CAT 2024 Quant Slot 2 Paper

Ans: 2160

Explanation:
Let the purchase price of Bina be $B_{cost}$.
Let the selling price of Bina be $B_{sell} = 4860$.
Bina incurs a 19% loss, so her selling price is 100% – 19% = 81% of her purchase price.
$B_{sell} = 0.81 \times B_{cost}$
$4860 = 0.81 \times B_{cost}$
$B_{cost} = \frac{4860}{0.81} = \frac{486000}{81} = 6000$.
So, Bina’s purchase price was Rs. 6000.

Shyam’s purchase price is Bina’s selling price, which is Rs. 4860.
Let Shyam’s selling price be $S_{sell}$.
Let Hari’s purchase price be $H_{cost}$, which is equal to Shyam’s selling price, $S_{sell}$.

We are given that if Bina had sold the product to Shyam at Hari’s purchase price, she would have obtained a 17% profit.
This means that Hari’s purchase price ($H_{cost}$) is 17% more than Bina’s purchase price ($B_{cost}$).
$H_{cost} = B_{cost} + 0.17 \times B_{cost}$
$H_{cost} = 1.17 \times B_{cost}$
$H_{cost} = 1.17 \times 6000$
$H_{cost} = 7020$.

Since Shyam sells the product to Hari, Shyam’s selling price is Hari’s purchase price.
$S_{sell} = H_{cost} = 7020$.

Shyam’s purchase price is Rs. 4860.
Shyam’s selling price is Rs. 7020.
The profit made by Shyam is $S_{sell} – S_{cost}$.
Shyam’s profit = $7020 – 4860 = 2160$.

Final_Answer:2160

Ans: 2

Explanation:Let the vertices of the triangle be A = (1, 2), B = (7, 2), and C = (1, 10).
We first calculate the lengths of the sides of the triangle.
Side AB: The y-coordinates are the same (2), so this is a horizontal line. The length is the absolute difference of the x-coordinates: $|7 – 1| = 6$.
Side AC: The x-coordinates are the same (1), so this is a vertical line. The length is the absolute difference of the y-coordinates: $|10 – 2| = 8$.
Side BC: We use the distance formula: $\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$
$BC = \sqrt{(7 – 1)^2 + (2 – 10)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

The lengths of the sides are 6, 8, and 10. We notice that $6^2 + 8^2 = 36 + 64 = 100 = 10^2$. This means the triangle is a right-angled triangle, with the right angle at vertex A (where the sides of length 6 and 8 meet).

The area of a right-angled triangle is (1/2) * base * height.
Area = (1/2) * 6 * 8 = 24.

The semi-perimeter (s) of the triangle is half the sum of its side lengths.
$s = (6 + 8 + 10) / 2 = 24 / 2 = 12$.

The radius of the incircle (r) of a triangle can be found using the formula: Area = r * s.
So, $r = Area / s$.
$r = 24 / 12 = 2$.

Alternatively, for a right-angled triangle with legs ‘a’ and ‘b’ and hypotenuse ‘c’, the inradius is given by $r = (a + b – c) / 2$.
Here, $a = 6$, $b = 8$, and $c = 10$.
$r = (6 + 8 – 10) / 2 = (14 – 10) / 2 = 4 / 2 = 2$.

Final_Answer:2

Ans: 340

Let the total initial quantity of all fruits be denoted by ‘X’.
Suppose the initial count of mangoes is represented by ‘m’ and apples by ‘p’.
The initial quantity of mangoes is 40% of X, which can be expressed as 2X/5.
The sum of all fruits sold comprises mangoes, apples, and bananas.
This total can be written as: (2X/10) + 96 + (4p/10) = X/2 (as per the problem statement).
Simplifying the equation: X/5 + 96 + 2p/5 = X/2.
Rearranging to solve for X: X = (4p + 960) / 3.
This can also be written as (4p/3) + 320.
For the term (4p/3) to yield an integer, ‘p’ must be a multiple of 3.
Additionally, for (4p/10) to represent an integer count of apples sold, ‘p’ must be a multiple of 5.
The smallest positive integer value for ‘p’ that satisfies both divisibility requirements (by 3 and 5) is 15.
Substituting this smallest value back into the expression for X:
X = (4 * 15 / 3) + 320
X = (60 / 3) + 320
X = 20 + 320
X = 340.
Hence, the value 340 is the correct solution.

