To find the time when the trains meet, we first need to calculate the duration of travel for each train and their respective speeds. Since the distance between stations M and N is unknown, we can assume it to be 'D'. Then, we can determine the relative speed of the trains when they are moving towards each other and calculate the time taken to meet.
Correct Option: B) 11:55 a.m.
To find the smallest number that should be added to 4519 so that the sum is completely divisible by 30, we need to determine the remainder when 4519 is divided by 30. The difference between the divisor (30) and this remainder will be the smallest number to add.
Divide 4519 by 30:
We perform the division:
\[ 4519 \div 30 \]Let's do long division:
So, we can write:
\[ 4519 = 30 \times 150 + 19 \]The quotient is \(150\) and the remainder is \(19\).
Determine the number to add:
For the sum to be completely divisible by 30, the remainder must be 0. Currently, the remainder is \(19\). To make the remainder 0, we need to add a number to 4519 such that the new remainder is 0 when divided by 30.
The smallest number to add will be the difference between the divisor (30) and the current remainder (19).
\[ \text{Number to add} = \text{Divisor} - \text{Remainder} \] \[ \text{Number to add} = 30 - 19 \] \[ \text{Number to add} = 11 \]Verify the result:
If we add 11 to 4519, we get:
\[ 4519 + 11 = 4530 \]Now, let's check if 4530 is divisible by 30:
\[ 4530 \div 30 = 151 \]Since 4530 is completely divisible by 30, 11 is the smallest number that needs to be added.
D) 11 is the correct answer because when 11 is added to 4519, the sum becomes 4530, which is perfectly divisible by 30.
We are given two conditions: the product of two positive numbers and the relationship between them. We will use these conditions to set up equations, solve for the individual numbers, and then find their sum.
Let the second number be \(x\).
According to the problem, the first number is five times the second number. So, the first number is \(5x\).
The product of the two numbers is 2880. Therefore:
\[ (5x) \times (x) = 2880 \]Divide both sides by 5:
\[ x^2 = \frac{2880}{5} \] \[ x^2 = 576 \]Take the square root of both sides. Since the numbers are positive, we take the positive root:
\[ x = \sqrt{576} \] \[ x = 24 \]The second number is \(x = 24\).
The first number is \(5x = 5 \times 24 = 120\).
Sum \( = \text{First number} + \text{Second number} \)
\[ \text{Sum} = 120 + 24 \] \[ \text{Sum} = 144 \]Correct Option: D) 144
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. Then, we can subtract the number of literate males from the total literate people to find the number of literate females. Finally, divide the number of literate females by the total number of females and multiply by 100 to get the percentage.
B) 14% is the correct answer. The calculations show that 14% of females are literate.
We are given the HCF and LCM of two numbers, as well as their difference. We will use the fundamental property that the product of two numbers is equal to the product of their HCF and LCM. Then, we will test the given options to find the pair of numbers that satisfies all conditions.
Let the two numbers be \(a\) and \(b\).
We are given:
The fundamental relationship between two numbers, their HCF, and LCM is:
\[ a \times b = \text{HCF} \times \text{LCM} \]Substitute the given values into the formula:
\[ a \times b = 6 \times 36 \]\[ a \times b = 216 \]Now we have two equations:
We can solve these equations or check the given options against these conditions.
Let's check the options:
C) 6 and 36 — This pair satisfies both conditions: their difference is \(36 - 6 = 30\), and their product is \(6 \times 36 = 216\), which is equal to HCF \(\times\) LCM (\(6 \times 36 = 216\)).
The problem involves simple interest calculations. We are given the total investment, two different interest rates, and the condition that the interest earned from both investments is equal after a certain period. We need to find the amount invested at one of the rates. We will use the simple interest formula \(I = \frac{P \times R \times T}{100}\) and set up an equation based on the equal interest condition.
Let the total investment be \(P_{\text{total}} = \text{₹}20,900\).
Let the amount invested at 10% per annum be \(P_1\). Then the amount 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 part is \(R_1 = 10\%\) per annum.
The rate of interest for the second part is \(R_2 = 9\%\) per annum.
The time period for both investments is \(T = 5\) years.
The simple interest earned from the first investment is \(I_1 = \frac{P_1 \times R_1 \times T}{100}\).
\[ I_1 = \frac{P_1 \times 10 \times 5}{100} = \frac{50 P_1}{100} = \frac{P_1}{2} \]The simple interest earned from the second investment is \(I_2 = \frac{P_2 \times R_2 \times T}{100}\).
\[ I_2 = \frac{(20,900 - P_1) \times 9 \times 5}{100} = \frac{45 (20,900 - P_1)}{100} \]According to the problem, the interests earned from the two investments are equal: \(I_1 = I_2\).
\[ \frac{P_1}{2} = \frac{45 (20,900 - P_1)}{100} \]Now, solve for \(P_1\):
\[ 100 P_1 = 2 \times 45 (20,900 - P_1) \]\[ 100 P_1 = 90 (20,900 - P_1) \]\[ 100 P_1 = 90 \times 20,900 - 90 P_1 \]\[ 100 P_1 + 90 P_1 = 90 \times 20,900 \]\[ 190 P_1 = 90 \times 20,900 \]\[ P_1 = \frac{90 \times 20,900}{190} \]\[ P_1 = \frac{9 \times 20,900}{19} \]\[ P_1 = 9 \times 1,100 \]\[ P_1 = 9,900 \]The sum invested at 10% per annum is ₹9,900.
