CAT 2025 Quant Slot 2 Paper

Ans: B

Explanation:Let the given equation be
$9^{x^{2}+2x-3}-4(3^{x^{2}+2x-2})+27=0$
We can rewrite the terms using the properties of exponents.
$9^{x^{2}+2x-3} = (3^2)^{x^{2}+2x-3} = 3^{2(x^{2}+2x-3)} = 3^{2x^{2}+4x-6}$
Also,
$3^{x^{2}+2x-2} = 3^{x^{2}+2x-3+1} = 3^{x^{2}+2x-3} \cdot 3^1 = 3 \cdot 3^{x^{2}+2x-3}$
Let $y = 3^{x^{2}+2x-3}$. Then the equation can be rewritten in terms of $y$.
We have $9^{x^{2}+2x-3} = (3^2)^{x^{2}+2x-3} = (3^{x^{2}+2x-3})^2 = y^2$.
And $3^{x^{2}+2x-2} = 3^{x^{2}+2x-3+1} = 3^{x^{2}+2x-3} \cdot 3^1 = 3y$.
Substituting these into the original equation:
$y^2 – 4(3y) + 27 = 0$
$y^2 – 12y + 27 = 0$
This is a quadratic equation in $y$. We can factor it:
$(y-3)(y-9) = 0$
So, the possible values for $y$ are $y=3$ or $y=9$.

Now we substitute back $y = 3^{x^{2}+2x-3}$:

Case 1: $y=3$
$3^{x^{2}+2x-3} = 3$
$3^{x^{2}+2x-3} = 3^1$
Equating the exponents:
$x^{2}+2x-3 = 1$
$x^{2}+2x-4 = 0$
Let the roots of this quadratic equation be $x_1$ and $x_2$. By Vieta’s formulas, the product of the roots is $x_1 x_2 = \frac{-4}{1} = -4$.

Case 2: $y=9$
$3^{x^{2}+2x-3} = 9$
$3^{x^{2}+2x-3} = 3^2$
Equating the exponents:
$x^{2}+2x-3 = 2$
$x^{2}+2x-5 = 0$
Let the roots of this quadratic equation be $x_3$ and $x_4$. By Vieta’s formulas, the product of the roots is $x_3 x_4 = \frac{-5}{1} = -5$.

The problem asks for the product of all possible values of $x$. The possible values of $x$ are the roots of $x^{2}+2x-4 = 0$ and $x^{2}+2x-5 = 0$.
The product of all possible values of $x$ is $(x_1 x_2) \cdot (x_3 x_4) = (-4) \cdot (-5) = 20$.

To verify that these quadratic equations yield real roots, we can check the discriminant.
For $x^{2}+2x-4 = 0$, the discriminant is $\Delta = b^2 – 4ac = 2^2 – 4(1)(-4) = 4 + 16 = 20 > 0$. So, there are two distinct real roots.
For $x^{2}+2x-5 = 0$, the discriminant is $\Delta = b^2 – 4ac = 2^2 – 4(1)(-5) = 4 + 20 = 24 > 0$. So, there are two distinct real roots.
Thus, there are four distinct real values of $x$.

The product of all possible values of $x$ is $(-4) \times (-5) = 20$.

The final answer is $\boxed{20}$.
Correct_Option:B

Ans: 49

Explanation:Let $S_i$ be the number of copies sold on the $i$-th day.

The average number of copies sold per day in the initial seven days is 60.
So, the total number of copies sold in the first seven days is $7 \times 60 = 420$.
This can be written as: $S_1 + S_2 + S_3 + S_4 + S_5 + S_6 + S_7 = 420$.

The average number of copies sold per day in the initial eight days is 63.
So, the total number of copies sold in the first eight days is $8 \times 63 = 504$.
This can be written as: $S_1 + S_2 + S_3 + S_4 + S_5 + S_6 + S_7 + S_8 = 504$.

From the above two equations, we can find the number of copies sold on the eighth day ($S_8$):
$S_8 = (S_1 + … + S_8) – (S_1 + … + S_7) = 504 – 420 = 84$.
So, $S_8 = 84$.

On the ninth day, she sells 11 copies less than the eighth day.
So, $S_9 = S_8 – 11 = 84 – 11 = 73$.
So, $S_9 = 73$.

The average number of copies sold per day from the second day to the ninth day is 66.
This means the sum of copies sold from day 2 to day 9 divided by the number of days (which is 8) is 66.
So, $S_2 + S_3 + S_4 + S_5 + S_6 + S_7 + S_8 + S_9 = 8 \times 66 = 528$.

We know the sum of copies sold from day 1 to day 7 is 420:
$S_1 + S_2 + S_3 + S_4 + S_5 + S_6 + S_7 = 420$.

We can rewrite the sum from day 2 to day 9 as:
$(S_1 + S_2 + … + S_7 + S_8 + S_9) – S_1 = 528$.

We also know that $S_1 + S_2 + … + S_7 = 420$, and $S_8 = 84$, $S_9 = 73$.
Let’s consider the sum of copies sold from the first day to the ninth day.
The sum of copies sold from day 2 to day 9 is 528.
So, $S_2 + S_3 + S_4 + S_5 + S_6 + S_7 + S_8 + S_9 = 528$.

We can express the sum of the first seven days as:
$(S_2 + S_3 + S_4 + S_5 + S_6 + S_7) + S_1 = 420$.

We also know the sum from day 2 to day 9:
$(S_2 + S_3 + S_4 + S_5 + S_6 + S_7) + S_8 + S_9 = 528$.

Substitute the sum from day 2 to day 7 into the second equation:
$(420 – S_1) + S_8 + S_9 = 528$.

Now, substitute the values of $S_8$ and $S_9$:
$(420 – S_1) + 84 + 73 = 528$.
$420 – S_1 + 157 = 528$.
$577 – S_1 = 528$.
$S_1 = 577 – 528$.
$S_1 = 49$.

