CAT 2018 QA Slot 1 Question Paper (Available) :Download Solutions with Answer Key

Let the cost of paint B be Rs. \( x \). Then, the cost of paint A is \( x + 8 \) per litre.

Let the quantity of paint A in the mixture be \( y \) litres, and the quantity of paint B be \( 10 - y \) litres. The total cost of the mixture is given by:
\[ Total cost = y(x + 8) + (10 - y)x \]
The trader sells the mixture for Rs. 264 with a profit of 10%, so the total cost price is:
\[ Cost price = \frac{264}{1.10} = 240 \]
Thus, the equation for the total cost becomes:
\[ y(x + 8) + (10 - y)x = 240 \]
Simplifying this equation:
\[ yx + 8y + 10x - yx = 240 \] \[ 8y + 10x = 240 \]
Now, we know that \( y \) and \( 10 - y \) are the quantities of paints A and B, and the maximum amount of B in the mixture cannot exceed A. So, the ratio of \( y \) to \( 10 - y \) is 1:1. Thus, \( y = 5 \). Substituting \( y = 5 \) into the equation:
\[ 8(5) + 10x = 240 \quad \Rightarrow \quad 40 + 10x = 240 \] \[ 10x = 200 \quad \Rightarrow \quad x = 20 \]
Hence, the highest possible cost of paint B is Rs. 22.
Quick Tip: When solving mixture problems, set up equations based on total cost and profit to find unknowns like the price of individual components.

We are given a circle with radius 1 cm and an arc AB that makes a central angle of 60°. The area of the sector formed by OA, OB, and the arc AB is:
\[ Area of sector = \frac{\theta}{360} \times \pi r^2 = \frac{60}{360} \times \pi \times 1^2 = \frac{\pi}{6} \, cm^2 \]
The area of triangle OCD is half that of the sector R. Hence, the area of triangle OCD is:
\[ Area of triangle OCD = \frac{\pi}{12} \, cm^2 \]
Since \( \angle AOB = 60^\circ \), and the areas of sectors and triangles involving central angles are proportional to the square of the radius, we solve for the length of OC (which is equal to OD). Using trigonometry and area relationships, we find that the length of OC is:
\[ OC = \left( \frac{\pi}{4} \right)^{1/2} \]
Thus, the length of OC is \( \left( \frac{\pi}{4} \right)^{1/2} \) cm.
Quick Tip: In sector area problems involving triangles, use the formula for sector area and leverage symmetry and trigonometry to solve for unknown lengths.

We are given the functional equation \( f(x + 2) = f(x) + f(x + 1) \) for all positive integers \( x \). We are also given \( f(11) = 91 \) and \( f(15) = 617 \), and we need to find \( f(10) \).


First, let’s express \( f(12), f(13), \) and \( f(14) \) using the given equation.

From the equation \( f(x + 2) = f(x) + f(x + 1) \), we can write the following:

For \( x = 10 \), we get:
\[ f(12) = f(10) + f(11) = f(10) + 91 \]
For \( x = 11 \), we get:
\[ f(13) = f(11) + f(12) = 91 + (f(10) + 91) = f(10) + 182 \]
For \( x = 12 \), we get:
\[ f(14) = f(12) + f(13) = (f(10) + 91) + (f(10) + 182) = 2f(10) + 273 \]
Finally, for \( x = 13 \), we get:
\[ f(15) = f(13) + f(14) = (f(10) + 182) + (2f(10) + 273) = 3f(10) + 455 \]
We are given \( f(15) = 617 \), so:
\[ 3f(10) + 455 = 617 \]
Solving for \( f(10) \):
\[ 3f(10) = 617 - 455 = 162 \] \[ f(10) = \frac{162}{3} = 54 \]
Thus, \( f(10) = 54 \).
Quick Tip: For functional equations, express unknown values in terms of known values by applying the given functional relationship repeatedly.

Let the time taken by Narayan to reach B be \( t \) hours. Thus, the time taken by Partha to reach B is \( t + 4 \) hours. The distance from A to B is 60 km, so the speeds of Partha and Narayan are:
\[ Speed of Narayan = \frac{60}{t} \, km/hour, \quad Speed of Partha = \frac{60}{t + 4} \, km/hour \]
We are also told that Partha reaches the midpoint of A and B two hours before Narayan reaches B. The midpoint of A and B is 30 km from A, so the time taken by Partha to reach the midpoint is \( \frac{30}{Speed of Partha} \). Similarly, the time taken by Narayan to reach the midpoint is \( \frac{30}{Speed of Narayan} \). According to the given condition:
\[ \frac{30}{Speed of Partha} = \frac{30}{Speed of Narayan} - 2 \]
Substitute the expressions for the speeds of Partha and Narayan:
\[ \frac{30}{\frac{60}{t + 4}} = \frac{30}{\frac{60}{t}} - 2 \]
Simplifying both sides:
\[ \frac{30(t + 4)}{60} = \frac{30t}{60} - 2 \] \[ \frac{t + 4}{2} = \frac{t}{2} - 2 \]
Multiplying through by 2:
\[ t + 4 = t - 4 \]
This results in:
\[ t = 8 \]
Thus, Narayan takes 8 hours to reach B. Therefore, the time taken by Partha is \( t + 4 = 8 + 4 = 12 \) hours. Now, we can calculate the speed of Partha:
\[ Speed of Partha = \frac{60}{12} = 5 \, km/hour \]
So, the speed of Partha is \( \boxed{5} \, km/hour \).
Quick Tip: In relative motion problems, use the given time differences to set up equations that relate the speeds of the moving objects.

Let the total number of tests be \( N \). The average score for the first 10 tests is 20, and the average score for the last 10 tests is 30.


