To find the LCM (Least Common Multiple) of decimal numbers, we first convert them into fractions. Then, we find the LCM of the numerators and the HCF (Highest Common Factor) of the denominators. The LCM of fractions is given by the formula:
\[ \text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}} \]Alternatively, we can make the number of decimal places equal and then find the LCM of the resulting integers, finally adjusting the decimal point.
Convert decimals to fractions:
The given numbers are \(6.3\) and \(0.07\).
\[ 6.3 = \frac{63}{10} \] \[ 0.07 = \frac{7}{100} \]Find the LCM of the numerators:
The numerators are \(63\) and \(7\).
We know that \(63 = 9 \times 7\). Since \(7\) is a factor of \(63\), the LCM of \(63\) and \(7\) is \(63\).
\[ \text{LCM}(63, 7) = 63 \]Find the HCF of the denominators:
The denominators are \(10\) and \(100\).
We know that \(100 = 10 \times 10\). Since \(10\) is a factor of \(100\), the HCF of \(10\) and \(100\) is \(10\).
\[ \text{HCF}(10, 100) = 10 \]Calculate the LCM of the given numbers:
Using the formula for LCM of fractions:
\[ \text{LCM}(6.3, 0.07) = \frac{\text{LCM}(63, 7)}{\text{HCF}(10, 100)} = \frac{63}{10} = 6.3 \]Make the number of decimal places equal:
The numbers are \(6.3\) and \(0.07\). The maximum number of decimal places is two (in \(0.07\)).
Rewrite \(6.3\) as \(6.30\).
Now consider the numbers as integers by removing the decimal point: \(630\) and \(7\).
Find the LCM of the integers:
We need to find \(\text{LCM}(630, 7)\).
Since \(630 = 90 \times 7\), \(7\) is a factor of \(630\). Therefore, the LCM of \(630\) and \(7\) is \(630\).
\[ \text{LCM}(630, 7) = 630 \]Place the decimal point:
Since we made the numbers have two decimal places initially, the final LCM should also have two decimal places.
So, \(630\) becomes \(6.30\) or \(6.3\).
Both methods yield the same result.
D) \(6.3\). The LCM of \(6.3\) and \(0.07\) is \(6.3\).
We need to find the amount invested at 10% per annum. We are given the total investment, the two interest rates, the time period, and that the simple interests earned from both investments are equal. We will use the simple interest formula \(I = \frac{P \times R \times T}{100}\) to set up an equation and solve for the unknown principal amounts.
Let the total investment be \(P_{\text{total}} = \text{₹}20,900\).
Let the sum invested at 10% per annum be \(P_1\). Then the sum invested at 9% per annum will be \(P_2 = P_{\text{total}} - P_1 = 20,900 - P_1\).
The rate of interest for the first investment is \(R_1 = 10\%\) per annum.
The rate of interest for the second investment is \(R_2 = 9\%\) per annum.
The time period for both investments is \(T = 8\) years.
The simple interest earned from the first investment is \(I_1 = \frac{P_1 \times R_1 \times T}{100}\).
\[ I_1 = \frac{P_1 \times 10 \times 8}{100} = \frac{80 P_1}{100} = \frac{4 P_1}{5} \]The simple interest earned from the second investment is \(I_2 = \frac{P_2 \times R_2 \times T}{100}\).
\[ I_2 = \frac{(20,900 - P_1) \times 9 \times 8}{100} = \frac{72 (20,900 - P_1)}{100} = \frac{18 (20,900 - P_1)}{25} \]According to the problem, the interests earned from the two investments are equal: \(I_1 = I_2\).
\[ \frac{4 P_1}{5} = \frac{18 (20,900 - P_1)}{25} \]To solve for \(P_1\), we can cross-multiply or multiply both sides by the least common multiple of the denominators (which is 25).
\[ 25 \times \frac{4 P_1}{5} = 25 \times \frac{18 (20,900 - P_1)}{25} \]\[ 5 \times 4 P_1 = 18 (20,900 - P_1) \]\[ 20 P_1 = 18 \times 20,900 - 18 P_1 \]\[ 20 P_1 + 18 P_1 = 18 \times 20,900 \]\[ 38 P_1 = 18 \times 20,900 \]\[ P_1 = \frac{18 \times 20,900}{38} \]Now, perform the calculation:
\[ P_1 = \frac{18 \times 20,900}{38} = \frac{9 \times 20,900}{19} \]\[ P_1 = 9 \times \frac{20,900}{19} \]\[ P_1 = 9 \times 1,100 \]\[ P_1 = 9,900 \]Thus, the sum invested at 10% per annum is ₹9,900.
