Let the number be 'abc'. Then,&nbsp;2&lt;a×&nbsp;b×&nbsp;c&lt;72&lt;a×&nbsp;b×&nbsp;c&lt;7. The product can be 3,4,5,6.We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of&nbsp;12&nbsp;numbers fulfilling the criteria.We can factories 4 as 2*2 and the combination 2,2,1 can be used to form&nbsp;3&nbsp;more distinct numbers.&nbsp;We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form&nbsp;6&nbsp;additional distinct numbers.&nbsp;Thus a total of 12 + 3 + 6 =&nbsp;21&nbsp;such numbers can be formed.

Let ‘r’ be the root of the function. It follows that $f(r) = 0$. We can represent this as $f(r) = f\{5 - (5 - r)\}$Based on the relation: $f(5 - x) = f(5 + x)$; $f(r) = f\{5 - (5 - r)\} = f\{5 + (5 - r)\}$∴ $f(r) = f(10 - r)$Thus, every root ‘r’ is associated with another root ‘(10-r)’ [these form a pair]. For even distinct roots, in this case four, let us assume the roots to be as follows: $r1$, $(10 - r1)$, $r2$, $(10 - r2)$The sum of these roots = $r1 + (10 - r1) + r2 + (10 - r2) = 20$Hence, Option D is the correct answer.

Let their individual Amounts be equal after ’t’ years. Let their initial investments amount to $A_V$ and $A_J$;&nbsp;$A_V$ = $10,000(1+\frac{5t}{100})$ and $A_J$ = $8,000(1+\frac{10(t-2)}{100})$Equating both: $10,000(1+\frac{5t}{100})$ = $8,000(1+\frac{10(t-2)}{100})$On simplifying both sides, we get: $15t$ = $180$ ; $t$ = $12$

Let the total distance be ‘D’ km and the speed of the train be ’s’ kmph. The time taken to cover D at speed d is ’t’ hours. Based on the information: on equating the distance, we get$s \times t = \frac{s}{3} \times (t + \frac{1}{2})$On solving we acquire the value of $t = \frac{1}{4}$ or 15 mins. We understand that during the return journey, the first 5 minutes are spent traveling at speed ’s’ {distance traveled in terms of s = $\frac{s}{12}$}. Remaining distance in terms of ’s’ = $\frac{s}{4} - \frac{s}{12}$ = $\frac{s}{6}$&nbsp;The rest 4 minutes of stoppage added to this initial 5 minutes amounts to a total of 9 minutes. The train has to complete the rest of the journey in $15 - 9 = 6$ mins or {$\frac{1}{10}$ hours}. Thus, let ‘x’ kmph be the new value of speed. Based on the above, we get $\frac{s}{6x} = \frac{1}{10}$ $or$ $x = \frac{10s}{6}$Since the increase in speed needs to be calculated: $\frac{(\frac{10s}{6} - s)}{s} \times 100 = \frac{200}{3} \approx 67$% increase.Hence, Option B is the correct answer.

$\frac{\log 5}{2 \log 2} = \frac{\log y}{2 \log 2} \cdot \frac{\log 5}{2 \log 6}$log 36 = log y;&nbsp;∴ y = 36

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/6.image_3ehaJXH.png"> The graphs of&nbsp;$2^x+2^{−x} \text{ and } 2−(x−2)^2$&nbsp;never intersect. So, number of solutions=0.Alternate method:We notice that the minimum value of the term in the LHS will be greater than or equal to&nbsp;2 {at x=0;&nbsp;LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x=2; RHS = 2}. The values of x&nbsp;at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.

Let the speed of Car 2 be ‘x’ kmph and the time taken by the two cars to meet be ’t’ hours. In ’t’ hours, Car 1 travels (60 × t) km while Car 2 travels (x × t) km. It is given that the time taken by Car 1 to travel (x × t) km is 45 minutes or (3/4) hours.∴ $\frac{(x × t)}{60} = \frac34$ or $t = \frac{180}{4x}$ ….(i) Similarly, the time taken by Car 2 to travel (60 × t) km is 20 minutes or (1/3) hours. ∴ $\frac{(60 × t)}{x} = \frac13$ or $t = \frac{x}{180}$ ….(ii) Equating the values in (i) and (ii), and solving for x: ∴ $\frac{180}{4x} = \frac{x}{180} ⟶ x = 90kmph$

GivenA+(B+C)/2=5 =&gt; 2A+B+C=10....(i)(A+C)/2 +B=7 =&gt; A+2B+C=14.....(ii)(i)-(ii)=&gt; B-A=4 =&gt; B=4+A.Given, A, B, C are positive integersIf A=1=&gt;B=5 =&gt; C=3If A=2=&gt;B=6 =&gt; C=0 but this is invalid as C is positive.Similarly if A&gt;2 C will be negative and cases are not valid.Hence, A+B=6.

