Linear Equations in one Variable

Q1: Multiple Choice Problems

i. The sum of three consecutive even numbers is 90; the middle number is:

Step 1: Let the three consecutive even numbers be \(x-2, x, x+2\)
Step 2: Sum of numbers = 90 → \((x-2) + x + (x+2) = 90\)
\(3x = 90 → x = 30\)
Answer: b. 30

ii. The length of a rectangle is 8 m more than its breadth. If the perimeter of the rectangle is 40 m, the length of the rectangle is:

Step 1: Let breadth = \(x\) → length = \(x + 8\)
Perimeter = \(2(l + b) = 40 → 2((x+8)+x) = 40\)
\(2(2x + 8) = 40 → 4x + 16 = 40\)
Step 2: Solve for x:
\(4x = 24 → x = 6\)
Length = \(x + 8 = 6 + 8 = 14\)
Answer: c. 14 m

iii. Two whole numbers are in the ratio 5 : 3. If their difference is 18; the numbers are:

Step 1: Let the numbers be \(5x\) and \(3x\)
Difference = \(5x – 3x = 2x = 18 → x = 9\)
Numbers = \(5x = 45, 3x = 27\)
Answer: c. 45 and 27

iv. Three more than twice a number is equal to four less than the number. The number is:

Step 1: Let the number = \(x\)
Equation: \(2x + 3 = x – 4\)
Step 2: Solve for x:
\(2x – x = -4 – 3 → x = -7\)
Answer: b. -7

v. The sum of two numbers is 70 and their difference is 16, the numbers are:

Step 1: Let numbers be \(x\) and \(y\)
Equations: \(x + y = 70\), \(x – y = 16\)
Step 2: Add both equations:
\((x+y) + (x-y) = 70 + 16 → 2x = 86 → x = 43\)
Then, \(y = 70 – x = 70 – 43 = 27\)
Answer: a. 43 and 27

Q2: Fifteen less than 4 times a is 9. Find the number.

Step 1: Translate the statement into an equation:
“Fifteen less than 4 times a” → \(4a – 15\)
Equation: \(4a – 15 = 9\)
Step 2: Solve for \(a\):
\(4a = 9 + 15 → 4a = 24\)
\(a = \frac{24}{4} = 6\)
Answer: a = 6

Q3: If Megha’s age is increased by three times her age, result is 60 years. Find her age.

Step 1: Let Megha’s age be \(x\) years.
“Increased by three times her age” → \(x + 3x = 4x\)
Equation: \(4x = 60\)
Step 2: Solve for \(x\):
\(x = \frac{60}{4} = 15\)
Answer: Megha’s age = 15 years

Q4: 28 is 12 less than 4 times a number. Find the number.

Step 1: Let the number be \(x\).
“12 less than 4 times a number” → \(4x – 12\)
Equation: \(4x – 12 = 28\)
Step 2: Solve for \(x\):
\(4x = 28 + 12 → 4x = 40\)
\(x = \frac{40}{4} = 10\)
Answer: Number = 10

Q5: Five less than 3 times a number is -20. Find the number.

Step 1: Let the number be \(x\).
“Five less than 3 times a number” → \(3x – 5\)
Equation: \(3x – 5 = -20\)
Step 2: Solve for \(x\):
\(3x = -20 + 5 → 3x = -15\)
\(x = \frac{-15}{3} = -5\)
Answer: Number = -5

Q6: Fifteen more than 3 times Neetu’s age is the same as 4 times her age. How old is she?

Step 1: Let Neetu’s age be \(x\) years.
“Fifteen more than 3 times her age” → \(3x + 15\)
Equation: \(3x + 15 = 4x\)
Step 2: Solve for \(x\):
\(15 = 4x – 3x → 15 = x\)
Answer: Neetu’s age = 15 years

Q7: A number decreased by 30 is the same as 14 decreased by 3 times the number. Find the number.

Step 1: Let the number be \(x\).
“A number decreased by 30” → \(x – 30\)
“14 decreased by 3 times the number” → \(14 – 3x\)
Equation: \(x – 30 = 14 – 3x\)
Step 2: Solve for \(x\):
\(x + 3x = 14 + 30 → 4x = 44 → x = \frac{44}{4} = 11\)
Answer: Number = 11

Q8: A’s salary is same as 4 times B’s salary. If together they earn ₹3,750 a month, find the salary of each.

Step 1: Let B’s salary = ₹\(x\)
Then A’s salary = ₹\(4x\)
Together: \(4x + x = 3750\)
Step 2: Solve for \(x\):
\(5x = 3750 → x = \frac{3750}{5} = 750\)
Step 3: Find A’s salary:
A’s salary = \(4x = 4 × 750 = 3000\)
Answer: A’s salary = ₹3,000, B’s salary = ₹750

Q9: Separate 178 into two parts so that first part is 8 less than twice the second part.

