IPM Indore 2019 Question Paper

Let, the principal be $$P$$, rate be $$r\%$$

So, C.I. earned in 2 years = $$P\left(1+\dfrac{r}{100}\right)^2-P$$

Also, S.I. earned in 2 years = $$\dfrac{P\times\ r\times\ 2}{100}$$

So, $$P\left(1+\dfrac{r}{100}\right)^2-P$$ = $$2\times\ $$ $$\dfrac{P\times\ r\times\ 2}{100}$$

or, $$\left(1+\dfrac{r}{100}\right)^2-1$$ = $$2\times\ $$ $$\dfrac{\ r\times\ 2}{100}$$

let, $$\dfrac{r}{100}=x$$

so, $$\left(1+x\right)^2-1=4x$$

or, $$1+2x+x^2-1=4x$$

or, $$x^2=2x$$

so, $$x=2$$

or, $$\dfrac{r}{100}=2$$

so, rate is $$200\%$$ per annum

$$21^n=3^n\times\ 7^n$$

Now for the maximum value of $$n$$ so that $$21^n$$ divides 50!, we can say we need to maximum value of n such that both $$3^n$$ and $$7^n$$ will divide 50!

Now, the exponent of prime p in n! is given by the formula,

$$E\left(p\right)=\left[\dfrac{n}{p}\right]+\left[\dfrac{n}{p^2}\right]+......$$

So, the exponent of 7 in 50! = $$=\left[\dfrac{50}{7}\right]+\left[\dfrac{50}{7\times\ 7}\right]+......$$ = $$7+1=8$$

Clearly 3 will have an exponent greater than 8 as 3 is less than 7

So, the maximum value of the natural number $$n$$ for which $$21^{n}$$ divides 50! is = 8

We know,

29≡2 (mod 9)

So, $$29^{29}$$≡$$2^{29}$$ (mod 9)

Now, $$2^6$$≡1 (mod 9)

so, $$2^{29}$$ ≡ $$\left(2^6\right)^4\cdot2^5$$ (mod 9) ≡ $$1\times2^5$$ (mod 9)

Now, $$\left(29^{29}\right)^{29}$$ ≡ $$\left(2^{29}\right)^{29}$$ (mod 9) ≡ $$\left(2^5\right)^{29}$$ (mod 9) ≡ $$2^{145}$$ (mod 9)

Now, $$2^{145}$$ ≡ $$\left(2^6\right)^{24}\cdot2$$ (mod 9) ≡ $$1\times\ 2$$ (mod 9)

So, remainder will be 2

Sum of the given digits = 4+5+3+0 =12

in option (a), sum of digits = 9+6 = 15

in option (b), sum of digits = 7+8 = 15

in option (c), sum of digits = 4+2 = 6

in option (d), sum of digits = 5+4 = 9

Now, for option (d), the overall sum of digits will be 12+9=21 which is not divisible by 9

So, this option can be eliminated.

Now among 453096, 453078 and 453042, all the numbers are divisible by both 6 and 9.

So, we need to find which number is divisible by 7

Now, $$453096=45\times\ 10^4+30\times\ 10^2+96$$

Now, $$10^2$$≡2(mod 7) and $$10^4$$≡4(mod 7)

So, 453096(mod 7)≡($$45\times\ 4+30\times\ 2+96$$)(mod 7)≡336(mod 7)≡0(mod 7) as 336 is divisible by 7

So, 453096 is divisible by 7

We can do similar operation in case of 453078 and 453042

453078(mod 7)≡($$45\times\ 4+30\times\ 2+78$$)(mod 7)≡318(mod 7)≡3(mod 7) so it is not divisible by 7

453042(mod 7)≡($$45\times\ 4+30\times\ 2+42$$)(mod 7)≡282(mod 7)≡2(mod 7) so it is not divisible by 7

So, option A is the correct answer

Let the total number of boys in the class be 100.

so, number of boys who like cricket = 70

similarly, number of boys who like football = 50

also, number of boys who like both cricket and football = $$x$$

also, let the number of students who like none of the sports be $$n$$

So, we can make a Venn diagram like:

So, $$\left(70-x\right)+x+\left(50-x\right)+n=100$$

or, $$x=n+20$$

Now, the minimum value of $$n$$ can be zero

So, minimum value of $$x$$ is 20

Also, $$50-x\ge0$$,or, $$x\le50$$

So, the maximum value of $$x$$ can be 50

So, $$20 \leq x \leq 50$$

Number of students playing all three sports = 5

So, number of students who like only cricket and football will be 10-5=5

number of students who like only football and hockey will be 8-5=3

Let, $$x$$ be the number of students who like only cricket and hockey.

