# Exercise : 5

1. A square carpet with an area 169 m2 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?
(a) 180 m2
(b) 164 m2
(c) 152 m2
(d) 143 m2
(e) None of these

#### View Ans & Explanation

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 m2

2. A picture 30" × 20" has a frame 2½" wide. The area of the picture is approximately how many times the area of the frame?
(a) 4
(b) 212
(c) 2
(d) 5
(e) None of these

#### View Ans & Explanation

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

3. A rectangular plot 15 m × 10 m, has a path of grass outside it. If the area of grassy pathway is 54 m2, find the width of the path.
(a) 4 m
(b) 3 m
(c) 2 m
(d) 1 m
(e) None of these

#### View Ans & Explanation

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 m2

and Area of rectangular plot (with path)

= 150 + 54 = 204 m2

Hence, (15 + 2W) × (10 + 2W) = 204

⇒ 4W2 + 50W – 54 = 0

⇒ 2W2 + 25W – 27 = 0

⇒ (W – 2)(W + 27) = 0

Thus W = 2 or –27

∴ with of the path = 2 m

4. If the area of a circle decreases by 36%, then the radius of a circle decreases by
(a) 20%
(b) 18%
(c) 36%
(d) 64%
(e) None of these

#### View Ans & Explanation

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%

5. The floor of a rectangular room is 15 m long and 12 m wide. The room is surrounded by a verandah of width 2 m on all its sides. The area of the verandah is :
(a) 124 m2
(b) 120 m2
(c) 108 m2
(d) 58 m2
(e) None of these

#### View Ans & Explanation

Ans.a

Area of the outer rectangle = 19 × 16 = 304 m2 Area of the inner rectangle = 15 × 12 = 180 m2

Required area = (304 – 180) = 124 m2

6. A typist uses a paper 12" by 5" length wise and leaves a margin of 1" at the top and the bottom and a margin of ½" on either side. What fractional part of the paper is available to him for typing ?
(a) 23
(b) 12
(c) 13
(d) 57
(e) None of these

#### View Ans & Explanation

Ans.a

Area of paper = 12 × 5 = 60 sq. inch

Area of typing part = (12 - 1 × 2) × (5 - 12 × 2)

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

∴ Required fraction = 4060 = 23

7. A rectangular lawn 70 m × 30 m has two roads each 5 metres wide, running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the cost of gravelling the road at the rate of 4 per square metre.
(a) 2,000
(b) 1,800
(c) 1,900
(d) 1,700
(e) None of these

#### View Ans & Explanation

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 m2

∴ Cost of gravelling the road

= 475 × 4 = 1900

8. A circular grass lawn of 35 metres in radius has a path 7 metres wide running around it on the outside. Find the area of path.
(a) 1694 m2
(b) 1700 m2
(c) 1598 m2
(d) 1500 m2
(e) None of these

#### View Ans & Explanation

Ans.a

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

∴ Area = πr2 = π(35)2

Radius of a circular grass lawn (with path)

= 35 + 7 = 42 m

∴ Area = πr2 = π(42)2

∴ Area of path = π(42)2 – π(35)2

= π(422 – 352) = π( 42 + 35)(42 – 35)

9. A cylindrical bucket of height 36 cm and radius 21 cm is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed, the height of the heap being 12 cm. The radius of the heap at the base is :
(a) 63 cm
(b) 53 cm
(c) 56 cm
(d) 66 cm
(e) None of these

#### View Ans & Explanation

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, 13πr2 × 12 = π(21)2 × 36

or r2 = (21)2 × 9 or r = 21 × 3 = 63

10. The radius of the wheel of a bus is 70 cms and the speed of the bus is 66 km/h, then the r.p.m. (revolutions per minutes)of the wheel is
(a) 200
(b) 250
(c) 300
(d) 330
(e) None of these

#### View Ans & Explanation

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

= 6660 × 1000 × 100 cms

Distance covered by one revolution of wheel

= circumference of wheel

= 440 cm

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