x − 4 = 3

9m = 81

2a + 4 = 0

3 − y = 4

17p – 2 = 49

Adding 2 to both sides

∴ 17p – 2 + 2 = 49 + 2

∴ 17p = 51

Dividing both sides by 17

\(\frac{17p}{17}\) = \(\frac{51}{17}\)

∴ p = 3

Hence, p = 3 is the solution of the equation.

2m + 7 = 9

Subtracting 7 from both sides

∴ 2m + 7 – 7 = 9 − 7

∴ 2m = 2

Dividing both sides by 2

\(\frac{2m}{2}\) = \(\frac{2}{2}\)

∴ m = 1

Hence, m = 1 is the solution of the equation.

3x + 12 = 2x − 4

Subtracting 12 from both sides

∴ 3x + 12 – 12 = 2x – 4 − 12

∴ 3x = 2x − 16

Subtracting 2x from both sides

3x − 2x = 2x −16 − 2x

∴ x = − 16

Hence, x = − 16 is the solution of the equation.

5(x − 3) = 3(x + 2)

∴ 5x – 15 = 3x + 6

Adding 15 to both sides

∴ 5x – 15 + 15 = 3x + 6 + 15

∴ 5x = 3x + 21

Subtracting 3x from both sides

∴ 5x − 3x = 3x + 21 − 3x

∴ 2x = 21

Dividing both sides by 2

\(\frac{2x}{2}\) = \(\frac{21}{2}\)

∴ x = \(\frac{21}{2}\)

Hence, x = \(\frac{21}{2}\) is the solution of the equation.

\(\frac{9x}{8}\) + 1 = 10

Multiplying both sides by 8

∴ \(\frac{9x}{8}\) × 8 + 1 × 8 = 10 × 8

∴ 9x + 8 = 80

Subtracting 8 from both sides

∴ 9x + 8 – 8 = 80 − 8

∴ 9x = 72

Dividing both sides by 9

\(\frac{9x}{9}\) = \(\frac{72}{9}\)

∴ x = 8

Hence, x = 8 is the solution of the equation.

\(\frac{y}{7}+\frac{y-4}{3}\) = 2

Multiplying both the sides by 21,

\(\frac{y}{7}×21+\frac{y-4}{3}×21\) = 2 × 21

∴ 3y + 7(y − 4) = 42

∴ 3y + 7y – 28 = 42

Adding 28 to both the sides,

10y – 28 + 28 = 42 + 28

∴ 10y = 70

Dividing both the sides by 10,

\(\frac{10y}{10}\) = \(\frac{70}{10}\)

∴ y = 7.

Hence, y = 7 is the solution of the equation.

13x − 5 = \(\frac{3}{2}\)

Multiplying both sides by 2

2(13x − 5) = 2 × \(\frac{3}{2}\)

∴ 26x – 10 = 3

Adding 10 on both sides

26x – 10 + 10 = 3 + 10

26x = 13

Dividing both sides by 26

\(\frac{26x}{26}\) = \(\frac{13}{26}\)

x = \(\frac{1}{2}\)

Hence, x = \(\frac{1}{2}\)  is the solution of the equation.

3(y + 8) = 10(y − 4) + 8.

3y + 24 = 10y – 40 + 8

∴  3y + 24 = 10y − 32

Subtracting 24 from both the sides,

3y + 24 – 24 = 10y – 32 −24

∴  3y = 10y − 56

Subtracting 3y from both the sides,

3y−3y = 10y – 56 − 3y

∴  0 = 7y − 56 i.e. 7y – 56 = 0

Adding 56 to both the sides,

7y – 56 + 56 = 0 + 56

∴  7y = 56

Dividing both the sides by 7,

\(\frac{7y}{7}\) = \(\frac{56}{7}\)

y = 8

Hence, y = 8 is the solution of the equation.

