CAT 2025 Quant Slot 3 Paper

Ans: B

Explanation:Let the marked price per unit be $M$.
The cost price per unit is Rs. 240.

Monthly sales:
January: 120 units
February: 135 units
March: 150 units
April: 165 units

Discounts:
January: 20%
February: 10%
March: 5%
April: 0%

Selling price per unit in each month:
January selling price ($SP_J$) = $M \times (1 – 0.20) = 0.80M$
February selling price ($SP_F$) = $M \times (1 – 0.10) = 0.90M$
March selling price ($SP_M$) = $M \times (1 – 0.05) = 0.95M$
April selling price ($SP_A$) = $M \times (1 – 0.00) = M$

Profit per unit in each month:
Profit per unit in January ($P_J$) = $SP_J – CP = 0.80M – 240$
Profit per unit in February ($P_F$) = $SP_F – CP = 0.90M – 240$
Profit per unit in March ($P_M$) = $SP_M – CP = 0.95M – 240$
Profit per unit in April ($P_A$) = $SP_A – CP = M – 240$

Total profit from January to April = Rs. 138825

Total profit = (Sales in Jan * Profit per unit in Jan) + (Sales in Feb * Profit per unit in Feb) + (Sales in Mar * Profit per unit in Mar) + (Sales in Apr * Profit per unit in Apr)

$138825 = 120(0.80M – 240) + 135(0.90M – 240) + 150(0.95M – 240) + 165(M – 240)$

$138825 = (120 \times 0.80M – 120 \times 240) + (135 \times 0.90M – 135 \times 240) + (150 \times 0.95M – 150 \times 240) + (165M – 165 \times 240)$

$138825 = (96M – 28800) + (121.5M – 32400) + (142.5M – 36000) + (165M – 39600)$

Combine the terms with $M$:
$96M + 121.5M + 142.5M + 165M = (96 + 121.5 + 142.5 + 165)M = 525M$

Combine the constant terms:
$-28800 – 32400 – 36000 – 39600 = -(28800 + 32400 + 36000 + 39600) = -136800$

So, the equation becomes:
$138825 = 525M – 136800$

Add 136800 to both sides:
$138825 + 136800 = 525M$
$275625 = 525M$

Now, solve for $M$:
$M = \frac{275625}{525}$

Let’s perform the division:
$275625 \div 525 = 525$

So, the marked price per unit is Rs. 525.

Let’s check the options:
Option A: 520
Option B: 525
Option C: 510
Option D: 515

The calculated marked price matches Option B.

Correct_Option:B

Ans: B

Explanation:
Let the productivity of each member in team A, B, and C be $p_A$, $p_B$, and $p_C$ respectively.
The number of members in team A, B, and C are 5, 8, and 10 respectively.

The total productivity of team A is $5p_A$.
The total productivity of team B is $8p_B$.
The total productivity of team C is $10p_C$.

Team A can complete the job in 40 hours, so the total work W = $5p_A \times 40 = 200p_A$.
Team B can complete the job in 50 hours, so the total work W = $8p_B \times 50 = 400p_B$.
Team C can complete the job in 4 hours, so the total work W = $10p_C \times 4 = 40p_C$.

Equating the work, we get:
$200p_A = 400p_B = 40p_C$

From $200p_A = 400p_B$, we get $p_A = 2p_B$.
From $400p_B = 40p_C$, we get $10p_B = p_C$.

Let’s express the productivity of each member in terms of $p_B$.
$p_A = 2p_B$
$p_B = p_B$
$p_C = 10p_B$

The total work W can be expressed in terms of $p_B$. Let’s use the work done by team B.
W = $400p_B$.

Two members from team A, three members from team B, and one member from team C together start the job.
The combined productivity of these members is $2p_A + 3p_B + 1p_C$.
Substitute the equivalent productivities in terms of $p_B$:
Combined productivity = $2(2p_B) + 3p_B + 1(10p_B) = 4p_B + 3p_B + 10p_B = 17p_B$.

These members work for 23 hours.
Work done in the first 23 hours = $17p_B \times 23 = 391p_B$.

The remaining work = W – Work done = $400p_B – 391p_B = 9p_B$.

After 23 hours, the member from team C leaves.
The remaining members are two from team A and three from team B.
Their combined productivity is $2p_A + 3p_B = 2(2p_B) + 3p_B = 4p_B + 3p_B = 7p_B$.

The job needs to be finished in the next 1 hour.
Let $x$ be the number of additional members from team B required to finish the remaining work in 1 hour.
The new productivity will be $7p_B + xp_B = (7+x)p_B$.

The work to be done in the next 1 hour is $9p_B$.
So, $(7+x)p_B \times 1 = 9p_B$.
$7+x = 9$
$x = 9 – 7$
$x = 2$.

Therefore, 2 additional members from team B are required.

