All members of a club went to Mumbai and stayed in a hotel. On the first day, 80% went for shopping and 50% went for sightseeing, whereas 10% took rest in the hotel. Which of the following conclusion(s) can be drawn from the above data?
1. 40% members went for shopping as well as sightseeing.
2. 20% members went for only shopping.
Select the correct answer using the code given below:
We are given percentages of club members who engaged in different activities and some who rested. We need to determine the percentage of members who engaged in both shopping and sightseeing, and the percentage who went for only shopping. This problem can be solved using the principle of inclusion-exclusion for sets or by using Venn diagrams.
Let \(S\) be the set of members who went for shopping and \(T\) be the set of members who went for sightseeing.
We are given:
The total percentage of members who engaged in at least one activity (shopping or sightseeing) or rested must be \(100\%\). The members who rested did not go for shopping or sightseeing. So, the percentage of members who went for either shopping or sightseeing or both is \(100\% - P(\text{Rest})\).
\[ P(S \cup T) = 100\% - 10\% = 90\% \]Now, we can use the principle of inclusion-exclusion to find the percentage of members who went for both shopping and sightseeing, \(P(S \cap T)\):
\[ P(S \cup T) = P(S) + P(T) - P(S \cap T) \]\[ 90\% = 80\% + 50\% - P(S \cap T) \]\[ 90\% = 130\% - P(S \cap T) \]\[ P(S \cap T) = 130\% - 90\% = 40\% \]Conclusion 1: 40% members went for shopping as well as sightseeing. This statement is correct.
Next, let's find the percentage of members who went for only shopping. This is \(P(S \text{ only}) = P(S) - P(S \cap T)\).
\[ P(S \text{ only}) = 80\% - 40\% = 40\% \]Conclusion 2: 20% members went for only shopping. This statement is incorrect, as we calculated 40%.
A) 1 only. Based on our calculations, only conclusion 1 is correct.
In a school, 60% students play cricket. A student who does not play cricket, plays football. Every football player has got a two-wheeler. Which of the following conclusions cannot be drawn from the above data?
1. 60% of the students do not have two-wheelers.
2. No cricketer has a two-wheeler.
3. Cricket players do not play football.
Select the correct answer using the code given below:
1 and 2 only
2 and 3 only
1 and 3 only
1, 2 and 3
The problem provides several statements about students, their sports, and possessions. We need to analyze each given conclusion based on these statements to determine which conclusions cannot be drawn from the data. This requires careful interpretation of the given conditions and identifying what is explicitly stated versus what is merely possible or not mentioned.
All three conclusions (1, 2, and 3) cannot be drawn from the given data.
Correct Option: D) 1, 2 and 3
We need to find the number of two-digit natural numbers such that the ratio of the number to the number formed by reversing its digits is 4:7. We will represent a two-digit number algebraically, form an equation based on the given ratio, and then find the possible pairs of digits that satisfy the condition.
Represent the two-digit number:
Let the two-digit number be \(10a + b\), where \(a\) is the tens digit and \(b\) is the units digit. Since it's a two-digit number, \(a\) must be a non-zero digit, so \(1 \le a \le 9\). The units digit \(b\) can be any digit from 0 to 9, so \(0 \le b \le 9\).
Represent the number with reversed digits:
The number formed by reversing its digits will be \(10b + a\). For this to be a two-digit number, \(b\) must also be a non-zero digit, so \(1 \le b \le 9\). (If \(b=0\), the reversed number would be a single-digit number, which is not implied by the problem statement of "reversing its digits" in the context of a two-digit number.)
Set up the ratio equation:
The problem states that the ratio of the original number to the reversed number is 4:7. So, we can write the equation:
Solve the equation for \(a\) and \(b\):
Cross-multiply to simplify the equation:
Find possible pairs of digits \((a, b)\):
We need to find integer values for \(a\) and \(b\) that satisfy \(2a = b\), with the conditions \(1 \le a \le 9\) and \(1 \le b \le 9\).
