This problem involves calculating the combined work rate of two taps, one filling and one emptying, and then using that net rate to find the time required to fill a specific volume. We will determine the individual rates, find the net filling rate, and then calculate the time needed for twice the tank's volume.
Calculate the filling rate of the first tap:
The first tap fills the cistern in 9 hours. So, in 1 hour, it fills \(\frac{1}{9}\) of the tank.
Calculate the emptying rate of the second tap:
The second tap empties the full tank in 90 hours. So, in 1 hour, it empties \(\frac{1}{90}\) of the tank.
Calculate the net filling rate when both taps are open:
When both taps are open, the net rate at which the tank fills is the difference between the filling rate and the emptying rate.
To subtract these fractions, find a common denominator, which is 90.
\[ \text{Net rate} = \frac{10}{90} - \frac{1}{90} \]\[ \text{Net rate} = \frac{10 - 1}{90} = \frac{9}{90} = \frac{1}{10} \text{ tank/hour} \]Calculate the time to fill one tank:
If the net rate is \(\frac{1}{10}\) tank per hour, it means it takes 10 hours to fill 1 tank.
Calculate the time to fill twice the volume of the tank:
The question asks for the time to fill twice the volume of the tank. Since it takes 10 hours to fill one tank, it will take twice that time to fill two tanks.
D) 20
A) 10 — This would be the time taken to fill one tank, not twice the volume, with both taps open.
B) 5 — This might result from an incorrect calculation, perhaps by dividing the time to fill one tank by 2, or by incorrectly combining the rates. For example, if one mistakenly thought the net rate was \(\frac{1}{5}\) tank/hour, which is not the case.
C) 15 — This value does not correspond to any logical step in the calculation. It might arise from an arithmetic error or an incorrect interpretation of the problem.
The Periplus of the Erythraean Sea was written in the:
1st century AD
2nd century BC
5th century AD
3rd century AD
The question asks about the period when the ancient Greek text "The Periplus of the Erythraean Sea" was written. This requires knowledge of historical timelines related to ancient maritime trade and literature.
A) 1st century AD is the correct answer because historical evidence places the authorship of "The Periplus of the Erythraean Sea" within this period, specifically around the mid-1st century AD.
To find the third proportional of two numbers, say 'a' and 'b', we assume the third proportional to be 'c'. The relationship for a third proportional is \( a : b :: b : c \), which can be written as a ratio \[\frac{a}{b} = \frac{b}{c}\]. We then solve for 'c'.
Correct Option: D) \( 72y^{2} \)
The problem presents a pattern in word triads. We need to identify the logical rule governing the transformation from the first word to the second, and from the second to the third, within each given triad. Then, we apply this same logic to the options to find the one that follows the established pattern.
D) COPE - ECPO - CEOP
Both transformations follow the established logic.
This is a logical reasoning problem involving the arrangement of people on different floors of a building. We need to deduce the exact floor on which each person lives based on the given clues. We will use a step-by-step elimination and placement method to determine the occupant of floor number 4.
Identify the floors: There are 6 floors, numbered 1 (lowermost) to 6 (topmost).
| Floor No. | Person |
|---|---|
| 6 | |
| 5 | |
| 4 | |
| 3 | |
| 2 | |
| 1 |
Clue 1: Only three people live between Y and Z, where Y lives below Z.
This means there are 3 floors between Y and Z. Possible arrangements:
Clue 2: P lives on an odd-numbered floor and immediately above Y.
Let's combine this with Clue 1.
So, we have:
| Floor No. | Person |
|---|---|
| 6 | Z |
| 5 | |
| 4 | |
| 3 | P |
| 2 | Y |
| 1 |
Clue 3: O lives on one of the floors below M and on one of the floors above N.
This implies the order M > O > N (M is above O, O is above N). The remaining floors are 1, 4, and 5. The people remaining are M, O, N.