Correct_Option: B

Ans: 700

Explanation:Let S be the number of science students, A be the number of arts students, and C be the number of commerce students.
The total number of students is 1500. So, S + A + C = 1500.
The fees for science, arts, and commerce are Rs 1100, Rs 1000, and Rs 800, respectively.
The total fee paid by all students is Rs 15,50,000. So, 1100S + 1000A + 800C = 15,50,000.
We can simplify this equation by dividing by 100: 11S + 10A + 8C = 15500.

We are also given that the number of science students is not more than the number of arts students, which means S ≤ A.

We have a system of two linear equations with three variables:
1) S + A + C = 1500
2) 11S + 10A + 8C = 15500

We want to maximize S, subject to S ≤ A.

From equation (1), we can express C as C = 1500 – S – A.
Substitute this expression for C into equation (2):
11S + 10A + 8(1500 – S – A) = 15500
11S + 10A + 12000 – 8S – 8A = 15500
3S + 2A + 12000 = 15500
3S + 2A = 15500 – 12000
3S + 2A = 3500

Now we have a single equation relating S and A: 3S + 2A = 3500.
We also have the constraint S ≤ A.

From 3S + 2A = 3500, we can express A in terms of S:
2A = 3500 – 3S
A = (3500 – 3S) / 2

Now substitute this expression for A into the inequality S ≤ A:
S ≤ (3500 – 3S) / 2
Multiply both sides by 2:
2S ≤ 3500 – 3S
Add 3S to both sides:
2S + 3S ≤ 3500
5S ≤ 3500
Divide by 5:
S ≤ 700

We also need to ensure that A and C are non-negative integers.
A = (3500 – 3S) / 2. For A to be an integer, (3500 – 3S) must be even. Since 3500 is even, 3S must be even, which means S must be even.
C = 1500 – S – A.

Since S ≤ 700 and S must be even, the maximum possible value for S is 700.
Let’s check if S = 700 is valid.
If S = 700:
A = (3500 – 3 * 700) / 2 = (3500 – 2100) / 2 = 1400 / 2 = 700.
In this case, S = 700 and A = 700. The condition S ≤ A is satisfied (700 ≤ 700).
Now calculate C:
C = 1500 – S – A = 1500 – 700 – 700 = 100.
All values S=700, A=700, C=100 are non-negative integers.
Let’s check the total fee:
1100 * 700 + 1000 * 700 + 800 * 100 = 770000 + 700000 + 80000 = 1550000.
This matches the given total fee.

Therefore, the maximum possible number of science students is 700.

Final_Answer:700

Ans: 65

Explanation:Let the arithmetic progression be denoted by $a_1, a_2, a_3, \dots$ with the first term $a$ and common difference $d$.
The $n$-th term of an arithmetic progression is given by $a_n = a + (n-1)d$.

We are given that the sum of the fourth, seventh, and tenth terms is 99.
The fourth term is $a_4 = a + (4-1)d = a + 3d$.
The seventh term is $a_7 = a + (7-1)d = a + 6d$.
The tenth term is $a_{10} = a + (10-1)d = a + 9d$.

So, $a_4 + a_7 + a_{10} = (a + 3d) + (a + 6d) + (a + 9d) = 3a + 18d$.
We are given that this sum is 99.
$3a + 18d = 99$
Dividing by 3, we get:
$a + 6d = 33$ (Equation 1)
Notice that $a + 6d$ is the seventh term, $a_7$. So, $a_7 = 33$.

We are also given that the sum of the first fourteen terms is 497.
The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}(2a + (n-1)d)$.
For $n=14$, $S_{14} = \frac{14}{2}(2a + (14-1)d) = 7(2a + 13d)$.
We are given $S_{14} = 497$.
$7(2a + 13d) = 497$
Divide by 7:
$2a + 13d = \frac{497}{7} = 71$ (Equation 2)

Now we have a system of two linear equations with two variables $a$ and $d$:
1) $a + 6d = 33$
2) $2a + 13d = 71$

From Equation 1, we can express $a$ in terms of $d$:
$a = 33 – 6d$

Substitute this expression for $a$ into Equation 2:
$2(33 – 6d) + 13d = 71$
$66 – 12d + 13d = 71$
$66 + d = 71$
$d = 71 – 66$
$d = 5$

Now substitute the value of $d$ back into the expression for $a$:
$a = 33 – 6(5)$
$a = 33 – 30$
$a = 3$

So, the first term is $a=3$ and the common difference is $d=5$.

We need to find the sum of the first five terms, $S_5$.
Using the formula $S_n = \frac{n}{2}(2a + (n-1)d)$ with $n=5$:
$S_5 = \frac{5}{2}(2a + (5-1)d)$
$S_5 = \frac{5}{2}(2a + 4d)$
Substitute the values of $a=3$ and $d=5$:
$S_5 = \frac{5}{2}(2(3) + 4(5))$
$S_5 = \frac{5}{2}(6 + 20)$
$S_5 = \frac{5}{2}(26)$
$S_5 = 5 \times 13$
$S_5 = 65$

Alternatively, we can list the first five terms and sum them:
$a_1 = 3$
$a_2 = 3 + 5 = 8$
$a_3 = 8 + 5 = 13$
$a_4 = 13 + 5 = 18$
$a_5 = 18 + 5 = 23$
$S_5 = 3 + 8 + 13 + 18 + 23 = 65$.