C) 9,900
Simplify the following:
3((\( \frac{4}{3} \))x² − 30x + 18) − 4(x² + 8x − 14)
To simplify the given algebraic expression, we will first distribute the constants into their respective parentheses. Then, we will combine like terms (terms with \(x^2\), terms with \(x\), and constant terms) to arrive at the simplified expression.
B) \(-122x + 110\) is the result obtained after distributing the constants and combining like terms.
To find the greatest ratio among the given options, we need to convert each ratio into its decimal equivalent. The ratio that yields the largest decimal value will be the greatest.
Comparing the calculated decimal values:
The largest value is \(0.7843\).
Correct Option: A) 40 : 51
The decimal equivalent of \(40 : 51\) is approximately \(0.7843\), which is the greatest among all the given options.
Sakshi bought a toy for ₹456, decorated it at a cost of ₹20 and sold the toy for ₹612. Find the loss or profit percentage.
\( \frac{210}{7} \) % loss
\( \frac{200}{7} \) % profit
200 % profit
\( \frac{200}{7} \) % loss
To find the profit or loss percentage, we first need to calculate the total cost price (CP) of the toy, which includes the purchase price and the decoration cost. Then, we compare this total CP with the selling price (SP) to determine if there was a profit or a loss. Finally, we calculate the percentage using the formula: \[\text{Percentage} = \frac{|\text{Profit/Loss}|}{\text{CP}} \times 100\].
The cost price of the toy is the sum of the price Sakshi bought it for and the cost she spent on decorating it.
\[ \text{CP} = \text{Purchase Price} + \text{Decoration Cost} \]\[ \text{CP} = \text{₹}456 + \text{₹}20 \]\[ \text{CP} = \text{₹}476 \]The selling price of the toy is given as ₹612.
\[ \text{SP} = \text{₹}612 \]Since the Selling Price (SP) is greater than the Cost Price (CP), there is a profit.
\[ \text{Profit} = \text{SP} - \text{CP} \]\[ \text{Profit} = \text{₹}612 - \text{₹}476 \]\[ \text{Profit} = \text{₹}136 \]The profit percentage is calculated using the formula:
\[ \text{Profit Percentage} = \frac{\text{Profit}}{\text{CP}} \times 100 \]\[ \text{Profit Percentage} = \frac{136}{476} \times 100 \]\[ \text{Profit Percentage} = \frac{13600}{476} \]Both 13600 and 476 are divisible by 4:
\[ \frac{13600 \div 4}{476 \div 4} = \frac{3400}{119} \]\[ \text{Profit Percentage} = \frac{3400}{119} \% \]Notice that \(119 = 7 \times 17\). Let's check if 3400 is divisible by 17 or 7.
\(3400 \div 17 = 200\)
So, we can simplify the fraction:
\[ \frac{3400}{119} = \frac{200 \times 17}{7 \times 17} = \frac{200}{7} \]\[ \text{Profit Percentage} = \frac{200}{7} \% \]Since the SP > CP, it is a profit.
Correct Option: B) \( \frac{200}{7} \% \) profit
If the interest earned during the 2nd year on a certain sum is ₹6,786, and the rate of interest is 20% 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 will use the formula for compound interest and the concept that the interest for any year is calculated on the amount accumulated at the end of the previous year.
Interest for the first year (\(CI_1\)) is calculated on the principal \(P\):
\[ CI_1 = P \times \frac{R}{100} = P \times \frac{20}{100} = 0.2P \]The amount at the end of the first year (\(A_1\)) will be:
\[ A_1 = P + CI_1 = P + 0.2P = 1.2P \]The interest for the second year (\(CI_2\)) is calculated on the amount at the end of the first year (\(A_1\)):
\[ CI_2 = A_1 \times \frac{R}{100} = (1.2P) \times \frac{20}{100} = 1.2P \times 0.2 = 0.24P \]We are given that the interest earned during the second year is ₹6,786. So,
\[ 0.24P = 6786 \]Now, we solve for \(P\):
\[ P = \frac{6786}{0.24} \]To simplify the division, we can write 0.24 as \(\frac{24}{100}\):
\[ P = \frac{6786}{\frac{24}{100}} = \frac{6786 \times 100}{24} \]Perform the division:
\[ P = \frac{678600}{24} \]Divide 678600 by 24:
\[ 678600 \div 24 = 28275 \]So, the sum (principal) is ₹28,275.
A) ₹28,275 is the correct sum. This is derived by correctly calculating the interest for the second year based on the amount accumulated after the first year and then solving for the principal.