To solve this problem, we need to track Harit's movements by breaking down each segment of his journey into horizontal (East-West) and vertical (North-South) components. We will then calculate his final displacement from the starting point Y to the stopping point Z. Finally, we will determine the shortest distance and direction needed to return to Y from Z.
Initial Position: Harit starts at Point Y. Let's assume Y is at coordinates \((0,0)\).
First Movement: Drives 47 km towards the East.
Second Movement: Takes a right turn (South), drives 23 km.
Third Movement: Turns right (West), drives 26 km.
Fourth Movement: Takes a right turn (North), drives 52 km.
Fifth Movement: Takes a left turn (West), drives 21 km.
Sixth Movement: Turns left (South), drives 11 km to stop at Point Z.
Final Position of Z relative to Y: Point Y is at \((0,0)\) and Point Z is at \((0, 18)\).
Distance and Direction from Z to Y:
Correct Option: B) 18 km towards the south
After all the movements, Harit is at Point Z, which is 18 km North of his starting point Y. To return to Y from Z, he needs to travel 18 km in the opposite direction, which is South.
We are given three numbers in a specific ratio and their Least Common Multiple (LCM). We need to find their Highest Common Factor (HCF). The key property to use here is that if numbers are in a ratio \(a:b:c\), they can be represented as \(hx\), \(hy\), \(hz\) where \(h\) is their HCF and \(x, y, z\) are coprime. In this case, the ratio is \(1:11:19\), and these numbers are already coprime. So, the numbers can be written as \(h \times 1\), \(h \times 11\), \(h \times 19\).
Let the three numbers be \(N_1\), \(N_2\), and \(N_3\).
Given that their ratio is \(1:11:19\).
Let their HCF be \(H\).
Then the numbers can be expressed as:
\[ N_1 = H \times 1 \] \[ N_2 = H \times 11 \] \[ N_3 = H \times 19 \]Here, \(1\), \(11\), and \(19\) are prime numbers and thus are coprime to each other.
The LCM of these three numbers \(N_1, N_2, N_3\) is given by the product of their HCF and the remaining coprime factors.
So, \(\text{LCM}(N_1, N_2, N_3) = H \times 1 \times 11 \times 19\).
This simplifies to \(\text{LCM} = H \times 11 \times 19\).
We are given that the LCM is \(8778\).
So, we can set up the equation:
\[ H \times 11 \times 19 = 8778 \]Calculate the product of \(11\) and \(19\):
\[ 11 \times 19 = 209 \]Substitute this value back into the equation:
\[ H \times 209 = 8778 \]Solve for \(H\):
\[ H = \frac{8778}{209} \]Perform the division:
To simplify the division, we can try dividing \(8778\) by \(11\) first:
\[ \frac{8778}{11} = 798 \]Now, divide \(798\) by \(19\):
\[ \frac{798}{19} = 42 \]So, \(H = 42\).
A) The HCF of the three numbers is \(42\).
The mode of a set of observations is the value that appears most frequently in the data set. To find the mode, we need to count the occurrences of each distinct observation and identify the one with the highest frequency.
Correct Option: C) 27
The question asks about the nutrient deficiency most directly linked to reduced oxygen-carrying capacity of blood. This requires knowledge of the components of blood responsible for oxygen transport and the nutrients essential for their formation.
D) Iron β Iron is an essential component of hemoglobin, the protein in red blood cells responsible for oxygen transport. A deficiency in iron directly impairs the production of hemoglobin, leading to reduced oxygen-carrying capacity of the blood, a condition known as anemia.
The problem asks us to find the frequency of a sound wave given its period. Frequency and period are inversely related to each other. We will use the fundamental formula connecting these two quantities to solve the problem.
Correct Option: D) 100 Hz is the correct frequency for a sound wave with a period of 0.01 seconds.
The problem involves similar triangles and their areas. When a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally, creating a smaller triangle similar to the original one. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. We will use this property to find the ratio of the area of \(\Delta BDE\) to the area of \(\Delta ABC\), and then use it to find the ratio of the area of \(\Delta BDE\) to the trapezium ADEC.