The number of copies sold on the first day of the book launch is 49.

Let’s verify the calculations.
Sum of first 7 days = 420.
Sum of first 8 days = 504. $S_8 = 504 – 420 = 84$.
$S_9 = 84 – 11 = 73$.
Average from day 2 to day 9 is 66.
Sum from day 2 to day 9 = $8 \times 66 = 528$.
Sum of first 7 days = $S_1 + S_2 + … + S_7 = 420$.
Sum of day 2 to day 9 = $(S_1 + S_2 + … + S_7 + S_8 + S_9) – S_1$.
$528 = (420 + 84 + 73) – S_1$.
$528 = 577 – S_1$.
$S_1 = 577 – 528 = 49$.

Final_Answer:49

Ans: A

Explanation:We want to find the set of all real values of x for which $x^2 – |x+9| + x > 0$.
We need to consider two cases based on the absolute value term $|x+9|$.

Case 1: $x+9 \ge 0$, which means $x \ge -9$.
In this case, $|x+9| = x+9$. The inequality becomes:
$x^2 – (x+9) + x > 0$
$x^2 – x – 9 + x > 0$
$x^2 – 9 > 0$
$(x-3)(x+3) > 0$
This inequality holds when $x < -3$ or $x > 3$.
Since we are in the case where $x \ge -9$, we need to find the intersection of $x \ge -9$ and ($x < -3$ or $x > 3$).
The intersection is $[-9, -3) \cup (3, \infty)$.

Case 2: $x+9 < 0$, which means $x < -9$.
In this case, $|x+9| = -(x+9)$. The inequality becomes:
$x^2 – (-(x+9)) + x > 0$
$x^2 + x + 9 + x > 0$
$x^2 + 2x + 9 > 0$
To determine when this quadratic is positive, we can look at its discriminant. The discriminant is $\Delta = b^2 – 4ac = (2)^2 – 4(1)(9) = 4 – 36 = -32$.
Since the discriminant is negative and the leading coefficient (1) is positive, the quadratic $x^2 + 2x + 9$ is always positive for all real values of x.
Since we are in the case where $x < -9$, the inequality $x^2 + 2x + 9 > 0$ is satisfied for all $x < -9$.
The intersection of $x < -9$ and ($x^2 + 2x + 9 > 0$) is $(-\infty, -9)$.

Now, we need to combine the solutions from both cases. The solution set is the union of the solutions from Case 1 and Case 2:
$(-\infty, -9) \cup [-9, -3) \cup (3, \infty)$
The union of $(-\infty, -9)$ and $[-9, -3)$ is $(-\infty, -3)$.
Therefore, the complete solution set is $(-\infty, -3) \cup (3, \infty)$.

Comparing this solution with the given options:
Option A: $(-\infty,-3)\cup (3,\infty)$
Option B: $(-\infty,-9)\cup (3,\infty)$
Option C: $(-9,-3)\cup (3,\infty)$
Option D: $(-\infty,-9)\cup (9,\infty)$

Our calculated solution matches Option A.

Final check of the boundaries:
If $x=-3$, $x^2-|x+9|+x = (-3)^2 – |-3+9| + (-3) = 9 – |6| – 3 = 9 – 6 – 3 = 0$. So $x=-3$ is not included.
If $x=3$, $x^2-|x+9|+x = (3)^2 – |3+9| + (3) = 9 – |12| + 3 = 9 – 12 + 3 = 0$. So $x=3$ is not included.
If $x=-9$, $x^2-|x+9|+x = (-9)^2 – |-9+9| + (-9) = 81 – |0| – 9 = 81 – 0 – 9 = 72 > 0$. So $x=-9$ is included in our solution set.
Let’s re-examine the union: $(-\infty, -9) \cup [-9, -3) \cup (3, \infty)$.
The union of $(-\infty, -9)$ and $[-9, -3)$ is indeed $(-\infty, -3)$.
So the final solution is $(-\infty, -3) \cup (3, \infty)$.

Correct_Option:A

Ans: C

Let the list price be denoted by $L$ and the initial discount percentage be $x$. The purchase price is $1650$. A profit of $20\%$ implies the sale price is:
$SP = 1650 \times 1.20 = 1980$
With discount $x$, we have:
$L(1-x) = 1980$
If the discount is doubled (i.e., $2x$), the sale price becomes $L(1-2x)$, and the profit is $110$. Therefore:
$L(1-2x) – 1650 = 110 \implies L(1-2x) = 1760$
Subtracting the two derived equations:
$L(1-x) – L(1-2x) = 1980 – 1760$
$Lx = 220 \implies L = \frac{220}{x}$
Substitute this expression for $L$ into the first equation:
$\frac{220}{x}(1-x) = 1980 \implies \frac{1-x}{x} = \frac{1980}{220} \implies \frac{1-x}{x} = 9$
$1-x = 9x \implies 1 = 10x \implies x = 0.1$
So, the initial discount percentage is $10\%$, and the list price is:
$L = \frac{220}{0.1} = 2200$
Now, let the new discount percentage be $y$ such that the profit percentage is equal to the discount percentage. The equation for this condition is:
$\frac{2200(1-y) – 1650}{1650} = y$
Rearranging and simplifying:
$2200(1-y) – 1650 = 1650y$
$2200 – 2200y – 1650 = 1650y$
$550 = 3850y$
$y = \frac{550}{3850} \approx 0.142857$
Thus, the required discount percentage is approximately $14\%$.

Ans: 17

Explanation:We are given the equation $(m+2n)(2m+n)=27$, where m and n are integers. We need to find the maximum possible value of $2m-3n$.

Since m and n are integers, m+2n and 2m+n must also be integers. The factors of 27 are:
(1, 27), (3, 9), (9, 3), (27, 1), (-1, -27), (-3, -9), (-9, -3), (-27, -1).

