Simple Equations

Given equation is ax + b = 0

⇒ ax = –b [transporting b to RHS]

⇒ [on dividing both side by b]

Given is ax = b

On dividing the equation by a, we get

Now, if a and b are positive integers, then the solution of the equation is also positive number as division of two positive integers is also a positive number.

Dividing both sides of the equation by the same non–zero number is allowed in a given equation, divisor of any number by zero is not allowed as set division of number by zero is not defined.

Let us solve the equations:

(a). Given equation is 2x + 6 = 0

⇒ 2x = –6 [transporting 6 to RHS]

⇒ [Dividing both sides by 2]

⇒ x = –3 (integer)

(b). Given equation is 3x – 5 = 0

⇒ 3x = 5 [transporting 5 to RHS]

⇒ [Dividing both sides by 3]

⇒ (fraction)

(c). Given equation is 5x – 8 = x + 4

2⇒ 5x = x + 4 + 8 [transporting 8 to RHS]

⇒ 5x = x + 12

⇒ 5x – x = 12 [transporting x to LHS]

⇒ 4x = 12

⇒

⇒ x = 3 (integer)

(d). Given equation is 4x + 7 = x + 2

⇒ 4x = x + 2 – 7 [transporting 8 to RHS]

⇒ 4x = x – 5

⇒ 4x – x = –5 [transporting x to LHS]

⇒ 3x = –5

⇒

⇒ (neither a positive fraction nor an integer)

(a) Given equation is 5y – 3 = –18

⇒ 5y = – 18 + 3 [transposing 3 to RHS]

⇒ 5y = –15

⇒ [dividing both sides by 5]

⇒ y = –5 (integer)

(b) Given equation is 3x – 9 = 0

⇒ 3x = 9 [Transposing 9 to RHS]

⇒ [Dividing both sides by 3]

⇒ x = 3 (integer)

(c) Given equation is 3z + 8 = 3 + z

On transposing z and 8 to LHS and RHS respectively, we get

⇒ 3z – z = 3 – 8

⇒ 2z = –5

⇒ [dividing both sides by 2]

⇒ (neither a positive fraction nor an integer)

(d) Given equation is 9y + 8 = 4y – 7

On transposing 4y and 8 to LHS and RHS respectively, we get

⇒ 9y – 4y = –7 – 8

⇒ 5y = –15

⇒ [Dividing both sides by 5]

⇒ y = – 3 (integer)

Given equation is 7x + 4 = 25

⇒ 7x = 25 – 4[Transposing 4 to RHS]

⇒ 7x = 21

On dividing the above equation by 7, we get

x = 3

Hence, the solution of the given equation is 3.

Given 3x + 7 = – 20

⇒ 3x = –20 – 7

⇒ 3x = –27

On dividing the above equation by 3, we get

⇒ x = –9

Hence, the solution of the given equation is –9.

It is given that both the expression is equal. So, the equation is,

⇒ y – 15 = 2y + 1

⇒ y – 2y = 1 + 15

⇒ –y = 16

Multiplying both sides by (–1), we get

y = –16

Given equation is k + 7 = 16

On transposing 7 to RHS, we get

⇒ k = 16 – 7 = 9

Put the value of k in the equation (8k – 72), we get

⇒ 8(9) – 72

⇒ 72 – 72 = 0

Given equation is 43m = 0.086

On dividing the given equation by 43, we get

⇒

If we remove the decimal, we get 1000 in denominator

⇒

The given statement means x is 7 more than 3

So, the equation is x – 7 = 3

We can write it as x – 3 = 7

(a) Given equation is 4x + 1 = 2

⇒ 4x = 2 – 1

⇒

⇒

(b) Given equation is 3 – x = 8

⇒ –x = 8 – 3

⇒ –x = 5

⇒ x = –5

(c) Given equation is x – 5 = 3

⇒ x = 3 + 5

⇒ x = 8

(d) Given equation is 3 + x = 8

⇒ x = 8 – 3

⇒ x = 5

(a) Given equation is x + 3 = 1

⇒ x = 1 – 3

⇒ x = –2

(b) Given equation is 8 + 2x = 3

⇒ 2x = 3 – 8

⇒ 2x = –5

⇒

(c) Given equation is 10 + 3x = 1

⇒ 3x = 1 – 10

⇒ 3x = –9

⇒

⇒ x = –3

(d) Given equation is 2x + 1 = 3

⇒ 2x = 3 – 1

⇒ 2x = 2

⇒

⇒ x = 1

We have, x = 0

On multiplying both the sides by 3, we get

⇒ 3 × x = 3 × 0

⇒ 3x = 0

On adding (–1) both the sides, we get

⇒ 3x + (–1) = 0 + (–1)

