To find the percentage of literate females, we need to calculate the total number of females, the total number of literate people, and the number of literate males. From these, we can deduce the number of literate females and then express it as a percentage of the total female population.
Given total population = 376000
Number of males = 230000
\[ \text{Number of females} = \text{Total population} - \text{Number of males} \]\[ \text{Number of females} = 376000 - 230000 = 146000 \]11% of the total population is literate.
\[ \text{Total literate people} = 11\% \times 376000 \]\[ \text{Total literate people} = \frac{11}{100} \times 376000 = 11 \times 3760 = 41360 \]11% of males are literate.
\[ \text{Number of literate males} = 11\% \times 230000 \]\[ \text{Number of literate males} = \frac{11}{100} \times 230000 = 11 \times 2300 = 25300 \]The total number of literate people is the sum of literate males and literate females.
\[ \text{Number of literate females} = \text{Total literate people} - \text{Number of literate males} \]\[ \text{Number of literate females} = 41360 - 25300 = 16060 \]Percentage of literate females is the number of literate females divided by the total number of females, multiplied by 100.
\[ \text{Percentage of literate females} = \frac{\text{Number of literate females}}{\text{Total number of females}} \times 100\% \]\[ \text{Percentage of literate females} = \frac{16060}{146000} \times 100\% \]\[ \text{Percentage of literate females} = \frac{1606}{14600} \times 100\% \]\[ \text{Percentage of literate females} = \frac{1606}{146} \% \]\[ \text{Percentage of literate females} = 11\% \]D) 11% is the correct answer. The calculations show that 11% of females are literate.
This problem involves calculating the capital contributions of two partners, Praveen and Hemant, based on their investment periods and profit shares. We will use the principle that profit is distributed in proportion to the product of capital invested and the time for which it is invested. We will set up equations based on the given information and solve for the unknown capital.
Let Hemant's capital be \(H\) and Praveen's capital be \(P\).
Given that Praveen invests ₹16,000 more than Hemant, we have:
\[ P = H + 16000 \quad (1) \]Praveen invests for 7 months, and Hemant invests for 6 months.
The ratio of their profits is equal to the ratio of (capital \(\times\) time):
\[ \frac{\text{Praveen's Profit}}{\text{Hemant's Profit}} = \frac{P \times 7}{H \times 6} \]The total profit is ₹7,049. Praveen's share is ₹1,007 more than Hemant's share.
Let Praveen's profit be \(P_p\) and Hemant's profit be \(P_h\).
\[ P_p + P_h = 7049 \]\[ P_p = P_h + 1007 \]Substitute the second equation into the first:
\[ (P_h + 1007) + P_h = 7049 \]\[ 2P_h + 1007 = 7049 \]\[ 2P_h = 7049 - 1007 \]\[ 2P_h = 6042 \]\[ P_h = \frac{6042}{2} \]\[ P_h = 3021 \]Now, find Praveen's profit:
\[ P_p = P_h + 1007 \]\[ P_p = 3021 + 1007 \]\[ P_p = 4028 \]Now we have the profit shares. Use the profit ratio equation:
\[ \frac{P_p}{P_h} = \frac{P \times 7}{H \times 6} \]\[ \frac{4028}{3021} = \frac{7P}{6H} \]Simplify the fraction \(\frac{4028}{3021}\). Both are divisible by 13. \(4028 = 13 \times 310\) and \(3021 = 13 \times 232.38\). Let's check for other common factors. Let's try dividing by 7. \(4028/7 = 575.4\). Let's try dividing by 6. \(3021/6 = 503.5\). Let's try dividing by 13. \(4028/13 = 310\). \(3021/13 = 232.38\). This is incorrect. Let's recheck the calculation of \(P_h\) and \(P_p\).
The sum of profits is \(P_p + P_h = 7049\). The difference is \(P_p - P_h = 1007\).
Adding the two equations: \(2P_p = 7049 + 1007 = 8056 \implies P_p = 4028\).
Subtracting the two equations: \(2P_h = 7049 - 1007 = 6042 \implies P_h = 3021\).
The profit shares are correct.
