CAT 2024 Quant Slot 3 Paper

Ans: C

The total student population is 250. The proportion of girls falls within the range of 44% to 60%.
This translates to the following range for the number of girls:
$0.44 \times 250 \le \text{Girls} \le 0.6 \times 250$
$110 \le \text{Girls} \le 150$

Condition 1:
Given a total of B boys and G girls, where $B+G=250$.
If 50% of boys and 80% of girls participate in swimming, the total number of swimming participants is $0.5B + 0.8G$.

Condition 2:
If 70% of boys and 60% of girls participate in running, the total number of running participants is $0.7B + 0.6G$.

The aggregate count of enrollments for both swimming and running is the sum of participants in each activity:
$(0.7B + 0.6G) + (0.5B + 0.8G) = 1.2B + 1.4G$

Employing the principle of inclusion-exclusion, let ‘I’ represent the number of students enrolled in exactly one activity, and ‘II’ represent the number of students enrolled in both activities.
We know that the total number of students is the sum of those in one activity and those in two:
$I + II = 250 = B + G$
The sum of the individual activity counts is equal to those in one activity plus twice those in both:
$I + 2II = 1.2B + 1.4G$

Subtracting the first equation from the second yields:
$II = (1.2B + 1.4G) – (B + G)$
$II = 0.2B + 0.4G$
This can be rewritten as:
$II = 0.2(B + 2G)$
Substituting $B = 250 – G$:
$II = 0.2((250 – G) + 2G)$
$II = 0.2(250 + G)$

Given that the number of girls (G) is between 110 and 150, we can determine the range for ‘II’:
The maximum value of ‘II’ occurs when G is maximum (150):
$\text{Maximum } II = 0.2(250 + 150) = 0.2(400) = 80$
The minimum value of ‘II’ occurs when G is minimum (110):
$\text{Minimum } II = 0.2(250 + 110) = 0.2(360) = 72$

Ans: B

Explanation:We are given the equation $(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}$, where a and b are natural numbers.
Let’s expand the left side of the equation:
$(a + b\sqrt{3})^2 = a^2 + 2(a)(b\sqrt{3}) + (b\sqrt{3})^2$
$(a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + b^2(3)$
$(a + b\sqrt{3})^2 = a^2 + 3b^2 + 2ab\sqrt{3}$

Now, we equate this to the right side of the given equation:
$a^2 + 3b^2 + 2ab\sqrt{3} = 52 + 30\sqrt{3}$

Since a and b are natural numbers, we can equate the rational and irrational parts of the equation:
1. Rational part: $a^2 + 3b^2 = 52$
2. Irrational part: $2ab\sqrt{3} = 30\sqrt{3}$

From the irrational part, we can divide both sides by $\sqrt{3}$:
$2ab = 30$
Dividing by 2, we get:
$ab = 15$

Now we need to find natural numbers a and b that satisfy both $ab = 15$ and $a^2 + 3b^2 = 52$.
The pairs of natural numbers (a, b) such that $ab = 15$ are:
(1, 15), (3, 5), (5, 3), (15, 1).

Let’s test each pair in the equation $a^2 + 3b^2 = 52$:

Case 1: a = 1, b = 15
$a^2 + 3b^2 = 1^2 + 3(15^2) = 1 + 3(225) = 1 + 675 = 676 \neq 52$

Case 2: a = 3, b = 5
$a^2 + 3b^2 = 3^2 + 3(5^2) = 9 + 3(25) = 9 + 75 = 84 \neq 52$

Case 3: a = 5, b = 3
$a^2 + 3b^2 = 5^2 + 3(3^2) = 25 + 3(9) = 25 + 27 = 52$
This pair satisfies both equations.

Case 4: a = 15, b = 1
$a^2 + 3b^2 = 15^2 + 3(1^2) = 225 + 3(1) = 225 + 3 = 228 \neq 52$

So, the natural numbers are $a = 5$ and $b = 3$.
We are asked to find $a + b$.
$a + b = 5 + 3 = 8$

The value of $a+b$ is 8.

The options are:
Option A: 7
Option B: 8
Option C: 9
Option D: 10

Our calculated value matches Option B.

Correct_Option:B

Ans: 70

Explanation:Let the three distinct real numbers in ascending order be $a$, $b$, and $c$.
The average of these three numbers is 28.
So, $\frac{a+b+c}{3} = 28$
$a+b+c = 28 \times 3 = 84$ (Equation 1)

The smallest number is increased by 7, so the new smallest number is $a+7$.
The largest number is reduced by 10, so the new largest number is $c-10$.
The middle number remains $b$.
The order of the numbers remains unchanged. This means:
$a+7 < b < c-10$

The new arithmetic mean is 2 more than the middle number.
The new arithmetic mean is $\frac{(a+7) + b + (c-10)}{3}$.
So, $\frac{a+b+c-3}{3} = b+2$
$a+b+c-3 = 3(b+2)$
$a+b+c-3 = 3b+6$
Substitute $a+b+c=84$ from Equation 1:
$84 – 3 = 3b+6$
$81 = 3b+6$
$81 – 6 = 3b$
$75 = 3b$
$b = \frac{75}{3} = 25$

Now we know the middle number is 25.
From Equation 1, $a+25+c = 84$
$a+c = 84 – 25$
$a+c = 59$ (Equation 2)

The difference between the largest and the smallest numbers becomes 64. This refers to the new set of numbers.
The new largest number is $c-10$.
The new smallest number is $a+7$.
So, $(c-10) – (a+7) = 64$
$c-10-a-7 = 64$
$c-a-17 = 64$
$c-a = 64+17$
$c-a = 81$ (Equation 3)

