Area
1.The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

Explanation:

Solution:
Perimeter = Distance covered in 8 min. =
$\left(\frac{12000}{60}×8\right)m=1600\phantom{\rule{thinmathspace}{0ex}}m$
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Therefore Length = 480 m and Breadth = 320 m.
Therefore Area = (480 x 320) m2 = 153600 m2.

2.An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

Explanation:

Solution:
100 cm is read as 102 cm.
∴ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]

= (102 + 100) x (102 - 100)

= 404 cm2.
∴ Percentage error =
$\left(\frac{404}{100×100}×100\right)\mathrm{%}=4.04\mathrm{%}$

3.The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

Explanation:

Solution:

4.A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

Explanation:

Solution:
Area of the park = (60 x 40) m2 = 2400 m2.
Area of the lawn = 2109 m2.
∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x + 40x - x2 = 291
x2 - 100x + 291 = 0
⇒ (x - 97)(x - 3) = 0
x = 3.

5.The diagonal of the floor of a rectangular closet is 7 1/2 feet. The shorter side of the closet is 4 1/2 feet. What is the area of the closet in square feet?

Explanation:

Solution:
$\begin{array}{rl}& \text{Outer}\phantom{\rule{thinmathspace}{0ex}}\text{Side}\\ & =\sqrt{{\left(\frac{15}{2}\right)}^{2}-{\left(\frac{9}{2}\right)}^{2}}ft\\ & =\sqrt{\frac{225}{4}-\frac{81}{4}ft}\\ & =\sqrt{\frac{144}{4}ft}\\ & =6ft\\ & \therefore \text{Area}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{closet}=\left(6×4.5\right)sq.\phantom{\rule{thinmathspace}{0ex}}ft=27\phantom{\rule{thinmathspace}{0ex}}sq.\phantom{\rule{thinmathspace}{0ex}}ft.\end{array}$

6.A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{original}\phantom{\rule{thinmathspace}{0ex}}\text{length}=x\phantom{\rule{thinmathspace}{0ex}}\text{and}\\ & \text{original}\phantom{\rule{thinmathspace}{0ex}}\text{breadth}=y\\ & \text{Decrease}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{area}\\ & =xy-\left(\frac{80}{100}x×\frac{90}{100}y\right)\\ & =\left(xy-\frac{18}{25}xy\right)\\ & =\frac{7}{25}xy\\ & \therefore \text{Decrease}\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}=\\ & \left(\frac{7}{25}xy×\frac{1}{xy}×100\right)\mathrm{%}=28\mathrm{%}\end{array}$

7.A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

Explanation:

Solution:
Let the side of the square(ABCD) be x metres.

AC = √2x = (1.41x) m.
Saving on 2x metres = (0.59x) m.

8.What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

Explanation:

Solution:
$\begin{array}{rl}& \text{Length}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{largest}\phantom{\rule{thinmathspace}{0ex}}\text{tile}=\\ & \text{H}\text{.C}\text{.F}\text{.}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}1517\phantom{\rule{thinmathspace}{0ex}}cm\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}902\phantom{\rule{thinmathspace}{0ex}}cm=41\phantom{\rule{thinmathspace}{0ex}}cm\\ & \text{Area}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{each}\phantom{\rule{thinmathspace}{0ex}}\text{tile}=\left(41×41\right)c{m}^{2}\\ & \therefore \text{Required}\phantom{\rule{thinmathspace}{0ex}}\text{number}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{tiles}\\ & =\left(\frac{1517×902}{41×41}\right)\\ & =814\end{array}$

9.The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

Explanation:

Solution:
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
∴ Area = (l x b) = (63 x 40) m2 = 2520 m2.

10.The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{original}\phantom{\rule{thinmathspace}{0ex}}\text{length}=x\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\\ & \text{Original}\phantom{\rule{thinmathspace}{0ex}}\text{breadth}=y\\ & \text{Original}\phantom{\rule{thinmathspace}{0ex}}\text{area}=xy\\ & \text{New}\phantom{\rule{thinmathspace}{0ex}}\text{length}=\frac{x}{2}\\ & \text{New}\phantom{\rule{thinmathspace}{0ex}}\text{breadth}=3y\\ & \text{New}\phantom{\rule{thinmathspace}{0ex}}\text{area}=\left(\frac{x}{2}×3y\right)=\frac{3}{2}xy\\ & \therefore \text{Increase}\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}=\left(\frac{1}{2}xy×\frac{1}{xy}×100\right)\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=50\mathrm{%}\end{array}$

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