CBSE Board 2020 Solution for Math for Class 10 Section A part 2

CBSE Board 2020

Mathematics

Question Paper Solution

Section A Part 2

1 Mark Questions

Question 11: In following figure, the angles of depressions from the observing positions O1 and O2 respectively of the object are ……., ……………..

figures of triangles

Answer: O1 30° and O2 45°


Question 12: In &Detla; ABC, Ab = `6sqrt3` cm, AC = 12 cm and BC = 6 cm, then ∠B = ?

Answer: 90°

OR

Two triangles are similar if their corresponding sides are ………………

Answer: In the same ratio

Question 13: In following figure, the length of PB = ?

concentric circles

Answer: 4 cm (Note: 3, 4 and 5 make Pythagorean triplet)

Question 14: In following figure, MN || BC and AM : MB = 1 : 2, then

`text(arΔAMN)/text(arΔABC)` = ?

triangles midpoint theorem

Answer: 1 : 9 (Ratio of areas is square of ratio of sides)


Question 15: The value of sin 32° cos 58° + cos 32° sin 58° is …………..

Answer: 1

Explanation: sin 32° cos 58° + cos 32° sin 58°

= sin 32° × cos(90°-32°) + cos 32° × sin(90°-32°)

= sin2 32° + cos2 32° = 1

Question 16: A die is thrown once. What is the probability of getting a prime number?

Answer: P(E) = `3/6=1/2`

There are three primer numbers below 6, i.e. 2, 3 and 5

Question 17: If a number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3, then find the probability of of `x^2 &let; 4`

Answer: `3/7`

OR

What is the probability that a randomly taken leap year has 52 Sundays?

Answer: `5/7`

Explanation: There are 366 days in a leap year. So, a leap year has 52 weeks and 2 days. These extra 2 days can be any of the following combinations:

Sunday-Monday, Monday-Tuesday, Tuseday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday, Saturday-Sunday

So, out of 7 possible combinations, two have Sundays which means there will be more than 52 Sundays on these two occasions. Rest of the occasions will have exactly 52 Sundays.


Question 18: If sin A + sin2 = 1, then find the value of the expression (cos2A + cos4A).

Answer: 1

Explanation: sin A + sin2 = 1

Or, sin2A = 1 – sin A ……………(1)

Now, cos2A + cos4A

= 1 – sin2A + (cos2A)2

= 1 – sin2A + (1 – sin2A)2

Substituting the value of sin2A from equation 1, we get:

1 – (1 – sin A) + (1 – (1 – sin A)2

= 1 – 1 + sin A + (1 – 1 + sin A )2

= sin A + sin2A = 1

Question 19: Find the area of the sector of a circle of radius 6 cm whose central angle is 30°. (Take π = 3.14).

Answer: Area of sector `=πr^2×θ/(360°)`

`=3.14×6^2×(30°)/(360°)`

`=3.14×36×1/(12)=3.14×3=9.42` sq cm

Question 20: Find the class marks of the classes 20 – 50 and 35 – 60.

Answer: 35 and 47.5



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