Ans: 14

The initial segment of the equation can be transformed as follows:

Start with:
$ \frac{\frac{1}{3}\log_2\left(a+b\right)}{\log_2c} $

Apply the rule $ m \log_x y = \log_x y^m $ to the numerator:
$ \frac{\log_2\left(a+b\right)^{\frac{1}{3}}}{\log_2c} $

Utilize the change of base formula for logarithms, $\frac{\log_x y}{\log_x z} = \log_z y$:
$ \log_c\left(a+b\right)^{\frac{1}{3}} $

Applying the same sequence of operations to the second term yields:
$ \log_c\left(a-b\right)^{\frac{1}{3}} $

The left side of the equation thus becomes:
$ \log_c\left(a+b\right)^{\frac{1}{3}}+\log_c\left(a-b\right)^{\frac{1}{3}} $

Using the property $\log_x y + \log_x z = \log_x (yz)$:
$ \log_c\left[\left(a+b\right)^{\frac{1}{3}}\left(a-b\right)^{\frac{1}{3}}\right] $

Combine the terms within the exponent:
$ \log_c\left[\left(a+b\right)\left(a-b\right)\right]^{\frac{1}{3}} $

Simplify the expression inside the brackets:
$ \log_c\left[\left(a^2-b^2\right)\right]^{\frac{1}{3}} $

This expression is stated to be equal to $\frac{2}{3}$.

By the definition of logarithms, if $\log_N M = x$, then $N^x = M$. Applying this here:
$ c^{\frac{2}{3}}=\left(a^2-b^2\right)^{\frac{1}{3}} $

Cube both sides of the equation:
$ \left(c^{\frac{2}{3}}\right)^3 = \left[\left(a^2-b^2\right)^{\frac{1}{3}}\right]^3 $
$ c^2 = a^2-b^2 $

Rearrange to solve for $a^2$:
$ a^2 = b^2+c^2 $

We are given that $b \le 10$ and $c \le 10$, and we aim to maximize $a^2$.
This relationship resembles the Pythagorean theorem. To maximize $a^2$, we should maximize $b$ and $c$. Setting $b=10$ and $c=10$ gives $a^2 = 10^2 + 10^2 = 100 + 100 = 200$.

However, we need to find an integer value for $a$. We look for the largest perfect square less than 200. This is $196$, which is $14^2$.
Therefore, the maximum integer value for $a$ is 14.

Ans: A

Explanation:The given functional equation is $f(xy) = f(x)f(y) + f(x) + f(y)$.
We can add 1 to both sides of the equation to get:
$f(xy) + 1 = f(x)f(y) + f(x) + f(y) + 1$
$f(xy) + 1 = (f(x) + 1)(f(y) + 1)$

Let $g(x) = f(x) + 1$. Then the functional equation becomes:
$g(xy) = g(x)g(y)$
This means that $g$ is a completely multiplicative function.

We are given that $f(p) = 1$ for every prime number $p$.
So, $g(p) = f(p) + 1 = 1 + 1 = 2$ for every prime number $p$.

Now we need to find the value of $f(160000)$.
First, let’s find the prime factorization of 160000.
$160000 = 16 \times 10000 = 2^4 \times (10^2)^2 = 2^4 \times 10^4 = 2^4 \times (2 \times 5)^4 = 2^4 \times 2^4 \times 5^4 = 2^8 \times 5^4$.

Now we can find $g(160000)$:
Since $g$ is completely multiplicative, $g(160000) = g(2^8 \times 5^4) = g(2^8) \times g(5^4)$.
Also, $g(a^n) = (g(a))^n$.
So, $g(2^8) = (g(2))^8$ and $g(5^4) = (g(5))^4$.

Since 2 and 5 are prime numbers, we have $g(2) = 2$ and $g(5) = 2$.
Therefore, $g(2^8) = 2^8$ and $g(5^4) = 2^4$.

Now, $g(160000) = g(2^8) \times g(5^4) = 2^8 \times 2^4 = 2^{8+4} = 2^{12}$.

We have $g(160000) = 2^{12}$.
We know that $g(x) = f(x) + 1$, so $f(x) = g(x) – 1$.
Therefore, $f(160000) = g(160000) – 1 = 2^{12} – 1$.

Let’s calculate $2^{12}$:
$2^{10} = 1024$
$2^{11} = 1024 \times 2 = 2048$
$2^{12} = 2048 \times 2 = 4096$

So, $f(160000) = 4096 – 1 = 4095$.

Comparing this value with the given options:
Option A: 4095
Option B: 8191
Option C: 2047
Option D: 1023

The calculated value matches Option A.

