UPSC CISF-AC-EXE

The smaller pipe can fill the tank in 12 minutes.
Rate of filling by the second pipe = V / 12 units per minute.
Normalized rate of the second pipe = 1/12 tank per minute.

When both pipes are opened simultaneously, their rates of filling are added together.
Combined rate of both pipes = Rate of pipe 1 + Rate of pipe 2
Combined rate = 1/4 + 1/12 tanks per minute.
To add the fractions, find a common denominator, which is 12.
1/4 = 3/12.
Combined rate = 3/12 + 1/12 = 4/12 tanks per minute.
Combined rate = 1/3 tank per minute.

The time taken to fill the tank is the reciprocal of the combined rate (since Rate * Time = Work Done, and Work Done = 1 tank).
Time taken = 1 / (Combined rate)
Time taken = 1 / (1/3) minutes.
Time taken = 3 minutes.

B is 60% more efficient than A.
B’s efficiency = A’s efficiency + 60% of A’s efficiency
B’s efficiency = A’s efficiency * (1 + 60/100) = A’s efficiency * (1 + 0.60) = 1.60 * A’s efficiency.
B’s efficiency = 1.60 * (W / 12) units/day.
B’s efficiency = (1.6 / 12) * W units/day
B’s efficiency = (16 / 120) * W units/day
B’s efficiency = (2 / 15) * W units/day.

Let the number of days taken by B to finish the same work be D days.
Work done by B = B’s efficiency * Number of days
W = (2W / 15) * D

To find D, we can divide both sides by W (assuming W > 0, which is true for work):
1 = (2 / 15) * D
D = 15 / 2
D = 7.5 days.
7.5 days is equal to 7½ days.

We want to find the probability that the sum of the outcomes is 7.
Let (d1, d2) represent the outcome of the first and second die, respectively. The possible pairs (d1, d2) that sum up to 7 are:
(1, 6)
(2, 5)
(3, 4)
(4, 3)
(5, 2)
(6, 1)

There are 6 favorable outcomes.

The probability of an event is calculated as:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes)

Probability of the sum being 7 = 6 / 36.
Simplifying the fraction:
Probability = 1 / 6.

We are given that the third term of the GP is 4.
$T_3 = ar^2 = 4$.

We need to find the product of the first five terms of the GP.
Product P = $T_1 \times T_2 \times T_3 \times T_4 \times T_5$
P = $a \times (ar) \times (ar^2) \times (ar^3) \times (ar^4)$

Let’s group the ‘a’ terms and the ‘r’ terms:
P = $(a \times a \times a \times a \times a) \times (r^0 \times r^1 \times r^2 \times r^3 \times r^4)$
P = $a^5 \times r^{(0+1+2+3+4)}$
P = $a^5 \times r^{10}$

We can rewrite this expression by grouping $(ar^2)$ terms:
P = $a^5 \times r^{10} = (a^5 \times r^{10/5 \times 5}) = (a^5 \times r^{2 \times 5})$
P = $(a \times r^2)^5$
P = $(ar^2)^5$

We know that $ar^2 = 4$.
Substitute the value of $ar^2$ into the expression for P:
P = $(4)^5 = 4^5$.

The second equation `+ # – & = 12` appears to have a leading ‘+’ which is likely a typo. Given the context of the other two equations, the most probable intended form for the second equation is either `# – & = 12` or `* + # – & = 12`.

Let’s analyze the system assuming the second equation is `# – & = 12`:
1) @ + * = 16
2) # – & = 12
3) @ – & = 16

From equation (3), we can express @ in terms of &: @ = 16 + &.
From equation (2), we can express # in terms of &: # = 12 + &.
Substitute @ = 16 + & into equation (1):
(16 + &) + * = 16
16 + & + * = 16
& + * = 0
This implies * = -&.

The relationships between the variables are:
@ = 16 + &
* = -&
# = 12 + &

This system has one degree of freedom, parameterized by &. There are infinitely many solutions unless there are unstated constraints. Let’s check which option leads to a valid solution within this framework:

A) @ = 9:
If @ = 9, then from @ = 16 + &, we get 9 = 16 + & => & = 9 – 16 = -7.
Now find * and # using & = -7:
* = -& = -(-7) = 7.
# = 12 + & = 12 + (-7) = 5.
Let’s check if the values (@=9, *=7, #=5, &=-7) satisfy the original equations:
1) @ + * = 9 + 7 = 16 (Correct)
2) # – & = 5 – (-7) = 5 + 7 = 12 (Correct, assuming this form of eq 2)
3) @ – & = 9 – (-7) = 9 + 7 = 16 (Correct)
So, the solution (@=9, *=7, #=5, &=-7) is valid if the second equation is `# – & = 12`. In this solution, the statement @=9 is true.

