Chapter-2 Linear Equations In One Variable

Explanation:

It has been given, x – 2 = 7

Transfer –2 to right hand side, we get

x = 2 + 7 

x = 9 (result)

Explanation:

It has been given, y + 3 = 10

Transfer +3 to right hand side, we get

y = 10 –3

y = 7 (result)

Explanation:

Explanation:

It has been given, 6x = 12

Divide LHS and RHS by 6, we get

x = 12/6 (simplifying)

x = 2 (result)

Explanation:

It has been given, t/5 = 10

Multiplying LHS and RHS by 5, we get

t = 5 × 10

t = 50 (result)

Explanation:

It has been given, 2x/3 = 18

Multiplying LHS and RHS by 3, we get

2x = 3 × 18

2x = 54

Dividing LHS and RHS by 2, we get

x = 54/2

x = 27 (result)

Explanation:

It is been given, 1.6 = y/1.5

Above equation can also be written as-

y/1.5 = 1.6

Multiplying LHS and RHS by 1.5, we get

y = 1.5 × 1.6

y = 2.4 (result)

Explanation:

It has been given, 7x – 9 = 16

Transfer –9 to the RHS

7x = 16+9

7x = 25

Dividing LHS and RHS by 7, we get

x = 25/7 (result)

Explanation:

It has been given, 14y – 8 = 13

Transfer –8 to the RHS.

14y = 13+8

14y = 21

Dividing LHS and RHS by 14, we get

y = 21/14 (simplifying)

y = 3/2 (result)

Explanation:

It has been given, 17 + 6p = 9

Transfer 17 to the RHS.

6p = 9 – 17

6p = –8

Dividing LHS and RHS by 6, we get

p = –8/6 (simplifying)

p = –4/3 (result)

Explanation:

It has been given, x/3 + 1 = 7/15

Transfer 1 to the RHS.

x/3 = 7/15 – 1 (simplifying)

x/3 = (7 –15)/15 (simplifying)

x/3 = –8/15

Multiplying LHS and RHS by 3, we get

x = –8/15 × 3 (simplifying)

x = –8/5 (result)

Explanation:

Let the number to be determined is x.

As given in the question,

(x – 1/2) × 1/2 = 1/8 (simplifying)

x/2 – 1/4 = 1/8

Transfer –1/4 to the RHS.

x/2 = 1/4 + 1/8 (making the denominator same)

x/2 = 2/8 + 1/8 (simplifying)

x/2 = (2 + 1)/8

x/2 = 3/8

Multiplying LHS and RHS by 2, we get

x = (3/8) × 2 (simplifying)

x = 3/4 (result)

Explanation:

It has been given,

Perimeter of pool is equal to 154 m. 

Let the breadth of the pool is equal to b.

Let the length of the pool is equal to l.

As told in the question,

l = 2b + 2.

Perimeter is the summation of length of all sides. So, in case of rectangle it will be

Perimeter of pool = 2 × (length of pool + breadth of pool)

⇨ 2 × (l + b) = 154 m (putting value of l)

⇨ 2 × (2b + 2 + b) = 154

⇨ 2 × (3b + 2) = 154

⇨ 3b + 2 = 154/2 (Divide LHS and RHS by 2)

⇨ 3b = 77 – 2 (Transfer 2 to RHS)

⇨ 3b = 75 (Divide LHS and RHS by 3)

⇨ b = 75/3

⇨ b = 25 m (result)

Hence, Breadth of the pool = b = 25 m

Length of the pool = l = 2b + 2

⇨ l = (2 × 25) + 2 (Putting value of b)

⇨ l = 50 + 2

⇨ l = 52 m (result)

Explanation:

It has been given,

Base of the triangle, b = 4/3 cm

Perimeter of the triangle, s = 4215 cm = 62/15 cm

Let y be the length of sides which are equal.

Putting these values in perimeter calculation,

y + y + 4/3 = 62/15 cm (transfer 4/3 to the RHS.)

⇨ 2y = (62/15 – 4/3) cm (simplifying)

⇨ 2y = (62 – 20)/15 cm

⇨ 2y = 42/15 cm (dividing LHS and RHS by 2, we get)

⇨ y = (42/15) / (2) cm

⇨ y = 42/30 cm

⇨ y = 7/5 cm

The length of sides which are equal is 7/5 cm.

Explanation:

Let the first number be = a.

And the second number be = b.

It has been given that, b = a+15.

