To find the overall average runs, we need to calculate the total runs scored in all matches and divide it by the total number of matches played. The total runs for each set of matches can be found by multiplying the average runs by the number of matches.
Given that the average runs in 23 matches is 43.
\[ \text{Total runs (first 23 matches)} = \text{Average runs} \times \text{Number of matches} \] \[ \text{Total runs (first 23 matches)} = 43 \times 23 = 989 \]
Given that the average runs in the next 10 matches is 16.
\[ \text{Total runs (next 10 matches)} = \text{Average runs} \times \text{Number of matches} \] \[ \text{Total runs (next 10 matches)} = 16 \times 10 = 160 \]
\[ \text{Total runs (all 33 matches)} = \text{Total runs (first 23 matches)} + \text{Total runs (next 10 matches)} \] \[ \text{Total runs (all 33 matches)} = 989 + 160 = 1149 \]
\[ \text{Total matches} = 23 + 10 = 33 \]
\[ \text{Overall average runs} = \frac{\text{Total runs (all 33 matches)}}{\text{Total matches}} \] \[ \text{Overall average runs} = \frac{1149}{33} \approx 34.8181... \]
Rounding 34.8181... to two decimal places gives 34.82.
Correct Option: B) The overall average runs scored by the batsman in all 33 matches is approximately 34.82.
The value of \(\sqrt{196}\) + \(\sqrt{0.0361}\) − \(\sqrt{7.29}\) is _____.
To solve this problem, we need to calculate the square root of each term individually and then perform the addition and subtraction operations as indicated. We will convert decimal numbers into fractions to simplify finding their square roots.
We know that \(14 \times 14 = 196\).
\[ \sqrt{196} = 14 \]First, convert the decimal to a fraction:
\[ 0.0361 = \frac{361}{10000} \]\[ \sqrt{0.0361} = \sqrt{\frac{361}{10000}} = \frac{\sqrt{361}}{\sqrt{10000}} \]\[ \sqrt{361} = 19 \quad \text{and} \quad \sqrt{10000} = 100 \]\[ \sqrt{0.0361} = \frac{19}{100} = 0.19 \]First, convert the decimal to a fraction:
\[ 7.29 = \frac{729}{100} \]\[ \sqrt{7.29} = \sqrt{\frac{729}{100}} = \frac{\sqrt{729}}{\sqrt{100}} \]\[ \sqrt{729} = 27 \quad \text{and} \quad \sqrt{100} = 10 \]\[ \sqrt{7.29} = \frac{27}{10} = 2.7 \]The expression is \(\sqrt{196} + \sqrt{0.0361} - \sqrt{7.29}\).
\[ 14 + 0.19 - 2.7 \]Correct Option: A) 11.49
The problem involves calculating the lateral surface area of a cuboid formed by joining five identical cubes end to end. First, we need to find the side length of a single cube from its given volume. Then, we determine the dimensions of the resulting cuboid. Finally, we apply the formula for the lateral surface area of a cuboid.
The volume of a cube is given by the formula \(V = a^3\), where \(a\) is the side length.
Given volume \(V = 39304 \text{ cm}^3\).
\[ a^3 = 39304 \]\[ a = \sqrt[3]{39304} \]To find the cube root, we can use prime factorization or estimation. We know that \(30^3 = 27000\) and \(40^3 = 64000\), so \(a\) is between 30 and 40. The last digit of 39304 is 4, so the last digit of its cube root must be 4 (since \(4^3 = 64\)). Thus, \(a\) must be 34.\[ a = 34 \text{ cm} \]Five cubes, each with side length \(a = 34 \text{ cm}\), are joined end to end.