Let’s check other options against the relationships @ = 16 + &, * = -&, # = 12 + &:
B) * = 7:
If * = 7, then from * = -&, we get 7 = -& => & = -7.
This gives the same value for & as option A, leading to the same solution (@=9, *=7, #=5, &=-7). In this solution, * = 7 is also true.

Since both A and B are true in the same valid solution set under this interpretation, there is likely an issue with the question, as only one option should be correct.

However, if we assume the intended correct answer is A, it implies that the system must have a solution where @=9, and this solution is the one the question is probing. Given the ambiguity, and that option A provides a specific value for one variable which leads to a consistent (though not unique system-wise) solution, we proceed with the derivation showing that @=9 is possible. The calculation above shows that if @=9, then &=-7, *=7, and #=5, which satisfies the system (with the second equation interpreted as `# – & = 12`).

Let’s also consider the alternative interpretation where the second equation is `* + # – & = 12`:
1) @ + * = 16
2) * + # – & = 12
3) @ – & = 16
From (1) and (3), subtracting (3) from (1) gives: (@ + *) – (@ – &) = 16 – 16 => * + & = 0 => * = -&.
Substitute * = -& into (1): @ + (-&) = 16 => @ – & = 16, which is (3).
Substitute * = -& into (2): (-&) + # – & = 12 => # – 2& = 12 => # = 12 + 2&.
Relationships: @ = 16 + &, * = -&, # = 12 + 2&.
A) @ = 9: 9 = 16 + & => & = -7. (@=9, *=7, #=12 + 2(-7) = 12 – 14 = -2, &=-7). Solution (@=9, *=7, #=-2, &=-7). This solution satisfies all three equations, assuming the form `* + # – & = 12` for the second one. In this solution, @=9 is true.
B) * = 7: 7 = -& => & = -7. (@=9, *=7, #=-2, &=-7). * = 7 is also true in this solution.

Again, options A and B are simultaneously true. Assuming option A is the intended correct answer, the explanation demonstrates that @=9 is a value that fits a valid solution derived from a plausible interpretation of the equations.

I sold this shirt to a friend at a loss of 10% on the price I bought it for (the CP).
The friend paid ₹729-00. This is the selling price (SP).
The loss is calculated on the CP.
Loss = 10% of CP = 0.10 * CP.
Selling Price (SP) = CP – Loss = CP – 0.10 * CP = 0.90 * CP.

We are given that the friend paid ₹729, so SP = ₹729.
729 = 0.90 * CP
To find CP, divide 729 by 0.90:
CP = 729 / 0.9 = 7290 / 9 = 810.
So, the cost price (the price I bought the shirt for) was ₹810.

Now, we need to find the undiscounted price U, knowing that CP = 0.90U.
810 = 0.90 * U
To find U, divide 810 by 0.90:
U = 810 / 0.9 = 8100 / 9 = 900.
The undiscounted price of the shirt was ₹900.

Let’s verify:
Undiscounted price = ₹900.
10% discount = 10% of 900 = 0.10 * 900 = ₹90.
Price after discount (CP) = 900 – 90 = ₹810. (This is the price I bought it for).
Sold at a loss of 10% on CP:
Loss amount = 10% of 810 = 0.10 * 810 = ₹81.
Selling Price (SP) = CP – Loss amount = 810 – 81 = ₹729.
This matches the amount the friend paid.

Let’s analyse the relationship between n and f(n).
Notice the outputs 0, 0, 2, 0, 2, ?, 2, 5.
The inputs are 1, 2, 3, 4, 5, 6, 7, 8.
f(1)=0, f(2)=0, f(4)=0. These inputs (1, 2, 4) are powers of 2 ($1=2^0$, $2=2^1$, $4=2^2$).
f(3)=2, f(5)=2, f(7)=2. These inputs (3, 5, 7) are primes greater than 2.
f(8)=5. This input (8) is $2^3$.
f(6)=x. This input (6) is not a power of 2 and not a prime.