Final_Answer:65

Ans: 444

Let Ankita’s pace while walking be represented by $5x$. Consequently, her speed when running, which is $40\%$ faster than her walking pace, can be expressed as $1.4 \times 5x = 7x$.
The proportional relationship between her walking and running speeds is thus $5:7$. Consequently, the ratio of the durations Ankita requires to traverse a set distance while walking and running will be $7:5$.
Ankita’s journey from B to C on foot takes 3 hours and 30 minutes, equivalent to $3.5$ hours. In the alternate situation, when Ankita covers the distance from B to C at her running pace, her time will decrease proportionally to the inverse of the speed ratio. The duration of her run from B to C will be $\frac{3.5}{7} \times 5 = 2.5$ hours.
Her run from A to B requires 3 hours and 30 minutes, or $3.5$ hours. When Ankita undertakes the journey from A to B at her walking pace in the second scenario, her time will increase proportionally to the inverse of the speed ratio. The duration of her walk from A to B will be $\frac{3.5}{5} \times 7 = 4.9$ hours.
Therefore, the aggregate time for Ankita in the second scenario is $4.9 + 2.5 = 7.4$ hours. This translates to $7.4 \times 60 = 444$ minutes.

Ans: A

Explanation:Let the 4-digit number be represented as $abcd$, where $a$, $b$, $c$, and $d$ are the digits in the thousands, hundreds, tens, and units places, respectively.
We are given the following information:
1. The number is a 4-digit number greater than 1000, which means $a$ can range from 1 to 9, and $b, c, d$ can range from 0 to 9.
2. Sum of the digits in the thousands, hundreds, and tens places is 15: $a + b + c = 15$
3. Sum of the digits in the hundreds, tens, and units places is 16: $b + c + d = 16$
4. The digit in the tens place is 6 more than the digit in the units place: $c = d + 6$

From equation (4), we can determine the possible values for $c$ and $d$. Since $c$ and $d$ are digits from 0 to 9:
If $d = 0$, then $c = 0 + 6 = 6$. This is valid.
If $d = 1$, then $c = 1 + 6 = 7$. This is valid.
If $d = 2$, then $c = 2 + 6 = 8$. This is valid.
If $d = 3$, then $c = 3 + 6 = 9$. This is valid.
If $d > 3$, then $c > 9$, which is not possible for a digit.
So, the possible pairs of $(c, d)$ are $(6, 0), (7, 1), (8, 2), (9, 3)$.

Now, let’s use equation (3) to find the value of $b$ for each pair of $(c, d)$: $b = 16 – c – d$.
Case 1: $(c, d) = (6, 0)$. Then $b = 16 – 6 – 0 = 10$. This is not possible as $b$ must be a single digit.
Case 2: $(c, d) = (7, 1)$. Then $b = 16 – 7 – 1 = 8$. This is valid.
Case 3: $(c, d) = (8, 2)$. Then $b = 16 – 8 – 2 = 6$. This is valid.
Case 4: $(c, d) = (9, 3)$. Then $b = 16 – 9 – 3 = 4$. This is valid.
So, the possible sets of $(b, c, d)$ are $(8, 7, 1), (6, 8, 2), (4, 9, 3)$.

Now, let’s use equation (2) to find the value of $a$ for each set of $(b, c)$: $a = 15 – b – c$. Remember that $a$ must be between 1 and 9.
Case 2a: $(b, c, d) = (8, 7, 1)$. Then $a = 15 – 8 – 7 = 0$. This is not possible as the number is a 4-digit number greater than 1000, so $a$ cannot be 0.
Case 3a: $(b, c, d) = (6, 8, 2)$. Then $a = 15 – 6 – 8 = 1$. This is valid. The number is 1682.
Case 4a: $(b, c, d) = (4, 9, 3)$. Then $a = 15 – 4 – 9 = 2$. This is valid. The number is 2493.

So, the only possible 4-digit numbers that satisfy all the conditions are 1682 and 2493.

The largest possible value of the number is 2493.
The smallest possible value of the number is 1682.

The difference between the largest and smallest possible value of the number is $2493 – 1682 = 811$.

Correct_Option: A

Ans: B

Let Rahul’s usual travel pace be represented by $x$ kilometers per hour. If his typical journey takes 6 hours (from 5 pm to 11 pm), the overall distance covered is $6 \times x = 6x$ kilometers.

Let $y$ kilometers be the distance covered before Rahul takes a break in both situations.

