CAT 2020 Question Paper Slot 2

Now 23 students choose maths as one of their subject.

This means (MPC)+ (MC) + (PC)=23 where MPC denotes students who choose all the three subjects maths, physics and chemistry and so on.

So MC + PM =5 Similarly we have PC+ MP =7 

We have to find the smallest number of students choosing chemistry

For that in the first equation let PM=5 and MC=0. In the second equation this PC=2

Hence minimum number of students choosing chemistry will be (18+2)=20 Since 18 students chose all the three subjects.

Let x(1) be the least number and x(10) be the largest number. Now from the condition given in the question , we can say that

x(2)+x(3)+x(4)+........x(10)= 47*9=423...................(1)

Similarly x(1)+x(2)+x(3)+x(4)................+x(9)= 42*9=378...............(2)

Subtracting both the equations we get x(10)-x(1)=45

Now, the sum of the 10 observations from equation (1) is 423+x(1)

Now the minimum value of x(10) will be 47 and the minimum value of x(1) will be 2 . Hence minimum average 425/10=42.5

Maximum value of x(1) is 42. Hence maximum average will be 465/10=46.5

Hence difference in average will be 46.5-42.5=4 which is the correct answer

Let the speed of Ram be v(r) and the speed of Rahim be v(h) respectively. Let them meet after time "t" from the beginning.

Hence Ram will cover v(r)(t) during that time and Rahim will cover v(h)t respectively.

Now after meeting Ram reaches his destination in 1 min i.e. Ram covered v(h)t in 1 minute or v(r)(1)= v(h)(t)

Similarly Rahim reaches his destination in 4 min i.e. Rahim covered v(r)t in 4 minutes or v(h)(4)= v(r)(t)

Dividing both the equations we get $$\frac{v\left(r\right)}{4v\left(h\right)}=\frac{v\left(h\right)}{v\left(r\right)}\ or\ \frac{v\left(r\right)}{v\left(h\right)}=2$$ Hence the ratio is 2.

To complete one round Ram takes 100m/15kmph and Rahim takes 20m/5kmph

They meet for the first time after L.C.M of (100m/15kmph , 20m/5kmph) = 100m/5kmph=20m/kmph.

Distance traveled by Ram =20m/kmph * 15kmph =300m.

So, he must have ran 300/100=3 rounds.

Note:

CAT gave both 2 and 3 as correct answers because of the word 'before'. 

Now we know that the perpendicular from the centre to a chord bisects the chord. Hence at the point of intersection of tangent, the chord will be divided into two parts of 3 cm each. As you can clearly see in the diagram, a right angled triangle is formed there.

Hence   $$\left(r+1\right)^2=r^2+9$$ or $$r^{2\ }+1+2r\ =\ r^2+9\ or\ 2r=8\ or\ r=4cm$$

Hence the radius of the larger circle is 5cm and diameter is 10cm.

Let the CP of the each toy be "x". CP of 12 toys will be "12x". Now the shopkeeper made a 10% profit on CP. This means that 

12x(1.1)= 2112 or x=160 . Hence the CP of each toy is ₹160.

Now let the SP of each toy be "m". Now he sold 8 toys at 20% discount. This means that 8m(0.8) or 6.4m

He sold 4 toys at an additional 25% discount. 4m(0.8)(0.75)=2.4m  Now 6.4m+2.4m=8.8m=2112 or m=240

Hence CP= 160 and SP=240. Hence profit percentage is 50%.

$$\left(x^2-5x+7\right)^{x+1}=1$$

There can be a solution when $$\left(x^2-5x+7\right)=1$$ or $$x^2-5x\ +6=0$$

or x=3 and x=2

There can also be a solution when x+1 = 0 or x=-1

Hence three possible solutions exist.

We have 2x + 5y = 99 or $$x=\frac{\left(99-5y\right)}{2}$$

Now $$x\ge\ y\ \ge\ -20$$ ; So $$\frac{\left(99-5y\right)}{2}\ge\ y\ ;\ 99\ge7y\ or\ y\le\ \approx\ 14$$

So $$-20\le y\le14$$. Now for this range of "y", we have to find all the integral values of "x". As the coefficient of "x" is 2,

then (99 - 5y) must be even, which will happen when "y" is odd. However, there are only 17 odd values of "y" be -20 and 14.

Hence the number of possible values is 17.

Based on the question: AD, CE and BF are the three altitudes of the triangle. It has been stated that {GD+GE+GF = s}

Now since the triangle is equilateral, let the length of each side be "a". So area of triangle will be

$$\frac{1}{2}\times\ GD\times\ a+\ \frac{1}{2}\times\ GE\times\ a\ +\frac{1}{2}\times\ GF\times\ a=\frac{\sqrt{\ 3}}{4}a^2$$

Now $$GD+GE+GF=\frac{\sqrt{\ 3}a}{2}$$ or $$s=\frac{\sqrt{\ 3}a}{2}\ or\ a=\frac{2s}{\sqrt{\ 3}}$$

Given the area of the equilateral triangle = $$\ \frac{\sqrt{3}}{4}a^2\ $$ ; substituting the value of 'a' from above, we get the area {in terms 's'}= $$\frac{s^2}{\sqrt{3}}$$

Let the first term of the GP be "a" . Now from the question we can show that

$$ar^{m-1}=\frac{3}{4}$$    $$ar^{n-1}=12$$

Dividing both the equations we get $$r^{m-1-n+1}=\frac{1}{16}\ or\ r^{m-n}=16^{-1\ }or\ r^{n-m}=16$$

So for the minimum possible value we take Now give minimum possible value to "r" i.e -4 and n-m=2

Hence minimum possible value of r+n-m=-4+2=-2