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
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.
Answer (Detailed Solution Below)
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 \({x}^3+\frac{1}{{x}^3}\).
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 2 Detailed Solution
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.
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
Answer (Detailed Solution Below)
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:
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 5 Detailed Solution
Let Yash’s age = , father’s age =, Suraj’s age = .
-
Given .
-
When Yash is as old as his father now (), time passed = .
-
At that time, Yash’s age = , Suraj’s age =
. -
Given .
-
Also, .
-
From these, solve:
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.
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 6 Detailed Solution
Download Solution PDFGIVEN :
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 = 182If 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?
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 7 Detailed Solution
Download Solution PDFLet 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?
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 8 Detailed Solution
Download Solution PDFGiven:
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.
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 9 Detailed Solution
Download Solution PDFCalculation:
Let the lottery worth be Rs. x
Winning amount = \(\frac{x}{3}\)
Donated amount \(= 6000 = \frac{x}{3} \times \frac{1}{6}\)
∴ x = Rs. 108000Ashish 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.
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 10 Detailed Solution
Download Solution PDFGiven:
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.
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 11 Detailed Solution
Download Solution PDFGiven:
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?
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 12 Detailed Solution
Download Solution PDFGiven:
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?
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 13 Detailed Solution
Download Solution PDFGiven 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).
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 14 Detailed Solution
Download Solution PDFGIVEN:
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 = 7Find the 13/16 worth of land whose 9/7 worth be Rs. 10116.
Answer (Detailed Solution Below)
Linear Equation in 1 Variable Question 15 Detailed Solution
Download Solution PDFLet the total worth of land is Rs. X
⇒ Given, 9x/7 = 10116
⇒ So, x = 7868
∴ 13/16 of 7868 = Rs. 6392.75