Now, let's simplify the ratio \(\frac{4028}{3021}\). Both are divisible by 13. \(4028 = 13 \times 310\) is incorrect. \(4028 \div 13 = 310\) is incorrect. \(4028 \div 13 = 309.84\). Let's recheck the division. \(4028 \div 1 = 4028\). \(3021 \div 1 = 3021\). Let's try to find the greatest common divisor (GCD).
Let's check if \(4028\) is divisible by \(7\). \(4028 \div 7 = 575.4\). No.
Let's check if \(3021\) is divisible by \(6\). No, it's an odd number.
Let's use the ratio directly:
\[ \frac{4028}{3021} = \frac{7P}{6H} \]\[ 4028 \times 6H = 3021 \times 7P \]\[ 24168H = 21147P \]\[ \frac{H}{P} = \frac{21147}{24168} \]\[ \frac{H}{P} = \frac{7 \times 3021}{6 \times 4028} \]\[ \frac{H}{P} = \frac{7 \times (3 \times 1007)}{6 \times (4 \times 1007)} \]\[ \frac{H}{P} = \frac{7 \times 3}{6 \times 4} \]\[ \frac{H}{P} = \frac{21}{24} \]\[ \frac{H}{P} = \frac{7}{8} \]So, \(8H = 7P\).
We have two equations:
1) \(P = H + 16000\)
2) \(8H = 7P\)
Substitute (1) into (2):
\[ 8H = 7(H + 16000) \]\[ 8H = 7H + 7 \times 16000 \]\[ 8H - 7H = 112000 \]\[ H = 112000 \]Now find Praveen's capital \(P\):
\[ P = H + 16000 \]\[ P = 112000 + 16000 \]\[ P = 128000 \]The capital contributed by Praveen is ₹1,28,000.
D) ₹1,28,000
The angle made by a point on the ground and the top of a building with ground is 30°. If the point is x units distance away from the bottom of the building, then the height of the building (in units) is:
\(\frac{x}{\sqrt{2}}\)
\( \frac{x}{3} \)
\( \frac{x}{2} \)
\(\frac{x}{\sqrt{3}}\)
This problem involves trigonometry, specifically the tangent function, to relate the angle of elevation, the distance from the base of the building, and the height of the building. We will form a right-angled triangle and use the definition of the tangent of an angle.
D) The height of the building is \(\frac{x}{\sqrt{3}}\) units, which matches our calculation.
The problem involves calculating simple interest for two different investments from a total sum, where the interests earned are equal. We will use the simple interest formula \(I = \frac{P \cdot R \cdot T}{100}\) and set up an equation based on the condition of equal interests to find the amount invested at 6%.
Let the total sum invested be \(P = \text{₹}21,100\).
Let the sum invested at 6% per annum be \(P_1\).
Then, the sum invested at 14% per annum will be \(P_2 = P - P_1 = (21,100 - P_1)\).
The rate of interest for the first part is \(R_1 = 6\%\) per annum.
The rate of interest for the second part is \(R_2 = 14\%\) per annum.
The time period for both investments is \(T = 3\) years.
The formula for simple interest is \(I = \frac{P \cdot R \cdot T}{100}\).
Interest from the first investment (\(I_1\)):
\[ I_1 = \frac{P_1 \cdot R_1 \cdot T}{100} = \frac{P_1 \cdot 6 \cdot 3}{100} = \frac{18 P_1}{100} \]Interest from the second investment (\(I_2\)):
\[ I_2 = \frac{P_2 \cdot R_2 \cdot T}{100} = \frac{(21,100 - P_1) \cdot 14 \cdot 3}{100} = \frac{42 (21,100 - P_1)}{100} \]According to the problem, the interests earned from the two investments are equal:
\[ I_1 = I_2 \]\[ \frac{18 P_1}{100} = \frac{42 (21,100 - P_1)}{100} \]Multiply both sides by 100 to eliminate the denominator:
\[ 18 P_1 = 42 (21,100 - P_1) \]Divide both sides by the common factor, 6:
\[ 3 P_1 = 7 (21,100 - P_1) \]Distribute 7 on the right side:
\[ 3 P_1 = 7 \cdot 21,100 - 7 P_1 \]\[ 3 P_1 = 147,700 - 7 P_1 \]Add \(7 P_1\) to both sides:
\[ 3 P_1 + 7 P_1 = 147,700 \]\[ 10 P_1 = 147,700 \]Solve for \(P_1\):
\[ P_1 = \frac{147,700}{10} \]\[ P_1 = 14,770 \]Therefore, the sum invested at the rate of 6% per annum is ₹14,770.
