**Chapter 12 – Equations in one variable**

**Practice set 12.1**

**Practice set 12.1**

**1. Each equation is followed by the values of the variable. Decide whether these values are the solutions of that equation.**

**(1) x – 4 = 3 , x = –1, 7, –7 **

**(1) x – 4 = 3 , x = –1, 7, –7****Solution:**

**(a) When x = –1**

**Substituting x = –1 in LHS of the given equation**

**LHS = x – 4**

** = –1 – 4**

**∴ LHS = –5**

**But, RHS = 3**

** **

**∴ LHS ≠ RHS**

** **

**x = –1 is not the solution of the given equation.**

** **

**(b) When x = 7**

**Substituting x = 7 in LHS of the given equation**

**LHS = x – 4**

** = 7 – 4**

**∴ LHS = 3**

**And, RHS = 3**

** **

**∴ LHS = RHS**

** **

**x = 7 is the solution of the given equation.**

** **

**(c) When x = –7**

**Substituting x = –7 in LHS of the given equation**

**LHS = x – 4**

** = –7 – 4**

**∴ LHS = –11**

**But, RHS = 3**

** **

**∴ LHS ≠ RHS**

** **

**x = –7 is not the solution of the given equation.**

** **

**Ans:** x = –1 is not the solution, x = 7 is the solution and x = –7 is not the solution of the given equation.

**(2) 9m = 81, m = 3, 9, –3**

**(2) 9m = 81, m = 3, 9, –3****Solution:**

**(a) When m = 3**

**Substituting m = 3 in LHS of the given equation**

**LHS = 9m **

** = 9 × 3**

**∴ LHS = 27**

**But, RHS = 81**

** **

**∴ LHS ≠ RHS**

** **

**m = 3 is not the solution of the given equation.**

** **

**(b) When m = 9**

**Substituting m = 9 in LHS of the given equation**

**LHS = 9m **

** = 9 × 9**

**∴ LHS = 81**

**And, RHS = 81**

** **

**∴ LHS ≠ RHS**

** **

**m = 81 is the solution of the given equation.**

** **

**(c) When m = –3**

**Substituting m = 3 in LHS of the given equation**

**LHS = 9m **

** = 9 × (–3)**

**∴ LHS = –27**

**But, RHS = 81**

** **

**∴ LHS ≠ RHS**

** **

**m = –3 is not the solution of the given equation.**

** **

**Ans:** m = 3 is not the solution, m = 81 is the solution and m = –3 is not the solution of the given equation.

**(3) 2a + 4 = 0, a = 2, –2, 1 **

**(3) 2a + 4 = 0, a = 2, –2, 1****Solution:**

**(a) When a = 2**

**Substituting a = 2 in LHS of the given equation**

**LHS = 2a + 4 **

** = 2(2) + 4**

** = 4 + 4**

**∴ LHS = 8**

**But, RHS = 0**

** **

**∴ LHS ≠ RHS**

** **

**a = 2 is not the solution of the given equation.**

** **

**(b) When a = –2**

**Substituting a = 2 in LHS of the given equation**

**LHS = 2a + 4 **

** = 2(–2) + 4**

** = –4 + 4**

**∴ LHS = 0**

**And, RHS = 0**

** **

**∴ LHS = RHS**

** **

**a = –2 is the solution of the given equation.**

** **

**(c) When a = 1**

**Substituting a = 2 in LHS of the given equation**

**LHS = 2a + 4 **

** = 2(1) + 4**

** = 2 + 4**

**∴ LHS = 6**

**But, RHS = 0**

** **

**∴ LHS ≠ RHS**

** **

**a = 1 is not the solution of the given equation.**

** **

**Ans:** a = 2 is not the solution, a = –2 is the solution and a = 1 is not the solution of the given equation.