$x = 2^{12(7 + 4\sqrt3)}$$x^{7/2} = 2^{42(7 + 4\sqrt3)}$$x^{2\sqrt3} = 2^{24\sqrt3(7 + 4\sqrt3)}$$\frac{x^{7/2}}{x^{2\sqrt3}} = 2^{(7 + 4\sqrt3)(42 - 24\sqrt3)} = 2^{(7 + 4\sqrt3)(7 - 4\sqrt3)6} = 2^6$Hence correct option is C.

The four digit even numbers will be of form:1100, 1122, 1144 ... 1188, 2200, 2222, 2244 ... 9900, 9922, 9944, 9966, 9988Their sum 'S' will be (1100+1100+22+1100+44+1100+66+1100+88)+(2200+2200+22+2200+44+...)....+(9900+9900+22+9900+44+9900+66+9900+88)=&gt; S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)....+9900*5+(22+44+66+88)=&gt; S=5*1100(1+2+3+...9)+9(22+44+66+88)=&gt;S=5*1100*9*10/2 + 9*11*20Total number of numbers are 9*5=45.'. Mean will be S/45 = 5*1100+44=5544.Option A

Let $a = x + \frac{1}{x}$So, the given equation is $a^2 - 3a + 2 = 0$So, $a$ can be either $2$ or $1$.If $a = 1$, $x + \frac{1}{x} = 1$ and it has no real roots.If $a = 2$, $x + \frac{1}{x} = 2$ and it has exactly one real root which is $x = 1$.So, the total number of distinct real roots of the given equation is 1

Let the price of desktop and laptop be x,y respectively.Given,x+y=50000...(i)1.2x+0.9y=50000(1.02)=51000...(ii)(ii)-0.9(i) gives0.3x=6000=&gt; x=20000.

$x_0 = \max(x_1, x_2, \ldots, x_{12})$$x_0$ will be minimum if $(x_1, x_2, \ldots, x_{12})$ are close to each other.$\frac{100}{12} = 8.33$Therefore, $(\max(x_1, x_2, \ldots, x_{12}))$ will be minimum if $((x_1, x_2, \ldots, x_{12})$ = (9, 9, 9, 9, 8, 8, 8, 8, 8, 8, 8, 8)).Hence, Option D is correct.

Let the length of the train be $l$ kms and speed be $s$ kmph.&nbsp;&nbsp;Based on the two scenarios presented, we obtain:$\frac{l}{s-2} = \frac{90}{3600}$ ...(i)&nbsp;&nbsp;$\frac{l}{s-4} = \frac{100}{3600}$ ...(ii)On dividing (ii) by (i) and simplifying we acquire the value of $s$ as $22$ kmph.&nbsp;&nbsp;Substituting this value in (i), we have $l = \frac{90}{3600} \times 20$ kms. {keeping it in km and hours for convenience}Since we need to find $\frac{l}{s}$, let this be equal to $x$.&nbsp;&nbsp;Then, $x = 90 \times \frac{20}{22} = 81.81 \approx 82$ seconds.

$(x^2 - 7x + 11)^{(x^2 - 13x + 42)} = 1$This holds if&nbsp;&nbsp;$(x^2 - 13x + 42) = 0$&nbsp;&nbsp;or $(x^2 - 7x + 11) = 1$&nbsp;&nbsp;or $(x^2 - 7x + 11) = -1$ and $(x^2 - 13x + 42)$ is even.For $x = 6,7$, the value $(x^2 - 13x + 42) = 0$.$(x^2 - 7x + 11) = 1$ for $x = 5,2$.$(x^2 - 7x + 11) = -1$ for $x = 3,4$ and for $x = 3$ or $4$, $(x^2 - 13x + 42)$ is even.∴ $\{2,3,4,5,6,7\}$ is the solution set of $x$.x can take 6 values.

