IPM Indore 2019 Question Paper

Let $$\beta=\alpha r$$, $$\gamma=\alpha r^2$$, and $$\delta=\alpha r^3$$, where r is the common difference.

$$x^{2} - x + p = 0$$

=> $$\alpha+\beta=1$$ => $$\alpha(1+r)=1\rightarrow1$$

and, $$\alpha\beta=p$$

$$x^{2}- 4x + q = 0$$

=> $$\gamma+\delta=1$$ => $$\alpha r^2(1+r)=4\rightarrow2$$

and, $$\gamma\delta=q$$

Dividing eq. 1 and eq. 2 - 

=> $$r^2=4$$ => $$r=\pm2$$

Case-1: If r = 2

=> $$\alpha(1+r)=1$$ => $$\alpha(1+2)=1$$ => $$\alpha = \dfrac{1}{3}$$

=> $$\beta=\alpha r=\dfrac{2}{3}$$

=> $$\gamma=\alpha r^2=\dfrac{4}{3}$$

=> $$\delta=\alpha r^3=\dfrac{8}{3}$$

=> $$\alpha\beta=p$$ => $$p=\left(\dfrac{1}{3}\right)\left(\dfrac{2}{3}\right)=\dfrac{2}{9}$$, but it is given that p and q are integers. Thus, this case is not valid.

Case-2: If r = -2

=> $$\alpha(1+r)=1$$ => $$\alpha(1-2)=1$$ => $$\alpha = -1$$

=> $$\beta=\alpha r=2$$

=> $$\gamma=\alpha r^2=-4$$

=> $$\delta=\alpha r^3=8$$

=> $$\alpha\beta=p$$ => $$p=(-1)(2)=-2$$

=> $$\gamma\delta=q$$ => $$q=(-4)(8)=-32$$

=> $$p+q=(-2)+(-32) = -34$$

$$(1 + x - 2x^{2})^{6} = A_{0} + \sum_{r = 1}^{12} A_{r}X^{r}$$

Substitute x = 1 

$$(1 + 1 - 2(1)^{2})^{6} = A_{0} + A_{1}(1^1)+A_{2}(1^2)+A_{3}(1^3)+A_{4}(1^4)+......+A_{12}(1^12)$$

$$(0)^{6} = A_{0} + A_{1}+A_{2}+A_{3}+A_{4}+......+A_{12}$$

$$0 = A_{0} + A_{1}+A_{2}+A_{3}+A_{4}+......+A_{12}\rightarrow1$$

Substitute x = -1

$$(1 + (-1) - 2(-1)^{2})^{6} = A_{0} + A_{1}(-1)^1+A_{2}(-1)^2+A_{3}(-1)^3+A_{4}(-1)^4+......+A_{12}(-1)^12$$

$$(-2)^{6} = A_{0} - A_{1}+A_{2}-A_{3}+A_{4}+......+A_{12}$$

$$64 = A_{0} - A_{1}+A_{2}-A_{3}+A_{4}+......+A_{12}\rightarrow2$$

Adding eq. 1 and eq. 2

$$64 =2( A_{0}+A_{2}+A_{4}+A_{6}+A_{8}+A_{10}+A_{12})$$

$$32 =A_{0}+A_{2}+A_{4}+A_{6}+A_{8}+A_{10}+A_{12}$$     

{To calculate the value of $$A_{0}$$, substitute the original eq. with x=0 => $$(1 + 0 - 2(0)^{2})^{6} = A_{0}$$ =>$$A_{0}=1$$}

$$32 =1+A_{2}+A_{4}+A_{6}+A_{8}+A_{10}+A_{12}$$

$$A_{2}+A_{4}+A_{6}+A_{8}+A_{10}+A_{12}=31$$

The first common term in both the arithmetic progressions is 5. The common difference will be the LCM of the common difference of the two AP. Thus, the common difference = LCM(2,3) = 6

Also, the last common term should be less than or equal to 101.

=> $$a=5$$, $$d=6$$, and $$a_n\le101$$

=> $$a_n\le101$$

=> $$a+(n-1)d\le101$$

=> $$5+(n-1)6\le101$$

=> $$(n-1)6\le96$$

=> $$(n-1)\le16$$

=> $$n\le17$$

Thus, the number of common terms is 17. 