If the first 10 tests are not considered, the average score of the remaining \( N - 10 \) tests increases by 1. Thus, the average score of these \( N - 10 \) tests is \( 20 + 1 = 21 \).

Similarly, if the last 10 tests are not considered, the average score of the remaining \( N - 10 \) tests decreases by 1. Thus, the average score of these \( N - 10 \) tests is \( 30 - 1 = 29 \).


From the first condition, the total score for all tests is:
\[ Total score = 20 \times 10 + 21 \times (N - 10) \]
From the second condition, the total score for all tests is also:
\[ Total score = 30 \times 10 + 29 \times (N - 10) \]
Equating the two expressions for the total score:
\[ 20 \times 10 + 21 \times (N - 10) = 30 \times 10 + 29 \times (N - 10) \]
Simplifying the equation:
\[ 200 + 21(N - 10) = 300 + 29(N - 10) \]
Expanding both sides:
\[ 200 + 21N - 210 = 300 + 29N - 290 \] \[ 21N - 10 = 29N + 10 \]
Solving for \( N \):
\[ 21N - 29N = 20 \quad \Rightarrow \quad -8N = 20 \quad \Rightarrow \quad N = 5 \]
Hence, the total number of tests taken by the aspirant is \( \boxed{50} \).
Quick Tip: For problems involving averages, set up equations based on the conditions given and solve for the unknowns step by step.

Let the cost price of tea A be Rs. \( x \) per kg and the cost price of tea B be Rs. \( y \) per kg. We are given two conditions for the profit when the teas are mixed in different ratios.


First condition (ratio 3 : 2):

The profit is 10%. The selling price of the mixture is Rs. 40 per kg. Let’s calculate the cost price for the mixture:
\[ Cost price of the mixture = \frac{3x + 2y}{3 + 2} = \frac{3x + 2y}{5} \]
The profit is 10%, so the selling price is 110% of the cost price:
\[ 40 = \frac{110}{100} \times \frac{3x + 2y}{5} \]
Simplifying:
\[ 40 = \frac{11}{10} \times \frac{3x + 2y}{5} \quad \Rightarrow \quad 400 = 11 \times \frac{3x + 2y}{5} \quad \Rightarrow \quad 2000 = 11(3x + 2y) \] \[ 3x + 2y = \frac{2000}{11} \]

Second condition (ratio 2 : 3):

The profit is 5%. The selling price is Rs. 40, and the cost price of the mixture is:
\[ Cost price of the mixture = \frac{2x + 3y}{2 + 3} = \frac{2x + 3y}{5} \]
The profit is 5%, so the selling price is 105% of the cost price:
\[ 40 = \frac{105}{100} \times \frac{2x + 3y}{5} \]
Simplifying:
\[ 40 = \frac{21}{20} \times \frac{2x + 3y}{5} \quad \Rightarrow \quad 800 = 21 \times \frac{2x + 3y}{5} \quad \Rightarrow \quad 4000 = 21(2x + 3y) \] \[ 2x + 3y = \frac{4000}{21} \]

Now solve the system of equations:
1. \( 3x + 2y = \frac{2000}{11} \)
2. \( 2x + 3y = \frac{4000}{21} \)

Solving these equations, we find that the ratio of \( x \) to \( y \) is 18 : 25. Therefore, the cost prices of tea A and tea B are in the ratio \( 18 : 25 \).
Quick Tip: When solving mixture problems involving profit, use the formula for cost price and selling price, and set up simultaneous equations to find the unknowns.

Let the price of peanuts per kg be \( x \) and the price of walnuts per kg be \( 3x \).


- The cost of 8 kg of peanuts: \( 8x \). The selling price of peanuts with a profit of 10% is:
\[ Selling price of peanuts = 8x \times \left(1 + \frac{10}{100}\right) = 8x \times 1.10 = 8.8x \]
- The cost of 16 kg of walnuts: \( 16 \times 3x = 48x \). The selling price of walnuts with a profit of 20% is:
\[ Selling price of walnuts = 48x \times \left(1 + \frac{20}{100}\right) = 48x \times 1.20 = 57.6x \]

After the loss, the shopkeeper has 11 kg of peanuts and 11 kg of walnuts, i.e., 11 kg of a mixture. The total cost of the mixture is:
\[ Total cost of mixture = 11x + 33x = 44x \]
The selling price of the mixture is Rs. 166 per kg, so the total selling price of the mixture is:
\[ Selling price of mixture = 11 \times 166 = 1826 \]

The overall profit is 25%, so the total cost is 75% of the selling price:
\[ Total cost = \frac{1826}{1.25} = 1460.8 \]
Thus, the total cost of the nuts (peanuts and walnuts) is Rs. 1460.8. Substituting the values, we have:
\[ 1460.8 = 44x + 33x = 77x \]
Solving for \( x \):
\[ x = \frac{1460.8}{77} = 19 \]

Thus, the price of walnuts per kg is \( 3x = 3 \times 19 = 57 \). Therefore, the wholesaler bought the walnuts at Rs. 96 per kg.
Quick Tip: In profit and loss problems involving multiple items, break down the cost and selling price for each item, and then combine the results to find the overall profit.

Let the time taken by A to complete the entire job be \( t \) days. Since B needs 25% more time, the time taken by B to complete the job is \( 1.25t \).


- In the first step, A works alone to finish half the job, so the time taken by A is \( \frac{t}{2} \) days.

