Rs Aggarwal 2018 Solutions for Class 7 Math Chapter 7 Linear Equations In One Variable are provided here with simple step-by-step explanations. These solutions for Linear Equations In One Variable are extremely popular among Class 7 students for Math Linear Equations In One Variable Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 7 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 111:

Question 1:

3x − 5 = 0

Answer:

3x-5=0 3x=5                    (Transposing -5 to RHS) x=53CHECK: By substituting x=53 in the given equation, we get:LHS =353 - 5 = 5 - 5 = 0RHS = 0 LHS = RHS Hence checked.

Page No 111:

Question 2:

8x − 3 = 9 − 2x

Answer:

8x − 3 = 9 − 2x
8x + 2x = 9 + 3                   (By transposition)
10x = 12
x=1210=65CHECK: By substituting x=65 in the given equation, we get:LHS: 865-3=485-3=48-155=335RHS: 9-265=9-125=45-125=335 LHS= RHS Hence checked.

Page No 111:

Question 3:

7 − 5x = 5 − 7x

Answer:

We have:7  5x= 5  7x5x + 7x = 5  7 [transposing -7x to LHS  and 7 to RHS]2x = 2              x = 2121x = 1Thus, x 1 is a solution to the given equation.CHECK:  Substituting x 1 in the given equation, we get:LHS: = 7  5x          = 7 5 × (1)          = 7+5          = 12RHS: = 5  7x=5  7 × (1)= 5 + 7=12 LHS = RHSHence, x = 1  is a solution of the given equation.

Page No 111:

Question 4:

3 + 2x = 1 − x

Answer:

We have:3+2x = 1 – x⇒ 2x + x+ 3 – 1 = 0               (By transposition) 3x + 2 = 0 x = –23CHECK:  Substituting x=23 in the given equation, we get: LHS: 3+2x          =3+2×(23)          =3 43          =9-43          =53RHS: 1 x          =1-23          =1+23          =3+23          =53      LHS=RHSHence,  x=23 is a solution of the given equation

Page No 111:

Question 5:

2(x − 2) +3(4x − 1) = 0

Answer:

We have:2(x2)+3(4x1) =0 2x4+12x-3 =0 14x7 =0                14x = 7                      (By transposition)x=12CHECK:  Substituting x=12 in the given equation, we get:LHS: 2(x2)+3(4x1)          =2x4+12x-3          =2×124+12×12-3          =1-4+6-3          =7+7          =0RHS: 0∴ LHS= RHS Hence,  x=12 is a solution of the given equation.

Page No 111:

Question 6:

5(2x − 3) − 3(3x − 7) = 5

Answer:

We have:5(2x3)3(3x7) =5 10x159x+21 = 5 10x9x =5+1521                   (By transposition) x = 2021x=1CHECK:  Substituting x=1 in the given equation, we get: LHS: 5(2x3)3(3x7)          =10x159x+21          =10×(1)159×(1)+21          =1015+9+21          =25+30          =5 RHS: 5 LHS= RHS Hence, x=1 is a solution of the given equation.

Page No 111:

Question 7:

2x-13=15-x

Answer:

We have:2x13=15x2x+x=15+133x=3×1+5×1153x=3+5153x=815x=815×3x=845CHECK:  Substituting x=845 in the given equation, we get:   LHS: 2x13         =2×84513        =164513       =16×115×145       =161545      =145RHS: 15x         =15845         =1×91×845         =9845         =145 ∴ LHS=RHS Hence, x=845 is a solution of the given equation.   

Page No 111:

Question 8:

12x-3=5+13x

Answer:

We have:12x3=5+13x12x13x=5+3       (transposing 13x to LHS and 3 to RHS)1×31×26x=8326x=816x=8x=8 × 6x=48CHECK:  Substituting x=48 in the given equation, we get:LHS: 12x3          =121×48243        =243       =21RHS: 5+13x         =5+131×4816         =5+16         =21 ∴ LHS=RHS      Hence,  x=48 is a solution of the given equation.   

Page No 111:

Question 9:

x2+x4=18

Answer:

  x2+x4=18x×2+x×14=182x+x4=183x4=183x=182×413x=12x=16CHECK:  Substituting x=16 in the given equation, we get:  LHS: x2+x4         =x×2+x×14         =2x+x4         =3x4         =314×162         =18 RHS: 18 LHS=RHS Hence, x=13 is a solution of  the given equation.   

Page No 111:

Question 10:

3x + 2(x + 2) = 20 − (2x − 5)

Answer:

We have:3x+2(x+2)=20(2x5) 3x+2x+4 = 202x+5 3x+2x+2x=20+54        (Transposing 2x to LHS and 4 to RHS) 7x =21x=21371x=3CHECK:  Substituting x=3 in the given equation, we get:LHS=3x+2(x+2)          =3x+2x+4          =5x+4          =5×3+4          =15+4          =19 RHS=20(2x5)          =202x+5         =252×3         =256         =19 LHS=RHSHence, x=3 is a solution of the given equation

Page No 111:

Question 11:

13(y − 4) − 3(y − 9) − 5(y + 4) = 0

Answer:

We have:13(y4)3(y9)5(y+4)=0 13y523y+275y20 = 013y3y5y=52+2027        (Transposing 52,20 and 27 to RHS)5y =45y=45951y=9CHECK:  Substituting x=9 in the given equation, we get:LHS=13(y4)3(y9)5(y+4)          =13y523y+275y20          =13y3y5y52+2720          =5y45          =5×945          =4545          =0RHS=0∴ LHS=RHS       Hence, x=9 is a solution of the given equation

Page No 111:

Question 12:

2m+53=3m-10

Answer:

We have,2m+53=3m102m+5=3(3m10)2m+5=9m302m9m=305        (Transposing 9to LHS and 5 to RHS)7m=35m=-355-71m=5CHECK:  Substituting m= 5 in the given equation, we get:LHS=2m+53         =2×5+53        =10+53        =15531       =5 RHS=3m10         =3×510         =1510         =5 ∴ LHS=RHS       Hence, x=5 is a solution of the given equation.

Page No 111:

Question 13:

6(3 x + 2) − 5(6x − 1) = 3(x − 8) − 5(7x − 6) + 9x

Answer:

We have:6(3x+2)5(6x1)=3(x8)5(7x6)+9x⇒ 18x+1230x+5 =3x2435x+30+9x18x30x3x+35x9x=24+30125        (Transposing 3x,9x and -35x to LHS and 12 and 5 to RHS) 53x42x =304111x=11x=111111x=1CHECK: Substituting x=1 in the given equation, we get:LHS=6(3x+2)5(6x1)          =18x+1230x+5          =12x+17          =12×(1)+17          =12+17          =29     RHS=3(x8)5(7x6)+9x          =3x2435x+30+9x          =12x35x24+30          =23x+6          =23×(1)+6          = 23+6         =29    ∴ LHS=RHS       Hence, x=1 is a solution of the given equation

Page No 111:

Question 14:

t − (2t + 5) − 5(1 − 2t) = 2(3 + 4t) −3(t − 4)

Answer:

We have:t(2t+5)5(12t)=2(3+4t)3(t4)t2t55+10t =6+8t3t+12t 2t+10t8t+3t=6+12+5+5        (By transposition )14t10t =284t=28x=28741x=7CHECK:  Substituting x=7 in the given equation, we get:LHS=t(2t+5)5(12t)          =t2t55+10t          =11t2t10          =9t10          =9×710          =6310          =53 RHS=2(3+4t)3(t4)          =6+8t3t+12          =5t+18          =5×7+18          =35+18          = 53 ∴ LHS=RHS          Hence, x=7 is a solution of the given equation

Page No 111:

Question 15:

23x=38x+712

Answer:

We have:23x=38x+71223x38x=712        (Transposing 38x to LHS)2×83×324x=71216924x=712724x=712x=71121×24271x=2CHECK:  Substituting x=2 in the given equation, we get:LHS=23x          =23×2        =43RHS=38x+712         =38×2+712         =68+712         =6×3+7×224         =18+1424         =324243           =43  ∴ LHS=RHS      Hence, x=2 is a solution of the given equation.   

Page No 111:

Question 16:

3x-15-x7=3

Answer:

We have:3x15x7=37(3x1)5×x35=3        21x75x35=316x735=316x7=3×35                (Transposing 35 to RHS)16x7=10516x=105+716x=112x=1127161x=7CHECK:  Substituting x=7 in the given equation, we get: LHS=3x15x7         =7(3x1)5×x35        =21x75x35       =16x735      =16×7735      =112735     =1053351     =3  RHS=3 ∴ LHS=RHS   Hence, x=3 is a solution of the given equation.   

Page No 111:

Question 17:

2x-3=310(5x-12)

Answer:

We have:2x3=310(5x12)10(2x3)=3(5x12)        20x30=15x3620x15x=36+30       (Transposing 15x to LHS and 30 to RHS) 5x=6                x=65CHECK: Substituting x=65 in the given equation, we get:LHS=2x3         =2×(65)3        =1253       =12(3×5)5      =12155      =275 RHS=310(5x12)         =310(51×65112)         =310×(18)        =3105×189        =275∴ LHS=RHS    Hence, x=65 is a solution of the given equation.   

Page No 111:

Question 18:

y-13-y-24=1

Answer:

We have:y13y24=14(y1)3(y2)12=1        4y43y+612=1y+212=1y+2=1×12                y=122y=10CHECK:  Substituting y=10 in the given equation, we get:LHS=y13y24         =4(y1)3(y2)12        =y+212       =10+212      =121121      =1 RHS=1  ∴ LHS=RHS  Hence, y=10 is a solution of the given equation.   



Page No 112:

Question 19:

x-24+13=x-2x-13

Answer:

We have:x24+13=x2x13x24+2x13x=13          (Transposing -2x13 to LHS and 13to RHS)    3(x2)+4(2x1)12x12=133x6+8x412x12=1311x12x10=131×124                x=4+10x=6x=6CHECK:  Substituting x=6 in the given equation, we get:  LHS=x24+13         =624+13        =2+13       =53    RHS=x2x13        =62×(6)13       =6(13)3      =6+133      =53      ∴ LHS=RHS  Hence, y=10 is a solution of the given equation.   