D) 9,900 is the correct amount invested at 10% per annum.
By how much is 65% of 65 greater than \( \frac{2}{5} \) of 55?
To solve this problem, we need to calculate two values: first, 65% of 65, and second, \( \frac{2}{5} \) of 55. After calculating both values, we will find the difference between the first value and the second value.
Calculate 65% of 65:
To find a percentage of a number, convert the percentage to a decimal or a fraction and then multiply by the number.
\[ 65\% \text{ of } 65 = \frac{65}{100} \times 65 \] \[ = \frac{4225}{100} \] \[ = 42.25 \]Calculate \( \frac{2}{5} \) of 55:
To find a fraction of a number, multiply the fraction by the number.
\[ \frac{2}{5} \text{ of } 55 = \frac{2}{5} \times 55 \] \[ = 2 \times \frac{55}{5} \] \[ = 2 \times 11 \] \[ = 22 \]Find the difference:
The question asks by how much the first value is greater than the second value. So, we subtract the second value from the first value.
\[ \text{Difference} = 42.25 - 22 \] \[ = 20.25 \]Thus, 65% of 65 is greater than \( \frac{2}{5} \) of 55 by 20.25.
D) 20.25 is the correct answer, as calculated above.
To determine which number the given number 648679059 is divisible by, we will apply the divisibility rules for each of the options: 10, 11, 3, and 8.
Based on these checks, only 3 divides 648679059.
C) 3 — The sum of the digits of 648679059 is \(6+4+8+6+7+9+0+5+9 = 54\). Since 54 is divisible by 3 (\(54 \div 3 = 18\)), the number 648679059 is divisible by 3.
Find the value of: 57-9 ÷ 574 × 57-10
The problem involves simplifying an expression with the same base and different exponents, using the laws of exponents. Specifically, we will use the rules for division \(\left(a^m \div a^n = a^{m-n}\right)\) and multiplication \(\left(a^m \times a^n = a^{m+n}\right)\).
Correct Option: C) \(57^{-23}\)
To find the ratio of the surface areas, we need to first define the formula for the surface area of a sphere (assuming the balloon is spherical). Then, we will express the initial and final radii and calculate their respective surface areas. Finally, we will find the ratio of the final surface area to the initial surface area.
Let the initial radius of the balloon be \(r_1\).
According to the problem, the balloon's radius increases to 5 times itself. So, the final radius \(r_2\) will be:
\[ r_2 = 5r_1 \]The surface area (\(A\)) of a sphere with radius \(r\) is given by:
\[ A = 4\pi r^2 \]Using the initial radius \(r_1\), the initial surface area \(A_1\) is:
\[ A_1 = 4\pi r_1^2 \]Using the final radius \(r_2 = 5r_1\), the final surface area \(A_2\) is:
\[ A_2 = 4\pi (5r_1)^2 \] \[ A_2 = 4\pi (25r_1^2) \] \[ A_2 = 100\pi r_1^2 \]The required ratio is \(A_2 : A_1\):
\[ \frac{A_2}{A_1} = \frac{100\pi r_1^2}{4\pi r_1^2} \]Cancel out the common terms \(4\pi r_1^2\):
\[ \frac{A_2}{A_1} = \frac{100}{4} \] \[ \frac{A_2}{A_1} = 25 \]So, the ratio is \(25 : 1\).
Correct Option: C) 25 : 1
The problem asks us to find the third proportional of two given numbers. The concept of third proportional relates three numbers in a continuous proportion. If three numbers \(a\), \(b\), and \(x\) are in continuous proportion, then the ratio of the first to the second is equal to the ratio of the second to the third. This can be written as \(a:b = b:x\), or equivalently, \[\frac{a}{b} = \frac{b}{x}\].
