# Exercise : 1

(b) 26 metres

(c) 34.5 metres

(d) Cannot be determined

(e) None of these

**Ans.e**

Let the breadth of the rectangular field be ‘x’ m. Then,length of the field will be

\(x + \frac{x \times 15}{100} = \frac{23x}{20}\)Now, \(x \times \frac{23x}{20} = 460\)

or, 23x^{2} = 460 × 20

or, x^{2} = 20 × 20

or, x = 20 m

(b) 1700

(c) 3440

(d) Cannot be determined

(e) None of these

**Ans.c**

Let l and b be the length and breadth of rectangular plot respectively.

∴ According to the question,we have

2(l + b) = 340 ⇒ l + b = 170

Now, (l + 2) and (b + 2) be the length and breadth of plot with boundary.

∴ Required area = (l + 2) (b + 2) – lb

= lb + 2l + 2b + 4 – lb

= 2(l + b) + 4 = 344

∴ Required cost = 344 × 10 = 3440

(b) 900

(c) 1800

(d) 360

(e) None of these

**Ans.c**

Area of cube

= 6× (side)^{2} = 6 × 10 × 10 = 600 square feet.

Cost to paint outside of the cube = ^{600}⁄_{20} × 60

= 1800

(b) 32

(c) 60

(d) Data inadequate

(e) None of these

**Ans.a**

Capacity of the tank = 20 × 13.5= 270 litres

When the capacity of each bucket = 9 litres, then the required no. of buckets

= ^{270}⁄_{9} = 30

(b) 80

(c) 60

(d) 40

(e) None of these

**Ans.c**

Let the length of the rectangular hall be ‘x’ m, then the breadth of the rectangular hall = ^{2x}⁄_{3} m.

Area of hall = ^{2x}⁄_{3} × x = ^{2x2}⁄_{3}

or, ^{2x2}⁄_{3} = 2400 or x = 60 m

^{2}

(b) 225 cm

^{2}

(c) 81 cm

^{2}

(d) 100 cm

^{2}

(e) None of these

**Ans.d**

Let the original side of the square = x cm

\(\frac{x + 5}{x} = \frac{3}{2}\) or 2x + 10 = 3x\ x = 10 cm

\ original area = (10)^{2} = 100 cm^{2}

(b) T > C > S

(c) T > S > C

(d) C > S > T

(e) None of these

**Ans.d**

Let the perimeter of each be a.

Then, side of the equilateral triangle = ^{a}⁄_{3}; side of square = ^{a}⁄_{4}; radius of the circle = ^{a}⁄_{2π}.

∴ \(T = \frac{\sqrt{3}}{4} \times \left(\frac{a}{2} \right)^{2} = \frac{\sqrt{3}a^{2}}{36}\); S = (^{a}⁄_{4})^{2} = ^{a2}⁄_{16};

C = π × (^{a}⁄_{2π})^{2} = ^{a2}⁄_{4π} = ^{7a2}⁄_{88}

So, C > S > T.

(b) 2.8 metres

(c) 28 metres

(d) 14 metres

(e) none of these

**Ans.e**

Capacity (volume) of a cylindrical tank = πr^{2}h

(Here r = radius and h = height of the tank)

Now, from the question, 246.4 × 0.001 = ^{22}⁄_{7} × r^{2} × 4

^{3}= 0.001 m

^{3}]

or, \(\frac{0.2464 \times 7}{22 \times 4}\) = r^{2}

or,r = 0.14 m

or,diameter = 2r = 0.28 m

(b) 81 sq.cm

(c) 120 sq.cm

(d) 225 sq.cm

(e) None of these

**Ans.e**

Let original length of each side = x cm.

Then, its area = (x^{2})cm^{2}.

Length of rectangle formed = (x + 5) cm

and its breadth = x cm.

∴ \(\frac{x + 5}{x} = \frac{3}{2}\) ⇔ 2x + 10 = 3x ⇔ x = 10

∴ Original length of each side = 10 cm and its area = 100 cm^{2}

(b) 150

(c) 280

(d) Data inadequate

(e) None of these

**Ans.e**

Let the length and breadth be l and b respectively.

^{l}⁄_{b} = ^{3}⁄_{2} or l = ^{3}⁄_{2}b .....(i)

From eq. (i)

10.5b – l 0b = 6 or, 0.5b = 3 or, b = 6 and l = 9

Area = l × b = 6 × 9 = 54 m^{2}