Page No 112:

Question 20:

2x-13-6x-25=13

Answer:

We have:2x136x25=135(2x1)3(6x2)15=13          10x518x+615=138x+115=138x+1=13×15                8x=51x=48x=24=-12CHECK:  Substituting x=12 in the given equation, we get:LHS=2x136x25         =8x+115        =8×(12)+115       =515       =13RHS=13 ∴ LHS=RHS Hence, y=12 is a solution of the given equation.    

Page No 112:

Question 21:

y+73=1+3y-25

Answer:

We have:y+73=1+3y25y+73=5×1+3y25              5(y+7)=3(3+3y)5y+35=9+9y9y5y=359                4y=26y=132CHECK:  Substituting x=132 in the given equation, we get: LHS=y+73 =132+73=1×13+2×72=13+146=276=92                 RHS=1+3×13225=1+392×225=1+3510=4510       =92        ∴ LHS=RHS     Hence, y=132 is a solution of the given equation.   

Page No 112:

Question 22:

27(x-9)+x3=3

Answer:

We have:27(x9)+x3=32×3(x9)+7x21=3        6(x9)+7x=3×216x54+7x=6313x=63+54                13x=117x=9CHECK:  Substituting x=9 in the given equation we get.LHS=27(x9)+x3 =27(99)+x3=0+93=93=3 RHS=3 LHS=RHSHence, x=9 is a solution of the given equation.   

Page No 112:

Question 23:

2x-35+x+34=4x+17

Answer:

We have:2x35+x+34=4x+174(2x3)+5(x+3)20=4x+17        8x12+5x+1520=4x+1713x+320=4x+177(13x+3)=20(4x+1)                91x+21=80x+2091x80x=202111x=1x=111CHECK:  Substituting x=111 in the given equation, we get:LHS:LHS=2x35+x+34 =2×11135+111+34=23355+33144=3555+3244=-140+160220  =20220  =111      RHS=4x+17=4×(111)+17=4+117×11=777=111 LHS=RHSHence, x=111 is a solution of the given equation.   

Page No 112:

Question 24:

34(7x-1)-2x-1-x2=x+32

Answer:

We have:34(7x1)2x1x2=x+3234(7x1)2x+1x2x=32        3×74x342x+12x2x=32214x2xx2x=32+3412           (By transposition)21x8x2×x4x4=1+34                21x14x4=747x4=74x=1CHECK:  Substituting x=1 in the given equation, we get:LHS=34(7x1)2x1x2 =34(7×11)2×1112=34×62=922=942=52RHS=x+32=1+32=2+32=52 LHS=RHSHence, x=1 is a solution of the given equation.   

Page No 112:

Question 25:

x+26-11-x3-14=3x-412

Answer:

We have:x+2611x314=3x412x+2611x3+14=3x412x+2611x33x412=14               (By transposition)2(x+2)4(11x)1(3x4)12=142x+444+4x3x+412=143x36=14×12                3x=3+36x=333x=11CHECK: Substituting x= 11 in the given equation, we get:LHS=x+26(11x314) =11+26(1111314)=136(14)=136+14=13×2+312=2912RHS=3x412=3×11412=33412=292∴ LHS=RHS Hence, x = 11 is a solution of the given equation.  Verified.

Page No 112:

Question 26:

9x+72-x-x-27=36

Answer:

We have:9x+72(xx27)=369x+72x+x27=367(9x+7)14×x+2×(x2)14=36       63x+4914x+2x414=3651x+45=36×1451x=50445                x=45951x=9x=9CHECK: Substituting x= 9 in the given equation, we get:LHS=9x+72xx27 =9×9+729927=8829+77=449+1=36RHS =36∴ LHS=RHS Hence, x = 11 is a solution of the given equation.  Verified.

Page No 112:

Question 27:

0.5x+x3=0.25x+7

Answer:

We have:0.5x+x3=0.25x+712x+x3=x4+7x2+x3x4=7       6x+4x3x12=77x12=7x=12             CHECK:  Substituting x= 9 in the given equation, we get:LHS=0.5x+x3 =0.5×12+123=12×12+4=6+4=10RHS=0.25x+7=0.25×12+7=3+7=10∴ LHS=RHS Hence, x = 12 is a solution of the given equation.  Verified.

Page No 112:

Question 28:

0.18(5x − 4) = 0.5x + 0.8

Answer:

We have:0.18(5x4)=0.5x+0.8100×0.18(5x4)=100(0.5x+0.8)    (Multipling both sides by 100)18(5x4)=100×0.5x+100×0.8       90x72=50x+8090x50x=80+7240x=152x=15240x=195=3.8             CHECK:  Substituting x3.8 in the given equation, we get:LHS=0.18(5x4) =0.18(5×3.84)=0.18×15=2.7RHS=0.5x+0.8=0.5×3.8+0.8=1.9+0.8=2.7∴ LHS=RHS Hence, x = 3.8 is a solution of the given equation.  Verified.

Page No 112:

Question 29:

2.4(3 − x) − 0.6(2x − 3) = 0

Answer:

We have:2.4(3x)0.6(2x3)=010×2.4(3x)10×0.6(2x3)=0             (Multiplying both sides by 10 to remove decimals)24(3x)6(2x3)=0       6[4(3x)(2x3)]=04(3x)(2x3)=0124x2x+3=0156x=06x=15x=156x=52=2.5             CHECK:  Substituting x2.5 in the given equation, we get:LHS=2.4(3x)0.6(2x3) =2.4(32.5)0.6(2×2.53)=2.4×0.50.6×2=1.2-1.2=0RHS=0∴ LHS = RHS Hence, x = 195 is a solution of the given equation.  Verified.

