# Exercise : 5

^{2}must have 2 metres cut-off one of its edges in order to be a perfect fit for a rectangular room. What is the area of rectangular room?

^{2}

(b) 164 m

^{2}

(c) 152 m

^{2}

(d) 143 m

^{2}

(e) None of these

**Ans.d**

Side of square carpet = \(\sqrt{Area} = \sqrt{169} = 13 \; m\)

After cutting of one side,

Measure of one side = 13 – 2 = 11 m

and other side = 13 m (remain same)

∴ Area of rectangular room = 13 × 11 = 143 m^{2}

(b) 2

^{1}⁄

_{2}

(c) 2

(d) 5

(e) None of these

**Ans.a**

Length of frame = 30 + 2.5 × 2= 35 inch

Breadth of frame = 20 + 2.5 × 2 = 25 inch

Now, area of picture = 30 × 20 = 600 sq. inch

Area of frame = (35 × 2.5) + (25 × 2.5)

∴ Area of frame = 875 – 600 = 275 sq. inch

^{2}, find the width of the path.

(b) 3 m

(c) 2 m

(d) 1 m

(e) None of these

**Ans.c**

Let the width of the path = W m

then, length of plot with path = (15 + 2W) m

and breadth of plot with path = (10 + 2W) m

Therefore, Area of rectangular plot (without path)

= 15 × 10 = 150 m^{2}

and Area of rectangular plot (with path)

= 150 + 54 = 204 m^{2}

Hence, (15 + 2W) × (10 + 2W) = 204

⇒ 4W^{2} + 50W – 54 = 0

⇒ 2W^{2} + 25W – 27 = 0

⇒ (W – 2)(W + 27) = 0

Thus W = 2 or –27

∴ with of the path = 2 m

(b) 18%

(c) 36%

(d) 64%

(e) None of these

**Ans.a**

If area of a circle decreased by x % then the radius of a circle decreases by

\(\left(100 - 10\sqrt{100 - x} \right)\% = \left(100 - 10\sqrt{100 - 36} \right)\%\)= \(\left(100 - 10\sqrt{64} \right)\%\)

= 100 - 80 = 20%

^{2}

(b) 120 m

^{2}

(c) 108 m

^{2}

(d) 58 m

^{2}

(e) None of these

**Ans.a**

Area of the outer rectangle = 19 × 16 = 304 m^{2}

Area of the inner rectangle = 15 × 12 = 180 m^{2}

Required area = (304 – 180) = 124 m^{2}

^{2}⁄

_{3}

(b)

^{1}⁄

_{2}

(c)

^{1}⁄

_{3}

(d)

^{5}⁄

_{7}

(e) None of these

**Ans.a**

Area of paper = 12 × 5 = 60 sq. inch

Area of typing part = (12 - 1 × 2) × (5 - ^{1}⁄_{2} × 2)

= (12 – 2) × (5 – 1) = (10 × 4) sq. inch

∴ Required fraction = ^{40}⁄_{60} = ^{2}⁄_{3}

(b) 1,800

(c) 1,900

(d) 1,700

(e) None of these

**Ans.c**

Total area of road

= Area of road which parallel to length + Area of road which parallel to breadth – overlapped road

= 70 × 5 + 30 × 5 – 5 × 5

= 350 + 150 – 25

= 500 – 25 = 475 m^{2}

∴ Cost of gravelling the road

= 475 × 4 = 1900

^{2}

(b) 1700 m

^{2}

(c) 1598 m

^{2}

(d) 1500 m

^{2}

(e) None of these

**Ans.a**

Radius of a circular grass lawn (without path) = 35 m

∴ Area = πr^{2} = π(35)^{2}

Radius of a circular grass lawn (with path)

= 35 + 7 = 42 m

∴ Area = πr^{2} = π(42)^{2}

∴ Area of path = π(42)^{2} – π(35)^{2}

= π(42^{2} – 35^{2}) = π( 42 + 35)(42 – 35)

(b) 53 cm

(c) 56 cm

(d) 66 cm

(e) None of these

**Ans.a**

Volume of the bucket = volume of the sand emptied

Volume of sand = π(21)^{2} × 36

Let r be the radius of the conical heap.

Then, ^{1}⁄_{3}πr^{2} × 12 = π(21)^{2} × 36

or r^{2} = (21)^{2} × 9 or r = 21 × 3 = 63

(b) 250

(c) 300

(d) 330

(e) None of these

**Ans.b**

Radius of the wheel of bus = 70 cm. Then,

circumference of wheel = 2πr = 140π = 440 cm

Distance covered by bus in 1 minute

= ^{66}⁄_{60} × 1000 × 100 cms

Distance covered by one revolution of wheel

= circumference of wheel

= 440 cm

∴ Revolutions per minute = \(\frac{6600000}{60 \times 440}\) = 250