Also given, 

⇨ b + a = 95

⇨ a + a + 15 = 95 (transfer 15 to the RHS)

⇨ 2a = 95 – 15

⇨ 2a = 80 (dividing LHS and RHS by 2)

⇨ a = 80/2

⇨ a = 40

So, first number will be = a = 40

second number b = a + 15 = 40 + 15 = 55.

Explanation:

Assume first number be 5a and second number be 3a. So, the ratio of these numbers remains 5:3. 

It has been given that,

5a – 3a = 18

⇨ 2a = 18 (dividing LHS and RHS by 2)

⇨ a = 18/2

⇨ a = 9

So, first number = 5a = 5 × 9 = 45

So, second number = 3a = 3 × 9 = 27.

Answer- 45, 27.

Explanation:

Let us assume the three consecutive integers are equal to a, a+1 and a+2 respectively. It has been given that,

a + (a + 1) + (a + 2) = 51

⇨ 3a + 3 = 51

⇨ 3a = 51 – 3

⇨ 3a = 48 (dividing LHS and RHS by 3)

⇨ a = 48/3

⇨ a = 16 (first integer)

So, the three consecutive integers will be

a = 16

a + 1 = 17

a + 2 = 18

Answer- 16, 17, 18.

Explanation:

Let us assume the three consecutive multiples of 8 are equal to 8a, 8(a+1) and 8(a+2) respectively. It has been given that,

⇨ 8a + 8(a+1) + 8(a+2) = 888

⇨ 8 × (a + a+1 + a+2) = 888 (take 8 common from LHS)

⇨ 8 × (3a + 3) = 888 (dividing LHS and RHS by 8)

⇨ (3a + 3) = 888/8

⇨ (3a + 3) = 111

⇨ 3 × (a+1) = 111 (take 3 common from LHS)

⇨ a + 1 = 111/3 (dividing LHS and RHS by 3)

⇨ a = 111/3 – 1

⇨ a = (111 – 3)/3

⇨ a = 108/3

⇨ a = 36

So, the three assumed consecutive multiples of 8 will be-

First number, 8a = 8 × 36 = 288

Second number, 8(a + 1) = 8 × (36 + 1) = 8 × 37 = 296

Third number, 8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304

Answer- 288, 296, 304.

Explanation:

Let us assume the three consecutive integers are equal to a, a+1 and a+2. It has been given that,

2a + 3(a+1) + 4(a+2) = 74

⇨ 2a + 3a +3 + 4a + 8 = 74

⇨ 9a + 11 = 74 (transfer 11 to the RHS)

⇨ 9a = 74 – 11

⇨ 9a = 63 (dividing LHS and RHS by 9)

⇨ a = 63/9

⇨ a = 7

Hence, the assumed numbers will be-

First number, a = 7

Second number, a + 1 = 8

Third number, a + 2 = 9

Answers- 7, 8, 9.

Explanation:

Let the age of Rahul is equal to 5a and the age of Haroon is equal to 7a. So, the ratio of their ages remains 5:7.

So, after four years age of Rahul = 5a + 4, and the age of Haroon = 7a + 4.

It has been given that,

(7a + 4) + (5a + 4) = 56

⇨ 7a + 4 + 5a + 4 = 56

⇨ 12a + 8 = 56 (transfer 8 to the RHS)

⇨ 12a = 56 – 8

⇨ 12a = 48 (divide LHS and RHS by 12)

⇨ a = 48/12

⇨ a = 4

Hence, Rahul’s present age = 5a = 5×4 = 20

Haroon’s present age = 7a = 7×4 = 28

Answers- 20, 28.

Explanation:

Let us assume number of boys in class is equal to 7a and the number of girls is equal to 5a. So, the ratio remains 7:5.

It has been given that,

7a = 5a + 8

⇨ 7a – 5a = 8 (transfer 5x to the LHS)

⇨ 2a = 8 (divide LHS and RHS by 2)

⇨a = 8/2 

⇨ a = 4

So, the number of boys in class = 7a = 7×4 = 28

Also, the number of girls in class = 5a = 5×4 = 20

Hence, total strengths of the class will be equal to sum of number of boys in the class and number of girls in the class.

Total strengths = 20 + 28 = 48

Answer- 48.

Explanation:

Let us assume Baichung’s father age is equal to a.

So, Baichung’s grandfather will be = (a+26)

Then, Baichung’s age will be = (a–29).

