Linear Equation in 1 Variable MCQ Quiz - Objective Question with Answer for Linear Equation in 1 Variable - Download Free PDF

Last updated on Jun 3, 2025

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Latest Linear Equation in 1 Variable MCQ Objective Questions

Linear Equation in 1 Variable Question 1:

1800 chocolates were distributed among the students of a class. Each student got twice as many chocolates as the number of students in the class. Calculate the number of students in the class.

  1. 30
  2. 40
  3. 60
  4. 90
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 30

Linear Equation in 1 Variable Question 1 Detailed Solution

Deatialed solution:-

Let the number of students in the class be ‘x’

No. of chocolates received by each student = 2x

Total number of chocolates = 2x × x

⇒ 2x2 = 1800

⇒ x2 = 900

⇒ x = √900 = 30

∴ There are 30 students in the class.

Linear Equation in 1 Variable Question 2:

If \({x}+\frac{1}{{x}}=1\), then find the value of \({x}^3+\frac{1}{{x}^3}\).

  1. -2
  2. -1
  3. 0
  4. 2
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : -2

Linear Equation in 1 Variable Question 2 Detailed Solution

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Given:

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x + 1/x = 1

Formula used:

x2 + 1/x2 = (x + 1/x)2 - 2

x3 + 1/x3 = (x + 1/x) × (x2 + 1/x2) - (x + 1/x)

Calculation:

1. Find x2 + 1/x2:

⇒ x2 + 1/x2 = (x + 1/x)2 - 2

⇒ x2 + 1/x2 = 12 - 2

⇒ x2 + 1/x2 = 1 - 2

⇒ x2 + 1/x2 = -1

2. Find x3 + 1/x3:

⇒ x3 + 1/x3 = (x + 1/x) × (x2 + 1/x2) - (x + 1/x)

⇒ x3 + 1/x3 = 1 × (-1) - 1

⇒ x3 + 1/x3 = -1 - 1

⇒ x3 + 1/x3 = -2

∴ The value of x3 + 1/x3 is -2.

Linear Equation in 1 Variable Question 3:

Find the value of x:

\(\frac{2}{5}(x)+\frac{3}{10}(x)-\frac{3}{5}(x)\) = 479

  1. 4890 
  2. 5190 
  3. 4790 
  4. 4690
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 4790 

Linear Equation in 1 Variable Question 3 Detailed Solution

Find the value of x:

\(\frac{2}{5}(x) + \frac{3}{10}(x) - \frac{3}{5}(x) = 479\)

Calculation:

Combining the terms with a common denominator:

\(\frac{2}{5}(x) - \frac{3}{5}(x) + \frac{3}{10}(x)\)

\(\frac{2x - 3x}{5} + \frac{3x}{10}\)

\(\frac{-x}{5} + \frac{3x}{10}\)

Finding a common denominator for the fractions:

\(\frac{-2x}{10} + \frac{3x}{10}\)

\(\frac{-2x + 3x}{10}\)

\(\frac{x}{10} = 479\)

Multiplying both sides by 10:

479 ×  10

x = 4790

The value of x is 4790.

Linear Equation in 1 Variable Question 4:

Arun is three years older than Varun. Eight years ago, \(\frac{1}{6}\)th of Arun's age exceeded \(\frac{2}{5}\)th of Varun's age by 6 years. If the present age of Varun is x years, then the value of x can be determined by
solving the equation:

  1. \(\rm \frac{3}{5}(x+5)-\frac{5}{6}(x-8)=6 \)
  2. \(\rm \frac{3}{5}(x+3)-\frac{5}{6}(x-8)=6 \)
  3. \(\frac{1}{6}\times (x - 5) - \frac{2}{5}\times (x - 8) =6\)
  4. \(\rm \frac{5}{6}(x-8)-\frac{3}{5}(x-5)=6 \)

Answer (Detailed Solution Below)

Option 3 : \(\frac{1}{6}\times (x - 5) - \frac{2}{5}\times (x - 8) =6\)

Linear Equation in 1 Variable Question 4 Detailed Solution

Given: 

Arun is three years older than Varun

Eight years ago \(\frac{5}{6}th\) of Arun's age exceeded \(\frac{3}{5}th\) of Varun's age by 6 years.