We can set up systems of equations for each pair of factors:

Case 1:
m + 2n = 1
2m + n = 27
Multiply the second equation by 2: 4m + 2n = 54.
Subtract the first equation from this: (4m + 2n) – (m + 2n) = 54 – 1 => 3m = 53. m = 53/3, which is not an integer.

Case 2:
m + 2n = 3
2m + n = 9
Multiply the second equation by 2: 4m + 2n = 18.
Subtract the first equation from this: (4m + 2n) – (m + 2n) = 18 – 3 => 3m = 15. m = 5.
Substitute m=5 into the first equation: 5 + 2n = 3 => 2n = -2 => n = -1.
Check with the second equation: 2(5) + (-1) = 10 – 1 = 9. This pair (m=5, n=-1) is valid.
For this pair, $2m-3n = 2(5) – 3(-1) = 10 + 3 = 13$.

Case 3:
m + 2n = 9
2m + n = 3
Multiply the second equation by 2: 4m + 2n = 6.
Subtract this from the first equation: (m + 2n) – (4m + 2n) = 9 – 6 => -3m = 3. m = -1.
Substitute m=-1 into the second equation: 2(-1) + n = 3 => -2 + n = 3 => n = 5.
Check with the first equation: -1 + 2(5) = -1 + 10 = 9. This pair (m=-1, n=5) is valid.
For this pair, $2m-3n = 2(-1) – 3(5) = -2 – 15 = -17$.

Case 4:
m + 2n = 27
2m + n = 1
Multiply the second equation by 2: 4m + 2n = 2.
Subtract this from the first equation: (m + 2n) – (4m + 2n) = 27 – 2 => -3m = 25. m = -25/3, not an integer.

Case 5:
m + 2n = -1
2m + n = -27
Multiply the second equation by 2: 4m + 2n = -54.
Subtract the first equation from this: (4m + 2n) – (m + 2n) = -54 – (-1) => 3m = -53. m = -53/3, not an integer.

Case 6:
m + 2n = -3
2m + n = -9
Multiply the second equation by 2: 4m + 2n = -18.
Subtract the first equation from this: (4m + 2n) – (m + 2n) = -18 – (-3) => 3m = -15. m = -5.
Substitute m=-5 into the first equation: -5 + 2n = -3 => 2n = 2 => n = 1.
Check with the second equation: 2(-5) + 1 = -10 + 1 = -9. This pair (m=-5, n=1) is valid.
For this pair, $2m-3n = 2(-5) – 3(1) = -10 – 3 = -13$.

Case 7:
m + 2n = -9
2m + n = -3
Multiply the second equation by 2: 4m + 2n = -6.
Subtract this from the first equation: (m + 2n) – (4m + 2n) = -9 – (-6) => -3m = -3. m = 1.
Substitute m=1 into the second equation: 2(1) + n = -3 => 2 + n = -3 => n = -5.
Check with the first equation: 1 + 2(-5) = 1 – 10 = -9. This pair (m=1, n=-5) is valid.
For this pair, $2m-3n = 2(1) – 3(-5) = 2 + 15 = 17$.

Case 8:
m + 2n = -27
2m + n = -1
Multiply the second equation by 2: 4m + 2n = -2.
Subtract this from the first equation: (m + 2n) – (4m + 2n) = -27 – (-2) => -3m = -25. m = 25/3, not an integer.

The possible values of $2m-3n$ are 13, -17, -13, and 17.
The maximum possible value of $2m-3n$ is 17.

Final_Answer:17

Ans: 25

Explanation:Let the given number be N.
$N = (625)^{65} \times (128)^{36}$
We can rewrite the bases in terms of their prime factors:
$625 = 5^4$
$128 = 2^7$
Substituting these into the expression for N:
$N = (5^4)^{65} \times (2^7)^{36}$
Using the exponent rule $(a^m)^n = a^{m \times n}$:
$N = 5^{4 \times 65} \times 2^{7 \times 36}$
$N = 5^{260} \times 2^{252}$
To find the sum of the digits, we need to express N in the form $a \times 10^k$. We can do this by pairing up powers of 2 and 5 to form powers of 10.
$N = 5^{260} \times 2^{252} = 5^{252} \times 5^{260-252} \times 2^{252}$
$N = 5^{252} \times 5^8 \times 2^{252}$
$N = (5 \times 2)^{252} \times 5^8$
$N = 10^{252} \times 5^8$
Now, let’s calculate $5^8$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
$5^5 = 3125$
$5^6 = 15625$
$5^7 = 78125$
$5^8 = 390625$
So, N can be written as:
$N = 390625 \times 10^{252}$
This means that the number N is 390625 followed by 252 zeros.
The digits of N are 3, 9, 0, 6, 2, 5, followed by 252 zeros.
The sum of the digits of N is the sum of these non-zero digits plus the sum of the zeros.
Sum of digits = 3 + 9 + 0 + 6 + 2 + 5 + (252 * 0)
Sum of digits = 3 + 9 + 6 + 2 + 5
Sum of digits = 12 + 6 + 2 + 5
Sum of digits = 18 + 2 + 5
Sum of digits = 20 + 5
Sum of digits = 25

Final_Answer:25

Ans: A

Explanation:Let the two quadratic equations be
$3x^{2}-5x+p=0 \quad \cdots (1)$
$2x^{2}-2x+q=0 \quad \cdots (2)$
Let the common root be $\alpha$.
Since $\alpha$ is a common root, it satisfies both equations:
$3\alpha^{2}-5\alpha+p=0 \quad \cdots (3)$
$2\alpha^{2}-2\alpha+q=0 \quad \cdots (4)$
Multiply equation (3) by 2 and equation (4) by 3:
$6\alpha^{2}-10\alpha+2p=0 \quad \cdots (5)$
$6\alpha^{2}-6\alpha+3q=0 \quad \cdots (6)$
Subtract equation (5) from equation (6):
$(6\alpha^{2}-6\alpha+3q) – (6\alpha^{2}-10\alpha+2p) = 0$
$6\alpha^{2}-6\alpha+3q – 6\alpha^{2}+10\alpha-2p = 0$
$4\alpha + 3q – 2p = 0$
$4\alpha = 2p – 3q$
$\alpha = \frac{2p – 3q}{4}$