\(\frac{x-9}{x-5}\) = \(\frac{5}{9}\)

∴ 7(x − 9) = 5(x − 5)                    ….(Cross multiplying)

∴ 7x – 63 = 5x − 25

∴ 7x − 63 + 63 = 5x – 25 + 63      ... [Adding 63 on both sides]

∴ 7x = 5x + 38

∴ 7x − 5x = 5x + 38 − 5x              ... [Subtracting 5x from both sides]

∴ 2x = 38

\(\frac{2x}{2}\) = \(\frac{38}{2}\)

x = 19

Hence, x = 19 is the solution of the equation.

\(\frac{y-4}{3}\) + 3y = 4

Multiplying both sides by 3

\(3(\frac{y-4}{3})\)+ 3(3y) = 4 × 3

∴ y – 4 + 9y = 12

∴ 10y − 4 = 12

Adding 4 to both sides

∴ 10y – 4 + 4 = 12 + 4

∴ 10y = 16

Dividing both sides by 10

\(\frac{10y}{10}\) = \(\frac{16}{10}\)

∴ y = \(\frac{8}{5}\)

Hence, y = \(\frac{8}{5}\) is the solution of the equation.

\(\frac{b+(b+1)+(b+2)}{4}\) = 21

Multiplying both sides by 4

∴ 4 × \(\frac{b+(b+1)+(b+2)}{4}\)  = 4 × 21

∴ b + (b + 1) + (b + 2) = 84

∴ 3b + 3 = 84

Subtracting 3 from both sides

∴ 3b + 3 − 3 = 84 − 3

∴ 3b = 81

Dividing both sides by 3

\(\frac{3b}{3}\) = \(\frac{81}{3}\)

∴ b = 27

Hence, b = 27 is the solution of the equation.

Let the present age of the son be x years.

Age of the mother = (x + 25) years

After 8 years,

Age of son = (x + 8) years

Age of mother = (x + 25 + 8) years = (x + 33) years

From the given information,

\(\frac{son\,sge}{mother\,age}\) = \(\frac{4}{9}\)

∴ \(\frac{x+8}{x+33}\) = \(\frac{4}{9}\)

∴ 9(x + 8) = 4(x + 33)

∴ 9x + 72 = 4x + 132

∴ 9x − 4x = 132 − 72

∴ 5x = 60

∴ x = 60/5

∴ x = 12

Hence, the present age of the son is 12 years.

Let the numerator of the fraction be x.

∴ Denominator = x + 12

∴ The fraction = \(\frac{x}{x+12}\)

If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction obtained is

From the given information,

\(\frac{x-2}{x+12+7}\) = \(\frac{1}{2}\)

2(x − 2) = 1(x + 12 + 7)

2x – 4 = x + 19

2x − x = 19 + 4

∴ x = 23

∴ Numerator of the fraction is 23.

∴ Denominator of the fraction is 23 + 12 = 35.

Hence, the fraction is \(\frac{23}{35}\)

The ratio of the weights of copper and zinc in brass is 13 : 7.

Let the weight of copper be 13x g.

Then the weight of zinc is 7x g.

From the given condition,

13x + 7x = 700

∴ 20x = 700

∴ x = 700/20

∴ x = 35.

The weight of zinc is 7x g.

∴ the weight of zinc = 7 × 35 = 245 g

The weight of zinc in a brass utensil is 245 g.

The difference between the consecutive whole numbers is 1.

Let the three consecutive whole numbers be x, x + 1 and x + 2.

Now, x + (x + 1) + (x + 2) > 45

∴ 3x + 3 > 45

∴ 3x > 45 − 3

∴ 3x > 42

∴ x > 14

∴ x = 15, x + 1 = 15 + 1 = 16 and x + 2 = 15 + 2 = 17

OR

x + x + 1 + x + 2 < 54

3x + 3 < 54

∴ 3x < 54 − 3

∴ 3x < 51

∴ x < 17

∴ x = 16,

x + 1 = 16 + 1 = 17,

x + 2 = 16 + 2 = 18

The three consecutive whole numbers are 15, 16, 17 or 16, 17, 18.

Let the digit at the units place be x.

The digit at the tens place is twice the digit at the units place.

∴ the digit at the tens place is 2x.