Correct_Option:A

Ans: B

Explanation:Given the quadratic equation $3x^2 + \lambda x – 1 = 0$.
Let the roots of this equation be $\alpha$ and $\beta$.
From Vieta’s formulas, we have:
Sum of roots: $\alpha + \beta = -\frac{\lambda}{3}$
Product of roots: $\alpha \beta = \frac{-1}{3}$

We are given the condition $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15$.
We can rewrite the left side of this equation:
$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} = \frac{(\alpha + \beta)^2 – 2\alpha \beta}{(\alpha \beta)^2}$

Substitute the values of $\alpha + \beta$ and $\alpha \beta$ from Vieta’s formulas into this expression:
$\frac{(-\frac{\lambda}{3})^2 – 2(\frac{-1}{3})}{(\frac{-1}{3})^2} = 15$
$\frac{\frac{\lambda^2}{9} + \frac{2}{3}}{\frac{1}{9}} = 15$
Multiply the numerator by 9:
$(\frac{\lambda^2}{9} + \frac{2}{3}) \times 9 = 15$
$\lambda^2 + 6 = 15$
$\lambda^2 = 15 – 6$
$\lambda^2 = 9$

Now we need to find the value of $(\alpha^3 + \beta^3)^2$.
We know the identity $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 – \alpha \beta + \beta^2)$.
We can also write $\alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha \beta$.
So, $\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 – 2\alpha \beta – \alpha \beta) = (\alpha + \beta)((\alpha + \beta)^2 – 3\alpha \beta)$.

Substitute the values of $\alpha + \beta$ and $\alpha \beta$:
$\alpha^3 + \beta^3 = (-\frac{\lambda}{3})((-\frac{\lambda}{3})^2 – 3(\frac{-1}{3}))$
$\alpha^3 + \beta^3 = (-\frac{\lambda}{3})(\frac{\lambda^2}{9} + 1)$

We found that $\lambda^2 = 9$.
So, $\alpha^3 + \beta^3 = (-\frac{\lambda}{3})(\frac{9}{9} + 1)$
$\alpha^3 + \beta^3 = (-\frac{\lambda}{3})(1 + 1)$
$\alpha^3 + \beta^3 = -\frac{2\lambda}{3}$

Now we need to find $(\alpha^3 + \beta^3)^2$:
$(\alpha^3 + \beta^3)^2 = (-\frac{2\lambda}{3})^2 = \frac{4\lambda^2}{9}$

Substitute $\lambda^2 = 9$:
$(\alpha^3 + \beta^3)^2 = \frac{4 \times 9}{9} = 4$

The value of $(\alpha^3 + \beta^3)^2$ is 4.

The options are:
Option A: 16
Option B: 4
Option C: 1
Option D: 9

Our calculated value is 4, which matches Option B.

Correct_Option:B

Ans: C

Explanation:Let R be the present age of Rajesh and G be the present age of Garima.
Let the age of Rajesh at that time be $R_1$ and the age of Garima at that time be $G_1$.
The problem states that when Rajesh’s age was the same as the present age of Garima, i.e., $R_1 = G$.
At that time, the ratio of their ages was 3 : 2. So, $\frac{R_1}{G_1} = \frac{3}{2}$.
Since $R_1 = G$, we can substitute this into the ratio equation: $\frac{G}{G_1} = \frac{3}{2}$.
This means $2G = 3G_1$, or $G_1 = \frac{2}{3}G$.

The difference in their ages is constant. The difference in their ages is $R – G$.
The difference in their ages at the time $R_1$ and $G_1$ was $R_1 – G_1$.
So, $R – G = R_1 – G_1$.
Substitute $R_1 = G$ and $G_1 = \frac{2}{3}G$:
$R – G = G – \frac{2}{3}G$
$R – G = \frac{1}{3}G$
$R = G + \frac{1}{3}G$
$R = \frac{4}{3}G$

Now consider the second part of the problem. When Garima’s age becomes the same as the present age of Rajesh, i.e., Garima’s age is R.
Let Garima’s age at that future time be $G_2 = R$.
The time elapsed for Garima to reach age R is $G_2 – G = R – G$.
The same amount of time will have passed for Rajesh. So, Rajesh’s age at that future time will be $R_2 = R + (R – G)$.
We know that $R = \frac{4}{3}G$. Substitute this into the expression for $R_2$:
$R_2 = \frac{4}{3}G + (\frac{4}{3}G – G)$
$R_2 = \frac{4}{3}G + \frac{1}{3}G$
$R_2 = \frac{5}{3}G$

The ratio of the ages of Rajesh and Garima at that future time will be $\frac{R_2}{G_2}$.
We have $R_2 = \frac{5}{3}G$ and $G_2 = R = \frac{4}{3}G$.
The ratio is $\frac{R_2}{G_2} = \frac{\frac{5}{3}G}{\frac{4}{3}G} = \frac{5/3}{4/3} = \frac{5}{4}$.
So, the ratio of the ages of Rajesh and Garima will become 5 : 4.