Let's test values for \(a\):
Count the number of such pairs:
The valid pairs \((a, b)\) are (1, 2), (2, 4), (3, 6), and (4, 8). There are 4 such pairs.
Correct Option: B) 4
We are given two pieces of information: the difference in marks between A and B, and the percentage difference in marks between B and A. We can set up algebraic equations based on these statements to find the marks scored by B.
Correct Option: B) 380
This problem involves calculating relative distances covered by three contestants (X, Y, and Z) in a race. We need to determine the head start Y would give to Z in a 1000 m race, based on their performance against X. The key is to find the ratio of their speeds or, equivalently, the distance covered by one when the other covers a certain distance.
In a 1000 m race, X gives Y a start of 40 m. This means when X runs 1000 m, Y runs \((1000 - 40) = 960\) m.
So, the ratio of distances covered by X to Y is:
\[ \frac{\text{Distance by X}}{\text{Distance by Y}} = \frac{1000}{960} \]In a 1000 m race, X gives Z a start of 64 m. This means when X runs 1000 m, Z runs \((1000 - 64) = 936\) m.
So, the ratio of distances covered by X to Z is:
\[ \frac{\text{Distance by X}}{\text{Distance by Z}} = \frac{1000}{936} \]We want to find out how much distance Y covers when Z covers a certain distance, or vice versa. We can use the ratios found in steps 1 and 2.
We have:
\[ \frac{\text{Distance by Y}}{\text{Distance by Z}} = \frac{\text{Distance by Y}}{\text{Distance by X}} \times \frac{\text{Distance by X}}{\text{Distance by Z}} \]\[ \frac{\text{Distance by Y}}{\text{Distance by Z}} = \frac{960}{1000} \times \frac{1000}{936} \]\[ \frac{\text{Distance by Y}}{\text{Distance by Z}} = \frac{960}{936} \]If Y runs 960 m, Z runs 936 m. We need to find out how much Z runs when Y runs 1000 m.
Let \(\text{D}_Y\) be the distance covered by Y and \(\text{D}_Z\) be the distance covered by Z.
When \(\text{D}_Y = 1000\) m, we need to find \(\text{D}_Z\).
\[ \frac{1000}{\text{D}_Z} = \frac{960}{936} \]\[ \text{D}_Z = \frac{1000 \times 936}{960} \]\[ \text{D}_Z = \frac{1000 \times 39}{40} \]\[ \text{D}_Z = 25 \times 39 \]\[ \text{D}_Z = 975 \text{ m} \]When Y completes 1000 m, Z completes 975 m. Therefore, Y gives Z a start of:
\[ 1000 \text{ m} - 975 \text{ m} = 25 \text{ m} \]B) 25 — As calculated, Y will give Z a start of 25 metres in a 1000 m race.
The problem provides two inequalities involving variables \(x\) and \(y\). We need to determine which of the given options is always true based on these conditions. We will analyze each option by manipulating the given inequalities.
We are given two conditions:
We need to find a relationship between \(x\) and \(y\) that is always true. Let's try to find the range of \((y-x)\).
From \(x \ge 25\), we can multiply by \(-1\) and reverse the inequality sign:
\[ -x \le -25 \]Now, we have two inequalities:
Add these two inequalities:
\[ y + (-x) \le 40 + (-25) \]\[ y - x \le 15 \]This inequality, \(y - x \le 15\), must always be true based on the given conditions.
C) \((y-x)\) is less than or equal to 15
As derived in the step-by-step analysis, by adding the inequality \(y \le 40\) and \(-x \le -25\), we get \(y - x \le 15\). This statement is always correct.
This is not always correct. For example, if \(x = 25\) and \(y = 40\), then \(x < y\). If \(x = 30\) and \(y = 20\), then \(x > y\). Since \(x\) can be 25 and \(y\) can be 40, it's possible for \(x\) to be less than \(y\). Thus, this is not always true.