⇒ 3x – 1 = –1

We have, x = 7

On multiplying both the sides by 7, we get

⇒ 7 × x = 7 × 7

⇒ 7x = 49

On adding (–1) both sides, we get

7x + (–1) = 49 + (–1)

⇒ 7x – 1 = 49 – 1

⇒ 7x – 1 = 48

Given

On multiplying both sides by 2, we get

⇒ x = 3 × 2 = 6

Put x = 6 in the equation 3x + 2, we get

⇒ 3(6) + 2

⇒ 18 + 2 = 20

Let us put the values given in the options in equation –6 + x = –12

(a) Put x = 2

⇒ –6 + 2 = –12

⇒ –4 = –12

⇒ LHS ≠ RHS

(b) Put x = 6

⇒ –6 + 6 = –12

⇒ 0 = –12

⇒ LHS ≠ RHS

(c) Put x = –6

⇒ –6 – 6 = –12

⇒ –12 = –12

⇒ LHS = RHS

(d) Put x = –2

⇒ –6 – 2 = –12

⇒ –8 = –12

⇒ LHS ≠ RHS

Transposition means shifting one term from one side of an equation to another side with a change of sign.

Given, the sum of two numbers is 60 and difference is 30.

(a) If the smaller number is x, then the other number is (60 – x), since the sum of both number is 60.

(b) Given, one number = x [from (a)]

Then, other number = (60 – x)

∴ Difference = (60 – x) – x = 60 – 2x

(c) We are given that difference between two numbers is 30.

So, the equation formed is 60 – 2x = 30

⇒ –2x = 30 – 60

⇒ –2x = –30

⇒ 2x = 30

(d) Let us solve the equation for x,

2x = 30

On dividing the above equation by 2, we get

⇒ 4x = 60

⇒

⇒ x = 15

(e) The numbers are x and (60 – x)

Now, put the value of x, we get

First number = 15

Second number = 60 – 15 = 45

(a) We are given that one number is twice the other.

If smaller number is x, then the other number is 2x.

(b) We are given that sum of two number is 81.so, the equation will be x + 2x = 81

(c) Now, solve the equation for x.

⇒ x + 2x = 81

⇒ 3x = 81

⇒

⇒ x = 27

Hence, the solution of the equation is 27.

(d) The two numbers are x = 27 and 2x = 2 × 27 = 54.

(a) If Palak gets x marks, then Abha gets twice the marks as that of Palak, i.e. 2x,

(b) Two times of Abha’s marks and three times Palak’s marks make 280.

So, the equation formed is 4x + 3x = 280

(c) Solve the equation for x,

4x + 3x = 280

⇒ 7x = 280

⇒

⇒ x = 40

Hence, the solution of the equation is 40.

(d) Marks obtained by Abha are 2x, i.e. 2 × 40 = 80

(a) It is given that the length of the rectangle is two times its breadth.

∴ Length = 2x cm.

(b) Perimeter of rectangle = 2(length + Breadth) = 2(2x + x)

(c) As we are given that perimeter of rectangle is 60 cm.

So, the equation formed is 2(2x + x) = 60

⇒ 2(3x) = 60

⇒ 6x = 60

(d) On dividing the equation by 6, we get

⇒ x = 10

Hence, the solution of the equation is 10.

Let the number of coins of ₹5 = x

Then, number of coins of ₹2 = x

(a) Number of coins of ₹5 = x

So, worth of ₹5 of x coins = ₹5 × x = ₹5x

(b) Similarly, the worth of 12 of x coins = ₹2x

(c) The equation formed is 5x + 2x = 70

(d) 5x + 2x = 70

⇒ 7x = 70

⇒ x = 10

There are 10 ₹5 coins and 10 ₹2 coins.

Given, number of prizes = 30

Total prize money = ₹52000, 1^st and 2^nd prizes are worth ₹2000 and ₹1000, respectively.

(a) 1^st prize is x in number, the number of 2^nd prizes are (30 – x), because total number of prizes are 30.

