To find the percentage of literate females, we first need to determine the total number of females, the total number of literate people, and the number of literate males. From these, we can calculate the number of literate females and then express it as a percentage of the total female population.
Total Population: Given as 328000.
Number of Males: Given as 164000.
Number of Females:
\[ \text{Number of Females} = \text{Total Population} - \text{Number of Males} \] \[ \text{Number of Females} = 328000 - 164000 = 164000 \]Total Literate Population: 24% of the total population.
\[ \text{Total Literate Population} = 0.24 \times 328000 \] \[ \text{Total Literate Population} = 78720 \]Number of Literate Males: 34% of the male population.
\[ \text{Number of Literate Males} = 0.34 \times 164000 \] \[ \text{Number of Literate Males} = 55760 \]Number of Literate Females:
\[ \text{Number of Literate Females} = \text{Total Literate Population} - \text{Number of Literate Males} \] \[ \text{Number of Literate Females} = 78720 - 55760 = 22960 \]Percentage of Literate Females:
\[ \text{Percentage of Literate Females} = \left( \frac{\text{Number of Literate Females}}{\text{Number of Females}} \right) \times 100 \] \[ \text{Percentage of Literate Females} = \left( \frac{22960}{164000} \right) \times 100 \] \[ \text{Percentage of Literate Females} = 0.14 \times 100 = 14\% \]Correct Option: D) 14%
To solve this problem, we need to apply the divisibility rule for 11. A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
The given 6-digit number is N75M52.
Starting from the rightmost digit (units place) as the 1st place:
According to the rule, \( |S_{\text{odd}} - S_{\text{even}}| \) must be 0 or a multiple of 11. So, we set the difference to 0 or \(\pm 11\), etc.
\[ (9 + M) - (10 + N) = 0 \quad \text{or} \quad \pm 11 \quad \text{or} \quad \pm 22 \quad \dots \]Since M and N are single digits (0-9), the value of \(M - N\) will be between \(0-9 = -9\) and \(9-0 = 9\).
Therefore, the only possible relation is \(M - N = 1\).
Correct Option: B) M - N = 1
We are given that a number, when decreased by 6%, results in 3760. We need to find the original number. We can set up an equation where the original number is represented by a variable, and then solve for that variable.
Let the original number be \(x\).
When the number is decreased by 6%, it means we are taking away 6% of \(x\) from \(x\). The percentage decrease can be written as \(6\%\) of \(x\), which is \(\frac{6}{100}x\).
So, the number after being decreased by 6% is \(x - \frac{6}{100}x\).
We are given that this decreased number is 3760. So, we can set up the equation:
\[ x - \frac{6}{100}x = 3760 \]To simplify the left side of the equation, we can write \(x\) as \(\frac{100}{100}x\):
\[ \frac{100}{100}x - \frac{6}{100}x = 3760 \] \[ \frac{100 - 6}{100}x = 3760 \] \[ \frac{94}{100}x = 3760 \]Now, to find \(x\), we need to isolate it. Multiply both sides by \(\frac{100}{94}\):
\[ x = 3760 \times \frac{100}{94} \]Perform the division:
We can see that \(3760\) is a multiple of \(94\). Let's divide \(3760\) by \(94\):
\[ 3760 \div 94 = 40 \]Substitute this back into the equation for \(x\):
\[ x = 40 \times 100 \] \[ x = 4000 \]Thus, the original number is 4000.
Correct Option: B) 4000
To find the third proportional of two numbers, say 'a' and 'b', we set up a proportion where 'b' is the mean proportional. The relationship is \(a : b :: b : x\), where \(x\) is the third proportional. This can be written as a fraction: \[\frac{a}{b} = \frac{b}{x}\].
Thus, the value of \(x\) is 400.
Correct Option: B) 400
The problem asks us to find the greater of two consecutive natural numbers whose product is 552. We can solve this by setting up an algebraic equation or by using estimation and checking the options.
A) 24 — If the greater number is 24, the smaller number is 23. Their product is \(23 \times 24 = 552\), which satisfies the condition given in the problem.
To determine which number divides 6,30,605, we can use divisibility rules for each of the given options. Divisibility rules provide a quick way to check if a number is divisible by another number without performing long division.
Sum of digits \( = 6 + 3 + 0 + 6 + 0 + 5 = 20 \).
