The problem involves a discount on the price of sugar, which allows a person to buy more quantity for the same amount of money. We need to find the original and discounted prices per kg. We can set up equations based on the total amount spent and the quantity of sugar bought before and after the discount.
Let the original price of sugar be \(P\) per kg.
The total amount spent is ₹225.
The quantity of sugar that could be bought originally for ₹225 is:
\[ \text{Original Quantity} = \frac{225}{P} \text{ kg} \]A discount of 10% is given on the sugar. So, the discounted price per kg is:
\[ \text{Discounted Price} = P - (0.10 \times P) = 0.9P \text{ per kg} \]With the discounted price, the man can buy 25 kg more sugar for ₹225. So, the new quantity of sugar is:
\[ \text{New Quantity} = \frac{225}{0.9P} \text{ kg} \]According to the problem, the new quantity is 25 kg more than the original quantity:
\[ \frac{225}{0.9P} - \frac{225}{P} = 25 \]Now, we solve this equation for \(P\):
\[ 225 \left( \frac{1}{0.9P} - \frac{1}{P} \right) = 25 \]\[ 225 \left( \frac{1}{0.9P} - \frac{0.9}{0.9P} \right) = 25 \]\[ 225 \left( \frac{1 - 0.9}{0.9P} \right) = 25 \]\[ 225 \left( \frac{0.1}{0.9P} \right) = 25 \]\[ 225 \left( \frac{1}{9P} \right) = 25 \]\[ \frac{225}{9P} = 25 \]\[ \frac{25}{P} = 25 \]\[ P = \frac{25}{25} \]\[ P = 1 \]So, the original price of sugar is ₹1 per kg.
Now, calculate the discounted price:
\[ \text{Discounted Price} = 0.9P = 0.9 \times 1 = \text{₹}0.90 \text{ per kg} \]₹0.90 is equal to 90 paise.
Correct Option: D) 90p, ₹1
To find the least number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, we first find the Least Common Multiple (LCM) of these divisors. Then, we add 1 to the LCM to satisfy the remainder condition. Finally, we check which of these numbers is perfectly divisible by 7.
Find the LCM of 2, 3, 4, 5, and 6:
Formulate the general number:
The numbers which leave a remainder of 1 when divided by 2, 3, 4, 5, and 6 are of the form \(60k + 1\), where \(k\) is a positive integer.
Test for divisibility by 7:
We need to find the smallest value of \(k\) such that \(60k + 1\) is perfectly divisible by 7. We can write this as:
\[60k + 1 \equiv 0 \pmod{7}\]Since \(60 \equiv 4 \pmod{7}\) (because \(60 = 8 \times 7 + 4\)), we can substitute this into the congruence:
\[4k + 1 \equiv 0 \pmod{7}\]Now, we test values for \(k\):
So, the least value for \(k\) is 5.
Calculate the number:
Substitute \(k=5\) into the expression \(60k + 1\):
\[\text{Number} = 60(5) + 1 = 300 + 1 = 301\]Correct Option: D) 301
When 301 is divided by 2, 3, 4, 5, or 6, it leaves a remainder of 1. When 301 is divided by 7, \(301 \div 7 = 43\) with no remainder.
The problem asks us to find out how many more boys need to be added to a line to reach a target number of boys. First, we need to determine the current total number of boys in the line using the given positional information. Once we have the current total, we can subtract it from the target total to find the number of boys to be added.
D) 14. As calculated, 14 boys need to be added to the line to reach a total of 29 boys.
What will be the value after simplifying this continued fraction?
\(\frac{1}{2+\frac{1}{3+\frac{1}{1+\frac{1}{4}}}}\)
\( \frac{43}{5} \)
\( \frac{43}{19} \)
\( \frac{5}{19} \)
\( \frac{19}{43} \)
To simplify a continued fraction, we start from the innermost fraction and work our way outwards. We will perform addition and division operations step-by-step.
Identify the innermost fraction and simplify it:
\[ 1 + \frac{1}{4} = \frac{4}{4} + \frac{1}{4} = \frac{4+1}{4} = \frac{5}{4} \]Substitute this value back into the expression:
\[ \frac{1}{2+\frac{1}{3+\frac{1}{\frac{5}{4}}}} \]Simplify the next layer: \[\frac{1}{\frac{5}{4}}\] is the reciprocal of \(\frac{5}{4}\), which is \(\frac{4}{5}\).
