2019

We are given that the cost remains the same (C₁ = C₂) while the distance is increased to 4 times (D₂ = 4D₁). We need to find the new volume (V₂) in terms of the original volume (V₁).

Using the formula:
C₁ = k * √D₁ * V₁
C₂ = k * √D₂ * V₂

Since C₁ = C₂, we have:
k * √D₁ * V₁ = k * √D₂ * V₂
√D₁ * V₁ = √D₂ * V₂

Substitute D₂ = 4D₁:
√D₁ * V₁ = √(4D₁) * V₂
√D₁ * V₁ = 2√D₁ * V₂

Assuming D₁ > 0, we can divide both sides by √D₁:
V₁ = 2 * V₂

Solving for V₂:
V₂ = V₁ / 2

This means the new volume V₂ is half of the original volume V₁, which is 50%.

Speed against the stream (upstream speed) = Speed in still water – Speed of stream = v – u.
We are given that the speed against the stream is 4.5 km/hr.
v – u = 4.5
Substitute the value of v:
6 – u = 4.5
u = 6 – 4.5 = 1.5 km/hr.
The speed of the river stream is 1.5 km/hr.

We need to find the speed if he rows along the stream (downstream speed) with the same effort. The effort remaining the same means his speed relative to the water is still v.
Speed along the stream (downstream speed) = Speed in still water + Speed of stream = v + u.
Substitute the values of v and u:
Downstream speed = 6 + 1.5 = 7.5 km/hr.

The question asks for the speed in one of the given options. 7.5 km/hr is equivalent to 7 and a half km/hr.
7.5 = 7 + 0.5 = 7 + 1/2 = 7½ km/hr.

The wheel makes 500 revolutions.
The total distance traveled is the distance per revolution multiplied by the number of revolutions.
Total distance = Circumference $\times$ Number of revolutions
Total distance = $(\pi \times 1.26) \times 500$ meters.

We can use the approximation $\pi \approx \frac{22}{7}$ or $\pi \approx 3.14$. Since the diameter 1.26 is divisible by 7 ($1.26 = 0.18 \times 7$), using $\pi \approx \frac{22}{7}$ will likely give a precise answer or one of the options.
$1.26 / 7 = 0.18$.
Total distance = $\frac{22}{7} \times 1.26 \times 500$
Total distance = $22 \times (1.26 / 7) \times 500$
Total distance = $22 \times 0.18 \times 500$
Total distance = $22 \times (0.18 \times 500)$
$0.18 \times 500 = 18 \times 5 = 90$.
Total distance = $22 \times 90$.
Total distance = $1980$ meters.

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.