In the first instance, a 20-minute break, equivalent to $\frac{1}{3}$ of an hour, means his actual travel time is $6 – \frac{1}{3} = \frac{17}{3}$ hours. The equation representing this scenario is:
$\dfrac{y}{x} + \dfrac{6x-y}{x+3} = \dfrac{17}{3}$
$\Rightarrow \dfrac{6x^2+3y}{x^2+3x} = \dfrac{17}{3}$
$\Rightarrow 18x^2 + 9y = 17x^2 + 51x$
$\Rightarrow x^2 = 51x – 9y$ …..(1)

In the second instance, a break of 20 minutes plus an additional 10 minutes, totaling 30 minutes or $\frac{1}{2}$ of an hour, means his actual travel time is $6 – \frac{1}{2} = \frac{11}{2}$ hours. The equation for this situation is:
$\dfrac{y}{x} + \dfrac{6x-y}{x+5} = \dfrac{11}{2}$
$\Rightarrow \dfrac{6x^2+5y}{x^2+5x} = \dfrac{11}{2}$
$\Rightarrow 12x^2 + 10y = 11x^2 + 55x$
$\Rightarrow x^2 = 55x – 10y$ …..(2)

Equating expressions for $x^2$ from equations (1) and (2):
$55x – 10y = 51x – 9y$
$4x = y$

Substituting the value of $y$ into equation (1):
$x^2 = 51x – 36x$
Since $x$ represents speed and must be positive, $x = 15$. Thus, the correct option is B.

Ans: A

Explanation:Let the rate of water flow through pipes A, B, and C be $4x$, $9x$, and $36x$ respectively, where $x$ is a constant.
The rate of pipe A is $4x$.
Pipe A can fill the empty tank completely in 15 hours.
The volume of the tank can be represented as the rate multiplied by the time.
Volume of the tank = Rate of pipe A * Time taken by pipe A
Volume of the tank = $(4x) * 15$ hours = $60x$ tank-units-hours.

If all three pipes A, B, and C are used simultaneously, their combined rate is the sum of their individual rates:
Combined rate = Rate of A + Rate of B + Rate of C
Combined rate = $4x + 9x + 36x = 49x$ tank-units-hours.

Let $T$ be the time required to fill the tank completely when all three pipes are used simultaneously.
Volume of the tank = Combined rate * $T$
$60x = (49x) * T$

To find $T$, we can divide both sides by $49x$:
$T = \frac{60x}{49x} = \frac{60}{49}$ hours.

The question asks for the time in minutes. To convert hours to minutes, we multiply by 60:
Time in minutes = $\frac{60}{49} * 60$ minutes
Time in minutes = $\frac{3600}{49}$ minutes.

Now, we need to calculate the value of $\frac{3600}{49}$ and find the nearest option.
$\frac{3600}{49} \approx 73.469$ minutes.

Let’s check the given options:
Option A: 73
Option B: 78
Option C: 76
Option D: 71

The calculated time is approximately 73.469 minutes. The nearest option is 73 minutes.

Correct_Option:A

Ans: D

Explanation:Let $y = x^2 + 3x$. Then the given function $f(x)$ can be written as $f(x) = y(y+2)$.
The equation is $\sqrt{f(x)+1} = 9701$.
Squaring both sides, we get $f(x)+1 = 9701^2$.
Substituting the expression for $f(x)$, we have $y(y+2)+1 = 9701^2$.
Expanding this, we get $y^2 + 2y + 1 = 9701^2$.
This can be written as $(y+1)^2 = 9701^2$.
Taking the square root of both sides, we get $y+1 = \pm 9701$.
So, we have two cases:
Case 1: $y+1 = 9701$
$y = 9701 – 1$
$y = 9700$
Substituting back $y = x^2 + 3x$, we have $x^2 + 3x = 9700$.
$x^2 + 3x – 9700 = 0$.
This is a quadratic equation. Let the roots be $x_1$ and $x_2$. By Vieta’s formulas, the sum of the roots is $x_1 + x_2 = -\frac{3}{1} = -3$.

Case 2: $y+1 = -9701$
$y = -9701 – 1$
$y = -9702$
Substituting back $y = x^2 + 3x$, we have $x^2 + 3x = -9702$.
$x^2 + 3x + 9702 = 0$.
This is a quadratic equation. To check if the roots are real, we calculate the discriminant $\Delta = b^2 – 4ac$.
Here, $a=1$, $b=3$, $c=9702$.
$\Delta = 3^2 – 4(1)(9702) = 9 – 38808 = -38799$.
Since the discriminant is negative, the roots of this quadratic equation are complex and not real.

Therefore, the only real roots come from the equation $x^2 + 3x – 9700 = 0$. The sum of these real roots is -3.

We need to ensure that for the real roots found, $f(x)+1 \geq 0$.
In Case 1, $y=9700$, so $f(x)+1 = y(y+2)+1 = 9700(9700+2)+1 = 9700(9702)+1 > 0$. So the square root is well-defined.
In Case 2, $y=-9702$, so $f(x)+1 = y(y+2)+1 = -9702(-9702+2)+1 = -9702(-9700)+1 > 0$. However, the roots are complex.

The question asks for the sum of all real roots of the equation $\sqrt{f(x)+1}= 9701$.
From Case 1, the sum of the real roots of $x^2 + 3x – 9700 = 0$ is -3.