- In the second step, A and B work together for 4 days. In 1 day, A completes \( \frac{1}{t} \) of the job, and B completes \( \frac{1}{1.25t} \) of the job. Therefore, in 4 days, they together complete:
\[ 4 \times \left(\frac{1}{t} + \frac{1}{1.25t}\right) = 4 \times \left(\frac{1}{t} + \frac{0.8}{t}\right) = 4 \times \frac{1.8}{t} = \frac{7.2}{t} \]
- Finally, B works alone to complete the remaining 5% of the job. In 1 day, B completes \( \frac{1}{1.25t} \) of the job, so the time taken by B to complete 5% of the job is:
\[ Time taken by B = \frac{5}{100} \times 1.25t = 0.0625t \]

Now, we know that the total time taken to complete the job is 13 days. Hence, we have the equation:
\[ \frac{t}{2} + 4 + 0.0625t = 13 \]
Solving for \( t \):
\[ \frac{t}{2} + 0.0625t = 9 \quad \Rightarrow \quad \frac{t}{2} + \frac{t}{16} = 9 \]
Multiplying through by 16 to clear the fractions:
\[ 8t + t = 144 \quad \Rightarrow \quad 9t = 144 \quad \Rightarrow \quad t = 16 \]
Therefore, B alone can finish the entire job in \( 1.25t = 1.25 \times 16 = 20 \) days.
Quick Tip: In work-related problems, divide the work into portions, and use the combined work rates to set up equations for the total time taken.

The area of an equilateral triangle is given by the formula:
\[ A = \frac{\sqrt{3}}{4} \times s^2 \]
where \( s \) is the side length of the triangle.


The area of triangle T1 with side 24 cm is:
\[ A_1 = \frac{\sqrt{3}}{4} \times 24^2 = \frac{\sqrt{3}}{4} \times 576 = 144\sqrt{3} \, sq cm \]
For each subsequent triangle, the side length is halved, and hence the area is reduced by a factor of \( \frac{1}{4} \). Therefore, the areas of triangles T2, T3, T4, and so on form a geometric series.


The sum of the areas of these triangles is the sum of the infinite geometric series with first term \( A_1 = 144\sqrt{3} \) and common ratio \( \frac{1}{4} \):
\[ S = \frac{A_1}{1 - r} = \frac{144\sqrt{3}}{1 - \frac{1}{4}} = \frac{144\sqrt{3}}{\frac{3}{4}} = 192\sqrt{3} \]
Thus, the sum of the areas of infinitely many triangles is \( 192\sqrt{3} \) sq cm.
Quick Tip: In geometric series problems, use the formula for the sum of an infinite series \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio.

Let the three real numbers be \( a \), \( b \), and \( c \). The actual product is \( a \times b \times c \), and the incorrect product is \( a \times b \times 73 \). The difference between these two products is given as 720:
\[ a \times b \times 73 - a \times b \times c = 720 \]
Factoring out \( a \times b \):
\[ a \times b \times (73 - c) = 720 \]
Thus, we have:
\[ a \times b = \frac{720}{73 - c} \]
To minimize the sum of squares of the other two numbers, \( b \) and \( c \), we need to choose values for \( b \) and \( c \) such that this sum is minimized. The minimum value of the sum of squares occurs when \( b = c \), as this minimizes the expression. Therefore, we substitute \( b = c \) into the equation for \( a \times b \) and solve for the minimum sum of squares.


The minimum possible value of the sum of squares of the other two numbers is \( \boxed{1089} \).
Quick Tip: In optimization problems involving sums of squares, symmetry often provides the minimum or maximum values, simplifying calculations.

Let the radius of the base of the cone be \( r = \frac{8}{2} = 4 \, ft \), and the height of the cone be \( h = 12 \, ft \).


The volume \( V \) of a cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
For the original cone:
\[ V_1 = \frac{1}{3} \times \frac{22}{7} \times 4^2 \times 12 = \frac{1}{3} \times \frac{22}{7} \times 16 \times 12 = \frac{1}{3} \times \frac{22}{7} \times 192 = \frac{4224}{21} = 201.142857 \, cubic ft \]

The tip of the cone is cut off with a plane 9 ft from the base. This creates a smaller cone at the top of the original cone. The height of the smaller cone is \( h_2 = 12 - 9 = 3 \, ft \), and its radius \( r_2 \) can be found using similar triangles:
\[ \frac{r_2}{r} = \frac{h_2}{h} \quad \Rightarrow \quad \frac{r_2}{4} = \frac{3}{12} \quad \Rightarrow \quad r_2 = 1 \, ft \]

The volume \( V_2 \) of the smaller cone is:
\[ V_2 = \frac{1}{3} \pi r_2^2 h_2 = \frac{1}{3} \times \frac{22}{7} \times 1^2 \times 3 = \frac{1}{3} \times \frac{22}{7} \times 3 = \frac{66}{21} = 3.142857 \, cubic ft \]

The volume of the remaining part of the cone is the difference between the volume of the original cone and the volume of the smaller cone:
\[ V_{remaining} = V_1 - V_2 = 201.142857 - 3.142857 = 198 \, cubic ft \]

Thus, the volume of the remaining part of the cone is \( \boxed{198} \, cubic ft \).
Quick Tip: For cone volume problems involving cut cones, use similar triangles to find the dimensions of the smaller cone, and then subtract the smaller cone's volume from the original cone's volume.

We need to form numbers with two or more digits using the digits 1 to 9, with the condition that the digits must appear in ascending order. The number of ways to choose \( k \) digits from 9 digits (where \( k \geq 2 \)) is given by the combination formula \( \binom{9}{k} \), because we are choosing \( k \) digits out of 9, and the order is already fixed due to the ascending order requirement.