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/16.image_D2ggvra.png">The area of the region contained by the lines&nbsp;∣ x ∣ − y ≤ 1, y ≥ 0 and&nbsp;y ≤ 1 is the white region.Total area = Area of rectangle + 2 * Area of triangle =&nbsp;2+($\frac12$ ×&nbsp;2 ×&nbsp;1)&nbsp;=3Hence, 3 is the correct answer.

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/17.image_tnbquxg.png">Let the base radius be 3r.Height of upper cone is 9 so, by symmetry radius of upper cone will be r.Volume of frustum=$\fracπ3(9r^2⋅27−r^2⋅9)$Volume of upper cone =&nbsp;$\fracπ3⋅r^2⋅9$Difference=&nbsp;$\fracπ3⋅9⋅r^2⋅25 = 225 = \frac{\pi}{3} ⋅ r^2 = 1$Volume of larger cone =&nbsp;$\fracπ3⋅9r^2⋅27=243$

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/18.image_yBmetC1.png">Let the length of radius be ‘$r$’.From the above diagram,$x^2 +r^2 = 6^2$&nbsp;....(i)$(10 - x)^2 + r^2 = 8^2$ ....(ii)Subtracting (i) from (ii), we get:&nbsp;$x = 3.6$ =&gt;&nbsp;$r^2 = 36 - (3.6)^2$&nbsp;= $23.04$ ==&gt; $r^2 = 36 - (3.6)^2$ = $23.04$Area of circle =&nbsp;$\pi r^2 = 23.04 \pi$&nbsp;Area of rhombus = 1/2*d1*d2 = 1/2*12*16 = 96..’. Ratio of areas = 23.04$\pi$ / 96 = $\frac{6 \pi}{25}$

The difference in the time take to traverse the same distance&nbsp;′d′&nbsp;at two different speeds is 35 minutes. Equating this:&nbsp;$\frac{d}8$ − $\frac{d}{15}$&nbsp;=&nbsp;$\frac{35}{60}$On solving, we obtain&nbsp;d=10kms Let&nbsp;x kmph&nbsp;be the speed at which Amal needs to travel to reach the office in 50 minutes; then$\frac{10}{x}=\frac{50}{60}$&nbsp;or&nbsp;x&nbsp;=&nbsp;12&nbsp;kmph​Hence, Option A is the correct answer.

Since&nbsp;c&lt;9, we can have the following viable combinations for&nbsp;b×&nbsp;c&nbsp;=96&nbsp;(given our objective is to minimize the sum):48×&nbsp;2&nbsp;;&nbsp;32×3; 24×&nbsp;4&nbsp;;&nbsp;16×6;&nbsp;12×8&nbsp;Similarly, we can factorize&nbsp;a×&nbsp;b&nbsp;=432&nbsp;into its factors. On close observation, we notice that&nbsp;18×24&nbsp;and&nbsp;24&nbsp;×&nbsp;4&nbsp;corresponding to&nbsp;a×b&nbsp;and&nbsp;b×c&nbsp;respectively&nbsp;together render us with the least value of the sum of&nbsp;a+b+c&nbsp;=&nbsp;18+24+4&nbsp;=46Hence, Option C is the correct answer.

$2^{Y^2 (\log_3 5)} = 5^{Y^2 (\log_3 2)}$Given,&nbsp;&nbsp;$5^{Y^2 (\log_3 2)} = 5^{(\log_2 3)}$⇒ $Y^2 (\log_3 2) = (\log_2 3)$&nbsp;&nbsp;⇒ $Y^2 = (\log_2 3)^2$⇒ $Y = (-\log_2 3)$ or $(\log_2 3)$Since $Y$ is a negative number,&nbsp;&nbsp;$Y = (-\log_2 3) = (\log_2 \tfrac{1}{3})$

