Seven businessmen, G, H, I, J, K, L and M, travelled in the same month on different dates 4th, 8th, 11th, 14th, 17th, 22nd and 28th (but not necessarily in the same order).
L travelled the last. K travelled just after I but before J.
K travelled on the 11th.
Only two people travelled between K and M.
Exactly one person travelled between G and K, and K travelled after G.
On which date did M travel?
This is a logical reasoning problem involving sequencing. We need to determine the travel dates of seven businessmen based on the given clues. We will use a table or a list to keep track of the dates and the corresponding individuals as we deduce them.
List the dates: The travel dates are 4th, 8th, 11th, 14th, 17th, 22nd, and 28th.
Use the definite clues first:
Current arrangement:
| Date | Person |
|---|---|
| 4th | |
| 8th | |
| 11th | K |
| 14th | |
| 17th | |
| 22nd | |
| 28th | L |
Incorporate relative positions:
Current arrangement:
| Date | Person |
|---|---|
| 4th | |
| 8th | I |
| 11th | K |
| 14th | J (or later) |
| 17th | |
| 22nd | |
| 28th | L |
Place M using the gap clue:
Current arrangement:
| Date | Person |
|---|---|
| 4th | |
| 8th | I |
| 11th | K |
| 14th | |
| 17th | |
| 22nd | M |
| 28th | L |
Place G using the gap clue:
Current arrangement:
| Date | Person |
|---|---|
| 4th | G |
| 8th | I |
| 11th | K |
| 14th | |
| 17th | |
| 22nd | M |
| 28th | L |
Fill the remaining spots:
The question asks: "On which date did M travel?"
From our deductions, M travelled on the 22nd.
A) 22nd is the correct date for M's travel, as determined in Step 4.
Find the value of
3-5 × 4-5 × 5-5.
\(\frac{1}{30^5}\)
\(\frac{1}{50^5}\)
\(\frac{1}{60^5}\)
\(\frac{1}{40^5}\)
The problem requires us to simplify an expression involving negative exponents. We will use the properties of exponents, specifically the rule \(a^{-n} = \frac{1}{a^n}\) and the rule \((ab)^n = a^n b^n\), to solve this problem.
C) The simplified value of the expression is \(\frac{1}{60^5}\).
The question asks about the opponent India defeated to win the bronze medal in men's hockey at the Tokyo Olympics in 2021. This is a factual question related to sports history and current affairs.
The Indian men's hockey team defeated Germany by a score of 5-4 in a thrilling bronze medal match at the Tokyo Olympics 2020 (held in 2021 due to the pandemic). This victory marked India's first Olympic medal in hockey in 41 years.
A) Germany is the correct answer. The Indian men's hockey team defeated Germany 5-4 to secure the bronze medal at the Tokyo Olympics 2020 (held in 2021).
The question asks to identify the Act that extended the provisions of Part IX of the Constitution, which deals with Panchayats, to the Scheduled Areas. This requires knowledge of specific legislation related to tribal self-governance in India.
Therefore, the PESA Act 1996 was specifically designed for this purpose.
Correct Option: D) PESA Act 1996
The question asks to identify the specific Indian Institute of Science Education and Research (IISER) campus whose scientists successfully extracted over 95% of uranium from natural seawater in an experiment conducted in 2022. This is a factual recall question related to recent scientific advancements in India.
B) Pune — Scientists from IISER Pune were responsible for this breakthrough in uranium extraction from seawater in 2022.
This problem involves properties of parallel lines and transversals, specifically the angles formed and their bisectors. We need to determine the shape formed by the internal bisectors of two pairs of interior angles.
Let the two parallel lines be \(L_1\) and \(L_2\), and let \(T\) be the transversal intersecting them.
Consider two interior angles on the same side of the transversal. Let these be \(\angle A\) and \(\angle B\). Since \(L_1 \parallel L_2\), the sum of these interior angles is \(180^\circ\). So, \(\angle A + \angle B = 180^\circ\).
Let \(AP\) be the bisector of \(\angle A\) and \(BP\) be the bisector of \(\angle B\). These bisectors meet at a point \(P\).
In the triangle formed by these bisectors and the transversal (let's call it \(\triangle APB\)), we have:
\[ \angle PAB = \frac{1}{2} \angle A \]\[ \angle PBA = \frac{1}{2} \angle B \]The sum of angles in \(\triangle APB\) is \(180^\circ\):
\[ \angle APB + \angle PAB + \angle PBA = 180^\circ \]\[ \angle APB + \frac{1}{2} \angle A + \frac{1}{2} \angle B = 180^\circ \]\[ \angle APB + \frac{1}{2} (\angle A + \angle B) = 180^\circ \]Substitute \(\angle A + \angle B = 180^\circ\):
\[ \angle APB + \frac{1}{2} (180^\circ) = 180^\circ \]\[ \angle APB + 90^\circ = 180^\circ \]\[ \angle APB = 90^\circ \]This means that the bisectors of any two consecutive interior angles are perpendicular to each other. Therefore, the angle formed by the intersection of these bisectors is a right angle.
Now, consider the four interior angles formed by the transversal. Let them be \(\angle 1, \angle 2, \angle 3, \angle 4\). Let the bisectors of these angles be \(b_1, b_2, b_3, b_4\).