Let the roots of equation (1) be $\alpha$ and $\beta$. From Vieta’s formulas for equation (1):
Sum of roots: $\alpha + \beta = \frac{-(-5)}{3} = \frac{5}{3}$
Product of roots: $\alpha \beta = \frac{p}{3}$

Let the roots of equation (2) be $\alpha$ and $\gamma$. From Vieta’s formulas for equation (2):
Sum of roots: $\alpha + \gamma = \frac{-(-2)}{2} = \frac{2}{2} = 1$
Product of roots: $\alpha \gamma = \frac{q}{2}$

We are asked to find the sum of the other roots, which is $\beta + \gamma$.
From the sum of roots for equation (1), we have $\beta = \frac{5}{3} – \alpha$.
From the sum of roots for equation (2), we have $\gamma = 1 – \alpha$.

Now, we find the sum $\beta + \gamma$:
$\beta + \gamma = \left(\frac{5}{3} – \alpha\right) + (1 – \alpha)$
$\beta + \gamma = \frac{5}{3} + 1 – 2\alpha$
$\beta + \gamma = \frac{5}{3} + \frac{3}{3} – 2\alpha$
$\beta + \gamma = \frac{8}{3} – 2\alpha$
Substitute the value of $\alpha = \frac{2p – 3q}{4}$:
$\beta + \gamma = \frac{8}{3} – 2\left(\frac{2p – 3q}{4}\right)$
$\beta + \gamma = \frac{8}{3} – \frac{2p – 3q}{2}$
$\beta + \gamma = \frac{8}{3} – \left(\frac{2p}{2} – \frac{3q}{2}\right)$
$\beta + \gamma = \frac{8}{3} – \left(p – \frac{3}{2}q\right)$
$\beta + \gamma = \frac{8}{3} – p + \frac{3}{2}q$

Comparing this with the given options:
Option A: $\frac{8}{3}-p+\frac{3}{2}q$

Correct_Option:A

Ans: A

Given that $64 = 8^2$ and $512 = 8^3$. The equation is:
$\log_{64}{x^{2}+\log_{8}{\sqrt{y}+3\log_{512}{(\sqrt{y}z)}}}=4$

We utilize the logarithmic property: $\log_{b^m} a^n = \frac{n}{m} \log_b a$.
Substituting this into the equation:
$\log_{8^2}{x^{2}+\log_{8}{\sqrt{y}+3\log_{8^3}{(\sqrt{y}z)}}}=4$

Applying the property to the terms:
$\log_{8}{x}+\log_{8}{\sqrt{y}+\log_{8}{(\sqrt{y}z)}}=4$

Combining the logarithmic terms using the property $\log_b A + \log_b B = \log_b (AB)$:
$\log_8(x \cdot \sqrt{y} \cdot (\sqrt{y}z)) = 4$
$\log_8(xyz) = 4$

Converting the logarithmic equation to exponential form:
$xyz = 8^4$
$xyz = (2^3)^4$
$xyz = 2^{12}$

Now, employing the Arithmetic Mean – Geometric Mean (AM-GM) inequality, which states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean:
$\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$

Substituting the value of $xyz$:
$\frac{x+y+z}{3} \ge \sqrt[3]{2^{12}}$
$\frac{x+y+z}{3} \ge 2^{12/3}$
$\frac{x+y+z}{3} \ge 2^4$
$\frac{x+y+z}{3} \ge 16$

Multiplying both sides by 3:
$x+y+z \ge 48$

Ans: 8

Explanation:Let $v_R$ be Rita’s speed in still water and $v_S$ be Sneha’s speed in still water.
We are given $v_R = 5$ km/h and $v_S = 6$ km/h.
Let $v$ be the velocity of the river.

For Sneha:
Speed downstream = $v_S + v = 6 + v$
Speed upstream = $v_S – v = 6 – v$

Time taken to row 14 km upstream = $\frac{14}{6-v}$ hours.
Time taken to row 14 km downstream = $\frac{14}{6+v}$ hours.

We are given that Sneha takes 48 minutes more to row 14 km upstream than to row the same distance downstream.
48 minutes = $\frac{48}{60}$ hours = $\frac{4}{5}$ hours.

So, $\frac{14}{6-v} – \frac{14}{6+v} = \frac{4}{5}$
Multiply by $(6-v)(6+v)$ on both sides:
$14(6+v) – 14(6-v) = \frac{4}{5}(36 – v^2)$
$84 + 14v – 84 + 14v = \frac{4}{5}(36 – v^2)$
$28v = \frac{4}{5}(36 – v^2)$
Multiply by 5:
$140v = 4(36 – v^2)$
$140v = 144 – 4v^2$
$4v^2 + 140v – 144 = 0$
Divide by 4:
$v^2 + 35v – 36 = 0$
Factor the quadratic equation:
$(v+36)(v-1) = 0$
Since the speed of the river cannot be negative, $v = 1$ km/h.

Now for Rita:
Rita’s speed in still water is $v_R = 5$ km/h.
The river velocity is $v = 1$ km/h.

When Rita rows upstream, her speed is $v_R – v = 5 – 1 = 4$ km/h.
When Rita rows downstream, her speed is $v_R + v = 5 + 1 = 6$ km/h.

Let the distance Rita rows upstream be $d$ km.
Then the distance Rita rows downstream is also $d$ km.
Total distance covered by Rita is $2d$ km.

