 CBSE Board 2020 Solution for Math for Class 10 Section B

# CBSE Board 2020

## Mathematics

### Question Paper Solution

#### Section B

##### 2 Marks Questions

Question 21: A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper. The following were the answers given by the students.

2x+3, 3x^2+7x+2, 4x^3+3x^2+2, x^3sqrt(3x)+7, 7x+sqrt7, 5x^3-7x+2, 2x^2+3-5/x, 5x-1/2, ax^3+bax^2+cx+d, x+1/x

1. How many of the above ten, are not polynomials?
2. How many of the above ten are quadratic polynomials?

Question 22: A child has a die whose six faces show the letters as shown here: A, B, C, D, E, A

The die is thrown once. What is the probability of getting (i) A, (ii) D?

Answer: (i) 1/3 (ii) 1/6

Question 23: In following figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

text(arΔABC)/text(areΔDBC)-(AO)/(DO) Answer: Let us draw altitudes AM and DN on BC; respectively from A and D text(ar ABC)/text(ar DBC)=(1/2xxBCxxAM)/(1/2xxBCxxDN)

=(AM)/(DN)

In ΔAMO and ΔDNO;

∠ AMO = ∠ DNO (Right angle)

∠ AOM = ∠ DON (Opposite angles)

Hence; ΔAMO ∼ ΔDNO

Hence;

(AM)/(DN)=(AO)/(DO)

Or, text(ar ABC)/text(ar DBC)=(AO)/(DO)

OR

In following figure, if AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2 So, AB2 - BD2 = AC2 - CD2

Or, AB2 - BD2 + CD2 = AC2

Or, AB2 + CD2 = BD2 + AC2 proved

Question 24: Prove that 1+(text(cot)^2α)/(1+text(cosecα))=text(cosecα)

Answer: LHS =1+(text(cot)^2α)/(1+text(cosecα))

=1+(text(cosecα+1)text(cosecα-1))/(1+text(cosecα))

=1+text(cosecα)-1=text(cosecα) proved

OR

Show that tan4θ + tan2θ = sec4θ - sec2θ

Answer: RHS = sec4θ - sec2θ

= (1 + tan2θ)2 - (1 + tan2θ)

= 1 + tan4θ + 2 tan2θ - 1 – tan2θ

= tan4θ + tan2θ = LHS proved

Question 25: Find the mode of the following frequency distribution:

 Class Frequency 15-20 20-25 25-30 30-35 35-40 40-45 3 8 9 10 3 2

Answer: Here, modal class is 30 – 35, l = 30, h = 5, f1 = 10, f0 = 9, f2 = 3

Mode can be calculated as follows:

=l+(f_1-f_0)/(2f_1-f_0-f_2)× x

=30+(10-9)/(2×10-9-3)× 5

=30+1/8×5

=(240+5)/8=(245)/8=30.625

Question 26: From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of the same height and same base radius is removed. Find the volume of the remaining solid.

Answer: Volume of cylinder =πr^2h

=(22)/7×6^2×14=1584 cm3

Volume of cone =1/3×πr^2h

=(1584)/3=528 cm3

So, volume of remaining solid =1584-528=1056 cm3