JIPMAT 2021 Question Paper

Let the length of first train be $$Y$$ metres & length of bridge be "X" metres.
So, now the length of second train will be = (Y-100) metres.

First train speed = $$\dfrac{\ 450}{18}$$ = 25 m/s

Second train speed = $$\dfrac{\ 225}{18}$$ = 12.5 m/s

Time taken by first train = $$\dfrac{\ Y\ +X}{25}=36$$

It implies, (Y+X) = 25(36).

Time taken by Second train = $$\dfrac{\ Y\ +X-100}{12.5}$$ = $$\dfrac{25\left(36\right)-100}{12.5}$$ = 64 seconds

Let the no of passengers at start be "N" .

At first station it drops 1/3rd and takes 280 more : 
                   Now the passengers = $$N(1-\dfrac{\ 1}{3})+280$$ = 2N/3 + 280.

At second station, it drops 1/2 part of new total and takes 12 more : 
                  Now the passengers count after 2nd station will be = N/3 + 140 + 12 = N/3 + 152.

It is given that this value {N/3 + 152} = 248.
So, N = 96(3) = 288

In 100 gm of mixture , alcohol= 35 gm & water = 65 gm.

So, if 25 gm water is added to this mixture : Alcohol = 35 gm
                                                                   Water = 65 + 25 = 90 gm.

Therefore, the percentage of alcohol in new mixture = $$\dfrac{\ 35}{125}\times100\ \%$$ = 28 %

Hollow cylinder Pipe Volume = $$\pi\ \left(R^2-r^2\right)H$$ 

In this case, H = 4 m, R = 6 cm, r = 5 cm.
                   H = 4 m, R = 0.06 m, r = 0.05 m.

So , the volume = $$\dfrac{\ 22}{7}\left(4\right)\left(\ \frac{\ 36-25}{10^4}\right)$$ = $$0.0138\ m^3$$

As one cubic metre of iron weighs 7.8 kg : 
   Total weight of iron of volume $$0.0138\ m^3$$ = 7.8(0.0138) kg = 0.1077 kg

As there is no answer, NTA has dropped this question !!

$$(7^{95}-7^{58})$$ = N

The unit digit of {$$7^1,7^2,7^3,7^4$$} = {7, 9, 3, 1} . And this cycle gets repeated for every 4 powers ..

So if the exponent is in the form of :
1. form of {4k} : unit digit is 1
2. form of {4k + 1} : unit digit is 7
3. form of {4k + 2} : unit digit is 9
4. form of {4k + 3} : unit digit is 3

Now, 95 = 4k + 3 , so the unit digit would be 3 .
         58 = 4k + 2 , so the unit digit would be 9.

Therefore, for this $$(7^{95}-7^{58})$$ = {Number with unit digit 3 - Number with unit Digit 9} 
                                                           = {Number with unit digit of (13-9)}
                                                           = 4.

The $$m^{th}$$ term of an AP : [a + (m-1)(d)] = $$\dfrac{1}{n}$$

The $$n^{th}$$ term of an AP : [a + (n-1)(d)] = $$\dfrac{1}{m}$$

Subtracting the two equations :$$(d)(m-n)=\dfrac{(m-n)\ }{mn}$$ 
                                                 $$d=\dfrac{\ 1}{mn}$$

Substituting "d" value in any one of the equation, we get : $$a=\dfrac{\ 1}{mn}$$

So, the $$mn^{th}$$ term of AP : [a + (mn-1)(d)] = $$\dfrac{\ 1}{mn}+\dfrac{\ \left(mn-1\right)}{mn}$$ = 1

Let the invested money be "P" at a rate of R % in compound interest :

After two years the value becomes =$$P(1+R/100)^2$$ = 1.44 P
                                   1 + R/100 = 1.2 
                                    R = 20 %.

If same amount of P is invested in simple interest, then value of money after 't' years = $$P\left(1+\frac{\ R\left(t\right)}{100}\right)$$ = 2(P)
                                                                 1 + R(t)/100 = 2 
                                                                  R(t) = 100
Therefore, t = 100/R = 100/20 = 5 years.

As BD is a tangent , we can say OD is perpendicular to BE i.e <ODB = 90

$$Sin\ \theta\ =\dfrac{\ OD}{OB}=\dfrac{\ 8}{13}$$

<BOD = $$90-\theta\ $$ 

Since, <BOD + <DOA = 180 degrees
this can imply us that <DOA = 180 - ($$90-\theta\ $$)
                                          = $$90+\theta\ $$

In triangle DOA, by applying cosine rule formula : 
                 cos ($$90+\theta\ $$) = $$\ \dfrac{\ OA^2+OD^2-AD^2}{2\left(OA\right)\left(OD\right)}$$

                 $$-\sin\theta\ $$ = $$\dfrac{\ 13^2+8^2-AD^2}{2\left(13\right)\left(8\right)}$$

                    $$\dfrac{\ -8}{13}$$ = $$\dfrac{\ 13^2+8^2-AD^2}{2\left(13\right)\left(8\right)}$$

Therefore, AD = 19.

Given that BC = a, AB = c, AC = b, CD = p
CD is perpendicular to AB & <ACB = 90 degrees.

Let <DBC = <ABC = $$\theta\ $$

$$Sin\ \theta\ =\dfrac{\ CD}{CB}=\dfrac{\ AC}{AB}$$

                     = $$\dfrac{\ p}{a}=\dfrac{\ b}{c}$$

Therefore, pc = ab.

$$(2\sin^{2}\theta+3\cos^{2}\theta)$$
= $$2\left[\sin^2\left(\theta\ \right)+\cos^2\left(\theta\ \right)\right]+\cos^2\left(\theta\ \right)$$
= $$2+\cos^2\theta\ $$

The min value occurs at $$\cos\theta\ =0$$ .
Therefore, the minimum value = 2 .