Correct_Option: D

Ans: C

We are given the expression $y= \dfrac{2x-3}{2x^{2}+4x-6}$.
Rearranging this equation to express it in terms of $x$ yields:
$2yx^2 + 4yx – 6y = 2x – 3$
This can be further simplified to:
$2yx^2 + (4y-2)x – 6y + 3 = 0$
Let’s refer to this as Equation (1).

Equation (1) represents a quadratic equation in the variable $x$, and since $x$ is a real number, the discriminant of this quadratic must be non-negative.
Applying the discriminant condition, we get:
$(4y-2)^2 + 4(2y)(-(6y-3)) \geq 0$
Expanding and simplifying the inequality:
$(16y^2 – 16y + 4) + 8y(3-6y) \geq 0$
$16y^2 – 16y + 4 + 24y – 48y^2 \geq 0$
$-32y^2 + 8y + 4 \geq 0$
Dividing by -4 and reversing the inequality sign:
$8y^2 – 2y – 1 \leq 0$

To find the roots of the quadratic $8y^2 – 2y – 1 = 0$, we use the quadratic formula:
$y = \dfrac{-(-2) \pm \sqrt{(-2)^2 – 4(8)(-1)}}{2(8)}$
$y = \dfrac{2 \pm \sqrt{4 + 32}}{16}$
$y = \dfrac{2 \pm \sqrt{36}}{16}$
$y = \dfrac{2 \pm 6}{16}$
The roots are $y = \dfrac{2+6}{16} = \dfrac{8}{16} = \dfrac{1}{2}$ and $y = \dfrac{2-6}{16} = \dfrac{-4}{16} = -\dfrac{1}{4}$.

Since the coefficient of $y^2$ in $8y^2 – 2y – 1$ is positive (8), the quadratic expression is less than or equal to zero between its roots.
Therefore, the possible values of $y$, which represent the range of the function $f(x)$, are:
$\left[-\dfrac{1}{4}, \dfrac{1}{2}\right]$

Option C is the correct answer.

Ans: B

The expression $10^{50} + 10^{25} – 123$ can be rewritten as $10^{50} + (10^{25}-123)$.

Let’s examine the pattern of subtracting 123 from powers of 10:
$10^3 – 123 = 1000 – 123 = 877$
$10^4 – 123 = 10000 – 123 = 9877$
$10^5 – 123 = 100000 – 123 = 99877$
$10^6 – 123 = 1000000 – 123 = 999877$

For an expression of the form $10^n – 123$, which has $n$ digits in total:
The digit $7$ appears $2$ times.
The digit $8$ appears $1$ time.
The digit $9$ appears $(n-3)$ times.

Now, consider the addition to $10^{50}$. The addition of $10^{50}$ to the result of $(10^{25}-123)$ will effectively shift the digits of $(10^{25}-123)$ and add a $1$ as the leftmost digit of the final sum. The digits originating from $(10^{25}-123)$ that are now part of the larger number will contribute their values, but any new digits introduced by the addition of $10^{50}$ to the left of these will be zeros.

Therefore, the sum of the digits of the final result is calculated as follows:
The contribution from the digits of $(10^{25}-123)$ is:
Digits $7$: $7 \times 2 = 14$
Digit $8$: $1 \times 8 = 8$
Digits $9$: $(25-3) \times 9 = 22 \times 9 = 198$
The contribution from the new leading digits due to the addition of $10^{50}$:
The leftmost digit will be $1$.
All other newly introduced digits will be $0$. Their sum contribution is $25 \times 0 = 0$.

Total sum of digits = $14 + 8 + 198 + 0 + 1 = 221$

The correct option is B.

Ans: 126

To be published

Ans: A

Given the expression $\left(x+\dfrac{1}{x}\right)^2$, we can expand it as follows:
$\left(x+\dfrac{1}{x}\right)^2 = x^2+\dfrac{1}{x^2} + 2$
Since we are provided that $x^2+\dfrac{1}{x^2} = 25$, we substitute this value into the expanded form:
$25 + 2 = 27$
Thus, we have:
$\left(x+\dfrac{1}{x}\right)^2 = 27$
Taking the square root of both sides, we get:
$\left(x+\dfrac{1}{x}\right) = \sqrt{27} = 3\sqrt{3}$

Next, consider the expansion of $\left(x+\dfrac{1}{x}\right)^3$:
$\left(x+\dfrac{1}{x}\right)^3 = x^3 + \dfrac{1}{x^3} + 3\left(x+\dfrac{1}{x}\right)$
We can rearrange this to solve for $x^3 + \dfrac{1}{x^3}$:
$x^3 + \dfrac{1}{x^3} = \left(x+\dfrac{1}{x}\right)^3 – 3\left(x+\dfrac{1}{x}\right)$
Substitute the value of $\left(x+\dfrac{1}{x}\right) = 3\sqrt{3}$:
$x^3 + \dfrac{1}{x^3} = (3\sqrt{3})^3 – 3(3\sqrt{3})$
$x^3 + \dfrac{1}{x^3} = 27 \times 3\sqrt{3} – 9\sqrt{3}$
$x^3 + \dfrac{1}{x^3} = 81\sqrt{3} – 9\sqrt{3} = 72\sqrt{3}$

