A gardener increased the area of his rectangular garden by increasing its length by 40% and decreasing its width by 20%. The area of the new garden
UPSC-2014
has increased by 20%.
has increased by 12%.
has increased by 8%.
is exactly the same as the old area.
To find the percentage change in the area of the rectangular garden, we will first define the original length, width, and area. Then, we will calculate the new length and new width based on the given percentage changes. Finally, we will calculate the new area and determine the percentage increase or decrease compared to the original area.
Define original dimensions and area:
Let the original length of the garden be \(L\) and the original width be \(W\).
The original area of the garden, \(A_{\text{original}}\), is given by:
Calculate new length:
The length is increased by 40%. So, the new length, \(L_{\text{new}}\), will be:
Calculate new width:
The width is decreased by 20%. So, the new width, \(W_{\text{new}}\), will be:
Calculate new area:
The new area of the garden, \(A_{\text{new}}\), will be:
Calculate percentage change in area:
The percentage change in area is given by:
Correct Option: B) has increased by 12%.
A cubical vessel of side 1 m is filled completely with water. How many millilitres of water is contained in it (neglect thickness of the (vessel)?
UPSC-2021
To find the amount of water in millilitres, we first need to calculate the volume of the cubical vessel in cubic metres. Then, we convert this volume to cubic centimetres, and finally to millilitres, using the standard conversion factor that \(1 \text{ cm}^3 = 1 \text{ mL}\).
Calculate the volume of the cubical vessel:
The side of the cubical vessel is given as \(1 \text{ m}\).
The volume of a cube is given by the formula: \(\text{Volume} = (\text{side})^3\).
So, \(\text{Volume} = (1 \text{ m})^3 = 1 \text{ m}^3\).
Convert cubic metres to cubic centimetres:
We know that \(1 \text{ m} = 100 \text{ cm}\).
Therefore, \(1 \text{ m}^3 = (100 \text{ cm})^3 = 100 \times 100 \times 100 \text{ cm}^3 = 1,000,000 \text{ cm}^3\).
Convert cubic centimetres to millilitres:
We know that \(1 \text{ cm}^3 = 1 \text{ mL}\).
So, \(1,000,000 \text{ cm}^3 = 1,000,000 \text{ mL}\).
Thus, the cubical vessel contains \(1,000,000 \text{ mL}\) of water.
D) 1000000 — The volume of the cube is \(1 \text{ m}^3\), which converts to \(1,000,000 \text{ cm}^3\). Since \(1 \text{ cm}^3 = 1 \text{ mL}\), the vessel contains \(1,000,000 \text{ mL}\) of water.
The village having a population of 4000 requires 150 litres of the water per head per day. It has a tank measuring 20 m x 15 m x 6 m. The water of this tank will last for
UPSC-2011
To determine how long the water in the tank will last, we need to calculate the total water required by the village per day and the total volume of water the tank can hold. Then, we divide the total tank volume by the daily water requirement.
Population of the village = 4000 people
Water required per head per day = 150 litres
\[ \text{Total daily water requirement} = \text{Population} \times \text{Water per head per day} \]\[ \text{Total daily water requirement} = 4000 \times 150 \text{ litres} \]\[ \text{Total daily water requirement} = 600000 \text{ litres} \]The tank is a cuboid with dimensions:
Length (\(L\)) = 20 m
Breadth (\(B\)) = 15 m
Height (\(H\)) = 6 m
\[ \text{Volume of tank} = L \times B \times H \]\[ \text{Volume of tank} = 20 \text{ m} \times 15 \text{ m} \times 6 \text{ m} \]\[ \text{Volume of tank} = 1800 \text{ cubic metres} \]We know that \(1 \text{ cubic metre} = 1000 \text{ litres}\).
\[ \text{Volume of tank in litres} = 1800 \text{ m}^3 \times 1000 \frac{\text{litres}}{\text{m}^3} \]\[ \text{Volume of tank in litres} = 1800000 \text{ litres} \]Correct Option: B) 3 days
Twelve equal squares are placed to fit in a rectangle of diagonal 5 cm. There are three rows containing four squares each. No gaps are left between adjacent squares. What is the area of each square?
UPSC-2018
The problem describes a rectangle formed by 12 equal squares arranged in three rows of four squares each. We are given the diagonal length of this larger rectangle and need to find the area of each individual square. We will use the properties of squares and rectangles, specifically the Pythagorean theorem, to solve this problem.
Let the side length of each small square be \(s\) cm.
The arrangement consists of three rows and four columns of squares. Therefore, the width of the large rectangle will be the sum of the side lengths of four squares, and the height of the large rectangle will be the sum of the side lengths of three squares.
Width of the rectangle (\(W\)) = \(4 \times s = 4s\).
Height of the rectangle (\(H\)) = \(3 \times s = 3s\).
The diagonal of a rectangle (\(D\)) is related to its width and height by the Pythagorean theorem:
\[ D^2 = W^2 + H^2 \]We are given that the diagonal \(D = 5\) cm. Substitute the expressions for \(W\) and \(H\) in terms of \(s\) into the Pythagorean theorem:
\[ 5^2 = (4s)^2 + (3s)^2 \]Calculate the squares:
\[ 25 = 16s^2 + 9s^2 \]Combine the terms on the right side:
\[ 25 = 25s^2 \]Solve for \(s^2\):
\[ s^2 = \frac{25}{25} \]\[ s^2 = 1 \]The area of each square is \(s^2\).
Therefore, the area of each square is \(1\) sq cm.
C) 1 sq cm is the correct answer. The side length of each square is 1 cm, and thus its area is \(1^2 = 1\) sq cm.
A cylindrical overhead tank of radius 2 m and height 7 m is to be filled from an underground tank of size 5.5 m x 4 m x 6 m. How much portion of the underground tank is still filled with water after filling the overhead tank completely?
UPSC-2016
To solve this problem, we need to calculate the volume of the cylindrical overhead tank and the volume of the underground tank. Then, we'll determine how much water is transferred from the underground tank to the overhead tank. Finally, we'll find the remaining volume in the underground tank and express it as a fraction of its original volume.
A) 1/3 is the correct answer, as calculated in the steps above.
AB is a vertical trunk of a huge tree with A being the point where the base of the trunk touches the ground. Due to a cyclone, the trunk has been broken at C which is at a height of 12 meters, broken part is partially attached to the vertical portion of the trunk at C. If the end of the broken part B touches the ground at D which is at a distance of 5 meters from A, then the original height of the trunk is:
UPSC-2016
This problem can be solved by visualizing the situation as a right-angled triangle. When the tree breaks, the broken part forms the hypotenuse, the standing part forms one leg, and the distance from the base of the tree to where the broken part touches the ground forms the other leg. We can then use the Pythagorean theorem to find the length of the broken part and subsequently the original height of the tree.
Let A be the base of the tree, C be the point where it broke, and D be the point where the top of the broken part (original B) touches the ground. The standing part of the trunk is AC, the distance on the ground is AD, and the broken part is CD. Triangle ADC is a right-angled triangle with the right angle at A (since the trunk is vertical to the ground).
According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse (CD) is equal to the sum of the squares of the other two sides (AC and AD).
\[ CD^2 = AC^2 + AD^2 \]The original height of the trunk (AB) was the sum of the standing part (AC) and the broken part (CB). Since the broken part CB is now CD (it's the same piece of the trunk), the original height is:
\[ \text{Original Height} = AC + CD \]\[ \text{Original Height} = 12 \text{ m} + 13 \text{ m} \]\[ \text{Original Height} = 25 \text{ m} \]Correct Option: B) 25 m