C) 14,770 is the correct answer, as calculated by setting the simple interests from both investments equal and solving for the principal amount invested at 6%.
The problem asks us to find the length of the equal sides of an isosceles triangle given its perimeter and the length of its base. We will use the definition of the perimeter of a triangle and the property of an isosceles triangle having two equal sides.
Understand the properties of an isosceles triangle: An isosceles triangle has two sides of equal length. Let these equal sides be denoted by \(s\).
Recall the formula for the perimeter of a triangle: The perimeter of any triangle is the sum of the lengths of its three sides.
Set up the equation: Given that the perimeter of the isosceles triangle is 184 cm and the base is 44 cm, we can write the equation for the perimeter:
\[ \text{Perimeter} = \text{Side 1} + \text{Side 2} + \text{Base} \]Substitute the known values into the equation. Since two sides are equal (\(s\)), the equation becomes:
\[ 184 = s + s + 44 \]\[ 184 = 2s + 44 \]Solve for \(s\):
Subtract 44 from both sides of the equation:
\[ 184 - 44 = 2s \]\[ 140 = 2s \]Divide by 2 to find the value of \(s\):
\[ s = \frac{140}{2} \]\[ s = 70 \]Therefore, the length of each of the equal sides is 70 cm.
D) 70 cm is the correct length for the equal sides, as calculated.
If the interest earned during the 2nd year on a certain sum is ₹2,508, and the rate of interest is 20% per annum compounded annually, then the sum is:
We are given the interest earned in the second year on a sum compounded annually and the rate of interest. We need to find the principal sum. We can use the formula for compound interest or calculate the interest year by year. Since the interest for the second year is given, we can work backward to find the principal.
Interest for the first year (\(CI_1\)) is calculated on the principal \(P\):
\[ CI_1 = P \times \frac{R}{100} = P \times \frac{20}{100} = 0.2P \]Amount at the end of the first year (\(A_1\)) will be the principal for the second year:
\[ A_1 = P + CI_1 = P + 0.2P = 1.2P \]The interest earned during the second year (\(CI_2\)) is calculated on the amount at the end of the first year (\(A_1\)):
\[ CI_2 = A_1 \times \frac{R}{100} \]We are given that \(CI_2 = \text{₹}2,508\). Substitute the values into the equation:
\[ 2508 = (1.2P) \times \frac{20}{100} \]\[ 2508 = 1.2P \times 0.2 \]\[ 2508 = 0.24P \]Now, solve for \(P\):
\[ P = \frac{2508}{0.24} \]\[ P = \frac{2508 \times 100}{24} \]\[ P = \frac{250800}{24} \]Perform the division:
\[ P = 10450 \]C) ₹10,450. This is the principal sum calculated using the given interest for the second year and the rate of interest.
To simplify the given algebraic expression, we will use the distributive property to expand the terms, then combine like terms by adding or subtracting their coefficients.
Correct Option: A) The simplified expression is \(6x^2 - 9x + 8\).
This problem involves calculating cost prices and selling prices with given profit/loss percentages. Since there is neither an overall loss nor gain, the total selling price must equal the total cost price. We can set up equations based on the given percentages and solve for the unknown cost price.
The cost price of the first bicycle is ₹816.
C) 816 — The calculation shows that the cost price of the first bicycle is ₹816.
We are given two scenarios involving the cost of pens and pencils, which allows us to set up a system of two linear equations with two variables. We will define variables for the original cost of a pen and a pencil, form equations based on the given information, solve the system of equations to find the original costs, and then calculate the cost of 9 pens and 6 pencils.
Define variables:
Let the original cost of one pen be \(P\) (in ₹).
Let the original cost of one pencil be \(C\) (in ₹).