Now we have a system of two linear equations with two variables, $a$ and $c$:
Equation 2: $a+c = 59$
Equation 3: $c-a = 81$

Add Equation 2 and Equation 3:
$(a+c) + (c-a) = 59 + 81$
$2c = 140$
$c = \frac{140}{2} = 70$

Substitute the value of $c$ into Equation 2:
$a+70 = 59$
$a = 59 – 70$
$a = -11$

The original three numbers are $a=-11$, $b=25$, and $c=70$.
Let’s check if they are distinct real numbers. Yes, they are.
Let’s check the conditions:
Average: $\frac{-11+25+70}{3} = \frac{84}{3} = 28$. Correct.

New numbers:
Smallest: $a+7 = -11+7 = -4$
Middle: $b = 25$
Largest: $c-10 = 70-10 = 60$
New order: $-4 < 25 < 60$. The order remains unchanged.

New arithmetic mean: $\frac{-4+25+60}{3} = \frac{81}{3} = 27$.
Middle number is 25. New mean is 27, which is $25+2$. Correct.

Difference between the largest and smallest numbers in the new set: $60 – (-4) = 60+4 = 64$. Correct.

The question asks for the largest number in the original set of three numbers, which is $c$.
$c=70$.

Final_Answer:70

Ans: C

Explanation:We need to find the remainder when $10^{68}$ is divided by 13. We can use modular arithmetic.
We observe the powers of 10 modulo 13:
$10^1 \equiv 10 \pmod{13}$
$10^2 \equiv 100 \equiv 13 \times 7 + 9 \equiv 9 \pmod{13}$
$10^3 \equiv 10 \times 9 \equiv 90 \equiv 13 \times 6 + 12 \equiv 12 \equiv -1 \pmod{13}$
$10^4 \equiv 10 \times 12 \equiv 120 \equiv 13 \times 9 + 3 \equiv 3 \pmod{13}$
$10^5 \equiv 10 \times 3 \equiv 30 \equiv 13 \times 2 + 4 \equiv 4 \pmod{13}$
$10^6 \equiv 10 \times 4 \equiv 40 \equiv 13 \times 3 + 1 \equiv 1 \pmod{13}$

Since $10^6 \equiv 1 \pmod{13}$, the remainders of powers of 10 when divided by 13 repeat in a cycle of length 6.
We need to find the remainder of $10^{68}$ when divided by 13. We can use the exponent 68 and divide it by the length of the cycle, which is 6.
$68 \div 6$
$68 = 6 \times 11 + 2$
So, $68 \equiv 2 \pmod{6}$.
Therefore, $10^{68} \equiv 10^{6 \times 11 + 2} \equiv (10^6)^{11} \times 10^2 \pmod{13}$.
Since $10^6 \equiv 1 \pmod{13}$, we have:
$10^{68} \equiv 1^{11} \times 10^2 \pmod{13}$
$10^{68} \equiv 1 \times 10^2 \pmod{13}$
$10^{68} \equiv 100 \pmod{13}$
As calculated before, $100 \equiv 9 \pmod{13}$.
Thus, the remainder when $10^{68}$ is divided by 13 is 9.

Alternatively, 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, $p = 13$ (which is prime) and $a = 10$ (which is not divisible by 13).
So, $10^{13-1} \equiv 10^{12} \equiv 1 \pmod{13}$.
Now we need to find the remainder of $10^{68}$ when divided by 13.
We divide the exponent 68 by 12:
$68 \div 12$
$68 = 12 \times 5 + 8$
So, $68 \equiv 8 \pmod{12}$.
Therefore, $10^{68} \equiv 10^{12 \times 5 + 8} \equiv (10^{12})^5 \times 10^8 \pmod{13}$.
Since $10^{12} \equiv 1 \pmod{13}$:
$10^{68} \equiv 1^5 \times 10^8 \pmod{13}$
$10^{68} \equiv 10^8 \pmod{13}$.
Now we need to find $10^8 \pmod{13}$.
From our earlier calculations:
$10^6 \equiv 1 \pmod{13}$
$10^8 \equiv 10^6 \times 10^2 \pmod{13}$
$10^8 \equiv 1 \times 10^2 \pmod{13}$
$10^8 \equiv 100 \pmod{13}$
$100 \equiv 9 \pmod{13}$.
Thus, the remainder is 9.

Correct_Option:C

Ans: B

Suppose an individual, Sam, can finish a task in 20 days. Another individual, Mohit, works at double Sam’s efficiency, implying Mohit would complete the task in 10 days. Mohit’s efficiency is thrice that of Ayna, meaning Ayna would take 30 days to complete the same task.

Let’s represent the total effort required for the task as 60 units. This assignment allows us to calculate the daily output of Mohit, Sam, and Ayna as 6, 3, and 2 units, respectively.

In the first day, Sam and Mohit collaborate, accomplishing 9 units.
On the second day, Sam and Ayna work together, completing 5 units.
On the third day, Mohit and Ayna team up, finishing 8 units.

This pattern results in 22 units of work completed over a 3-day period.
After two such 3-day cycles, 44 units of work are done, leaving 16 units remaining (60 – 44 = 16).

On the seventh day (the start of the next cycle), Sam and Mohit work, contributing 9 units, leaving 7 units.
On the eighth day, Sam and Ayna work, completing 5 units, leaving 2 units.
On the ninth day, Ayna and Mohit will finish the remaining 2 units.

