Olympiad Test: Linear Equations In One Variable - Class 8 MCQ

Let the speed of boat in still water be x km/hour. 
∴ (x + 2) 4 = (x – 2) 5
⇒  4x + 8 = 5x - 10
⇒ x = 18 km/hr

Let Mohan’s age = 9x
and Sohan’s age = 7x
10, years ago,
Mohan’s age = 9x – 10
Sohan’s age = 7x – 10

⇒ 45x – 50 = 49x – 70
⇒ 4x = 20 ⇒ x = 5
∴ Mohan’s age = 9 × 5 = 45
and Sohan’s age = 7 × 5 = 35
Hence difference = 45 – 35 = 10 years

Let the one number be x.
Other number = 360 – x
Now 65% of x = 85% of (360 – x)

⇒ 65 x + 85x = 85 × 360
⇒ 150x = 85 × 360
⇒ 

Let the no. of workers in starting be x.
∴ 70 × x = (x – 20) 80
⇒ 70 x = 80x – 20 × 80
⇒ 10x = 20 × 80
⇒ 

Let the numbers be
11x, 11(x + 1), 11(x + 2)
Hence 11x + 11(x + 1) + 11(x + 2) = 363
⇒ 11x + 11x + 11 + 11x + 22 = 363
⇒ 33x + 33 = 363
⇒ 
Greatest multiple = 11(x + 2)
= 11 (10 + 2) = 11 × 12 = 132

Let Son’s age = x years
∴ Arun’s age = 3x.
Now 10 years ago, son’s age = x – 10.
Given Arun’s age = 3x – 10
⇒ 3x -10 = 5 (x – 10)
⇒ 3x – 10 = 5x – 50
2x = 40 ⇒ x = 20 years
Arun’s  age = 3 × 20 = 60 years
Sum of Arun’s and Son’s age = 20 + 60
= 80 years

Let the number be x.

⇒ 8x - 20 = 3x
⇒ 5x = 20
⇒ x = 4

Let the unit’s digit be x.
Ten’s digit = x + 3.Original number =10(x + 3) + 1(x)
= 10x + 30 + x
= 11x + 30
After interchange, resulting number
= 10 (x) + 1 (x + 3)
= 10x + x + 3
= 11x + 3
∴ 11x + 3 + 11x + 30 = 143
⇒ 22x + 33 = 143
⇒ 22x = 143 – 33
⇒ 22x = 110
⇒ 
Original number = 11x + 3 = 11 × 5 + 3
= 55 + 3 = 58

 Let the age of grand son be x years.
Grand father’s age = 10x.
⇒ 10x = 54 + x
⇒ 9x = 54
⇒ x = 6
Grand father’s age = 10 × 6 = 60 years
Difference of their present ages = 60 – 6 = 54.

Initial Area of the Triangle

The area of the triangle is calculated as:

Area = 12 × b × 53 b = 56 b²

New Area with Changes

Now, with the new base and altitude, the area becomes:

New Area = 12 × (b - 2) × ( 53 b + 4 )

Setting the New Area Equal to the Original

We equate the new area to the original area:

(b - 2) × ( 53 b + 4) = 53 b²

Simplification

After simplifying, we solve for b:

12b - 10b - 24 = 0
2b = 24
b = 12 cm

Finding the Altitude

The altitude is:

Altitude = 53 × 12 = 20 cm

Final Answer: The base is 12 cm and the altitude is 20 cm.

Let the area of the field ploughed daily be x hectares.
Area of the field = 18x
⇒ 18x = (x + 16) 12
⇒ 18x = 12x + 16 × 12
⇒ 6x = 16 × 12
⇒ 
​Area of the field = 18 × 32 = 576 hectares

Let the unit’s digit be x.
Ten’s digit = 9 – x
Number = 10 (9 – x) + 1(x)
= 90 – 10x + x= 90 – 9x
90 – 9x – 9 = x(10) + 1(9 – x)
⇒ 81 – 9x = 10x + 9 – x
⇒ 81 – 9x = 9x + 9
⇒ 18x = 72 ⇒ x = 4
Number = 90 – 9 × 4 = 90 – 36 = 54

⇒ 5x + 30 – 30 + 15x = 30x – 10
⇒ 20x + 10 = 30x
⇒ 10x = 10 ⇒ x = 1

⇒ 53x - 41 = 38x + 34
⇒ 15x = 34 + 41
⇒ 

⇒ (x – n) (m – n) = (x + n)(m + n)
⇒ mx – mn – nx + n² = mx + mn + nx + n²
⇒ - mn – nx = mn + nx
⇒ 2nx = -2mn
⇒ 

Let the cost price of the shirt be Rs. x.
∴ x + 7% of x = 1498
⇒ 
⇒ 107x 1498 x 100
⇒ 

Let number of deer be x.
∴ 
⇒ 
⇒ 4x + 3x 72 = 8x
⇒  x = 72 

Let Raju’s age = 5x.
∴ Rajan ‘s age = 8x
∴ 5x + 5 = 8x – 4
⇒ 3x = 9 ⇒ x = 3
Raju’s age = 5 × = 15 years

Let the number be x.

⇒ 8x = 16
⇒ x = 2

Let length of total Journey be x km.


⇒ 39x + 320 = 40x
⇒ x = 320