Mensuration Model Questions Set 2 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

We have P (A ∪ B) = 0.7 and P (A ∩ B) = 0.2

Now, P(A ∪ B) = P(A) + P(B) - P (A ∩ B)

⇒P(A) + P(B) = 0.9⇒1-P($\ov{A}$) + 1 - P($\ov{B}$) = 0.9

⇒P($\ov{A}$) + P($\ov{B}$) = 1.1

Answer: (b)

(i) Miss C is taken

(1) B included ⇒A excluded ⇒$^4C_1 . ^4C_2$ = 24

(2) B excluded ⇒$^4C_1 . ^5C_3$ = 40

(ii) Miss C is not taken

⇒B does not comes ; $^4C_2 . ^5C_3$ = 60⇒Total = 124

Answer: (c)

Let $E_1$ be the event that exactly two players scored more than 50 runs then P$(E_1) = 1/2 × 1/3 × 3/4 × 9/{10} + 1/2 × 2/3 × 1/4 × 9/{10} + 1/2 × 2/3 × 3/4 × 1/{10} + 1/2 × 1/3 × 1/4 × 9/{10} + 1/2 × 1/3 × 3/4 × 1/{10} + 1/2 × 2/3 × 1/4 × 1/{10} = {65}/{240}$

Let $E_2$ be the event that A and B scored more than 50 runs, then P$(E_1 ∩ E_2) = 1/2 × 1/3 × 3/4 × 9/{10} = {27}/{240}$

∴ Desired probability

= $P(E_2/E_1) = {P(E_1 ∩ E_2)}/{P(E_1)} = {27}/{65}$

Answer: (b)

We know that when two triangles are similar then ratio of their areas is equal to square of corresponding sides.

${\text"area of ΔABC"}/{\text"area of ΔXYZ"} = {AB^2}/{XY^2} ⇒ {32}/{60.5} = {AB^2}/{(7.7)^2}$

⇒ ${32 × 59.29}/{60.5} = AB^2 ⇒ 31.36 = AB^2$

∴ AB = $√{31.36}$ = 5.6 cm

Answer: (a)

Since we are given that a and c are co-prime i.e. HCF of a and c is 1, therefore we can say that a definitely divides d exactly.

So, a is a factor of d.

Answer: (b)

Let Radius of circle be r

Case I (Square)

side = x

$x^2 + x^2 = (2r)^2$

$2x^2 = 4r^2$

x = $√2$r

Case II (equilateral triangle)

y cos30° = h

y × $√3/2$ = h

r = $2/3 × √3/2$ y

r = $y/√3$

[if h= 3 r = 2]

$y/{2r} = √3/2$

y = ${2 √3r}/2 = √3r$

$x/y = {√2 r}/{√3 r} = √{2/3}$