Final Arrangement:
| Floor No. | Person |
|---|---|
| 6 | Z |
| 5 | M |
| 4 | O |
| 3 | P |
| 2 | Y |
| 1 | N |
Answer the question: Who lives on floor numbered 4? From the final arrangement, O lives on floor 4.
D) O lives on floor number 4, as determined by the logical deduction from the given clues.
This problem involves tracking Mr. O's movements using cardinal directions and distances. We need to find his final position relative to his starting point (Point L) and then determine the shortest distance and direction required to return to Point L.
Now, let's calculate the net displacement from Point L to Point M:
So, Point M is 7 km to the east of Point L. To reach Point L again from Point M, Mr. O must drive 7 km towards the west.
C) 7 km towards the west — As calculated, Mr. O ends up 7 km to the east of his starting point L. Therefore, to return to L, he must travel 7 km towards the west.
We are given a quadrilateral PQRS where the angle bisectors of \(\angle R\) and \(\angle S\) meet at point T. We know \(\angle STR = 32.5^\circ\) and the ratio of \(\angle P\) to \(\angle Q\) is 4:9. We need to find the difference between \(\angle Q\) and \(\angle P\).
We will use the property that the sum of angles in a triangle is \(180^\circ\) and the sum of angles in a quadrilateral is \(360^\circ\). We will also use the formula for the angle formed by the bisectors of two adjacent angles in a quadrilateral.
In \(\triangle STR\), we know that the sum of angles is \(180^\circ\):
\[ \angle STR + \angle TSR + \angle TRS = 180^\circ \]
Given \(\angle STR = 32.5^\circ\), we have:
\[ 32.5^\circ + \angle TSR + \angle TRS = 180^\circ \]
\[ \angle TSR + \angle TRS = 180^\circ - 32.5^\circ = 147.5^\circ \]
Since ST is the bisector of \(\angle S\) and RT is the bisector of \(\angle R\):
\[ \angle TSR = \frac{1}{2} \angle S \]
\[ \angle TRS = \frac{1}{2} \angle R \]
Substituting these into the equation from Step 1:
\[ \frac{1}{2} \angle S + \frac{1}{2} \angle R = 147.5^\circ \]
\[ \frac{1}{2} (\angle S + \angle R) = 147.5^\circ \]
\[ \angle S + \angle R = 2 \times 147.5^\circ = 295^\circ \]
The sum of angles in a quadrilateral PQRS is \(360^\circ\):
\[ \angle P + \angle Q + \angle R + \angle S = 360^\circ \]
Substitute the value of \((\angle S + \angle R)\) from Step 2:
\[ \angle P + \angle Q + 295^\circ = 360^\circ \]
\[ \angle P + \angle Q = 360^\circ - 295^\circ = 65^\circ \]
We are given that the ratio of \(\angle P\) to \(\angle Q\) is 4:9. Let \(\angle P = 4x\) and \(\angle Q = 9x\).
Substitute these into the equation from Step 3:
\[ 4x + 9x = 65^\circ \]
\[ 13x = 65^\circ \]
\[ x = \frac{65^\circ}{13} = 5^\circ \]
Now, calculate the measures of \(\angle P\) and \(\angle Q\):
\[ \angle P = 4x = 4 \times 5^\circ = 20^\circ \]
\[ \angle Q = 9x = 9 \times 5^\circ = 45^\circ \]
Finally, find the difference between \(\angle Q\) and \(\angle P\):
\[ \angle Q - \angle P = 45^\circ - 20^\circ = 25^\circ \]
C) 25°
The question asks about the Central Water Commission's (CWC) collaboration for 'Snowmelt Runoff Forecasting in Himalayan River Basin'. This requires identifying the specific organization that works with CWC on remote sensing and hydrological forecasting, especially for snowmelt in the Himalayas.
B) National Remote Sensing Centre — The Central Water Commission (CWC) has collaborated with the National Remote Sensing Centre (NRSC), ISRO, for snowmelt runoff forecasting in Himalayan river basins. NRSC provides satellite-based snow cover information which is crucial for these forecasts.
