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Question Detail
The area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field ?
- 18 meter
- 20 meter
- 22 meter
- 25 meter
Answer: Option B
Explanation:
Let breadth =x metres.
Then length =(115x/100)metres.
\begin{aligned}
=x*\frac{115x}{100}= 460\\
x^2=(460 x 100/115) \\
x^2=400 \\
x= 20 \\
\end{aligned}
1. 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}
2. If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ?
- Area is quadrupled
- Area is tripled
- Area is doubles
- Area become half
Answer: Option A
Explanation:
\begin{aligned}
2\pi R1 = 4 \pi \\
=> R1 = 2 \\
2\pi R2 = 8 \pi \\
=> R2 = 4 \\
\text{Original Area =} 4\pi * 2^2 \\
= 16 \pi \\
\text{New Area =} 4\pi * 4^2 \\
= 64 \pi
\end{aligned}
So the area quadruples.
3. 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 = 30-2b
\begin{aligned}
Area = 100m^2 \\
=> l \times b = 100 \\
=> b(30-2b) = 100 \\
b^2 - 15b + 50 = 0 \\
=>(b-10)(b-5)=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
4. The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.
- 4:25
- 4:15
- 3:25
- 3:15
Answer: Option A
Explanation:
Let the diagonals of the squares be 2x and 5x.
Then ratio of their areas will be
\begin{aligned}
\text{Area of square} = \frac{1}{2}*{Diagonal}^2 \\
\frac{1}{2}*{2x}^2:\frac{1}{2}*{5x}^2 \\
4x^2:25x^2 = 4:25
\end{aligned}
5. The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares .
- 22 cm
- 24 cm
- 26 cm
- 28 cm
Answer: Option B
Explanation:
We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm
Area of third Square = 10*10 - 8*8
= 36 cm
So side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cm
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