JIPMAT 2021 Question Paper

It is given conical base area = 346.5 
So, $$\pi\left(r\right)^2\ $$ = 346.5
This gives radius = 10.5 

Slant length of cone = $$\sqrt{h^2+r^2\ }$$
                               = $$\sqrt{14^2+10.5^2\ }\ =\ 17.5$$ 

Surface area of cone = $$\pi\ rl$$ = (10.5)(17.5)(3.14) = 576.97

Therefore, the length of canvas = 576.97/1.1 = 524.52

Team has 11 players with average of 28 years age : 
$$X_1+X_2+......X_{11}$$ = 28(11)

Group I : $$X_1+X_2+X_3$$ = 25(3)
Group II : $$X_4+X_5+X_6$$ = 28(3)
Group III : $$X_7+X_8+X_9$$ = 3(30)

Let the non included Youngest players age be $$X_{10}$$ & Captain age be $$X_{11}$$.
Hence, it is given that , $$X_{11}=11+X_{10}$$.

Hence, 75 + 90 + 84 + $$X_{10}$$ + 11 +  $$X_{10}$$ = 28(11)
it implies, $$X_{10}=24$$

Therefore, the Captain age ($$X_{11}$$) = 11+$$X_{10}$$. = 11 + 24 = 35

Angle between hands of a clock at H:M  = $$|\dfrac{\ 11}{2}M-30H|$$ degrees.

So, at 5:15 pm ,the angle = $$|\frac{\ 11}{2}\left(15\right)-30\left(5\right)|\ =\left|-67.5\right|\ =\ 67.5\ $$

Let the Rent be 'K' per annum . Now it is said that man purchases at single go the property at the value of 40(K).

So, the investment = 40(K) 
Rate of returns = $$\ \dfrac{\ Annual Returns\ }{Investment}\times100$$
                       = $$\dfrac{\ K}{40K}\times100$$
                       =  2.5 % 

Rhombus has all sides equal and their diagonals bisect each other perpendicularly : 
So, Length of each side of rhombus = $$\sqrt{\dfrac{\ d_1^2+d_2^2}{4}}$$
                                                     = $$\sqrt{\dfrac{\ 20^2+15^2}{4}}$$
                                                     = 25.

Therefore, perimeter of rhombus = 4(l) = 4x25 = 100

HCF of $$\left(\frac{9}{10},\frac{12}{25},\frac{18}{35},\frac{21}{40} \right)$$ = $$\dfrac{\ HCF\left(9,12,18,21\right)}{LCM\left(10,25,35,40\right)}$$. 

Hence, the value = $$\dfrac{3\ }{1400}$$

The first term = 15
Second = (15x1) + 1 = 16
Third = (16x2) + 2 = 34
Fourth = (34x3) + 3 = 105
Fifth = (105x4) + 4 = 424
Sixth = (424x5) + 5 = 2125
Seventh = (2125x6) + 6 = 12756
 
Hence, the order of sequence = {15, 16, 34, 105, 424, 2125, 12756}

Therefore the odd term is 2124.

If the total length of pole = l
The initial part which is in mud = $$\dfrac{\ 4}{7}\left(l\right)$$

As it is said that $$\dfrac{\ 1}{3}$$rd part is pulled out of mud, hence the remaining part which lies inside mud = $$(l)\left(\dfrac{\ 4}{7}-\ \dfrac{\ 1}{3}\right)\ =\ \dfrac{\ 5l}{21}$$

Therefore, $$\dfrac{\ 5l}{21}$$ = 250 cm

Therefore, the value of L = 1050 cm

Given information : 
Capital of A = 12,000
Capital of B = 20,000
                                Hence, the ratio of investments for A is to B = 3:5

Total profit given as = 9600

As "A" is managing the business , he is being paid = (0.1)(9600) = 960
So, the remaining profit = 9600-960 = 8640

This remaining has to be shared in proportion of investment : A's share = $$\dfrac{\ 3}{8}\left(8640\right)$$ = 3240
                                                                                             B's share = $$\dfrac{\ 5}{8}\left(8640\right)$$ = 5400

Therefore, total earning of A = 3240 + 960 = 4200

Let the filling capacity of pump be "X" $$m^3$$ .
It is given emptying capacity of pump = X + 10 
Total volume of the tank = 2400 

Time taken to fill the tank = $$\dfrac{2400\ }{X}$$

Time taken to empty the tank = $$\dfrac{\ 2400}{X\ +10}$$

Now, it is given that : $$\dfrac{\ 2400}{X\ +10}$$ = $$\dfrac{2400\ }{X}$$ - 8
 
So , this gives $$\dfrac{\ 2400(10)}{X\left(X+10\right)}=8$$

therefore, 3000 = X(X+10)

Hence, X = 50 or X = -60.
But X cannot be negative as it is filling capacity which is +ve.

Therefore, answer will be 50 $$m^3$$