Step 1: Let the second part = \(x\)
Then the first part = \(2x – 8\)
Equation: \( (2x – 8) + x = 178 \)
Step 2: Solve for \(x\):
\(3x – 8 = 178 → 3x = 178 + 8 → 3x = 186 → x = \frac{186}{3} = 62\)
Step 3: Find the first part:
First part = \(2x – 8 = 2 × 62 – 8 = 124 – 8 = 116\)
Answer: First part = 116, Second part = 62

Q10: Six more than one-fourth of a number is two-fifth the number. Find the number.

Step 1: Let the number be \(x\).
“Six more than one-fourth of a number” → \(\frac{x}{4} + 6\)
Equation: \(\frac{x}{4} + 6 = \frac{2x}{5}\)
Step 2: Eliminate denominators (LCM of 4 and 5 = 20):
Multiply both sides by 20: \(20 × (\frac{x}{4} + 6) = 20 × \frac{2x}{5}\)
\(5x + 120 = 8x\)
Step 3: Solve for \(x\):
\(120 = 8x – 5x → 120 = 3x → x = \frac{120}{3} = 40\)
Answer: Number = 40

Q11: The length of a rectangle is twice its width. If its perimeter is 54 cm, find its length.

Step 1: Let the width = \(x\) cm.
Then length = \(2x\) cm
Perimeter of rectangle = \(2(\text{length + width}) = 54\)
Equation: \(2(2x + x) = 54\)
Step 2: Simplify:
\(2(3x) = 54 → 6x = 54 → x = \frac{54}{6} = 9\)
Step 3: Find the length:
Length = \(2x = 2 × 9 = 18\) cm
Answer: Length = 18 cm

Q12: A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm, the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of original rectangle.

Step 1: Let the width of the original rectangle = \(x\) cm.
Then the length = \(2x – 5\) cm
Step 2: Length decreased by 5 → \(2x – 5 – 5 = 2x – 10\) cm
Width increased by 2 → \(x + 2\) cm
Perimeter of resulting rectangle = 74 cm → \(2(\text{length + width}) = 74\)
Equation: \(2((2x – 10) + (x + 2)) = 74\)
Step 3: Simplify the equation:
\(2(2x – 10 + x + 2) = 74 → 2(3x – 8) = 74 → 6x – 16 = 74\)
\(6x = 74 + 16 → 6x = 90 → x = \frac{90}{6} = 15\)
Step 4: Find the length:
Length = \(2x – 5 = 2 × 15 – 5 = 30 – 5 = 25\) cm
Answer: Original width = 15 cm, Original length = 25 cm

Q13: The sum of three consecutive odd numbers is 57. Find the numbers.

Step 1: Let the first odd number = \(x\)
Then the second odd number = \(x + 2\)
Third odd number = \(x + 4\)
Equation: \(x + (x + 2) + (x + 4) = 57\)
Step 2: Simplify the equation:
\(3x + 6 = 57 → 3x = 57 – 6 → 3x = 51 → x = \frac{51}{3} = 17\)
Step 3: Find the other two numbers:
Second number = \(x + 2 = 17 + 2 = 19\)
Third number = \(x + 4 = 17 + 4 = 21\)
Answer: The numbers are 17, 19, 21

Q14: A man’s is three times that of his son and in twelve years he will be twice as old as his son would be. What are their present ages?

Step 1: Let the son’s present age = \(x\) years.
Then the man’s present age = \(3x\) years.
Step 2: In 12 years:
Son’s age = \(x + 12\)
Man’s age = \(3x + 12\)
Equation: \(3x + 12 = 2(x + 12)\)
Step 3: Simplify the equation:
\(3x + 12 = 2x + 24 → 3x – 2x = 24 – 12 → x = 12\)
Step 4: Find the man’s age:
Man’s age = \(3x = 3 × 12 = 36\)
Answer: Son’s age = 12 years, Man’s age = 36 years

Q15: A man is 42 years old and his son is 12 years old. In how many years will the age of the son be half age of the man at that time?

Step 1: Let the number of years after which the son’s age will be half of the man’s age = \(x\)
Man’s age after \(x\) years = \(42 + x\)
Son’s age after \(x\) years = \(12 + x\)
Equation: \(12 + x = \frac{1}{2}(42 + x)\)
Step 2: Eliminate the fraction:
\(2(12 + x) = 42 + x → 24 + 2x = 42 + x\)
Step 3: Solve for \(x\):
\(2x – x = 42 – 24 → x = 18\)
Answer: After 18 years, the son’s age will be half of the man’s age.

Q16: A man completed a trip of 136 km in 8 hours. Some parts of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.

