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

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

Given:

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.

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.

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

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

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 =
   $s+(f-y)$.

4. Given $f=5s$.

5. Also,   $s + (f - y) = y + 12$.

6. From these, solve:

   
   y = 3s - 6 \quad \text{and} \quad (3s - 6) + 5s = 90
   $$8s=96⟹s=12$$

Answer: Suraj is 12 years old.

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.

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

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

Calculation:

Let the lottery worth be Rs. x

Winning amount = \(\frac{x}{3}\)

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

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

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.

Given:

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

Given y² = y + 7

Multiplying LHS and RHS by y

⇒ y³ = y² + 7y

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

GIVEN:

x² - 7x + 1 = 0

CALCULATION:

x² - 7x + 1 = 0

Dividing by x:

⇒ x - 7 + 1/x = 0

⇒ x + 1/x = 7

Let the total worth of land is Rs. X

⇒ Given, 9x/7 = 10116

⇒ So, x = 7868