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Rohan borrowed some money at 10% per annum for first 6 years, 5% per annum for next three years 13% per annum for the period after 9 years. If the interest paid by him at the end of 12 year is Rs 22800, then find how much did he borrowed.
Solution Let's denote the amount borrowed by Rohan as \(P\) (the principal amount). Given: - Interest rate for the first 6 years: 10% per annum - Interest rate for the next 3 years: 5% per annum - Interest rate for the period after 9 years: 13% per annum - Total interest paid after 12 years: Rs 2280Read more
Solution
Let’s denote the amount borrowed by Rohan as \(P\) (the principal amount).
Given:
– Interest rate for the first 6 years: 10% per annum
– Interest rate for the next 3 years: 5% per annum
– Interest rate for the period after 9 years: 13% per annum
– Total interest paid after 12 years: Rs 22800
Step 1: Calculate the interest for each period
– Interest for the first 6 years: \(P \times 10\% \times 6 = 0.6P\)
– Interest for the next 3 years: \(P \times 5\% \times 3 = 0.15P\)
– Interest for the last 3 years: \(P \times 13\% \times 3 = 0.39P\)
Step 2: Calculate the total interest paid
Total interest paid = Interest for the first 6 years + Interest for the next 3 years + Interest for the last 3 years
\[ 22800 = 0.6P + 0.15P + 0.39P \]
\[ 22800 = 1.14P \]
Step 3: Find the principal amount
\[ P = \frac{22800}{1.14} \]
\[ P = 20000 \]
Conclusion
Rohan borrowed Rs 20,000.
See lessIn an examination the marks of Anil was 28.57% less than that of Barun’s marks and Barun’s marks was 11.11% less than that of Chandan’s marks. If the difference between the marks obtained by Anil and Chandan is 80.5 then find the marks obtained by Barun?
Solution Let's denote the marks obtained by Anil, Barun, and Chandan as \(A\), \(B\), and \(C\) respectively. Given: - Anil's marks are 28.57% less than Barun's marks: \(A = B - 0.2857B = 0.7143B\). - Barun's marks are 11.11% less than Chandan's marks: \(B = C - 0.1111C = 0.8889C\). - The differenceRead more
Solution
Let’s denote the marks obtained by Anil, Barun, and Chandan as \(A\), \(B\), and \(C\) respectively.
Given:
– Anil’s marks are 28.57% less than Barun’s marks: \(A = B – 0.2857B = 0.7143B\).
– Barun’s marks are 11.11% less than Chandan’s marks: \(B = C – 0.1111C = 0.8889C\).
– The difference between the marks obtained by Anil and Chandan is 80.5: \(C – A = 80.5\).
Step 1: Express Anil’s marks in terms of Chandan’s marks
Using the relation between Anil’s and Barun’s marks:
\[ A = 0.7143B \]
Using the relation between Barun’s and Chandan’s marks:
\[ B = 0.8889C \]
So, Anil’s marks in terms of Chandan’s marks:
\[ A = 0.7143 \times 0.8889C = 0.6351C \]
Step 2: Use the difference between Anil’s and Chandan’s marks
\[ C – A = 80.5 \]
\[ C – 0.6351C = 80.5 \]
\[ 0.3649C = 80.5 \]
\[ C = \frac{80.5}{0.3649} \]
\[ C = 220.5 \]
Step 3: Find Barun’s marks
Using the relation between Barun’s and Chandan’s marks:
\[ B = 0.8889C \]
\[ B = 0.8889 \times 220.5 \]
\[ B = 196 \]
Conclusion
The marks obtained by Barun are 196.
See lessThree cubes of metal whose edges are in the ratio 3 : 4 : 5, are melted and one cube is formed. If the diagonal of the cube is 12√3 cm, then find the edge of the largest among three cubes.
Solution Given: - The ratio of the edges of the three cubes is 3:4:5. - The diagonal of the new cube formed by melting the three cubes is \(12\sqrt{3}\) cm. Step 1: Find the edge of the new cube The diagonal of a cube is related to its edge (\(a\)) by the formula: \[ \text{Diagonal} = a\sqrt{3} \] SRead more
Solution
Given:
– The ratio of the edges of the three cubes is 3:4:5.
– The diagonal of the new cube formed by melting the three cubes is \(12\sqrt{3}\) cm.