Considering Sam’s contribution: Sam participated on days 1, 2, 4, 5, 7, and 8. He worked for a total of 6 days, contributing 3 units per day, for a cumulative output of 18 units of work (6 days * 3 units/day).

The proportion of the total work completed by Sam is calculated as $\frac{18}{60}$, which simplifies to $\frac{3}{10}$.
Consequently, Option B represents the correct solution.

Ans: D

Explanation:Let $r$ be the radius of the circular plot and $O$ be the center. Let the chord be $AB$. We are given that the length of the chord $AB = 10\sqrt{3}$ meters. The angle subtended by the chord at the center is $\angle AOB = 120^\circ$.

We can find the radius $r$ using the law of cosines in triangle $AOB$:
$AB^2 = OA^2 + OB^2 – 2(OA)(OB)\cos(\angle AOB)$
$(10\sqrt{3})^2 = r^2 + r^2 – 2(r)(r)\cos(120^\circ)$
$100 \times 3 = 2r^2 – 2r^2(-\frac{1}{2})$
$300 = 2r^2 + r^2$
$300 = 3r^2$
$r^2 = 100$
$r = 10$ meters.

The area of the sector $AOB$ is given by:
Area of sector $= \frac{\theta}{360^\circ} \pi r^2$
Area of sector $= \frac{120^\circ}{360^\circ} \pi (10)^2$
Area of sector $= \frac{1}{3} \pi (100)$
Area of sector $= \frac{100\pi}{3}$ square meters.

The area of triangle $AOB$ is given by:
Area of triangle $= \frac{1}{2} OA \times OB \sin(\angle AOB)$
Area of triangle $= \frac{1}{2} r \times r \sin(120^\circ)$
Area of triangle $= \frac{1}{2} (10)(10) \frac{\sqrt{3}}{2}$
Area of triangle $= \frac{100\sqrt{3}}{4}$
Area of triangle $= 25\sqrt{3}$ square meters.

The area of the smaller region is the area of the minor segment, which is the area of the sector $AOB$ minus the area of triangle $AOB$.
Area of smaller region = Area of sector $AOB$ – Area of triangle $AOB$
Area of smaller region $= \frac{100\pi}{3} – 25\sqrt{3}$

We need to match this result with the given options. Let’s factor out 25 from our result:
Area of smaller region $= 25\left(\frac{4\pi}{3} – \sqrt{3}\right)$

Comparing this with the given options:
Option A: $20\left(\cfrac{4 \pi}{3} + \sqrt{3}\right)$
Option B: $25\left(\cfrac{4 \pi}{3} + \sqrt{3}\right)$
Option C: $20\left(\cfrac{4 \pi}{3} – \sqrt{3}\right)$
Option D: $25\left(\cfrac{4 \pi}{3} – \sqrt{3}\right)$

Our calculated area matches Option D.

The larger region is the area of the circle minus the area of the smaller region.
Area of circle = $\pi r^2 = \pi (10)^2 = 100\pi$.
Area of larger region = $100\pi – (\frac{100\pi}{3} – 25\sqrt{3}) = \frac{300\pi – 100\pi}{3} + 25\sqrt{3} = \frac{200\pi}{3} + 25\sqrt{3}$.
Since $\frac{100\pi}{3} \approx 104.7$ and $25\sqrt{3} \approx 43.3$, the area of the smaller region is approximately $104.7 – 43.3 = 61.4$.
The area of the larger region is approximately $100\pi – 61.4 \approx 314.16 – 61.4 = 252.76$.
Clearly, $\frac{100\pi}{3} – 25\sqrt{3}$ is the area of the smaller region.

Correct_Option:D

Ans: A

Observing the terms in the series, we find that the 3rd term is 0, the 4th term is -1/3, and the 5th term is 0.
It is apparent that all terms at odd positions are zero. Since the question specifically asks for the sum of only the even-positioned terms, we can disregard the odd-positioned terms.
The 6th term is -1/5.
The sequence of even-positioned terms forms a Harmonic Progression (HP): -1, -1/3, -1/5, -1/7, and so on.
The sum requested is the sum of the reciprocals of these terms. This means we need to sum the following sequence: -1, -3, -5, -7, continuing for 1012 terms.
This sequence is an Arithmetic Progression (AP). The sum of this AP can be calculated as:
$ \frac{[-(2 \times 1) + (1012 – 1)(-2)]}{2} \times 1012 $
This evaluates to:
$ -1012 \times 1012 = -1024144 $
Thus, the correct option is A.

Ans: 2

To be published

Ans: 9

Explanation:Let P be the principal amount invested, which is Rs 4000.
Let r be the annual rate of interest.
The value of the investment after t years, compounded annually, is given by $A(t) = P(1 + r)^t$.

We are given that the ratio of the value of the investment after 3 years to the value of the investment after 5 years is 25 : 36.
So, $\frac{A(3)}{A(5)} = \frac{P(1 + r)^3}{P(1 + r)^5} = \frac{25}{36}$.

Simplifying the equation:
$\frac{(1 + r)^3}{(1 + r)^5} = \frac{1}{(1 + r)^2} = \frac{25}{36}$.

Taking the reciprocal of both sides:
$(1 + r)^2 = \frac{36}{25}$.

Taking the square root of both sides:
$1 + r = \sqrt{\frac{36}{25}} = \frac{6}{5}$.
Since the interest rate must be positive, we take the positive square root.

