JIPMAT 2022 Question Paper

The value of the principal after n years in a simple interest scheme is given by $$=P\left[1+\dfrac{RT}{100}\right]$$, where R is the rate of interest and T is the time period.

It is given that the sum becomes double in 10 years.

$$P\left[1+\dfrac{R\left(10\right)}{100}\right]=2P$$

$$1+\dfrac{R}{10}=2$$

$$R=10\%$$

P, Q, and R started the business together and kept their investment for two years. 

Thus, their profit-sharing ratios will be in the ratio of their capital investment. Therefore, it is $$1:2:1$$

Now, the net overall profit is Rs 48000.

Thus, Q's share will be = $$\dfrac{2}{4}\times48000=24000$$ Rs

(A)
$$6x^2+x-12=0$$
$$6x^2+9x-8x-12=0$$
$$3x(2x+3)-4(2x+3)=0$$
$$(3x-4)(2x+3)=0$$
$$x=\dfrac{4}{3}\ or\ x=-\dfrac{3}{2}$$
(A) - (IV)

(D)
$$2x^2-5x-3=0$$
$$2x^2-6x+x-3=0$$
$$2x(x-3)+1(x-3)=0$$
$$(2x+1)(x-3)=0$$
$$x=\dfrac{-1}{2}\ or\ x=3$$
(D) - (III)

Thus, the correct answer is option B. 

NOTE:- Equation (B) and Equation (C) were incorrectly typed in the actual examination as well by NTA. 

Let the number of males be x and the number of females be y.

$$x+y=5000\rightarrow1$$

If the number of males were to increase by 5% and that of females by 3% annually, the population would grow to 5202 at the end of the year. 

$$1.05x+1.03y=5202\rightarrow2$$

Multiply eq. 1 by 1.05, and subtract eq. 2 from it-

$$1.05y-1.03y=5000(1.05)-5202$$

$$0.02y=48$$

$$y=2400$$

Thus, $$x=2600$$

Therefore, the number of males is 2600, and the number of females is 2400. 

A earns $$33\dfrac{1}{3}$$% more than B. Let the amount earned by B be $$3x$$

Then the amount earned by A $$=3x\left(1+\dfrac{1}{3}\right)=4x$$

The percentage B earns less than A will be $$=\dfrac{4x-3x}{4x}\times100=25\%$$

(A)
$$\sin^{2} (41^{\circ}) + \sin^{2} (49^{\circ})$$
$$\sin^{2} (41^{\circ}) + \sin^{2} ((90-41)^{\circ})$$
$$\sin^{2} (41^{\circ}) + \cos^{2} (49^{\circ})$$
$$=1$$
This is TRUE.

(B)
$$\sin^{2} (60^{\circ}) - 2\tan (45^{\circ}) - \cos^{2}(30^{\circ})$$
$$\left(\dfrac{\sqrt{3}}{2}\right)^2-2\left(1\right)-\left(\dfrac{\sqrt{3}}{2}\right)^2$$
$$=-2$$
This is FALSE

(C)
$$\sin^{2} (\theta) + \dfrac{1}{1 + \tan^{2} (\theta)} $$
$$\sin^{2} (\theta) + \dfrac{1}{\sec^{2} (\theta)}$$
$$\sin^{2} (\theta) + \cos^{2} (\theta)$$
$$=1$$
This is TRUE

Thus, statement A and statement C are TRUE. 

$$f(x)=x^{4}+10x^{3}+25x^{2}+15x+m $$ is exactly divisible by $$x+7$$. Thus, $$x+7$$ is a factor of $$f(x)$$

Therefore, using the remainder theorem, $$f(-7)=0$$

$$(-7)^{4}+10(-7)^{3}+25(-7)^{2}+15(-7)+m=0$$ 

$$2401-3430+1225-105+m=0$$

$$-1134+1225+m=0$$

$$m=-91$$

Let the radius of the circular playground be x. The area of a circular playground is 22176 $$cm^{2}$$.

$$\dfrac{22}{7}\times r^2=22176$$

$$r=84$$ cm

$$r=0.84m$$

Cost of fencing = Rs 50 per meter

Total cost of fencing = Circumference of playground * Cost of fencing

Total cost of fencing $$=2\times\dfrac{22}{7}\times0.84\times50$$

Total cost of fencing $$=264$$ Rs

$$\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}.......$$

$$\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+....$$

$$\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+....+\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$$

$$1-\dfrac{1}{n+1}$$

$$\dfrac{n}{n+1}$$

A can do 1/4 of the work in 3 days. This means that he can do the entire work in 12 days. 

B can do 1/6 of the same work in 4 days. This means that he can do the entire work in 24 days. 

Thus, A's efficiency will be 1/12 and B's efficiency will be 1/24, which shows that A is twice as efficient as B.

Now, the wages are divided in the ratio of their efficiency, and efficiency of A : efficiency of B = 2:1

Thus, A's payment will be $$\dfrac{2}{3}\times180=120$$ Rs