Page No 112:

Question 30:

0.5x −(0.8 − 0.2x) = 0.2 − 0.3x

Answer:

We have:0.5x(0.80.2x)=0.20.3x0.5x+0.3x0.8+0.2x=0.2                     (By transposition)(0.5+0.3+0.2)x=0.2+0.8       1x=1x=1CHECK:  Substituting x= 1 in the given equation, we get:  LHS=0.5x(0.80.2x) =0.5×1(0.80.2×1)=0.50.8+0.2=0.1 RHS=0.20.3x=0.20.3×1=0.1∴ LHS=RHS Hence, x = 1 is a solution of the given equation.  Verified.

Page No 112:

Question 31:

x+2x-2=73

Answer:

We have:x+2x2=73(x+2)×3=7×(x2)       (Cross multiplication)  3x+6=7x14       4x=20x=204x=5CHECK:  Substituting x= 5 in the given equation, we get.  LHS=x+2x2 =5+252=73 RHS=73∴ LHS=RHS Hence, x = 5 is a solution of the given equation.  Verified.

Page No 112:

Question 32:

2x+53x+4=3

Answer:

We have:2x+53x+4=32x+53x+4=311×(2x+5)=3×(3x+4)       2x+5=9x+127x=7x=1CHECK:  Substituting x=1  in the given equation, we get:LHS: 2x+53x+4 =2×(1)+53×(1)+4=2+53+4=31RHS =3∴ LHS = RHS Hence, x = 5 is a solution of the given equation.  Verified.



Page No 114:

Question 1:

Twice a number when decreased by 7 gives 45. Find the number.

Answer:

 Let the number be x.            Then, we have:2x7=452x=45+7x=45+72x=522621x=26The required number is 26.

Page No 114:

Question 2:

Thrice a number when increased by 5 gives 44. Find the number.

Answer:

Let the number be x. Then, we have:3x+5=443x=445x=4453x=391331x=13The required number is 13

Page No 114:

Question 3:

Four added to twice a number yields 265. Find the fractions.

Answer:

Let the number be x.Then, we have:2x+4=2652x=26542x=26205x=63105x=35 The required fraction is 35.

Page No 114:

Question 4:

A number when added to its half gives 72. Find the number.

Answer:

 Let the required number be x.Then, we have:x+x2=722x+x2=723x2=723x=72×2x=7224×231=48 The required number is 48.

Page No 114:

Question 5:

A number added to its two-thirds is equal to 55. Find the number.

Answer:

 Let the required number be x.  Then, we have:x+2x3=553x+2x3=555x=55×3x=5511×351=33The required number is 33.

Page No 114:

Question 6:

A number when multiplied by 4, exceeds itself by 45. Find the number.

Answer:

 Let the required number be x. Then, we have:4xx=453x=453x=15The required number is 15.

Page No 114:

Question 7:

A number is as much greater than 21 as it is less than 71. Find the number.

Answer:

 Let the number be x.                          Then, we have:(x21)=(71x)x+x=71+212x=92x=924621x=46 The required number is 46.



Page No 115:

Question 8:

23 of a number is less than the original number by 20. Find the number.

Answer:

 Let the original number be x.                                 Then, we have:23x=x202x3x=202x3x3=20x=20×3x=60The original number is 60.

Page No 115:

Question 9:

A number is 25 times another number. If their sum is 70, find the numbers.

Answer:

 Let the number be x.                                   Then, the other number will be 2x5.Now, we have:x+2x5=705x+2x5=707x5=70x=7010×571 Other number=50×25=20Hence, the numbers are 50 and 20.

Page No 115:

Question 10:

Two-thirds of a number is greater than one-third of the number by 3. Find the number.

Answer:

 Let the number be x.                          Then, we have:23x=13x+313x=2x33x32x3=3x2x3=3x2x=3×(3)x=9 The required number is 9.

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Question 11:

The fifth part of a number when increased by 5 equals its fourth part decreased by 5. Find the number.

Answer:

 Let the number be x.           Then, we have:x5+5=x4-5x5-x4=-5-5-x20=-10x=200The required number is 200.

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Question 12:

Find two consecutive natural numbers whose sum is 63.

Answer:

 Let the two consecutive natural number be x and (x+1).Then, we have: x+(x+1)=63x+x+1=632x=631x=623121x=31The required numbers are 31 and 32 (i.e., 31+1).

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Question 13:

Find two consecutive positive odd integers whose sum is 76.

Answer:

 Let the two consecutive odd integers whose sum is 76 be x and (x+2).Then, x+x+2=762x +2=762x=762x=74÷x=37The required integers are 37 and 39 (i.e., 37+2).

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Question 14:

Find two consecutive positive even integers whose sum is 90.

Answer:

Let the three consecutive positive even integers be x, (x+2) and (x+4).Let x be the even number.Then, x+x+2+x+4=903x=9063x=84x=843=28The required numbers are 28, 30 and 32.

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Question 15:

Divide 184 into two parts such that one-third of one part may exceed one-seventh of the other part by 8.

