# Exercise : 2

(b) 90 cc

(c) 110 cc

(d) 105 cc

(e) 100 cc

**Ans.b**

Edge of the cube = \(\sqrt[3]{343}\) = 7 cm

∴ Radius of cone = 3.5 cm

height = 7 cm

volume of cone = ^{1}⁄_{3}πr^{2}h

^{1}⁄_{3}πr^{2}h = ^{1}⁄_{3} × ^{22}⁄_{7} × (3.5)^{2} × 7 = ^{1}⁄_{3} × 22 × 12.25 ≈ 90 sec

(b) 6% increase

(c) 5% decrease

(d) 4% decrease

(e) None of these

**Ans.d**

Percentage change = x - y - ^{xy}⁄_{100}

= 20 – 20 – \(\frac{20 \times 20}{100} = -4\% = 4\% \;\; decrease\)

(b) 1260 sq.metres

(c) 1280 sq.metres

(d) 1380 sq.metres

(e) None of these

**Ans.b**

Let the length and breadth be 7x and 5x respectively.

Then, P = 2(7x + 5x) = 144 ⇒ x = 6

Area = 7 × 6 × 5 × 6 = 1260 sq.m.

(b) 24 cm

(c) 22 cm

(d) Data inadequate

(e) None of these

**Ans.d**

P = 2(l + b) = L + B + h = L + b + 12.

Data inadequate.

^{2}) sq cm, where B is less than 8 cm. What is the breadth of that rectangle?

(b) 10 cm

(c) 8 cm

(d) Data inadequate

(e) None of these

**Ans.a**

Diagonal^{2} = 64 + B^{2} or, 10^{2} = 64 + 6^{2}

(b) 16% increase

(c) 8% decrease

(d) 16% decrease

(e) None of these

**Ans.d**

Regd effect = \(\left | 40 - 40 - \frac{40 \times 40}{100} \right |\% = -16\%\)

i.e., the area will decrease by 16%

(b) 6.16

(c) 122.66

(d) Data inadequate

(e) None of these

**Ans.a**

Req. area = π[(17.5 + 1.4)^{2} – (17.5)^{2}]

= ^{22}⁄_{7} × (36.4 × 1.4)[since a^{2} - b^{2} = (a + b)(a - b)]

= 22 × 36.4 × 0.2 = 160.16 sq m

(b) 120

(c) 50

(d) Data inadequate

(e) None of these

**Ans.e**

Perimeter of the rectangular plot = [(b + 20) + b] × 2

= \(\frac{5300}{26.5} = 200\)

∴ (2b + 20)2 = 200

⇒ b = 40

⇒ l = 40 + 20 = 60 m

(b) 68.5 sq.in

(c) 64.5 sq.in

(d) 66.5 sq.in

(e) None of these

**Ans.e**

Required area = 6 × 12 – \(\left \{ 2 \times \pi\left ( \frac{2}{2} \right )^{2} + \pi\left ( \frac{1}{2} \right )^{2} \right \}\)

= 72 - (2π + ^{π}⁄_{4}) = 72 - ^{9π}⁄_{4} = 72 - ^{9}⁄_{4} × ^{22}⁄_{7}

= 72 - (^{99}⁄_{14}) = 7.07 = 64.94 sq in.

(b) 60 m

(c) 4800 m

(d) Data inadequate

(e) None of these

**Ans.b**

a^{2} = 45 × 40 = 1800

∴ a = \(\sqrt{1800} = 30\sqrt{2}\)

∴ Diagonal of the square = √2 a = √2 × 30√2

= 30 × 2 = 60 m