IPM Indore 2020 Question Paper

Given number is $$10^{19}$$ , this can be presented as $$2^{19}\times\ \ 5^{19}$$ .

so, total no of factors = (19+1)(19+1) = $$20^2$$.

To calculate no of factors which are multiples of $$10^{15}$$ , lets write $$10^{19}$$ = $$10^{15}\times\ 2^4\times\ 5^4$$ . This implies no of factors which are multiple of $$10^{15}$$ are equal to no of factors of $$2^4\times\ 5^4$$ . 

Therefore, total no of factors which are multiple of $$10^{15}$$ are :

$$2^4=\left\{2^0,2^1,...2^4\right\}$$

$$5^4=\left\{5^0,5^1,...5^4\right\}$$

Therefore , it equal to = (4+1)x(4+1) = 25

The probability = $$\dfrac{\ 25}{20^2}\ =\ \ \dfrac{\ 1}{16}$$

Since the sides are three consecutive integers, let the sides be n, n+1, and n+2, where n > 0.

The sum of the two sides is greater than the third side.

=> $$n+n+1>n+2$$

=> $$n>1\rightarrow1$$

The perimeter of the triangle is at most 100.

=> $$n+n+1+n+2\le100$$

=> $$3n+3\le100$$

=> $$n\le32.33\rightarrow2$$

Also, the given triangle is an acute angled triangle.

=> $$\left(n+2\right)^2< n^2+\left(n+1\right)^2$$

=> $$n^2+4n+4< n^2+n^2+2n+1$$

=> $$n^2-2n-3>0$$

=> $$\left(n-3\right)\left(n+1\right)>0$$

=> $$n<-1\ or\ n>3\rightarrow3$$

Thus, comparing eq. 1, 2 and 3, we will get -

=> $$3<n\le32.33$$

The values of n are - $$n=4,5,6,7,......32$$. Thus, there are 29 values possible. 

The straight line passing through the point M (-5,4), and the portion of it between the axes is divided by the point M into two halves. This means that M is the midpoint of the line segment joining the intercepts of X-axis and Y-axis

=> $$\dfrac{a+0}{2}=-5$$ => $$a=-10$$

=> $$\dfrac{0+b}{2}=4$$ => $$b=8$$

To get the equation of the line, we will use the line-intercept form.

=> $$\dfrac{x}{a}+\dfrac{y}{b}=1$$

=> $$\dfrac{x}{-10}+\dfrac{y}{8}=1$$

=> $$10y-8x=80$$

ALTERNATE SOLUTION

Check in which of the following equations the point (-5,4) lies on the line by putting the x-coordinate as -5, and the y-coordinate as 4. There is only one equation that will satisfy this condition. 

$$\cos^{2}\dfrac{\pi}{8} + \cos^{2}\dfrac{3\pi}{8} + \cos^{2}\dfrac{5\pi}{8} + \cos^{2}\dfrac{7\pi}{8}$$

$$\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\left(\pi-\dfrac{3\pi}{8}\right)+\cos^2\left(\pi-\dfrac{\pi}{8}\right)$$

We know that, $$\cos\left(\pi-\theta\ \right)=-\cos\left(\theta\right)$$, therefore -

$$\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\dfrac{\pi}{8}$$

$$2\cos^2\dfrac{\pi}{8}+2\cos^2\dfrac{3\pi}{8}$$

$$2\cos^2\left(\dfrac{\pi}{2}-\dfrac{3\pi}{8}\right)+2\cos^2\dfrac{3\pi}{8}$$

We know that, $$\cos\left(\dfrac{\pi}{2}-\theta\ \right)=\sin\left(\theta\right)$$, therefore - 

$$2\sin^2\dfrac{3\pi}{8}+2\cos^2\dfrac{3\pi}{8}$$

$$2\left(\sin^2\dfrac{3\pi}{8}+\cos^2\dfrac{3\pi}{8}\right)$$

$$2\left(1\right)=2$$

$$\dfrac{1}{1^{2}} + \dfrac{1}{2^{2}} + \dfrac{1}{3^{2}} + $$...... upto $$ \infty = \dfrac{\pi^{2}}{6}$$

the above equation can be written as :
[$$\dfrac{1}{1^{2}} + \dfrac{1}{3^{2}} + \dfrac{1}{5^{2}} + $$......$$upto\ \infty\ $$] + [$$\dfrac{1}{2^{2}} + \dfrac{1}{4^{2}} + $$......$$upto\ \infty$$] = $$\dfrac{\pi^{2}}{6}$$

Now lets assume $$\dfrac{1}{1^{2}} + \dfrac{1}{3^{2}} + \dfrac{1}{5^{2}} + $$......$$upto\ \infty\ $$ = X.

And for the second half of the equation lets take $$\ \dfrac{\ 1}{2^2}$$ as common :

X + $$\ \dfrac{\ 1}{2^2}$$[$$\dfrac{1}{1^{2}} + \dfrac{1}{2^{2}} + \dfrac{1}{3^{2}} + $$......$$upto\ \infty\ $$ ] = $$\dfrac{\pi^{2}}{6}$$

X + $$\dfrac{\ 1}{4}$$[$$\dfrac{\pi^{2}}{6}$$] = $$\dfrac{\pi^{2}}{6}$$

Therefore, X = [$$\dfrac{\pi^{2}}{6}$$][$$\dfrac{\ 3}{4}$$] = $$\dfrac{\pi^{2}}{8}$$

Let A be the event that the die shows a five. Let B be the event that the die does not show a five. And let E be the event that the person reports that the die shows a five.

