UPSC CISF-AC-EXE Archives

41. Consider the following statements and conclusions: Statement I : Som

Consider the following statements and conclusions:

Statement I : Some P are Q.
Statement II : Every R is Q.

Conclusion I : Some P are R.
Conclusion II : Every R is P.
Which of the above conclusions can be certainly drawn?

Conclusion I only

Conclusion II only

Both conclusions I and II

Neither conclusion I nor II

This question was previously asked in

UPSC CISF-AC-EXE – 2024

Download PDF Attempt Online

Based on the given statements, neither Conclusion I nor Conclusion II can be certainly drawn.

Statements:
I: Some P are Q.
II: Every R is Q.
Conclusion I: Some P are R.
Conclusion II: Every R is P.
Statement II (Every R is Q) means that the set R is a subset of the set Q. Statement I (Some P are Q) means that the set P has at least one element in common with the set Q.

Let’s analyze the conclusions:
Conclusion I: Some P are R. This means the intersection of sets P and R is not empty.
Conclusion II: Every R is P. This means the set R is a subset of the set P.

Consider a scenario where the set P overlaps with Q, but the overlap region is entirely outside R. In this case, “Some P are Q” and “Every R is Q” hold true, but “Some P are R” is false, and “Every R is P” is false.
Since there exists a valid interpretation of the statements where both conclusions are false, neither conclusion can be certainly drawn from the given statements.

42. $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ is equal to

$\frac{4}{1+\sqrt{2}+\sqrt{3}}$ is equal to

$sqrt{2}-sqrt{3}+2$

$sqrt{2}+sqrt{3}+2$

$sqrt{2}+sqrt{6}+2$

$sqrt{2}-sqrt{6}+2$

This question was previously asked in

UPSC CISF-AC-EXE – 2024

Download PDF Attempt Online

The correct answer is $\sqrt{2}-\sqrt{6}+2$.

We need to simplify the expression $\frac{4}{1+\sqrt{2}+\sqrt{3}}$. This involves rationalizing the denominator, which contains multiple surds.
We can group the terms in the denominator, for instance, as $(1+\sqrt{2}) + \sqrt{3}$. We multiply the numerator and denominator by the conjugate of this expression, which is $(1+\sqrt{2}) – \sqrt{3}$.
Expression = $\frac{4}{(1+\sqrt{2})+\sqrt{3}} \times \frac{(1+\sqrt{2})-\sqrt{3}}{(1+\sqrt{2})-\sqrt{3}}$
Numerator = $4(1+\sqrt{2}-\sqrt{3})$.
Denominator = $((1+\sqrt{2})+\sqrt{3})((1+\sqrt{2})-\sqrt{3})$. This is in the form $(a+b)(a-b) = a^2 – b^2$, where $a = (1+\sqrt{2})$ and $b = \sqrt{3}$.
Denominator = $(1+\sqrt{2})^2 – (\sqrt{3})^2$.
$(1+\sqrt{2})^2 = 1^2 + 2(1)(\sqrt{2}) + (\sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}$.
$(\sqrt{3})^2 = 3$.
Denominator = $(3 + 2\sqrt{2}) – 3 = 2\sqrt{2}$.
So the expression becomes $\frac{4(1+\sqrt{2}-\sqrt{3})}{2\sqrt{2}}$.
We can simplify by dividing 4 by 2: $\frac{2(1+\sqrt{2}-\sqrt{3})}{\sqrt{2}}$.
Now, we need to rationalize the denominator $\sqrt{2}$ by multiplying the numerator and denominator by $\frac{\sqrt{2}}{\sqrt{2}}$.
Expression = $\frac{2(1+\sqrt{2}-\sqrt{3})}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
Expression = $\frac{2\sqrt{2}(1+\sqrt{2}-\sqrt{3})}{2}$.
Cancel the 2 in the numerator and denominator:
Expression = $\sqrt{2}(1+\sqrt{2}-\sqrt{3})$.
Now, distribute $\sqrt{2}$:
Expression = $\sqrt{2} \times 1 + \sqrt{2} \times \sqrt{2} – \sqrt{2} \times \sqrt{3}$
Expression = $\sqrt{2} + 2 – \sqrt{6}$.
Rearranging the terms to match the options, we get $2 + \sqrt{2} – \sqrt{6}$, which is the same as $\sqrt{2}-\sqrt{6}+2$.

