NCERT Solutions for Chapter 4 Simple Equations Class 7 Maths
Book Solutions1
Complete the last column of the table:
|
S. No. |
Equation |
Value |
Say, whether the Equation is satisfied. (Yes / No) |
|
1 |
x+3=0 |
x=3 |
Β |
|
2 |
x+3=0 |
x=0 |
Β |
|
3 |
x+3=0 |
x=β3 |
Β |
|
4 |
xβ7=1 |
x=7 |
Β |
|
5 |
xβ7=1 |
x=8 |
Β |
|
6 |
5x=25 |
x=0 |
Β |
|
7 |
5x=25 |
x=5 |
Β |
|
8 |
5x=25 |
x=β5 |
Β |
|
9 |
m/3=2 |
m=β6 |
Β |
|
10 |
m/3=2 |
m=0 |
Β |
|
11 |
m/3=2 |
m=6 |
Β |
Answer
|
S. No. |
Equation |
Value |
Say, whether the Equation is satisfied. (Yes / No) |
|
1 |
x+3=0 |
x=3 |
No |
|
2 |
x+3=0 |
x=0 |
No |
|
3 |
x+3=0 |
x=β3 |
Yes |
|
4 |
xβ7=1 |
x=7 |
No |
|
5 |
xβ7=1 |
x=8 |
Yes |
|
6 |
5x=25 |
x=0 |
No |
|
7 |
5x=25 |
x=5 |
Yes |
|
8 |
5x=25 |
x=β5 |
No |
|
9 |
m/3=2 |
m=β6 |
No |
|
10 |
m/3=2 |
m=0 |
No |
|
11 |
m/3=2 |
m=6 |
Yes |
Exercise 4.1
Page Number 81
2
Check whether the value given in the brackets is a solution to the given equation or not:
(a) n+5=19(n=1)
(b) 7n+5=19(n=β2)
(c) 7n+5=19(n=2)
(d) 4pβ3=13(p=1)
(e) 4pβ3=13(p=β4)
(f) 4pβ3=13(p=0)
Answer
(a) n+5=19(n=1)
Putting n=1 in L.H.S.,
1 + 5 = 6
β΅ L.H.S. β R.H.S.,
β΄ n=1 is not the solution of given equation.
(b) 7n+5=19(n=β2)
Putting n=β2 in L.H.S.,
7(β2)+5=β14+5=β9
β΅ L.H.S. β R.H.S.,
β΄ n=β2 is not the solution of given equation.
(c) 7n+5=19(n=2)
Putting n=2 in L.H.S.,
7(2)+5=14+5=19
β΅ L.H.S. = R.H.S.,
β΄ n=2is the solution of given equation.
(d) 4pβ3=13(p=1)
Putting p=1 in L.H.S.,
4(1)β3=4β3=1
β΅ L.H.S. β R.H.S.,
β΄ p=1 is not the solution of given equation.
(e) 4pβ3=13(p=β4)
Putting p=β4 in L.H.S.,
4(β4)β3=β16β3=β19
β΅ L.H.S. β R.H.S.,
β΄ p=β4 is not the solution of given equation.
(f) 4pβ3=13(p=0)
Putting p=0 in L.H.S.,
4(0)β3=0β3=β3
β΅ L.H.S. β R.H.S.,
β΄ p=0 is not the solution of given equation.
Exercise 4.1
Page Number 81
3
Solve the following equations by trial and error method:
(i) 5p+2=17
(ii) 3mβ14=4
Answer
(i) 5p+2=17
Putting p=β3 in L.H.S. 5(β3)+2 = β15+2=β13
β΅β13β 17 Therefore, p=β3 is not the solution.
Putting p=β2 in L.H.S. 5(β2)+2=β10+2=β8
β΅β8β 17 Therefore, p=β2 is not the solution.
Putting p=β1 in L.H.S. 5(β1)+2=β5+2=β3
β΅β3β 17 Therefore, p=β1 is not the solution.
Putting p=0 in L.H.S. 5(0)+2=0+2=2
β΅ 2β 17 Therefore, p=0 is not the solution.
Putting p=1 in L.H.S. 5(1)+2=5+2=7
β΅7β 17 Therefore, p=1 is not the solution.
Putting p=2 in L.H.S. 5(2)+2=10+2=12
β΅12β 17 Therefore, p=2 is not the solution.
Putting p=3 in L.H.S. 5(3)+2=15+2=17
β΅17=17Therefore, p=3 is the solution.