Time taken to row upstream = $\frac{d}{4}$ hours.
Time taken to row downstream = $\frac{d}{6}$ hours.

Total time taken by Rita = $\frac{d}{4} + \frac{d}{6}$ hours.
We are given that the total time taken is 100 minutes.
100 minutes = $\frac{100}{60}$ hours = $\frac{5}{3}$ hours.

So, $\frac{d}{4} + \frac{d}{6} = \frac{5}{3}$
Find a common denominator for the left side:
$\frac{3d + 2d}{12} = \frac{5}{3}$
$\frac{5d}{12} = \frac{5}{3}$
Multiply both sides by 12:
$5d = \frac{5}{3} \times 12$
$5d = 5 \times 4$
$5d = 20$
$d = 4$ km.

The total distance Rita will cover is $2d$.
Total distance = $2 \times 4 = 8$ km.

Final_Answer:8

Ans: 12

The objective is to find the smallest possible value of $3a+2b+c$.
Observe that the coefficient of $c$ is the smallest among the terms.
The given condition is $3ac=8(a+b)$.
Since $a$, $b$, and $c$ are positive integers, the product $ac$ must be divisible by 8.
Let’s consider possible values for $a$ and $c$ that satisfy this divisibility, and then determine $b$.

Scenario 1: If $a = 1$ and $c = 8$, then $3(1)(8) = 8(1+b)$, which simplifies to $24 = 8+8b$. Solving for $b$ gives $16 = 8b$, so $b = 2$.
In this case, the expression $3a+2b+c$ evaluates to $3(1) + 2(2) + 8 = 3 + 4 + 8 = 15$.

Scenario 2: If $a = 2$ and $c = 4$, then $3(2)(4) = 8(2+b)$, which simplifies to $24 = 16+8b$. Solving for $b$ gives $8 = 8b$, so $b = 1$.
In this case, the expression $3a+2b+c$ evaluates to $3(2) + 2(1) + 4 = 6 + 2 + 4 = 12$.

Comparing the values obtained, 12 is the smaller result.
Therefore, the minimum value is 12.

Ans: A

Explanation:To find the domain of $h(x) = f(g(x)) + g(f(x))$, we need to consider the domains of $f(x)$, $g(x)$, $f(g(x))$, and $g(f(x))$.

The domain of $f(x) = \frac{x}{(2x-1)}$ is all real numbers except where the denominator is zero, i.e., $2x-1 = 0 \implies x = \frac{1}{2}$.
So, Domain of $f = \mathbb{R} \setminus \{\frac{1}{2}\}$.

The domain of $g(x) = \frac{x}{(x-1)}$ is all real numbers except where the denominator is zero, i.e., $x-1 = 0 \implies x = 1$.
So, Domain of $g = \mathbb{R} \setminus \{1\}$.

Now, let’s find the expression for $f(g(x))$:
$f(g(x)) = f\left(\frac{x}{x-1}\right) = \frac{\frac{x}{x-1}}{2\left(\frac{x}{x-1}\right)-1}$
$f(g(x)) = \frac{\frac{x}{x-1}}{\frac{2x}{x-1}-1} = \frac{\frac{x}{x-1}}{\frac{2x – (x-1)}{x-1}} = \frac{\frac{x}{x-1}}{\frac{x+1}{x-1}} = \frac{x}{x+1}$

For $f(g(x))$ to be defined:
1. $g(x)$ must be defined. So, $x \neq 1$.
2. The denominator of $f(g(x))$ cannot be zero. So, $x+1 \neq 0 \implies x \neq -1$.
Therefore, the domain of $f(g(x))$ is $\mathbb{R} \setminus \{1, -1\}$.

Next, let’s find the expression for $g(f(x))$:
$g(f(x)) = g\left(\frac{x}{2x-1}\right) = \frac{\frac{x}{2x-1}}{\frac{x}{2x-1}-1}$
$g(f(x)) = \frac{\frac{x}{2x-1}}{\frac{x – (2x-1)}{2x-1}} = \frac{\frac{x}{2x-1}}{\frac{-x+1}{2x-1}} = \frac{x}{-x+1}$

For $g(f(x))$ to be defined:
1. $f(x)$ must be defined. So, $x \neq \frac{1}{2}$.
2. The denominator of $g(f(x))$ cannot be zero. So, $-x+1 \neq 0 \implies x \neq 1$.
Therefore, the domain of $g(f(x))$ is $\mathbb{R} \setminus \{\frac{1}{2}, 1\}$.

The domain of $h(x) = f(g(x)) + g(f(x))$ is the intersection of the domains of $f(g(x))$ and $g(f(x))$.
This means $x$ must not be in the set $\{1, -1\}$ (from $f(g(x))$) and $x$ must not be in the set $\{\frac{1}{2}, 1\}$ (from $g(f(x))$).

Combining these restrictions, $x$ cannot be $1$, $-1$, or $\frac{1}{2}$.
Thus, the domain of $h(x)$ is all real numbers except $-1, \frac{1}{2}, 1$.

Comparing this with the given options:
Option A: $-1, \frac{1}{2}, \text{and } 1$

Correct_Option:A

Ans: D

Explanation:Let P be the principal amount of the loan, which is Rs 1000.
Let $I_1$ be the first installment paid at the end of the first year, which is Rs 530.
Let $I_2$ be the second installment paid at the end of the second year, which is Rs 594.
Let r be the annual rate of interest in percentage.

The present value of the installments must be equal to the principal amount.
The present value of the first installment is $I_1$ discounted back by one year. So, it is $\frac{I_1}{(1+r)}$.
The present value of the second installment is $I_2$ discounted back by two years. So, it is $\frac{I_2}{(1+r)^2}$.