Step 1: Find the edge of the new cube
The diagonal of a cube is related to its edge (\(a\)) by the formula:
\[ \text{Diagonal} = a\sqrt{3} \]
So, the edge of the new cube is:
\[ a = \frac{\text{Diagonal}}{\sqrt{3}} = \frac{12\sqrt{3}}{\sqrt{3}} = 12 \text{ cm} \]
Step 2: Find the volume of the new cube
The volume of the new cube is:
\[ V = a^3 = 12^3 = 1728 \text{ cm}^3 \]
Step 3: Find the edge of the largest original cube
Let the edges of the three original cubes be \(3x\), \(4x\), and \(5x\) respectively. The volume of the largest cube is:
\[ V_{\text{largest}} = (5x)^3 = 125x^3 \]
The total volume of the three cubes is equal to the volume of the new cube:
\[ 3^3x^3 + 4^3x^3 + 5^3x^3 = 1728 \]
\[ 27x^3 + 64x^3 + 125x^3 = 1728 \]
\[ 216x^3 = 1728 \]
\[ x^3 = 8 \]
\[ x = 2 \]
So, the edge of the largest cube is:
\[ 5x = 5 \times 2 = 10 \text{ cm} \]
Conclusion
The edge of the largest among the three cubes is 10 cm.
See lessIf \(a+b+c=0\) then the value of \[ \frac{a^2}{a^2-b c}+\frac{b^2}{b^2-c a}+\frac{c^2}{c^2-a b} \]
Solution Given: \[ a + b + c = 0 \] We need to find the value of: \[ \frac{a^2}{a^2 - bc} + \frac{b^2}{b^2 - ca} + \frac{c^2}{c^2 - ab} \] Since \(a + b + c = 0\), we can write \(a = -(b + c)\). Step 1: Substitute \(a = -(b + c)\) \[ \frac{(-b - c)^2}{(-b - c)^2 - bc} + \frac{b^2}{b^2 - c(-b - c)} +Read more
Solution
Given:
\[ a + b + c = 0 \]
We need to find the value of:
\[ \frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab} \]
Since \(a + b + c = 0\), we can write \(a = -(b + c)\).
Step 1: Substitute \(a = -(b + c)\)
\[ \frac{(-b – c)^2}{(-b – c)^2 – bc} + \frac{b^2}{b^2 – c(-b – c)} + \frac{c^2}{c^2 – b(-b – c)} \]
Step 2: Simplify
\[ \frac{(b + c)^2}{(b + c)^2 – bc} + \frac{b^2}{b^2 + bc – c^2} + \frac{c^2}{c^2 + bc – b^2} \]
Step 3: Simplify further
\[ \frac{(b + c)^2}{b^2 + c^2 + 2bc – bc} + \frac{b^2}{b^2 + c^2 + bc} + \frac{c^2}{b^2 + c^2 + bc} \]
\[ \frac{(b + c)^2}{b^2 + c^2 + bc} + \frac{b^2}{b^2 + c^2 + bc} + \frac{c^2}{b^2 + c^2 + bc} \]
Step 4: Combine the fractions
\[ \frac{(b + c)^2 + b^2 + c^2}{b^2 + c^2 + bc} = \frac{b^2 + c^2 + 2bc + b^2 + c^2}{b^2 + c^2 + bc} \]
\[ = \frac{2b^2 + 2c^2 + 2bc}{b^2 + c^2 + bc} \]
\[ = 2 \frac{b^2 + c^2 + bc}{b^2 + c^2 + bc} \]
\[ = 2 \]
Conclusion
The value of \(\frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab}\) is 2.
See lessOne flies a kite with a thread 180 meter long. If the thread of the kite makes an angle of 60° with the horizontal line, then the height of the kite from the ground (assuming thread to be in straight line) is
Solution Given: - The length of the kite's thread is 180 meters. - The angle between the thread and the horizontal line is \(60^\circ\). We can use trigonometry to find the height of the kite from the ground. The height (\(h\)) can be found using the sine function: \[ \sin \theta = \frac{\text{OpposRead more
Solution
Given:
– The length of the kite’s thread is 180 meters.
– The angle between the thread and the horizontal line is \(60^\circ\).
We can use trigonometry to find the height of the kite from the ground. The height (\(h\)) can be found using the sine function:
\[ \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} \]
In this case, the opposite side is the height of the kite (\(h\)), and the hypotenuse is the length of the thread (180 meters):
\[ \sin 60^\circ = \frac{h}{180} \]
We know that \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), so:
\[ \frac{\sqrt{3}}{2} = \frac{h}{180} \]
Solving for \(h\):
\[ h = 180 \times \frac{\sqrt{3}}{2} \]
\[ h = 90\sqrt{3} \]
The height of the kite from the ground is \(90\sqrt{3}\) meters.