Now, let’s find $x^4 + \dfrac{1}{x^4}$. We square the expression for $x^2+\dfrac{1}{x^2}$:
$\left(x^2+\dfrac{1}{x^2}\right)^2 = x^4 + \dfrac{1}{x^4} + 2$
We are given $x^2+\dfrac{1}{x^2} = 25$. Substituting this value:
$(25)^2 = x^4 + \dfrac{1}{x^4} + 2$
$625 = x^4 + \dfrac{1}{x^4} + 2$
Rearranging to solve for $x^4 + \dfrac{1}{x^4}$:
$x^4 + \dfrac{1}{x^4} = 625 – 2 = 623$

Finally, to find $x^7+\dfrac{1}{x^7}$, we multiply the expressions for $x^4 + \dfrac{1}{x^4}$ and $x^3 + \dfrac{1}{x^3}$:
$\left(x^4+\dfrac{1}{x^4}\right)\left(x^3+\dfrac{1}{x^3}\right) = x^7+\dfrac{1}{x^7} + x\left(\dfrac{1}{x^3}\right) + \dfrac{1}{x^4}(x^3)$
$\left(x^4+\dfrac{1}{x^4}\right)\left(x^3+\dfrac{1}{x^3}\right) = x^7+\dfrac{1}{x^7} + \dfrac{1}{x^2} + \dfrac{1}{x}$
This can be rearranged as:
$\left(x^4+\dfrac{1}{x^4}\right)\left(x^3+\dfrac{1}{x^3}\right) = x^7+\dfrac{1}{x^7} + \left(x + \dfrac{1}{x}\right)$
Now, substitute the calculated values: $x^4 + \dfrac{1}{x^4} = 623$, $x^3 + \dfrac{1}{x^3} = 72\sqrt{3}$, and $x + \dfrac{1}{x} = 3\sqrt{3}$:
$(623)(72\sqrt{3}) = x^7+\dfrac{1}{x^7} + 3\sqrt{3}$
$44856\sqrt{3} = x^7+\dfrac{1}{x^7} + 3\sqrt{3}$
Solving for $x^7+\dfrac{1}{x^7}$:
$x^7+\dfrac{1}{x^7} = 44856\sqrt{3} – 3\sqrt{3} = 44853\sqrt{3}$

Option A is the correct answer.

Ans: B

To be published

Ans: D

For a logarithm to be defined, its base must be positive and not equal to 1. Therefore, $x$ must be greater than 3 and cannot be 4. Additionally, the argument of a logarithm must be positive, so $x^2 – 9 > 0$, which implies $x > 3$.

The given equation can be expressed as:
$\log_{x-3}{(x^{2}-9)}-\log_{x-3}{(x+1)} = 2$

Using the logarithm property $\log_b M – \log_b N = \log_b \frac{M}{N}$:
$\log_{x-3}{\dfrac{x^2-9}{x+1}} = 2$

Converting the logarithmic equation to an exponential form:
$\dfrac{x^2-9}{x+1} = (x-3)^2$

Factor the numerator:
$\dfrac{(x+3)(x-3)}{x+1} = (x-3)^2$

Assuming $x \neq 3$ (which is already covered by $x>3$), we can divide both sides by $(x-3)$:
$\dfrac{x+3}{x+1} = x-3$

Cross-multiply:
$x+3 = (x-3)(x+1)$

Expand the right side:
$x+3 = x^2 – 3x + x – 3$
$x+3 = x^2 – 2x – 3$

Rearrange into a quadratic equation:
$x^2 – 2x – x – 3 – 3 = 0$
$x^2 – 3x – 6 = 0$

The roots of this quadratic equation can be found using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$x = \dfrac{-(-3) \pm \sqrt{(-3)^2 – 4(1)(-6)}}{2(1)}$
$x = \dfrac{3 \pm \sqrt{9 + 24}}{2}$
$x = \dfrac{3 \pm \sqrt{33}}{2}$

Considering the constraints established earlier ($x > 3$), the negative root is not valid. The positive root is $\dfrac{3+ \sqrt{33}}{2}$. This value is greater than 3, and since $\sqrt{33}$ is between 5 and 6, the value is approximately $(3+5.something)/2$, which is greater than 4. Thus, it satisfies the conditions.

The correct answer is option D.

Ans: C

Explanation:Let $S_M$ be the total salary of 5 managers and $S_E$ be the total salary of 25 engineers.
The total number of employees is $5 + 25 = 30$.
The average salary of all 30 employees is 60000 rupees.
So, the total salary of all 30 employees is $30 \times 60000 = 1800000$ rupees.
This means $S_M + S_E = 1800000$.