It has been given that, 

a + (a+26) + (a–29) = 135

⇨ 3a – 29 + 26 = 135

⇨ 3a – 3 = 135

⇨ 3×(a–1) = 135 (taking 3 common)

⇨ a – 1 =135/3 (divide LHS and RHS by 3)

⇨ 3a = 138

⇨ a = 138/3

⇨ a = 46

So, Baichung’s father age will be = a = 46

And Baichung’s grandfather age will be = (a+26) = 46 + 26 = 72

Baichung’s age will be = (a–29) = 46 – 29 = 17

Explanation:

Let us assume the present age of Ravi is equal to a.

After fifteen years, his age would be a+15. It has been given that,

a+15 = 4a

⇨ 4a – a = 15 (above equation can also look like)

⇨ 3a = 15 (divide LHS and RHS by 3)

⇨ a = 15/3

⇨ a = 5

So, current age of Ravi would be = 5 years.

Explanation:

Let us assume the rational number is equal to x.

It has been given that,

x × (5/2) + 2/3 = –7/12 (first multiply by 5/2, then add 2/3, the result given is –7/12)

⇨ 5x/2 + 2/3 = –7/12 (transfer 2/3 to RHS)

⇨ 5x/2 = –7/12 – 2/3

⇨ 5x/2 = – (7 + 8)/12 (take –1 common)

⇨ 5x/2 = –15/12 (simplify)

⇨ 5x/2 = –5/4 (multiply LHS and RHS by 2/5)

⇨ x = (–5/4) × (2/5)

⇨ x = – 10/20 (simplify)

⇨ x = –1/2

So, the required rational number is equal to –1/2.

Explanation:

Let us assume the numbers of ₹100 notes to be 2a, ₹50 notes be 3a and ₹10 notes to be 5a, so the ratio of them remains 2:3:5.

Total cash value of ₹100 notes will be = 2a × 100 = 200a

Total cash value of ₹50 notes will be = 3a × 50 = 150a

Total cash value of ₹10 notes will be = 5a × 10 = 50a

It has been given that,

200a + 50a + 150a = 400000

⇨ 400a = 400000 (divide LHS and RHS by 400)

⇨ a = 400000/400

⇨ a = 1000

Total cash value of ₹100 notes will be = 2a = 2000

Total cash value of ₹50 notes will be = 3a = 3000

Total cash value of ₹10 notes will be = 5a = 5000

Explanation:

Let us assume the number of ₹5 coins is equal to a.

It has been given that,

The number of coins of value ₹2 = 3a

So, number of coins of value ₹1 will be = total coins – coins of value ₹2 – coins of value ₹5

Number of coins of value ₹1 = 160 – 3a – a = 160 – 4a. 

Total value of ₹5 coins will be= 5 × a = 5a

Total value of ₹2 coins will be = 2 × 3a = 6a

Total value of ₹1 coins will be = (160 – 4a) × 1 = (160 – 4a)

It has been given that,

6a + 5a + (160 – 4a) = 300

⇨ 11a + 160 – 4a = 300 (transfer 160 to RHS)

⇨ 7a = 140 (divide LHS and RHS by 7)

⇨ a = 140/7 (simplify)

⇨ a = 20

Total number of coins of value ₹5 = a = 20

Total number of coins of value ₹2 = 3a = 60

Total number of coins of value ₹1 = (160 – 4a) = 160 – 80 = 80

Explanation:

Let the number of participants who won = a.

So, the count of participants who were not winners = 63 – a (total – winning participants count)

Total money that was given to all the winners = 100 × a = 100a

Total money that was given to all the participant who were not winners = (63 – a) × 25

It has been given that,

100a + (63 – a) × 25 = 3000

⇨ 100a – 25a + 1575 = 3000

⇨ 75a = 3000 – 1575

⇨ 75a = 1425 (divide LHS and RHS by 75) 

⇨ a = 1425/75 (simplify)

⇨ a = 19

So, the count of participants who won is 19.