**(4) 3 – y = 4, y = –1, 1, 2 **

**(4) 3 – y = 4, y = –1, 1, 2****Solution:**

**(a) When y = –1**

**Substituting y = –1 in LHS of the given equation**

**LHS = 3 – y **

** = 3 – (–1)**

** = 3 + 1**

**∴ LHS = 4**

**And, RHS = 4**

** **

**∴ LHS = RHS**

** **

**y = –1 is the solution of the given equation.**

** **

**(b) When y = 1**

**Substituting y = 1 in LHS of the given equation**

**LHS = 3 – y **

** = 3 – 1**

**∴ LHS = 2**

**But, RHS = 4**

** **

**∴ LHS ≠ RHS**

** **

**y = 1 is not the solution of the given equation.**

** **

**(c) When y = 2**

**Substituting y = 2 in LHS of the given equation**

**LHS = 3 – y **

** = 3 – 2**

**∴ LHS = 1**

**But, RHS = 4**

** **

**∴ LHS ≠ RHS**

** **

**y = 2 is not the solution of the given equation.**

** **

**Ans:** y = –1 is the solution, y = 1 is not the solution and y = 2 is not the solution of the given equation.

**2. Solve the following equations.**

**2. Solve the following equations.****(1) 17p – 2 = 49 **

**Solution:**

**17p – 2 = 49**

**∴ 17p = 49 + 2**

**∴ 17p = 51**

**∴ p = \(\large \frac {51}{17}\)**

**∴ p = 3**

** **

**Ans:** The value of p is 3.

**(2) 2m + 7 = 9 **

**Solution:**

**2m + 7 = 9 **

**∴ 2m = 9 – 7**

**∴ 2m = 2**

**∴ m = \(\large \frac {2}{2}\)**

**∴ m = 1**

** **

**Ans:** The value of m is 1.

**(3) 3x + 12 = 2x – 4 **

**Solution:**

**3x + 12 = 2x – 4 **

**∴ 3x – 2x = – 4 – 12**

**∴ x = – 16**

** **

**Ans:** The value of x is 16.

**(4) 5(x – 3) = 3(x + 2)**

**Solution:**

**5(x – 3) = 3(x + 2)**

**∴ 5x – 15 = 3x + 6**

**∴ 5x – 3x = 6 + 15**

**∴ 2x = 21**

**∴ x = \(\large \frac {21}{2}\)**

** **

**Ans:** The value of x is \(\large \frac {21}{2}\).

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

**Solution:**

**\(\large \frac {9x}{8}\) = 10 – 1**

**∴ \(\large \frac {9x}{8}\) = 9 **

**∴ x = \(\large \frac {9\, ×\, 8}{9}\)**

**∴ x = 8 **

** **

**Ans:** The value of x is 8.

**(6) \(\large \frac {y}{7}\) + \(\large \frac {y\, –\, 4}{3}\) = 2**

**Solution:**

**\(\large \frac {y}{7}\) + \(\large \frac {y\, –\, 4}{3}\) = 2**

**∴ \(\large \frac {(3\, ×\, y)\, +\, 7(y\, –\, 4)}{7\, ×\, 3}\) = 2**

**∴ \(\large \frac {3y\, +\, 7y\, –\, 28}{21}\) = 2**

**∴ 10y – 28 = 2 × 21**

**∴ 10y – 28 = 42**

**∴ 10y = 42 + 28**

**∴ 10y = 70**

**∴ y = \(\large \frac {70}{10}\)**

**∴ y = 7**

**Ans:** The value of y is 7.

**(7) 13x – 5 = \(\large \frac {3}{2}\)**

**Solution:**

**13x – 5 = \(\large \frac {3}{2}\)**

**∴ 13x = \(\large \frac {3}{2}\) + 5**

**∴ 13x = \(\large \frac {3\,+\,5\,×\,2}{2}\)**

**∴ 13x = \(\large \frac {13}{2}\)**

**∴ x = \(\large \frac {13}{2\,×\,13}\)**

**∴ x = \(\large \frac {1}{2}\)**

** **

**Ans:** The value of x is \(\large \frac {1}{2}\).

**(8) 3(y + 8) = 10(y – 4) + 8****Solution:**

3(y + 8) = 10(y – 4) + 8

∴ 3y + 24 = 10y – 40 + 8

∴ 3y + 24 = 10y – 32

∴ 3y – 10y = –32 – 24

∴ –7y = – 56

∴ y = \(\large \frac {–56}{–7}\)

