Test: Linear Equations in One Variable - 2 - SAT MCQ

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

Total students = 100

According to the question

Every students study 8 subjects, then total subjects = 8 × 100 = 800

If every subject is studied by 10 students, then number of subjects = 800/10 = 80

Let the total worth of land is Rs. X

⇒ Given, 9x/7 = 10116

⇒ So, x = 7868

∴ 13/16 of 7868 = Rs. 6392.75

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. 

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]

Let's denote the total lottery amount as xxx.

Jane receives 1/3​ of the total lottery amount, so the amount she gets is x/3​.

Jane donates Rs. 6000, and this amount is given as 1/6​ of what she received. Therefore, we can set up the equation:

​​​​​​

Now, solve for x:

To find x, multiply both sides by 18:

x=6000 × 18 = 108000

So, the total lottery amount is Rs. 108000.

Thus, the correct answer is (d) 108000.

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.

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:

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

Calculation:

Given that, 5x = 2

Divide the equation by 5

⇒ (5x/5) = (2/5)

⇒ x = 2/5

∴ The value of x is 2/5

Calculation:

Given that

5x - 8 = 7 

⇒ 5x = 7 + 8

⇒ 5x = 15

⇒ x =3

∴ The required value is 3.

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 

⇒ -22x – 26 = 8 - 56x

⇒ 56x - 22x = 26 + 8

⇒ 34x = 34

∴ x = 1

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