The present age of Varun x years

Calculation: 

Since Arun is 3 years older than Varun: 

The present age of Arun x +3 

Eight years ago:

Arun's age was (x + 3) - 8 = x - 5 years

Varun's age was x - 8 years

According to the problem, Eight years ago \(\frac{1}{6}th\) of Arun's age exceeded \(\frac{2}{5}th\) of Varun's age by 6 years.

⇒ \(\frac{1}{6}\times (x - 5) - \frac{2}{5}\times (x - 8) =6\)

This matches the third option.

∴ The correct answer is Option (3): \(\frac{1}{6}\times (x - 5) - \frac{2}{5}\times (x - 8) =6\)

Linear Equation in 1 Variable Question 5:

The sum of the ages of Yash and his father is 90 years. When Yash is as old as his father's present age, he will be five times as old as his son Suraj's present age. Suraj will be 12 years older than Yash's present age, when Yash is as old as his father at present. How old is Suraj at present?

  1. 15 years
  2. 18 years
  3. 12 years
  4. 16 years

Answer (Detailed Solution Below)

Option 3 : 12 years

Linear Equation in 1 Variable Question 5 Detailed Solution

Let Yash’s age =
y
, father’s age =f, Suraj’s age = .

  1. Given
    y + f = 90
    .

  2. When Yash is as old as his father now (f), time passed =
    f - y
    .

  3. At that time, Yash’s age =
    f
    , Suraj’s age =
    .

  4. Given .

  5. Also,   s + (f - y) = y + 12.

  6. From these, solve:


    y = 3s - 6 \quad \text{and} \quad (3s - 6) + 5s = 90

Answer: Suraj is 12 years old.

Top Linear Equation in 1 Variable MCQ Objective Questions

Find the product of two consecutive numbers where four times the first number is 10 more than thrice the second number.

  1. 210
  2. 182
  3. 306
  4. 156

Answer (Detailed Solution Below)

Option 2 : 182

Linear Equation in 1 Variable Question 6 Detailed Solution

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GIVEN :

Four times the first number is 10 more than thrice the second number.

CALCULATION :

Suppose the numbers are ‘a’ and ‘a + 1’.

According to the question :

4a = 3 × (a + 1) + 10

⇒ a = 13

Hence, the numbers are 13 and 14.

∴ Product = 13 × 14 = 182

If 80% of my age 6 years ago is the same as 60% of my age after 10 years. What is the product of digits of my present age?

  1. 24
  2. 20
  3. 30
  4. 15

Answer (Detailed Solution Below)

Option 2 : 20

Linear Equation in 1 Variable Question 7 Detailed Solution

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Let the present age be x

⇒ Six years ago my age will be = (x - 6)

⇒ Age after 10 years = (x + 10)

According to the question

⇒ 80% of (x – 6) = 60% of (x + 10)

⇒ 4x – 24 = 3x + 30

⇒ x = 54

∴ Product of digits = 5 × 4 = 20

Rajeev was to earn Rs. 500 and a free holiday for seven weeks’ work. He worked for only 5 weeks and earned Rs. 50 and a free holiday. What was the value of the holiday?

  1. Rs. 1,075
  2. Rs. 1,850
  3. Rs. 1,550
  4. Rs. 1,675

Answer (Detailed Solution Below)

Option 1 : Rs. 1,075

Linear Equation in 1 Variable Question 8 Detailed Solution

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Given:

500 + 1 holiday = 7 weeks      ----(1)

50 + 1 holiday = 5 weeks      ----(2)

⇒ From (1) - (2)

450 = 2 weeks

⇒ 1 week = 225 Rs.

From (1),

⇒ 500 + 1 holiday = 7 × 225

⇒ 1 holiday = 1575 - 500

⇒ 1 holiday = 1075 Rs.

Jane won a lottery and gets 1/3rd of the winning amount and donates Rs. 6000 which is 1/6th  of amount he got, find how much the lottery was worth.

  1. 36000
  2. 18000
  3. 54000
  4. 108000

Answer (Detailed Solution Below)

Option 4 : 108000

Linear Equation in 1 Variable Question 9 Detailed Solution

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Calculation:

Let the lottery worth be Rs. x

Winning amount = \(\frac{x}{3}\)

Donated amount \(= 6000 = \frac{x}{3} \times \frac{1}{6}\)

∴ x = Rs. 108000 

Ashish purchased 50 kg of grains for Rs. 710 from a grocery shop. The price of rice was Rs. 12 per kilogram and that of wheat is Rs. 17 per kilogram. Find the money he spent on rice.