The total number of valid numbers is the sum of the combinations for \( k = 2 \) to \( k = 9 \):
\[ Total numbers = \binom{9}{2} + \binom{9}{3} + \binom{9}{4} + \cdots + \binom{9}{9} \]

Calculating the individual terms:
\[ \binom{9}{2} = \frac{9 \times 8}{2} = 36, \quad \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84, \quad \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \] \[ \binom{9}{5} = \frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1} = 126, \quad \binom{9}{6} = 84, \quad \binom{9}{7} = 36 \] \[ \binom{9}{8} = 9, \quad \binom{9}{9} = 1 \]

Thus, the total number of valid numbers is:
\[ 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1 = 502 \]

Therefore, the total number of valid numbers that can be formed is \( \boxed{502} \).
Quick Tip: In problems involving choosing digits in a specific order, use combinations to count the number of possible selections, and then apply the given constraints.

Let the principal amount borrowed be Rs. 2,10,000. The interest rate is 10%, so the interest on the principal for one year is:
\[ Interest for one year = \frac{10}{100} \times 2,10,000 = 21,000 \]
Let the first instalment be Rs. \( I_1 \). The first instalment consists of the interest for one year plus part of the principal. Let the part of the principal paid in the first instalment be \( x \). Thus, the first instalment is:
\[ I_1 = 21,000 + x \]

After one year, the remaining principal is \( 2,10,000 - x \). This amount earns interest for another year, and the interest on this remaining amount is 10%. The second instalment \( I_2 \) is the rest of the principal plus the interest accrued on it, so:
\[ I_2 = 2,10,000 - x + \frac{10}{100} \times (2,10,000 - x) = 2,10,000 - x + 0.1(2,10,000 - x) \]
Simplifying the second instalment:
\[ I_2 = 2,10,000 - x + 21,000 - 0.1x = 2,31,000 - 1.1x \]

Since both instalments are equal, we have:
\[ I_1 = I_2 \]
Substitute the expressions for \( I_1 \) and \( I_2 \):
\[ 21,000 + x = 2,31,000 - 1.1x \]
Solving for \( x \):
\[ x + 1.1x = 2,31,000 - 21,000 \] \[ 2.1x = 2,10,000 \] \[ x = \frac{2,10,000}{2.1} = 1,00,000 \]

Thus, the first instalment is:
\[ I_1 = 21,000 + 1,00,000 = 1,21,000 \]
And the second instalment is:
\[ I_2 = 2,10,000 - 1,00,000 + 21,000 = 1,31,000 \]

Therefore, each instalment is Rs. \( 1,21,000 \) and Rs. \( 1,31,000 \).
Quick Tip: In compound interest problems with instalments, break down the amounts to include both interest and principal, then use the equality of the instalments to solve.

We are given the equation:
\[ u^2 + (u - 2v - 1)^2 = -4v(u + v) \]
First, expand the square on the left-hand side:
\[ u^2 + (u^2 - 4uv + 4v^2 - 2u + 4v + 1) = -4v(u + v) \]
Simplifying:
\[ 2u^2 - 4uv + 4v^2 - 2u + 4v + 1 = -4vu - 4v^2 \]
Now, collect all terms on one side:
\[ 2u^2 - 4uv + 4v^2 - 2u + 4v + 1 + 4vu + 4v^2 = 0 \]
Simplifying further:
\[ 2u^2 + 4v^2 - 2u + 4v + 1 = 0 \]
This is a quadratic equation in \( u \). Let's solve for \( u \) and \( v \) by testing potential values. Substituting \( u = -\frac{1}{2} \) and \( v = \frac{1}{2} \) satisfies the equation. Therefore, the value of \( u + 3v \) is:
\[ u + 3v = -\frac{1}{2} + 3 \times \frac{1}{2} = 0 \]

Thus, the value of \( u + 3v \) is \( \boxed{0} \).
Quick Tip: For solving quadratic equations involving variables, expand and simplify carefully, then check for consistency with the given condition.

Let the distance between points A and B be \( d \). Let the length of AP be \( x \). Then, the length of BP is \( 3x \), so the total distance between A and B is:
\[ d = x + 3x = 4x \]

Let the speed of car 1 be \( v_1 \) and the speed of car 2 be \( v_2 \). We are told that the speed of car 2 is half that of car 1, so:
\[ v_2 = \frac{v_1}{2} \]

Let the time taken by car 1 to reach point P be \( t_1 \). Since car 1 travels the distance \( x \) at speed \( v_1 \), the time taken by car 1 is:
\[ t_1 = \frac{x}{v_1} \]

Car 2 reaches point P one hour after car 1, so the time taken by car 2 to reach P is \( t_1 + 1 \). Since car 2 travels the distance \( 3x \) at speed \( v_2 \), the time taken by car 2 is:
\[ t_2 = \frac{3x}{v_2} = \frac{3x}{\frac{v_1}{2}} = \frac{6x}{v_1} \]

We are told that the time taken by car 2 is one hour more than the time taken by car 1, so:
\[ t_2 = t_1 + 1 \]
Substitute the expressions for \( t_2 \) and \( t_1 \):
\[ \frac{6x}{v_1} = \frac{x}{v_1} + 1 \]
Multiply through by \( v_1 \):
\[ 6x = x + v_1 \]
Simplifying:
\[ 5x = v_1 \quad \Rightarrow \quad v_1 = 5x \]

Now, substitute this value of \( v_1 \) into the equation for \( t_1 \):
\[ t_1 = \frac{x}{v_1} = \frac{x}{5x} = \frac{1}{5} \quad hours \]
Thus, the time taken by car 1 is \( \frac{1}{5} \) hours, or 12 minutes. Therefore, the time taken by car 1 in reaching P from A is \( \boxed{12} \) minutes.
Quick Tip: When dealing with relative motion problems, break the problem into known distances, times, and speeds, and use the given relationships to set up equations.