We can make a Venn diagram and fill it like:

So, $$\left(30-x\right)+x+\left(12-x\right)+25=65$$

or, $$42-x=40$$

or, $$x=2$$

So, number of students who like both cricket and hockey = $$x+5=2+5=7$$

$$\dfrac{x-3}{x^2+3x+2}\ge1$$

or, $$\dfrac{x-3}{x^2+3x+2}-1\ge0$$

or, $$\dfrac{x-3-x^2-3x-2}{x^2+3x+2}\ge0$$

or, $$\dfrac{-x^2-2x-5}{x^2+3x+2}\ge0$$

or, $$\left(x^2+2x+5\right)\left(x^2+3x+2\right)\le0$$

Now, $$\left(x^2+2x+5\right)=\left(x+1\right)^2+4>0$$ always

So, $$x^2+3x+2\le0$$

or, $$\left(x+2\right)\left(x+1\right)\le0$$

or, $$-2\le x\le-1$$

So, $$x$$ lies in the interval $$(-2,-1)$$

So, $$a=-2$$ and $$b=-1$$

So, $$b-a=-1-(-2)=1$$

We know, 

$$\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)$$We know, 

so, $$\left(a+b+c\right)^2=1+2\left(ab+bc+ac\right)$$

or, $$ab+bc+ac=\dfrac{\left(a+b+c\right)^2-1}{2}$$

Now, the least value of a square can be 0

so, the least value of $$ab+bc+ac=\dfrac{0-1}{2}=-\dfrac{1}{2}$$

Also, we know, $$A.M.\ge G.M.$$

so, $$\dfrac{a^2+b^2}{2}\ge ab$$

or, $$a^2+b^2\ge2ab$$

similarly, $$b^2+c^2\ge2bc$$

similarly, $$a^2+c^2\ge2ac$$

Adding the above three inequalities we get, $$2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)$$

or, $$a^2+b^2+c^2\ge ab+bc+ac$$

or, $$1\ge ab+bc+ac$$

So, the maximum value of $$ab+bc+ac$$ is 1

so, $$ab+bc+ac\in\ \left[-\dfrac{1}{2},1\right]$$

Given, $$\log_{2} \dfrac{3x - 1}{2 - x} < 1$$

So, $$\dfrac{3x-1}{2-x}>0$$ and $$\dfrac{3x-1}{2-x}<2$$

Solving the first inequality:

$$\dfrac{3x-1}{2-x}>0$$

or, $$\left(3x-1\right)\left(x-2\right)<0$$

or, $$\dfrac{1}{3}$$ $$<$$ $$x$$ $$<$$ $$2$$

Also, $$\dfrac{3x-1}{2-x}<2$$

or, $$\dfrac{3x-1}{2-x}-2<0$$

or, $$\dfrac{3x-1-4+2x}{2-x}<0$$

or, $$\dfrac{5x-5}{2-x}<0$$

or, $$\left(x-1\right)\left(x-2\right)>0$$

so, $$x<1$$ and $$x>2$$

So, overall we can say, $$\dfrac{1}{3}$$ $$<$$ $$x$$ $$<$$ $$1$$

So, option A is the correct answer

Given,

$$0.7^{2x^{2}-3x+4} < 0.343$$

or, $$0.7^{2x^2-3x+4}<0.7^3$$

since $$0.7$$ is a fractional value so, $$2x^2-3x+4>3$$

or, $$2x^2-3x+1>0$$

or, $$2x^2-2x-x+1>0$$

or, $$(2x-1)(x-1)>0$$

or, $$x<\dfrac{1}{2}$$ and $$x>1$$

So, option D is the correct answer.