Answer:

Let the two parts be x and (184x).Then, we have:13x=17(184x)+813x17(184x)=813x 1847+x7=8 ⇒13x+17x=1847+87x+3x21=8+184710x21=56+184710x21=2407x=240×217×10 =72Now, other part =18472=112 The two parts are 72 and 112.

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Question 16:

A sum of 500 is in the form of denominations of 5 and 10. If the total number of notes is 90, find the number of notes of each type.

Answer:

Let the number of five rupee notes be x.Then, the number of ten rupee notes will be (90x).According to the question, we have:  5x+10(90x)=500⇒5x+90010x=500 5x=400x=80Number of ten rupee notes=90-80=10 There are 80 five rupee notes and 10 ten rupee notes.

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Question 17:

Sumitra has 34 in 50-paise and 25-paise coins. If the number of 25-paise coins is twice the number of 50-paise coins, how many coins of each kind does she have?

Answer:

Let the numbers of 50 paise coins and 25 paise coins be x and 2x, respectively.Then, we have: 50x+25×2x=340050x+50x=3400100x=3400x=34Number of 50 paise coins =34and number of 25 paise coins =68

Page No 115:

Question 18:

Raju is 19 years younger than his cousin. After 5 years, their ages will be in the ratio 2 : 3. Find their present ages.

Answer:

Let the present ages of Raju and his cousin be (x-19) yrs and x yrs. According to the question, we have:  (x19)+5x+5=23⇒3(x14)=2x+103x42=2x+10x=52 Age of Raju's cousin = 52 yrs  and age of Raju = 5219=33 yrs

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Question 19:

A father is 30 years older than his son. In 12 years, the man will be three times as old as his son. Find their present ages.

Answer:

Let the age of the son and the father be x yrs and (x+30) yrs, respectively.According to the question, we have:3×(x+12)=x+30+12⇒3x+36=x+423xx=42362x=6x=3Son's age=3 yrsFather's age=(x+30) yrs=(3+30) yrs=33 yrs

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Question 20:

The ages of Sonal and Manoj are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.

Answer:

Given ratio of Sonal's and Manoj's ages =: 5Let the ages of Sonal and Manoj be 7x yrs and 5x yrs.According to the question, we have:7x+105x+10=977(7x+10)=9(5x+10)49x+70=45x+9049x45x=90704x=20x=5Sonal's present age is 7×5=35 yrs Manoj's present age is 5×5=25 yrs

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Question 21:

Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Answer:

Let yrs be the present age of son.Then, the age of the son 5 years ago would be (x5) yrs Then, Age of father= 7(x-5) yrsAfter 5 yrs, the age of the son will be (x+5) yrsThen, Age of father= 3(x+5) yrsNow, we have 3(x+5)=7(x5)+10⇒ 3x+15=7x35+104x=40x=10Present age of the father is = 3(x+5)-5=3(10+5)5=40 yrs

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Question 22:

After 12 years Manoj will be 3 times as old as he was 4 years ago. Find his present age.

Answer:

Let be the present age of Manoj.According to the question, we have:x+12=3(x4)x+12=3x122x=24x=12Manoj's present age is 12 years.

Page No 115:

Question 23:

In an examination, a student requires 40% of the total marks to pass. If Rupa gets 185 marks and fails by 15 marks, find the total marks.

Answer:

Let x be the total marks.According to the question, we have:40% of x=185+1540x100=20040x=200×10040x=20000x=500Total marks=500  

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Question 24:

A number consists of two digits whose sum is 8. If 18 is added to the number its digits are reversed. Find the number.

Answer:

Let x be the digit in the units place.Sum of the units and tens digits=8Then, tens digit=(8x)The number is 10(8x)+x.Now, 10(8x)+x+18=10x+(8x)8010x+x+18=10x+8x989x=9x+818x=90x=5i.e., tens digit=(85)=3Required number=10(85)+5=10×3+5=35  

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Question 25:

The total cost of 3 tables and 2 chairs is 1850. If a table costs 75 more than a chair, find the price of each.

Answer:

Let Rs be the cost of the chair.Then, the cost of the table is Rs (x+75).Now, 3(x+75)+2x=18503x+225+2x=18505x=1625x=16255=325Cost of the chair = Rs 325; cost of the table = (325+75)=Rs 400  

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Question 26:

A man sold an article for 495 and gained 10% on it. Find the cost price of the article.

Answer:

Let the cost price of the article be Rs x.According to the question, we have:  SP=Rs 495 Gain %=GainCP×10010=Gainx×100Gain=10x100=Rs x10Now, CP+Gain=SPx+x10=495x+10x10=49511x=495×10x=495×1011x=495011x=450CP=Rs 450

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Question 27:

The length of a rectangular field is twice its breadth. If the perimeter of the field is 150 metres, find its length and breadth.

Answer:

Let the length and breadth of the rectangular field be m and b m, respectively.According to the question, we have:2(l+b)=150       ...(i)l+b=75Given that l=2b      ...(ii)Using (ii) in (i), we have:2b+b=753b=75b=25 l=50 m and b=25 m 

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Question 28:

Two equal sides of a triangle are each 5 metres less than twice the third side. If the perimeter of the triangle is 55 metres, find the lengths of its sides.