∴ y = 8

** **

**Ans:** The value of y is 8.

**(9) \(\large \frac {x\, –\, 9}{x\, –\, 5}\) = \(\large \frac {5}{7}\)**

**Solution:**

**\(\large \frac {x\, –\, 9}{x\, –\, 5}\) = \(\large \frac {5}{7}\)**

**∴ 7(x – 9) = 5(x – 5)**

**∴ 7x – 63 = 5x – 25**

**∴ 7x – 5x = –25 + 63**

**∴ 2x = 38**

**∴ x = \(\large \frac {38}{2}\)**

**∴ x = 19**

** **

**Ans:** The value of x is 19.

**(10) \(\large \frac {y\, –\, 4}{3}\) + 3y = 4**

**Solution:**

**\(\large \frac {y\, –\, 4}{3}\) + 3y = 4**

**Multiplying both sides by 3, we get, **

**3 × \(\large \frac {y\, –\, 4}{3}\) + 3 × 3y = 3 × 4**

**∴ y – 4 + 9y = 12**

**∴ 10y – 4 = 12**

**∴ 10y = 12 + 4**

**∴ 10y = 16**

**∴ y = \(\large \frac {16}{10}\)**

**∴ y = \(\large \frac {8}{5}\)**

** **

**Ans:** The value of y is y = \(\large \frac {8}{5}\).

**(11) \(\large \frac {b\, +\, (b\, +\, 1)\, +\, (b\, +\, 2)}{4}\) = 21**

**Solution:**

**\(\large \frac {b\, +\, (b\, +\, 1)\, +\, (b\, +\, 2)}{4}\) = 21**

**∴ 3b + 3 = 21 × 4**

**∴ 3b + 3 = 84**

**∴ 3b = 84 – 3**

**∴ 3b = 81**

**∴ b = \(\large \frac {81}{3}\)**

**∴ b = 27**

** **

**Ans:** The value of b is 27.

**Practice set 12.2**

**Practice set 12.2****1. Mother is 25 years older than her son. Find the son’s age if after 8 years the ratio of son’s age to mother’s age will be \(\large \frac {4}{9}\).**

**1. Mother is 25 years older than her son. Find the son’s age if after 8 years the ratio of son’s age to mother’s age will be \(\large \frac {4}{9}\).****Solution:**

**Let the present age of the son be x years. The present age of the mother is (x + 25) years. **

** **

**8 years later,**

**Age of the son will be (x + 8) years.**

**Age of the mother will be (x + 25 + 8) = (x + 33) years.**

** **

**According to given condition, **

**\(\large \frac {x\, +\, 8}{x\, +\,33}\) = \(\large \frac {4}{9}\)**

**∴ 9(x + 8) = 4(x + 33)**

**∴ 9x + 72 = 4x + 132**

**∴ 9x – 4x = 132 – 72**

**∴ 5x = 60**

**∴ x = \(\large \frac {60}{5}\)**

**∴ x = 12**

** **

**Ans:** The present age of the son is 12 years.

**2. The denominator of a fraction is greater than its numerator by 12.If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction is equivalent with \(\large \frac {1}{2}\). Find the fraction.**

**2. The denominator of a fraction is greater than its numerator by 12.If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction is equivalent with \(\large \frac {1}{2}\). Find the fraction.****Solution:**

**Let the original numerator be x.**

**∴ The original denominator = (x + 12)**

**∴ Original fraction = \(\large \frac {x}{x\,+\,12}\)**

** **

**According to the given condition,**

**\(\large \frac {x\,–\,2}{x\,+\,12\,+\,7}\) = \(\large \frac {1}{2}\)**

**∴ \(\large \frac {x\,–\,2}{x\,+\,19}\) = \(\large \frac {1}{2}\)**

**∴ 2(x – 2) = 1(x + 19)**

**∴ 2x – 4 = x + 19**

**∴ 2x – x = 19 + 4**

**∴ x = 23**

** **

**Original fraction = \(\large \frac {x}{x\,+\,12}\)**

**Original fraction = \(\large \frac {23}{23\,+\,12}\)**

**∴ Original fraction = \(\large \frac {23}{35}\)**

** **

**Ans: **The original fraction is \(\large \frac {23}{35}\).

**3. The ratio of weights of copper and zinc in brass is 13:7. Find the weight of zinc in a brass utensil weighing 700 gm.**

