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CBSE Class 10  >  Class 10 Test  >  The Complete SAT Course  >  Test: Linear Equations in One Variable - 2 - Class 10 MCQ

Linear Equations in One Variable - 2 - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Linear Equations in One Variable - 2 (15 Questions)

You can prepare effectively for Class 10 The Complete SAT Course with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Linear Equations in One Variable - 2". These 15 questions have been designed by the experts with the latest curriculum of Class 10 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 25 minutes
  • - Number of Questions: 15

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Test: Linear Equations in One Variable - 2 - Question 1
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The sum of three consecutive number is 126. Find the highest number?

Detailed Solution: Question 1

Let the three consecutive number be x, x + 1, x + 2

⇒ x + (x + 1) + (x + 2) = 126

⇒ 3x + 3 = 126

⇒ 3x = 123

⇒ x = 41

∴ Highest number = x + 2 = 41 + 2 = 43

Test: Linear Equations in One Variable - 2 - Question 2
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In a group of 100 students, each student offers exactly 8 subjects, and exactly 10 students take each subject. Find the total number of different subjects.

Detailed Solution: Question 2

Step 1 — Count student–subject pairs: each of the 100 students offers 8 subjects, so total pairs = 100 × 8 = 800.

Step 2 — Relate to subjects: each subject is taken by 10 students, so if S is the number of subjects, total pairs = 10 × S.

Step 3 — Equate and solve: 10 × S = 800, hence S = 800 ÷ 10 = 80.

Answer: a) Exactly 80

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Test: Linear Equations in One Variable - 2 - Question 3
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Find the 5/8 worth of land whose 3/4 worth be Rs. 10116.

Detailed Solution: Question 3

Test: Linear Equations in One Variable - 2 - Question 4
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A total of 324 notes comprising of Rs. 20 and Rs. 50 denominations make a sum of Rs. 12450. The number of Rs. 20 notes is

Detailed Solution: Question 4

Given:

Total number of notes of Rs. 20 and Rs. 50 = 324

Total Sum = Rs. 12450

Concept used:

Value of note × Number of total notes = Total amount

Calculation:

Let the number of Rs. 20 notes be x.

No. of Rs. 50 notes = 324 – x

⇒ 20 × x + 50 × (324 – x) = 12450 

⇒ 20x + 16200 – 50x = 12450

⇒ -30x = 12450 – 16200

⇒ 30x = 3750

⇒ x = 125

∴ The number of Rs. 20 notes is 125. 

Test: Linear Equations in One Variable - 2 - Question 5
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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?

Detailed Solution: Question 5

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 

Test: Linear Equations in One Variable - 2 - Question 6
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If y2 = y + 7, then what is the value of y3?

Detailed Solution: Question 6

Given y2 = y + 7

Multiplying LHS and RHS by y

⇒ y3 = y2 + 7y

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

Test: Linear Equations in One Variable - 2 - Question 7
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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.

Detailed Solution: Question 7

Step 1: Define Variables

  • Let W represent the total winning amount.
  • Jane receives 1/3 of the winning amount: (1/3)W.

Step 2: Calculate the Donation

  • Jane donates Rs. 6000, which is 1/6 of the amount she got.
  • Mathematically, this can be expressed as:

    (1/6) × (1/3)W = 6000

Step 3: Set Up the Equation

  • Simplify the equation:

    (1/6) × (1/3)W = 6000

  • (1/18)W = 6000

Step 4: Solve for W

  • Multiply both sides of the equation by 18 to isolate W:

    W = 6000 × 18

  • W = 108,000

Step 5: Final Answer

  • The total winning amount Jane won is Rs. 108,000.

Test: Linear Equations in One Variable - 2 - Question 8
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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?

Detailed Solution: Question 8

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.

Test: Linear Equations in One Variable - 2 - Question 9
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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?

Detailed Solution: Question 9

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

Test: Linear Equations in One Variable - 2 - Question 10
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Find the product of two consecutive numbers where four times the first number is 10 more than thrice the second number.

Detailed Solution: Question 10

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

Test: Linear Equations in One Variable - 2 - Question 11
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A woman had only 50 paisa and 1 rupee coins in her purse. The total number of coins were 120 and the total amount was Rs.100. So the number of 1 rupee coins are_______.

Detailed Solution: Question 11

Total number of coin = 120 

Total amount = Rs 100 

Calculation 

Let the number of coins of 50 p and Re 1 be x and y 

According to question 

x + y = 120       ----(1)

x/2 + y = 100      ----(2) 

Subtract equation (2) 

By solving (1) and (2) from equation (1)

⇒ x - x/2 = 120 - 100

⇒ x/2 = 20

We get x = 40 and y = 80 

So number of coins of Rs 1 = 80 

∴ The required answer is 80 

Test: Linear Equations in One Variable - 2 - Question 12
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Find x:

Detailed Solution: Question 12

⇒ -22x – 26 = 8 - 56x

⇒ 56x - 22x = 26 + 8

⇒ 34x = 34

∴ x = 1

Test: Linear Equations in One Variable - 2 - Question 13
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A is 5 years older than B and 10 years younger than C. If the sum of ages of A, B and C together after 5 years is 80 years. Find the present age of B.

Detailed Solution: Question 13

Let the age of B be x years

So, the age of A = (x + 5)

And the age of C = (x + 5 +10) = (x + 15)

Now, age of B after 5 years = (x + 5)

Age of A after 5 years = (x + 5) + 5 = (x + 10)

 Age of C after 5 years = (x + 15) + 5 = (x + 20)

according to question,

sum of their ages after 5 years = 80

So, (x + 5) + (x + 10) + (x + 20) = 80

⇒ 3x + 35 = 80 

⇒ 3x = 45

So, x = 15

∴ present age of B = 15 years

Test: Linear Equations in One Variable - 2 - Question 14
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Solve:  3x/4 – 7/4 = 5x + 12

Detailed Solution: Question 14

  • 3x/4 - 7/4 = 5x + 12
  • 3x - 7 = 4(5x +12)
  • 3x - 7 = 20 x + 48
  • -17 x = 55
  • x = -55/17

Test: Linear Equations in One Variable - 2 - Question 15
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In 10 years, a father will be twice as old as his son. Five years ago, the father was three times as old as his son. Find their current ages.

Detailed Solution: Question 15

  • Father's current age = F
  • Son's current age = S

Step 1: Translate the first statement into an equation.

F + 10 = 2(S + 10) Simplify: F + 10 = 2S + 20 F = 2S + 10 (Equation 1)

Step 2: Translate the second statement into an equation.

F - 5 = 3(S - 5)
Simplify:
F - 5 = 3S - 15 F = 3S - 10 (Equation 2)

Step 3: Set Equation 1 equal to Equation 2.

2S + 10 = 3S - 10
Subtract 2S from both sides: 10 = S - 10 Add 10 to both sides: S = 20

Step 4: Find Father’s age using Equation 1.

F = 2(20) + 10 F = 50

Answer:

  • Father's current age = 50 years
  • Son's current age = 20 years

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About this Class 10 Practice Test

This test contains 15 MCQs on Linear Equations in One Variable - 2 with detailed solutions for each question. It is part of the The Complete SAT Course course on EduRev, designed to align with the current Class 10 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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