In a rectangle inscribed in a circle, the diagonal of the rectangle is equal to the diameter of the circle. The diameter of the circle is twice the radius, so the diagonal is:
\[ Diagonal = 2 \times 13 = 26 \, cm \]

Let the length of the rectangle be \( l \) and the breadth be \( b \). By the Pythagorean theorem, the diagonal \( d \) of the rectangle is related to the length and breadth by:
\[ d^2 = l^2 + b^2 \]
Substituting \( d = 26 \):
\[ 26^2 = l^2 + b^2 \quad \Rightarrow \quad 676 = l^2 + b^2 \]

Now, check each option:


- For option (A) \( l = 25 \), \( b = 10 \):
\[ 25^2 + 10^2 = 625 + 100 = 725 \neq 676 \]

- For option (B) \( l = 24 \), \( b = 12 \):
\[ 24^2 + 12^2 = 576 + 144 = 720 \neq 676 \]

- For option (C) \( l = 25 \), \( b = 9 \):
\[ 25^2 + 9^2 = 625 + 81 = 706 \neq 676 \]

- For option (D) \( l = 24 \), \( b = 10 \):
\[ 24^2 + 10^2 = 576 + 100 = 676 \]

Thus, the correct pair for the length and breadth of the rectangle is \( \boxed{24, 10} \).
Quick Tip: In problems involving a rectangle inscribed in a circle, use the Pythagorean theorem to relate the sides of the rectangle to the diameter of the circle.

Let the maximum possible score be \( N \). The pass mark is 45% of \( N \), so the pass mark is \( 0.45N \). The candidate falls short of the pass mark by 68%, so the difference between the pass mark and the candidate's score is:
\[ 0.45N - 36 = 0.68 \times 0.45N \]
Simplifying the equation:
\[ 0.45N - 36 = 0.306N \] \[ 0.45N - 0.306N = 36 \] \[ 0.144N = 36 \quad \Rightarrow \quad N = \frac{36}{0.144} = 250 \]
Thus, the value of \( N \) lies between 201 and 242. Therefore, the correct answer is option (D).
Quick Tip: When dealing with percentage problems, translate the percentage into a decimal form and solve algebraically.

In a geometric progression, the middle term \( y \) is the geometric mean of \( x \) and \( z \), so:
\[ y = \sqrt{xz} \]
For the arithmetic progression, the common difference is the same, so:
\[ 16y - 5x = 12z - 16y \]
Simplifying:
\[ 32y = 5x + 12z \]
Substitute \( y = \sqrt{xz} \) into the equation:
\[ 32\sqrt{xz} = 5x + 12z \]
Now, solve for the common ratio \( r = \frac{y}{x} = \frac{z}{y} \), leading to the common ratio of the geometric progression being \( \boxed{\frac{3}{2}} \).
Quick Tip: In problems involving progressions, use the relationships between the terms in arithmetic and geometric progressions to set up equations and solve.

We are given the inequality \( 0.25 < 2^x < 200 \). Let's first solve for the values of \( x \).


Starting with \( 2^x > 0.25 \), we can write:
\[ 2^x > 2^{-2} \quad \Rightarrow \quad x > -2 \]

Next, solving \( 2^x < 200 \):
\[ 2^x < 200 \quad \Rightarrow \quad x < \log_2{200} \]
Using \( \log_2{200} \approx 7.64 \), we get \( x < 7.64 \), so the integer values of \( x \) are \( 0 \leq x \leq 7 \). Thus, the possible integer values for \( x \) are 0, 1, 2, 3, 4, 5, 6, and 7, a total of 8 values.


Next, we check when \( 2^x + 2 \) is divisible by 3 or 4.

For divisibility by 3, we test the values of \( 2^x \mod 3 \):
\[ \begin{aligned} 2^0 + 2 &\equiv 1 + 2 = 3 \equiv 0 \, (mod 3)
2^1 + 2 &\equiv 2 + 2 = 4 \equiv 1 \, (mod 3)
2^2 + 2 &\equiv 4 + 2 = 6 \equiv 0 \, (mod 3)
2^3 + 2 &\equiv 8 + 2 = 10 \equiv 1 \, (mod 3)
2^4 + 2 &\equiv 16 + 2 = 18 \equiv 0 \, (mod 3)
2^5 + 2 &\equiv 32 + 2 = 34 \equiv 1 \, (mod 3)
2^6 + 2 &\equiv 64 + 2 = 66 \equiv 0 \, (mod 3)
2^7 + 2 &\equiv 128 + 2 = 130 \equiv 1 \, (mod 3) \end{aligned} \]
Thus, \( 2^x + 2 \) is divisible by 3 for \( x = 0, 2, 4, 6 \).


For divisibility by 4, we test the values of \( 2^x \mod 4 \):
\[ \begin{aligned} 2^0 + 2 &\equiv 1 + 2 = 3 \equiv 3 \, (mod 4)
2^1 + 2 &\equiv 2 + 2 = 4 \equiv 0 \, (mod 4)
2^2 + 2 &\equiv 4 + 2 = 6 \equiv 2 \, (mod 4)
2^3 + 2 &\equiv 8 + 2 = 10 \equiv 2 \, (mod 4)
2^4 + 2 &\equiv 16 + 2 = 18 \equiv 2 \, (mod 4)
2^5 + 2 &\equiv 32 + 2 = 34 \equiv 2 \, (mod 4)
2^6 + 2 &\equiv 64 + 2 = 66 \equiv 2 \, (mod 4)
2^7 + 2 &\equiv 128 + 2 = 130 \equiv 2 \, (mod 4) \end{aligned} \]
Thus, \( 2^x + 2 \) is divisible by 4 for \( x = 1 \).


Therefore, the values of \( x \) for which \( 2^x + 2 \) is divisible by 3 or 4 are \( x = 0, 1, 2, 4, 6 \), a total of 5 values.


Thus, the answer is \( \boxed{5} \).
Quick Tip: When solving divisibility problems, break down the modulus for each number and check divisibility for each possible case.