Correct_Option:C

Ans: A

To be published

Ans: A

Explanation:ABCD is a trapezium with AB || CD. AD and BC intersect at E.
Given AB = 2 cm, CD = 1 cm.
Perimeter of ABCD = AB + BC + CD + AD = 6 cm.
So, 2 + BC + 1 + AD = 6.
BC + AD = 6 – 2 – 1 = 3 cm.

Since AB || CD, by the property of similar triangles, $\triangle AEB \sim \triangle DEC$.
The ratio of corresponding sides is equal:
$\frac{AE}{DE} = \frac{BE}{CE} = \frac{AB}{CD}$
$\frac{AE}{DE} = \frac{BE}{CE} = \frac{2}{1}$

From $\frac{AE}{DE} = 2$, we have AE = 2 * DE.
Also, AE = AD + DE.
So, AD + DE = 2 * DE, which implies AD = DE.

From $\frac{BE}{CE} = 2$, we have BE = 2 * CE.
Also, BE = BC + CE.
So, BC + CE = 2 * CE, which implies BC = CE.

Now we know AD = DE and BC = CE.
From BC + AD = 3 cm, we substitute BC with CE and AD with DE.
So, CE + DE = 3 cm.

We need to find the perimeter of $\triangle AEB$.
Perimeter of $\triangle AEB$ = AE + BE + AB.
We know AB = 2 cm.

From AE = AD + DE and AD = DE, we have AE = DE + DE = 2 * DE.
From BE = BC + CE and BC = CE, we have BE = CE + CE = 2 * CE.

So, Perimeter of $\triangle AEB$ = 2 * DE + 2 * CE + 2.
Perimeter of $\triangle AEB$ = 2 * (DE + CE) + 2.
We know that DE + CE = 3 cm.
Perimeter of $\triangle AEB$ = 2 * (3) + 2.
Perimeter of $\triangle AEB$ = 6 + 2.
Perimeter of $\triangle AEB$ = 8 cm.

Therefore, the perimeter of $\triangle AEB$ is 8 cm.

The final answer is $\boxed{8}$.
Correct_Option: A

Ans: A

We can categorize the list into four segments: the initial 10 items as ‘a’, the subsequent 10 as ‘b’, the next 10 as ‘c’, and the final 10 as ‘d’.

Based on the provided relationships, we can establish the following algebraic expressions:
$ \frac{a+b+c}{3} = 40,000 $
$ \frac{b+c+d}{3} = 60,000 $
$ \frac{a+d}{2} = 50,000 $

Combining the first two expressions yields:
$ a + 2(b+c) + d = 300,000 $

Substituting the value of $a+d$ from the third expression ($a+d = 100,000$) into the combined equation allows us to determine $b+c$:
$ 100,000 + 2(b+c) = 300,000 $
$ 2(b+c) = 200,000 $
$ b+c = 100,000 $

Now, using the value of $b+c$ in the first and second original expressions, we can solve for ‘a’ and ‘d’:
For the first expression: $a + 100,000 = 3 \times 40,000 \implies a = 120,000 – 100,000 = 20,000$.
For the second expression: $100,000 + d = 3 \times 60,000 \implies d = 180,000 – 100,000 = 80,000$.

We are informed that the average of the first 10 items experiences a 100% increase, meaning it changes from $20,000$ to $40,000$.
The average of the last 10 items sees a 200% increase, changing from $80,000$ to $240,000$.

The sum of all four segments is $40,000 + 100,000 + 240,000 = 380,000$.
The average across these four segments is $\frac{380,000}{4} = 95,000$.

Thus, Option A represents the correct solution.

Ans: 139

Let the individual work outputs of Amal, Vimal, and Sunil be represented by A, V, and S, respectively.
Assume the total task to be completed is denoted by T.
The problem provides the following relationships:
150A + 150V = T      …(1)
100V + 100S = T      …(2)
75A + 135V + 45S = T    …(3)

Summing equations (1) and (2) yields: 150A + 250V + 100S = 2T   …(4)
Multiplying equation (3) by 2 gives: 150A + 270V + 100S = 2T  …(5)

Subtracting equation (5) from equation (4) results in 10S = 20V, which simplifies to S = 2V.
Substituting S = 2V into equation (2) allows us to express the total work T in terms of V: T = 100V + 100(2V) = 300V.
Now, substituting T = 300V into equation (1) gives 150A + 150V = 300V, leading to 150A = 150V, or A = V.
Thus, the daily work rates of Amal, Vimal, and Sunil are equivalent to V, V, and 2V units per day, respectively.

