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. $2\%$ |

| Discuss |

Answer: Option A

Explanation:

**Solution 1**

Percentage error in calculated area

$=\left(2+2+\dfrac{2×2}{100}\right)\%=4.04\%$

(This formula is explained in detail here)

**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 $=100×100=10000$

Calculated area of the square $=102×102=10404$

Error $=10404-10000=404$

Percentage error $=\dfrac{\text{error}}{\text{actual value}}×100$

$=\dfrac{404}{10000}×100 = 4.04\%$

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 |

| Discuss |

Answer: Option D

Explanation:

Please refer the diagram given above.

Area of the park $=60×40=2400$ m^{2}

Given that area of the lawn $=2109$ m^{2}

∴ Total area of the cross roads $=2400-2109=291$ m^{2}

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 $(x^2)$ two times (part $2$ and part $5$), we need to subtract the intersecting area of $(x^2)$ 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 $=291$ m^{2}

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. $26\%$ |

| Discuss |

Answer: Option B

Explanation:

**Solution 1**

percentage change in area

$=\left(-20-10+\dfrac{20×10}{100}\right)\%=-28\%$

i.e., area is decreased by $28\%$

(This formula is explained in detail here)

**Solution 2**

Let original length $=10$

original breadth $=10$

Then, original area $=10×10=100$

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 $=8×9=72$

Decrease in area

= original area - new area

$=100-72=28$

Percentage decrease in area

$=\dfrac{\text{decrease in area}}{\text{original area}}×100\\=\dfrac{28}{100}×100=28\%$

**Solution 3**

Let original length $=l,$

original breadth $=b$

Then, original area $=lb$

Lost $20\%$ of length

⇒ New length $=l×\dfrac{80}{100}=0.8l$

Lost $10\%$ of breadth

⇒ New breadth $=b×\dfrac{90}{100}=0.9b$

New area $=0.8l×0.9b=0.72lb$

Decrease in area

= original area - new area

$=lb-0.72lb=0.28lb$

Percentage decrease in area

$=\dfrac{\text{decrease in area}}{\text{original area}}×100\\=\dfrac{0.28lb}{lb}×100=28\%$

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 |

| Discuss |

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

$=\left(-50+200-\dfrac{50×200}{100}\right)\%=50\%$

i.e., area is increased by $50\%$

(This formula is explained in detail here)

**Solution 2**

Let original length $=10$

original breadth $=10$

Then, original area $=10×10=100$

Length is halved

⇒ New length $=\dfrac{10}{2}=5$

breadth is tripled.

⇒ New breadth $=10×3=30$

New area $=5×30=150$

Increase in area

= new area - original area

$=150-100=50$

Percentage increase in area

$=\dfrac{\text{increase in area}}{\text{original area}}×100\\=\dfrac{50}{100}×100=50\%$

**Solution 3**

Let original length $=l,$

original breadth $=b$

Then, original area $=lb$

Length is halved

⇒ New length $=\dfrac{l}{2}$

breadth is tripled

⇒ New breadth $=3b$

New area $=\dfrac{l}{2}×3b=\dfrac{3lb}{2}$

Increase in area

= new area - original area

$=\dfrac{3lb}{2}-lb=\dfrac{lb}{2}$

Percentage increase in area

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

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. $25\%$ |

| Discuss |

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

$=\dfrac{0.59}{2}×100=0.59×50 \approx 30 \%$

**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

$=\dfrac{0.59x}{2x}×100=0.59×50 \approx 30\%$

^{2}

so area of larger square = $\dfrac{1}{2}×4^2=\dfrac{1}{2}×16=8$

now area of the smaller square $=\dfrac{1}{2}$(area of larger square)$=\dfrac{1}{2}×8=4$

Area of the larger square $=2\sqrt{2}×2\sqrt{2}=8$

Area of smaller square $=\dfrac{8}{2}=4$

Side of the smaller square $=\sqrt{4}=2$

Diagonal of the smaller square $=2\sqrt{2}$

^{2}= 2464

22/7 * r

^{2}= 2464

r

^{2}=2464 * 7/22= 784

r = 28

diameter = 2r = 56 m

root(30^2+24^2+18^2)

= root(1800)

= 30 * root(2) m

2 * pi * r = 44

2 * (22/7) * r = 44

r = 7

Area = pi * r

^{2}= (22/7) * 7

^{2}= 154

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