Total area = 5 parts <img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/22.q22_circle_area.png">Let ABCD be the rectangle with length $2l$ and breadth $2b$ respectively.Area of the circle i.e. area of painted region $= \pi b^{2}$.Given, $4lb - \pi b^{2} = \frac{2}{3}\pi b^{2}$.$\Rightarrow 4lb = \frac{5}{3}\pi b^{2}$.$\Rightarrow l = \frac{5\pi}{12} b$.Given, $4lb = 135$&nbsp;&nbsp;$\Rightarrow 4 \cdot \frac{5\pi}{12} b^{2} = 135$&nbsp;&nbsp;$\Rightarrow b = \frac{9}{\sqrt{\pi}}$.$\Rightarrow l = \frac{15}{4}\sqrt{\pi}$.Perimeter of rectangle $= 4(l + b)$&nbsp;&nbsp;$= 4\left( \frac{15}{4}\sqrt{\pi} + \frac{9}{\sqrt{\pi}} \right)$&nbsp;&nbsp;$= 3\sqrt{\pi}\left(5 + \frac{12}{\pi}\right)$.Hence option A is correct.

cat 2020 Complete Paper Solution | Quant Slot 1

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cat 2020 Complete Paper Solution | Quant Slot 1

Question 1.

<div><p>How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?</p></div>

Question 2.

<div><p>If f(5 + x) = f(5 - x) for every real x and f(x) = 0 has four distinct real roots, then the sum of the roots is</p></div>

Question 3.

<div><p>Veeru invested Rs 10000 at 5% simple annual interest, and exactly after two years, Joy invested Rs 8000 at 10% simple annual interest. How many years after Veeru&rsquo;s investment, will their balances, i.e., principal plus accumulated interest, be equal?</p></div>

Question 4.

Question 5.

<div><p>If $log_{4}$ 5 = ($log_{4}$ y) ($log_{6}$ &radic;5), then y equals</p></div>

Question 6.

<div><p>The number of real-valued solutions of the equation $2^x$ + $2^{-x}$ = 2 - $(x - 2)^2$ is</p></div>

Question 7.

Question 8.

<div><p>Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is</p></div>

Question 9.

<div><p>If x = $(4096)^{7+4&radic;3}$, then which of the following equals 64?</p></div>

Question 10.

<div><p>The mean of all 4 digit even natural numbers of the form 'aabb', where a&gt;0, is</p></div>

Question 11.

<div><p>The number of distinct real roots of the equation $(x + \frac{1}{x})^{2}$ - 3(x + $\frac{1}{x}$ ) + 2 = 0 equals</p></div>

Question 12.

<div><p>A person spent Rs 50000 to purchase a desktop computer and a laptop computer. He sold the desktop at 20% profit and the laptop at 10% loss. If overall he made a 2% profit then the purchase price, in rupees, of the desktop is</p></div>

Question 13.

<div><p>Among 100 students, $x_{1}$ have birthdays in January, $x_{2}$ have birthdays in February, and so on. If $x_{0}$ = max($x_{1}$, $x_{2}$, ..., $x_{12}$), then the smallest possible value of $x_{0}$ is</p></div>

Question 14.

<div><p>Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to</p></div>

Question 15.

<div><p>How many distinct positive integer-valued solutions exist to the equation $(x^{2} - 7x + 11)^{(x^{2} - 13x + 42)}$ = 1?</p></div>

Question 16.

<div><p>The area of the region satisfying the inequalities |x| - y &le; 1, y &ge; 0, and y &le; 1 is</p></div>

Question 17.

<div><p>A solid right circular cone of height 27 cm is cut into 2 pieces along a plane parallel to it's base at a height of 18 cm from the base. If the difference in the volume of the two pieces is 225 cc, the volume, in cc, of the original cone is</p></div>

Question 18.

<div><p>A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of the circle to the area of the rhombus is</p></div>

Question 19.

<div><p>Leaving home at the same time, Amal reaches office at 10:15 am if he travels at 8kmph, and at 9:40 am if he travels at 15kmph. Leaving home at 9:10 am, at what speed, in kmph, must he travel so as to reach office exactly at 10:00 am?</p></div>

Question 20.

<div><p>If a, b and c are positive integers such that ab = 432, bc = 96 and c &lt; 9, then the smallest possible value of a + b + c is</p></div>

Question 21.

<div><p>If y is a negative number such that $2^{y^{2}log_{3}5}$ = $5^{log_{2}3}$, then y equals</p></div>

Question 22.

<div><p>On a rectangular metal sheet of area 135 sq in, a circle is painted such that the circle touches opposite two sides. If the area of the sheet left unpainted is two-thirds of the painted area then the perimeter of the rectangle in inches is</p></div>