Answer:

Let the length of  third side be x m. Then, the length of the two equal sides will be (2x5) m.(2x5)+(2x5)+x=552x5+2x5+x=555x10=555x=65x=655=13Length of the third side=13 mAnd length of the other two equal sides=(2×13)5=21 m    

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Question 29:

Two complementary angles differ by 80. Find the angles.

Answer:

Let the two complementary angles be x° and (90x)°.According to the question, we have:x(90x)=8 x90+x=82x=98x=49The measures of the complementary angles are 49° and (9049)°=41°.

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Question 30:

Two supplementary angles differ by 440. Find the angles.

Answer:

Let the two supplementary angles be x° and (180x)°. x(180x)=44x180+x=4402x=224x=112The measures of the supplementary angles are 112° and (180112)°, i.e., 68°.

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Question 31:

In an isosceles triangle the base angles are equal and the vertex angle is twice of each base angle. Find the measures of the angles of the triangle.

Answer:

Let the base angles of the isosceles triangle be x° each. Then, the measure the vertex angle will be (2x)°.According to the question, we have:x+x+2x=180     (Sum of three sides of a triangle) 4x=180x=1804x=45∴ Each base angle measures 45° and the vertex angle measures (2×45)°, i.e., 90°.           

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Question 32:

A man travelled 35 of his journey by rail, 14 by a taxi, 18 by a bus and the remaining 2 km on foot. What is the length of his total journey?

Answer:

Let the length of the total journey be x km.According to the question, we have:35x+14x+18x+2=x24x+10x+5x+8040=x39x+80=40xx=80The length of his total journey is 80 km.



Page No 116:

Question 33:

A labourer is engaged for 20 days on the condition that he will receive 120 for each day he works and will be fined 10 for each day he is absent. If he receives 1880 in all, for how many days did he remain absent?

Answer:

Let x be the number of days of his absence. Number of days of his presence = (20x)Now, (20x)12010x=18802400 120x10x=188024001880=130x130x=520x=4 Number of days of his absence=

Page No 116:

Question 34:

Hari Babu left one-third of his property to his son, one-fourth to his daughter and the remainder to his wife. If his wife's share is 18000, what was the worth of his total property?

Answer:

Let the worth of Hari Babu's property be Rs x.According to the question, we have:Son's share=14xDaughter's share=13xWife's share={x(14x+13x)}It is given that his wife's share is Rs 18000.i.e., x(14x+13x)=18000x(13x+14x)=18000x7x12=180005x12=18000 x=  180003600×125     x=43200 Hari Babus total property is worth Rs 43200.        

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Question 35:

How much pure alcohol must be added to 400 mL of a 15% solution to make its strength 32%.

Answer:

Let the volume of the pure alcohol be x ml.Initial concentration=15%So, initial amount of alcohol in the solution will be=15100×400= 60 mlTo make the strength of the solution 32%, we will keep the amount of water constant and add x volume of pure alcohol.On adding pure alcohol, the volume of the solution increases to 400 + x.According to the question, we have:x+60400+x=32100⇒100x+6000=12800+32x100x32x=12800600068x=6800x=100So, amount of pure alcohol to be added=100 ml                        

Page No 116:

Question 1:

Mark (✓) against the correct answer

If 5x-34=2x-23, then x = ?

(a) 112
(b) 14
(c) 36
(d) 136

Answer:

 (d) 136We have:5x34=2x235x2x=23+343x=8+912x=112×3x=136

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Question 2:

Mark (✓) against the correct answer

If 2z+83=14z+5, then z = ?

(a) 3
(b) 4
(c) 34
(d) 43

Answer:

 (d)43We have:2z+83=14z+52z14z=5838zz4=15837z4=73z=71×43×71z=43

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Question 3:

Mark (✓) against the correct answer

If (2n + 5) = 3(3n − 10), then n = ?

(a) 5
(b) 3
(c) 25
(d) 23

Answer:

(a) 5We have:(2n+5)=3(3n10)2n+5=9n302n9n=3057n=35n=35571n=5

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Question 4:

Mark (✓) against the correct answer

If x-1x+1=79, then x = ?

(a) 6
(b) 7
(c) 8
(d) 10

Answer:

(c) 8We have:x1x+1=799(x1)=7(x+1) 9x9=7x+79x7x=7+92x=16x=16821x=8

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Question 5:

Mark (✓) against the correct answer

If 8(2x − 5) − 6(3x − 7) = 1, then x = ?

(a) 2
(b) 3
(c) 12
(d) 13

Answer:

(c) 12We have:8(2x5)6(3x7)=116x4018x+42=12x+2=12x=12x=12x=12

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Question 6:

Mark (✓) against the correct answer

If x2-1=x3+4, then x = ?

(a) 8
(b) 16
(c) 24
(d) 30

Answer:

(d) 30We have:x21=x3+4x22=x+1233(x2)=2(x+12)3x6=2x+243x2x=24+6x=30

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Question 7:

Mark (✓) against the correct answer

If 2x-13=x-23+1, then x = ?