Step 1: Let the distance covered at 18 km/hr = \(x\) km
Then the distance covered at 15 km/hr = \(136 – x\) km
Step 2: Time = Distance ÷ Speed
Total time = 8 hours → \(\frac{136 – x}{15} + \frac{x}{18} = 8\)
Step 3: Eliminate denominators (LCM of 15 and 18 = 90):
\(\frac{136 – x}{15} × 90 + \frac{x}{18} × 90 = 8 × 90\)
\(6(136 – x) + 5x = 720\)
\(816 – 6x + 5x = 720 → -x + 816 = 720 → -x = -96 → x = 96\)
Answer: Distance covered at 18 km/hr = 96 km

Q17: The difference of two numbers is 3 and the difference of their squares is 69. Find the numbers.

Step 1: Let the numbers be \(x\) and \(y\) such that \(x > y\)
Given: \(x – y = 3\)
Also, \(x^2 – y^2 = 69\)
Step 2: Factorize the difference of squares:
\(x^2 – y^2 = (x – y)(x + y) = 69\)
Substitute \(x – y = 3\):
\(3(x + y) = 69 → x + y = \frac{69}{3} = 23\)
Step 3: Solve for \(x\) and \(y\):
\(x – y = 3\)
\(x + y = 23\)
Add both equations: \(2x = 26 → x = 13\)
Then \(y = 23 – 13 = 10\)
Answer: The numbers are 13 and 10

Q18: Two consecutive natural numbers are such that one-fourth of the smaller exceeds one-fifth of me greater by 1. Find the numbers.

Step 1: Let the smaller number = \(x\)
Then the next consecutive number = \(x + 1\)
Equation: \(\frac{x}{4} – \frac{x + 1}{5} = 1\)
Step 2: Eliminate denominators (LCM of 4 and 5 = 20):
\(20 × \left(\frac{x}{4} – \frac{x + 1}{5}\right) = 20 × 1\)
\(5x – 4(x + 1) = 20\)
\(5x – 4x – 4 = 20 → x – 4 = 20 → x = 24\)
Step 3: Find the consecutive number:
Next number = \(x + 1 = 24 + 1 = 25\)
Answer: The numbers are 24 and 25

Q19: Three consecutive whole numbers are such that if they be divided by 5, 3 and 4 respectively, sum of the quotients is 40. Find the numbers.

Step 1: Let the numbers be \(x\), \(x+1\), \(x+2\)
Given condition: \(\frac{x}{5} + \frac{x+1}{3} + \frac{x+2}{4} = 40\)
Step 2: Eliminate denominators (LCM of 5, 3, 4 = 60):
\(60 × \left(\frac{x}{5} + \frac{x+1}{3} + \frac{x+2}{4}\right) = 60 × 40\)
\(12x + 20(x+1) + 15(x+2) = 2400\)
\(12x + 20x + 20 + 15x + 30 = 2400\)
\(47x + 50 = 2400\)
Step 3: Solve for \(x\):
\(47x = 2400 – 50 → 47x = 2350 → x = \frac{2350}{47} = 50\)
Step 4: Find the other two numbers:
Second number = \(x+1 = 50 + 1 = 51\)
Third number = \(x+2 = 50 + 2 = 52\)
Answer: The numbers are 50, 51, 52

Q20: If the same number be added to numbers 5, 11, 15 and 31. the resulting numbers are in proportion. Find the number.

Step 1: Let the number to be added = \(x\)
Numbers after adding \(x\): \(5+x\), \(11+x\), \(15+x\), \(31+x\)
They are in proportion: \(\frac{5+x}{11+x} = \frac{15+x}{31+x}\)
Step 2: Cross multiply:
\((5 + x)(31 + x) = (11 + x)(15 + x)\)
Expand both sides:
\(5×31 + 5x + 31x + x^2 = 11×15 + 11x + 15x + x^2\)
\(155 + 36x + x^2 = 165 + 26x + x^2\)
Step 3: Simplify:
\(36x – 26x + 155 – 165 = 0 → 10x – 10 = 0 → 10x = 10 → x = 1\)
Answer: The number to be added is 1

Q21: The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.

Step 1: Let the son’s present age = \(x\) years
Then the man’s present age = \(2x\) years
Step 2: Eight years hence:
Son’s age = \(x + 8\)
Man’s age = \(2x + 8\)
Given ratio: \(\frac{2x + 8}{x + 8} = \frac{7}{4}\)
Step 3: Cross multiply:
\(4(2x + 8) = 7(x + 8) → 8x + 32 = 7x + 56\)
Step 4: Solve for \(x\):
\(8x – 7x = 56 – 32 → x = 24\)
Step 5: Find the man’s age:
Man’s age = \(2x = 2 × 24 = 48\)
Answer: Son’s present age = 24 years, Man’s present age = 48 years