The number = 10 × 2x + x = 20x + x = 21x             ... (1)

The number obtained by interchanging the digits.

x at tens place and 2x at units place

∴ the new number = 10x + 2x = 12x                       ... (2)

From the given condition,

21x + 12x = 66                    ….[From (1) and (2)]

∴ 33x = 66

x = 66/33

∴ x = 2

The number is 21x = 21 x 2 = 42

The original number is 42.

Let the number of ₹ 100 tickets sold be x.

The number of tickets of ₹ 200 sold was 20 more than the number of tickets of ₹ 100 sold.

∴ the number of ₹ 200 tickets sold is (x+ 20)

The amount received by the sale of ticket is 100x + 200(x + 20).

This amount is ₹ 37,000.

∴ 100x + 200(x + 20) = 37000

∴ x + 2(x + 20) = 370          ... (Dividing both the sides by 100)

∴ x + 2x + 40 = 370

∴ 3x = 370 − 40

∴ 3x = 330

∴ x = 330/3

∴ x = 110.

The number of ₹ 100 tickets sold

Let the three consecutive natural numbers be x, x + 1 and .x + 2.

x is the smallest and .x + 2 is the greatest.

From the given condition, 5 times the smallest = 5.x and four times the greatest = 4(x + 2).

5x is greater than 4(x + 2) by 9.

∴ 5x = 4(x + 2) + 9

∴ 5x = 4x + 8 + 9

∴ 5x − 4x = 17

∴ x = 17.

x + 1 = 17 + 1 = 18 and x + 2 = 17 + 2 = 19.

The three consecutive natural numbers are 17, 18 and 19 respectively.

Let the cost price of the bicycle for Raju = ₹ x

The selling price of bicycle for Raju = Cost price of bicycle + Profit

= x + 8% on cost price

= x + \(\frac{8}{100}\) × x

= x + \(\frac{8x}{100}\)

∴ The selling price of the bicycle for Raju = Purchase price of bicycle for amit

∴ Purchase price of bicycle for Amit = x + \(\frac{8x}{100}\)       ... (i)

Now selling price of bicycle for Amit = purchase price of bicycle for amit + Reparing

= x + \(\frac{8x}{100}\) + 54            …..from (i) and given

∴ Selling price of bicycle for Amit = purchase price of bicycle for Nikhil

∴ Purchase price of bicycle for Nikhil = x + \(\frac{8x}{100}\) + 54

∴ 1134 = x + \(\frac{8x}{100}\) + 54

x + \(\frac{8x}{100}\) = 1134 − 54

\(\frac{100x+8x}{100}\) = 1080

∴ 108x = 1080 × 100

∴ x = \(\frac{1080 × 100}{108}\)

∴ x = ₹ 1000

The cost price of bicycle for which Raju purchased = ₹ 1000

Runs scored in First match = 180

Runs scored in Second match = 257

Let's cricketer should score x runs in the 3rd match

Average of run in 3 matches = \(\frac{\text{sum of all the runs in 3 matches}}{\text{no of matches}}\)

230 = \(\frac{180+257+x}{3}\)

230 = \(\frac{437+x}{3}\)

230 × 3 = 437 + x

∴ x = 690 − 437

∴ x = 253

The cricketer should score 253 runs in the 3rd match.

Let Viru's present age be x years.

Sudhir's present age = (5 + 3x) years

Anil's present age = \(\frac{5+3x}{2}\) years

∴ From the given information,

(5 + 3x) + (x) : 3\((\frac{5+3x}{2})\) = 5:6

\(\frac{5+4x}{3(\frac{5+3x}{2})}\) = \(\frac{5}{6}\)

\(\frac{2(5+4x)}{3(5+3x)}\) = \(\frac{5}{6}\)

\(\frac{10+8x}{15+9x}\) = \(\frac{5}{6}\)

6(10 + 8x) = 5(15 + 9x)        ….(cross multiplying)

60 + 48x = 75 + 45x

48x − 45x = 75 – 60

3x = 15

∴ x = 5

Viru's age = 5 years