IPM Indore 2020 Question Paper

We are given:

The product of the quotient and remainder is 24:  $$Q\times R=24$$
The sum of the quotient and remainder is 10:   $$Q+R=10$$

Solving above two equations, we get {Q,R} as {6,4} or {4,6} . But now the given divisor is 5, the remainder should be less than 5, i.e remainder only possibility is to be 4 .

Hence, Q = 6, R = 4.

Now, Dividend = (Divisor×Quotient) + Remainder = 5(6) + 4 = 34.        

Therefore, Dividend = 34 .

The shortest distance would be the, distance between the given point P(-4,3) and the point of intersection of circle and line joining centre of circle and P(-4,3).

The effective shortest distance = Distance between {Centre, P} - Radius of circle .

As the given circle is having centre at origin (0,0) and given point P(-4,3) : 

Distance b/w Centre and $$P=\sqrt{\ \left(0+4\right)^2+\left(0+3\right)^2}$$ = 5                                           

Given radius of circle = 1.

Therefore, shortest distance is 5-1 = 4.

$$\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+....=\dfrac{\frac{1}{4}}{1-\frac{1}{2}}=\dfrac{\frac{1}{4}}{\frac{1}{2}}=\dfrac{1}{2}$$  (Applying the formula for infinite GP)

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=0.04^{\log_{\sqrt{5}}\left(\frac{1}{2}\right)}$$

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=5^{\left(-2\right)\log_{\sqrt{5}}\left(\frac{1}{2}\right)}$$

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=5^{-\dfrac{2}{\frac{1}{2}}\log_5\left(2\right)^{-1}}$$

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=5^{4\log_52}$$

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=2^{4\log_55}$$

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=16$$

$$\begin{vmatrix}a & a^2 & a^3-1 \\b & b^2 & b^3-1 \\ c & c^2 & c^3-1 \end{vmatrix}=0$$

$$\begin{vmatrix}a & a^2 & a^3 \\b & b^2 & b^3 \\ c & c^2 & c^3 \end{vmatrix}+\begin{vmatrix}a & a^2 & -1 \\b & b^2 & -1 \\ c & c^2 & -1 \end{vmatrix}=0$$

Taking -1 common from the second matrix of the third row.

$$\begin{vmatrix}a & a^2 & a^3 \\b & b^2 & b^3 \\ c & c^2 & c^3 \end{vmatrix}-\begin{vmatrix}a & a^2 & 1 \\b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix}=0$$

Taking a,b,c common from row 1, row 2 and row 3 respectively from the first matrix.

$$abc\begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}-\begin{vmatrix}a & a^2 & 1 \\b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix}=0$$

Now, we will switch column 3 of the second matrix to column 1. Since with each switch the sign changes, and for this we have to switch the columns twice, thus the sign will remain the same as before.

$$abc\begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}-\begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}=0$$

$$(abc-1)\begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}=0$$

Performing row operations on matrix, R2 -> R2-R1, and R3 -> R3-R1.

$$(abc-1)\begin{vmatrix}1 & a & a^2 \\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2 \end{vmatrix}=0$$

$$\left(abc-1\right)\left[\left(b-a\right)\left(c^2-a^2\right)-\left(c-a\right)\left(b^2-a^2\right)\right]$$

$$\left(abc-1\right)\left(b-a\right)\left(c-a\right)\left(c+a-b-a\right)$$

$$\left(abc-1\right)\left(b-a\right)\left(c-a\right)\left(c-b\right)$$

Since a,b,c are distinct values, thus (b-a), (c-a), and (c-b) can never be 0, therefore, only (abc-1) can be = 0.

=> $$abc-1=0$$

=> $$abc=1$$

$$f(x)=|3-x|+|2+x|+|5-x|$$

We will check the inflection points, because any set of values less than the least inflection value point, or any values more than the highest inflection point will give a higher value than it's minimum

Inflection Points are = -2, 3, 5

At x = -2 => $$f(x)=|3-(-2)|+|2+(-2)|+|5-(-2)|=12$$

At x = 3 => $$f(x)=|3-3|+|2+3|+|5-3|=7$$

At x = 5 => $$f(x)=|3-5|+|2+5|+|5-5|=9$$

Thus, the minimum value of f(x) = 7, at x = 3. 