Therefore, we have the equation:
$P = \frac{I_1}{(1+r)} + \frac{I_2}{(1+r)^2}$

Substitute the given values:
$1000 = \frac{530}{(1+r)} + \frac{594}{(1+r)^2}$

To solve for r, we can let $x = \frac{1}{(1+r)}$. The equation becomes:
$1000 = 530x + 594x^2$

Rearrange the equation into a quadratic form:
$594x^2 + 530x – 1000 = 0$

We can test the given options for r to find the correct value.

If r = 10% = 0.10, then $1+r = 1.10$.
$x = \frac{1}{1.10} \approx 0.90909$
Check if $594(0.90909)^2 + 530(0.90909) – 1000 = 0$
$594(0.82644) + 530(0.90909) – 1000$
$491.11 + 481.81 – 1000 = 972.92 – 1000 = -27.08 \neq 0$

If r = 11% = 0.11, then $1+r = 1.11$.
$x = \frac{1}{1.11} \approx 0.90090$
Check if $594(0.90090)^2 + 530(0.90090) – 1000 = 0$
$594(0.81162) + 530(0.90090) – 1000$
$482.12 + 477.47 – 1000 = 959.59 – 1000 = -40.41 \neq 0$

If r = 9% = 0.09, then $1+r = 1.09$.
$x = \frac{1}{1.09} \approx 0.91743$
Check if $594(0.91743)^2 + 530(0.91743) – 1000 = 0$
$594(0.84167) + 530(0.91743) – 1000$
$500.23 + 486.24 – 1000 = 986.47 – 1000 = -13.53 \neq 0$

If r = 8% = 0.08, then $1+r = 1.08$.
$x = \frac{1}{1.08} \approx 0.92593$
Check if $594(0.92593)^2 + 530(0.92593) – 1000 = 0$
$594(0.85735) + 530(0.92593) – 1000$
$509.44 + 490.74 – 1000 = 1000.18 – 1000 = 0.18 \approx 0$

Let’s re-check the calculation for r=8% more precisely.
$1+r = 1.08$
$x = \frac{1}{1.08}$
$1000 = 530 \times \frac{1}{1.08} + 594 \times \frac{1}{(1.08)^2}$
$1000 = \frac{530}{1.08} + \frac{594}{1.1664}$
$1000 = 490.7407… + 509.2592…$
$1000 = 1000$

The rate of interest is 8%.

Correct_Option:D

Ans: 80

To be published

Ans: D

Any factor of the given number will be of the structure $2^a*3^b*5^c*7^d$ where the exponents satisfy $0\le a\le6,\ 0\le b\le5,\ 0\le c\le3,\ 0\le d\le2$.
Since the factors must be of the form 3r+1, they cannot be multiples of 3, which implies b must be 0.
Let’s consider the remainders when the prime factors are divided by 3:
$2 \pmod 3 = 2$
$5 \pmod 3 = 2$
$7 \pmod 3 = 1$
Therefore, a factor of the form $2^a5^c7^d$ will have a remainder modulo 3 as:
$2^a5^c7^d \equiv 2^a \cdot 2^c \cdot 1^d \pmod 3$
$ \equiv 2^{a+c} \pmod 3$
For $2^k$ to be of the form 3r+1, k must be an even number. Thus, we require the sum of the exponents a+c to be even.
The possible values for ‘a’ are $0, 1, 2, 3, 4, 5, 6$. Among these, there are 4 even values (0, 2, 4, 6) and 3 odd values (1, 3, 5).
The possible values for ‘c’ are $0, 1, 2, 3$. Among these, there are 2 even values (0, 2) and 2 odd values (1, 3).
The number of pairs (a,c) such that (a+c) is even occurs in two scenarios:
1. ‘a’ is even and ‘c’ is even: $4 \text{ (choices for a)} \times 2 \text{ (choices for c)} = 8$
2. ‘a’ is odd and ‘c’ is odd: $3 \text{ (choices for a)} \times 2 \text{ (choices for c)} = 6$
The total number of pairs (a,c) for which a+c is even is $8 + 6 = 14$.
For each of these 14 pairs, there are 3 possible choices for the exponent ‘d’ (0, 1, or 2).
Therefore, the total count of divisors in the form (3r+1) is $14 \times 3 = 42$.

Ans: B

To be published

Ans: 2

Consider the expression: $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$
This expression represents the sum of squared differences. Its value will always be non-negative.
To achieve the absolute minimum value of zero, all the terms $(a-b)$, $(a-c)$, and $(a-d)$ would need to be zero. This implies that $b$, $c$, and $d$ must all be equal to $a$.
However, the problem states that $a$, $b$, $c$, and $d$ are integers whose sum is 46. If $a=b=c=d$, then their sum would be $4a$. For the sum to be 46, $4a=46$, which means $a = 46/4 = 11.5$. Since $a$ must be an integer, this scenario is not possible.
Therefore, we cannot make all the terms equal to zero. We need to find integer values for $a$, $b$, $c$, and $d$ that minimize the expression.
The sum of the four integers is 46. To minimize the sum of squares of differences from a central value ‘a’, the values of $b$, $c$, and $d$ should be as close to ‘a’ as possible. Since we cannot make them all equal, we look for integers that are clustered around the average value of $46/4 = 11.5$.
The closest possible integer set for $b$, $c$, and $d$ that are near the average and sum to $46-a$ would be integers around 11 and 12.
If we consider the average value of $11.5$, the closest integer distribution for four numbers summing to 46 would be two 11s and two 12s. For example, if $b=11, c=11, d=12$, then $a$ could be chosen to minimize the expression.
Let’s try setting $a$ to one of these values, say $a=12$. Then, $b$ and $c$ could be $11$, and $d$ could be $12$ (this fits the criteria as the set of numbers could be {12, 11, 11, 12} which sums to 46).
In this case, the expression evaluates to:
$(12-11)^{2}+(12-11)^{2}+(12-12)^{2} = (1)^2 + (1)^2 + (0)^2 = 1 + 1 + 0 = 2$
Alternatively, if we set $a=11$, and let $b=12, c=12, d=11$:
$(11-12)^{2}+(11-12)^{2}+(11-11)^{2} = (-1)^2 + (-1)^2 + (0)^2 = 1 + 1 + 0 = 2$
Thus, the minimum value attainable is 2.