Identify similar triangles:
Given that in \(\Delta ABC\), \(DE \parallel AC\). This implies that \(\Delta BDE\) is similar to \(\Delta BAC\) (by AA similarity criterion, as \(\angle BDE = \angle BAC\) and \(\angle BED = \angle BCA\) (corresponding angles), and \(\angle B\) is common).
Determine the ratio of corresponding sides:
We are given \(BD = 10 \text{ cm}\) and \(AD = 17 \text{ cm}\).
The side \(AB\) of \(\Delta BAC\) is \(AB = AD + DB = 17 + 10 = 27 \text{ cm}\).
The ratio of corresponding sides of \(\Delta BDE\) and \(\Delta BAC\) is:
\[ \frac{BD}{BA} = \frac{10}{27} \]Calculate the ratio of areas of similar triangles:
For similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides.
\[ \frac{\text{Area}(\Delta BDE)}{\text{Area}(\Delta BAC)} = \left(\frac{BD}{BA}\right)^2 = \left(\frac{10}{27}\right)^2 = \frac{100}{729} \]Express the area of trapezium ADEC:
The area of trapezium ADEC can be found by subtracting the area of \(\Delta BDE\) from the area of \(\Delta BAC\).
\[ \text{Area}(\text{trapezium ADEC}) = \text{Area}(\Delta BAC) - \text{Area}(\Delta BDE) \]Find the required ratio:
Let \(\text{Area}(\Delta BDE) = 100k\) and \(\text{Area}(\Delta BAC) = 729k\) for some constant \(k\).
Then, \(\text{Area}(\text{trapezium ADEC}) = 729k - 100k = 629k\).
The ratio of the area of \(\Delta BDE\) to the trapezium ADEC is:
\[ \frac{\text{Area}(\Delta BDE)}{\text{Area}(\text{trapezium ADEC})} = \frac{100k}{629k} = \frac{100}{629} \]D) 100 : 629
The problem involves simple interest. We are given the amount after 3 years, the rate of interest, and the time. We need to find the principal sum first, and then calculate the simple interest for 2 years on that sum at the same rate.
Let the principal sum be \(P\). The rate of interest \(R = 20\%\) per annum. The time \(T_1 = 3\) years. The amount after 3 years \(A_1 = \text{βΉ}8,640\).
The formula for simple interest is \(SI = \frac{P \times R \times T}{100}\).
The amount \(A\) is given by \(A = P + SI\). Substituting the formula for \(SI\), we get:
\[ A = P + \frac{P \times R \times T}{100} = P \left(1 + \frac{R \times T}{100}\right) \]Using the given values for the first scenario:
\[ 8640 = P \left(1 + \frac{20 \times 3}{100}\right) \] \[ 8640 = P \left(1 + \frac{60}{100}\right) \] \[ 8640 = P \left(1 + 0.6\right) \] \[ 8640 = P \times 1.6 \]Now, solve for \(P\):
\[ P = \frac{8640}{1.6} \] \[ P = \frac{86400}{16} \] \[ P = 5400 \]So, the principal sum is \(\text{βΉ}5,400\).
Next, we need to find the simple interest on this sum for \(T_2 = 2\) years at the same rate \(R = 20\%\).
\[ SI_2 = \frac{P \times R \times T_2}{100} \] \[ SI_2 = \frac{5400 \times 20 \times 2}{100} \]Calculate the simple interest:
\[ SI_2 = 54 \times 20 \times 2 \] \[ SI_2 = 54 \times 40 \] \[ SI_2 = 2160 \]The simple interest is \(\text{βΉ}2,160\).
Correct Option: B) 2,160
The question asks about the constitutional requirement for the President to consult specific authorities when appointing judges to the Supreme Court. This falls under the Indian Constitution's provisions regarding the judiciary.
Correct Option: B) The Chief Justice of India
The question asks to identify the Chairperson of the National Commission for Women (NCW) appointed in October 2024. This is a current affairs question requiring knowledge of recent appointments to significant government bodies.
As per recent announcements, Vijaya Kishore Rahatkar was appointed as the Chairperson of the National Commission for Women in October 2024.
B) Vijaya Kishore Rahatkar was appointed as the Chairperson of the National Commission for Women (NCW) in October 2024. This is a direct factual recall from current events.