JIPMAT 2022 Question Paper

We can see that AB = 5m, and BC = 12m

We need to find the distance between the top of the two poles, which is equal to AC.

$$AC^2=AB^2+BC^2$$

$$AC^2=5^2+12^2$$

$$AC^2=169$$

$$AC=13\ m$$

The value of a flat worth ₹500,000 is depreciating at the rate of 10% per annum. Thus, at the end of the first year, the value of the flat will be = 500000(1 - 0.1) = 500000(0.9).

Thus, the value of flat at the end of the second year will be = 500000(0.9)(1 - 0.1) = 500000(0.9)(0.9)

Therefore, the value of the flat at the end of n years will be $$=500000\left(0.9\right)^n$$

This value is given as 364500.

$$500000\left(0.9\right)^n=364500$$

$$0.9^n=0.729$$

$$n=3$$ years

30 litres of a mixture of milk and water contains milk and water in the ratio 7:3. Therefore, the amount of milk = 21 litres and the amount of water = 9 litres in the solution.

Let us assume we added 'x' litres of water. Now, we want the resultant solution to have 40% water in it. 

$$9+x=\dfrac{40}{100}\times\left(30+x\right)$$

$$45+5x=60+2x$$

$$3x=15$$

$$x=5$$

Thus, 5 litres of water must be added. 

Statement - I
$$392=2^3\times7^2$$
Thus, the sum of exponents of prime factors of 392 is (3 + 2) = 5
This statement is TRUE. 

Statement - II
$$\dfrac{13}{2^3\times5}=\dfrac{13}{40}=0.325$$
This  will terminate after 3 decimal places.
This statement is FALSE.

$$P(A)=p$$
$$P(B)=2p$$

P(Exactly one of A or B) = $$\dfrac{5}{9}$$. This means that either A happens and B does not happen, or B happens and A does not happen. Therefore - 

$$P\left(\overline{A}\right)P\left(B\right)+P\left(\overline{B}\right)P\left(A\right)=\dfrac{5}{9}$$

$$(1-p)(2p)+(1-2p)(p)=\dfrac{5}{9}$$

$$3p-4p^2=\dfrac{5}{9}$$

$$36p^2-27p+5=0$$

$$36p^2-12p-15p+5=0$$

$$12p\left(3p-1\right)-5\left(3p-1\right)=0$$

$$\left(12p-5\right)\left(3p-1\right)=0$$

$$p=\dfrac{5}{12}\ or\ p=\dfrac{1}{3}$$

Thus, both statements are true. 

Let the speed of boat in still water be B, and the speed of stream be R.

A motor boat can travel 30 km upstream and 28 km downstream in 7 hours.

$$\dfrac{30}{B-R}+\dfrac{28}{B+R}=7$$

It can travel 21 km upstream and return in 5 hours.

$$\dfrac{21}{B-R}+\dfrac{21}{B+R}=5$$

Let $$\dfrac{1}{B-R}=x$$ and $$\dfrac{1}{B+R}=y$$

$$30x+28y=7\rightarrow1$$
$$21x+21y=5\rightarrow2$$

Multiply eq. 1 by 3 and eq. 2 by 4, and then subtract -

$$90x-84x=21-20$$

$$x=\dfrac{1}{6}$$

Therefore, $$y=\dfrac{1}{14}$$

Thus, $$B+R=14$$ and $$B-R=6$$

$$B=10$$ kmph and $$R=4$$ kmph

Therefore, speed of boat is 10 km/h and speed of stream is 4 km/h. 

Length of the floor = 15m 17cm = 1517 cm

Width of the floor = 9m 2cm = 902 cm

The overall area of the floor $$=1517\times902$$

To minimise the number of tiles needed to pave the floor, we should calculate the HCF of both the length and width of the floor to determine the side length of each square tile. 

$$1517=41\times37$$

$$902=2\times11\times41$$

Thus, HCF(1517,902) = 41

The side length of each square tile is 41 cm.

Number of tiles = Area of Floor / Area of each tile

Number of tiles = $$\dfrac{1517\times902}{41\times41}=37\times22=814$$

Let the speed of B be v km/h. The speed of A is 4km/hr slower than that of B. Thus, the speed of A will be (v - 4) km/h. 

B, when it reaches Ambala, starts back via the same route without taking any rest and meets A, who was still 12km away from Ambala. The distance between Chandigarh and Ambala is 60 km. 

When A and B are meeting, A is 12 km away from Ambala, which means that he has covered 48 km. At that point in time, B has covered [60 + (60 - 48)] = 72 km. 

So we can say that in time T (let's assume), A has covered 48 km and B has covered 72 km. 

$$(v-4)\timesT=48$$
$$v\timesT=72$$

$$\dfrac{v-4}{v}=\dfrac{48}{72}$$

$$3v-12=2v$$

$$v=12$$

Speed of A = (v - 4) = (12 - 4) = 8 km/h

Let the other number be A, the LCM be L, and the HCF be H.

The LCM of two numbers is 12 times their HCF. Thus, $$L=12H$$

The sum of LCM and HCF is 402. Thus - 

$$L+H=403$$
$$13H=403$$
$$H=31$$
Therefore, $$L=12H=12\times31=372$$

We know that the product of two numbers is the product of their LCM and the HCF. 

$$93\times A=L\times H$$

$$93\times A=372\times31$$

$$A=124$$

Thus, the other number is 124. 

Let the son's age be X. Then, the father's age will be 6X.

Four years hence, the age of the father will be four times his sonʼs age.

6x + 4 = 4(X + 4) 

6X + 4 = 4X + 16

2X = 12

X = 6

Thus, the present age of son is 6 years, and the present agae of father is 36 years.