See lessIf \(\sin \alpha+(\operatorname{Sin} \alpha)^2=1\), then the value of \((\cos \alpha)^{12}+3(\cos \alpha)^{10}+3(\cos \alpha)^8+(\cos \alpha)^6-1\) is
Solution Given: \[ \sin \alpha + (\sin \alpha)^2 = 1 \] Step 1: Simplify the given equation \[ \sin \alpha = 1 - (\sin \alpha)^2 \] \[ \sin \alpha = (\cos \alpha)^2 \] (Since \((\sin \alpha)^2 + (\cos \alpha)^2 = 1\)) Step 2: Substitute \(\sin \alpha = (\cos \alpha)^2\) into the expression \[ (\cosRead more
Solution
Given:
\[ \sin \alpha + (\sin \alpha)^2 = 1 \]
Step 1: Simplify the given equation
\[ \sin \alpha = 1 – (\sin \alpha)^2 \]
\[ \sin \alpha = (\cos \alpha)^2 \] (Since \((\sin \alpha)^2 + (\cos \alpha)^2 = 1\))
Step 2: Substitute \(\sin \alpha = (\cos \alpha)^2\) into the expression
\[ (\cos \alpha)^{12} + 3(\cos \alpha)^{10} + 3(\cos \alpha)^8 + (\cos \alpha)^6 – 1 \]
\[ = \left((\cos \alpha)^4 + (\cos \alpha)^2\right)^3 – 1 \]
\[ = \left((\sin \alpha)^2 + (\cos \alpha)^2\right)^3 – 1 \]
\[ = 1^3 – 1 \]
\[ = 1 – 1 \]
\[ = 0 \]
Conclusion
The value of \((\cos \alpha)^{12} + 3(\cos \alpha)^{10} + 3(\cos \alpha)^8 + (\cos \alpha)^6 – 1\) is 0.
See lessFind the value of \[ \left(\frac{\sin 35^{\circ}}{\cos 55^{\circ}}\right)^2+\left(\frac{\cos 55^{\circ}}{\sin 35^{\circ}}\right)^2-2 \cos 30^{\circ} . \]
Solution Given: \[ \left(\frac{\sin 35^\circ}{\cos 55^\circ}\right)^2 + \left(\frac{\cos 55^\circ}{\sin 35^\circ}\right)^2 - 2\cos 30^\circ \] Step 1: Use the identity \(\cos(90^\circ - \theta) = \sin \theta\) \[ \left(\frac{\sin 35^\circ}{\cos 55^\circ}\right)^2 + \left(\frac{\cos 55^\circ}{\sin 35Read more
Solution
Given:
\[ \left(\frac{\sin 35^\circ}{\cos 55^\circ}\right)^2 + \left(\frac{\cos 55^\circ}{\sin 35^\circ}\right)^2 – 2\cos 30^\circ \]
Step 1: Use the identity \(\cos(90^\circ – \theta) = \sin \theta\)
\[ \left(\frac{\sin 35^\circ}{\cos 55^\circ}\right)^2 + \left(\frac{\cos 55^\circ}{\sin 35^\circ}\right)^2 – 2\cos 30^\circ \]
\[ = \left(\frac{\sin 35^\circ}{\sin(90^\circ – 35^\circ)}\right)^2 + \left(\frac{\sin(90^\circ – 55^\circ)}{\sin 35^\circ}\right)^2 – 2\cos 30^\circ \]
\[ = \left(\frac{\sin 35^\circ}{\sin 35^\circ}\right)^2 + \left(\frac{\sin 35^\circ}{\sin 35^\circ}\right)^2 – 2\cos 30^\circ \]
Step 2: Simplify
\[ = 1 + 1 – 2\left(\frac{\sqrt{3}}{2}\right) \]
\[ = 2 – \sqrt{3} \]
Conclusion
The value of the given expression is \(2 – \sqrt{3}\).
See lessIn \(\triangle A B C, \angle A=\angle B=60^{\circ}, A C=\sqrt{ } 34 \mathrm{~cm}\). The lines \(A D\) and \(B D\) intersect at \(D\) with \(\angle D=\) \(90^{\circ}\). If \(\mathrm{DB}=\mathbf{3 \mathrm { cm }}\), then the length of \(\mathrm{AD}\) is:
Solution Given: - In \(\triangle ABC\), \(\angle A = \angle B = 60^\circ\), \(AC = \sqrt{34} \text{ cm}\). - Lines \(AD\) and \(BD\) intersect at \(D\) with \(\angle D = 90^\circ\). - \(DB = 3 \text{ cm}\). Step 1: Determine the Type of Triangle Since \(\angle A = \angle B = 60^\circ\), \(\angle C\)Read more
Solution
Given:
– In \(\triangle ABC\), \(\angle A = \angle B = 60^\circ\), \(AC = \sqrt{34} \text{ cm}\).