Rationalizing a denominator with three terms involving square roots often requires multiplying by conjugates twice. In this case, grouping $(1+\sqrt{2})$ as a single term ‘a’ allowed us to use the difference of squares formula $(a+b)(a-b)=a^2-b^2$ once to eliminate one square root term from the denominator. The resulting denominator ($2\sqrt{2}$) still contained a square root, necessitating a second multiplication by $\sqrt{2}/\sqrt{2}$ to complete the rationalization. The goal is to make the denominator a rational number.

43. Which one among the following digits can never be at the units place i

Which one among the following digits can never be at the units place in $(273)^n$, where $n$ is a positive integer?

This question was previously asked in

UPSC CISF-AC-EXE – 2024

Download PDF Attempt Online

The correct answer is 5.

The units digit of a number raised to a power is determined solely by the units digit of the base number and the power. In this case, the base number is 273, and its units digit is 3. So, the units digit of $(273)^n$ is the same as the units digit of $3^n$, where n is a positive integer.
Let’s examine the pattern of the units digits of powers of 3:
$3^1$: Units digit is 3.
$3^2$: Units digit of $3 \times 3 = 9$.
$3^3$: Units digit of $9 \times 3 = 27$, which is 7.
$3^4$: Units digit of $7 \times 3 = 21$, which is 1.
$3^5$: Units digit of $1 \times 3 = 3$. (The pattern repeats)
$3^6$: Units digit of $3 \times 3 = 9$.
The cycle of the units digits of powers of 3 is 3, 9, 7, 1. This cycle has a length of 4.
For any positive integer n, the units digit of $3^n$ will be one of these four digits: 1, 3, 7, or 9.
Let’s check the given options:
A) 1: Possible (e.g., for $n=4$)
B) 3: Possible (e.g., for $n=1$)
C) 5: Not in the cycle {3, 9, 7, 1}
D) 7: Possible (e.g., for $n=3$)
Therefore, the digit 5 can never be at the units place in $(273)^n$ for any positive integer n.

The units digits of powers of any single digit follow a repeating pattern (a cycle). For example, powers of 2 have units digits 2, 4, 8, 6, 2, 4, 8, 6, … (cycle length 4). Powers of 7 have units digits 7, 9, 3, 1, 7, 9, 3, 1, … (cycle length 4). Powers of 0, 1, 5, 6 have a cycle length of 1 (always 0, 1, 5, 6 respectively). Powers of 4 and 9 have a cycle length of 2 (4, 6, 4, 6, … and 9, 1, 9, 1, …). The units digit of a number like 273 raised to a power is determined only by the units digit of 273, which is 3.

44. A man walks 7 km in the North direction. He then walks 4 km Eastward,

A man walks 7 km in the North direction. He then walks 4 km Eastward, and then 4 km Southward. How far and in which direction is he now from his original position?

5 km in South-East direction

5 km in North-East direction

15 km towards East direction

7 km towards North direction

This question was previously asked in

UPSC CISF-AC-EXE – 2024

Download PDF Attempt Online

The correct answer is 5 km in North-East direction.

We can solve this problem by tracking the man’s position relative to his starting point. Let’s assume the starting point is the origin (0,0) of a coordinate system, with North along the positive y-axis and East along the positive x-axis.
1. He walks 7 km in the North direction. His position is now (0, 7).
2. He then walks 4 km Eastward. His position changes by +4 in the x-direction. His new position is (0 + 4, 7) = (4, 7).
3. He then walks 4 km Southward. His position changes by -4 in the y-direction. His new position is (4, 7 – 4) = (4, 3).
His final position is (4, 3). His original position was (0, 0).
The displacement from the original position is a vector from (0,0) to (4,3).
The distance from the original position is the magnitude of this displacement vector, which can be calculated using the distance formula (or Pythagorean theorem):
Distance = $\sqrt{(\text{final x} – \text{initial x})^2 + (\text{final y} – \text{initial y})^2}$
Distance = $\sqrt{(4 – 0)^2 + (3 – 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ km.
The final position (4, 3) is in the positive x and positive y quadrant relative to the origin. In terms of directions, positive x is East and positive y is North. Therefore, the direction is North-East.

