# CBSE Class 10 Maths 2018 Question Paper Solution

###### Class 10 Board Paper Solution | 1 Mark Questions | Section A

Section A contains 6 questions of 1 mark each. Scroll down for explanatory answer and video solution to all six 1-mark questions. Questions appeared from Quadratic Equations, Real Numbers, Coordinate Geometry, Arithmetic Progressions, Trigonometry, and Triangles.

1. If x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0, then find the value of k.

Hint to solve this Quadratic Equations question

Because 3 is a root of the equation, substitute the value of x as 3 in the equation and solve to find the value of k.

2. What is the HCF of smallest prime number and the smallest composite number?

Hint to solve this Real Numbers problem

What is the smallest prime number? What is the smallest composite number?

Compute the HCF of these two numbers.

3. Find the distance of a point P(x, y) from the origin.

Hint to solve this Coordinate Geometry Question

Concept: The formula to compute the distance between two points whose coordinates are (x1, y1) and (x2, y2).
Approach: One of the points is the origin (0, 0). Let it be (x1, y1)
The second point is P(x, y). Let it be (x2, y2).
Substitute these values in the formula to find the distance between two points to find the answer.

4. In an AP, if the common difference (d) = - 4, and the seventh term (a7) is 4, then find the first term.

Hint to solve this CBSE 2018 1 mark Arithmetic Progressions problem

Approach: The nth term of an AP an = a1 + (n - 1)d, where a1 is the first term of the AP and 'd' is the common difference.
Given Data: a7 = 4 and d = (-4). So, n = 7.
Substitute: The above values in the equation to find the nth term an and find a1.

5. What is the value of (cos2 67° - sin2 23°)?

Hint to solve this CBSE 2018 1 mark Trigonometry question

Concept: Trigonometric values of complementary angles.
Step 1: Rewrite cos 67 as cos (90 - 23).
Step 2: Apply properties of trigonometric values of complementary angles and compute the answer.

6. Given ∆ ABC is similar to ∆ PQR. If $$frac{AB}{PQ} = \frac{1}{3}$, then find $\frac{ar ∆ ABC}{ar ∆ PQR}$ Hint to solve this CBSE 2018 maths 1 mark Triangles problem Concept / Theorem: Ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides. Given Data: Ratio of corresponding sides of the two similar triangles is given. Apply Theorem: Apply the theorem about the relation between the ratio of corresponding sides of two similar triangles and their areas to compute the answer. ### Video Solution and Explanatory Answers to CBSE 2018 Section A - 1 mark questions 1 to 6 Question 1: If x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0, then find the value of k. #### Q1 Video Explanation Scroll down further for explanatory answer text #### Explanatory Answer | Q1 2018 Board Paper x = 3 is one root of x2 - 2kx - 6 = 0 Because 3 is one of the roots of the equation, substituting x= 3 will satisfy the equation. 32 - 2 × k × 3 - 6 = 0 9 - 6k - 6 = 0 6k = 3 or k = $\frac{3}{6}$ = $\frac{1}{2}$ Question 2: What is the HCF of smallest prime number and the smallest composite number? #### Q2 Video Explanation Scroll down further for explanatory answer text #### Explanatory Answer | Q2 2018 Board Paper Smallest prime number is 2. Smallest composite number is 4. HCF$2, 4) = 2.

Question 3: Find the distance of a point P(x, y) from the origin.

#### Q3 Video Explanation

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#### Explanatory Answer | Q3 2018 Board Paper

Distance between two points A(x1, y1) and B(x2, y2) = $$sqrt{$x_2 - x_1$^2 + (y_2 - y_1)^2})
∴ distance between (x, y) and (0, 0) = $$sqrt{$x - 0$^2 + (y - 0)^2})
= $$sqrt{x^2 + y^2}$ Question 4: In an AP, if the common difference$d) = - 4, and the seventh term (a7) is 4, then find the first term.

#### Q4 Video Explanation

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#### Explanatory Answer | Q4 2018 Board Paper

Common difference, d = -4
7th term a7 = 4
Let a1 be the first term of this AP.

###### nth term of an AP an = a1 + (n - 1)d

∴ a7 = a1 + (7 - 1)d
4 = a1 + 6(-4)
4 = a1 - 24
a1 = 28

Question 5: What is the value of (cos2 67° - sin2 23°)?

#### Q5 Video Explanation

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#### Explanatory Answer | Q5 2018 Board Paper

The given trigonometric expression is cos267° – sin223°
The expression is of the form (a2 - b2)
(a2 - b2) = (a + b)(a - b)
∴ cos267° – sin223° = (cos 67° + sin 23°) (cos 67° - sin 23°) ... eqn(1)

Trigonometric Ratio of Complementary Angles: sin (90 - θ) = cos θ
Let us express sin 23° = sin(90° - 67°)
sin (90° - 67°) = cos 67°
Substitute sin 23° as cos 67° in eqn(1): (cos 67° + sin 23°) (cos 67° - sin 23°) = (cos 67° + cos 67°) (cos 67° - cos 67°)
(2 cos 67°) (0) = 0.

Question 6: Given ∆ ABC is similar to ∆ PQR. If $\frac{AB}{PQ} = \frac{1}{3}$, then find $\frac{ar ∆ ABC}{ar ∆ PQR}$

#### Q6 Video Explanation

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#### Explanatory Answer | Q6 2018 Board Paper

Theorem: Ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides.
$\frac{\text{Area of Δ ABC}}{\text{Area of Δ PQR}}$ = $\frac{\text{Square of corresponding side of ABC}}{\text{Square of corresponding side of PQR}}$ = $\frac{{AB}^2}{{PQ}^2}$
$\frac{{AB}^2}{{PQ}^2}$ = $\frac{1^2}{3^2}$ = $\frac{1}{9}$

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