Problems on Area - Solved Examples(Set 1)

Problems on Area - Solved Examples(Set 1)

1. An error $2\%$ in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?
A. $4.04\%$	B. $2.02\%$
C. $4\%$	D. $2\%$

answer with explanation

Answer: Option A
Explanation:
Solution 1

Percentage error in calculated area
$=(2+2+\frac{2\times 2}{100})\%=4.04\%$

(This formula is explained in detail here)

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Solution 2

Error $=2\%$ while measuring the side of a square.

Let correct value of the side of the square $=100$
Then, measured value $=100+2=102$(∵ $2$ is $2\%$ of $100$)

Correct area of the square $=100\times 100=10000$
Calculated area of the square $=102\times 102=10404$

Error $=10404-10000=404$

Percentage error $=\frac{\text{error}}{\text{actual value}}\times 100$
$=\frac{404}{10000}\times 100=4.04\%$

2. 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. The area of the lawn is $2109$ sq. m. what is the width of the road?
A. $5$ m	B. $4$ m
C. $2$ m	D. $3$ m

answer with explanation

Answer: Option D
Explanation:


Please refer the diagram given above.

Area of the park $=60\times 40=2400$ m²
Given that area of the lawn $=2109$ m²
∴ Total area of the cross roads $=2400-2109=291$ m²

Assume that the width of the cross roads $=x$

Then, total area of the cross roads
= Area of road $1$ + Area of road $2$ - (Common area of the cross roads)
$=60x+40x-x^2$

(Let's look in detail how we got the total area of the cross roads as $60x+40x-x^2$. As shown in the diagram, area of road $1$ $=60x$. This has the areas of the parts $1,2$ and $3$given in the diagram. Area of road $2$ $=40x$. This has the parts $4,5$ and $6$. You can see that there is an area which is intersecting (i.e. part $2$ and part $5$) and the intersection area$=x^2$.

Since $60x+40x$ covers the intersecting area $\left(x^2\right)$ two times (part $2$ and part $5$), we need to subtract the intersecting area of $\left(x^2\right)$ one time to get the total area. Hence total area of the cross roads $=60x+40x-x^2$

Now, we have
Total area of cross roads $=60x+40x-x^2$
But total area of the cross roads $=291$ m²

Hence,
$60x+40x-x^2=291\Rightarrow 100x-x^2=291\Rightarrow x^2-100x+291=0\Rightarrow (x-97)(x-3)=0$$\Rightarrow x=3$ ($x$ cannot be $97$ as the park is only $60$ m long and $40$ m wide)

3. A towel, when bleached, lost $20\%$ of its length and $10\%$ of its breadth. What is the percentage decrease in area?
A. $30\%$	B. $28\%$
C. $32\%$	D. $26\%$

answer with explanation

Answer: Option B
Explanation:
Solution 1

percentage change in area
$=(-20-10+\frac{20\times 10}{100})\%=-28\%$
i.e., area is decreased by $28\%$

(This formula is explained in detail here)

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Solution 2

Let original length $=10$
original breadth $=10$
Then, original area $=10\times 10=100$

Lost $20\%$ of length
⇒ New length $=10-2=8$(∵ $2$ is $20\%$ of $10$)

Lost $10\%$ of breadth
⇒ New breadth $=10-1=9$(∵ $1$ is $10\%$ of $10$)

New area $=8\times 9=72$

Decrease in area
= original area - new area
$=100-72=28$

Percentage decrease in area
$=\frac{\text{decrease in area}}{\text{original area}}\times 100=\frac{28}{100}\times 100=28\%$

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Solution 3

Let original length $=l,$
original breadth $=b$
Then, original area $=lb$

Lost $20\%$ of length
⇒ New length $=l\times \frac{80}{100}=0.8l$

Lost $10\%$ of breadth
⇒ New breadth $=b\times \frac{90}{100}=0.9b$

New area $=0.8l\times 0.9b=0.72lb$

Decrease in area
= original area - new area
$=lb-0.72lb=0.28lb$

Percentage decrease in area
$=\frac{\text{decrease in area}}{\text{original area}}\times 100=\frac{0.28lb}{lb}\times 100=28\%$

4. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?
A. $25\%$ increase	B. $25\%$ decrease
C. $50\%$ decrease	D. $50\%$ increase

answer with explanation

Answer: Option D
Explanation:
Solution 1

Length is halved.
i.e., length is decreased by $50\%$

Breadth is tripled
i.e., breadth is increased by $200\%$

Change in area
$=(-50+200-\frac{50\times 200}{100})\%=50\%$

i.e., area is increased by $50\%$

(This formula is explained in detail here)

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Solution 2

Let original length $=10$
original breadth $=10$
Then, original area $=10\times 10=100$

Length is halved
⇒ New length $=\frac{10}{2}=5$

breadth is tripled.
⇒ New breadth $=10\times 3=30$

New area $=5\times 30=150$

Increase in area
= new area - original area
$=150-100=50$

Percentage increase in area
$=\frac{\text{increase in area}}{\text{original area}}\times 100=\frac{50}{100}\times 100=50\%$

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Solution 3

Let original length $=l,$
original breadth $=b$
Then, original area $=lb$

Length is halved
⇒ New length $=\frac{l}{2}$

breadth is tripled
⇒ New breadth $=3b$

New area $=\frac{l}{2}\times 3b=\frac{3lb}{2}$

Increase in area
= new area - original area
$=\frac{3lb}{2}-lb=\frac{lb}{2}$

Percentage increase in area
$=\frac{\text{increase in area}}{\text{original area}}\times 100=\frac{\left(\frac{lb}{2}\right)}{lb}\times 100=\frac{1}{2}\times 100=50\%$

5. A person walked diagonally across a square plot. Approximately, what was the percent saved by not walking along the edges?
A. $35\%$	B. $30\%$
C. $20\%$	D. $25\%$

answer with explanation

Answer: Option B
Explanation:
Solution 1



Consider a square plot as shown above.
Let length of each side $=1$
Then, length of the diagonal $=\sqrt{1^2+1^2}=\sqrt{2}$

Distance travelled if walked along the edges
= BC + CD $=1+1=2$

Distance travelled if walked diagonally
= BD $=\sqrt{2}=1.41$

Distance saved $=2-1.41=0.59$

Percent distance saved
$=\frac{0.59}{2}\times 100=0.59\times 50\approx 30\%$

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Solution 2



Consider a square plot as shown above.
Let length of each side = $x$
Then, length of the diagonal = $\sqrt{x^2+x^2}=\sqrt{2}x$

Distance travelled if walked along the edges
= BC + CD = $x+x=2x$

Distance travelled if walked diagonally
= BD = $\sqrt{2}x=1.41x$

Distance saved $=2x-1.41x=0.59x$

Percent distance saved
$=\frac{0.59x}{2x}\times 100=0.59\times 50\approx 30\%$