The length of the cuboid will be the sum of the side lengths of the five cubes:
\[ L = 5 \times a = 5 \times 34 = 170 \text{ cm} \]The breadth (width) of the cuboid will be the side length of one cube:\[ B = a = 34 \text{ cm} \]The height of the cuboid will also be the side length of one cube:\[ H = a = 34 \text{ cm} \]The lateral surface area (LSA) of a cuboid is the sum of the areas of its four side faces, excluding the top and bottom faces. The formula for LSA is:
\[ \text{LSA} = 2(L+B)H \]Substitute the dimensions we found:\[ \text{LSA} = 2(170 + 34) \times 34 \]\[ \text{LSA} = 2(204) \times 34 \]\[ \text{LSA} = 408 \times 34 \]\[ \text{LSA} = 13872 \text{ cm}^2 \]A) \(13872 \text{ cm}^2\). This matches our calculated lateral surface area.
Pipe A can fill a tank in 18 hours, pipe B can fill the same tank in 28 hours and pipe C can fill the same tank in 11 hours. The time taken by them to fill the same tank if they operate together is:
10 \(\frac{247}{505}\) hours
7 \(\frac{247}{505}\) hours
11 \(\frac{247}{505}\) hours
5 \(\frac{247}{505}\) hours
To find the time taken by multiple pipes to fill a tank when operating together, we first determine the individual filling rates of each pipe. The rate is the reciprocal of the time taken to fill the tank. Then, we sum these individual rates to find the combined filling rate. Finally, the total time taken to fill the tank together is the reciprocal of the combined rate.
Correct Option: D) \(5 \frac{247}{505}\) hours
If P = tanθ + secθ, then the value of tanθ is:
\(\frac{P^2+1}{P}\)
\(\frac{P^2-1}{P+1}\)
\(\frac{P^2+1}{2P}\)
\(\frac{P^2-1}{2P}\)
The problem asks us to find the value of \(\tan\theta\) in terms of \(P\), given that \(P = \tan\theta + \sec\theta\). We will use the fundamental trigonometric identity relating \(\sec\theta\) and \(\tan\theta\).
We are given the equation:
\[ P = \tan\theta + \sec\theta \quad \ldots (1) \]Recall the fundamental trigonometric identity:
\[ \sec^2\theta - \tan^2\theta = 1 \]This identity can be factored as a difference of squares:
\[ (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1 \]Substitute the given value of \(P\) from equation (1) into this factored identity:
\[ (\sec\theta - \tan\theta)(P) = 1 \]From this, we can express \(\sec\theta - \tan\theta\) in terms of \(P\):
\[ \sec\theta - \tan\theta = \frac{1}{P} \quad \ldots (2) \]Now we have a system of two linear equations with \(\tan\theta\) and \(\sec\theta\):
\[ P = \tan\theta + \sec\theta \quad \ldots (1) \]\[ \frac{1}{P} = \sec\theta - \tan\theta \quad \ldots (2) \]To find \(\tan\theta\), subtract equation (2) from equation (1):
\[ (\tan\theta + \sec\theta) - (\sec\theta - \tan\theta) = P - \frac{1}{P} \]\[ \tan\theta + \sec\theta - \sec\theta + \tan\theta = P - \frac{1}{P} \]\[ 2\tan\theta = P - \frac{1}{P} \]Combine the terms on the right-hand side:
\[ 2\tan\theta = \frac{P^2 - 1}{P} \]Finally, solve for \(\tan\theta\):
\[ \tan\theta = \frac{P^2 - 1}{2P} \]D) \(\frac{P^2-1}{2P}\) is the correct expression for \(\tan\theta\).
This problem involves two trains moving towards each other. We need to find the time when they meet. The key is to calculate the time taken by each train, their respective speeds (or rates), and then use the concept of relative speed or the distance covered by each train until they meet.
Correct Option: D) 11:13 AM
To find the mode of a given set of data, we need to identify the number that appears most frequently in the dataset. The mode is a measure of central tendency.
Correct Option: D) 19
To find the number of diagonals in a regular polygon, we first need to determine the number of sides of the polygon. We can do this by using the given relationship between the sum of its interior angles and the sum of its exterior angles. Once the number of sides (\(n\)) is known, we can use the formula for the number of diagonals.