Now, we can find the annual rate of interest:
$r = \frac{6}{5} – 1 = \frac{6 – 5}{5} = \frac{1}{5}$.
So, the rate of interest is $\frac{1}{5}$ or 20%.

The value of the investment after t years is given by $A(t) = 4000 \left(\frac{6}{5}\right)^t$.

We need to find the minimum number of years required for the value of the investment to exceed Rs 20000.
So, we need to find the smallest integer t such that $A(t) > 20000$.
$4000 \left(\frac{6}{5}\right)^t > 20000$.

Divide both sides by 4000:
$\left(\frac{6}{5}\right)^t > \frac{20000}{4000}$.
$\left(\frac{6}{5}\right)^t > 5$.

Now, we can test values of t:
For t = 1: $\left(\frac{6}{5}\right)^1 = 1.2$ (not > 5)
For t = 2: $\left(\frac{6}{5}\right)^2 = \left(\frac{36}{25}\right) = 1.44$ (not > 5)
For t = 3: $\left(\frac{6}{5}\right)^3 = \frac{216}{125} = 1.728$ (not > 5)
For t = 4: $\left(\frac{6}{5}\right)^4 = \frac{1296}{625} = 2.0736$ (not > 5)
For t = 5: $\left(\frac{6}{5}\right)^5 = \frac{7776}{3125} = 2.48832$ (not > 5)
For t = 6: $\left(\frac{6}{5}\right)^6 = \left(\frac{6}{5}\right)^5 \times \frac{6}{5} = 2.48832 \times 1.2 = 2.985984$ (not > 5)
For t = 7: $\left(\frac{6}{5}\right)^7 = \left(\frac{6}{5}\right)^6 \times \frac{6}{5} = 2.985984 \times 1.2 = 3.5831808$ (not > 5)
For t = 8: $\left(\frac{6}{5}\right)^8 = \left(\frac{6}{5}\right)^7 \times \frac{6}{5} = 3.5831808 \times 1.2 = 4.29981696$ (not > 5)
For t = 9: $\left(\frac{6}{5}\right)^9 = \left(\frac{6}{5}\right)^8 \times \frac{6}{5} = 4.29981696 \times 1.2 = 5.159780352$ ( > 5)

Alternatively, we can use logarithms:
$t \log\left(\frac{6}{5}\right) > \log(5)$
$t (\log(6) – \log(5)) > \log(5)$
$t (0.7781 – 0.6990) > 0.6990$
$t (0.0791) > 0.6990$
$t > \frac{0.6990}{0.0791} \approx 8.836$.

Since t must be an integer, the minimum number of years required is 9.

Final_Answer:9

Ans: A

Let’s represent the expression $10^x$ as $y$.
Substituting this into the given equation, we have:
$y + \frac{4}{y} = \frac{81}{2}$
To solve for $y$, we can rearrange this into a quadratic equation:
$2y^2 – 81y + 8 = 0$
We are looking for the sum of the possible values of $x$. Let these values be $x_1$ and $x_2$.
These values of $x$ correspond to the logarithms of the possible values of $y$. Specifically, if $y_1$ and $y_2$ are the roots of the quadratic equation, then $x_1 = \log_{10} y_1$ and $x_2 = \log_{10} y_2$.
The sum of the possible values of $x$ is $x_1 + x_2$, which can be written as $\log_{10} y_1 + \log_{10} y_2$.
Using the properties of logarithms, this sum is equal to $\log_{10} (y_1 \times y_2)$.
From the quadratic equation $2y^2 – 81y + 8 = 0$, the product of the roots ($y_1 \times y_2$) is given by the constant term divided by the coefficient of the $y^2$ term, which is $\frac{8}{2} = 4$.
Therefore, the sum of the values of $x$ is $\log_{10} 4$.
This can be further simplified as $2 \log_{10} 2$.
Thus, the correct option is A.

Ans: A

Explanation:Let the distance travelled by the train be $D$ km and the original uniform speed be $S$ km per hour. Let the original time taken be $T$ hours.
We know that distance = speed × time, so $D = S \times T$.

According to the first condition:
If the speed had been 6 km per hour more, i.e., $S+6$, it would have needed 4 hours less, i.e., $T-4$.
So, $D = (S+6)(T-4)$
$ST = (S+6)(T-4)$
$ST = ST – 4S + 6T – 24$
$0 = -4S + 6T – 24$
$4S – 6T = -24$
Dividing by 2, we get:
$2S – 3T = -12$ (Equation 1)

According to the second condition:
If the speed had been 6 km per hour less, i.e., $S-6$, it would have needed 6 hours more, i.e., $T+6$.
So, $D = (S-6)(T+6)$
$ST = (S-6)(T+6)$
$ST = ST + 6S – 6T – 36$
$0 = 6S – 6T – 36$
$6S – 6T = 36$
Dividing by 6, we get:
$S – T = 6$ (Equation 2)

Now we have a system of two linear equations with two variables:
1) $2S – 3T = -12$
2) $S – T = 6$

From Equation 2, we can express $S$ in terms of $T$:
$S = T + 6$

Substitute this expression for $S$ into Equation 1:
$2(T+6) – 3T = -12$
$2T + 12 – 3T = -12$
$-T + 12 = -12$
$-T = -12 – 12$
$-T = -24$
$T = 24$ hours

Now substitute the value of $T$ back into the equation for $S$:
$S = T + 6$
$S = 24 + 6$
$S = 30$ km per hour

The distance travelled by the train is $D = S \times T$.
$D = 30 \times 24$
$D = 720$ km

Let’s check the conditions with the calculated values:
Original speed $S = 30$ km/h, original time $T = 24$ h, distance $D = 720$ km.