(ii) 3mβ14=4
Putting m=β2 in L.H.S. 3(β2)β14=β6β14=β20
β΅β20β 4 Therefore, m=β2 is not the solution.
Putting m=β1 in L.H.S. 3(β1)β14=β3β14=β17
β΅β17β 4 Therefore, m=β1 is not the solution.
Putting m=0 in L.H.S. 3(0)β14=0β14=β14
β΅β14β 4 Therefore, m=0 is not the solution.
Putting m=1 in L.H.S. 3(1)β14=3β14=β11
β΅β11β 4 Therefore, m=1 is not the solution.
Putting m=2 in L.H.S. 3(2)β14=6β14=β8
β΅β8β 4 Therefore, m=2is not the solution.
Putting m=3 in L.H.S. 3(3)β14=9β14=β5
β΅β5β 4 Therefore, m=3 is not the solution.
Putting m=4 in L.H.S. 3(4)β14=12β14=β2
β΅β2β 4 Therefore, m=4 is not the solution.
Putting m=5 in L.H.S. 3(5)β14=15β14=1
β΅1β 4 Therefore, m=5 is not the solution.
Putting m=6 in L.H.S. 3(6)β14=18β14=4
β΅4=4 Therefore, m=6m=6 is the solution.
Exercise 4.1
Page Number 81
4
Β Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times aa is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of tt is 15.
(vi) Seven times mm plus 7 gets you 7.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
Answer
(i) x+4=9
(ii) yβ2=8
(iii) 10a=70
(iv) b/5=6
(v) 3/4.t=15
(vi) 7m+7=77
(vii) x/4β4=4
(viii) 6yβ6=60
(ix) z/3+3=30
Exercise 4.1
Page Number 81
5
Write the following equations in statement form:
(i) p+4=15
(ii) mβ7=3
(iii) 2m=7
(iv) m/5=3
(v) 3m/5=6
(vi) 3p+4=25
(vii) 4pβ2=18
(viii) p/2+2=8
Answer
(i) The sum of numbers p and 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Two times m is 7.
(iv) The number m is divided by 5 gives 3.
(v) Three-fifth of the number m is 6.
(vi) Three times p plus 4 gets 25.
(vii) If you take away 2 from 4 times p, you get 18.
(viii) If you added 2 to half is p, you get 8.
Exercise 4.1
Page Number 81
6
Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Tale mm to be the number of Parmitβs marbles.)
(ii) Laxmiβs father is 49 years old. He is 4 years older than three times Laxmiβs age. (Take Laxmiβs age to be y years.)
(iii) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to beΒ l. )
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180β.)
Answer
(i) Let m be the number of Parmitβs marbles.
β΄ 5m+7=37
(ii) Let the age of Laxmi be y years.
β΄ 3y+4=49
(iii) Let the lowest score beΒ l.
β΄ 2l+7=87
(iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b
β΄ 2b+b+b=180β
β 4b=180βΒ [Angle sum property of a Ξ]
Exercise 4.1
Page Number 83
1
Give first the step you will use to separate the variable and then solve the equations:
(a) xβ1=0
(b) x+1=0
(c) xβ1=5
(d) x+6=2
(e) yβ4=β7
(f) yβ4=4
(g) y+4=4
(h) y+4=β4
Answer
(a) xβ1=0Β
β xβ1+1=0+1 [Adding 1 both sides]
β x=1
(b) x+1=0Β
β x+1β1=0β1 [Subtracting 1 both sides]