Let the cost of pens be A, and the cost of pencils be B.

During his first visit, he purchased pens and pencils in the ratio 2:3, and paid Rs. 86. Let's assume he purchased 2k pens and 3k pencils.

=> $$2kA+3kB=86\rightarrow1$$

During his second visit, he purchased four pens, and since he bought pens and pencils in a 4:1 ratio, he must have also bought one pencil. He paid Rs 112 for this.

=> $$4A+B=112$$ => $$B=112-4A$$

Substitute the value of B in eq. 1 -

=> $$k(2A+3(112-4A))=86$$

=> $$k(2A+336-12A)=86$$

=> $$k(336-10A)=86$$

k is a positive integer, and A is also a positive integer; therefore, (336-10A) must be a factor of 86, and thus (336-10A) will be equal to 1, 2, 43, or 86.

Only when (336-10A) = 86 => A = 25 will give an integer value of A, otherwise we will have a decimal value for A

=> A = 25 and B = 112 - 4A = 12

The amount he paid for pens during his second visit = 4 * 25 = 100 rs.

The product of the thousands digit and the unit digit is zero; thus, the unit digit will be 0 (because if the thousands digit is 0, then it won't be a four-digit number). Also, their difference is 7, thus the thousands digit is 7.

Number = 7ab0

The thousands digit is as much more than the units digit as the hundreds digit is more than the tens digit => a = b + 7

The product of the middle digits is 18 

=> ab = 18

=> b(b+7) = 18

=> b = 2 and a = 9

Therefore, the number is 7920. 

Let the number of students who passed in all three subjects be x.

$$M∪P∪C=M+P+C-\left(M∩P\right)-\left(M∩C\right)-\left(P∩C\right)+\left(M∩P∩C\right)$$

$$80=50+30+40-\left(M∩P\right)-\left(M∩C\right)-\left(P∩C\right)+x$$

$$80=120-\left(M∩P\right)-\left(M∩C\right)-\left(P∩C\right)+x$$

$$\left(M∩P\right)+\left(M∩C\right)+\left(P∩C\right)-40=x$$

The value of x will be maximum when maths and physics, physics and chemistry, and maths and chemistry values are maximum, which is equal to 20.

$$20+20+20-40=x$$

=> $$x=20$$

Lets assume two people to be "A" & "B" . It is told that they each had run 3Kms on a 300 m circular track , this implies each had run$$\dfrac{\ 3000\ m}{300\ m}\ =\ 10\ rounds$$ .

Speeds of A:B given as 3:2 . Now, this statement implies that if A takes 100 sec to finish 1 round i.e 300 m circular track, then B should be taking 150 sec to complete 1 round . { Time ratio and Speed ratio are inverse related as distance of each round is constant for both }.

The above deduction can say that if A finishes 3 rounds in 300 sec of time, then B will finish only 2 rounds in the same 300 sec of time. This implies for every 3 rounds which A makes , he will meet/cross B once in his journey.

Therefore, it is given 10 rounds of A, then he will be meeting B : $$\left[\dfrac{\ 10}{3}\right]$$ = 3 times.

As there are total 13 toys and 4 indistinguishable , hence 13-4 = 9 distinct toys are present.

Since the toys are indistinguishable , we can pick 5 combinations of (indistinguishable, distinct) : (0,4)/(1,3)/(2,2)/(3,1),/(4,0)

Hence, the question can be solved by focusing on no of ways to pick 'n' distinct balls out of 9 balls for n = {4,3,2,1,0}

Total possibilities = $$9C_4+9C_3+9C_2+9C_1+9C_0$$

                          =  126 + 84 + 36 + 9 + 1 = 256

Therefore, the answer is 256.