Condition 1: Speed $S+6 = 30+6 = 36$ km/h. Time $T-4 = 24-4 = 20$ h.
Distance $= 36 \times 20 = 720$ km. This matches.

Condition 2: Speed $S-6 = 30-6 = 24$ km/h. Time $T+6 = 24+6 = 30$ h.
Distance $= 24 \times 30 = 720$ km. This matches.

Therefore, the distance travelled by the train is 720 km.

Correct_Option:A

Ans: 2

Explanation:We are given the following equations:
1) $3^a = 4$
2) $4^b = 5$
3) $5^c = 6$
4) $6^d = 7$
5) $7^e = 8$
6) $8^f = 9$

We want to find the value of the product $abcdef$.

Let’s express the later terms in terms of the earlier terms by substituting the equations into each other.

From (1), we can write $a = \log_3 4$.
From (2), we can write $b = \log_4 5$.
From (3), we can write $c = \log_5 6$.
From (4), we can write $d = \log_6 7$.
From (5), we can write $e = \log_7 8$.
From (6), we can write $f = \log_8 9$.

Now, we can multiply these logarithmic values together:
$abcdef = (\log_3 4) (\log_4 5) (\log_5 6) (\log_6 7) (\log_7 8) (\log_8 9)$

We can use the change of base formula for logarithms, which states that $\log_x y = \frac{\log_k y}{\log_k x}$ for any valid base $k$. Let’s use base 10 or the natural logarithm for simplicity. However, a more direct way to simplify this product is to observe the telescoping nature of the logarithmic product.

Using the property $\log_x y \cdot \log_y z = \log_x z$:
$(\log_3 4) (\log_4 5) = \log_3 5$
$(\log_3 5) (\log_5 6) = \log_3 6$
$(\log_3 6) (\log_6 7) = \log_3 7$
$(\log_3 7) (\log_7 8) = \log_3 8$
$(\log_3 8) (\log_8 9) = \log_3 9$

So, $abcdef = \log_3 9$.

We know that $3^2 = 9$. Therefore, $\log_3 9 = 2$.

Alternatively, we can substitute the equations sequentially:
Start with $3^a = 4$.
Substitute this into $4^b = 5$: $(3^a)^b = 5 \implies 3^{ab} = 5$.
Substitute this into $5^c = 6$: $(3^{ab})^c = 6 \implies 3^{abc} = 6$.
Substitute this into $6^d = 7$: $(3^{abc})^d = 7 \implies 3^{abcd} = 7$.
Substitute this into $7^e = 8$: $(3^{abcd})^e = 8 \implies 3^{abcde} = 8$.
Substitute this into $8^f = 9$: $(3^{abcde})^f = 9 \implies 3^{abcdef} = 9$.

Since $3^2 = 9$, we can equate the exponents:
$abcdef = 2$.

Final_Answer:2

Ans: A

Explanation:Let the cost price of the product be CP.
Gopi marks a price on the product in order to make 20% profit. So, the marked price (MP) is such that if sold at MP, Gopi would get a 20% profit. This implies that MP is the selling price if there were no discount.
Let the marked price be MP.
Ravi gets a 10% discount on the marked price.
Discount amount = 10% of MP = 0.10 * MP.
We are given that Ravi saves Rs 15, which means the discount amount is Rs 15.
So, 0.10 * MP = 15.
MP = 15 / 0.10 = 150.
The marked price is Rs 150.
Gopi marks the price to make a 20% profit. This means that if the product was sold at the marked price, the profit would be 20% of the cost price.
Let CP be the cost price.
MP = CP + 20% of CP = CP + 0.20 * CP = 1.20 * CP.
We found that MP = 150.
So, 1.20 * CP = 150.
CP = 150 / 1.20 = 150 / (12/10) = 150 * (10/12) = 1500 / 12 = 125.
The cost price of the product is Rs 125.
Ravi gets a 10% discount on the marked price.
Selling Price (SP) to Ravi = MP – Discount = 150 – 15 = 135.
The profit made by Gopi by selling the product to Ravi is the difference between the selling price and the cost price.
Profit = SP – CP = 135 – 125 = 10.
The profit made by Gopi is Rs 10.

Let’s recheck the initial assumption about the marked price. The problem states “Gopi marks a price on a product in order to make 20% profit.” This phrasing usually implies that the intended selling price is the marked price, and that intended selling price should yield a 20% profit over the cost price.

Let CP be the Cost Price.
Gopi marks a price MP.
The intention is to make 20% profit, so if sold at MP, the profit would be 20% of CP.
So, MP = CP + 0.20 * CP = 1.20 * CP.

Ravi gets a 10% discount on the marked price.
Discount = 10% of MP = 0.10 * MP.
This discount amount is Rs 15.
0.10 * MP = 15
MP = 15 / 0.10 = 150.

Now, we can find the Cost Price using MP = 1.20 * CP.
150 = 1.20 * CP
CP = 150 / 1.20 = 125.

The selling price to Ravi is the marked price minus the discount.
SP = MP – Discount = 150 – 15 = 135.

The profit made by Gopi is the selling price minus the cost price.
Profit = SP – CP = 135 – 125 = 10.

Let’s consider an alternative interpretation where the marked price is set to achieve a 20% profit on the selling price, but this is less common. The standard interpretation is profit on cost price.

The initial interpretation aligns with standard business terminology.