Let the average salary of managers be $A_M$ and the average salary of engineers be $A_E$.
Then, $S_M = 5 \times A_M$ and $S_E = 25 \times A_E$.
So, $5 A_M + 25 A_E = 1800000$.

When each of the managers received a 20% salary increase, their new total salary becomes $S_M \times (1 + 0.20) = 1.20 S_M$.
The salary of the engineers remained unchanged, so their total salary is still $S_E$.
The new total salary of all 30 employees is $1.20 S_M + S_E$.

The average salary of all 30 employees would have increased by 5%.
The new average salary is $60000 \times (1 + 0.05) = 60000 \times 1.05 = 63000$ rupees.
The new total salary of all 30 employees is $30 \times 63000 = 1890000$ rupees.

So, $1.20 S_M + S_E = 1890000$.

We have a system of two linear equations:
1) $S_M + S_E = 1800000$
2) $1.20 S_M + S_E = 1890000$

Subtract equation (1) from equation (2):
$(1.20 S_M + S_E) – (S_M + S_E) = 1890000 – 1800000$
$0.20 S_M = 90000$
$S_M = \frac{90000}{0.20} = 90000 \times 5 = 450000$ rupees.

Now substitute the value of $S_M$ into equation (1):
$450000 + S_E = 1800000$
$S_E = 1800000 – 450000 = 1350000$ rupees.

We need to find the average salary of the engineers, which is $A_E = \frac{S_E}{25}$.
$A_E = \frac{1350000}{25}$

To calculate $\frac{1350000}{25}$:
$\frac{1350000}{25} = \frac{135 \times 10000}{25} = 135 \times \frac{10000}{25} = 135 \times 400$.
$135 \times 400 = 135 \times 4 \times 100 = 540 \times 100 = 54000$.

So, the average salary of the engineers is 54000 rupees.

Let’s verify the average salary of managers:
$A_M = \frac{S_M}{5} = \frac{450000}{5} = 90000$ rupees.
Initial average salary = $\frac{5 \times 90000 + 25 \times 54000}{30} = \frac{450000 + 1350000}{30} = \frac{1800000}{30} = 60000$. (Correct)

New salary of managers = $90000 \times 1.20 = 108000$.
New average salary = $\frac{5 \times 108000 + 25 \times 54000}{30} = \frac{540000 + 1350000}{30} = \frac{1890000}{30} = 63000$.
Increase in average salary = $63000 – 60000 = 3000$.
Percentage increase = $\frac{3000}{60000} \times 100\% = \frac{1}{20} \times 100\% = 5\%$. (Correct)

Correct_Option: C

Ans: A

To be published

Ans: 112

Explanation:We are given the equation:
$12^{12x}\times 4^{24x+12}\times 5^{2y}=8^{4z}\times 20 ^{12x} \times 243^{3x-6}$
First, we express all the bases in terms of their prime factors:
$12 = 2^2 \times 3$
$4 = 2^2$
$8 = 2^3$
$20 = 2^2 \times 5$
$243 = 3^5$

Substitute these into the equation:
$(2^2 \times 3)^{12x} \times (2^2)^{24x+12} \times 5^{2y} = (2^3)^{4z} \times (2^2 \times 5)^{12x} \times (3^5)^{3x-6}$

Now, apply the exponent rules $(a^m)^n = a^{mn}$ and $(ab)^m = a^m b^m$:
$(2^{2 \times 12x} \times 3^{12x}) \times 2^{2 \times (24x+12)} \times 5^{2y} = 2^{3 \times 4z} \times (2^{2 \times 12x} \times 5^{12x}) \times 3^{5 \times (3x-6)}$
$2^{24x} \times 3^{12x} \times 2^{48x+24} \times 5^{2y} = 2^{12z} \times 2^{24x} \times 5^{12x} \times 3^{15x-30}$

Combine the terms with the same base by adding their exponents:
$2^{24x + 48x + 24} \times 3^{12x} \times 5^{2y} = 2^{12z + 24x} \times 3^{15x-30} \times 5^{12x}$
$2^{72x + 24} \times 3^{12x} \times 5^{2y} = 2^{12z + 24x} \times 3^{15x-30} \times 5^{12x}$

For the equality to hold, the exponents of each prime base on both sides of the equation must be equal.

Equating the exponents of base 2:
$72x + 24 = 12z + 24x$
$72x – 24x + 24 = 12z$
$48x + 24 = 12z$
Divide by 12:
$4x + 2 = z$ (Equation 1)

Equating the exponents of base 3:
$12x = 15x – 30$
$30 = 15x – 12x$
$30 = 3x$
$x = 10$

Equating the exponents of base 5:
$2y = 12x$
Divide by 2:
$y = 6x$

Now substitute the value of x into Equation 1 to find z:
$z = 4(10) + 2$
$z = 40 + 2$
$z = 42$

Now substitute the value of x into the equation for y:
$y = 6(10)$
$y = 60$

We are given that x, y, and z are natural numbers. Our calculated values are x=10, y=60, and z=42, which are all natural numbers.

