An agricultural field is in the form of a rectangle having length X1 meters and breadth X2 meters (X1 and X2 are variable). If X1 + X2 = 40 meters, then the area of the agricultural field will not exceed which one of the following values?
UPSC-2016
The problem asks for the maximum possible area of a rectangular field given that the sum of its length and breadth is constant. This is a classic optimization problem that can be solved using algebraic inequalities or calculus. The area of a rectangle is maximized when it is a square.
Let the length of the rectangular field be \(X_1\) meters and the breadth be \(X_2\) meters.
The given condition is that the sum of the length and breadth is 40 meters:
\[ X_1 + X_2 = 40 \]The area of the agricultural field, denoted by \(A\), is given by the product of its length and breadth:
\[ A = X_1 \times X_2 \]We need to find the maximum possible value of \(A\). We can use the AM-GM (Arithmetic Mean - Geometric Mean) inequality, which states that for any non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. For two non-negative numbers \(a\) and \(b\), it is:
\[ \frac{a+b}{2} \ge \sqrt{ab} \]Applying the AM-GM inequality to \(X_1\) and \(X_2\):
\[ \frac{X_1 + X_2}{2} \ge \sqrt{X_1 X_2} \]Substitute the given sum \(X_1 + X_2 = 40\):
\[ \frac{40}{2} \ge \sqrt{X_1 X_2} \]\[ 20 \ge \sqrt{X_1 X_2} \]Square both sides of the inequality to find the maximum value of \(X_1 X_2\):
\[ (20)^2 \ge X_1 X_2 \]\[ 400 \ge X_1 X_2 \]This means that the area \(A = X_1 X_2\) will not exceed 400 square meters. The maximum area is achieved when \(X_1 = X_2\), which means the rectangle is a square. In this case, \(X_1 + X_2 = 40 \implies 2X_1 = 40 \implies X_1 = 20\) meters. So, \(X_1 = 20\) meters and \(X_2 = 20\) meters. The area would be \(20 \times 20 = 400\) square meters.
A) 400 sq m — As derived from the AM-GM inequality, the maximum area is 400 sq m, which occurs when the length and breadth are equal (i.e., when the field is a square with sides of 20 m each).
Let x, y be the volumes, m, n be the masses of two metallic cubes P and Q respectively. Each side of Q is two times that of P and mass of Q is two times that of. P. Let u=m/x and v=n/y. Which one of the following is correct?
UPSC-2020
This problem involves comparing the densities of two metallic cubes, P and Q, given their relative dimensions and masses. We need to express the relationship between \(u\) and \(v\), where \(u\) and \(v\) represent the densities of cubes P and Q, respectively.
Define variables for cube P:
Define variables for cube Q:
Substitute the expressions for \(n\) and \(y\) into the formula for \(v\):
\[ v = \frac{2m}{8s_P^3} \] \[ v = \frac{1}{4} \left( \frac{m}{s_P^3} \right) \]Relate \(v\) to \(u\):
We know that \(u = \frac{m}{s_P^3}\). Substitute this into the equation for \(v\):
\[ v = \frac{1}{4} u \]Rearrange the equation to match the options:
Multiply both sides by 4:
\[ 4v = u \]Or, \(u = 4v\).
A) u=4v is correct because our derivation shows that \(u = 4v\).
A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form of a square, then its area will be
The problem involves a wire that is first bent into a circle and then re-bent into a square. The key concept here is that the length of the wire remains constant regardless of its shape. Therefore, the perimeter of the circle will be equal to the perimeter of the square. We will first calculate the length of the wire (circumference of the circle) and then use this length to find the side of the square. Finally, we will calculate the area of the square.
Calculate the circumference of the circle:
Given the radius of the circle, \(r = 56 \text{ cm}\).
The circumference of a circle is given by the formula \(C = 2\pi r\).
\[ C = 2 \times \frac{22}{7} \times 56 \] \[ C = 2 \times 22 \times \frac{56}{7} \] \[ C = 2 \times 22 \times 8 \] \[ C = 44 \times 8 \] \[ C = 352 \text{ cm} \]So, the length of the wire is \(352 \text{ cm}\).
Calculate the side of the square:
When the wire is bent into a square, its length becomes the perimeter of the square.
Let the side of the square be \(s\).
The perimeter of a square is given by the formula \(P = 4s\).
Since the length of the wire is \(352 \text{ cm}\), we have:
\[ 4s = 352 \] \[ s = \frac{352}{4} \] \[ s = 88 \text{ cm} \]So, the side of the square is \(88 \text{ cm}\).
