To find the missing number in the series, we need to identify the pattern or rule governing the sequence of numbers. We will look at the differences between consecutive terms.
Thus, the missing number in the series is 35.
C) 35 — The pattern of differences between consecutive terms is \(-10, -9, -8, -7\). Following this pattern, the next difference should be \(-6\). Subtracting 6 from 41 gives 35.
The problem involves calculating the value of an unknown discount percentage given the marked price, the selling price, and one of the two successive discounts. We will apply the formula for successive discounts to find the price after the first discount and then use that to determine the second discount percentage.
The first discount is 10% on the Marked Price.
\[ \text{Price after } D_1 = \text{MP} - (\text{MP} \times \frac{D_1}{100}) \] \[ \text{Price after } D_1 = 18000 - (18000 \times \frac{10}{100}) \] \[ \text{Price after } D_1 = 18000 - 1800 \] \[ \text{Price after } D_1 = \text{₹}16,200 \]The second discount is applied to the price after the first discount to arrive at the Selling Price.
\[ \text{Second Discount Amount} = \text{Price after } D_1 - \text{SP} \] \[ \text{Second Discount Amount} = 16200 - 12960 \] \[ \text{Second Discount Amount} = \text{₹}3,240 \]The second discount percentage (x%) is calculated on the price after the first discount (₹16,200).
\[ x\% = \frac{\text{Second Discount Amount}}{\text{Price after } D_1} \times 100 \] \[ x = \frac{3240}{16200} \times 100 \] \[ x = \frac{324}{1620} \times 100 \] \[ x = \frac{1}{5} \times 100 \] \[ x = 20 \]Thus, the value of x is 20.
A) 20 is the correct value for the second discount percentage.
To solve this problem, we need to perform algebraic operations involving addition and subtraction of polynomials. The key steps are to first find the sum of the two given expressions and then subtract this sum from the third expression. We will combine like terms at each step.
The first two expressions are \(2x - 3y + 7z\) and \(4z - 5x\). Let's add them:
\[ (2x - 3y + 7z) + (4z - 5x) \]\[ = 2x - 5x - 3y + 7z + 4z \]\[ = (2 - 5)x - 3y + (7 + 4)z \]\[ = -3x - 3y + 11z \]The third expression is \(13x - z\). We need to subtract the sum \(-3x - 3y + 11z\) from \(13x - z\).
\[ (13x - z) - (-3x - 3y + 11z) \]Distribute the negative sign to each term inside the parenthesis:
\[ = 13x - z + 3x + 3y - 11z \]Group the terms with \(x\), \(y\), and \(z\) together:
\[ = (13x + 3x) + 3y + (-z - 11z) \]\[ = (13 + 3)x + 3y + (-1 - 11)z \]\[ = 16x + 3y - 12z \]D) \(16x + 3y - 12z\) is the correct result after performing the specified algebraic operations.
To find the combined ratio \(a : b : c\), we need to make the common term, which is 'b' in this case, equal in both given ratios. We will find the least common multiple (LCM) of the 'b' values from the individual ratios and then adjust the ratios accordingly.
Write down the given ratios:
\[ a : b = 3 : 4 \]\[ b : c = 2 : 3 \]Identify the common term: The common term in both ratios is \(b\). In the first ratio, \(b\) is 4. In the second ratio, \(b\) is 2.
Find the LCM of the values of the common term: The values of \(b\) are 4 and 2. The LCM of 4 and 2 is 4.
Adjust the ratios so that the common term has the LCM value:
The first ratio \(a : b = 3 : 4\) already has \(b = 4\). So, no change is needed for this ratio.
The second ratio \(b : c = 2 : 3\) needs to be adjusted so that \(b\) becomes 4. To do this, multiply both parts of the ratio by \(\frac{4}{2} = 2\):
\[ b : c = (2 \times 2) : (3 \times 2) = 4 : 6 \]Combine the adjusted ratios: Now we have:
\[ a : b = 3 : 4 \]\[ b : c = 4 : 6 \]Since the value of \(b\) is the same (4) in both ratios, we can combine them directly to get \(a : b : c\).\[ a : b : c = 3 : 4 : 6 \]C) 3 : 4 : 6 is the correct combined ratio.
The question asks for the specific dates during which the Department for Promotion of Industry and Internal Trade (DPIIT) organized the Startup India Innovation Week 2024. This is a factual question requiring knowledge of recent government initiatives and events.