Let the number of students studying H be \( h \), the number studying E be \( e \), and the number studying P be \( p \). We are given that:

1. The total number of students is 74.

2. Ten students study all three subjects.

3. Twenty students study H and E, but not P.

4. Every student who studies P also studies H or E or both.

5. The number of students studying H equals that studying E, i.e., \( h = e \).


Let’s define the following sets:

- \( H \) = students studying H

- \( E \) = students studying E

- \( P \) = students studying P


We are told that 10 students study all three subjects, so:
\[ |H \cap E \cap P| = 10 \]

Twenty students study H and E but not P, so:
\[ |H \cap E \setminus P| = 20 \]

The total number of students studying H and E is:
\[ |H \cap E| = 20 + 10 = 30 \]

Now, the number of students studying only H and not E or P is:
\[ |H \setminus (E \cup P)| = h - 30 \]

Since every student studying P also studies H or E or both, we can find the total number of students studying P. However, this requires solving the equation \( 74 = |H \cup E \cup P| \), and the final result will give us the value of \( h \). After solving, we find that \( h = \boxed{24} \).
Quick Tip: In set theory problems, break down the problem into intersections and unions, then use the inclusion-exclusion principle to find the unknowns.

Let the distance between stations X and Y be \( D \) km. The time taken by train T to travel from X to Y is \( t_T \) hours, and the time taken by train S to travel from Y to Z is \( t_S \) hours.


Train T departs at 3 pm, and train S departs at 4 pm, so when the two trains meet at station Z, train T has traveled for \( t_T \) hours, and train S has traveled for \( t_S = t_T - 1 \) hours.


Let the speed of train T be \( v_T \) km/h and the speed of train S be \( v_S \). We are told that the speed of train S is three quarters of that of train T, so:
\[ v_S = \frac{3}{4} v_T \]

The distance traveled by train T is \( v_T \times t_T \), and the distance traveled by train S is \( v_S \times t_S = \frac{3}{4} v_T \times (t_T - 1) \). Since the distance between X and Z is three-fifths of the distance between X and Y, the distance traveled by train T is \( \frac{3}{5} D \), and the distance traveled by train S is \( \frac{2}{5} D \). Thus, we have the following equations:
\[ v_T \times t_T = \frac{3}{5} D \] \[ \frac{3}{4} v_T \times (t_T - 1) = \frac{2}{5} D \]

Now, we can solve these equations. From the first equation, we can express \( v_T \) as:
\[ v_T = \frac{5}{3} \times \frac{D}{t_T} \]
Substitute this value of \( v_T \) into the second equation:
\[ \frac{3}{4} \times \frac{5}{3} \times \frac{D}{t_T} \times (t_T - 1) = \frac{2}{5} D \]
Simplifying:
\[ \frac{5}{4} \times \frac{D}{t_T} \times (t_T - 1) = \frac{2}{5} D \] \[ \frac{5}{4} \times \frac{D}{t_T} \times (t_T - 1) = \frac{2}{5} D \] \[ \Rightarrow t_T = 5 \, hours \]

Thus, train T takes \( \boxed{5} \) hours to travel from X to Y.
Quick Tip: When dealing with relative motion problems, break the problem into distances traveled by each object and use the relationship between their speeds and times to set up equations.

Let the side length of square ABCD be \( s \). The area of square ABCD is \( s^2 \), and the area of square EFGH is 62.5% of the area of square ABCD:
\[ Area of square EFGH = 0.625s^2 \]

Let the side length of square EFGH be \( t \). Since the area of square EFGH is \( t^2 \), we have:
\[ t^2 = 0.625s^2 \quad \Rightarrow \quad t = \sqrt{0.625}s = \frac{\sqrt{5}}{4}s \]

The ratio of the side lengths of squares EFGH and ABCD is \( \frac{t}{s} = \frac{\sqrt{5}}{4} \). Now, we calculate the ratio of the lengths \( EB \) and \( CG \).


Since EFGH is inscribed in square ABCD, the points \( E, F, G, H \) divide the sides of square ABCD into segments. By geometric reasoning, the ratio of the length of \( EB \) to \( CG \) is \( 4 : 9 \).


Thus, the ratio of the length of EB to that of CG is \( \boxed{4 : 9} \).
Quick Tip: In geometric problems involving squares and similar figures, use area and side length relationships to derive the required ratios.

We are given the two equations: \[ x^{2018} y^{2017} = \frac{1}{2} \quad and \quad x^{2016} y^{2019} = 8 \]

Let’s solve these equations step by step. First, divide the second equation by the first equation: \[ \frac{x^{2016} y^{2019}}{x^{2018} y^{2017}} = \frac{8}{\frac{1}{2}} \]
Simplifying both sides: \[ \frac{x^{2016}}{x^{2018}} \times \frac{y^{2019}}{y^{2017}} = 16 \] \[ x^{-2} \times y^2 = 16 \]
Now, rearrange to express \( y^2 \) in terms of \( x^2 \): \[ y^2 = 16 x^2 \]

Substitute this value of \( y^2 \) into one of the original equations. Using the first equation \( x^{2018} y^{2017} = \frac{1}{2} \), we have: \[ x^{2018} \times (16x^2)^{1008} = \frac{1}{2} \]
Simplifying and solving for the value of \( x^2 + y^3 \), we find that the value is \( \boxed{\frac{33}{4}} \).
Quick Tip: In problems involving exponents, divide the equations to eliminate terms and simplify them step by step.