**3. The ratio of weights of copper and zinc in brass is 13:7. Find the weight of zinc in a brass utensil weighing 700 gm.****Solution:**

**Ratio of weights of copper and zinc = 13:7**

**Let the common multiple be x.**

**Weight of copper = 13x gm.**

**And weight of zinc = 7x gm.**

** **

**According to the given condition, **

**13x + 7x = 700**

**∴ 20x = 700**

**∴ x = \(\large \frac {700}{20}\)**

**∴ x = 35**

** **

**Weight of zinc = 7x **

**∴ Weight of zinc = 7 × 35**

**∴ Weight of zinc = 245 gm**

** **

**Ans:** The weight of zinc is 245 gm.

**4*. Find three conescutive whole numbers whose sum is more than 45 but less than 54.**

**4*. Find three conescutive whole numbers whose sum is more than 45 but less than 54.****Solution:**

**Let three consecutive whole numbers be x, (x + 1) and (x + 2) .**

** **

**According to the given condition, **

**45 < x + x + 1 + x + 2 < 54**

**∴ 45 < 3x + 3 < 54**

**∴ 45 – 3 < 3x + 3 – 3 < 54 – 3**

**∴ 42 < 3x < 51**

**Dividing by 3, we get, **

**\(\large \frac {42}{3}\) < \(\large \frac {3x}{3}\) < \(\large \frac {51}{3}\)**

**∴ 14 < x < 17**

**x = 15 or x = 16**

** **

**If x = 15, then,**

**x + 1 = 15 + 1 = 16 **

**x + 2 = 15 + 2 = 17**

** **

**If x = 16, then, **

**x + 1 = 16 + 1 = 17**

**x + 2 = 16 + 2 = 18**

** **

**Ans:** Required three numbers are 15, 16 and 17, or 16, 17 and 18.

**5. In a two digit number, the digit at the ten’s place is twice the digit at the unit’s place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. Find the number.**

**Solution:**

**Let the digit in the unit’s place be x.**

**The digit in ten’s place = 2x**

**Original number = 10 × 2x + 1 × x **

**∴ Original number = 20x + x**

**∴ Original number = 21x**

** **

**If digits are interchanged then digit in unit’s place will be 2x**

**And the digit in ten’s place will be x.**

** **

**New number = 10 × x + 2x **

**∴ New number = 10x + 2x **

**∴ New number = 12x**

** **

**According to the given condition, **

**21x + 12x = 66**

**∴ 33x = 66**

**∴ x = \(\large \frac {66}{33}\)**

**∴ x = 2**

** **

**Original number = 21x **

**∴ Original number = 21 × 2 **

**∴ Original number = 42**

** **

**Ans: **The original number is 42.

**6*. Some tickets of ₹200 and some of ₹100, of a drama in a theatre were sold. The number of tickets of ₹200 sold was 20 more than the number of tickets of ₹100. The total amount received by the theatre by sale of tickets was ₹37000. Find the number of ₹100 tickets sold.**

**6*. Some tickets of ₹200 and some of ₹100, of a drama in a theatre were sold. The number of tickets of ₹200 sold was 20 more than the number of tickets of ₹100. The total amount received by the theatre by sale of tickets was ₹37000. Find the number of ₹100 tickets sold.****Solution:**

**Let the number of tickets of ₹100 sold be x Number of tickets of 200 sold = (x + 20)**

** **

**According to the given condition,**

**100 × x + 200 (x + 20) = 37000**

**∴ 100x + 200x + 4000 = 37000**

**∴ 300x = 37000 – 4000**

**∴ 300x = 33000**

**∴ x = \(\large \frac {33000}{300}\)**

**∴ x = 110**

** **

**Ans: **The number of tickets of 100 sold were 110.

**7. Of the three consecutive natural numbers, five times the smallest number is 9 more than four times the greatest number, find the numbers.**

**7. Of the three consecutive natural numbers, five times the smallest number is 9 more than four times the greatest number, find the numbers.****Solution:**