This contradicts our derivation. We found that \(y - x \le 15\). Therefore, \((y-x)\) cannot always be greater than 15. For example, if \(x=25\) and \(y=40\), then \(y-x = 15\), which is not greater than 15.
This is not always correct. While \(x \ge 25\) and \(y \le 40\), the sum \(x+y\) can be less than 65. For example, if \(x = 25\) and \(y = 20\) (which satisfies \(y \le 40\)), then \(x+y = 25+20 = 45\), which is not greater than or equal to 65. Therefore, this statement is not always true.
We need to find the age of Ena's father at the time of his marriage. We are given Ena's current age, the age difference between Ena and her mother, the age difference between her parents, and how many years after marriage Ena was born. We will work backward from Ena's current age to determine the ages of her mother and father, and then calculate the father's age at marriage.
Correct Option: B) 23 years
The problem asks us to calculate the discount percentage offered by a retailer. Rakesh initially had enough money to buy 8 mobile handsets. Due to a discount, he was able to buy 10 handsets with the same amount of money. We need to find the percentage discount on the price of each handset.
Let the original price of one mobile handset be \(P\).
The total amount of money Rakesh had was sufficient to buy 8 handsets at the original price. So, the total amount Rakesh had is:
\[ \text{Total Amount} = 8 \times P \]After the discount, Rakesh could buy 10 mobile handsets with the same total amount. Let the discounted price of one mobile handset be \(P'\).
So, the total amount Rakesh had can also be expressed as:
\[ \text{Total Amount} = 10 \times P' \]Equating the two expressions for the total amount:
\[ 8 \times P = 10 \times P' \]From this equation, we can find the relationship between the original price \(P\) and the discounted price \(P'\):
\[ P' = \frac{8}{10} P = \frac{4}{5} P \]Now, we need to calculate the discount percentage. The discount amount per handset is the original price minus the discounted price:
\[ \text{Discount Amount} = P - P' \]\[ \text{Discount Amount} = P - \frac{4}{5} P = \frac{1}{5} P \]The discount percentage is calculated as (Discount Amount / Original Price) \(\times\) 100%:
\[ \text{Discount Percentage} = \frac{\text{Discount Amount}}{\text{Original Price}} \times 100\% \]\[ \text{Discount Percentage} = \frac{\frac{1}{5} P}{P} \times 100\% \]\[ \text{Discount Percentage} = \frac{1}{5} \times 100\% \]\[ \text{Discount Percentage} = 20\% \]Correct Option: B) 20%
To find the corrected mean marks, we first need to calculate the original (incorrect) total sum of marks. Then, we adjust this total sum by subtracting the misread mark and adding the correct mark. Finally, we divide the corrected total sum by the total number of students to get the corrected mean.
Given that the average marks of 100 students are 40.
\[ \text{Initial Total Sum} = \text{Number of Students} \times \text{Average Marks} \] \[ \text{Initial Total Sum} = 100 \times 40 = 4000 \]Marks of one student were 53, but misread as 83.
The difference due to the error is the incorrect mark minus the correct mark.
\[ \text{Error Difference} = \text{Incorrect Mark} - \text{Correct Mark} \] \[ \text{Error Difference} = 83 - 53 = 30 \]Since the misread mark (83) was higher than the actual mark (53), the initial total sum was inflated by 30.
Subtract the error difference from the initial total sum.
\[ \text{Corrected Total Sum} = \text{Initial Total Sum} - \text{Error Difference} \] \[ \text{Corrected Total Sum} = 4000 - 30 = 3970 \]Divide the corrected total sum by the number of students.
\[ \text{Corrected Mean} = \frac{\text{Corrected Total Sum}}{\text{Number of Students}} \] \[ \text{Corrected Mean} = \frac{3970}{100} = 39.7 \]Correct Option: B) 39.7