Thus, the value of \(x\) is 15.
Correct Option: C) 15
To find the average of the first 'n' even numbers, we can use the formula for the average of an arithmetic progression, or a simpler direct formula for the average of the first 'n' even numbers. The first 'n' even numbers form an arithmetic progression.
Correct Option: D) 157
We need to find the list price of the fan given the profit earned, the percentage by which the list price is above the cost price, and the discount percentage offered. We will set up equations based on the given percentages and profit to find the cost price first, and then the list price.
Let the Cost Price (CP) of the fan be \(C\).
The shopkeeper lists the price of a fan at 25% above its cost price. So, the List Price (LP) is:
\[ \text{LP} = C + 0.25C = 1.25C \]A 16% discount is offered on the list price. The Selling Price (SP) will be:
\[ \text{SP} = \text{LP} - 0.16 \times \text{LP} = (1 - 0.16) \times \text{LP} = 0.84 \times \text{LP} \]Substitute the expression for LP from step 2 into the SP equation:
\[ \text{SP} = 0.84 \times (1.25C) \]\[ \text{SP} = 1.05C \]The profit earned is given as ₹64. Profit is calculated as \(\text{SP} - \text{CP}\).
\[ \text{Profit} = \text{SP} - C \]\[ 64 = 1.05C - C \]\[ 64 = 0.05C \]Now, calculate the Cost Price (C):
\[ C = \frac{64}{0.05} \]\[ C = \frac{64}{\frac{5}{100}} \]\[ C = \frac{64 \times 100}{5} \]\[ C = 64 \times 20 \]\[ C = 1280 \]Finally, calculate the List Price (LP) using the value of C:
\[ \text{LP} = 1.25C \]\[ \text{LP} = 1.25 \times 1280 \]\[ \text{LP} = \frac{5}{4} \times 1280 \]\[ \text{LP} = 5 \times 320 \]\[ \text{LP} = 1600 \]The list price of the fan is ₹1600.
B) 1,600
To solve this problem, we will represent the two consecutive even natural numbers algebraically. Then, we will form an equation based on the given condition that the sum of their squares is 2452. Solving this quadratic equation will give us the values of the numbers, and finally, we can find their sum.
Define the numbers: Let the first even natural number be \(x\). Since the numbers are consecutive even natural numbers, the next even natural number will be \(x+2\).
Formulate the equation: The problem states that the sum of the squares of these two numbers is 2452. So, we can write the equation:
\[ x^2 + (x+2)^2 = 2452 \]Expand and simplify the equation:
\[ x^2 + (x^2 + 4x + 4) = 2452 \] \[ 2x^2 + 4x + 4 = 2452 \]Subtract 2452 from both sides:
\[ 2x^2 + 4x + 4 - 2452 = 0 \] \[ 2x^2 + 4x - 2448 = 0 \]Divide by 2 to simplify:
\[ x^2 + 2x - 1224 = 0 \]Solve the quadratic equation: We can solve this quadratic equation by factoring or using the quadratic formula. Let's try factoring. We need two numbers that multiply to -1224 and add up to 2. These numbers are 36 and -34.
\[ x^2 + 36x - 34x - 1224 = 0 \] \[ x(x+36) - 34(x+36) = 0 \] \[ (x-34)(x+36) = 0 \]This gives two possible values for \(x\):
\[ x-34 = 0 \implies x = 34 \] \[ x+36 = 0 \implies x = -36 \]Choose the correct value for \(x\): Since the problem specifies natural numbers, \(x\) must be a positive even number. Therefore, \(x = 34\).
Find the two consecutive even numbers:
The first number is \(x = 34\).
The second number is \(x+2 = 34+2 = 36\).
Verify the condition: Let's check if the sum of their squares is 2452:
\[ 34^2 + 36^2 = 1156 + 1296 = 2452 \]The condition is satisfied.
Calculate the sum of the numbers:
Sum \( = 34 + 36 = 70\).
A) 70 is the sum of the two consecutive even natural numbers (34 and 36).