(b) Total value of prizes in terms of x are 2000x + 1000(30 – x)

(c) The equation formed is 1000x + 30000 = 52000

From (b), 2000x + 1000(30 – x) = 52000

⇒ 2000x + 30000 – 1000x = 54000

⇒ 1000x = 54000 – 30000

⇒ 1000x = 24000

(d) The solution of the equation is 22.

From (c), 1000x + 30000 = 52000.

⇒ 1000x = 52000 – 30000 = 22000

⇒ 1000x = 22000

⇒

Solve the given equation for z,

z + 3 = 5

⇒ z = 5 – 3

⇒ z = 2

Solve the given equation for x,

3x – 2 = 7

⇒ 3x = 7 + 2

⇒ 3x = 9

⇒

⇒ x = 3

Solve the given equation for x,

3x + 10 = 7

⇒ 3x = 7 – 10

⇒ 3x = – 3

⇒

⇒ x = – 1

Solve the given equation for x,

2x + 3 = 5

⇒ 2x = 5 – 3

⇒ 2x = 2

⇒

⇒ x = 1

Solve the equation for x,

4x – 1 = 8

⇒ 4x = 8 + 1

⇒ 4x = 9

⇒

Solve the equation for x,

4x + 5 = –7

⇒ 4x = –7 – 5

⇒ 4x = –12

⇒

⇒ x = –3

Since, the value of x is not natural number, hence the equation ha no solution in natural numbers.

Solve the equation for x,

x – 5 = –5

⇒ x = –5 + 5

⇒ x = 0

Since, natural numbers do not contain zero, hence, the equation has no solution.

Solve the equation for x,

x + 8 = 12 – 4

⇒ x + 8 = 8

⇒ x = 8 – 8

⇒ x = 0

Since, natural numbers do not contain zero, hence, the equation has no solution.

Let the number be x.

Now, 5 is added to 3 times the number, i.e. 5 + 3x

It is same as 7 is subtracted from 4 times the number, i.e. Ax – 7.

So, the equation formed is 5 + 3x = 4x – 7

Solve the equation for x,

x + 7 = 10

⇒ x = 10 – 7

⇒ x = 3

Given that, 3x = 12

⇒

⇒ x = 4

⇒ x – 0 = 4 – 0 = 4

Given that, 2x = 2

⇒

⇒ x = 1

⇒ x – 1 = 1 – 1 = 0

Given that,

⇒ x = 12 [multiplying both sides by 2]

∴ 12 – (–3) = 15

Hence, x – (–3) = 15

Solve the equation for x,

⇒ x + 15 = 19

⇒ x = 19 – 15 [transposing 15 to RHS]

⇒ x = 4

The solution of the equation x + 15 = 19 is 4.

Finding the value of a variable in a linear equation that satisfies the equation is called a root of the equation.

Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the sign of the term.

Given that

Multiplying 5 both sides

⇒ 9x = 18

On dividing both sides by 9

⇒

⇒ x = 2

Given that, 3 – x = –4

⇒ –x = –4 – 3

⇒ –x = –7

⇒ x = 7 [multiplying by (–1)]

Given that,

⇒

⇒ x = 0

Given that,

⇒

⇒ –x = 0

⇒ x = 0 [multiplying both sides by (–1)]

Let the number be x.

Then, the equation will be x – 10 = 65

Now, solving the equation for x,

x = 65 + 10

⇒ x = 75

Hence, the number is 75.

Let the number be x.

If it is increased by 20, it becomes (x + 20),

So, the equation formed is x + 20 = 45

⇒ x = 45 – 20 [transposing 20 to RHS]

⇒ x = 25

Hence, the number is 25.

Let the number be x.

84 – x = 12

⇒ –x = 12 – 84 [transposing 84 to RHS]

⇒ –x = –72

⇒ x = 72 [ multiplying both sides by (–1)]

Hence, the number is 72.

Given that,

⇒

⇒

True

Solve the equation for x, 3x + 2 = 17

⇒ 3x = 17 – 2 [transposing 2 to RHS]

⇒ 3x = 15

⇒ [dividing both sides by 3]

⇒ x = 5

False

Solve the equation for x,

⇒ 4x – 1 = 8

⇒ 4x = 8 + 1 [transposing 1 to RHS]

⇒ 4x = 9

⇒ [dividing both sides by 4]

False

Given equation is 4x – 5 = 7

⇒ 4x = 7 + 5 [transposing 5 to RHS]

⇒ 4x = 12

⇒

⇒ x = 3

False

Let the number be x.