Ans: A

Explanation:Let L_income and L_expenditure be the income and expenditure of Lakshmi, respectively.
Let M_income and M_expenditure be the income and expenditure of Meenakshi, respectively.

We are given the following ratios:
1. Ratio of expenditures of Lakshmi and Meenakshi:
L_expenditure : M_expenditure = 2 : 3
Let L_expenditure = 2x and M_expenditure = 3x

2. Ratio of income of Lakshmi to expenditure of Meenakshi:
L_income : M_expenditure = 6 : 7
We know M_expenditure = 3x.
So, L_income : 3x = 6 : 7
L_income = (6/7) * 3x = 18x/7

3. Excess of income over expenditure is saved.
Lakshmi’s saving (L_saving) = L_income – L_expenditure
Meenakshi’s saving (M_saving) = M_income – M_expenditure

4. Ratio of their savings is 4 : 9:
L_saving : M_saving = 4 : 9

Now, let’s express L_saving in terms of x:
L_saving = L_income – L_expenditure
L_saving = (18x/7) – 2x
L_saving = (18x – 14x) / 7
L_saving = 4x/7

From the ratio of savings, we have:
(4x/7) : M_saving = 4 : 9
M_saving = (9/4) * (4x/7)
M_saving = 9x/7

Now we can find Meenakshi’s income (M_income):
M_saving = M_income – M_expenditure
9x/7 = M_income – 3x
M_income = (9x/7) + 3x
M_income = (9x + 21x) / 7
M_income = 30x/7

We need to find the ratio of their incomes:
L_income : M_income
= (18x/7) : (30x/7)

We can cancel out x/7 from both sides:
= 18 : 30

Now, simplify the ratio by dividing both numbers by their greatest common divisor, which is 6:
= 18/6 : 30/6
= 3 : 5

Therefore, the ratio of their incomes is 3 : 5.

Comparing this with the given options:
Option A: 3:5
Option B: 5:6
Option C: 2:1
Option D: 7:8

The calculated ratio matches Option A.

Correct_Option:A

Ans: B

Explanation:Let the first term of the decreasing infinite geometric progression be $a$ and the common ratio be $r$. Since it is a decreasing progression, we have $0 < r < 1$ (as terms are positive from the given equations).
The terms of the geometric progression are $a_1 = a$, $a_2 = ar$, $a_3 = ar^2$.

Given the sum of the first three terms:
$a_1 + a_2 + a_3 = 52$
$a + ar + ar^2 = 52$
$a(1 + r + r^2) = 52$ (Equation 1)

Given the sum of pairwise products of the first three terms:
$a_1a_2 + a_2a_3 + a_3a_1 = 624$
$(a)(ar) + (ar)(ar^2) + (ar^2)(a) = 624$
$a^2r + a^2r^3 + a^2r^2 = 624$
$a^2r(1 + r^2 + r) = 624$
$a^2r(1 + r + r^2) = 624$ (Equation 2)

Now we have a system of two equations:
1) $a(1 + r + r^2) = 52$
2) $a^2r(1 + r + r^2) = 624$

Divide Equation 2 by Equation 1:
$\frac{a^2r(1 + r + r^2)}{a(1 + r + r^2)} = \frac{624}{52}$
$ar = \frac{624}{52}$

To simplify the fraction $\frac{624}{52}$:
$624 \div 52 = 12$
So, $ar = 12$.

This means the second term $a_2 = 12$.

Substitute $ar = 12$ into Equation 1.
From Equation 1, $a(1 + r + r^2) = 52$.
We can rewrite $a(1 + r + r^2)$ as $a + ar + ar^2 = 52$.
We know $ar = 12$.
So, $a + 12 + ar^2 = 52$.
$a + ar^2 = 52 – 12$
$a + ar^2 = 40$.

We also know $ar = 12$, so $a = \frac{12}{r}$.
Substitute this into $a + ar^2 = 40$:
$\frac{12}{r} + \left(\frac{12}{r}\right)r^2 = 40$
$\frac{12}{r} + 12r = 40$

Multiply by $r$ to clear the denominator:
$12 + 12r^2 = 40r$
$12r^2 – 40r + 12 = 0$

Divide by 4 to simplify the quadratic equation:
$3r^2 – 10r + 3 = 0$

Factor the quadratic equation:
We need two numbers that multiply to $3 \times 3 = 9$ and add to $-10$. These numbers are $-1$ and $-9$.
$3r^2 – 9r – r + 3 = 0$
$3r(r – 3) – 1(r – 3) = 0$
$(3r – 1)(r – 3) = 0$

This gives two possible values for $r$:
$3r – 1 = 0 \implies r = \frac{1}{3}$
$r – 3 = 0 \implies r = 3$

Since the geometric progression is decreasing, the common ratio $r$ must be between 0 and 1 ($0 < r < 1$). Therefore, we choose $r = \frac{1}{3}$.

Now we find the first term $a$ using $ar = 12$:
$a \times \frac{1}{3} = 12$
$a = 12 \times 3$
$a = 36$

So, the first term $a_1 = 36$ and the common ratio $r = \frac{1}{3}$.
The sum of an infinite geometric progression is given by the formula $S = \frac{a}{1-r}$, provided $|r| < 1$.

In this case, $a = 36$ and $r = \frac{1}{3}$.
$S = \frac{36}{1 – \frac{1}{3}}$
$S = \frac{36}{\frac{3}{3} – \frac{1}{3}}$
$S = \frac{36}{\frac{2}{3}}$
$S = 36 \times \frac{3}{2}$
$S = 18 \times 3$
$S = 54$

Let’s check if the terms are indeed decreasing.
$a_1 = 36$
$a_2 = 36 \times \frac{1}{3} = 12$
$a_3 = 12 \times \frac{1}{3} = 4$
$36 > 12 > 4$, so it is a decreasing progression.