The profit made by Gopi is Rs 10.

Correct_Option:A

Ans: 30

Let the initial quantity of milk in the container be denoted by $M$ and the initial quantity of water be denoted by $W$.
We are given that the total volume of the mixture is 300 units, so $M + W = 300$.

The problem describes a process: “a portion of this mixture was removed, with the removed volume being double the initial volume of water added, and the container was then replenished with water.”

The initial volume of water is $W$. If twice this amount, $2W$, is removed from the container, the fraction of the total mixture that is removed is $\frac{2W}{M+W}$.

Consequently, the quantity of milk remaining in the mixture after this removal will be $M \times \left(1 – \frac{2W}{M+W}\right)$.

We are told that this remaining quantity of milk constitutes 72% of the total original volume (300 units). So, the remaining milk is $0.72 \times 300 = 216$ units.

Therefore, we can set up the equation:
$M \times \left(1 – \frac{2W}{M+W}\right) = 216$

Substitute $M+W = 300$:
$M \times \left(1 – \frac{2W}{300}\right) = 216$
$M \times \left(\frac{300 – 2W}{300}\right) = 216$

Now, substitute $M = 300 – W$:
$(300 – W) \times \left(\frac{300 – 2W}{300}\right) = 216$

Multiply both sides by 300:
$(300 – W)(300 – 2W) = 216 \times 300$
$(300 – W)(300 – 2W) = 64800$

Expand the left side of the equation:
$90000 – 600W – 300W + 2W^2 = 64800$
$2W^2 – 900W + 90000 = 64800$
$2W^2 – 900W + 90000 – 64800 = 0$
$2W^2 – 900W + 25200 = 0$

Divide the entire equation by 2 to simplify:
$W^2 – 450W + 12600 = 0$

Factorize this quadratic equation. We need two numbers that multiply to 12600 and add up to -450. These numbers are -30 and -420.
$(W – 30)(W – 420) = 0$

This gives two possible values for $W$: $W = 30$ or $W = 420$.

Given that the total capacity of the container is 300 units, the initial amount of water ($W$) cannot exceed 300. Therefore, $W$ must be 30.

Hence, the initial quantity of water in the container was 30 units.

Ans: D

Explanation:Let the side length of the regular octagon be $s = 6$ cm.
A regular octagon can be inscribed in a circle. The interior angle of a regular octagon is $(8-2) \times 180^\circ / 8 = 6 \times 180^\circ / 8 = 6 \times 22.5^\circ = 135^\circ$.
Consider the vertices of the octagon. The vertices A, C, E, and G form a square. We need to find the area of this square. The side length of the square ACEG is the distance between consecutive vertices A and C (or C and E, E and G, G and A).

Let’s consider the triangle ABC. AB = BC = $s = 6$. The angle ABC is $135^\circ$.
We can use the law of cosines to find the length of AC, which is the side of the square ACEG.
$AC^2 = AB^2 + BC^2 – 2(AB)(BC)\cos(135^\circ)$
$AC^2 = 6^2 + 6^2 – 2(6)(6)\cos(135^\circ)$
$AC^2 = 36 + 36 – 72 \times (-\frac{\sqrt{2}}{2})$
$AC^2 = 72 + 36\sqrt{2}$
$AC^2 = 36(2 + \sqrt{2})$

The area of the square ACEG is the square of its side length, which is $AC^2$.
Area of square ACEG = $AC^2 = 36(2 + \sqrt{2})$ sq. cm.

Alternatively, we can consider the octagon as a square with four triangles cut off from its corners.
Let the side length of the regular octagon be $s$.
The vertices of the square ACEG can be visualized. The distance between alternate vertices forms the sides of the square.
Consider vertex A and C. The angle subtended by the side AB at the center of the circumscribing circle is $360^\circ/8 = 45^\circ$.
Let R be the radius of the circumscribing circle.
Using the law of cosines on triangle OAB (where O is the center), $s^2 = R^2 + R^2 – 2R^2 \cos(45^\circ) = 2R^2(1 – \frac{\sqrt{2}}{2}) = R^2(2 – \sqrt{2})$.
So, $R^2 = \frac{s^2}{2 – \sqrt{2}} = \frac{s^2(2 + \sqrt{2})}{(2 – \sqrt{2})(2 + \sqrt{2})} = \frac{s^2(2 + \sqrt{2})}{4 – 2} = \frac{s^2(2 + \sqrt{2})}{2}$.
For $s=6$, $R^2 = \frac{36(2 + \sqrt{2})}{2} = 18(2 + \sqrt{2})$.

The angle subtended by AC at the center O is $2 \times 45^\circ = 90^\circ$.
So, triangle AOC is an isosceles right-angled triangle with OA = OC = R.
The length of AC is $AC = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$.
$AC^2 = 2R^2$.
Substituting the value of $R^2$:
$AC^2 = 2 \times \frac{s^2(2 + \sqrt{2})}{2} = s^2(2 + \sqrt{2})$.
With $s=6$:
$AC^2 = 6^2(2 + \sqrt{2}) = 36(2 + \sqrt{2})$.
The area of the square ACEG is $AC^2 = 36(2 + \sqrt{2})$ sq. cm.

Correct_Option:D

Ans: 8

The absolute value function yields only non-negative results. Considering only integer values for x and y significantly limits the potential solutions.