One third of a number added to itself gives 10

Then, the equation formed is

True

Solve the equation for x,

⇒ 8x = 7 + 5 [transposing 5 to RHS]

⇒ 8x = 12

⇒

False

Solve the equation for x,

4x – 7 = 11

⇒ 4x = 11 + 7 [transposing 7 to RHS]

⇒ 4x = 18

⇒

False

Given that, x = 9

Put the value of x in the equation, we get

⇒

⇒

⇒

⇒ 19 = y

(i). (c) Given equation is x + 5 = 9

⇒ x = 9 – 5

⇒ x = 4

(ii). (e) Given equation is x – 7 = 4

⇒ x = 4 + 7

⇒ x = 11

(iii). (f) Given equation is

⇒

⇒ x = –60

(iv). (d) Given equation is 5x = 30

⇒

⇒ x = 6

(v). (b) given equation is 3y = 5

⇒

⇒

Let the number be x.

13 is subtracted from twice of a number i.e., 2x – 13 and it results 3.

So, the equation formed is 2x – 13 = 3.

Let the number be x.

Then, ^th of the number

Now, is 5 less than x.

So, the equation formed is

Let the number be x.

Then, ^rd of number

So, the equation formed is 6x = 10 + x

Let the number be x.

Then, 6 times of a number = 6x

So, the equation formed is 6x = 10 + x.

Let the number be x.

Then, 10 is subtracted from and its result is 4

So, the equation formed is .

Subtract 5 from p i.e. p – 5 and its result is 2. Hence, the equation formed is p – 5 = 2.

Let the number be x.

Five times a number increased by 7 is 27

⇒ 5x + 7 = 27

Let the age of Sohan be x yr. Then, the age of Mohan is (x + 3) yr.

∴ Sum of their ages = 43

So, the equation formed is x + (x + 3) = 43

Let the number be x.

Then, 1 is subtracted from a number and the difference is multiplied by

It gives result 7.

So, the equation formed is

Let the number be x.

Then, x is divided by 2 and increased by 5 i.e. and gives result 9.

So, the equation formed is .

Let the number be x.

Then, sum of twice of a number and 4 gives result 18. Hence, 2x + 4 = 18 is the equation.

Let x yr be the age of Amit.

Then, age of Sohan Lal = 4x yr

According to the question,

4x – x = 27

⇒ 3x = 27

⇒

Let x be a number, then another number will be x + 12

According to the question,

x + x + 12 = 72

⇒ 2x + 12 = 72

⇒ 2x = 72 – 12

⇒ 2x = 60

⇒ x = 30

Hence, the numbers are 30 and (30 + 12) i.e. 30 and 42.

Let the number be x.

Then, seven times of this number = 7x and thirteen times of this number = 13x

According to the question,

⇒ 13x – 7x = 12

⇒ 6x = 12

⇒ x = 2 [dividing both sides by 6]

Hence, the number is 2.

Let the interest received by Karim be ₹x, then interest received by Ramesh will be ₹ (x – 30). So, the interest received by both will be ₹ (x + x – 30)

According to the question, x + x – 30 = 70

⇒ 2x – 30 = 70

⇒ 2x = 70 + 30

⇒ 2x = 100

⇒ x = 50

So, interest received by Ramesh = ₹ (x – 30) = ₹ (50 – 30) = ₹20.

Let ₹x be the amount donated in a relief fund by Subramaniam. Then, the amount donated by Naidu will be ₹ (x + 125)

According to the question,

⇒ x + x + 125 = 975

⇒ 2x + 125 = 975

⇒ 2x = 975 – 125

⇒ 2x = 850

⇒ x = ₹425

Hence, the amount of money donated by Subramaniam is ₹ 425

Let x be the number of boys in the school. Then, the number of girls in the school will be x + 50

According to the question,

⇒ x + (x + 50) = 1070

⇒ 2x + 50 = 1070

⇒ 2x = 1070 – 50

⇒ 2x = 1020

⇒ x = 510

So, the number of boys in the school = 510

∴ Number of girls in the school = 510 + 50 = 560

Let the number be x.

It is given that two times this number increased by 5 equals 9

∴ 2x + 5 = 9

⇒ 2x = 9 – 5

⇒ 2x = 4

⇒ x = 2

Hence, the required number is 2.

Let the required number be x.

