JIPMAT 2022 Question Paper

Let the present ages of Amisha and Nimisha be A and N, respectively. 

Three years ago, the ratio of the ages of Amisha and Namisha was 8:9.

$$\dfrac{A-3}{N-3}=\dfrac{8}{9}$$

$$9A-27=8N-24$$

$$9A-8N=3\rightarrow1$$

Three years from now, the ratio will become 11:12.

$$\dfrac{A+3}{N+3}=\dfrac{11}{12}$$

$$12A+36=11N+33$$

$$11N-12A=3\rightarrow2$$

Multiply eq. 1 by 4 and eq. 2 by 3, and then adding them up - 

$$33N-32N=12+9$$

$$N=21$$

Therefore, $$A=19$$

The present age of Amisha is 19 years. 

Let the radius of the inner circle be x. If we join the four centres of the outer circle, then it will be a square as the radius of each circle is equal, thus, the side length of the square will be 2r. 

Now, the innermost circle is touching all four circles and the radius of all the circles is equal, thus the length of the line joining the diagonally opposite points will be = (r + x + x + r) = (2r + 2x)

Also, the length of the diagonal is $$\sqrt{\left(2r\right)^2+\left(2r\right)^2}=2\sqrt{2}r$$

Therefore - 

$$2r+2x=2\sqrt{2}r$$

$$x=\left(\sqrt{2}-1\right)r$$

$$\sin(\alpha)$$ and $$\cos(\alpha)$$ are the roots of the equation $$ax^{2}+bx+c=0$$. Therefore - 

$$\sin\alpha\cos\alpha=\dfrac{c}{a}$$  and 

$$\sin\alpha+\cos\alpha=\dfrac{-b}{a}$$

Squaring both sides -

$$\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha=\dfrac{b^2}{a^2}$$

$$1+\dfrac{2c}{a}=\dfrac{b^2}{a^2}$$

$$b^2=a^2\left(1+\dfrac{2c}{a}\right)$$

$$b^2=a^2+2ac$$

It is given that $$ab+bc+ca=0$$

The mean of $$a^{2}$$, $$b^{2}$$ and $$c^{2}$$ is $$KM^{2}$$

$$a^2+b^2+c^2=3kM^2$$

The mean of a, b and c is $$M$$.

$$a+b+c=3M$$

Squaring both sides - 

$$a^2+b^2+c^2+2(ab+bc+ca)=9M^2$$

$$3kM^2+2(0)=9M^2$$

$$k=3$$

(A) 
First five prime numbers are {2,3,5,7,11}
Mean = {2+3+5+7+11} / 5 = 5.6 
(A) - (III)

(B) 
All factors of 24 are {1,2,3,4,6,8,12,24}
Mean = (1+2+3+4+6+8+12+24) / 8 = 60 / 8 = 7.5
(B) - (II)

(C)
First 6 multiples of 4 = {4,8,12,16,20,24}
Mean = (4+8+12+16+20+24) / 6 = 14
(C) - (IV)

(D)
Mean = [(x-5y) + (x-3y) + (x-y) + (x+y) + (x+3y) + (x+5y)] / 6 = 12
6x / 6 = 12
x = 12
(D) - (I)

$$\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{8-2\sqrt{15}}}+\dfrac{\sqrt{11+2\sqrt{30}}}{\sqrt{6}-\sqrt{5}}$$

$$\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}+\dfrac{\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}}{\sqrt{6}-\sqrt{5}}$$

$$\dfrac{\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)}+\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}$$

$$\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\times\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}\times\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}$$

$$\dfrac{8+2\sqrt{15}}{2}+11+2\sqrt{30}$$

$$15+\sqrt{15}+2\sqrt{30}$$

$$aXb=2a- 3b+ab$$

$$3X5=2\times3-3\times5+3\times5=6$$

$$5X3=2\times5-3\times3+5\times3=16$$

$$3X5+5X3=6+16=22$$

$$\dfrac{325\times 325 \times 325 + 175 +175 +175}{325\times 325 -325 \times 175+175 \times 175}$$

Let $$a=325$$ and $$b=175$$

$$\dfrac{a\times a\times a+ b+b+b}{a\times a-a\times b+b\times b}$$

$$\dfrac{a^3+b^3}{a^2-ab+b^2}$$

$$\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2-ab+b^2}$$

$$a+b$$

Substituting the values of a and b

$$325+175$$

$$500$$

Statement I
$$AB^2+BC^2=(6\sqrt{3})^2+6^2=108+36=144$$
$$AC^2=12^2=144$$
$$AB^2+BC^2=AC^2$$
ABC is a right-angled triangle, with a right angle at B. 
Statement I is true. 

Statement II
$$AC^2+BC^2=AC^2+AC^2=2AC^2$$
It is given that $$AB^2=2AC^2$$
$$AB^2=AC^2+BC^2$$
ABC is a right-angled triangle, right-angled at C
Statement II is true

Statement I
$$sin(\theta) =\dfrac{5}{13}=\dfrac{P}{H}$$
We can assume P = 5k, and H = 13k
$$P^2+B^2=H^2$$
$$25k^2+B^2=169k^2$$
$$B=12k$$
$$tan(\theta)=\dfrac{P}{B}=\dfrac{5}{12}$$
Statement I is true

Statement II
$$cot(\theta) =\dfrac{12}{5}=\dfrac{B}{P}$$
We can assume P = 5k, and H = 12k
$$P^2+B^2=H^2$$
$$25k^2+144k^2=H^2$$
$$H=13k$$
$$sin(\theta)=\dfrac{P}{H}=\dfrac{5}{13}$$
Statement II is false