We can obtain a sum of 2 in two primary ways:
1. **2 + 0:** This occurs when one of the terms within the absolute value is zero.
* **Case 1: The second term is zero.** This means x = y. In this scenario, |2x| = 2, which implies x can be 1 or -1. This gives us the pairs (1, 1) and (-1, -1).
* **Case 2: The first term is zero.** This means x = -y. In this scenario, |x – (-x)| = 2, which simplifies to |2x| = 2. Again, x can be 1 or -1. This yields the pairs (1, -1) and (-1, 1).

2. **1 + 1:** This occurs when each term within the absolute value contributes 1.
* **Case 3: If y = 0.** Then |x| + |x| = 2, which simplifies to 2|x| = 2. This means x can be 1 or -1. This provides the pairs (1, 0) and (-1, 0).
* **Case 4: If x = 0.** Similarly, |0| + |y| = 2, which implies |y| = 2. This means y can be 1 or -1. This gives us the pairs (0, 1) and (0, -1).

In total, there are 8 distinct pairs of (x, y) that satisfy the given condition.

Ans: C

Explanation:We are given the functional equation $f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$ for any non-zero real number $x$.
We are asked to find the sum of all possible values of $x$ for which $f(x) = 3$.

First, let’s find an expression for $f(x)$.
From the given equation:
1) $f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$

Now, let’s replace $x$ with $\frac{1}{x}$ in the given equation. Since $x$ is a non-zero real number, $\frac{1}{x}$ is also a non-zero real number.
$f \left(\cfrac{1}{x}\right) + 2f \left(\cfrac{1}{\frac{1}{x}}\right) = 3 \left(\cfrac{1}{x}\right)$
$f \left(\cfrac{1}{x}\right) + 2f(x) = \cfrac{3}{x}$

Now we have a system of two linear equations with $f(x)$ and $f \left(\cfrac{1}{x}\right)$:
1) $f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$
2) $2f(x) + f \left(\cfrac{1}{x}\right) = \cfrac{3}{x}$

To eliminate $f \left(\cfrac{1}{x}\right)$, we multiply equation (2) by 2:
$4f(x) + 2f \left(\cfrac{1}{x}\right) = \cfrac{6}{x}$ (Equation 3)

Now, subtract equation (1) from equation (3):
$(4f(x) + 2f \left(\cfrac{1}{x}\right)) – (f(x) + 2f \left(\cfrac{1}{x}\right)) = \cfrac{6}{x} – 3x$
$3f(x) = \cfrac{6}{x} – 3x$

Divide by 3:
$f(x) = \cfrac{2}{x} – x$

Now we need to find the values of $x$ for which $f(x) = 3$.
Substitute the expression for $f(x)$:
$\cfrac{2}{x} – x = 3$

To solve for $x$, we can multiply the entire equation by $x$ (since $x$ is non-zero):
$2 – x^2 = 3x$

Rearrange the terms to form a quadratic equation:
$x^2 + 3x – 2 = 0$

We need to find the sum of all possible values of $x$ for which this quadratic equation holds. For a quadratic equation of the form $ax^2 + bx + c = 0$, the sum of the roots is given by $-\frac{b}{a}$.

In our equation, $a = 1$, $b = 3$, and $c = -2$.
The sum of the roots is $-\frac{3}{1} = -3$.

Let’s check if the roots are real and non-zero. The discriminant of the quadratic equation $x^2 + 3x – 2 = 0$ is $\Delta = b^2 – 4ac = 3^2 – 4(1)(-2) = 9 + 8 = 17$. Since the discriminant is positive, there are two distinct real roots.
The roots are $x = \frac{-3 \pm \sqrt{17}}{2}$. Both of these roots are non-zero.
Therefore, the sum of all possible values of $x$ for which $f(x) = 3$ is -3.

Correct_Option: C

Ans: B

To determine when there are no solutions, we require the lines to be parallel. This occurs when the ratio of coefficients of x and y are equal, but not equal to the ratio of the constant terms. Thus, the condition for parallel lines is:
$\frac{p}{3}=-\frac{4}{k}\ne\ \frac{2}{a}$

Let’s evaluate the provided options based on this condition:

Option A: From the first and last parts of our relation, we deduce that $ap \ne 6$ is required for the lines to be parallel. This option imposes no constraints on this relationship, rendering it irrelevant.

Option C: This option represents the inverse of our desired condition. If this were true, the lines would never be parallel.

Option D: Considering the first and second parts of the relation, we need $pk = -12$, which can be rewritten as $pk – 12 = 0$. Therefore, this statement does not align with our requirements.

Option B: Examining the second and third parts of the relation, we see that we must avoid the case where $k = -2a$, or in other words, we must avoid $k + 2a = 0$.

Thus, Option B presents a condition that is essential for the lines to be parallel and have no solution.

Therefore, Option B is the correct answer.

Ans: B

Explanation:
Let $R_w$ and $R_m$ be the cultivation area of wheat and mustard in Rajesh’s land, respectively.
Let $V_w$ and $V_m$ be the cultivation area of wheat and mustard in Vimal’s land, respectively.

Rajesh owns 20 hectares of land, so $R_w + R_m = 20$.
Vimal owns 30 hectares of land, so $V_w + V_m = 30$.

The cultivation area of wheat and mustard in Vimal’s land are in the ratio 5 : 3.
So, $V_w : V_m = 5 : 3$.
We can write $V_w = 5x$ and $V_m = 3x$.
Since $V_w + V_m = 30$, we have $5x + 3x = 30$.
$8x = 30$
$x = \frac{30}{8} = \frac{15}{4}$.
Therefore, $V_w = 5 \times \frac{15}{4} = \frac{75}{4}$ hectares and $V_m = 3 \times \frac{15}{4} = \frac{45}{4}$ hectares.