The question involves work cycles. To simplify calculations, we can consider a time frame that is the least common multiple (LCM) of the periods mentioned. In this scenario, the LCM of the given periods is 6.

Over a 6-day period:
Amal completes 6 * V = 6V units of work.
Vimal completes 3 * V = 3V units of work.
Sunil completes 2 * 2V = 4V units of work.

Therefore, in one 6-day cycle, a total of 6V + 3V + 4V = 13V units of work are accomplished.
To determine the number of full cycles to complete the total work T (which is 300V), we divide the total work by the work done per cycle: 300V / 13V = 23 with a remainder.
This means that 23 full cycles are completed, accounting for 23 * 13V = 299V units of work.
The duration of these 23 cycles is 23 * 6 days = 138 days.

The remaining 1V unit of work will be completed on the subsequent day.
Consequently, the total time required is 138 days + 1 day = 139 days.

Ans: A

To be published

Ans: D

Let the sum contributed by Sunil be denoted by ‘S’.
The sum obtained by Anil after 6 years, assuming a compounded interest rate of 4% per annum, compounded semi-annually, can be calculated as:
$22000\left(1+\frac{4}{2\times\ 100}\right)^{6\times\ 2}=22000\left(1.02\right)^{12}$
The sum obtained by Sunil after 5 years, assuming the same compounded interest rate, would be:
$S\left(1.02\right)^{10}$
In the sixth year, Sunil then applies a simple interest rate of 10% to the sum he possesses at the end of 5 years. The interest earned during this year is:
$S\left(1.02\right)^{10}\times\ 0.1$
Thus, the total sum held by Sunil at the conclusion of 6 years is:
$S\left(1.02\right)^{10}\times\ \left(1+0.1\right)$
By setting the final sums held by Sunil and Anil equal, we establish the following equation:
$S\left(1.02\right)^{10}\times\ \left(1.1\right)=22000\left(1.02\right)^{12}$
Solving for S:
$S=\frac{22000\left(1.02\right)^2}{1.1}=20808$
Consequently, the correct option is D.

Ans: A

Let’s denote the typical travel duration of the bus as $t$.

Based on the first scenario (bus traveling at 60 km/h), the total distance covered can be expressed as $60(t + 3.5)$.

The second scenario indicates that the bus covered two-thirds of its usual journey distance in one-third of its usual travel time. This implies that the remaining one-third of the distance was covered in the remaining two-thirds of the usual travel time.

Specifically, if $\frac{1}{3}$ of the distance is covered in $\frac{2}{3}t$ time, then the speed during this segment is $\frac{\frac{1}{3} \text{distance}}{\frac{2}{3}t} = \frac{1}{2} \times \frac{\text{distance}}{t}$. Since the usual speed is $\frac{\text{distance}}{t}$, this speed is half of the usual speed.

We are given that this speed is 40 km/h. Therefore, half of the usual speed is 40 km/h, which means the usual speed of the bus is $2 \times 40 = 80$ km/hr.

Now, using the relationship from the first scenario, we have $60(t + 3.5) = 80t$.
Solving for $t$:
$60t + 210 = 80t$
$210 = 20t$
$t = \frac{210}{20} = 10.5$ hours.

Thus, the bus typically completes its journey in 10.5 hours.

If the journey begins at 9:00 AM, it will conclude at 7:30 PM.
Consequently, Option A is the correct choice.

Ans: D

Explanation:Given the equation $m^n = 2^{25} \times 3^{40}$, where $m$ and $n$ are natural numbers and $n > 1$.
We need to express the right-hand side in the form of a power. To do this, we can use the property of exponents $(a^x)^y = a^{xy}$.
We can rewrite the exponents of the prime factors 2 and 3 so that they are divisible by $n$.
Let’s find the greatest common divisor (GCD) of the exponents 25 and 40.
GCD(25, 40) = 5.
This suggests that $n$ could be a factor of 5. Since $n$ is a natural number and $n > 1$, $n$ can be 5.
If $n=5$, we can rewrite the equation as:
$m^5 = 2^{25} \times 3^{40}$
$m^5 = (2^{25/5})^5 \times (3^{40/5})^5$
$m^5 = (2^5)^5 \times (3^8)^5$
Using the property $(a \times b)^x = a^x \times b^x$, we can combine the terms:
$m^5 = (2^5 \times 3^8)^5$
Therefore, $m = 2^5 \times 3^8$.