Let’s hypothesize a rule based on the observed pattern:
– If n is a power of 2, $n=2^k$: the output depends on k? f(1)=f($2^0$)=0, f(2)=f($2^1$)=0, f(4)=f($2^2$)=0, f(8)=f($2^3$)=5.
– If n is a prime greater than 2: f(n)=2. (Matches f(3), f(5), f(7)).
– If n is composite and not a power of 2: f(n)=? (Only n=6 in the given range).

Let’s refine the rule based on inputs being powers of 2 vs not powers of 2.
– If n is a power of 2: $n=2^k$. f(1)=0, f(2)=0, f(4)=0, f(8)=5. This still doesn’t follow a simple rule like k, k+c, k^2, etc.
– If n is NOT a power of 2: f(3)=2, f(5)=2, f(7)=2. This strongly suggests f(n)=2 for n which are not powers of 2, with a potential exception for n=8.

Let’s assume the rule is:
f(n) = 2 if n is not a power of 2.
f(n) = 0 if n is a power of 2, except for n=8.
f(8) = 5.

Let’s test this rule:
f(1) = f($2^0$). Power of 2, not 8. Rule says 0. Matches.
f(2) = f($2^1$). Power of 2, not 8. Rule says 0. Matches.
f(3). Not a power of 2. Rule says 2. Matches.
f(4) = f($2^2$). Power of 2, not 8. Rule says 0. Matches.
f(5). Not a power of 2. Rule says 2. Matches.
f(6). Not a power of 2 ($6 = 2 \times 3$). Rule says 2. So x=2.
f(7). Not a power of 2. Rule says 2. Matches.
f(8) = f($2^3$). This is the special case $n=8$. Rule says 5. Matches.

This rule consistently explains the sequence values 0, 0, 2, 0, 2, ?, 2, 5, and predicts x=2.
The number 6 is not a power of 2 ($1, 2, 4, 8, 16, …$). According to the rule, f(6) should be 2.
Therefore, x = 2.

Let’s check the numbers between 200 and 230:
– 201: Divisible by 3 (sum of digits 2+0+1=3). Not prime.
– 202: Divisible by 2. Not prime.
– 203: $203 = 7 \times 29$. Not prime.
– 204: Divisible by 2. Not prime.
– 205: Divisible by 5. Not prime.
– 206: Divisible by 2. Not prime.
– 207: Divisible by 3 (sum of digits 2+0+7=9). Not prime.
– 208: Divisible by 2. Not prime.
– 209: $209 = 11 \times 19$. Not prime.
– 210: Divisible by 10. Not prime.
– 211: Check divisibility by primes 2, 3, 5, 7, 11, 13. Not divisible by any of these. $14^2 = 196$, $15^2 = 225$. $\sqrt{211} \approx 14.5$. Primes to check up to 13. 211 is prime.
– 212: Divisible by 2. Not prime.
– 213: Divisible by 3 (sum of digits 2+1+3=6). Not prime.
– 214: Divisible by 2. Not prime.
– 215: Divisible by 5. Not prime.
– 216: Divisible by 2. Not prime.
– 217: $217 = 7 \times 31$. Not prime.
– 218: Divisible by 2. Not prime.
– 219: Divisible by 3 (sum of digits 2+1+9=12). Not prime.
– 220: Divisible by 10. Not prime.
– 221: $221 = 13 \times 17$. Not prime.
– 222: Divisible by 2. Not prime.
– 223: Check divisibility by primes 2, 3, 5, 7, 11, 13. Not divisible by any of these. $\sqrt{223} \approx 14.9$. Primes to check up to 13. 223 is prime.
– 224: Divisible by 2. Not prime.
– 225: Divisible by 5. Not prime.
– 226: Divisible by 2. Not prime.
– 227: Check divisibility by primes 2, 3, 5, 7, 11, 13. Not divisible by any of these. $\sqrt{227} \approx 15.07$. Primes to check up to 13. 227 is prime.
– 228: Divisible by 2. Not prime.
– 229: Check divisibility by primes 2, 3, 5, 7, 11, 13. Not divisible by any of these. $\sqrt{229} \approx 15.13$. Primes to check up to 13. 229 is prime.

The prime numbers between 200 and 230 are 211, 223, 227, and 229.
There are 4 prime numbers in this range.