β x=β1
(c) xβ1=5Β
β xβ1+1=5+1 [Adding 1 both sides]
β x=6
(d) x+6=2Β
β x+6β6=2β6 [Subtracting 6 both sides]
β x=β4
(e) yβ4=β7
β yβ4+4=β7+4 [Adding 4 both sides]
β y=β3
(f) yβ4=4Β
β yβ4+4=4+4 [Adding 4 both sides]
β y=8
(g) y+4=4Β
β y+4β4=4β4 [Subtracting 4 both sides]
β y=0
(h) y+4=β4
β y+4β4=β4β4 [Subtracting 4 both sides]
β y=β8
Exercise 4.2
Page Number 86
2
Give first the step you will use to separate the variable and then solve the equations
(a) 3l=42
(b) b/2=6
(c) p/7=4
(d) 4x=25
(e) 8y=36
(f) z/3=54
(g) a/5=7/15
(h) 20t=β10
Answer
(a) 3l=42Β
β 3l/3=42/3 [Dividing both sides by 3]
βΒ l=14
(b) b/2=6Β
β b/2Γ2=6Γ2 [Multiplying both sides by 2]
β b=12
(c) p/7=4Β
β p/7Γ7=4Γ7[Multiplying both sides by 7]
β p=28
(d) 4x=25
β 4x4=25/4 [Dividing both sides by 4]
β x=25/4
(e) 8y=36
β 8y/8=36/8 [Dividing both sides by 8]
β y=9/2
(f) z/3=54
β z/3Γ3=5/4Γ3 [Multiplying both sides by 3]
β z=15/4
(g) a/5=7/15Β
β a/5Γ5=7/15Γ5Β [Multiplying both sides by 5]
β a=73
(h) 20t=β10
β 20t/20=β10/20 [Dividing both sides by 20]
β t=β1/2
Exercise 4.2
Page Number 86
3
Give first the step you will use to separate the variable and then solve the equations
(a) 3nβ2=46
(b) 5m+7=17
(c) 20p/3=40
(d) 3p/10=6
Answer
(a) 3nβ2=46
Step I:Β 3nβ2+2=46+2=48
[Adding 2 both sides]
Step II:Β 3n/3=48/3
βn=16 [Dividing both sides by 3]
(b) 5m+7=17
Step I:Β 5m+7β7=17β7=17β7Β
β 5m=10 [Subtracting 7 both sides]
Step II:Β 5m/5=10/5
βm=2 [Dividing both sides by 5]
(c) 20p/3=40
Step I:Β 20p/3Γ3=40Γ3Β
β 20p=120 [Multiplying both sides by 3]
Step II:Β 20p/20=120/20
β p=6 [Dividing both sides by 20]
(d) 3p/10=6
Step I:Β 3p/10Γ10=6Γ10
β 3p=60 [Multiplying both sides by 10]
Step II:Β 3p/3=60/3
β p=20 [Dividing both sides by 3]
Exercise 4.2
Page Number 86
4
Solve the following equation:
(a) 10p=100
(b) 10p+10=100
(c) p/4=5
(d) βp/3=5
(e) 3p/4=6
(f) 3s=β9
(g) 3s+12=0
(h) 3s=0
(i) 2q=6
(j) 2qβ6=0
(k) 2q+6=0
(l) 2q+6=12
Answer
(a) 10p=100
β 10p/10=100/10 [Dividing both sides by 10]
β p=10
(b) 10p+10=100
β 10p+10β10=100β10 [Subtracting both sides 10]
β 10p=90 β 10p/10=90/10 [Dividing both sides by 10]
β p=9
(c) p/4=5
β p/4Γ4=5Γ4 [Multiplying both sides by 4]
β p=20
(d) βp/3=5
β βp/3Γ(β3)=5Γ(β3) [Multiplying both sides by β 3]
β p=β15
(e) 3p/4=6
β 3p/4Γ4=6Γ4 [Multiplying both sides by 4]
β 3p=24β 3p/3=24/3 [Dividing both sides by 3]
β p=8
(f) 3s=β9
β 3s/3=β9/3 [Dividing both sides by 3]
β s=β3
(g) 3s+12=0
β 3s+12β12=0β12 [Subtracting both sides 10]
β 3s=β12 β 3s/3=β12/3 [Dividing both sides by 3]
β s=β4
(h) 3s=0
β 3s/3=0/3 [Dividing both sides by 3]
β s=0
(i) 2q=6
β 2q/2=6/2 [Dividing both sides by 2]