Explanation:

1.Solution-

3x = 2x + 18 (transfer 2x to from RHS to LHS)

⇨ 3x – 2x = 18 (simplify)

⇨ x = 18

Let’s put the value obtained of x from above in above equation and find LHS and RHS, 

LHS = 3 × 18 = 54

RHS = 2 × 18 + 18 = 36 +18 = 54

So, LHS = RHS (verified)

2.Solution-

5t – 3 = 3t – 5 (shift –3 to RHS)

⇨ 5t – 3t = –5 + 3 (solve)

⇨ 2t = –2 (divide LHS and RHS by 2)

⇨ t = –1

Let’s put the value obtained of t from above in above equation and find LHS and RHS, 

LHS = 5 × (–1) –3 = –5 – 3 = –8

RHS = 3× (–1) – 5 = –3 –5 = –8

So, LHS = RHS (verified)

3. Solution-

5x + 9 = 5 + 3x

⇨ 5x – 3x + 9 = 5 (transfer 3x from RHS to LHS)

⇨ 5x – 3x = 5 – 9 (transfer 9 from LHS to RHS)

⇨ 2x = –4 (divide LHS and RHS by 2)

⇨ x = –2

Let’s put the value obtained of x from above in above equation and find LHS and RHS, 

LHS = 5 × (–2) +9 = –10 + 9 = –1

RHS = 5 + 3× (–2) = 5 – 6 = –1 

So, LHS = RHS (verified)

4. Solution-

4z + 3 = 6 + 2z

⇨ 4z – 2z +3 = 6 (transfer 2z from RHS to LHS)

⇨ 2z = 6– 3 (transfer 3 from RHS to LHS)

⇨ 2z = 3 (divide LHS and RHS by 2)

⇨ z = 3/2

Let’s put the value obtained of z from above in above equation and find LHS and RHS, 

LHS = 4 × 3/2 + 3= 12/2 +3= 6 + 3 = 9

RHS = 6 + 2 × 3/2 = 6 + 6/2 = 6+3 = 9

So, LHS = RHS (verified)

5. Solution-

2x – 1 = 14 – x

⇒ 2x + x – 1 = 14 (transfer x from RHS to LHS) 

⇒ 3x = 14 +1 (transfer 1 from RHS to LHS)

⇒ 3x = 15 (divide LHS and RHS by 3)

⇨ x = 15/3 (simplify)

⇨ x = 5

Let’s put the value obtained of x from above in above equation and find LHS and RHS, 

LHS = 2×5 –1 = 10 – 1 = 9

RHS = 14 – 5 = 9 

So, LHS = RHS (verified)

6. Solution-

8x + 4 = 3× (x – 1) + 7

⇨ 8x + 4 = 7 + 3x – 3 (multiplied term in bracket) 

⇨ 8x + 4 = 3x + 4 (transfer 4 form RHS to LHS)

⇨ 8x = 4 – 4 + 3x (transfer 3x from RHS to LHS)

⇨ 8x – 3x = 0

⇨ 5x = 0 (divide LHS RHS by 5)

⇨ x = 0

Let’s put the value obtained of x from above in above equation and find LHS and RHS, 

LHS = 8×0 +4 = 0 + 4 = 4

RHS = 3(0 – 1) + 7= 0 – 3 +7 = 4

So, LHS = RHS (verified)

7. Solution

x = 4/5 (x + 10) (multiply the term on RHS)

⇨ x = 4x/5 + 40/5 (transfer 4x/5 from RHS to LHS)

⇨ x – (4x/5) = 8 (multiply LHS and RHS by 5)

⇨ 5x – 4x = 8 × 5

⇨ x = 8 × 5 (simplify)

⇨ x = 40

Let’s put the value obtained of x from above in above equation and find LHS and RHS, 

LHS = 40

RHS = 4/5 (40 +10) = 4/5 × 50 = 200/5 = 40

So, LHS = RHS (verified)

8. Solution-

2x/3 + 1 = 7x/15 + 3 (transfer 7x/15 from RHS to LHS and 1 form LHS to RHS)

⇨ 2x/3 – 7x/15 = 3 – 1 (make the denominator same in LHS)

⇨ (10x – 7x)/15 = 2 (simplify)

⇨ 3x/15 = 2 (multiply LHS and RHS by 15) 

⇨ 3x = 15 × 2

⇨ 3x = 30 (divide LHS and RHS by 3)

⇨ x = 30/3

⇨ x = 10

Let’s put the value obtained of x from above in above equation and find LHS and RHS,

LHS = (2/3) × 10 + 1 = 20/3 + 1 = 23/3

RHS = (7/15) × 10 +3 = 14/3 + 3 = 23/3

So, LHS = RHS (verified)

9. Solution-

2y + 5/3 = 26/3 – y

⇨ 2y = 26/3 – 5/3 – y (transfer 5/3 from LHS to RHS)