Probability that ace will appear in 1st turn $$f\left(1\right)=\dfrac{4}{52}=\dfrac{1}{13}$$

Probability that ace will appear in 2nd turn $$f\left(2\right)=\dfrac{48}{52}\times \dfrac{4}{51}=\dfrac{1}{13}\times \dfrac{48}{51}$$

Probability that ace will appear in 3rd turn $$f\left(3\right)=\dfrac{48}{52}\times \dfrac{47}{51}\times \dfrac{4}{50}=\dfrac{1}{13}\times \dfrac{48}{51} \times \dfrac{47}{50}$$

$$f\left(1\right)=\dfrac{1}{13}$$

$$f\left(2\right)=\dfrac{1}{13}\times \dfrac{48}{51}=f(1)\times (number<1)$$ => $$f(2)<f(1)$$

$$f\left(3\right)==\dfrac{1}{13}\times \dfrac{48}{51} \times \dfrac{47}{50}=f(2)\times (number<1)$$ => $$f(3)<f(2)$$

$$\dfrac{1}{13} > f(2) > f(3)$$

A die is thrown three times and the sum of the three numbers is found to be 15. This is possible only when the combinations are - (3,6,6), (4,5,6), and (5,5,5)

When the combination is (3,6,6), there are 3 cases possible
When the combination is (4,5,6), there are 6 cases possible
When the combination is (5,5,5), there is only 1 possible case.

Thus, there are a total of 10 cases when the sum of the numbers is 15.

We need to find the probability that the first throw was a four. If the first throw was a four, then the second throw must be either 5 or 6, and the third throw must be either 6 or 5, respectively. Thus, there are 2 cases possible.

Probability that the sum is 15, and the first throw is 4 = $$\dfrac{2}{10}=\dfrac{1}{5}$$

Let us assume there are a total of 3X families. Since the proportion of families with 2,4, and 6 members is roughly equal, we will assume there are X families of each size. 

=> Total number of people in village = 2X + 4X + 6X = 12X

Probability that the person selected is from a 2-membered family = $$\dfrac{2X}{12X}=\dfrac{1}{6}$$

Probability that the person selected is from a 4-membered family = $$\dfrac{4X}{12X}=\dfrac{1}{3}$$

Probability that the person selected is from a 6-membered family = $$\dfrac{6X}{12X}=\dfrac{1}{2}$$

The average family size will be given by summing up the products of number of family members and the probability of that membered family being chosen.

=> Average family size = $$\dfrac{\frac{1}{6}\times2+\frac{1}{3}\times4+\frac{1}{2}\times6}{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=\dfrac{14}{3}=4.667$$

$$(\log_{3}30)^{−1}+(\log_{4}900)^{−1}+(\log_{5}30)^{−1}$$

$$\dfrac{1}{\log_330}+\dfrac{1}{\log_4900}+\dfrac{1}{\log_530}$$

$$\dfrac{1}{\log_330}+\dfrac{1}{\log_{2^2}{30^2}}+\dfrac{1}{\log_530}$$

$$\dfrac{1}{\log_330}+\dfrac{1}{\frac{2}{2}\log_230}+\dfrac{1}{\log_530}$$

$$\dfrac{1}{\log_330}+\dfrac{1}{\log_230}+\dfrac{1}{\log_530}$$

$$\log_{30}3+\log_{30}2+\log_{30}5$$

$$\log_{30}\left(3\times2\times5\right)$$

$$\log_{30}\left(30\right)=1$$

If $$\log_ax<\log_ay$$, then both $$x$$ and $$y$$ are greater than 0, and - 

For $$0<a<1$$ => $$x>y$$

For $$a>1$$ => $$x<y$$

We will replace x by f(x) and y by g(x).

If $$\log_af(x)<\log_ag(x)$$, then both $$f(x)$$ and $$g(x)$$ are greater than 0, and -

For $$0<a<1$$ => $$f(x)>g(x)>0$$

For $$a>1$$ => $$0<f(x)<g(x)$$

Let the sides of the three cubes be a, b, and c. It is known that the sum of their surface areas is 564 $$cm^{2}$$.

$$6\left(a^2+b^2+c^2\right)=564$$

$$\left(a^2+b^2+c^2\right)=94$$

Now, it is given that the sides of the cubes are integers, thus, (a,b,c) can be either (9,2,3) or (6,7,3)

Therefore, the sum of volumes can be either $$9^3+2^3+3^3=764\ cm^3$$ or it can be $$6^3+7^3+3^3=586\ cm^3$$

$$2^{300}=\left(2^3\right)^{100}=8^{100}$$

$$3^{200}=\left(3^2\right)^{100}=9^{100}$$

$$2^{100}+3^{100}<6^{100}$$   (Because if $$a+b<c$$, then $$a^x+b^x$$, given a,b,c, and x are greater than 1)

$$4^{100}$$

Thus, among all these, $$9^{100}$$ is highest and thus, $$3^{200}$$ is the greatest value.