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 within each pair and then identify the one that deviates from this rule.
Option A: IE – NG
Option B: QM – VO
Option C: GC – LF
Option D: DZ – IB
Let's re-examine the pattern based on the differences found:
C) GC – LF. This pair does not follow the pattern of (+5, +2) for the letter positions. Instead, it follows (+5, +3).
This problem involves substituting given symbols with mathematical operators and then evaluating the expression according to the order of operations (BODMAS/PEMDAS).
Correct Option: A) 41
To solve this problem, we first need to determine the cost price (CP) of the orange. We are given the selling price (SP) and the percentage loss. Once we have the cost price, we can calculate the new selling price required to achieve the desired profit percentage.
Calculate the Cost Price (CP):
The vendor sells oranges at ₹44 each and incurs a loss of 78%. This means the selling price (₹44) represents \((100 - 78)\%\) of the cost price.
\[ \text{SP} = \text{CP} \times \left(1 - \frac{\text{Loss percentage}}{100}\right) \] \[ 44 = \text{CP} \times \left(1 - \frac{78}{100}\right) \] \[ 44 = \text{CP} \times \left(\frac{100 - 78}{100}\right) \] \[ 44 = \text{CP} \times \frac{22}{100} \]Now, solve for CP:
\[ \text{CP} = \frac{44 \times 100}{22} \] \[ \text{CP} = 2 \times 100 \] \[ \text{CP} = \text{₹}200 \]Calculate the New Selling Price for 78% Profit:
Now that we know the cost price is ₹200, we want to find the selling price (New SP) that yields a 78% profit.
\[ \text{New SP} = \text{CP} \times \left(1 + \frac{\text{Profit percentage}}{100}\right) \] \[ \text{New SP} = 200 \times \left(1 + \frac{78}{100}\right) \] \[ \text{New SP} = 200 \times \left(\frac{100 + 78}{100}\right) \] \[ \text{New SP} = 200 \times \frac{178}{100} \] \[ \text{New SP} = 2 \times 178 \] \[ \text{New SP} = \text{₹}356 \]Therefore, the vendor should sell each orange at ₹356 to make a profit of 78%.
Correct Option: A) 356
This question tests knowledge of recent corporate partnerships, specifically in the context of sports sponsorships. To answer correctly, one needs to recall which bank entered into an exclusive banking partnership with Sunrisers Hyderabad (SRH) around March 2025.
Correct Option: C) City Union Bank
We need to find the annual rate of interest when the interest is compounded half-yearly. We are given the principal amount, the final amount, and the time period. We will use the compound interest formula, adjusting for half-yearly compounding, to solve for the rate.
Since the interest is compounded half-yearly, the number of compounding periods in 1 year is 2. Therefore:
The formula for compound interest is:
\[ A = P \left(1 + \frac{r}{100}\right)^n \]We need to find the cube root of \(\frac{19683}{15625}\).
We know that \(15625 = 25^3\).
To find the cube root of 19683, we can test values. Since \(20^3 = 8000\) and \(30^3 = 27000\), the cube root is between 20 and 30. The last digit is 3, so the cube root must end in 7 (since \(7^3 = 343\)). So, let's try 27.
\(27^3 = 27 \times 27 \times 27 = 729 \times 27 = 19683\).
So, we have:
\[ \left(\frac{27}{25}\right)^3 = \left(1 + \frac{R}{200}\right)^3 \]Taking the cube root on both sides:
\[ \frac{27}{25} = 1 + \frac{R}{200} \]\[ \frac{27}{25} - 1 = \frac{R}{200} \]\[ \frac{27 - 25}{25} = \frac{R}{200} \]\[ \frac{2}{25} = \frac{R}{200} \]\[ R = \frac{2}{25} \times 200 \]\[ R = 2 \times 8 \]\[ R = 16\% \]C) 16% is the correct annual rate of interest.
This problem involves calculating the time lost due to stoppages for a train. We can determine the time taken to cover the distance without stoppages and the time taken with stoppages. The difference between these two times will give us the total time spent on stoppages.
Given:
The formula for time is \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\).
\[ \text{Time without stoppages} = \frac{D}{S_1} = \frac{630 \text{ km}}{70 \text{ km/hr}} = 9 \text{ hours} \]Given:
Using the same formula:
\[ \text{Time with stoppages} = \frac{D}{S_2} = \frac{630 \text{ km}}{30 \text{ km/hr}} = 21 \text{ hours} \]The total time for stoppages is the difference between the time taken with stoppages and the time taken without stoppages.
\[ \text{Time for stoppages} = \text{Time with stoppages} - \text{Time without stoppages} \]\[ \text{Time for stoppages} = 21 \text{ hours} - 9 \text{ hours} = 12 \text{ hours} \]B) 12 — The total time taken for stoppages on the route is 12 hours, as calculated by finding the difference between the time taken with and without stoppages.
The problem asks us to identify symbols in a given series that are immediately preceded by a number and immediately followed by a number. We need to scan the series from left to right and apply this condition to each symbol.
A) 1 is the correct answer because only one symbol, the second '@', satisfies the condition of being immediately preceded by a number and immediately followed by a number (1 @ 7).
To find the greatest ratio among the given options, we need to convert each ratio into its decimal equivalent or simplify them to their lowest terms and then compare. Converting to decimals is often the most straightforward method for comparison.
Correct Option: B) The ratio \(20 : 36\) simplifies to \(\frac{5}{9}\), which is approximately \(0.5555\). This is the largest decimal value among all the options.
The mean proportional between (21 + 16\(\sqrt{2}\)) and (16 − 7\(\sqrt{2}\)) is:
17\(\sqrt{7}\)
\(\sqrt{112+109\sqrt{2}}\)
\(\sqrt{109+112\sqrt{2}}\)
4\(\sqrt{7}\)
To find the mean proportional between two numbers, say \(a\) and \(b\), we use the formula \(\sqrt{ab}\). We will substitute the given expressions into this formula and simplify.
Correct Option: B) \(\sqrt{112+109\sqrt{2}}\)