Section - Mathematics
The angles of elevation of the top of a tower from two points X and Y (on opposite sides of the tower) at distances b2 m and a2 m, respectively, from the base and in the same straight line with it are complementary. The height (in m) of the tower is _____.
ab\( \frac{1}{2} \)
(ab)\( \frac{1}{2} \)
ba\( \frac{1}{2} \)
ab
This problem involves trigonometry, specifically the concept of angles of elevation and complementary angles. We will use the tangent function to relate the height of the tower to the distances from its base and the angles of elevation. Since the angles are complementary, we can express one in terms of the other, which will help us find the height of the tower.
Let the height of the tower be \(h\) meters.
Let the two points be X and Y. They are on opposite sides of the tower and in the same straight line with its base.
The distance of point X from the base of the tower is \(b^2\) m. Let the angle of elevation from X be \(\theta\).
The distance of point Y from the base of the tower is \(a^2\) m. Since the angles of elevation are complementary, the angle of elevation from Y will be \(90^\circ - \theta\).
From the right-angled triangle formed by the tower, its base, and point X, we have:
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{b^2} \quad \ldots (1) \]From the right-angled triangle formed by the tower, its base, and point Y, we have:
\[ \tan (90^\circ - \theta) = \frac{h}{a^2} \]Using the trigonometric identity \(\tan (90^\circ - \theta) = \cot \theta\), we get:
\[ \cot \theta = \frac{h}{a^2} \quad \ldots (2) \]We know that \(\cot \theta = \frac{1}{\tan \theta}\). So, from equation (2), we can write:
\[ \frac{1}{\tan \theta} = \frac{h}{a^2} \]\[ \tan \theta = \frac{a^2}{h} \quad \ldots (3) \]Now, equate equation (1) and equation (3):
\[ \frac{h}{b^2} = \frac{a^2}{h} \]Cross-multiply to solve for \(h\):
\[ h^2 = a^2 b^2 \]Take the square root of both sides to find \(h\):
\[ h = \sqrt{a^2 b^2} \]\[ h = ab \]The height of the tower is \(ab\) meters.
Correct Option: D) ab
To solve this problem, we need to follow the order of operations, often remembered by the acronym BODMAS/PEMDAS. This means we perform operations in the following order: Brackets (Parentheses), Orders (Exponents/Powers and Square Roots), Division and Multiplication (from left to right), Addition and Subtraction (from left to right).
First, evaluate the innermost bracket: \((3 - 6)\).
\[ 3 - 6 = -3 \]Next, substitute this value back into the expression: \([954 \div \{30 + 9 \times (-3)\}]\).
Inside the curly braces, perform the multiplication before addition: \(9 \times (-3)\).
\[ 9 \times (-3) = -27 \]Substitute this value back into the expression: \([954 \div \{30 + (-27)\}]\).
Now, perform the addition inside the curly braces: \(30 + (-27)\).
\[ 30 + (-27) = 30 - 27 = 3 \]Finally, perform the division: \(954 \div 3\).
\[ 954 \div 3 = 318 \]B) 318 is the correct answer, as calculated by following the order of operations.
The problem involves calculating the time taken by Nisha to travel from Delhi to Kanpur, given Prashant's speed and time, and Nisha's speed. Since both travel the same distance, we can first calculate the total distance using Prashant's speed and time. Then, we can use this distance and Nisha's speed to find the time taken by Nisha.
Correct Option: A) 39 hours
If the interest earned during the 2nd year on a certain sum is ₹2,352, and the rate of interest is 5% per annum compounded annually, then the sum is:
The problem involves compound interest. We are given the interest earned in the second year and the annual interest rate. We need to find the principal sum. We can use the formula for compound interest or calculate the interest year by year, considering that compound interest means interest is calculated on the principal plus accumulated interest from previous periods.
Correct Option: C) ₹44,800
To find the area of a triangle given its three side lengths, we can use Heron's formula. Alternatively, we can check if the triangle is a right-angled triangle using the Pythagorean theorem, as this often simplifies the area calculation.
Check for a Right-Angled Triangle:
The given side lengths are 82 cm, 18 cm, and 80 cm. Let's check if they satisfy the Pythagorean theorem \(a^2 + b^2 = c^2\), where \(c\) is the longest side.
Here, the longest side is 82 cm. So, we check if \(18^2 + 80^2 = 82^2\).
Since \(18^2 + 80^2 = 82^2\) (\(6724 = 6724\)), the triangle is a right-angled triangle. The sides 18 cm and 80 cm are the perpendicular sides (base and height).
Calculate the Area:
For a right-angled triangle, the area is given by the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]Using the perpendicular sides as base and height:
\[ \text{Area} = \frac{1}{2} \times 18 \text{ cm} \times 80 \text{ cm} \]\[ \text{Area} = 9 \times 80 \text{ cm}^2 \]\[ \text{Area} = 720 \text{ cm}^2 \]D) 720
The area of the triangle is 720 cm², as calculated by identifying it as a right-angled triangle and using the formula for its area.
By how much is 65% of 65 greater than \( \frac{1}{5} \) of 15?
To solve this problem, we need to calculate two values: first, 65% of 65, and second, \( \frac{1}{5} \) of 15. After calculating both values, we will find the difference between the first value and the second value to determine how much greater the first is than the second.