  1. Rs. 318
  2. Rs. 328
  3. Rs. 348
  4. Rs. 336
  5. Rs. 364

Answer (Detailed Solution Below)

Option 4 : Rs. 336

Linear Equation in 1 Variable Question 10 Detailed Solution

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Given:

Total grains = 50 kg, price = Rs. 710

Price of rice per kg = Rs. 12

Price of wheat per kg = Rs. 17

Calculation:

Let Ashish purchased x kg of rice

So, quantity of wheat he purchased = (50 – x)

Now,

⇒x × 12 + (50 – x) × 17 = 710

⇒ 12x + 850 – 17x = 710

⇒ 5x = 140

⇒ x = 28 kg

∴ Money spent on rice = 12 × 28 = Rs. 336

The monthly incomes of Ravi and Shiv are in the ratio 1 : 2 and their monthly expenditures are in the ratio 1 : 3. If each saves ₹4,000 per month, then find the monthly income of Shiv.

  1. ₹16,000
  2. ₹18,000
  3. ₹14,000
  4. ₹20,000

Answer (Detailed Solution Below)

Option 1 : ₹16,000

Linear Equation in 1 Variable Question 11 Detailed Solution

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Given:

Ratio of monthly incomes of Ravi and Shiv = 1 : 2

Ratio of monthly expenditures = 1 : 3

Savings per month each = Rs 4,000

Concept Used:

Income – Savings = Expenditure

Calculation:

⇒ Let the monthly incomes of Ravi and Shiv be x and 2x.

⇒ According to the question,

⇒ \(\frac{x-4000}{2x-4000}=\frac{1}{3}\)

⇒ 3(x – 4000) = 1(2x – 4000)

⇒ 3x – 12000 = 2x – 4000

⇒ 3x – 2x = 12000 – 4000

⇒ x = 8000

⇒ Income of Shiv = 2x = 2 × 8000 = Rs 16,000

Therefore, the monthly income of Shiv is Rs 16,000.

If 1/3rd of the first of the three consecutive odd numbers is 2 more than 1/5th of the third number, then the second number is?

  1. 21
  2. 23
  3. 25
  4. 19

Answer (Detailed Solution Below)

Option 2 : 23

Linear Equation in 1 Variable Question 12 Detailed Solution

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Given:

1/3rd of the first of the three consecutive odd numbers is 2 more than 1/5th of the third number.

Calculation:

Let, the three consecutive odd numbers be x, x + 2 and x + 4 respectively, then

According to the question,

x × (1/3) = (x + 4) × (1/5) + 2

⇒ x/3 = (x + 4 + 10)/5

⇒ 5x = 3x + 42

⇒ 5x – 3x = 42

⇒ 2x = 42

⇒ x = 21

∴ Second odd number is x + 2 = 21 + 2 = 23 

If y2 = y + 7, then what is the value of y3?

  1. 8y + 7
  2. y + 14
  3. y + 2
  4. 4y + 7

Answer (Detailed Solution Below)

Option 1 : 8y + 7

Linear Equation in 1 Variable Question 13 Detailed Solution

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Given y2 = y + 7

Multiplying LHS and RHS by y

⇒ y3 = y2 + 7y

∴ y3 = y + 7 + 7y = 8y + 7 [∵ y2 = y + 7]

If x2 - 7x + 1 = 0 then find the value of (x + 1/x).

  1. 3
  2. 7
  3. - 7
  4. - 3

Answer (Detailed Solution Below)

Option 2 : 7

Linear Equation in 1 Variable Question 14 Detailed Solution

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GIVEN:

x2 - 7x + 1 = 0

CALCULATION:

x2 - 7x + 1 = 0

Dividing by x:

⇒ x - 7 + 1/x = 0

⇒ x + 1/x = 7

∴ Value of x + 1/x = 7 

Find the 13/16 worth of land whose 9/7 worth be Rs. 10116.

  1. Rs. 6391.75
  2. Rs. 6394.75
  3. Rs. 6392.75
  4. Rs. 6302.75

Answer (Detailed Solution Below)

Option 3 : Rs. 6392.75

Linear Equation in 1 Variable Question 15 Detailed Solution

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Let the total worth of land is Rs. X

⇒ Given, 9x/7 = 10116

⇒ So, x = 7868

∴ 13/16 of 7868 = Rs. 6392.75
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