D) 10 January 2024 to 18 January 2024. The Department for Promotion of Industry and Internal Trade (DPIIT) organized the Startup India Innovation Week 2024 from January 10th to January 18th, 2024. This event aims to foster the Indian startup ecosystem and celebrate National Startup Day on January 16th.
The problem asks us to identify the letter-cluster pair that does not follow the same pattern as the others, based on the English alphabetical order. We need to find a consistent rule or relationship between the letters in each pair and then identify the option that deviates from this rule.
B) ZT – PG is the pair that does not belong to the group. In this pair, Z (26) to P (16) is a shift of -10 (or +16 cyclically), and T (20) to G (7) is a shift of -13 (or +13 cyclically). This is different from the +5 shift observed in the other options.
This problem involves setting up and solving a system of linear equations based on the given age relationships. We will define variables for the present ages of Anita and Sunita and then translate the given conditions into algebraic equations.
Define variables: Let Anita's present age be \(A\) years. Let Sunita's present age be \(S\) years.
Formulate the first equation: "The ages of Anita and Sunita differ by 30 years." This means the absolute difference between their ages is 30. Since Anita was 3 times as old as Sunita 5 years ago, Anita must be older than Sunita. So, we can write:
\[ A - S = 30 \quad \text{(Equation 1)} \]Formulate the second equation: "5 years ago, Anita was 3 times as old as Sunita."
According to the problem:
\[ A - 5 = 3(S - 5) \quad \text{(Equation 2)} \]Solve Equation 2:
\[ A - 5 = 3S - 15 \] \[ A = 3S - 15 + 5 \] \[ A = 3S - 10 \quad \text{(Equation 3)} \]Substitute Equation 3 into Equation 1: Now we have an expression for \(A\) in terms of \(S\). Substitute this into Equation 1:
\[ (3S - 10) - S = 30 \]Solve for \(S\):
\[ 2S - 10 = 30 \] \[ 2S = 30 + 10 \] \[ 2S = 40 \] \[ S = \frac{40}{2} \] \[ S = 20 \]So, Sunita's present age is 20 years.
(Optional) Find Anita's present age: Using Equation 1: \(A - 20 = 30 \implies A = 50\) years. Check with Equation 3: \(A = 3(20) - 10 = 60 - 10 = 50\) years. The ages are consistent.
Correct Option: C) 20 years
The question asks about the number of villages to be covered under the Pradhan Mantri Janjatiya Unnat Gram Abhiyan, as stated in the Union Budget 2024-25. This requires knowledge of recent government schemes and budget announcements.
Correct Option: B) 63,000
This problem involves decoding a blood relation expression using given symbols and then determining the relationship between two individuals. We will break down the given expression step-by-step, interpreting each symbol to build a family tree or identify relationships, and then trace the path from M to T.
Correct Option: D) Wife's father's mother's brother
The resistance of a conductor depends on its material, length, and cross-sectional area. This question asks how resistance changes when the length is tripled, keeping other factors (material and cross-sectional area) constant. We will use the formula for resistance to determine the relationship.
The formula for the resistance (\(R\)) of a conductor is given by:
\[ R = \rho \frac{L}{A} \]where:
From the formula, we can see that resistance (\(R\)) is directly proportional to the length (\(L\)) of the conductor, assuming resistivity (\(\rho\)) and cross-sectional area (\(A\)) remain constant. This can be written as:
\[ R \propto L \]Let the initial resistance be \(R_1\) and the initial length be \(L_1\). So, \(R_1 = \rho \frac{L_1}{A}\).
When the length of the conductor is tripled, the new length \(L_2\) will be \(3L_1\). The problem states that "other factors" (resistivity \(\rho\) and cross-sectional area \(A\)) are kept the same.
Now, calculate the new resistance \(R_2\) using the new length \(L_2\):
\[ R_2 = \rho \frac{L_2}{A} \]\[ R_2 = \rho \frac{3L_1}{A} \]\[ R_2 = 3 \left( \rho \frac{L_1}{A} \right) \]Substitute \(R_1 = \rho \frac{L_1}{A}\) into the equation:
\[ R_2 = 3 R_1 \]This shows that the new resistance \(R_2\) is three times the original resistance \(R_1\).
B) tripled — As derived, when the length of the conductor is tripled, its resistance also triples, assuming resistivity and cross-sectional area remain constant.