1. The line ‘p’ passes through two points (2, 4) and (-1, 1). Which of the following could be the equation of the line perpendicular to ‘p’?

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**Correct answer is option - 5**

**Explanation: **Slope of a line is given by (change in ‘y’) / (change in ‘x’). Therefore the slope of the line ‘p’ is (1 – 4)/ (-1 – 2) ==> slope = -3/-3 = 1.If two lines are perpendicular to each other, then the product of their slopes is -1.

Let the slope of the other line be = m.

Then 1 * m = -1 ==> m = -1.

Hence option d is the right answer since it is the only equation which has -1 slope.

2. Which of the following is equivalent to the expression, 169 – x^{2}?

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**Correct answer is option - 5**

**Explanation: **169 can also be written as 13^{2}.

This gives: 169 – x^{2} ==> 13^{2} – x^{2} and is in the form of a^{2} – b^{2} and can be expanded as (a + b) (a – b).

This implies:13^{2} – x^{2} = (13 + x) (13 – x).

3. If f(x) = 2x^{2} – 1 and f(2p) = 31, then which of the following could be the value of ‘p’?

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**Correct answer is option - 3**

**Explanation: **Given f(x) = 2x^{2} – 1, then f(2p) = 2* (2p)^{2} – 1 ==> f(2p) = 8p^{2} – 1 = 31.

This gives: 8p^{2} = 32 ==> p^{2} = 4 ==> p = √4 ==> p = ± 2.

Since p = -2 is not given in the options, hence the only answer is p = 2.

4. In a class there are ‘m’ students. If ‘p%’ among the ‘m’ number of students are members of the sports club, then which of the following expression represents the number of students not in the sports club?

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**Correct answer is option - 5**

**Explanation:**

In a class of ‘m’ students, number of students in the sports club = p% of ‘m’.

So this gives (p/100) * m ==> (mp)/ 100.

Therefore students not in the sports club = m – [(mp)/ 100] ==> (100m – mp)/ 100.

This gives: m(100 – p)/ 100 as the answer.

5. If xy = 4 and x – y = 6, then what is the value of x^{2} + y^{2}?

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**Correct answer is option - 3**

**Explanation: **Squaring the equation (x - y) = 6 on both sides, we get: (x – y)^{2} = 36.

Expanding the above equation we have: x^{2} – 2xy + y^{2} = 36 and given xy = 4.

This implies: x^{2} + y^{2} – 2(4) = 36 ==> x^{2} + y^{2} = 36 + 8 ==> x^{2} + y^{2} = 44.

6. The legs of a right angled triangle are of lengths 12m and 16m. What is the area of the circle in ‘m^{2}’ passing through all the three vertices of the triangle?

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**Correct answer is option - 1**

**Explanation: **Since the circle is passing through all the three vertices, hence the diameter of the circle is the hypotenuse of the right angled triangle.

If the hypotenuse is the diameter, only then it can have an angle of 90° on the circle.

Hence the hypotenuse of the triangle can be calculated using the Pythagorean Theorem.

This gives: (hypotenuse)^{2} = 12^{2} + 16^{2} ==> (hypotenuse)^{2} = 400 ==> hypotenuse = √400.

Therefore the hypotenuse = diameter of the circle = 20m è radius of the circle = 10m.

Area of the circle = π * (radius)^{2} = π * 100 ==> Area = 100π m^{2}.

7. Which of the following represents the domain of the function, f(x) = √(2x – 6)?

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**Correct answer is option - 3**

**Explanation: **Any square root function cannot have a negative number inside the square root if it needs to have only real solutions.

Hence we set the inequality like this: 2x – 6 ≥ 0.

This gives: 2x ≥ 6 ==> x ≥ 3 is the domain of the given function.