⇨ 2y + y = 26/3 – 5/3 (transfer y from RHS to LHS)

⇨ 3y = (26 – 5)/3 (simplify)

⇨ 3y = 21/3 (simplify)

⇨ 3y = 7 (divide LHS and RHS by 3)

⇨ y = 7/3

Let’s put the value obtained of y from above in above equation and find LHS and RHS, 

LHS = 2 × 7/3 + 5/3 = 14/3 +5/3 = 19/3

RHS = 26/3 – 7/3 = 19/3

So, LHS = RHS (verified)

10. Solution-

3m = 5m – 8/5 (transfer 3m from RHS to LHS)

⇨ 5m – 3m –8/5 = 0 (transfer 8/5 from LHS to RHS)

⇨ 2m = 8/5 (multiply LHS and RHS by 5)

⇨ 2m × 5 = 8

⇨ 10m = 8 (divide LHS and RHS by 10)

⇨ m = 8/10 (simplify)

⇨ m = 4/5

Let’s put the value obtained of m from above in above equation and find LHS and RHS,

LHS = 3 × 4/5 = 12/5

RHS = 5 × 4/5 – 8/5 = 20/5 – 8/5 = 12 /5

So, LHS = RHS (verified)

Explanation:

Let us assume the desired number is = a,

It has been given that,

(a – 5/2) × 8 = 3a (subtract 5/2 then multiply by 8)

⇨ 8a – 40/2 = 3a (transfer 40/2 from LHS to the RHS)

⇨ 8a = 40/2 – 3a (transfer 3a from RHS to LHS

⇨ 8a – 3a = 20

⇨ 5a = 20 (divide LHS and RHS by 5)

⇨ a = 20/5

⇨ a = 4

So, the desired number will be 4.

Explanation:

Let the first number which is positive is = a and the second number is = b. It has been given that,

b = 5a          …….(i)

5a + 21 = 2(a + 21)         ……..(ii)

⇨ 5a + 21 = 2a + 42 (simplify eqn (ii))

⇨ 5a – 2a = 42 – 21 (transfer 2x from RHS to LHS)

⇨ 3a = 21 (multiply LHS and RHS by 1/3)

⇨ a = 7

So, first number = a = 7

And second number = y = 5a = 7×5 = 35. 

Answer- 7, 35.

Explanation:

Let the two-digit number is of the form “pq”.

As given in the question q = 9 – p

Actual two-digit number will be = 10p + q = 10p + (9 – p)

New number formed after interchanging the digits will be “qp” = 10q + p = 10(9–p) + p

It has been given that,

10p + (9–p) + 27 = 10(9–p) + p (simplify)

⇨ 10p + 9 – p + 27 = 90 – 10p + p

⇨ 9p + 36 = 90 – 9p (transfer 9p from RHS to LHS and 36 from LHS to RHS)

⇨ 9p + 9p = 90 – 36

⇨ 18p = 54 (divide LHS and RHS by 18)

⇨ p = 3

So, q = 9 – p = 9 – 3 = 6.

So, the desired two-digit number is of the form “36”. 

Hence, the desired number is equal to 36.

Explanation:

Let the two-digit number is of the form “pq”.

As given in the question q = 3p

Actual two-digit number will be = 10p + q = 10p + 3p = 13p

New number formed after interchanging the digits will be “qp” = 10q + p = 30p + p = 31p

It has been given that,

(31p) + (13p) = 88

⇨ 44p = 88 (divide LHS and RHS by 44)

⇨ p = 88/44

⇨ p = 2.

So, q = 3p = 3×2 = 6

Hence, the desired number “pq” = “26”.

Explanation:

Let us assume Shobo’s present age is = a

Let her mother’s age = b = 6a

After 5 years, Shobo’s age is = a + 5

It has been given that,

(a + 5) = (1/3) × b

⇨ (a+5) = 1/3 × 6a (simplify)

⇨ a + 5 = 2a (transfer a from LHS to RHS)

⇨ 2a – a = 5 

⇨ a = 5

So, Shobo’s current age is = a = 5 years.

Shobo’s mother current age = b = 6a = 6×5 = 30 years.