Let \(n\) be the number of sides of the regular polygon.
The sum of the interior angles of a polygon with \(n\) sides is given by the formula:
\[ S_{\text{interior}} = (n-2) \times 180^\circ \]The sum of the exterior angles of any convex polygon is always \(360^\circ\).
\[ S_{\text{exterior}} = 360^\circ \]According to the problem statement, the sum of the interior angles is equal to four times the sum of its exterior angles:
\[ S_{\text{interior}} = 4 \times S_{\text{exterior}} \]Substitute the expressions for \(S_{\text{interior}}\) and \(S_{\text{exterior}}\) into the equation:
\[ (n-2) \times 180^\circ = 4 \times 360^\circ \]Divide both sides by \(180^\circ\):
\[ n-2 = \frac{4 \times 360^\circ}{180^\circ} \] \[ n-2 = 4 \times 2 \] \[ n-2 = 8 \]Add 2 to both sides:
\[ n = 8 + 2 \] \[ n = 10 \]So, the polygon is a decagon (a 10-sided polygon).
The number of diagonals in a polygon with \(n\) sides is given by the formula:
\[ D = \frac{n(n-3)}{2} \]Substitute \(n=10\) into the formula:
\[ D = \frac{10(10-3)}{2} \] \[ D = \frac{10 \times 7}{2} \] \[ D = \frac{70}{2} \] \[ D = 35 \]Therefore, the number of diagonals in the polygon is 35.
Correct Option: B) 35
To simplify the given algebraic expression, we need to apply the distributive property, combine like terms, and perform basic arithmetic operations.
First, distribute \(x\) into \((2x - 3)\) and \(2\) into \((x^2 - 4)\).
\[ x(2x - 3) = x \cdot 2x - x \cdot 3 = 2x^2 - 3x \] \[ 2(x^2 - 4) = 2 \cdot x^2 - 2 \cdot 4 = 2x^2 - 8 \]Now, substitute these expanded forms back into the original expression:
\[ (2x^2 - 3x) + (2x^2 - 8) + 16 \]Group the terms with \(x^2\), terms with \(x\), and constant terms together.
Combine the results from step 3:
\[ 4x^2 - 3x + 8 \]Correct Option: A) \(4x^2 - 3x + 8\)
To find the final selling price after successive discounts, we need to apply each discount one after the other to the marked price. First, calculate the price after the first discount. Then, apply the second discount to this new price.
Marked Price (MP) = ₹25,000
First Discount (\(D_1\)) = 15%
Second Discount (\(D_2\)) = 12%
The price after a 15% discount means the customer pays \(100\% - 15\% = 85\%\) of the marked price.
\[ \text{Price after } D_1 = \text{MP} \times \left(1 - \frac{D_1}{100}\right) \]\[ \text{Price after } D_1 = 25000 \times \left(1 - \frac{15}{100}\right) \]\[ \text{Price after } D_1 = 25000 \times \frac{85}{100} \]\[ \text{Price after } D_1 = 250 \times 85 \]\[ \text{Price after } D_1 = \text{₹}21,250 \]Now, the second discount of 12% is applied to the price after the first discount (₹21,250). This means the customer pays \(100\% - 12\% = 88\%\) of this intermediate price.
\[ \text{Final Selling Price} = (\text{Price after } D_1) \times \left(1 - \frac{D_2}{100}\right) \]\[ \text{Final Selling Price} = 21250 \times \left(1 - \frac{12}{100}\right) \]\[ \text{Final Selling Price} = 21250 \times \frac{88}{100} \]\[ \text{Final Selling Price} = 212.5 \times 88 \]\[ \text{Final Selling Price} = \text{₹}18,700 \]B) ₹18,700 is the final selling price after applying successive discounts of 15% and 12%.