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 ……., ……………..
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 = ?
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)` = ?
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