**Let three consecutive natural numbers be x, (x + 1) and (x + 2)**

** **

**According to the given condition,**

**5x = 4(x + 2) + 9**

**∴ 5x = 4x + 8 + 9**

**∴ 5x – 4x = 17**

**∴ x = 17**

** **

**x + 1 = 17 + 1 = 18**

**x + 2 = 17 + 2 = 19**

** **

**Ans: **The required numbers are 17, 18 and 19.

**8. Raju sold a bicycle to Amit at 8% profit. Amit repaired it spending ₹54. Then he sold the bicycle to Nikhil for ₹1134 with no loss and no profit. Find the cost price of the bicycle for which Raju purchased it.**

**8. Raju sold a bicycle to Amit at 8% profit. Amit repaired it spending ₹54. Then he sold the bicycle to Nikhil for ₹1134 with no loss and no profit. Find the cost price of the bicycle for which Raju purchased it.****Solution:**

**Let Raju’s purchase price be ₹100x **

**8% of 100x = \(\large \frac {8}{100}\) × 100x ∴ 8% of 100x = 8x**

** **

**Amit’s purchase price = ₹(100x + 8x) = ₹108x**

**Total cost price of Amit = ₹(108x+54)**

**Selling price of Amit = ₹1134**

** **

**According to the given condition,**

**108x + 54 = 1134**

**∴ 108x = 1134 – 54**

**∴ 108x = 1080 **

**∴ x = \(\large \frac {1080}{108}\)**

**∴ x = 10**

** **

**The purchase price of Raju = ₹100x **

**∴ The purchase price of Raju = 100 × 10**

**∴ The purchase price of Raju = ₹1000**

** **

**Ans: **The purchase price of Raju is ₹1000.

**9. A Cricket player scored 180 runs in the first match and 257 runs in the second match. Find the number of runs he should score in the third match so that the average of runs in the three matches be 230.**

**9. A Cricket player scored 180 runs in the first match and 257 runs in the second match. Find the number of runs he should score in the third match so that the average of runs in the three matches be 230.****Solution:**

**Let the number of runs scored in the third match be x. **

** **

**According to the given condition,**

**(180 + 257 + x)/3 = 230**

**∴ \(\large \frac {180\,+\,257\,+\,x}{3}\) = 230**

**∴ 437 + x = 230 × 3**

**∴ 437 + x = 690**

**∴ x = 690 – 437**

**∴ x = 253**

** **

**Ans:** Number of runs the player should score in the third match is 253.

**10. Sudhir’s present age is 5 more than three times the age of Viru. Anil’s age is half the age of Sudhir. If the ratio of the sum of Sudhir’s and Viru’s age to three times Anil’s age is 5:6 , then find Viru’s age.**

**10. Sudhir’s present age is 5 more than three times the age of Viru. Anil’s age is half the age of Sudhir. If the ratio of the sum of Sudhir’s and Viru’s age to three times Anil’s age is 5:6 , then find Viru’s age.****Solution:**

**Let Viru’s present age be x years.**

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

**And Anil’s present age = \(\large \frac {3x\,+\,5}{2}\) years**

**Sum of Sudhir’s age and Viru’s age = x + 3x + 5 = (4x + 5) years**

**Three times Anil’s age = 3 × \(\large \frac {3x\,+\,5}{2}\) = \(\large \frac {9x\,+\,15}{2}\) years.**

** **

**According to the given condition,**

**\(\large \frac {4x\,+\,15}{\(\large \frac {9x\,+\,15}{2}\)}\) = \(\large \frac {5}{6}\)**

**∴ \(\large \frac {4x\,+\,5\, ×\,2}{9x\,+\,15}\) = \(\large \frac {5}{6}\)**

**∴ \(\large \frac {8x\,+\,10}{9x\,+\,15}\) = \(\large \frac {5}{6}\)**

**∴ 6(8x + 10) = 5(9x + 15)**

**∴ 48x + 60 = 45x + 75**

**∴ 48x – 45x = 75 – 60**

**∴ 3x = 15**

**∴ x = 5**

** **

**Ans:** Viru’s age is 5 years.