The total cultivation area of wheat and mustard are in the ratio 11 : 9.
Let the total cultivation area of wheat be $W$ and the total cultivation area of mustard be $M$.
So, $W : M = 11 : 9$.
We can write $W = 11y$ and $M = 9y$.

The total cultivation area is the sum of the areas owned by Rajesh and Vimal.
Total area = 20 + 30 = 50 hectares.
So, $W + M = 50$.
$11y + 9y = 50$
$20y = 50$
$y = \frac{50}{20} = \frac{5}{2}$.
Therefore, $W = 11 \times \frac{5}{2} = \frac{55}{2}$ hectares and $M = 9 \times \frac{5}{2} = \frac{45}{2}$ hectares.

The total cultivation area of wheat is the sum of wheat cultivation in Rajesh’s and Vimal’s land: $W = R_w + V_w$.
$\frac{55}{2} = R_w + \frac{75}{4}$
$R_w = \frac{55}{2} – \frac{75}{4} = \frac{110}{4} – \frac{75}{4} = \frac{35}{4}$ hectares.

The total cultivation area of mustard is the sum of mustard cultivation in Rajesh’s and Vimal’s land: $M = R_m + V_m$.
$\frac{45}{2} = R_m + \frac{45}{4}$
$R_m = \frac{45}{2} – \frac{45}{4} = \frac{90}{4} – \frac{45}{4} = \frac{45}{4}$ hectares.

We need to find the ratio of cultivation area of wheat and mustard in the land owned by Rajesh, which is $R_w : R_m$.
$R_w : R_m = \frac{35}{4} : \frac{45}{4}$
Multiply both sides by 4:
$R_w : R_m = 35 : 45$
Divide both sides by 5:
$R_w : R_m = 7 : 9$.

Let’s check if $R_w + R_m = 20$:
$\frac{35}{4} + \frac{45}{4} = \frac{80}{4} = 20$. This is correct.

Correct_Option:B

Ans: 90

To be published

Ans: 378

We will analyze numbers with one, two, and three digits separately.

**One-digit numbers:**
There are nine such positive integers (1 through 9).

**Two-digit numbers:**
The first digit (tens place) has 9 possible choices (1 through 9). The second digit (units place) has 9 possible choices (0 through 9, excluding the digit already used in the tens place). This gives a total of $9 \times 9 = 81$ numbers.

**Three-digit numbers:**
The first digit (hundreds place) can be one of 1, 2, 3, or 4, providing 4 options. The second digit (tens place) has 9 possible choices (0 through 9, excluding the digit used in the hundreds place). The third digit (units place) has 8 possible choices (0 through 9, excluding the two digits already used in the hundreds and tens places). This results in $4 \times 9 \times 8 = 288$ numbers.

Summing these up: $9 + 81 + 288 = 378$.

Thus, 378 is the final count.

Ans: C

Explanation:Let the initial salary of Gopal be S.
Let the percentage of salary increase in the first increment be x%.
Let the percentage of salary increase in the second increment be y%.

According to the problem, the percentage of salary increase in the second increment was twice of that in the first increment.
So, y = 2x.

After the first increment, Gopal’s salary became S * (1 + x/100).
After the second increment, Gopal’s salary became S * (1 + x/100) * (1 + y/100).

We are given that after two successive increments, Gopal’s salary became 187.5% of his initial salary.
So, S * (1 + x/100) * (1 + y/100) = S * (187.5/100)
(1 + x/100) * (1 + y/100) = 1.875

Substitute y = 2x into the equation:
(1 + x/100) * (1 + 2x/100) = 1.875

Let’s rewrite 1.875 as a fraction:
1.875 = 1875/1000 = 375/200 = 75/40 = 15/8.

So, (1 + x/100) * (1 + 2x/100) = 15/8.

Let’s try the options given:
Option A: x = 30
y = 2*30 = 60
(1 + 30/100) * (1 + 60/100) = (1.30) * (1.60) = 2.08. This is not 1.875.

Option B: x = 27.5
y = 2*27.5 = 55
(1 + 27.5/100) * (1 + 55/100) = (1.275) * (1.55)
1.275 * 1.55 = 1.97625. This is not 1.875.

Option C: x = 25
y = 2*25 = 50
(1 + 25/100) * (1 + 50/100) = (1.25) * (1.50) = 1.875. This matches the given condition.

Option D: x = 20
y = 2*20 = 40
(1 + 20/100) * (1 + 40/100) = (1.20) * (1.40) = 1.68. This is not 1.875.

Alternatively, we can solve the quadratic equation:
(100 + x)/100 * (100 + 2x)/100 = 15/8
(100 + x)(100 + 2x) = 15/8 * 10000
10000 + 200x + 100x + 2x^2 = 15 * 1250
10000 + 300x + 2x^2 = 18750
2x^2 + 300x – 8750 = 0
Divide by 2:
x^2 + 150x – 4375 = 0

Using the quadratic formula: x = [-b ± sqrt(b^2 – 4ac)] / 2a
x = [-150 ± sqrt(150^2 – 4*1*(-4375))] / 2*1
x = [-150 ± sqrt(22500 + 17500)] / 2
x = [-150 ± sqrt(40000)] / 2
x = [-150 ± 200] / 2

Since percentage increase cannot be negative, we take the positive root:
x = (-150 + 200) / 2 = 50 / 2 = 25.

Correct_Option: C