Formulate the first equation:
The cost of 7 pens and 17 pencils is ₹92.
\[ 7P + 17C = 92 \quad \text{(Equation 1)} \]
Formulate the second equation based on changed costs:
The cost of a pen decreases by ₹0.50, so the new cost of a pen is \(P - 0.50\).
The cost of a pencil increases by ₹8, so the new cost of a pencil is \(C + 8\).
The cost of 8 pens and 14 pencils with these new prices is ₹186.
\[ 8(P - 0.50) + 14(C + 8) = 186 \]
Expand and simplify:
\[ 8P - 4 + 14C + 112 = 186 \]
\[ 8P + 14C + 108 = 186 \]
\[ 8P + 14C = 186 - 108 \]
\[ 8P + 14C = 78 \quad \text{(Equation 2)} \]
Solve the system of equations:
We have:
1) \(7P + 17C = 92\)
2) \(8P + 14C = 78\)
To eliminate one variable, let's multiply Equation 1 by 8 and Equation 2 by 7:
Multiply Equation 1 by 8: \(56P + 136C = 736 \quad \text{(Equation 3)}\)
Multiply Equation 2 by 7: \(56P + 98C = 546 \quad \text{(Equation 4)}\)
Subtract Equation 4 from Equation 3:
\[ (56P + 136C) - (56P + 98C) = 736 - 546 \]
\[ 38C = 190 \]
\[ C = \frac{190}{38} \]
\[ C = 5 \]
So, the original cost of one pencil is ₹5.
Substitute C back into Equation 1 to find P:
\[ 7P + 17(5) = 92 \]
\[ 7P + 85 = 92 \]
\[ 7P = 92 - 85 \]
\[ 7P = 7 \]
\[ P = 1 \]
So, the original cost of one pen is ₹1.
Calculate the original cost of 9 pens and 6 pencils:
Cost \( = 9P + 6C \)
Cost \( = 9(1) + 6(5) \)
Cost \( = 9 + 30 \)
Cost \( = 39 \)
The original cost of 9 pens and 6 pencils is ₹39.
Correct Option: D) ₹39
This problem involves concepts of speed, distance, and time. We are given information about two individuals, Pranay and Aman, covering a fixed distance. We need to set up equations based on the given conditions relating their speeds and times, and then solve for Pranay's original speed.
Let the distance be \(D = 110 \text{ km}\).
Let Pranay's original speed be \(S_P \text{ km/hr}\) and Aman's speed be \(S_A \text{ km/hr}\).
Let Pranay's original time be \(T_P\) and Aman's time be \(T_A\).
Using the formula \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\), we can write:
\(T_P = \frac{110}{S_P}\)
\(T_A = \frac{110}{S_A}\)
Condition 1: Pranay takes 6 hours more than Aman.
\(T_P = T_A + 6\)
Substituting the time expressions:
\[ \frac{110}{S_P} = \frac{110}{S_A} + 6 \quad \text{(Equation 1)} \]Condition 2: If Pranay doubles his speed, he would take 5 hours less than Aman.
Pranay's new speed is \(2S_P\).
Pranay's new time \(T_P' = \frac{110}{2S_P}\).
According to the condition:
\(T_P' = T_A - 5\)
Substituting the time expressions:
\[ \frac{110}{2S_P} = \frac{110}{S_A} - 5 \quad \text{(Equation 2)} \]We have a system of two equations. Let's simplify them.
From Equation 1:
\[ \frac{110}{S_A} = \frac{110}{S_P} - 6 \quad \text{(Equation 3)} \]Substitute Equation 3 into Equation 2:
\[ \frac{110}{2S_P} = \left( \frac{110}{S_P} - 6 \right) - 5 \]\[ \frac{110}{2S_P} = \frac{110}{S_P} - 11 \]Now, solve for \(S_P\). Multiply the entire equation by \(2S_P\) to eliminate denominators:
\[ 110 = 2S_P \left( \frac{110}{S_P} \right) - 11(2S_P) \]\[ 110 = 220 - 22S_P \]Rearrange the terms to solve for \(S_P\):
\[ 22S_P = 220 - 110 \]\[ 22S_P = 110 \]\[ S_P = \frac{110}{22} \]\[ S_P = 5 \text{ km/hr} \]Correct Option: B) 5 km/hr