– Lines \(AD\) and \(BD\) intersect at \(D\) with \(\angle D = 90^\circ\).
– \(DB = 3 \text{ cm}\).
Step 1: Determine the Type of Triangle
Since \(\angle A = \angle B = 60^\circ\), \(\angle C\) must also be \(60^\circ\) (as the sum of angles in a triangle is \(180^\circ\)). Therefore, \(\triangle ABC\) is an equilateral triangle.
Step 2: Find the Length of AB
Since \(\triangle ABC\) is equilateral, \(AB = AC = \sqrt{34} \text{ cm}\).
Step 3: Use Pythagoras’ Theorem
In \(\triangle ADB\), which is a right-angled triangle, we can use Pythagoras’ theorem:
\[ AB^2 = AD^2 + DB^2 \]
\[ (\sqrt{34})^2 = AD^2 + 3^2 \]
\[ 34 = AD^2 + 9 \]
\[ AD^2 = 34 – 9 \]
\[ AD^2 = 25 \]
\[ AD = \sqrt{25} \]
\[ AD = 5 \text{ cm} \]
Conclusion
The length of \(AD\) is 5 cm.
See lessA boat goes to a place and return back in 45 hours. It can go 10 km upstream in 1 hour and 20 km downstream in the same time. Find the total distance covered by the boat in the whole journey.
Solution Given: - The boat goes to a place and returns back in 45 hours. - The speed of the boat upstream is 10 km/hour. - The speed of the boat downstream is 20 km/hour. Let the distance between the starting point and the destination be \(x\) km. Time Taken for the Journey: - Time taken to go upstrRead more
Solution
Given:
– The boat goes to a place and returns back in 45 hours.
– The speed of the boat upstream is 10 km/hour.
– The speed of the boat downstream is 20 km/hour.
Let the distance between the starting point and the destination be \(x\) km.
Time Taken for the Journey:
– Time taken to go upstream (to the destination) = \(\frac{x}{10}\) hours
– Time taken to go downstream (return) = \(\frac{x}{20}\) hours
– Total time for the round trip = \(\frac{x}{10} + \frac{x}{20}\) hours
Given that the total time for the round trip is 45 hours, we can write:
\[ \frac{x}{10} + \frac{x}{20} = 45 \]
Multiplying all terms by 20 to clear the denominators:
\[ 2x + x = 900 \]
\[ 3x = 900 \]
\[ x = 300 \]
Total Distance Covered:
The total distance covered by the boat in the whole journey (going and returning) is:
\[ 2 \times x = 2 \times 300 = 600 \text{ km} \]
Conclusion
The total distance covered by the boat in the whole journey is 600 km.
See lessPQRSTU is a regular hexagon whose diagonals meet at point at O. Find the ratio of area of quadrilateral PQOU to the area of hexagon PQRSTU.
Solution In a regular hexagon, all sides are equal, and all internal angles are 120°. The diagonals of a regular hexagon divide it into six equilateral triangles. Let's denote the side length of the hexagon as \(a\). Area of the Hexagon PQRSTU: The area of an equilateral triangle with side length \(Read more
Solution
In a regular hexagon, all sides are equal, and all internal angles are 120°. The diagonals of a regular hexagon divide it into six equilateral triangles.
Let’s denote the side length of the hexagon as \(a\).
Area of the Hexagon PQRSTU:
The area of an equilateral triangle with side length \(a\) is given by:
\[ \text{Area of equilateral triangle} = \frac{\sqrt{3}}{4}a^2 \]
Since the hexagon is made up of six equilateral triangles, the area of the hexagon is:
\[ \text{Area of hexagon} = 6 \times \frac{\sqrt{3}}{4}a^2 = \frac{3\sqrt{3}}{2}a^2 \]
Area of Quadrilateral PQOU:
Quadrilateral PQOU is made up of two equilateral triangles, POQ and UOQ. Therefore, the area of quadrilateral PQOU is:
\[ \text{Area of quadrilateral PQOU} = 2 \times \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{2}a^2 \]
Ratio of Areas:
The ratio of the area of quadrilateral PQOU to the area of hexagon PQRSTU is:
\[ \text{Ratio} = \frac{\text{Area of quadrilateral PQOU}}{\text{Area of hexagon}} = \frac{\frac{\sqrt{3}}{2}a^2}{\frac{3\sqrt{3}}{2}a^2} = \frac{1}{3} \]
Conclusion
The ratio of the area of quadrilateral PQOU to the area of hexagon PQRSTU is 1:3.
See less