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

cat 2020 Complete Paper Solution | Slot 1

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cat 2020 Complete Paper Solution | 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>

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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>

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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>