The expression becomes:
\[ \frac{1}{2+\frac{1}{3+\frac{4}{5}}} \]Simplify the denominator of the next fraction:
\[ 3 + \frac{4}{5} = \frac{3 \times 5}{5} + \frac{4}{5} = \frac{15}{5} + \frac{4}{5} = \frac{15+4}{5} = \frac{19}{5} \]Substitute this value back into the expression:
\[ \frac{1}{2+\frac{1}{\frac{19}{5}}} \]Simplify the next layer: \[\frac{1}{\frac{19}{5}}\] is the reciprocal of \(\frac{19}{5}\), which is \(\frac{5}{19}\).
The expression becomes:
\[ \frac{1}{2+\frac{5}{19}} \]Simplify the final denominator:
\[ 2 + \frac{5}{19} = \frac{2 \times 19}{19} + \frac{5}{19} = \frac{38}{19} + \frac{5}{19} = \frac{38+5}{19} = \frac{43}{19} \]Substitute this value back into the expression:
\[ \frac{1}{\frac{43}{19}} \]Finally, simplify the entire expression:
The reciprocal of \(\frac{43}{19}\) is \(\frac{19}{43}\).
Correct Option: D) \( \frac{19}{43} \)
The question asks for the state where the Murudeshwar Temple, located on Kanduka Giri, is situated. This is a factual question related to Indian geography and famous religious sites.
D) Karnataka is the correct answer because Murudeshwar Temple, including Kanduka Giri, is located in the state of Karnataka.
This is a blood relation problem. We need to deduce the relationships between individuals based on the given statements. We will build a family tree or list out the relationships step-by-step to find how Jayant is related to the lady.
Correct Option: C) Husband
What is the value of \(\frac{\sqrt{5}+1}{\sqrt{5}-1}\)?
\(\frac{3-\sqrt{5}}{2}\)
\(\frac{\sqrt{5}}{2}\)
\(\frac{3\sqrt{5}}{2}\)
\(\frac{3+\sqrt{5}}{2}\)
To simplify the given expression, we need to rationalize the denominator. This is done by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{5}-1\) is \(\sqrt{5}+1\).
The given expression is:
\[ \frac{\sqrt{5}+1}{\sqrt{5}-1} \]Multiply the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{5}+1\):
\[ \frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}+1} \]Apply the algebraic identity \((a-b)(a+b) = a^2 - b^2\) to the denominator and \((a+b)^2 = a^2 + 2ab + b^2\) to the numerator.
For the numerator: \((\sqrt{5}+1)^2 = (\sqrt{5})^2 + 2(\sqrt{5})(1) + (1)^2 = 5 + 2\sqrt{5} + 1 = 6 + 2\sqrt{5}\)
For the denominator: \((\sqrt{5}-1)(\sqrt{5}+1) = (\sqrt{5})^2 - (1)^2 = 5 - 1 = 4\)
Substitute these simplified expressions back into the fraction:
\[ \frac{6 + 2\sqrt{5}}{4} \]Factor out 2 from the numerator:
\[ \frac{2(3 + \sqrt{5})}{4} \]Simplify the fraction by dividing the numerator and denominator by 2:
\[ \frac{3 + \sqrt{5}}{2} \]This matches option D.
D) \(\frac{3+\sqrt{5}}{2}\) is the correct simplification of the given expression after rationalizing the denominator.
Jaya and Hema can clean an apartment in 30 days. Hema and Sushma can do the same task in 40 days, while Jaya and Sushma take 60 days to do it. How many days will it take all three of them working together to clean the apartment?
15 \( \frac{2}{3} \)
\( \frac{2}{3} \)
6\( \frac{2}{3} \)
26 \( \frac{2}{3} \)
This problem can be solved using the concept of work and time. We will calculate the one-day work rate for each pair and then use these rates to find the combined one-day work rate of all three individuals. Finally, we will determine the total time required for all three to complete the task together.
D) 26 \( \frac{2}{3} \) is the correct answer, as calculated above.
To answer this question, one needs to have a clear understanding of the geographical boundaries of India and its neighboring country, Nepal. We will identify each Indian state that shares a border with Nepal.
Correct Option: A) 5. The five Indian states that share a boundary with Nepal are Uttarakhand, Uttar Pradesh, Bihar, West Bengal, and Sikkim.
The question asks for the name of India's first planned manned space flight. This is a factual question related to India's space program, specifically ISRO's human spaceflight mission.
Correct Option: A) Gaganyaan is the name of India's first planned manned space flight mission. It is an ambitious project by ISRO to demonstrate human spaceflight capability by sending a crew of three astronauts to an orbit of 400 km for a three-day mission and bringing them back safely to Earth by landing in Indian sea waters.