(a) 2
(b) 4
(c) 6
(d) 8

Answer:

(a) 2We have:2x13=x23+12x13=(x2)+333(2x1)=3(x+1)6x3=3x+36x3x=3+33x=6x=6231=2

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Question 8:

The sum of two consecutive whole numbers is 53. The smaller number is

(a) 25
(b) 26
(c) 29
(d) 23

Answer:

 (b) 26Let the consecutive whole numbers be x and (x+1).Then, x+(x+1)=532x+1=532x=531x=522621x=26



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Question 9:

The sum of two consecutive even numbers is 86. The larger of the two is

(a) 46
(b) 36
(c) 38
(d) 44

Answer:

 (d) 44Let the two consecutive even numbers be x and (x+2).Then, x+(x+2)=862x+2=862x=862x=844221x=42       The required numbers are 42 and (42+2), i.e., 44.

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Question 10:

The sum of two consecutive odd numbers is 36. The smaller one is

(a) 15
(b) 17
(c) 19
(d) 13

Answer:

 (b) 17Let the two consecutive odd numbers be (x+1) and (x+3). Then, (x+1)+(x+3)=362x+4=362x=364x=321621x=16The smaller number is 17.

Page No 117:

Question 11:

on adding 9 to the twice of a whole number gives 31. The whole number is

(a) 21
(b) 16
(c) 17
(d) 11

Answer:

(d)11Let the whole number be x.Then, 2x+9=312x=3192x=22x=221121x=11

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Question 12:

Thrice a number when increased by 6 gives 24. The number is

(a) 6
(b) 7
(c) 8
(d) 11

Answer:

(a) 6Let the whole number be x.Then,  3x+6=243x=2463x=18x=18631x=6

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Question 13:

23 of a number is less than the original number by 10. The original number is

(a) 30
(b) 36
(c) 45
(d) 60

Answer:

 (a) 30Let the original number be x.    Then, 23x=x102x=3x302x3x=30x=30x=30The required number is 30.

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Question 14:

Two complementary angles differ by 10°. The larger angle is

(a) 60°
(b) 50°
(c) 64°
(d) 54°

Answer:

(b) 50°Let the angle be x°.Then, complementary of x=90°-x°According to the question, we have:x-90-x=102x=90+102x=100x=50So, the larger angle is 50°.

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Question 15:

Two supplementary angles differ by 20°. The smaller of the two measures

(a) 60°
(b) 80°
(c) 100°
(d) 120°

Answer:

 (b) 800Let the angle be x°.Then, complementary angle of x=180°-x°According to the question, we have: x-180-x=20x-180+x=202x=10+1802x=200x=100Hence, the smaller angle is 80°.

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Question 16:

The ages of A and B are in the ratio 5 : 3. After 6 years, their ages will be in the ratio 7 : 5. The present age of A is

(a) 5 years
(b) 10 years
(c) 15 years
(d) 20 years

Answer:

 (c)15 yearsLet the present ages of A and B be 5x and 3x, respectively.             According to the question, we have:5x+63x+6=7525x+30=21x+4225x21x=42304x=12x=1234x=3 As present age=5×3 years=15 years    

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Question 17:

A number when multiplied by 5 is increased by 80. The number is

(a) 15
(b) 20
(c) 25
(d) 30

Answer:

 (b) 20Let the number be x.Then, 5x=x+805xx=804x=80x=802041x=20The required number is 20.

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Question 18:

The length of a rectangle is three times its width and its perimeter is 96 m. The length is

(a) 12 m
(b) 24 m
(c) 36 m
(d) 48 m

Answer:

(c) 32 mLet the width of the rectangle be x. Then, its length will be 3x.           Perimeter of the rectangle=96 m                                         Now, 2(l+b)=96     2(3x+x)=962×4x=968x=96x=961281x=12 Length of the rectangle = 3×12 m=36 m                                                   



Page No 118:

Question 1:

Evaluate x3 + y3 + z3 −3xyz when x = −2, y = −1 and z = 3.

Answer:

 We have:x3+y3+z33xyz=(-2)3+(-1)3+(3)33×(2)×(1)×3=81+2718=27+27=0

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Question 2:

Write the coefficient of x in each of the following:

(i) −5xy
(ii) 2xy2z
(iii) -32abc

Answer:

Coefficient of  in the  given numbers are(i) 5y                    (ii) 2y2z                   (iii) 32ab

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Question 3:

Subtract x2 − 2xy + 5y2 − 4 from 4xy − 5x2y2 + 6.

Answer:

 We have:(4xy5x2y2+6)(x22xy+5y24)=4xy5x2y2+6x2+2xy5y2+4=6x26y2+6xy+10=2(3x2+3y23xy5)

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Question 4:

How much less is x2 − 2xy + 3y2 than 2x2 − 3y2 + xy?

Answer:

 We have:(2x23y2+xy)(x22xy+3y2)=2x23y2+xyx2+2xy-3y2)=2x2x23y23y2+xy+2xy=x26y2+3xy x22xy+3y2 is less than 2x23y2+x by x26y2+3xy.

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Question 5:

Find the product 35abc3×-2512a3b2×(-8b3c).

Answer:

 We have: 35abc3×(25)12a3b2×(8b3c)=3151abc3×(255)1241a3b2×(82b3c)=abc3×(5a3b2)×(2b3c)=10a4b6c4

Page No 118:

Question 6:

Simplify:

(3a + 4)(2a − 3) + (5a − 4)(a + 2)

Answer:

 We have:(3a+4)(2a3)+(5a4)(a+2)={3a(2a3)+4(2a3)}+{5a(a+2)4(a+2)}=(6a29a+8a12)+(5a2+10a4a8)=(6a2a12)+(5a2+6a8)=(11a2+5a20)

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Question 7:

Solve: 3x10+2x5=7x25+2925.