β q=3
(j) 2qβ6=0
β 2qβ6+6=0+6 [Adding both sides 6]
β 2q=6 β 2q/2=6/2 [Dividing both sides by 2]
β q=3
(k) 2q+6=0
β 2q+6β6=0β6 [Subtracting both sides 6]
β 2q=β6β 2q/2=β6/2 [Dividing both sides by 2]
β q=β3
(l) 2q+6=12
β 2q+6β6=12β6 [Subtracting both sides 6]
β 2q=6β 2q/2=6/2 [Dividing both sides by 2]
β q=3
Exercise 4.2
Page Number 86
1
Solve the following equations:
(a) 2y+5/2=37/2
(b) 5t+28=10
(c) a/5+3=2
(d) q/4+7=5
(e) 5/2x=10
(f) 5/2x=25/4
(g) 7m+19/2=13
(h) 6z+10=β2
(i) 3l/2=2/3
(j) 2b/3β5=3
Answer
(a) 2y+5/2=37/2
β 2y=37/2β5/2 β 2y=37β5/2
β 2y=32/2β 2y=16/2
β y=8
(b) 5t+28=10
β 5t=10β28 β 5t=β18
β t=β18/5
(c) a/5+3=2
β a/5=2β3β a/5=β1
β a=β1Γ5
β a=β5
(d) q/4+7=5
β q/4=5β7β q/4=β2
β q=β2Γ4
β q=β8
(e) 5/2x=10
β 5x=10Γ2
β 5x=20
β x=20/5
β x=4
(f) 5/2x=25/4
β 5x=25/4Γ2
β 5x=25/2
β x=25/(2Γ5)
β x=5/2
(g) 7m+19/2=13
β 7m=13β19/2
β 7m=26β19/2
β 7m=7/2β m=7/(2Γ7)β m=12
(h) 6z+10=β2
β 6z=β2β10
β 6z=β12
β z=β12/6
β z=β2
(i) 3l/2=2/3
β 3l=2/3Γ2β 3l=4/3
βΒ l=4/3Γ3
βl=49
(j) 2b/3β5=3
β 2b/3=3+5
β 2b/3=8
β 2b=8Γ3
β 2b=24
β b=24/2
β b=1/2
Exercise 4.3
Page Number 89
2
Solve the following equations:
(a) 2(x+4)=12
(b) 3(nβ5)=21
(c) 3(nβ5)=β21
(d) 3β2(2βy)=7
(e) β4(2βx)=9
(f) 4(2βx)=9
(g) 4+5(pβ1)=34
(h) 34β5(pβ1)=4
Answer
(a) 2(x+4)=12
β x+4=12/2
β x+4=6
β x=6β4
β x=2
(b) n=7+5
β n=12
(c) n=β7+5
β n=β2
(d) 2βy=4β2
β 2βy=β2
β βy=β2β2
β βy=β4
β y=4
(e) 4x=9+8
β 4x=17
β x=174
(f) β4x=9β8
β β4x=1
β x=β14
(g) pβ1=30/5
β pβ1=6
β p=6+1
β p=7
(h) pβ1=β30/β5
β pβ1=6
β p=6+1
β p=7
Exercise 4.3
Page Number 89
3
Solve the following equations:
(a) 4=5(pβ2)
(b) β4=5(pβ2)
(c) β16=β5(2βp)
(d) 10=4+3(t+2)
(e) 28=4+3(t+5)
(f) 0=16+4(mβ6)
Answer
(a) 2(x+4)=12
β x+4=12/2
β x+4=6
β x=6β4
β x=2
(b) 3(nβ5)=21
β nβ5=21/3
β nβ5=7
β n=7+5
β n=12
(c) 3(nβ5)=β21
β nβ5=β21/3
β nβ5=β7
β n=β7+5
β n=β2
(d) 3β2(2βy)=7
β β2(2βy)=7β3
β β2(2βy)=4
β 2βy=4/β2
β 2βy=β2
β βy=β2β2
β βy=β4
β y=4
(e) β4(2βx)=9
β β4Γ2βxΓ(β4)=9
β β8+4x=9
β 4x=9+8
β 4x=17
β x=17/4
(f) 4(2βx)=9
β 4Γ2βxΓ(4)=9
β 8β4x=9
β β4x=9β8
β β4x=1
β x=β1/4
(g) 4+5(pβ1)=34
β 5(pβ1)=34β4
β 5(pβ1)=30
β pβ1=30/5
β pβ1=6
β p=6+1
β p=7p=7
(h) 34β5(pβ1)=4
β β5(pβ1)=4β34
β β5(pβ1)=β30
β pβ1=β30/β5
β pβ1=6
β p=6+1
β p=7p=7
Exercise 4.3
Page Number 89
4
(a) Construct 3 equations starting with x=2.
(b) Construct 3 equations starting with x=β2
Answer
(a) 3 equations starting with x=2.
(i) x=2
Multiplying both sides by 10, 10x=20
Adding 2 both sides 10x+2 =20+2 = 10x + 2 = 22
(ii) x=2
Multiplying both sides by 5 5x=10
Subtracting 3 from both sides 5xβ3=10β3 = 5xβ3=7
(iii) x=2
Dividing both sides by 5 x 5=2/5
(b) 3 equations starting with x=β2.