The internal bisectors of two pairs of interior angles will form a quadrilateral. As established, the angle formed by the bisectors of consecutive interior angles is \(90^\circ\). Since there are four pairs of consecutive interior angles (two on each side of the transversal), all four angles of the quadrilateral formed by these bisectors will be \(90^\circ\).
A quadrilateral with all four angles equal to \(90^\circ\) is a rectangle.
Correct Option: B) rectangle
A diver rowing at the speed of 3 km/h in still water takes double the time going 50 km upstream compared to going 50 km downstream. The speed of the stream is:
\( \frac{7}{3} \) km/h
1 km/h
\( \frac{4}{3} \) km/h
3 km/h
This problem involves concepts of relative speed in rivers (boats and streams). We are given the speed of the diver in still water and a relationship between the time taken to travel upstream and downstream for the same distance. We need to find the speed of the stream.
B) 1 km/h is the speed of the stream, as calculated from the given conditions.
This problem involves the concept of work and time, specifically inverse proportionality. When the amount of work is constant, the number of men and the time taken to complete the work are inversely proportional. This means that if the number of men increases, the time taken decreases, and vice versa. We can use the formula \(M_1 D_1 = M_2 D_2\), where \(M\) represents the number of men and \(D\) represents the number of days.
Identify the given values:
Apply the work-time formula:
The total work done is constant. Therefore, the product of the number of men and the number of days remains constant:
\[ M_1 D_1 = M_2 D_2 \]Substitute the known values into the formula:
\[ 16 \text{ men} \times 5 \text{ days} = M_2 \times 20 \text{ days} \]Calculate the total work units:
\[ 16 \times 5 = 80 \text{ work units} \]Solve for \(M_2\):
\[ 80 = M_2 \times 20 \] \[ M_2 = \frac{80}{20} \] \[ M_2 = 4 \]So, 4 men are required to complete the same work in 20 days.
B) 4 is the correct answer. As calculated, 4 men are needed to complete the work in 20 days, maintaining the same total work output.
A) 3 — This would imply that the work can be done by even fewer men in 20 days, which is incorrect as it would result in less total work (3 men * 20 days = 60 work units, which is less than 80 work units).
C) 6 — This would imply that more men are needed to complete the work in 20 days (6 men * 20 days = 120 work units, which is more than 80 work units). This is incorrect as it would mean the work is done faster or more work is done.
D) 5 — This would imply that 5 men can do the work in 20 days (5 men * 20 days = 100 work units, which is more than 80 work units). This is incorrect for the same reason as option C.
The problem asks us to evaluate the strength of two arguments in relation to a given statement. The statement proposes that all zoos should be banned except for those housing endangered species. We need to determine if each argument supports (strengthens) or opposes (weakens) this statement.
D) Both I and II strengthen the statement.
The cost price of 42 articles is equal to the selling price of 36 articles. What is the gain or loss made in the transaction?
14 \( \frac{1}{7} \) % gain
16 \( \frac{2}{3} \) % gain
14 \( \frac{1}{7} \) % loss
16 \( \frac{2}{3} \) % loss
The problem states that the cost price (CP) of 42 articles is equal to the selling price (SP) of 36 articles. We need to find the percentage gain or loss in this transaction. We can set up an equation based on the given information and then calculate the gain or loss percentage.
Let the Cost Price of one article be \( \text{CP} \) and the Selling Price of one article be \( \text{SP} \).
According to the problem statement, the cost price of 42 articles is equal to the selling price of 36 articles. We can write this as an equation:
\[ 42 \times \text{CP} = 36 \times \text{SP} \]Now, we need to find the relationship between \( \text{SP} \) and \( \text{CP} \). Let's rearrange the equation to express \( \frac{\text{SP}}{\text{CP}} \):
\[ \frac{\text{SP}}{\text{CP}} = \frac{42}{36} \]Simplify the fraction:
\[ \frac{\text{SP}}{\text{CP}} = \frac{7}{6} \]From this ratio, we can see that \( \text{SP} = \frac{7}{6} \times \text{CP} \). Since \( \text{SP} > \text{CP} \), there is a gain.
Calculate the gain:
\[ \text{Gain} = \text{SP} - \text{CP} \]\[ \text{Gain} = \frac{7}{6} \times \text{CP} - \text{CP} \]\[ \text{Gain} = \left( \frac{7}{6} - 1 \right) \times \text{CP} \]\[ \text{Gain} = \left( \frac{7-6}{6} \right) \times \text{CP} \]\[ \text{Gain} = \frac{1}{6} \times \text{CP} \]Calculate the gain percentage. The formula for gain percentage is:
\[ \text{Gain Percentage} = \left( \frac{\text{Gain}}{\text{CP}} \right) \times 100\% \]Substitute the value of Gain:
\[ \text{Gain Percentage} = \left( \frac{\frac{1}{6} \times \text{CP}}{\text{CP}} \right) \times 100\% \]\[ \text{Gain Percentage} = \frac{1}{6} \times 100\% \]\[ \text{Gain Percentage} = \frac{100}{6}\% \]\[ \text{Gain Percentage} = \frac{50}{3}\% \]Convert the improper fraction to a mixed number:
\[ \frac{50}{3}\% = 16 \frac{2}{3}\% \]Since \( \text{SP} > \text{CP} \), it is a gain.
B) 16 \( \frac{2}{3} \) % gain