Since 20 is not divisible by 3, 6,30,605 is not divisible by 3.
The last digit of 6,30,605 is 5.
Since the last digit is 5, 6,30,605 is divisible by 5.
Let's check by division or the rule. Using the rule:
\(63060 - (2 \times 5) = 63060 - 10 = 63050\)
\(6305 - (2 \times 0) = 6305\)
\(630 - (2 \times 5) = 630 - 10 = 620\)
\(62 - (2 \times 0) = 62\)
Since 62 is not divisible by 7, 6,30,605 is not divisible by 7.
For a number to be divisible by 14, it must be divisible by both 2 and 7. The last digit of 6,30,605 is 5, which is an odd number. Therefore, 6,30,605 is not divisible by 2, and consequently not divisible by 14.
B) 5 — The number 6,30,605 ends with the digit 5, which means it is divisible by 5.
The perimeter of any polygon is the sum of the lengths of its sides. For an equilateral triangle, all three sides are equal in length. Therefore, to find the perimeter, we simply multiply the length of one side by 3.
The perimeter of the equilateral triangle is \(39 \text{ cm}\).
Correct Option: B) 39 cm
To find the Least Common Multiple (LCM) of given numbers expressed in prime factor form, we identify all unique prime factors present in any of the numbers. For each prime factor, we take the highest power that appears in any of the given numbers. The product of these highest powers is the LCM.
The given numbers are:
First, we need to express the composite bases (8 and 16) as powers of their prime factors:
Now, substitute these into the given numbers:
Let's re-list them clearly:
The unique prime factors are 2, 3, and 13.
Multiply the highest powers of all unique prime factors:
\[ \text{LCM} = 2^{17} \times 3^3 \times 13^2 \]The options use 8 and 16. We know \(8 = 2^3\) and \(16 = 2^4\).
We need to express \(2^{17}\) using powers of 8 and 16. Notice that \(2^{17} = 2^9 \times 2^8 = (2^3)^3 \times (2^4)^2 = 8^3 \times 16^2\).
So, the LCM is \(3^3 \times 8^3 \times 13^2 \times 16^2\).
Correct Option: A) \(3^3 \times 8^3 \times 13^2 \times 16^2\)
To determine divisibility, we will use the divisibility rules for each given option. We need to check if the sum of the digits is divisible by 3, if the number is divisible by both 3 and 4 for 12, if the last digit is 0 or 5 for 5, and if the last three digits are divisible by 8 for 8.
A number is divisible by 3 if the sum of its digits is divisible by 3.
The given number is \(253118139\).
Sum of digits \( = 2 + 5 + 3 + 1 + 1 + 8 + 1 + 3 + 9 = 33 \).
Since \(33\) is divisible by \(3\) (\(33 \div 3 = 11\)), the number \(253118139\) is divisible by \(3\).
A number is divisible by 12 if it is divisible by both 3 and 4.
We already know the number is divisible by 3.
For divisibility by 4, the number formed by its last two digits must be divisible by 4.
The last two digits of \(253118139\) are \(39\).
Since \(39\) is not divisible by \(4\) (\(39 \div 4 = 9\) with a remainder of \(3\)), the number \(253118139\) is not divisible by \(4\). Therefore, it is not divisible by \(12\).
A number is divisible by 5 if its last digit is 0 or 5.
The last digit of \(253118139\) is \(9\).
Since the last digit is neither \(0\) nor \(5\), the number \(253118139\) is not divisible by \(5\).
A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
The last three digits of \(253118139\) are \(139\).
Let's divide \(139\) by \(8\):
\[ 139 \div 8 \]\[ 139 = 8 \times 17 + 3 \]\[ 139 \div 8 = 17 \text{ with a remainder of } 3 \]Since \(139\) is not divisible by \(8\), the number \(253118139\) is not divisible by \(8\).
Based on these checks, only option A (3) is a divisor of the given number.
A) 3 — The sum of the digits of \(253118139\) is \(33\), which is divisible by \(3\). Therefore, the number is divisible by \(3\).
To find the average of the first \(n\) even numbers, we can use the formula for the sum of an arithmetic progression or a direct formula for the average of the first \(n\) even numbers. The first \(n\) even numbers form an arithmetic progression with the first term \(a = 2\) and common difference \(d = 2\).