Complete the last column of the table.

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 = -2)

(c) 7n + 5 = 19 (n = 2)

(d) 4p - 3 = 13 (p = 1)

(e) 4p - 3 = 13 (p = - 4)

(f) 4p - 3 = 13 (p = 0)

(a) LHS 1+ 5 =6 &ne RHS

So it is not the solution

(b) LHS 7 * (-2)+5=--9 &ne RHS

So it is not the solution

(c) LHS 7 * (2) + 5=19 = RHS

So it is the solution

(d) LHS 4* 1 -3 =1 &ne RHS

So it is not the solution

(e) LHS 4* (-4) -3 =-19 &ne RHS

So it is not the solution

(f) LHS 4* 0 -3 =-3 &ne RHS

So it is not the solution

Solve the following equations by trial and error method:

(i) 5p + 2 = 17

(ii) 3m – 14 = 4

(i)

(ii)

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 a is 70.

(iv) The number b divided by 5 gives 6.

(v) Three-fourth of t is 15.

(vi) Seven times m plus 7 gets you 77.

(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.

(i) x + 4=9

(ii) y -2 =8

(iii) 10a=70

(iv) b/5=6

(v) 3/4 t=15

(vi) 7m + 7 =77

(vii) x/4 -4=4

(viii) 6y - 6=60

(ix) z/3 + 3 =30

Write the following equations in statement forms:

(i) p + 4 = 15

(ii) m – 7 = 3

(iii) 2m = 7

(iv)m/5= 3

(v)3m/5= 6

(vi) 3p + 4 = 25

(vii) 4p – 2 = 18

(viii)p/2 + 2 = 8

(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

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. (Take m to be the number of Parmit’s marbles.)

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)

(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be 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 180 degrees).

(i) 5m + 7 =37

(ii) 3y + 4=49

(iii) 2l + 7=87

(iv) Base angle =b

Vertex angle = 2b

Now sum of angles of a triangle is 180 degrees

b + b + 2b=180

3b=180

Give first the step you will use to separate the variable and then solve the equations:

(a) x-1=0

(b) x+1=0

(c) x-1=5

(d) x+6=2

(e) y-4=-7

(f) y-4=4

(g) y+4=4

(h) y+4=-4

(a) x-1=0

Adding 1 both sides

x-1+1=0+1

x=1

(b) x+1=0

Subtracting 1 both sides

x+1-1=0-1

x=-1

(c) x-1=5

Adding 1 both sides

x-1+1=5+1

x=6

(d) x+6=2

Subtracting 6 both sides

x+6-6=2-6

x=-4

(e) y-4=-7

Adding 4 both sides

y-4+4=-7+4

y=-3

(f) y-4=4

Adding 4 both sides

y-4+4=4+4

y=8

(g) y+4=4

Subtracting 4 both sides

y+4-4=4-4

y=0

(h) y+4=-4

Subtracting 4 both sides

y+4-4=-4-4

y=-8

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 = 5/4

(g) a/5 = 7/15

(h) 20t = -10

(a) 3l=42

Dividing both sides by 3

3l/3=42/3

l=14

(b) b/2=6

Multiplying both sides by 2

(b/2)*2=6*2

b=12

(c) p/7=4

Multiplying both sides by 7

(p/7)*7=4*7

p=28

(d) 4x=25

Dividing both sides by 4

x=25/4

(e) 8y=36

Dividing both sides by 8

8y/8=36/8

y=9/2

(f) z/3=5/4

Multiplying both sides by 3

(z/3)*3=(5/4)*3

z=15/4

(g) a/5=7/15

Multiplying both sides by 5

(a/5)*5=(7/15)*5

a=73

(h) 20t=-10

Dividing both sides by 20

20t/20=-10/20

t=-1/2

Give first the step you will use to separate the variable and then solve the equations

(a) 3n-2=46

(b) 5m+7=17

(c) 20p/3=40

(d) 3p/10=6

(a) 3n-2=46

Adding 2 both sides

3n-2+2=46+2=48

Dividing both sides by 3

3n/3=48/3

n=16

(b) 5m+7=17

Subtracting 7 both sides

5m+7-7=17-7=17-7

5m=10

Dividing both sides by 5

5m/5=10/5

m=2

(c) 20p/3=40

Multiplying both sides by 3

(20p/3)*3 = 40*3

20p = 120

Dividing both sides by 20

20p/20=120/20

p=6

(d) 3p/10=6

Multiplying both sides by 10

(3p/10)*10 = 6*10

3p=60

Dividing both sides by 3

3p/3 = 60/3

p=20

Solve the following equation:

(a) 10 p = 100

(b) 10p + 10 =100

(c) p/4 = 5

(d) -p/3 = 5

(e) 3p/4 = 6

(f) 3s = -9

(g) 3s + 12 = 0

(h) 3s = 0

(i) 2q = 6

(j) 2q-6 = 0

(k) 2q+6 = 0

(l) 2q+6 = 12

(a) 10p=100

Dividing both sides by 10

10p/10=100/10

p=10

(b) 10p+10=100

Subtracting both sides 10

10p+10-10 = 100-10

10p=90

Dividing both sides by 10

10p/10=90/10

p=9

(c) p/4=5

Multiplying both sides by 4

p/4 * 4 = 5*4

p = 20

(d) -p/3=5

Multiplying both sides by – 3

-p/3*(-3) = 5*(-3)

p = -15

(e) 3p/4 =6

Multiplying both sides by 4

(3p/4) * 4 = 6*4

3p = 24

Dividing both sides by 3

3p/3 = 24/3

p=8

(f) 3s = -9

Dividing both sides by 3

3s/3 = -9/3

s = -3

(g) 3s+12=0

Subtracting both sides 10

3s+12-12=0-12

3s=-12

Dividing both sides by 3

3s/3 = -12/3

s=-4

(h) 3s=0

Dividing both sides by 3

3s/3=0/3

s=0

(i) 2q = 6

Dividing both sides by 2

2q/2 = 6/2

q = 3

(j) 2q-6 = 0

Adding both sides 6

2q-6+6 = 0+6

2q = 6

Dividing both sides by 2

2q/2 = 6/2

q=3

(k) 2q+6 = 0

Subtracting both sides 6

2q+6-6 = 0-6

2q = -6

Dividing both sides by 2

2q/2 = -6/2

q = -3

(l) 2q+6 = 12

Subtracting both sides 6

2q+6-6 = 12-6

2q = 6

Dividing both sides by 2

2q/2 = 6/2

q = 3

Set up equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number; you get 60.

(b) One-fifth of a number minus 4 gives 3.

(c) If I take three-fourth of a number and add 3 to it, I get 21.

(d) When I subtracted 11 from twice a number, the result was 15.

(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.

(f) Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5, she will get 8.

(g) Anwer thinks of a number. If he takes away 7 from 5/2 of the number, the result is 11/2.

(a) Let the number be x.

According to the question

8x+4=60

Transposing 4 to other side

8x = 60-4

8x = 56

Dividing both sides by 8

x = 56/8

x = 7

(b) Let the number be y.

According to the question

y/5-4=3

Transposing 4 to other side

y/5 = 3+4

y/5 = 7

Multiplying both sides by 5

y = 7*5

y = 35

(c) Let the number be z.

According to the question

$\frac {3}{4} z+3=21$

Transposing 3 to other side

$\frac {3}{4} z=21 -3$

$\frac {3}{4} z=18$

Multiplying both sides by 4

3z = 18*4

3z = 72

Dividing both sides by 3

z = 72/3

z = 24

(d) Let the number be x

According to the question

2x-11=15

Transposing 11 to other side

2x = 15+11

2x = 26

Dividing both sides by 2

x = 26/2

x = 13

(e) Let the number be z.

According to the question

50-3z=8

Transposing 50 to other side

-3z = 8-50

-3z = -42

Dividing both sides by -3

z = -42/-3

z = 14

(f) Let the number be x.

According to the question,

$\frac {x+19}{5}=8$

Multiplying both sides by 5

x+19 = 8*5

x+19 = 40

x = 40-19

x = 21

(g) Let the number be x.

According to the question

$\frac {5}{2} x-7=\frac {11}{2}$

$\frac {5}{2}x = \frac {11}{2} + 7$

$\frac {5}{2} x = \frac {25}{2}$

Multiplying both sides by 2

5x = 25

5x = 25

Dividing both sides by 5

x = 25/5

x = 5

Solve the following:

(a) 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?

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°.)

(c) 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?

(a) Let the lowest marks be m.

According to the question

$2m+7=87$

$2m = 87-7$

$2m = 80$

Dividing both sides by 2

$m = 40$

Hence, the lowest score is 40.

(b) Let the base angle of the triangle be x.

Given, a=40° , b =x ,c=x

Angle sum property of a triangle

a+b+c=180

40+x+x=180

40 +2x = 180

2x= 180 - 40

2x = 140

Dividing both sides by 2

x = 140/2

x = 70

Therefore , the base angles of the isosceles triangle are 70° each.

(c) Let the score of Rahul be x runs then Sachin’s score will be 2x.

According to the question

x+2x =198

3x = 198

Dividing both sides by 3

x = 198/3

x = 66

Thus, Rahul’s score = 66 runs

And Sachin’s score = 2 * 66 = 132 runs.

Solve the following:

(i) 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?

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

(iii) 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?

(i) Let the number of marbles Parmit has be m.

According to the question

5m+7=37

Transposing 7 to other side

5m=37-7

5m=30

Dividing both sides by 5

m=30/5

m=6

Hence, Parmit has 6 marbles.

(ii) Let the age of Laxmi be y years.

Then her father’s age = (3y+4) years

According to question

3y+4=49

Transposing 4 to other side

3y=49-4

3y=45

Dividing both sides by 3

y=45/3

y=15

Therefore, the age of Laxmi is 15 years

.

(iii) Let the number of fruit trees be x.

Then the number of non-fruits tree = 3x+2

According to the question $x+3x+2 = 102$

4x+2 =102

Transposing 2 to other side

4x = 102-2

4x = 100

Dividing both sides by 4

x = 100/4

x = 25

Thus, the number of fruit trees are 25.

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!

Let the number be n.

According to the question

7n+50+40 = 300

7n+90 = 300

Transposing 90 to other side

7n = 300-90

7n = 210

Dividing both sides by 7

n = 210/7

n = 30

Thus, the required number is 30.

**NCERT Solutions and Worksheets**

Class 7 Maths Class 7 Science