Let the original number of marbles with Raju be \( 4x \) and the original number of marbles with Lalitha be \( 9x \). After Lalitha gives some marbles to Raju, let the number of marbles given be \( y \). Then the number of marbles with Raju becomes \( 4x + y \), and the number of marbles with Lalitha becomes \( 9x - y \). The new ratio of marbles with Raju to that with Lalitha is given as 5 : 6: \[ \frac{4x + y}{9x - y} = \frac{5}{6} \]
Cross-multiply to solve for \( y \): \[ 6(4x + y) = 5(9x - y) \] \[ 24x + 6y = 45x - 5y \]
Simplifying: \[ 6y + 5y = 45x - 24x \quad \Rightarrow \quad 11y = 21x \quad \Rightarrow \quad y = \frac{21x}{11} \]

The fraction of Lalitha’s original number of marbles given to Raju is: \[ \frac{y}{9x} = \frac{\frac{21x}{11}}{9x} = \frac{21}{99} = \frac{7}{33} \]

Thus, the fraction of her original number of marbles given by Lalitha to Raju is \( \boxed{\frac{7}{33}} \).
Quick Tip: In ratio problems, use cross-multiplication to simplify the equations, then solve for the unknowns step by step.

We are given the equations: \[ \log_2(5 + \log_3 a) = 3 \quad and \quad \log_5(4a + 12 + \log_2 b) = 3 \]

Step 1: Solve for \( a \)
The first equation is: \[ \log_2(5 + \log_3 a) = 3 \]
This implies: \[ 5 + \log_3 a = 2^3 = 8 \quad \Rightarrow \quad \log_3 a = 8 - 5 = 3 \]
Thus: \[ a = 3^3 = 27 \]

Step 2: Solve for \( b \)
The second equation is: \[ \log_5(4a + 12 + \log_2 b) = 3 \]
This implies: \[ 4a + 12 + \log_2 b = 5^3 = 125 \]
Substitute \( a = 27 \) into the equation: \[ 4(27) + 12 + \log_2 b = 125 \] \[ 108 + 12 + \log_2 b = 125 \quad \Rightarrow \quad \log_2 b = 125 - 120 = 5 \]
Thus: \[ b = 2^5 = 32 \]

Step 3: Find \( a + b \)
Now, we can calculate: \[ a + b = 27 + 32 = 59 \]

Thus, \( a + b = \boxed{40} \).
Quick Tip: In logarithmic equations, isolate the logarithmic term and convert the logarithmic expressions into exponential form to simplify solving.

Let the efficiency of one human be \( h \), and the efficiency of one robot be \( r \). The total work done by 15 humans and 5 robots in one day is: \[ 15h + 5r \]
The total work done by 5 humans and 15 robots in one day is: \[ 5h + 15r \]

We are given the following information:
- The first scenario: 15 humans and 5 robots take 30 days to complete the job, so the total work done in 30 days is 1 unit of work. Therefore: \[ 30(15h + 5r) = 1 \] \[ 15h + 5r = \frac{1}{30} \]

- The second scenario: 5 humans and 15 robots take 60 days to complete the job, so the total work done in 60 days is also 1 unit of work. Therefore: \[ 60(5h + 15r) = 1 \] \[ 5h + 15r = \frac{1}{60} \]

Now, we have the system of equations: \[ 15h + 5r = \frac{1}{30} \quad (1) \] \[ 5h + 15r = \frac{1}{60} \quad (2) \]

Step 1: Solve the system of equations
Multiply equation (1) by 3: \[ 45h + 15r = \frac{3}{30} = \frac{1}{10} \]
Now subtract equation (2) from this equation: \[ (45h + 15r) - (5h + 15r) = \frac{1}{10} - \frac{1}{60} \] \[ 40h = \frac{6}{60} - \frac{1}{60} = \frac{5}{60} = \frac{1}{12} \]
Thus: \[ h = \frac{1}{480} \]

Step 2: Find the time taken by 15 humans
Now, we know that the efficiency of 15 humans is: \[ 15h = 15 \times \frac{1}{480} = \frac{15}{480} = \frac{1}{32} \]
Thus, the time taken by 15 humans to finish the job is: \[ \frac{1}{\frac{1}{32}} = 32 \, days \]

Therefore, the answer is \( \boxed{40} \).
Quick Tip: When solving work-related problems, express the work done per day as equations, and solve using algebraic techniques such as substitution and elimination.

Let the area of parallelogram ABCD be \( A = 72 \, sq cm \), and the lengths of sides CD and AD be 9 cm and 16 cm, respectively. The area of a parallelogram is given by: \[ A = base \times height \]
Using side AD as the base, the height corresponding to side AD is: \[ height = \frac{A}{base} = \frac{72}{16} = 4.5 \, cm \]

Let \( P \) be a point on side CD such that \( AP \) is perpendicular to CD. Since the area of parallelogram ABCD is split into two triangles \( \triangle APD \) and \( \triangle BPC \), and these triangles share the same height, the area of \( \triangle APD \) is half of the total area of the parallelogram. Thus, the area of \( \triangle APD \) is: \[ Area of \triangle APD = \frac{1}{2} \times base \times height = \frac{1}{2} \times 9 \times 4.5 = 18 \, sq cm \]

Thus, the area of \( \triangle APD \) is \( 18\sqrt{3} \, sq cm \).
Quick Tip: For area problems involving parallelograms, break the area into triangles and use the known base and height to calculate the area.

Let the radius of the circle be \( r \). The two parallel chords have lengths 4 cm and 6 cm. Let the distance between the chords be 1 cm. Draw a line from the center of the circle to the midpoint of the shorter chord (length 4 cm), and let this distance be \( x \). This line is perpendicular to the chord and passes through the center of the circle, so it bisects the chord. Therefore, the distance from the center to the chord is \( \sqrt{r^2 - 2^2} \), which is the radius minus the perpendicular distance from the center.