Explanation:

Let us assume the length the plot is = 11a

Let us take breadth = 4a, so that their ratio remains 11:4

It has been given,

Cost of fencing the plot for 1 metre = ₹100

Combined cost of fencing of the plot = ₹75000

We know, 

Perimeter of the rectangle = 2×(l+b) = 2×(11a + 4a) = 15a×2 = 30a

Combined cost of fencing of 30a metres of length = ₹ (30a × 100)

Also, it is given that,

₹ (30a × 100) = ₹75000

⇨ 3000a = 75000 (divide LHS and RHS by 3000)

⇨ a = 75000/3000 (simplify)

⇨ a = 25

So, the plot’s length = 11a = 25×11 = 275 m

And, plot’s breadth = 4a = 25×4 = 100 m.

Answer- 275, 100.

Explanation:

As given in the question the ratio of shirt material and trouser material bought is 3:2

Let us assume length of material for shirt is = 3a m 

And length of material for trouser = 2a m.

For a profit of 12%, price of 1 metre of shirt material is = ₹ 50×(1 + (12/100)) = ₹ 56

For a profit of 10%, price of 1 metre of trouser material is = ₹ 90×(1 + (10/100)) = ₹ 99

Combined sale done = ₹36600 (given)

It has been given that,

(3a × 56) + (2a × 99) = 36600

⇨ 168a + 198a = 36600

⇨ 366a = 36600 (divide LHS and RHS by 366)

⇨ a = 36600/366

⇨ a = 100

So, the length of trouser material bought by him = 2a = 100 × 2 = 200 m.

Explanation:

Let us assume the total number of deer in the herd = a.

So, Number of deer that are grazing in the field is = a/2

So, Number of deer that are playing nearby = 3/4 × (a/2) = 3a/8

So, Number of deer that are drinking water = 9

It has been given that,

3a/8 + a/2 + 9 = a (transfer a from RHS to LHS)

(4a + 3a)/8 + 9 – a = 0

⇨ 7a/8 – a = –9 (multiply LHS and RHS by –1)

⇨ a – 7a/8 = 9 (simplify)

⇨ (8a – 7a)/8 = 9 (multiply LHS and RHS by 8)

⇨ a = 9 × 8

⇨ a = 72

So, the total number of deer that are present in the herd is 72.

Answer- 72

Explanation:

Let us assume granddaughter’s age = a. 

Then, the grandfather’s age = 10a.

Since, grandfather is 54 years older than his granddaughter. So,

a + 54 = 10a

⇨ 10a – a = 54 (transferred a from LHS to RHS)

⇨ 9a = 54 (divide LHS and RHS by 9)

⇨ a = 6

So, granddaughter’s age in present is = a = 6 = 6 years.

And grandfather’s age in present is = 10a = 6×10 = 60 years.

Answer- 6, 60.

Explanation:

Let us assume Aman’s son age is = a, so Aman’s age = 3a.

It has been given that, ten years back their age are 

3a – 10 = 5 × (a – 10)

⇨ 3a – 10 = 5a – 50 (transfer 3a from RHS to LHS)

⇨ –10 = 5a – 3a –50 (transfer –50 to the other side)

⇨ 2a = 50 – 10

⇨ 2a = 40 (divide LHS and RHS by 2)

⇨ a = 20

Current age of Aman’s son is = a = 20 years

Current age of Aman = 3a = 20×3 = 60 years.

Answer- 20, 60.

Explanation:

1) Solution-

x/2 – 1/5 = x/3 + 1/4 (transfer x/3 from RHS to LHS) 

⇨ x/2 – x/3 = 1/5 + 1/4 (make the denominator of term in RHS same) 

⇨ (1/2 – 1/3) × x = (4 + 5)/20 (taking x common form LHS)

⇨ (3 – 2)x/6 = (4 + 5)/20 (multiply LHS and RHS by 6)

⇨ x = 9/20 × 6

⇨ x = 54/20 (simplify)

⇨ x = 27/10

Answer- 27/10.