Now we calculate the values of $m$ and $n$:
$n = 5$
$m = 2^5 \times 3^8 = 32 \times (3^4)^2 = 32 \times (81)^2 = 32 \times 6561$
$m = 32 \times 6561 = 209952$

We need to find $m – n$.
$m – n = 209952 – 5 = 209947$.

Let’s consider if there are other possible values for $n$. $n$ must be a common divisor of 25 and 40, and $n>1$. The common divisors of 25 and 40 are 1 and 5. Since $n>1$, the only possible value for $n$ is 5.

If we tried to express $m^n = 2^{25} \times 3^{40}$ in other ways, say if $n$ was not a common divisor of 25 and 40, we would have fractional exponents for $m$, which is not allowed for natural numbers. For example, if $n=25$, $m = (2^{25} \times 3^{40})^{1/25} = 2 \times 3^{40/25} = 2 \times 3^{8/5}$, which is not a natural number.
Similarly, if $n=40$, $m = (2^{25} \times 3^{40})^{1/40} = 2^{25/40} \times 3 = 2^{5/8} \times 3$, which is not a natural number.
Thus, $n$ must be a common divisor of the exponents, and the largest possible $n$ that allows $m$ to be a natural number is the GCD of the exponents if the exponents are written in their lowest terms relative to $n$.

In this case, $n$ must be a common divisor of 25 and 40. The common divisors are 1 and 5. Since $n > 1$, $n$ must be 5.
$m^5 = 2^{25} \times 3^{40}$
$m = (2^{25} \times 3^{40})^{1/5} = 2^{25/5} \times 3^{40/5} = 2^5 \times 3^8$
$m = 32 \times 6561 = 209952$
$n = 5$
$m – n = 209952 – 5 = 209947$.

Correct_Option: D

Ans: A

Explanation:We want to find the remainder when $3^{333}$ is divided by 11. This is a problem of modular arithmetic. We can use Fermat’s Little Theorem, which states that if $p$ is a prime number, then for any integer $a$ not divisible by $p$, we have $a^{p-1} \equiv 1 \pmod{p}$.
In this case, $p=11$ (which is prime) and $a=3$ (which is not divisible by 11).
According to Fermat’s Little Theorem, $3^{11-1} \equiv 3^{10} \equiv 1 \pmod{11}$.

Now we need to find the remainder of $3^{333}$ when divided by 11. We can write the exponent 333 in terms of multiples of 10.
$333 = 10 \times 33 + 3$.

So, we can write $3^{333}$ as $3^{10 \times 33 + 3}$.
Using the properties of exponents, this is $(3^{10})^{33} \times 3^3$.

Now we can apply the modular arithmetic:
$3^{333} \equiv (3^{10})^{33} \times 3^3 \pmod{11}$.
Since $3^{10} \equiv 1 \pmod{11}$, we substitute this into the expression:
$3^{333} \equiv (1)^{33} \times 3^3 \pmod{11}$.
$3^{333} \equiv 1 \times 3^3 \pmod{11}$.
$3^{333} \equiv 3^3 \pmod{11}$.

Now, we calculate $3^3$:
$3^3 = 3 \times 3 \times 3 = 9 \times 3 = 27$.

So, we need to find the remainder of 27 when divided by 11.
$27 = 11 \times 2 + 5$.
The remainder is 5.

Therefore, $3^{333} \equiv 5 \pmod{11}$.

The remainder is 5.

Let’s check the options:
Option A: 5
Option B: 10
Option C: 1
Option D: 6

The calculated remainder is 5, which corresponds to Option A.

Correct_Option:A

Ans: 7

For this type of problem, the strategy is to aim for completing the squares.
Observe the $xy$ term. This suggests creating a squared expression that includes both x and y.
Given that there’s only one other term involving x, it should also be fully contained within this squared term.
Consider the expansion: $\left(2x-y\right)^{^2}=4x^2+y^2-4xy$
Substituting this into the provided equation leaves us with $\left(2x-y\right)^{^2}+3y^2+3-6y$
This expression can be rearranged as $\left(2x-y\right)^{^2}+3\left(y^2+1-2y\right)$
Further simplification leads to $\left(2x-y\right)^{^2}+3\left(y-1\right)^2=0$
Since the sum of two squared terms equals zero, the only way this can be true is if each squared term is individually zero.
From the second term, we deduce that $y=1$.
Substituting this value into the first term allows us to find $x = 1/2$.
Consequently, the value of $4x+5y$ is calculated as $2+5 = 7$.
Therefore, the correct result is 7.