We are given two positive numbers. We need to find these numbers using the given conditions: their difference and the result of dividing the greater by the smaller. Once we find the greater number, we will calculate the HCF of this greater number and 4225.
Let the two positive numbers be \(x\) and \(y\), where \(x > y\).
From the first condition, the numbers differ by 2134:
\[ x - y = 2134 \quad \text{(Equation 1)} \]From the second condition, when the greater number (\(x\)) is divided by the smaller number (\(y\)), the quotient is 2 and the remainder is 147. Using the division algorithm (Dividend = Divisor × Quotient + Remainder):
\[ x = 2y + 147 \quad \text{(Equation 2)} \]Substitute Equation 2 into Equation 1:
\[ (2y + 147) - y = 2134 \]\[ y + 147 = 2134 \]\[ y = 2134 - 147 \]\[ y = 1987 \]Now, find \(x\) using Equation 1:
\[ x = 2134 + y \]\[ x = 2134 + 1987 \]\[ x = 4121 \]The greater number is 4121. We need to find the HCF of 4121 and 4225.
We can use the Euclidean algorithm to find the HCF:
Divide 4225 by 4121:
\[ 4225 = 4121 \times 1 + 104 \]Divide 4121 by 104:
\[ 4121 = 104 \times 39 + 65 \]Divide 104 by 65:
\[ 104 = 65 \times 1 + 39 \]Divide 65 by 39:
\[ 65 = 39 \times 1 + 26 \]Divide 39 by 26:
\[ 39 = 26 \times 1 + 13 \]Divide 26 by 13:
\[ 26 = 13 \times 2 + 0 \]Since the remainder is 0, the HCF is the last non-zero remainder, which is 13.
Correct Option: D) 13
If \( \sin ^{4}\theta - \cos ^{4}\theta = \frac{1}{2} \), find the value of \( 2\sin ^{2}\theta - 1 \).
1
\( \frac{1}{2} \)
\(\frac{\sqrt{3}}{2}\)
\(\frac{1}{\sqrt{2}}\)
The problem requires us to find the value of an expression involving \(\sin^2\theta\) given an equation involving \(\sin^4\theta\) and \(\cos^4\theta\). We will use algebraic identities and fundamental trigonometric identities to simplify the given equation and then find the required value.
Start with the given equation:
\[ \sin ^{4}\theta - \cos ^{4}\theta = \frac{1}{2} \]Recognize that the left side is a difference of squares, \((a^2 - b^2) = (a-b)(a+b)\), where \(a = \sin^2\theta\) and \(b = \cos^2\theta\). So, we can factor it:
\[ (\sin ^{2}\theta - \cos ^{2}\theta)(\sin ^{2}\theta + \cos ^{2}\theta) = \frac{1}{2} \]Apply the fundamental trigonometric identity: \(\sin ^{2}\theta + \cos ^{2}\theta = 1\). Substitute this into the equation:
\[ (\sin ^{2}\theta - \cos ^{2}\theta)(1) = \frac{1}{2} \]\[ \sin ^{2}\theta - \cos ^{2}\theta = \frac{1}{2} \]The expression we need to find is \(2\sin ^{2}\theta - 1\). We can use the identity \(\cos ^{2}\theta = 1 - \sin ^{2}\theta\) to express the equation solely in terms of \(\sin ^{2}\theta\):
\[ \sin ^{2}\theta - (1 - \sin ^{2}\theta) = \frac{1}{2} \]Simplify the equation:
\[ \sin ^{2}\theta - 1 + \sin ^{2}\theta = \frac{1}{2} \]\[ 2\sin ^{2}\theta - 1 = \frac{1}{2} \]This is exactly the expression we needed to find.
B) The value of \(2\sin ^{2}\theta - 1\) is \(\frac{1}{2}\), as derived from the given equation using trigonometric identities.