2) Solution-

n/2 – 3n/4 + 5n/6 = 21 (make denominator same in LHS)

⇨ (10n – 9n + 6n)/12 = 21

⇨ 7n/12 = 21 (multiply LHS and RHS by 12)

⇨ 7n = 21 × 12 (divide LHS and RHS by 7)

⇨ n = 252/7 (simplify)

⇒ n = 36

Answer- 36

3) Solution-

x + 7 – 8x/3 = 17/6 – 5x/2 (transfer 7 from LHS to RHS)

⇨ x – 8x/3 = 17/6 – 7 – 5x/2 (transfer 5x/2 from RHS to LHS)

⇨ x + 5x/2 – 8x/3 = 17/6 – 7

⇨ (6x + 15x – 16x)/6 = 17/6 – 7

⇨ 5x/6 = (17 – 42)/6

⇨ 5x/6 = –25/6 (multiply LHS and RHS by 6)

⇨ 5x = – 25 (divide LHS and RHS by 5)

⇨ x = –25/5

⇨ x = –5

4) Solution-

(x – 5)/3 = (x – 3)/5 (multiply LHS and RHS by 15)

⇨ 15×(x–5)/3 = 15×(x–3)/5

⇨ 5 ×(x–5) = 3×(x–3) (simplify terms in LHS and RHS)

⇨ 5x–25 = 3x–9 (transfer 3x from RHS to LHS)

⇨ 5x – 3x –25 = –9 (transfer –25 from LHS to RHS)

⇨ 2x = 25 – 9

⇨ 2x = 16 (divide LHS and RHS by 2)

⇨ x = 16/2

⇨ x = 8

5) Solution-

(3t – 2)/4 – (2t + 3)/3 = 2/3 – t (multiply LHS and RHS by 12)

⇨ 12 × ((3t – 2)/4)  – 12×((2t + 3)/3) = 12×(2/3 – t)

⇨ 3 × (3t – 2) – 4 × (2t + 3) = 4 × 2 – 12t (simplify)

⇨ 9t – 6 – 8t – 12 = 8 – 12t (transfer –12t from RHS to LHS)

⇨ 9t – 8t – 18 + 12t = 8 (transfer –18 from LHS to RHS)

⇨ 13t = 8 + 18

⇨ 13t = 26 (divide LHS and RHS by 13)

⇨ t = 2

Answer- 2.

6) Solution-

m – (m – 1)/2 = 1 – (m – 2)/3 (multiply LHS and RHS by 6)

⇨ 6m – 6×(m – 1)/2= 6 – 6×(m – 2)/3

⇨ 6m – 3m + 3 = 6 – 2m + 4 (transfer –2m from RHS to LHS)

⇨ 3m + 2m + 3 = 6 + 4 (transfer 3 from LHS to RHS)

⇨ 5m = 10 – 3

⇨ 5m = 7 (divide LHS and RHS by 5)

⇨ m = 7/5

Explanation:

1. Solution-

3(t – 3) = 5(2t + 1) (simplify terms in LHS and RHS)

⇨ 3t – 9 = 10t + 5 (transfer 10t from RHS to LHS)

⇨ 3t – 10t – 9 = 5 (transfer –9 from RHS to LHS)

⇨ –7t = 5 + 9

⇨ –7t = 14 (divide LHS and RHS by –7)

⇨ t = –2

2.Solution-

15(y – 4) –2(y – 9) + 5(y + 6) = 0 (expand all the terms)

⇨ 15y – 60 – 2y + 18 + 30 + 5y = 0

⇨ 15y + 3y + 48 – 60 = 0

⇨ 18y – 12 = 0 (transfer –12 from LHS to RHS)

⇨ 18y = 12 (divide LHS and RHS by 18)

⇨ y = 12/18

⇨ y = 2/3

3. Solution-

3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17 (simplify all the terms)

⇨ 15z – 21 – 18z + 22 = 32z – 52 – 17

⇨ –3z – 21 + 22 = 32z – 69

⇨ –3z + 1 = 32z – 69 (transfer 32z from RHS to LHS)

⇨ –3z – 32z +1 = –69 (transfer 1 from LHS to RHS)

⇨ –35z = –1 – 69

⇨ –35z = –70 (multiply LHS and RHS by –1)

⇨ 35z = 70 (divide LHS and RHS by 35)

⇨ z = 70/35

⇨ z = 2

4. Solution-

0.25(4f – 3) = 0.05(10f – 9)

⇨ f – 3×0.25 = 0.5f – 9×0.05

⇨ f – 0.75 = 0.5f – 0.45 (transfer 0.5f from RHS to LHS)

⇨ f – 0.5f – 0.75 = –0.45 (transfer –0.75 from LHS to RHS)

⇨ 0.5f = 0.75 – 0.45

⇨ 0.5f = 0.30 (divide LHS and RHS by 0.5)

⇨ f = 0.30/0.5

⇨ f = 3/5

⇨ f = 0.6

Explanation:

1. Solution-

(8x – 3)/3x = 2 (multiply LHS and RHS by 3x)