Check the given conditions:
$a_1 + a_2 + a_3 = 36 + 12 + 4 = 52$ (Correct)
$a_1a_2 + a_2a_3 + a_3a_1 = (36)(12) + (12)(4) + (4)(36) = 432 + 48 + 144 = 624$ (Correct)

The sum of this geometric progression is 54.

Correct_Option:B

Ans: A

Let the cost of coffee be denoted by $C$ per kilogram and the cost of cocoa be denoted by $K$ per kilogram.
We are given information about two existing blends:
Blend 1: $0.16$ parts coffee and $0.84$ parts cocoa cost Rs $240$.
This can be represented as:
$0.16C + 0.84K = 240$

Blend 2: $0.36$ parts coffee and $0.64$ parts cocoa cost Rs $320$.
This can be represented as:
$0.36C + 0.64K = 320$

To eliminate the decimals, we multiply both equations by $100$:
$16C + 84K = 24000$ (Equation 1)
$36C + 64K = 32000$ (Equation 2)

Subtract Equation 1 from Equation 2:
$(36C + 64K) – (16C + 84K) = 32000 – 24000$
$36C – 16C + 64K – 84K = 8000$
$20C – 20K = 8000$
Dividing by $20$, we get:
$C – K = 400$
This implies $C = K + 400$.

Substitute this expression for $C$ into Equation 1:
$16(K + 400) + 84K = 24000$
$16K + 6400 + 84K = 24000$
$100K + 6400 = 24000$
$100K = 24000 – 6400$
$100K = 17600$
$K = \frac{17600}{100} = 176$

Now, find the value of $C$ using $C = K + 400$:
$C = 176 + 400 = 576$
So, coffee costs Rs $576$ per kilogram and cocoa costs Rs $176$ per kilogram.

We need to create a new blend that costs Rs $376$ per kilogram. Let $p$ be the fraction of coffee in this new blend. Then, the fraction of cocoa will be $(1-p)$. The cost of the new blend can be expressed as:
$p \cdot C + (1-p) \cdot K = 376$
$p \cdot 576 + (1-p) \cdot 176 = 376$

Now, solve for $p$:
$576p + 176 – 176p = 376$
$576p – 176p = 376 – 176$
$400p = 200$
$p = \frac{200}{400} = \frac{1}{2}$

So, the new mixture contains $p = \frac{1}{2}$ or $50\%$ coffee.
For a total of $10$ kilograms of this new mixture, the amount of coffee would be:
$10 \text{ kg} \times 0.5 = 5 \text{ kg}$

Ans: B

To be published

Ans: 340

Explanation:Let the efficiency of Chandan be $x$ units per day.
Then, the efficiency of Bipin is $2x$ units per day.
And the efficiency of Ankita is $2 \times (2x) = 4x$ units per day.

Let the total work to be done be $W$.
The total work done by all three together in one day is the sum of their efficiencies: $4x + 2x + x = 7x$ units per day.

Bipin leaves the job after 20 days.
In the first 20 days, all three worked together.
Work done in the first 20 days = (efficiency of Ankita + efficiency of Bipin + efficiency of Chandan) $\times$ 20
Work done in the first 20 days = $(4x + 2x + x) \times 20 = 7x \times 20 = 140x$ units.

The job was completed in 60 days.
This means that after Bipin left, the remaining work was completed by Ankita and Chandan in $60 – 20 = 40$ days.
The combined efficiency of Ankita and Chandan is $4x + x = 5x$ units per day.
Work done by Ankita and Chandan in the remaining 40 days = $(4x + x) \times 40 = 5x \times 40 = 200x$ units.

The total work $W$ is the sum of the work done in the first 20 days and the work done in the remaining 40 days.
$W = 140x + 200x = 340x$ units.

We need to find the number of days needed by Chandan to complete the job alone.
Let $D$ be the number of days Chandan needs to complete the job alone.
Work done by Chandan alone = Chandan’s efficiency $\times$ Number of days
$W = x \times D$

We have $W = 340x$.
So, $340x = x \times D$.
Dividing both sides by $x$ (since $x$ is an efficiency, $x > 0$), we get $D = 340$.

Therefore, Chandan needs 340 days to complete the job alone.

Final_Answer:340

Ans: C

Explanation:Let the total amount of money be T.
Pinu received 20% of the total amount, so Pinu’s share = 0.20T.
The remaining amount after Pinu’s share is T – 0.20T = 0.80T.
Meena received 40% of the remaining amount, so Meena’s share = 0.40 * (0.80T) = 0.32T.
Seema received 20% less than Pinu.
Pinu’s share = 0.20T.
Seema’s share = Pinu’s share – 0.20 * Pinu’s share
Seema’s share = 0.20T – 0.20 * (0.20T)
Seema’s share = 0.20T – 0.04T
Seema’s share = 0.16T.
The total amount distributed among Pinu, Meena, and Seema is:
0.20T (Pinu) + 0.32T (Meena) + 0.16T (Seema) = 0.68T.
The amount received by Rinu is the total amount minus the sum of amounts received by Pinu, Meena, and Seema.
Rinu’s share = T – 0.68T = 0.32T.
We need to find the ratio of the amounts received by Pinu and Rinu.
Ratio of Pinu to Rinu = Pinu’s share / Rinu’s share
Ratio = (0.20T) / (0.32T)
Ratio = 0.20 / 0.32
To simplify the ratio, we can multiply both the numerator and denominator by 100:
Ratio = 20 / 32
Divide both by their greatest common divisor, which is 4:
Ratio = 5 / 8
So, the ratio of amounts received by Pinu and Rinu is 5:8.

Correct_Option:C