Visualizing the path can be helpful. The movements are 7 km Up, then 4 km Right, then 4 km Down. The net movement is 4 km Right (East) and 7 – 4 = 3 km Up (North) from the starting point. These two movements form the legs of a right-angled triangle. The hypotenuse of this triangle is the direct distance from the start to the end point. The lengths of the legs are 4 km and 3 km. The hypotenuse is $\sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$ km. The final position is East of the start and North of the start, hence North-East.

45. If $x : y = 3 : 7$, then $\frac{6x^2 – 2xy + y^2}{xy}$ is equal to

If $x : y = 3 : 7$, then $\frac{6x^2 – 2xy + y^2}{xy}$ is equal to

$rac{62}{21}$

$rac{61}{21}$

$rac{63}{21}$

$rac{64}{21}$

This question was previously asked in

UPSC CISF-AC-EXE – 2024

Download PDF Attempt Online

The correct answer is $\frac{61}{21}$.

Given the ratio $x : y = 3 : 7$. This means that $\frac{x}{y} = \frac{3}{7}$.
We can express x and y in terms of a common variable. Let $x = 3k$ and $y = 7k$ for some non-zero constant k.
We need to evaluate the expression $\frac{6x^2 – 2xy + y^2}{xy}$.
Substitute $x = 3k$ and $y = 7k$ into the expression:
Numerator:
$6x^2 – 2xy + y^2 = 6(3k)^2 – 2(3k)(7k) + (7k)^2$
$= 6(9k^2) – 2(21k^2) + 49k^2$
$= 54k^2 – 42k^2 + 49k^2$
Combine the terms with $k^2$:
$= (54 – 42 + 49)k^2$
$= (12 + 49)k^2$
$= 61k^2$.
Denominator:
$xy = (3k)(7k) = 21k^2$.
Now, form the ratio:
$\frac{6x^2 – 2xy + y^2}{xy} = \frac{61k^2}{21k^2}$.
Since k is a non-zero constant, $k^2$ is also non-zero, so we can cancel $k^2$ from the numerator and the denominator.
The expression simplifies to $\frac{61}{21}$.

Alternatively, we can divide the numerator and the denominator of the expression by $y^2$ (assuming y is not zero, which it isn’t since the ratio is 3:7):
$\frac{6x^2 – 2xy + y^2}{xy} = \frac{\frac{6x^2}{y^2} – \frac{2xy}{y^2} + \frac{y^2}{y^2}}{\frac{xy}{y^2}} = \frac{6(\frac{x}{y})^2 – 2(\frac{x}{y}) + 1}{(\frac{x}{y})}$.
Substitute the value of $\frac{x}{y} = \frac{3}{7}$:
$= \frac{6(\frac{3}{7})^2 – 2(\frac{3}{7}) + 1}{(\frac{3}{7})} = \frac{6(\frac{9}{49}) – \frac{6}{7} + 1}{\frac{3}{7}}$
$= \frac{\frac{54}{49} – \frac{42}{49} + \frac{49}{49}}{\frac{3}{7}} = \frac{\frac{54 – 42 + 49}{49}}{\frac{3}{7}} = \frac{\frac{61}{49}}{\frac{3}{7}}$
$= \frac{61}{49} \times \frac{7}{3} = \frac{61}{7 \times 7} \times \frac{7}{3} = \frac{61}{7 \times 3} = \frac{61}{21}$.
Both methods yield the same result.

46. The average of seven consecutive natural numbers is 13. If the next th

The average of seven consecutive natural numbers is 13. If the next three consecutive natural numbers are also included, the new average will be

15.0

14.5

13.5

13.0

This question was previously asked in

UPSC CISF-AC-EXE – 2024

Download PDF Attempt Online

The correct answer is 14.5.