Calculate 65% of 65:
To find a percentage of a number, convert the percentage to a decimal or a fraction and then multiply it by the number.
\[ 65\% \text{ of } 65 = \frac{65}{100} \times 65 \] \[ = 0.65 \times 65 \] \[ = 42.25 \]Calculate \( \frac{1}{5} \) of 15:
To find a fraction of a number, multiply the fraction by the number.
\[ \frac{1}{5} \text{ of } 15 = \frac{1}{5} \times 15 \] \[ = \frac{15}{5} \] \[ = 3 \]Find the difference:
Now, subtract the second value from the first value to find out by how much 65% of 65 is greater than \( \frac{1}{5} \) of 15.
\[ \text{Difference} = 42.25 - 3 \] \[ = 39.25 \]Thus, 65% of 65 is 39.25 greater than \( \frac{1}{5} \) of 15.
Correct Option: D) 39.25
Jaya can do a certain piece of work in 30 days. Jaya and Sushma can together do the same work in 21 days, and Jaya, Sushma and Hina can do the same work together in 14 days. In how many days can Jaya and Hina do the same work together?
\( \frac{35}{2} \)
\( \frac{31}{3} \)
\( \frac{44}{2} \)
\( \frac{27}{4} \)
This problem involves work and time. We will use the concept that if a person can do a piece of work in \(N\) days, then their one day's work is \(1/N\) of the total work. We will calculate the individual work rates of Jaya, Sushma, and Hina, and then combine the rates of Jaya and Hina to find the time they take to complete the work together.
Calculate Jaya's one day's work:
Jaya can do the work in 30 days. So, Jaya's one day's work is:
Calculate Jaya and Sushma's combined one day's work:
Jaya and Sushma together can do the work in 21 days. So, their combined one day's work is:
Calculate Sushma's one day's work:
We can find Sushma's one day's work by subtracting Jaya's one day's work from their combined work:
Calculate Jaya, Sushma, and Hina's combined one day's work:
Jaya, Sushma, and Hina together can do the work in 14 days. So, their combined one day's work is:
Calculate Hina's one day's work:
We can find Hina's one day's work by subtracting Jaya and Sushma's combined one day's work from the total combined work of all three:
Calculate Jaya and Hina's combined one day's work:
Now, we add Jaya's one day's work and Hina's one day's work:
Calculate the number of days Jaya and Hina take together:
If their combined one day's work is \(\frac{2}{35}\), then they will take the reciprocal of this value to complete the work:
Correct Option: A) \( \frac{35}{2} \)
Jaya and Hina together can do the work in \( \frac{35}{2} \) days.
To find the final selling price after successive discounts, we need to apply each discount one after the other to the current price. First, calculate the price after the first discount. Then, apply the second discount to this new price.
Identify the marked price and discounts:
Calculate the price after the first discount:
A 10% discount means the item is sold at \(100\% - 10\% = 90\%\) of its marked price.
\[ \text{Price after 1st discount} = \text{MP} \times \left(1 - \frac{10}{100}\right) \] \[ = 5000 \times \frac{90}{100} \] \[ = 5000 \times 0.90 \] \[ = \text{₹}4,500 \]Calculate the final selling price after the second discount:
Now, apply the second discount of 5% to the price after the first discount (₹4,500). A 5% discount means the item is sold at \(100\% - 5\% = 95\%\) of this price.
\[ \text{Final Selling Price} = \text{Price after 1st discount} \times \left(1 - \frac{5}{100}\right) \] \[ = 4500 \times \frac{95}{100} \] \[ = 4500 \times 0.95 \] \[ = \text{₹}4,275 \]C) ₹4,275 is the final selling price after successive discounts of 10% and 5%.
This problem involves profit sharing in a partnership. The profit share of each partner is proportional to the product of their investment and the duration of their investment. We are given the total profit, the difference in investments, the investment durations, and the difference in their profit shares. We need to find the capital contributed by M.
Let N's investment be \(\text{₹}x\).
Then M's investment is \(\text{₹}(x + 46000)\).
M invests for \(t_M = 3\) months.
N invests for \(t_N = 4\) months.
Total profit \(P = \text{₹}1359\).
M's share is \(P_M\).
N's share is \(P_N\).
We are given \(P_M - P_N = \text{₹}453\).
We know \(P_M + P_N = P = \text{₹}1359\).
We have a system of two linear equations:
\[ P_M + P_N = 1359 \quad \text{(1)} \]
\[ P_M - P_N = 453 \quad \text{(2)} \]
Adding (1) and (2):
\[ 2P_M = 1359 + 453 = 1812 \]
\[ P_M = \frac{1812}{2} = \text{₹}906 \]
Substituting \(P_M\) into (1):
\[ 906 + P_N = 1359 \]
\[ P_N = 1359 - 906 = \text{₹}453 \]
The ratio of profit shares is equal to the ratio of (investment \(\times\) time):
\[ \frac{P_M}{P_N} = \frac{\text{M's investment} \times \text{M's time}}{\text{N's investment} \times \text{N's time}} \]
\[ \frac{906}{453} = \frac{(x + 46000) \times 3}{x \times 4} \]
\[ 2 = \frac{3(x + 46000)}{4x} \]
\[ 2 \times 4x = 3(x + 46000) \]
\[ 8x = 3x + 3 \times 46000 \]
\[ 8x - 3x = 138000 \]
\[ 5x = 138000 \]
\[ x = \frac{138000}{5} \]
\[ x = \text{₹}27600 \]
So, N's capital is \(\text{₹}27600\).
M's capital = \(x + 46000\)
M's capital = \(27600 + 46000\)
M's capital = \(\text{₹}73600\).
Correct Option: D) ₹73,600
The mean proportional (or geometric mean) between two numbers, say \(a\) and \(b\), is given by the square root of their product. If \(x\) is the mean proportional between \(a\) and \(b\), then the relationship is \(a:x :: x:b\), which implies \(x^2 = ab\), and therefore \(x = \sqrt{ab}\).
Correct Option: A) 60