Ans: 11

By raising both expressions to the power of two, we obtain:
$x+6\sqrt{\ 2}+x-6\sqrt{\ 2}-2\left(x^2-72\right)^{\frac{1}{2}}=8$
$x-\left(x^2-72\right)^{\frac{1}{2}}=4$
Relocating $x$ to the opposite side yields:
$-\left(x^2-72\right)^{\frac{1}{2}}=4-x$
Performing another squaring operation on both sides results in:
$x^2-72=16+x^2-8x$
$8x=88$
$x=11$
Hence, 11 represents the accurate solution.

Ans: 7

To be published

Ans: B

Explanation:We are given two equations:
1) $|x| + x + y = 15$
2) $x + |y| – y = 20$

We can analyze these equations by considering different cases for the signs of x and y.

Case 1: $x \ge 0$ and $y \ge 0$
Equation 1 becomes: $x + x + y = 15 \implies 2x + y = 15$
Equation 2 becomes: $x + y – y = 20 \implies x = 20$
Substitute $x=20$ into $2x+y=15$: $2(20) + y = 15 \implies 40 + y = 15 \implies y = -25$.
This contradicts our assumption that $y \ge 0$. So, this case is not possible.

Case 2: $x < 0$ and $y \ge 0$
Equation 1 becomes: $-x + x + y = 15 \implies y = 15$.
Equation 2 becomes: $x + y – y = 20 \implies x = 20$.
This contradicts our assumption that $x < 0$. So, this case is not possible.

Case 3: $x \ge 0$ and $y < 0$
Equation 1 becomes: $x + x + y = 15 \implies 2x + y = 15$.
Equation 2 becomes: $x + (-y) – y = 20 \implies x – 2y = 20$.
We have a system of two linear equations:
a) $2x + y = 15$
b) $x – 2y = 20$
From equation (a), $y = 15 – 2x$. Substitute this into equation (b):
$x – 2(15 – 2x) = 20$
$x – 30 + 4x = 20$
$5x = 50$
$x = 10$
Now substitute $x=10$ back into $y = 15 – 2x$:
$y = 15 – 2(10) = 15 – 20 = -5$.
This case satisfies our assumptions: $x=10 \ge 0$ and $y=-5 < 0$.
Now we need to find $x – y$:
$x – y = 10 – (-5) = 10 + 5 = 15$.

Case 4: $x < 0$ and $y < 0$
Equation 1 becomes: $-x + x + y = 15 \implies y = 15$.
This contradicts our assumption that $y < 0$. So, this case is not possible.

The only valid solution is $x = 10$ and $y = -5$.
Then, $x – y = 10 – (-5) = 15$.

Correct_Option: B

Ans: B

Explanation:Let A be the amount of acid and W be the amount of water in the original solution.
The total volume of the original solution is A + W.

Step 1: When 2 litres of water was added.
The new amount of acid is A.
The new amount of water is W + 2.
The new total volume is A + W + 2.
The new solution had 50% acid concentration.
So, A / (A + W + 2) = 50/100 = 1/2.
This gives us the equation: 2A = A + W + 2.
Simplifying this, we get: A = W + 2. (Equation 1)

Step 2: When 15 litres of acid was further added to this new solution.
The amount of acid in the new solution from Step 1 was A.
After adding 15 litres of acid, the amount of acid becomes A + 15.
The amount of water in the new solution from Step 1 was W + 2.
The final total volume is (A + 15) + (W + 2) = A + W + 17.
The final solution had 80% acid concentration.
So, (A + 15) / (A + W + 17) = 80/100 = 4/5.
This gives us the equation: 5(A + 15) = 4(A + W + 17).
Expanding this, we get: 5A + 75 = 4A + 4W + 68.
Simplifying this, we get: A + 7 = 4W. (Equation 2)

Step 3: Solve the system of equations.
We have two equations:
1) A = W + 2
2) A + 7 = 4W

Substitute Equation 1 into Equation 2:
(W + 2) + 7 = 4W
W + 9 = 4W
9 = 3W
W = 3 litres.

Now substitute the value of W back into Equation 1 to find A:
A = W + 2
A = 3 + 2
A = 5 litres.

Step 4: Find the ratio of water and acid in the original solution.
The original amount of water was W = 3 litres.
The original amount of acid was A = 5 litres.
The ratio of water to acid in the original solution is W : A = 3 : 5.