The average of seven consecutive natural numbers is given as 13.
In an arithmetic progression (which consecutive natural numbers are), the average is equal to the middle term.
For 7 consecutive numbers, the middle term is the 4th number.
So, the 4th number is 13.
Let the numbers be $n, n+1, n+2, n+3, n+4, n+5, n+6$. The 4th number is $n+3$.
$n+3 = 13 \implies n = 10$.
The seven consecutive natural numbers are 10, 11, 12, 13, 14, 15, 16.
The sum of these 7 numbers is $7 \times 13 = 91$.
The next three consecutive natural numbers are 17, 18, 19.
When these three numbers are included, the new set of numbers is 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. There are now 10 numbers.
The sum of the new set of numbers = Sum of original 7 numbers + Sum of next 3 numbers.
Sum of next 3 numbers = 17 + 18 + 19 = 54.
Total sum of 10 numbers = 91 + 54 = 145.
The new average = Total sum / Number of terms = 145 / 10 = 14.5.

For any series of consecutive numbers (or any arithmetic progression), adding new terms that are consecutive to the series will shift the average. If ‘k’ consecutive terms are added to an arithmetic progression with common difference ‘d’, the new average will be the original average plus $(k \times d)/2$. In this case, 3 consecutive natural numbers (d=1) were added to the original 7 numbers. The increase in average is $(3 \times 1)/2 = 1.5$. New average = Original average + Increase = 13 + 1.5 = 14.5. This shortcut works specifically because the numbers are consecutive and added sequentially to the original list.

47. Assume there to be some foodstuff that is sufficient for a team for $x

Assume there to be some foodstuff that is sufficient for a team for $x$ number of days. After 20 days, 25% of team members happen to quit. It is observed that the remaining foodstuff is sufficient to feed the remaining members for another $x$ number of days. What is the value of $x$?

This question was previously asked in

UPSC CISF-AC-EXE – 2024

Download PDF Attempt Online

The correct answer is 80.

This is a variation of a work/provision problem. The total amount of food is proportional to the number of team members multiplied by the number of days it lasts.
Let the initial number of team members be M.
The food is sufficient for M members for x days. Total food “units” can be represented as M * x.
After 20 days, the M members have consumed food for 20 days. The food consumed is M * 20 units.
The remaining food units = (Initial food) – (Food consumed) = M * x – M * 20 = M(x – 20).
After 20 days, 25% of the team members quit.
The remaining number of team members = M – 25% of M = M – 0.25M = 0.75M.
It is observed that the remaining food is sufficient to feed the remaining members (0.75M) for another x number of days.
So, the remaining food units can also be represented as (Remaining members) * (Number of additional days) = (0.75M) * x.
Equating the two expressions for the remaining food:
M(x – 20) = 0.75M * x
Since M represents the number of members, M must be greater than 0 (otherwise there is no team or food). We can divide both sides by M:
x – 20 = 0.75x
Now, solve for x:
x – 0.75x = 20
0.25x = 20
x = 20 / 0.25
x = 20 / (1/4)
x = 20 * 4
x = 80.

This type of problem relies on the principle that the total amount of work (or food) is directly proportional to the number of workers (or members) and the time they work (or are fed). If N workers can do a job in D days, the total work is N*D units. Similarly, if N members have food for D days, the total food is N*D units. When the number of workers or the duration changes, the total work or remaining work can be related using this principle.

48. A train with a length of 335 m is travelling at a speed of 90 km/h. Wh

A train with a length of 335 m is travelling at a speed of 90 km/h. What is the time taken by the train to pass a 290 m long platform?

25 s

23 s

20 s

30 s

This question was previously asked in

UPSC CISF-AC-EXE – 2024

Download PDF Attempt Online

The correct answer is 25 seconds.