It is given that 9 added to twice this number gives 13.

2x + 9 = 13

⇒ 2x = 13 – 9

⇒ 2x = 4

⇒

Hence, the required number is 2.

Let the number be x. then, one–third of the number

⇒

⇒

⇒

⇒ x = 6 [multiply by 3 both sides]

Hence, the required number is 6.

Let the Rama’s present age be x yr.

Then, Rama’s age after 25 yr = (x + 25) yr.

It is given that after 25 yr, Rama’s age will be 5 times his present age.

Therefore, the equation is x + 25 = 5x

⇒ 25 = 5x – x

⇒ 25 = 4x

⇒ [Divide by 4 both sides]

⇒

⇒

Hence, the present age of Rama is yr.

Let the Manoj’s present age be x yr.

Then, Manoj’s age after 20 yr = (x + 20) yr.

It is given that after 20 yr, Manoj’s age will be 5 times his present age.

Therefore, the equation is x + 20 = 5x

⇒ 20 = 5x – x

⇒ 20 = 4x

⇒ [Divide by 4 both sides]

⇒ 5 = x

Hence, the present age of Manoj is 5 yrs.

Let the age of my younger sister be x yr.

Then, her age after 3 yr = (x + 3) yr.

Also, her age, 3 yr ago = (x – 3) yr.

It is given that her present age is 3 times her age after 3 yr minus 3 times her age 3 yr ago.

Therefore, the equation is x = 3(x + 3) – 3(x – 3)

⇒ x = 3x + 9 – 3x + 9

⇒ x = 18 yr.

Hence, her present age is 18 yr.

Let x be the number. Then, half of the number is

According to the question,

⇒

⇒

⇒ x + 90 = 6x

⇒ 90 = 6x – x

⇒ 90 = 5x

⇒ [divide both side by 5]

⇒ x = 18.

Hence, the number is 18.

As per the given information in the question, total consumption of the cereals = 80kg

Let x be the consumption of rice.

Then, consumption of wheat = 4x

According to the question,

x + 4x = 80

⇒ 5x = 80

⇒

∴ Consumption of wheat = 4x = 4 × 16 = 64 kg.

Hence, the consumption rice and wheat are 16kg and 64 kg, respectively.

Let the number of two–rupee coin be y. Then, the number of one–rupee coins is 3y.

Total money by two –rupee coins = 2 × y = 2y

Total money by one –rupee coins = 1 × 3y = 3y

Total worth of coins = ₹120

So, the equation formed is 2y + 3y = 120

⇒ 5y = 120

⇒ [Divide by 5 both side]

⇒ y = 24

∴ Number of two–rupee coins = y = 24 and number of one–rupee coins = 3y = 3 × 24 = 72.

Let x be the number thought by Anamika.

If she multiplied it be 2, then the number will be 2x.

Also, added 5 to it and obtained 17 as result.

∴ 2x + 5 = 17

⇒ 2x = 17 – 5

⇒ 2x = 12

⇒

Hence, the number 6 is thought by Anamika.

Let x be the one of the number. Then, the other number is twice the first one = 2x

According to the question,

x + 2x = 12

⇒ 3x = 12

⇒ [Divide both side by 3]

⇒ x = 4

Hence, the numbers are x = 4 and 2x = 2 × 4 = 8.

Let one number be x. Then, the next two consecutive numbers will be x + 1 and x + 2. Sum of these three numbers = x + (x + 1) + (x + 2) = 3x + 3

According to the question, 3x + 3 = x + 5

⇒ 3x – x = 5 – 3

⇒ 2x = 2

⇒ [Dividing both sides by 2]

⇒ x = 1

∴ Hence, the numbers are 1, 1 + 1, 1 + 2, i.e. 1, 2, 3.

Let the required number be x. Then, x divided by 6

It is given that when is x is divided by 6, gives the quotient as 6.

So, we obtain the following equation

⇒ [multiply both sides by 6]

⇒ x = 36

Hence, the required number is 36.

As per given information in the question, the perimeter of a rectangle is 40m.

Let x be the breadth of the rectangle.

Then, length of the rectangle = 5x – 4

According to the question, 2x + 2(5x – 4) = 40

⇒ 2x + 10x – 8 = 40

⇒ 12x = 48

⇒

Hence, the length of the rectangle = 5x – 4 = (5 × 4) – 4 = 20 – 4 = 16m.

