Olympiad Test: Algebra - Free MCQ with solutions Grade 11 Mathematics

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 represents 1 subtracted from the product of 2 and n. This is accurately described by option A, which states that it is '1 subtracted from the product of n and 2.'

The solution correctly identifies the:

• Mathematical operation: subtraction
• Order of operations: multiplication before subtraction

The given equation is:

Let's solve it step by step:

1. Convert the mixed fractions into improper fractions:

   •  = 5/2
   •  = 7/2

   The equation now becomes: (2/3) * P - 5/2 = 7/2

2. Add 5/2 to both sides to isolate the term with P: (2/3) * P = 7/2 + 5/2 (2/3) * P = 12/2 (2/3) * P = 6

3. Multiply both sides by 3/2 to solve for P: P = (6) * (3/2) P = 18/2 P = 9

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