Probability of getting a five $$P\left(A\right)=\dfrac{1}{6}$$

Probability of not getting a five $$P\left(B\right)=\dfrac{5}{6}$$

Now, we will find the conditional probabilities.

The probability of the number being reported as five, and it is five (he is telling the truth) $$P(E|A)=\dfrac{4}{5}$$

The probability of the number being reported as five, but it is not five (he is telling the lie) $$P(E|B)=\dfrac{1}{5}$$

Now, we will apply Bayes' rule. Since events A and B are mutually exclusive -

=> $$P(A|E)=\dfrac{P(E|A)\cdot P\left(A\right)}{P(E|B)\cdot P\left(B\right)+P(E|A)\cdot P\left(A\right)}$$

=> $$P(A|E)=\dfrac{\frac{4}{5}\times\frac{1}{6}}{\frac{1}{5}\times\frac{5}{6}+\frac{4}{5}\times\frac{1}{6}}$$

=> $$P(A|E)=\dfrac{\frac{4}{30}}{\frac{9}{30}}$$

=> $$P(A|E)=\dfrac{4}{9}$$

$$\log_{5}\log_{8}(x^2 - 1) = 0$$

$$\log_{8}(x^2 - 1) = 1$$

$$(x^2 - 1) = 8$$

$$x^2=9$$

$$x=3\ or\ x=-3$$

$$\dfrac{1}{1+x} < 1 - x + x^{2}$$

$$\dfrac{1}{1+x}-\left(1-x+x^2\right)<0$$

$$\dfrac{1-\left(1+x\right)\left(1-x+x^2\right)}{1+x}<0$$

$$\dfrac{1-\left(1+x^3\right)}{1+x}<0$$

$$\dfrac{1-1-x^3}{1+x}<0$$

$$\dfrac{x^3}{1+x}>0$$

The solution for the above equation is $$x<-1\ or\ x>0$$

$$\dfrac{1}{1+x} > 1 - x + x^{2}$$

$$\dfrac{1}{1+x}-\left(1-x+x^2\right)>0$$

$$\dfrac{1-\left(1+x\right)\left(1-x+x^2\right)}{1+x}>0$$

$$\dfrac{1-\left(1+x^3\right)}{1+x}>0$$

$$\dfrac{1-1-x^3}{1+x}>0$$

$$\dfrac{x^3}{1+x}<0$$

The solution for the above equation is $$-1<x<0$$

Thus, $$\dfrac{1}{1+x} < 1 - x + x^{2}$$ is satisfied when $$x<-1\ or\ x>0$$

and, $$\dfrac{1}{1+x} > 1 - x + x^{2}$$ is satisfied when $$-1<x<0$$

Thus, statements (i) and (iv) are correct. 

Fifty litres of a mixture of milk and water contains 30% water. This means that in this mixture, there are 15 litres of water and 35 litres of milk.

Eighty litres of another mixture of milk and water that contains 20% water. This means that in this mixture, there are 16 litres of water and 64 litres of milk.

Now, both mixtures are combined. There, the total quantity of the mix is 130 litres, of which 31 litres is water and 89 litres is milk. 

Now, we need to find out how many litres of water should be added to the resulting mixture to obtain a final mixture that contains 25% water.

=> $$\dfrac{31+x}{130+x}=\dfrac{1}{4}$$

=> $$124+4x=130+x$$

=> $$3x=6$$

=> $$x=2$$ litres of water should be added to make the concentration of water 25%.

Let the time taken for the first worker to construct the wall alone be A hours. Let the time taken for the second worker to build the wall alone be B hours. Let the time taken for the third worker to construct the wall alone be C hours.

The first worker, working alone, can construct the wall twice as fast as the third worker.

=> $$\dfrac{1}{A}=2\left(\dfrac{1}{C}\right)$$ => $$\dfrac{1}{C}=\dfrac{1}{2A}$$

The first worker can complete the task an hour sooner than the second worker.

=> $$A=B-1$$ => $$B=A+1$$

The three workers working together need 1 hour to construct a wall.

=> $$\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=\frac{1}{1}$$

=> $$\dfrac{1}{A}+\dfrac{1}{A+1}+\dfrac{1}{2A}=1$$

=> $$\dfrac{2A+2+2A+A+1}{2A\left(A+1\right)}=1$$

=> $$5A+3=2A^2+2A$$

=> $$2A^2-3A-3=0$$

=> $$A=\dfrac{3\pm\sqrt{9-4\left(-3\right)\left(2\right)}}{2\left(2\right)}$$

=> $$A=\dfrac{3+\sqrt{33}}{4}$$          [Since the value of A cannot be negative]

The average time in hours taken by the three workers, when working alone, to construct the wall is = $$\dfrac{A+B+C}{3}$$

$$=\dfrac{\frac{3+\sqrt{33}}{4}+\frac{3+\sqrt{33}}{4}+1+2\left(\frac{3+\sqrt{33}}{4}\right)}{3}$$

$$=\dfrac{3+\sqrt{33}+1}{3}$$

$$=\dfrac{4+\sqrt{33}}{3}$$