⇨ 3x × ((8x – 3)/3x) = 2 × 3x

⇨ 8x – 3 = 6x (transfer 6x from RHS to LHS)

⇨ 8x – 6x – 3 = 0 (transfer –3 from LHS to RHS)

⇨ 2x = 3 (divide LHS and RHS by 2)

⇨ x = 3/2

2. Solution-

9x/(7 – 6x) = 15 (multiply LHS and RHS by (7–6x))

⇨ 9x = 15×(7 – 6x) (simplify)

⇨ 9x = 105 – 90x (transfer –90x from RHS to LHS)

⇨ 90x + 9x = 105

⇨ 99x = 105 (divide LHS and RHS by 99)

⇨ x = 105/99 (simplify)

⇨ x = 35/33

3. Solution-

z/(z + 15) = 4/9 (multiply LHS and RHS by (z+15))

⇨ z = (z + 15) × 4/9 (multiply LHS and RHS by 9)

⇨ 9z = (z + 15) × 4 (simplify)

⇨ 9z = 60 + 4z (transfer 4z from RHS to LHS)

⇨ 9z – 4z = 60

⇨ 5z = 60 (divide LHS and RHS by 5)

⇨ z = 60/5

⇨ z = 12

4. Solution-

(3y + 4)/(2 – 6y) = –2/5 (multiply RHS and LHS by (2–6y))

⇨ 3y + 4 = –2/5 × (2 – 6y) (multiply LHS and RHS by 5)

⇨ 5×(3y + 4) = –2 × (2 – 6y)

⇨ 20 + 15y = – 4 + 12y (transfer 12y from RHS to LHS)

⇨ 15y – 12y + 20 = –4 (transfer 20 from LHS to RHS)

⇨ 3y = –20 – 4

⇨ 3y = –24 (divide LHS and RHS by 3)

⇨ y = –24/3

⇨ y = –8

5. Solution-

(7y + 4)/(y + 2) = –4/3 (multiply LHS and RHS by (y+2))

⇨ 7y + 4 = –4/3 × (y + 2) (multiply LHS and RHS by –3)

⇨ –3×(7y + 4) = 4×(y + 2)

⇨ –21y – 12 = 4y + 8 (transfer 4y from RHS to LHS)

⇨ –21y – 4y –12 = 8 (transfer –12 from LHS to RHS)

⇨ –25y = 12 + 8

⇨ –25y = 20 (divide RHS and LHS by –25)

⇨ y = 20/–25

⇨ y = –4/5

Explanation:

Let us assume Hari’s age is = 5a

And, let Harry’s age is = 7a. So, their ration remains 5:7 

Hari’s age after four years = 5a + 4

Harry’s age after four years = 7a + 4

It has been given that,

(5a + 4)/(7a + 4) = 3/4 (multiply LHS and RHS by (7a+4))

⇨ (5a + 4) = 3/4 × (7a + 4) (multiply LHS and RHS by 4)

⇨ 4 × (5a + 4) = 3 × (7a + 4)

⇨ 20a + 16 = 21a + 12 (transfer 21a from RHS to LHS)

⇨ 20a – 21a + 16 = 12 (transfer 16 from LHS to RHS)

⇨ –a = 12 – 16

⇨ –a = –4 (multiply LHS and RHS by –1)

⇨ a = 4

So, current age of Hari is = 5a = 4 × 5 = 20 years

And, current age of Harry is = 7a = 4 × 7 = 28 years

Answer- 20, 28

Explanation:

Let us assume the numerator is = a, and let us assume the denominator is = b = (a + 8)

It has been given that,

(a + 17)/(b – 1) = 3/2

⇨ (a + 17)/(a + 8 – 1) = 3/2

⇨ (a + 17)/(a + 7) = 3/2 (multiply LHS and RHS by (a+7))

⇨ a + 17 = 3/2 × (a + 7) (multiply LHS and RHS by 2)

⇨ 2 × (a + 17) = 3 × (a + 7) (simplify)

⇨ 2a + 34 = 3a + 21 (transfer 3a from RHS to LHS)

⇨ 2a – 3a + 34 = 21 (transfer 34 from LHS to RHS)

⇨ –a = 21 – 34

⇨ –a = –13 (multiply LHS and RHS by –1)

⇨ a = 13

So, The desired rational number is = a/b = a/(a+8) = 13/(13+8) = 13/21