Olympiad Test: Algebra - Class 6 MCQ

In algebra, unknown quantities are represented by lower case letters.

Here are some key points about variables:

• Variables can be any letter, such as n, l, m, p, x, y, or z.
• They allow us to express relationships in various situations.
• Although variables represent numbers, their values are not fixed.
• We can perform operations like addition, subtraction, multiplication, and division on variables, just like with fixed numbers.
• Examples of expressions with variables include: x + 3, 2n, 5m, and 2y + 3.
• Variables help express general rules in geometry and arithmetic, such as a + b = b + a.
• An equation is a statement that a variable expression equals a fixed number, like x - 3 = 10.

Literals usually represent variables.

Here are some key points about variables:

• A variable can be represented by any letter, such as n, l, m, x, y, or z.
• Variables allow us to express relationships in practical situations.
• They can take on different values, meaning their value is not fixed.
• Operations like addition, subtraction, multiplication, and division can be performed on variables.
• For example, expressions like x + 3 or 2n involve variables.

Variables help express general rules in both geometry and arithmetic. For instance:

• The rule that the order of addition does not change the sum can be expressed as a + b = b + a.
• In this case, a and b represent any number.

An equation is a condition involving a variable, shown as an expression equal to a fixed number, such as x - 3 = 10.

from the figure it is confirmed that we need 2 matchstick minium for a 'L' shape. so matchsticks will be multiple of 2. so, we need 2n matchsticks

Let the number be represented by x.

The statement "half of a number added to 10 is 15" can be translated into an equation:

x / 2 + 10 = 15

Here's the breakdown:

• "Half of a number" means dividing the number (x) by 2, which gives x / 2.
• "Added to 10" means adding 10 to x / 2.
• The phrase "is 15" represents an equality, so we set the expression equal to 15.

Thus, the final equation is:

x / 2 + 10 = 15

⇒ Let the length of the room be l.

⇒ Therefore, thrice the length is represented as 3l.

∴ According to the statement provided, we can express this as:

⇒ 3l = 340

- The method of finding a solution by trying out various values for the variable is called the Trial and Error method.
- Trial and Error involves testing different possibilities until a satisfactory solution is found.
- This method is often used when there is no straightforward formula to solve a problem.
- It is a fundamental problem-solving technique, particularly useful in situations where other methods are impractical.
- It helps in learning and discovering solutions through experimentation and iteration.

Solution:

To solve the equation 6a = -144, follow these steps:

• Start with the equation: 6a = -144.
• To isolate a, divide both sides by 6:
• a = -144 / 6.
• Calculate the result:
• a = -24.

The value of a that satisfies the equation is -24.

Solution:

To solve the equation 6a = -30, follow these steps:

• Start with the equation: 6a = -30.
• To isolate a, divide both sides by 6:
• a = -30 / 6
• Simplifying gives: a = -5.

Thus, the value of a that satisfies the equation is -5.

Solution:

To find the value of x in the equation:

• Start with the equation: 3x/4 + 8 = 17.
• Subtract 8 from both sides:
• 3x/4 = 17 - 8
• 3x/4 = 9
• Next, multiply both sides by 4 to eliminate the fraction:
• 3x = 9 * 4
• 3x = 36
• Now, divide both sides by 3:
• x = 36 / 3
• x = 12

Thus, the value of x is 12.

Solution:

To solve the equation 6(2a - 1) + 8 = 14, follow these steps:

• Start with the equation: 6(2a - 1) + 8 = 14
• Distribute the 6: 12a - 6 + 8 = 14
• Simplify the left side: 12a + 2 = 14
• Subtract 2 from both sides: 12a = 12
• Divide both sides by 12: a = 1

The value of a that satisfies the equation is 1.

- Start by solving the equation 6x - 1 = 2x + 9.

• Subtract 2x from both sides:
• 6x - 2x - 1 = 9
• Simplify:
• 4x - 1 = 9
• Add 1 to both sides:
• 4x = 10
• Divide both sides by 4 to solve for x:
• x = 10/4 = 5/2

- Therefore, the correct answer is x = 5/2, matching option D.

Let Ramu's age be x years.

Then, Ramu's father's age is 3x years.

Given that the father's age is 45 years, we can set up the equation:

• 3x = 45

To find Ramu's age, divide both sides by 3:

• x = 45 / 3
• x = 15

Thus, Ramu's age is 15 years.

Let the number be x.

The equation can be set up as follows:

• x multiplied by 6 plus 12 equals 84.
• So, 6x + 12 = 84.

To solve for x:

• First, subtract 12 from both sides:
• 6x = 84 - 12
• 6x = 72

Next, divide both sides by 6:

• x = 72 / 6
• x = 12

Thus, the required number is x = 12.

Let the length of a side of the equilateral triangular garden be x m.

The formula for the perimeter of an equilateral triangle is:

• Perimeter = 3 × side length

Given that the perimeter is 66 m, we can set up the equation:

• 3x = 66

To find the length of one side, divide both sides by 3:

• x = 66 ÷ 3
• x = 22 m

Thus, the length of a side of the equilateral triangular garden is 22 m.

From the figure,x^∘ + 3x^∘ = 180^∘ (straight angle) 
⇒ 4x^∘ =180^∘
⇒x^∘ = 45^∘
∴ 3x^∘ = 3×45^∘ = 135^∘

Let the breadth of the rectangle be b units. Since the length is twice the breadth, we can express it as:

Length = 2b units

The area of the rectangle is given by:

• Area = breadth × length
• Area = b × 2b = 2b²

We know the area is 72 sq. m, so:

2b² = 72

Dividing both sides by 2 gives:

b² = 36

Taking the square root of both sides, we find:

b = 6 m

A regular hexagon has 6 equal sides. To find the perimeter, we use the formula:

• Perimeter = Number of sides × Length of one side
• For a hexagon, this becomes:
• Perimeter = 6 × s

Thus, the perimeter of a regular hexagon is 6s units.

2n - 1 is 1 subtracted from the product of 2 and n.

Let the number of 50 paise coins be x.

Then, the number of 25 paise coins is 5x.

According to the problem, the total number of coins is 120:

• x + 5x = 120
• 6x = 120
• x = 120 / 6 = 20

Thus, there are:

• 50 paise coins = 20
• 25 paise coins = 5 × 20 = 100

Now, we calculate the total amount:

• Amount from 50 paise coins = 20 × 50p = Rs. 10
• Amount from 25 paise coins = 100 × 25p = Rs. 25

Therefore, the total amount in the bank is:

Rs. (25 + 10) = Rs. 35