Let third side of an isosceles triangle be x. then, two other equal sides are twice.

So, both equal sides are 2x and 2x.

We know that, perimeter of a triangle is sum of all sides of the triangle.

According to the question,

x + 2x + 2x = 30

⇒ 5x = 30

On dividing both sides by 5, we get

⇒

⇒ x = 6cm

∴ Third side = x = 6cm

So, the other equal sides are 2x = 2 × 6 = 12cm and 12 cm.

Let the two consecutive multiples of 2 be 2x and 2x + 2.

According to the question,

2x + 2x + 2 = 18

⇒ 4x + 2 = 18

⇒ 4x = 18 – 2

⇒ 4x = 16

⇒ [dividing both sides by 4]

Hence, the required numbers are 2x = 2× 4 = 8 and 2x + 2 = 2 × 4 + 2 = 8 + 2 = 10.

Let one of the angle be x, then other will be x – 20

According to the question,

x + (x – 20) = 90° [∵ sum of the complementary angles is 90°]

⇒ 2x – 20 = 90°

⇒ 2x = 90° + 20

⇒ 2x = 110°

⇒

⇒ x = 55°

Hence, the required angles are 55° and (55 – 20) ° i.e. 55° and 35°

Let one part be x, then other part will be 2x as second part is twice the first part.

Since, 150 has been divided into above two parts.

According to the question,

x + 2x = 150

⇒ 3x = 150

⇒

⇒ x = 50

Hence, the first part is 50 and the second part is 2× 50 = 100.

As per given information in the equation, the total number of students in the class = 60.

Let x be the number of boys in the class.

Then, the number of girls in the class =

According to the question,

⇒

⇒

⇒

⇒ 4x = 60 × 3

⇒ 4x = 180

⇒

Hence, the number of boys in the class is 45 and the number of girls in the class is

Let the number be x.

The, two–third of this number and one–third of this number

According to the question,

⇒

⇒

⇒ x = 3 × 3

⇒ x = 9

Hence, the required number is 9.

Let the number be x. If we subtract 27 from x. i.e. (x – 27) and subtract x from 73 i.e. (73 – x), we get result. Therefore, we get the following equation:

x – 27 = 73 – x

⇒ x + x = 73 + 27

⇒ 2x = 100

⇒ [Dividing both sides by 2]

⇒ x = 50

Hence, the required by is 50.

Let his total journey length be x.

∴ Then, Distance travelled by train

Travelled by bus and travelled by car x.

∴ Total journey travelled by train, bus and car

∴ Remaining journey

According to the question, remaining journey is 3km.

∴

⇒ x = 3 × 60 = 180 km

Hence, the length of his total journey = 180km.

Let the number be x. then, twice of this number = 2x and half of this number

According to the question,

Multiplying both sides by 2, we get 4x + x = 48

⇒ 5x = 48

⇒ [Dividing both sides by 5]

⇒

⇒ x = 9.6

Hence, the required number is 9.6.

Let the number be x. Then, thrice the number = 3x and twice of this number = 2x.

If we decrease thrice of x by 5, we get (3x – 5).

According to the question,

(3x – 5) – 2x = 1

⇒ 3x – 5 –2x = 1

⇒ x – 5 = 1

⇒ x = 1 + 5

⇒ x = 6

Hence, the required number is 6.

Let x be the age of girl.

Then, age of her father = (x + 28) yr.

According to the question,

∴ x + (x + 28) = 50

⇒ 2x + 28 = 50

⇒ 2x = 50 – 28

⇒ 2x = 22

⇒

⇒ x = 11

Hence, the age of the girl is 11yr and age of her father’s age is (11 + 28) = 39yr.

Let x be the width of the rectangle. Then, length of the rectangle will be 2x.

∵ Perimeter of a rectangle = 2[Length + Width]

According to the question.

2 (x + 2x) = 180

⇒ 2 (3x) = 180

⇒ 6x = 180

⇒ [dividing both sides by 6]

⇒ x = 30

Hence, width of the rectangle is 30cm and length of the rectangle is 2 × 30 i.e. 60xm.

Let the cost of one pencil be ₹x.

Now, cost of such 7 pencils will be ₹7x and of 5 pencils will be ₹5x.

It is given that cost of 7 pencils is ₹6 more than cost of 5 pencils. Therefore, we get the following equation

7x – 5x = 6

⇒ 2x = 6

⇒ [dividing both sides by 2]

⇒ x = 3

Since, cost of one pencil is ₹3.