Calculate the area of the square:
The area of a square is given by the formula \(A = s^2\).
\[ A = (88)^2 \] \[ A = 88 \times 88 \] \[ A = 7744 \text{ cm}^2 \]C) The area of the square is \(7744 \text{ cm}^2\), which matches our calculation.
A typist uses a paper 12" by 5" length wise and leaves a margin of 1" at the top and the bottom and a margin of ½"on either side. What fractional part of the paper is available to him for typing ?
To find the fractional part of the paper available for typing, we need to calculate the area of the entire paper and the area available for typing. The area available for typing is found by subtracting the margins from the total dimensions of the paper.
Given dimensions of the paper are 12" by 5".
\[ \text{Length of paper} = 12 \text{ inches} \] \[ \text{Width of paper} = 5 \text{ inches} \] \[ \text{Total Area of paper} = \text{Length} \times \text{Width} = 12 \times 5 = 60 \text{ square inches} \]Margins are left at the top, bottom, and on either side.
New length available for typing:
\[ \text{Typing Length} = \text{Total Length} - (\text{Top Margin} + \text{Bottom Margin}) \] \[ \text{Typing Length} = 12 - (1 + 1) = 12 - 2 = 10 \text{ inches} \]New width available for typing:
\[ \text{Typing Width} = \text{Total Width} - (\text{Left Margin} + \text{Right Margin}) \] \[ \text{Typing Width} = 5 - \left(\frac{1}{2} + \frac{1}{2}\right) = 5 - 1 = 4 \text{ inches} \]Correct Option: A) 2/3
A rectangular tank measuring 5m × 4.5 m × 2.1 m is dug in the centre of the field measuring 13.5 m × 2.5 m. The earth dug out is spread evenly over the remaining portion of a field. How much is the level of the field raised ?
The problem involves calculating the volume of earth dug out from a rectangular tank and then spreading this earth evenly over the remaining area of a field. The key is to understand that the volume of earth dug out will be equal to the volume by which the remaining field's level is raised. We need to calculate the volume of the tank, the total area of the field, the area of the tank, and then the remaining area of the field. Finally, we can find the height by dividing the volume of earth by the remaining area.
Correct Option: C) 4.2 m
A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. What is the area of the circle ?
To find the area of the circle, we first need to determine its radius. The problem states that the circle and the rectangle have the same perimeter. We can calculate the perimeter of the rectangle using its given side lengths. Then, we equate this perimeter to the formula for the circumference of a circle to find the radius. Finally, we use the radius to calculate the area of the circle.
Calculate the perimeter of the rectangle:
The sides of the rectangle are given as \(l = 26 \text{ cm}\) and \(w = 18 \text{ cm}\).
The perimeter of a rectangle is given by the formula:
\[ P_{\text{rectangle}} = 2(l + w) \]Substitute the given values:
\[ P_{\text{rectangle}} = 2(26 \text{ cm} + 18 \text{ cm}) \] \[ P_{\text{rectangle}} = 2(44 \text{ cm}) \] \[ P_{\text{rectangle}} = 88 \text{ cm} \]Equate the perimeter of the rectangle to the circumference of the circle:
Given that the perimeter of the circle is the same as the perimeter of the rectangle, the circumference of the circle is \(88 \text{ cm}\).
The circumference of a circle is given by the formula:
\[ C = 2\pi r \]So, we have:
\[ 2\pi r = 88 \text{ cm} \]Calculate the radius of the circle:
Using the value of \(\pi = \frac{22}{7}\):
\[ 2 \times \frac{22}{7} \times r = 88 \] \[ \frac{44}{7} \times r = 88 \] \[ r = \frac{88 \times 7}{44} \] \[ r = 2 \times 7 \] \[ r = 14 \text{ cm} \]Calculate the area of the circle:
The area of a circle is given by the formula:
\[ A = \pi r^2 \]Substitute the value of \(r = 14 \text{ cm}\) and \(\pi = \frac{22}{7}\):
\[ A = \frac{22}{7} \times (14 \text{ cm})^2 \] \[ A = \frac{22}{7} \times 196 \text{ cm}^2 \] \[ A = 22 \times \frac{196}{7} \text{ cm}^2 \] \[ A = 22 \times 28 \text{ cm}^2 \] \[ A = 616 \text{ cm}^2 \]Correct Option: D) 616 cm 2