Let’s check the options:
Option A: 5 : 3 (Acid : Water)
Option B: 3 : 5 (Water : Acid)
Option C: 5 : 4 (Acid : Water)
Option D: 4 : 5 (Water : Acid)

The question asks for the ratio of water and acid in the original solution, which is Water : Acid.
Our calculated ratio is 3 : 5.

Correct_Option:B

Ans: B

Explanation:Let the given infinite series be denoted by $S$.
The series is given by:
$S = \cfrac{1}{5}\left(\cfrac{1}{5} – \cfrac{1}{7}\right) + \left(\cfrac{1}{5}\right)^2 \left(\left(\cfrac{1}{5}\right)^2 – \left(\cfrac{1}{7}\right)^2\right) + \left(\cfrac{1}{5}\right)^3 \left(\left(\cfrac{1}{5}\right)^3 – \left(\cfrac{1}{7}\right)^3\right) + \dots$
We can rewrite the terms of the series. Let $a = \frac{1}{5}$ and $b = \frac{1}{7}$. The series becomes:
$S = a(a-b) + a^2(a^2-b^2) + a^3(a^3-b^3) + \dots$
Expanding the terms:
$S = (a^2 – ab) + (a^4 – a^2b^2) + (a^6 – a^3b^3) + \dots$
We can split this into two separate infinite series:
$S = (a^2 + a^4 + a^6 + \dots) – (ab + a^2b^2 + a^3b^3 + \dots)$
The first part is a geometric series with first term $a^2$ and common ratio $a^2$. Since $a = \frac{1}{5}$, $a^2 = \frac{1}{25}$. The sum of this geometric series is:
$S_1 = \frac{a^2}{1-a^2} = \frac{\left(\frac{1}{5}\right)^2}{1-\left(\frac{1}{5}\right)^2} = \frac{\frac{1}{25}}{1-\frac{1}{25}} = \frac{\frac{1}{25}}{\frac{24}{25}} = \frac{1}{24}$
The second part is a geometric series with first term $ab$ and common ratio $ab$. Since $a = \frac{1}{5}$ and $b = \frac{1}{7}$, $ab = \frac{1}{5} \times \frac{1}{7} = \frac{1}{35}$. The sum of this geometric series is:
$S_2 = \frac{ab}{1-ab} = \frac{\frac{1}{35}}{1-\frac{1}{35}} = \frac{\frac{1}{35}}{\frac{34}{35}} = \frac{1}{34}$
Therefore, the sum of the given series is:
$S = S_1 – S_2 = \frac{1}{24} – \frac{1}{34}$
To subtract these fractions, we find a common denominator, which is $24 \times 34 = 816$.
$S = \frac{34}{816} – \frac{24}{816} = \frac{34-24}{816} = \frac{10}{816}$
We can simplify this fraction by dividing both the numerator and the denominator by 2:
$S = \frac{5}{408}$

Alternatively, we can rewrite the general term of the series as:
$T_n = \left(\frac{1}{5}\right)^n \left(\left(\frac{1}{5}\right)^n – \left(\frac{1}{7}\right)^n\right) = \left(\frac{1}{5}\right)^{2n} – \left(\frac{1}{5}\right)^n \left(\frac{1}{7}\right)^n = \left(\frac{1}{25}\right)^n – \left(\frac{1}{35}\right)^n$
The sum of the series is:
$S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \left(\frac{1}{25}\right)^n – \sum_{n=1}^{\infty} \left(\frac{1}{35}\right)^n$
These are two infinite geometric series.
The first series is $\sum_{n=1}^{\infty} \left(\frac{1}{25}\right)^n = \frac{\frac{1}{25}}{1-\frac{1}{25}} = \frac{\frac{1}{25}}{\frac{24}{25}} = \frac{1}{24}$.
The second series is $\sum_{n=1}^{\infty} \left(\frac{1}{35}\right)^n = \frac{\frac{1}{35}}{1-\frac{1}{35}} = \frac{\frac{1}{35}}{\frac{34}{35}} = \frac{1}{34}$.
So, $S = \frac{1}{24} – \frac{1}{34} = \frac{34 – 24}{24 \times 34} = \frac{10}{816} = \frac{5}{408}$.

Comparing with the given options:
Option A: $\cfrac{7}{816}$
Option B: $\cfrac{5}{408}$
Option C: $\cfrac{7}{408}$
Option D: $\cfrac{5}{816}$
Our calculated sum is $\frac{5}{408}$, which matches Option B.

The final answer is $\boxed{\text{\cfrac{5}{408}}}$.
Correct_Option:B