(i) x=β2
Multiplying both sides by 3 3x=β6
(ii) x=β2
Multiplying both sides by 3 3x=β6
Adding 7 to both sides 3x+7=β6+7 = 3x+7=1
(iii) x=β2
Multiplying both sides by 3 3x=β6
Adding 10 to both sides 3x+10=β6+10= 3x+10=4
Exercise 4.3
Page Number 89
1
Set up equations and solve them to find the unknown numbers in the following cases:
1. Add 4 to eight times a number; you get 60.
2Β One-fifth of a number minus 4 gives 3.
3. If I take three-fourth of a number and add 3 to it, I get 21.
4. When I subtracted 11 from twice a number, the result was 15.
5. Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
6. Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5, she will get 8.
7. Answer thinks of a number. If he takes away 7 from 5/2 of the number, the result is 11/2.
Answer
(a) Let the number be x.
According to the question, 8x+4=60
β 8x=60β4
β 8x=56
β x=56/8
β x=7
(b) Let the number be y.
According to the question, y/5β4=3
β y/5=3+4
β y/5=7
β y=7Γ5
β y=35
(c) Let the number be z.
According to the question, 3/4.z+3=21
β 3/4.z=21β3
β 3/4.z=18
β 3z=18Γ4
β 3z=72
β z=72/3
β z=24
(d) Let the number be x
According to the question, 2xβ11=15
β 2x=15+11
β 2x=26
β x=26/2
β x=13
(e) Let the number be m.
According to the question, 50β3m=8
β β3m=8β50
β β3m=β42
β m=β42/β3
β m=14
(f) Let the number be n.
According to the question, (n+190/5=8
β n+19=8Γ5
β n+19=40
β n=40β19
β n=21
(g) Let the number be x.
According to the question, 5/2xβ7=11/2
β 5/2x=11/2+7
β 5/2x=(11+14)/2
β 5/2x=25/2
β 5x=(25Γ2)/2
β 5x=25
β x=25/5
β x=5
Exercise 4.4
Page Number 91
2
Solve the following:
1. The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
2. In an isosceles triangle, the base angles are equal. The vertex angle is 40β. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180β.)
3. Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Answer
(a) Let the lowest marks be y.
According to the question, 2y+7=87
β 2y=87β7
β 2y=80
β y=80/2
β y=40
Thus, the lowest score is 40.
(b) Let the base angle of the triangle be b.
Given, a=40β,b=c
Since, a+b+c=180β [Angle sum property of a triangle]
β 40β+b+b=180β
Β 
β 40β+2b=180β
β 2b=180ββ40β
β 2b=140β
β b=140β/2
β b=70β
Thus, the base angles of the isosceles triangle are 70β each.
(c) Let the score of Rahul be xx runs and Sachinβs score is 2x.
According to the question, x+2x=198
β 3x=198
β x=198/3
β x=66
Thus, Rahulβs score = 66 runs
And Sachinβs score = 2 x 66 = 132 runs.
Exercise 4.4
Page Number 91
3
Solve the following:
1. Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
2. Laxmiβs father is 49 years old. He is 4 years older than three times Laxmiβs age. What is Laxmiβs age?
3. People of Sundergram planted a total of 102 trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted?
Answer
(i) Let the number of marbles Parmit has be m.m.
According to the question, 5m+7=37
β 5m=37β7
β 5m=30
β m=30/5
β m=6
Thus, Parmit has 6 marbles.
(ii) Let the age of Laxmi be yy years.
Then her fatherβs age = (3y+4) years
According to question, 3y+4=49
β 3y=49β4
β 3y=45
β y=45/3
β y=15
Thus, the age of Laxmi is 15 years.
(iii) Let the number of fruit trees be t.
Then the number of non-fruits tree = 3t+2
According to the question, t+3t+2=102
β 4t+2=102
β 4t=102β2
β 4t=100
β t=100/4
β t=25
Thus, the number of fruit trees are 25.
Exercise 4.4
Page Number 91
4
Solve the following riddle:
I am a number, Tell my identity!
Take me seven times over, And add a fifty!
To reach a triple century, You still need forty!
Answer
Let the number be n.
According to the question, 7n+50+40=300
β 7n+90=300
β 7n=300β90
β 7n=210
β n=210/7
β n=30
Thus, the required number is 30.
Exercise 4.4
Page Number 91