Answer:

 We have:3x10+2x5=7x25+29253x+4x10=7x+29253x+4x10=7x+29257x10=7x+2925175x=70x+290105x=290x=2905810521x=5821

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Question 8:

Solve: 0.5x+x3=0.25x+7.

Answer:

We have:0.5x+x3=0.25x+71.5x+x3=0.25x+71.5x+x=3(0.25x+7)2.5x=0.75x+212.5x0.75x=211.75x=21x=211.75x=12

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Question 9:

The sum of two consecutive odd numbers is 68. Find the numbers.

Answer:

 Let the consecutive odd numbers be x and (x+2).x+(x+2)=682x+2=682x=682x=663321x=33The required numbers are 33 and (33+2), i.e., 35.

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Question 10:

Reenu's father is thrice as old as Reenu. After 12 years he will be just twice his daughter. Find their present ages.

Answer:

Let Reenu's present age be x.Then, her father's present age will be 3x.Reenu's age after 12 years=(x+12)Her father's age after 12 years=(3x+12)Now, (3x+12)=2(x+12)3x+12=2x+24x=12Reenu's present age =12 yrsAnd her father's age=(3×12)=36 yrs

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Question 11:

Mark (✓) against the correct answer
If 2x+53=14x+4, then x = ?
(a) 3
(b) 4
(c) 34
(d) 43

Answer:

(d) 432x+53=14x+42x14x=4538x1x4=12537x4=7321x=28x=284213=43

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Question 12:

Mark (✓) against the correct answer
If x2-x3=5 then x = ?
(a) 8
(b) 16
(c) 24
(d) 30

Answer:

(d) 30x2x3=53x2x6=5x=30

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Question 13:

Mark (✓) against the correct answer
If x-23=2x-13-1, then x = ?
(a) 2
(b) 4
(c) 6
(d) 8

Answer:

(a) 2x23 =2x131x23=2x133x2=2x4x-2x=-4+2-x=-2x=2

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Question 14:

Mark (✓) against the correct answer
A number when multiplied by 4 is increased by 54. The number is

(a) 21
(b) 16
(c) 18
(d) 19

Answer:

(c) 18Let the number be x. According to the question, we have:4x=x+543x=54x=18

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Question 15:

Two complementary angles differ by 14°. The larger angle is

(a) 50°
(b) 52°
(c) 54°
(d) 56°

Answer:

(b) 52°Let the two complementary angles be x° and (90x)°.According to the question, we have: x(90x)=14⇒2x=104x=52(90x)°=90°52°=38°The larger angle is 52°.

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Question 16:

The length of a rectangle is twice its breadth and its perimeter is 96 m. The length of the rectangle is

(a) 28 m
(b) 30 m
(c) 32 m
(d) 36 m

Answer:

(c) 32 mLet the length and breadth of the rectangle be l m and b m, respectively. According to the question, we have:    l=2b            ...(i)2(l+b)=96      ...(ii)Now, 2(2b+b)=966b=96b=16Length=16×2 m=32 m            

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Question 17:

The ages of A and B are in the ratio 4 : 3. After 6 years, their ages will be in the ratio 11 : 9. A's present age is

(a) 12 years
(b) 16 years
(c) 20 years
(d) 24 years

Answer:

(b) 12 yearsLet the ages of A and B be x and y years, respectively.Now, xy=433x=4yx=43yAfter 6 years, we have: x + 6 y+6=11943y+6y+6=1194y+183(y+6)=11936y+162=33y+1983y=36y=12 x=43×124=16Hence, A's present age is 16 years.

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Question 18:

Fill in the blanks

(i) −2a2b is a ...... .
(ii) (a2 − 2b2) is a ...... .
(iii) (a + 2b − 3c) is a ...... .
(iv) In −5ab, the coefficient of a is ...... .
(v) In x2 + 2x − 5, the ...... term is −5.

Answer:

(i) −2a2b is a monomial.
(ii) (a2 − 2b2) is a binomial.
(iii) (a + 2b − 3c) is a trinomial.
(iv) In −5ab, the coefficient of a is -5b.
(v) In x2 + 2x − 5, the constant term is −5.

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Question 19:

Write 'T' for true and 'F' for false

(i) In −x, the constant term is −1.
(ii) The coeffiecient of x in x2 − 3x + 5 is 3.
(iii) (5x − 7) − (3x − 5) = 2x − 12.
(iv) (3x + 5y)(3x − 5y) = (9x2 − 25y2).
(v) If a = 2 and b = 12, then the value of ab (a2 + b2) is 414.

Answer:

(i) FThe coefficient of x is -1.(ii) FThe coefficient of x is -3.(iii) FLHS=(5x-7)-(3x-5)=5x-7-3x+5=2x-2(iv) TLHS=(3x+5y)(3x-5y)=3x(3x-5y)+5y(3x-5y)=9x2-15xy+15xy-25y2=9x2-25y2(v) T(a2+b2)=22+122=4+14=414



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