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Question Detail
One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
- \begin{aligned} 120m^2 \end{aligned}
- \begin{aligned} 130m^2 \end{aligned}
- \begin{aligned} 140m^2 \end{aligned}
- \begin{aligned} 150m^2 \end{aligned}
Answer: Option A
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
1. 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}
2. 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 ?
- 25%
- 26%
- 27%
- 28%
Answer: Option D
Explanation:
Let original length = x
and original width = y
Decrease in area will be
\begin{aligned}
= xy-\left( \frac{80x}{100}\times\frac{90y}{100}\right) \\
= \left(xy- \frac{18}{25}xy\right) \\
= \frac{7}{25}xy \\
\text{Decrease = }\left(\frac{7xy}{25xy} \times100\right) \% \\
= 28\%
\end{aligned}
3. 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
4. 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
5. Find the circumference of a circle, whose area is 24.64 meter sqaure
- 17.90m
- 17.80m
- 17.60m
- 17.40m
Answer: Option C
Explanation:
\begin{aligned}
\text{Area of Square =} \pi*r^2\\
=> \pi*r^2 = 24.64 \\
=> r^2 = \frac{24.64}{22}*7 \\
=> r^2 = 7.84 \\
=> r = \sqrt{7.84} \\
=> r = 2.8 \\
\text{Circumference =}2\pi*r\\
= 2*\frac{22}{7}*2.8 \\
= 17.60m
\end{aligned}
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