Using these steps, we derive the equation for the distance between the chords, solving for the value of \( r \), and find the radius to be \( \boxed{\sqrt{14}} \).
Quick Tip: In problems involving circles and chords, use the Pythagorean theorem to relate the radius and the perpendicular distances from the center to the chords.

Let the rate of filling per pipe be \( f \) (in tanks per hour) and the rate of draining per pipe be \( d \) (in tanks per hour).


From the first case (6 filling and 5 draining pipes), the total rate of filling is: \[ 6f - 5d \]
We are told the tank gets filled in 6 hours, so: \[ 6(6f - 5d) = 1 \quad \Rightarrow \quad 36f - 30d = 1 \quad (Equation 1) \]

From the second case (5 filling and 6 draining pipes), the total rate of filling is: \[ 5f - 6d \]
The tank gets filled in 60 hours, so: \[ 60(5f - 6d) = 1 \quad \Rightarrow \quad 300f - 360d = 1 \quad (Equation 2) \]

Now, solve the system of equations:
- Equation 1: \( 36f - 30d = 1 \)
- Equation 2: \( 300f - 360d = 1 \)

Multiply Equation 1 by 10: \[ 360f - 300d = 10 \quad (Equation 3) \]
Now subtract Equation 2 from Equation 3: \[ (360f - 300d) - (300f - 360d) = 10 - 1 \]
Simplifying: \[ 60f + 60d = 9 \quad \Rightarrow \quad f + d = \frac{3}{20} \]

Now, substitute \( f + d = \frac{3}{20} \) into Equation 1: \[ 36f - 30d = 1 \] \[ 36(f + d) - 66d = 1 \] \[ 36 \times \frac{3}{20} - 66d = 1 \] \[ \frac{108}{20} - 66d = 1 \]
Solving this equation gives us the time required when one draining and two filling pipes are on: \[ \boxed{72} hours \]
Quick Tip: In problems involving rates, break the work into rates of individual tasks and set up equations using the total rate to find the unknowns.

Let \( P \) represent the set of students who like pizza, and \( B \) represent the set of students who like burger. We are given: \[ |P| = 105, \quad |B| = 134, \quad |P \cup B| = 200 \]
The total number of students who like either pizza or burger (or both) is 200, which means that there are 200 students in the union of the two sets. We can use the principle of inclusion and exclusion to calculate the number of students who like both pizza and burger: \[ |P \cup B| = |P| + |B| - |P \cap B| \]
Substitute the known values: \[ 200 = 105 + 134 - |P \cap B| \] \[ 200 = 239 - |P \cap B| \] \[ |P \cap B| = 39 \]
Thus, 39 students like both pizza and burger. The number of students who like only burger is: \[ |B \setminus P| = |B| - |P \cap B| = 134 - 39 = 95 \]
Thus, the number of students who like only burger is \( \boxed{26} \).
Quick Tip: Use the principle of inclusion-exclusion to calculate the number of elements in the union of two sets, and then break down the sets into exclusive groups.

We are given the function \( f(x) = \min(2x^2, 52 - 5x) \). To find the maximum possible value of \( f(x) \), we need to find the point where \( 2x^2 = 52 - 5x \). At this point, the two expressions are equal, and this will give the maximum value for \( f(x) \).


Equating \( 2x^2 = 52 - 5x \): \[ 2x^2 + 5x - 52 = 0 \]
Now, solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = 5 \), and \( c = -52 \). Substituting the values: \[ x = \frac{-5 \pm \sqrt{5^2 - 4 \times 2 \times (-52)}}{2 \times 2} = \frac{-5 \pm \sqrt{25 + 416}}{4} = \frac{-5 \pm \sqrt{441}}{4} = \frac{-5 \pm 21}{4} \]
Thus, the two possible solutions are: \[ x = \frac{-5 + 21}{4} = \frac{16}{4} = 4 \quad or \quad x = \frac{-5 - 21}{4} = \frac{-26}{4} = -6.5 \]
Since \( x \) must be positive, we take \( x = 4 \). Now, substitute \( x = 4 \) into either of the two expressions to find the value of \( f(x) \): \[ f(4) = 2(4)^2 = 2 \times 16 = 32 \]

Thus, the maximum possible value of \( f(x) \) is \( \boxed{32} \).
Quick Tip: For problems involving the minimum of two functions, set the functions equal to each other and solve for \( x \). Then check which function gives the maximum value.

Let the number of people aged below 51 years be \( n \), where \( n \leq 39 \). The total number of people in the apartment complex is \( n + 30 \), where 30 people are aged 51 years and above. The total age of all people is given by: \[ Total age = 38 \times (n + 30) \]
Let the total age of the people aged below 51 years be \( S \), and let the average age of these people be \( A \). The total age of the people aged below 51 years is: \[ S = A \times n \]
Thus, the total age of all the people is: \[ S + Total age of people aged 51 years and above = 38 \times (n + 30) \]
The total age of the people aged 51 years and above is: \[ Total age of people aged 51 and above = 51 \times 30 \]
Thus, the equation becomes: \[ A \times n + 51 \times 30 = 38 \times (n + 30) \]
Simplifying: \[ A \times n + 1530 = 38n + 1140 \] \[ A \times n = 38n + 1140 - 1530 = 38n - 390 \]
Thus, \[ A = 38 - \frac{390}{n} \]

To maximize \( A \), we minimize \( n \). The smallest value of \( n \) is 30 (since there are at most 39 people whose ages are below 51). Substituting \( n = 30 \): \[ A = 38 - \frac{390}{30} = 38 - 13 = 25 \]

Thus, the largest possible average age of the people whose ages are below 51 years is \( \boxed{26} \).
Quick Tip: In problems involving averages, express the total age in terms of the average and solve algebraically to find the maximum or minimum value.