To pass a platform, a train must cover a distance equal to its own length plus the length of the platform.
Total distance to be covered = Length of train + Length of platform.
Total distance = 335 m + 290 m = 625 m.
The speed of the train is given in km/h, so we need to convert it to m/s to be consistent with the distance unit (meters) and time unit (seconds).
Speed = 90 km/h.
To convert km/h to m/s, multiply by the factor (1000 m / 3600 s) = 5/18.
Speed in m/s = 90 * (5/18) m/s = (90/18) * 5 m/s = 5 * 5 m/s = 25 m/s.
The time taken is calculated using the formula: Time = Distance / Speed.
Time = 625 m / 25 m/s.
Time = 25 seconds.

When a train crosses a stationary object with negligible length (like a pole or a man), the distance covered is just the length of the train. When it crosses an object with considerable length (like a bridge or a platform), the distance is the sum of the train’s length and the object’s length. The conversion factor 5/18 comes from $1 \text{ km} = 1000 \text{ m}$ and $1 \text{ hour} = 3600 \text{ seconds}$, so $1 \text{ km/h} = \frac{1000}{3600} \text{ m/s} = \frac{5}{18} \text{ m/s}$.

49. The ratio between the sale price and the cost price of an article is 7

The ratio between the sale price and the cost price of an article is 7:4. What is the ratio between the profit and the cost price of that article?

3:4

4:3

5:4

7:3

This question was previously asked in

UPSC CISF-AC-EXE – 2024

Download PDF Attempt Online

The correct answer is 3:4.

The ratio between the sale price (SP) and the cost price (CP) of an article is given as 7:4.
This means that for every 4 units of cost price, the sale price is 7 units.
We can represent this by saying SP = 7k and CP = 4k, where k is a positive constant.
Profit (P) is calculated as the difference between the sale price and the cost price:
Profit (P) = Sale Price (SP) – Cost Price (CP)
P = 7k – 4k = 3k.
The question asks for the ratio between the profit and the cost price.
Ratio = Profit : Cost Price = P : CP
Ratio = 3k : 4k.
Since k is a common factor in both terms of the ratio (and k is not zero), we can cancel k:
Ratio = 3 : 4.

Profit percentage is calculated as $(\text{Profit} / \text{Cost Price}) \times 100\%$. In this case, the profit percentage would be $(3k / 4k) \times 100\% = (3/4) \times 100\% = 75\%$. The ratio of SP to CP being greater than 1 indicates a profit; if it were less than 1, it would indicate a loss.

50. An amount is adequate to pay salaries to R for 20 days or to S for 30

An amount is adequate to pay salaries to R for 20 days or to S for 30 days. For how many days is the money enough to cover salaries of both of them together?

12 days

15 days

11 days

20 days

This question was previously asked in

UPSC CISF-AC-EXE – 2024

Download PDF Attempt Online

The correct answer is 12 days.

This is a work and time problem. We can determine the efficiency (or daily work rate) of R and S based on the number of days they take to complete the work (pay salaries).
Let the total amount of money (representing the total work) be M.
R can be paid for 20 days with M. So, R’s daily salary (work rate) = M/20 per day.
S can be paid for 30 days with M. So, S’s daily salary (work rate) = M/30 per day.
If both R and S are paid together, their combined daily salary is R’s daily salary + S’s daily salary = M/20 + M/30.
To find how many days (let’s call this D) the money M will last for both, we set M equal to the combined daily salary multiplied by D:
M = (M/20 + M/30) * D
Divide both sides by M (assuming M > 0):
1 = (1/20 + 1/30) * D
Find a common denominator for 1/20 and 1/30, which is 60:
1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12.
So, 1 = (1/12) * D
Multiply by 12 to find D:
D = 12 days.

This problem can also be viewed as calculating the time taken by two individuals to complete a task together, given the time each takes individually. If R takes ‘a’ days and S takes ‘b’ days, the time taken together is given by the formula $\frac{1}{\frac{1}{a} + \frac{1}{b}} = \frac{ab}{a+b}$. In this case, a=20 and b=30. Time together = $\frac{20 \times 30}{20 + 30} = \frac{600}{50} = 12$ days.