IPM Indore 2024 Question Paper

Let 'a' be first term, 'r' be common ratio of an infinite GP :
$$\ \dfrac{\ a}{1-r}=80$$ ..... $$1^{st}$$ Equation .
sum of first two terms = a + ar = a(1 + r) = 35.... Equation 2.

Substituting equation 1 in $$2^{nd}$$ equation : 80(1 - r)(1 + r) = 35
(1 - r)(1 + r) = $$1-r^2$$ = $$\dfrac{\ 7}{16}$$
hence,$$r=\pm\ \dfrac{\ 3}{4}$$ 
The possible values for (a,r) = (20, 3/4) or (140, -3/4)

Sum of first 'n' terms of GP = a($$\dfrac{\ 1-r^n}{1-r}$$) = 100

If we check for (a,r) = (20, 3/4):
It gives us that : $$1-\left(\dfrac{\ 3^n}{4^n}\right)$$ = $$\dfrac{\ 5}{4}$$ . But 1 minus some positive value can never be $$\dfrac{\ 5}{4}$$ which is greater than 1. Hence, this possibility is ignored.

Now, the only possibility is (a,r) = (140, -3/4) :

It gives us that : $$1-\dfrac{\ \left(-1\right)^n3^n}{4^n}=\dfrac{\ 5}{4}$$ 

This implies , $$\dfrac{-1}{4}=\dfrac{\left(-1\right)^n3^n}{4^n}$$

Therefore, closest integral value of n is 5.

$$a = \dfrac{(\log_7 4)(\log_7 5 - \log_7 2)}{\log_7 25(\log_7 8 - \log_7 4)}$$ 

$$a=\dfrac{\ 2\left(\log_72\right)\left(\log_72.5\right)}{2\left(\log_75\right)\left(\log_72\right)}$$

$$a=\dfrac{\ \log_72.5}{\log_75}=\log_52.5$$

Therefore ,$$5^a=2.5$$

Let, the number of girls be $$N$$.

Now, in the first option:

a.) $$\dfrac{48}{100}\times\ 27=12.96$$ and $$\dfrac{50}{100}\times\ 27=13.5$$

Now, $$12.96<N<13.5$$

So, we can have an integer value of $$N$$, $$N=13$$

So, option(a) can be a possible answer choice

So, automatically option (b) and (c) are eliminated as they are greater than 27

So, we need to check only option (d)

Now, $$\dfrac{48}{100}\times\ 25=12$$ and $$\dfrac{50}{100}\times\ 25=12.5$$

Now, $$12<N<12.5$$

In this range we can't have an integer value for $$N$$

So, option(d) is also eliminated

So the smallest value of $$N$$ is 27.

Total no of ways to arrange 8 people around a circular table is 7! ways.

let No of boys between G1 and G2 be X , G2 and G3 be Y , G3 and G1 be Z . Then as total boys are 5 , it implies X+Y+Z = 5.
But given at least one boy between any two girls i.e $$X\ \ge1,\ \ Y\ge1\ ,\ \ Z\ge1$$: 
(X,Y, Z) = (1,2,2) or (2,1,2) or (2,2,1) ~ 3 ways.
(X,Y, Z) = (1,3,1) or (3,1,1) or (1,1,3) ~ 3 ways.

No of ways of arranging 3 girls (G1, G2, G3) on a circular table is 2! ways = 2 ways . After arranging girls , we need to now place 5 boys in between them = (6 triplets) x 5! ways = 6! ways.

Total ways of arranging 8 members such that there is at least one boy b/w any two girls = 2x6! ways .

Therefore , probability is $$\dfrac{\ 2\times\ 6!}{7!}$$ = $$\dfrac{2}{7}$$