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Question Detail
The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
 32%
 34%
 42%
 44%
Answer: Option D
Explanation:
Let original length = x metres and original breadth = y metres.
\begin{aligned}
\text{Original area } = \text{xy } m^2 \\
\text{New Length }= \frac{120}{100}x = \frac{6}{5}x \\
\text{New Breadth }= \frac{120}{100}y = \frac{6}{5}y \\
=>\text{New Area }= \frac{6}{5}x * \frac{6}{5}y \\
=>\text{New Area }= \frac{36}{25}xy \\
\text{Area Difference} = \frac{36}{25}xy  xy \\
= \frac{11}{25}xy \\
Increase \% = \frac{Differnce}{Actual}*100 \\
= \frac{11xy}{25}*\frac{1}{xy}*100 = 44\%
\end{aligned}
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:
 \begin{aligned} 152600 m^2\end{aligned}
 \begin{aligned} 153500 m^2\end{aligned}
 \begin{aligned} 153600 m^2\end{aligned}
 \begin{aligned} 153800 m^2\end{aligned}
Answer: Option C
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
As per question length is 3x and width is 2x
We know perimeter of rectangle is 2(L+B)
So, 2(3x+2x) = 1600
=> x = 160
So Length = 160*3 = 480 meter
and Width = 160*2 = 320 meter
Finally, Area = length * breadth
= 480 * 320 = 153600
2. A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.
 10 m * 5 m
 15 m * 5 m
 20 m * 5 m
 25 m * 5 m
Answer: Option C
Explanation:
From the question, 2b+l = 30
=> l = 302b
\begin{aligned}
Area = 100m^2 \\
=> l \times b = 100 \\
=> b(302b) = 100 \\
b^2  15b + 50 = 0 \\
=>(b10)(b5)=0 \\
\end{aligned}
b = 10 or b = 5
when b = 10 then l = 10
when b = 5 then l = 20
Since the garden is rectangular so we will take value of breadth 5.
So its dimensions are 20 m * 5 m
3. What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs. 10 per square meter ?
 Rs. 3430
 Rs. 3440
 Rs. 3450
 Rs. 3460
Answer: Option B
Explanation:
In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)lb)
= 2(l+b)+4 = 340+4 = 344
So required cost will be = 344 * 10 = 3440
4. There is a plot which is rectangular in shape and it area is 460 square metres. What will be breadth of the plot if length of plot is 15% more than the breadth of plot ?
 8 mtr
 10 mtr
 15 mtr
 20 mtr
Answer: Option D
Explanation:
\begin{aligned} l*b = 460 m^2 \end{aligned} .. (i)
Lets suppose breadth of plot is = b rnLength will be,\begin{aligned}l=b*\frac{100+15}{100}\\=\frac{115b}{100}\end{aligned} .. (ii)
From (i) and (ii), we get :
\begin{aligned}
\frac{115b}{100}*b = 460 \\
b^2 = \frac{46000}{115} = 400 \\
=> b = \sqrt{400} = 20 meter
\end{aligned}
5. What are the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?
 714
 814
 850
 866
Answer: Option B
Explanation:
In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
Length of largest tile can be obtained from HCF of length and breadth.
So lets solve this,
Length of largest tile = HCF of (1517 cm and 902 cm)
= 41 cm
Required number of tiles =
\begin{aligned}
\frac{\text{Area of floor}}{\text{Area of tile}} \\
= \left(\frac{1517\times902}{41 \times 41}\right)\\
= 814
\end{aligned}
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