So, the cost of 10 pencils = 3 × 10 = ₹30.

Thus, she has to pay ₹30 for 10 pencils.

Let the number of students who got first division in year 2006 be x.

Since, the number of first division in year 2007 exceeded those in the year 2006 by 34, therefore the number of students who got first division in year 2007 will be (x + 34)

It is given that total number of students who got first division in years 2006 and 2007 is 3768.

According to the question,

x + (x + 34) = 3768

⇒ 2x + 34 = 3768

⇒ 2x = 3768 – 34

⇒ 2x = 3734

⇒ [dividing both sides by 2]

⇒ x = 1867

Hence, 1867 students got first division in year 2006.

Radha’s monthly salary and over–time = ₹17480

Let ₹ x be her monthly salary

The, overtime = ₹ (x – 10000)

∴, 17480 – x = x – 10000

⇒ 17480 + 10000 = x + x

⇒ 27480 = 2x

⇒

⇒ x = 13740

Hence, her monthly salary is ₹13740.

Given, one side of a square is 18x – 20 and adjacent side is 42 – 13x.

We know that, all the sides of a square are always equal.

∴ 18x – 20 = 42 – 13x

⇒ 18x + 13x = 42 + 20

⇒ 31x = 62

⇒

⇒ x = 2 units

Hence, the side of the square is (18 × 2) – 20 = 36 – 20 = 16 units.

(a) Given, IX – VI = V

According to the question we have to remove two matchsticks to make a valid equation.

Hence, X – V = V

⇒ 10 – 5 = 5

(b) Given, VI – IV = XI

According to the question, we have to move one matchstick to make a valid equation.

(i) VI + IV = X

⇒ 6 + 4 = 10

(ii) VI + V = XI

⇒ 6 + 5 = 11

We have, i + 69 = 70

⇒ i = 70 – 69

⇒ i = 1

and 8u = 6u + 8

⇒ 8u – 6u = 8

⇒ 2u = 8

⇒ [dividing both side by 2]

⇒ u = 4

We have, 4a = –5a + 45

⇒ 4a + 5a = 45

⇒ 9a = 45

⇒ [dividing both side by 9]

⇒ a = 5

and 4q + 5 = 17

⇒ 4q = 17 – 5

⇒ 4q = 12

⇒ [dividing both side by 4]

⇒ q = 3

We have, –5t – 60 = –70

⇒ –5t = –70 + 60

⇒ –5t = –10

⇒ [dividing both side by –5]

⇒ t = 2

and

⇒

⇒

⇒ 5p = 15 × 3

⇒ 5p = 45

⇒ [dividing both side by 5]

⇒ p = 9

We have, 3c = c + 12

⇒ 3c – c = 12

⇒ 2c = 12

⇒

⇒ c = 6

We have, 3(k + 1) = 24

⇒ [dividing both side by 3]

⇒ k + 1 = 8

⇒ k = 8 – 1

⇒ k = 7

By substituting the number for the letter, it equals, we get

Let the value of Δ and * be x and y, respectively and it is given that • = 3.

From (a), y + y + y + y + y = x + x + 3 + 3

⇒ 5y = 2x + 6

⇒ 5y – 2x = 6

⇒ 2x – 5y = –6 … (1)

From (b) x + x = y + y + 3 + 3

⇒ 2x = 2y + 6

⇒ 2x – 2y = 6

⇒ x – y = 3 [dividing both sides by 2] …. (2)

From (c), y + y + y + 3 + 3 + 3 = x + x + x

⇒ 3y + 9 = 3x

⇒ 3x – 3y = 9

⇒ x – y = 3 [Dividing both sides by 3] … (3)

From eq. (3) x – y = 3

⇒ x = y + 3

On putting x = y + 3 in Eq (1), we get

⇒ 2(y + 3) – 5y = –6

⇒ 2y + 6 – 5y = –6

⇒ –3y = –6 – 6

⇒ –3y = – 12

⇒ [Dividing both sides by –3]

⇒ y = 4

On putting y = 4 in eq (3) we get

x = y + 3

⇒ x = 4 + 3 = 7

∴ value of Δ = x = 7 and value of * = y = 4

Weight on LHS = 40 kg

Weight on RHS = 14 + 4 = 18kg

Weight should be equal,

Therefore, circle weight = 40 – 18 = 22kg