We need to find $x + y + z$:
$x + y + z = 10 + 60 + 42$
$x + y + z = 112$

Final_Answer:112

Ans: C

To be published

Ans: 16

Container P holds 60 litres of pure spirit, and Container Q holds 60 litres of pure water. Suppose $x$ litres are removed from container P. Container P now has $60-x$ litres of spirit. Container Q now contains 60 litres of water and $x$ litres of spirit.
After combining the contents of Q and then removing the same amount $x$, the spirit removed from Container Q would be ${\left(\dfrac{x}{60+x}\right)}$ of $x$, which equates to $\dfrac{x^2}{60+x}$ litres.
The initial total volume of 60 litres has been re-established in P after the substitution. The final quantity of spirit in Container P is:
$60 – x + \dfrac{x^2}{60+x} = \dfrac{(60+x)(60-x) + x^2}{60+x} = \dfrac{3600}{60+x}$
The total volume in container P is 60 litres, with the proportion of spirit to the total volume being $\dfrac{15}{15+4} = \dfrac{15}{19}$.
Consequently,
$\dfrac{\left(\dfrac{3600}{60+x}\right)}{60} = \dfrac{15}{19}$
$\Rightarrow \dfrac{60}{60+x} = \dfrac{15}{19}$
$\Rightarrow x = 16$
Hence, the amount transferred and replaced in both steps is 16 litres.

Ans: 272

Explanation:Let the initial number of coins in box A be $17x$ and in box B be $7x$.
The ratio of the number of coins in boxes A and B was 17:7.

After 108 coins were shifted from box A to box B:
Number of coins in box A becomes $17x – 108$.
Number of coins in box B becomes $7x + 108$.

The new ratio became 37:20.
So, we can write the equation:
$\frac{17x – 108}{7x + 108} = \frac{37}{20}$

Cross-multiply:
$20(17x – 108) = 37(7x + 108)$
$340x – 2160 = 259x + 3996$

Now, solve for x:
$340x – 259x = 3996 + 2160$
$81x = 6156$
$x = \frac{6156}{81}$
$x = \frac{684}{9}$
$x = 76$

Now, let’s find the current number of coins in boxes A and B.
Current number of coins in box A = $17x – 108 = 17(76) – 108 = 1292 – 108 = 1184$.
Current number of coins in box B = $7x + 108 = 7(76) + 108 = 532 + 108 = 640$.

We want to find the number of coins that needs to be shifted further from A to B to make the ratio 1:1. Let this number be $y$.
After shifting $y$ coins from A to B:
Number of coins in box A becomes $1184 – y$.
Number of coins in box B becomes $640 + y$.

For the ratio to be 1:1, the number of coins in both boxes must be equal:
$1184 – y = 640 + y$

Now, solve for y:
$1184 – 640 = y + y$
$544 = 2y$
$y = \frac{544}{2}$
$y = 272$

So, 272 coins need to be shifted further from A to B.

Final_Answer:272

Ans: 397

Given the equation $p+q+r = 900$, where $r$ is a perfect square and its value is greater than $150$ and less than $500$.
Let $q$ be a constant value, irrespective of any range.
The problem states that $0.3q \leq p \leq 0.7q$.
Adding $q$ to all parts of this inequality, we get $0.3q + q \leq p+q \leq 0.7q + q$, which simplifies to $1.3q \leq p+q \leq 1.7q$.
From the initial equation, we can express $p+q$ as $900-r$.
Substituting this into the inequality, we have $1.3q \leq 900-r \leq 1.7q$.
The bounds of $p$ are directly related to $q$. Therefore, to determine the extreme values of $p$, we need to find the extreme values of $r$.

To find the minimum possible value of $p$, we consider the lower bound of the inequality: $1.3q \leq 900-r$.
The minimum value of $p$ occurs when $1.3q$ is minimized. This happens when $q$ is minimized, which in turn occurs when $900-r$ is minimized. For $900-r$ to be minimized, $r$ must be maximized.
The largest perfect square between $150$ and $500$ is $484$ ($22^2$).
Setting $r = 484$, we have $1.3q = 900-484$, which gives $1.3q = 416$. Solving for $q$, we get $q = 320$.
With $q=320$, the minimum value of $p$ is $0.3 \times 320 = 96$.

To find the maximum possible value of $p$, we consider the upper bound of the inequality: $900-r \leq 1.7q$.
The maximum value of $p$ occurs when $0.7q$ is maximized. This happens when $q$ is maximized, which in turn occurs when $900-r$ is maximized. For $900-r$ to be maximized, $r$ must be minimized.
The smallest perfect square between $150$ and $500$ is $169$ ($13^2$).
Setting $r = 169$, we have $900-169 = 1.7q$, which gives $731 = 1.7q$. Solving for $q$, we get $q = 430$.
With $q=430$, the maximum value of $p$ is $0.7 \times 430 = 301$.

The sum of the maximum and minimum values of $p$ is $96 + 301 = 397$.