4 , 7 and 14 is =2×2×3 =28
24,7,1422,7,771,7,71,1,1
⇒7×1−6×4+3×228
⇒7−24+628
⇒13−2428
⇒−1128
(ii) 715+−925+−310
Sol :
⇒715−925+−310
L.C.M of 15 , 25 and 10 is =2×3×5×5 =150
215,25,10315,25,555,25,551,5,11,1,1⇒7×10−9×6−3×15150
⇒70−54−45150
⇒70−99150
⇒−29150
(iii) 113+2−9+−56
Sol :
Denominator can not be negative
⇒(3×1+13)+(2−9×−1−1)−56
⇒43−29−56
L.C.M of 3 , 9 and 6 is =2×3×3 =18
23,9,633,9,331,3,11,1,1⇒4×6−2×2−5×318
⇒24−4−1518
⇒24−1918
⇒518
(iv) 415+(−5310)+112
Sol :
⇒215−5310+32
L.C.M of 5 , 10 and 2 is =2×5 =10
25,10,255,5,11,1,1
⇒21×2−53×1+3×510
⇒42−53+1510
⇒57−5310
⇒410=25
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Question 4
Verify the following.
(i)
38+−23=−23+38
Sol :
They are equal to each other by commutative property
ALTERNATE METHOD
L.C.M of 8 and 3 is =2×2×2×3 =24
28,324,322,331,31,1
⇒3×3−2×824=−2×8+3×324
⇒9−1624=−16+924
⇒−724=−724
(ii) −711+15−22=15−22+−711
Sol :
They are equal to each other by commutative property
ab+cd=cd+ab (by commutative property)
ALTERNATE METHOD
−711−1522=−1522−711
[denominator cannot be negative]
LCM of 11 and 22 is 22.
−7×2−15×122=−15×1−7×222
−14−1522=−15−1422
−2922=−2922
(iii) −8+−11−12=−11−12+(−8)
Sol :
They are equal to each other by commutative property
ALTERNATE METHOD
−8−11−12=−11−12−8
−8+1112=1112−8
−8×12+11×112=11×1−8×1212
−96+1112=−96+1112
−10712=−10712
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Question 5
Verify that:
(i)
(−58+98)+138=−58+(98+138)
Sol :
They are equal to each other by associative property
ALTERNATE METHOD
(−5+98)+138=−58+(9+138)
48+138=−58+228
178=178
(ii) −20+(3−5+−7−10)=(−20+3−5)+−7−10
Sol :
They are equal to each other by associative property
ALTERNATE METHOD
(denominator can not be negative)
−20+(−35+710)=(−20−35)+710
[L.C.M of 5 and 10 is 10]
−20+(−3×2+7×110)=(−20×5−3×15)+710
−20+(−6+710)=(−100−35)+710
−20+110=−1035+710
[L.C.M of 5 and 10 is 10]
(−20×10+1×110)=(−103×2+7×110)
−200+110=−208+710
−19910=−19910
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Question 6
Find the additive inverse of each of the following.
(i) 1011
Sol :
⇒−(1011)
(ii) 0
Sol :
0=0
(iii) 218
Sol :
⇒−(218)
(iv) −11−8
Sol :
⇒−(⧸−11⧸−8)
⇒−(118)=−118
(v) −2613
Sol :
⇒−(−2613)=2613
(vi) 18−39
Sol :
⇒−(18−39)=1839
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Question 7
Arrange and simplify:
(i)
12+−35+32
Sol :
12+(−35)+32
=12+32−35
=1+32−35
=42−35
[L.C.M of 2 and 5 is 10]
=4×5−3×210
=20−610
=1410=75=125
(ii) 2817+3517+−1617+−2317
Sol :
=28+35−16−2317
=63−3917
=2417=1717
(iii) 23+−35+16+−815
Sol :
23+(−35)+16+(−815)
=23+16−35−815
L.C.M of 3 and 6 is 6
L.C.M of 5 and 15 is 15
=2×2+1×16+−3×3−8×115
=4+16+−9−815
L.C.M of 15 and 6 is 30
=56−1715
26,1533,1551,51,1
=5×5−17×230
=25−3430
=−930 or−310
(iv) 35+53+−115+−23
Sol :
35+53+−115+−23
=35−115+53−23
=3−115+5−23
=−85+33
L.C.M of 5 and 3 is 15
=−8×3+3×515
=−24+1515
=−915 or−35
(v) 43+35+−23+−115
Sol :
43+35+−23+−115
=43−23+35−115
=4−23+3−115
=23+−85
L.C.M of 5 and 3 is 15
=2×5−8×315
=−1415
(vi) 49+53+−45+79+−23+95
Sol :
49+53+−45+79+−23+95
=49+79+53+−23−45+95
=4+79+5−23+−4+95
=119+33+55
L.C.M of 9,3,5 is 45
=11×5+3×15+5×945
=55+45+4545
=14545 or299=329
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Question 8
Verify that -(-x)=x , when x= (i)76
(ii)−89
Sol :
(i) 76
⇒−(−76)=76
⇒76=76
(ii) −89
⇒−[−(−89)]=−89
⇒−[+89]=−89
⇒−89=−89
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Question 9
Verify that -(x+y)=(-x)+(-y) , when (i)x=34,y=67 (ii)x=−34,y=−67
Sol :
(i) x=34,y=87
−(34+67)=(−34)+(−67)
−(3×7+6×428)=−34−67
−(21+2428)=(−3×7−6×428)
−4528=−21−2428
−4528=−4528
(ii) x=−34,y=−67
-(x+y)=(-x)+(-y)
−[−34+−67]=[−(−34)]+[−(−67)]
−[−3×7−6×428]=34+67
−[−21−2428]=3×7+6×428
−[−4528]=21+2428
4528=4528
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Question 10
Subtract:
(i) 29 from 79
Sol :
=79−29
=7−29
=59
(ii) 89 from −56
Sol :
=−56−89
LCM of 6,9 is 18
212,926,933,931,31,1=−5×3−8×218
=−15−1618
=−3118
(iii) −811 from 322
Sol:
=322−−811
=322+811
222,111111,111,1
3×1+8×222
=3+1622
=1922
(iv) 3−4 from 45
Sol :
=45−(3−4×−1−1)
=45−−34
=45+34
LCM of 4,5 is 20
=4×4+3×520
=16+1520
=3120
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Question 11
The sum of two rational numbers is −1960 . If one of the numbers is −712 , find the other .
Sol :
Sum of two rational numbers is −1960
One of the number is −712 and Let the other be x
−712+x=−1960
x=−1960+712 (on transposing )
LCM of 60 and 12 is 60
=−19×1+7×560
=−19+3560=1660 or 415
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Question 12
What number should be subtracted from −1415 to get −130 ?
Sol :
let x be subtstaded from −1415 to get −130
−1415−x=−130
−1415+130=x
or
x=−1415+130
LCM of 15 and 30 is 30
x=−14×2+1×130
x=−28+130
=−2730
or−910
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Question 13
What should be subtracted from (34+13+25) to get 12 ?
Sol :
[34+13+25]−x=12
[34+13+25]−12=x
x=34+13+25−12
LCM of 4,3,5,2 is 60
x=3×15+1×20+2×12−1×3060
x=45+20+24−3060
=89−3060
=5960
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