IPMAT Rohtak 2019 Question Paper

Let us assume the speeds of trains A and B to be a km/hr and b km/hr, respectively.

The total distance that has to be travelled = 450 km

It is given that Train A takes 45 minutes more than Train B to travel 450 km.

Time taken by train A to travel 450 km = $$\dfrac{450}{a}$$

Time taken by train B to travel 450 km = $$\dfrac{450}{b}$$

We know that Train A takes 45 minutes more than Train B

$$\dfrac{450}{a}=\dfrac{45}{60}+\dfrac{450}{b}$$  ------equation (1)  (We have divided 45 by 60 to convert in hour)

Also, due to engine trouble, the speed of train B falls by a quarter. So it takes 30 minutes more than Train A to complete the same journey.

The time taken by A to complete the journey will remain the same.

Time taken by B to complete the journey will become = $$\dfrac{450}{\frac{3b}{4}}$$ (As it's speed falls by a quarter, new speed = 3b/4)

Thus, $$\dfrac{450}{\frac{3b}{4}}=\dfrac{450}{a}+\dfrac{30}{60}$$   -----equation (2)

=> Substituting the value of $$\dfrac{450}{a}$$ from equation (1) into (2)

=> $$\dfrac{450}{\frac{3b}{4}}=\dfrac{45}{60}+\dfrac{450}{b}+\dfrac{30}{60}$$

=>$$\dfrac{600}{b}=\dfrac{75}{60}+\dfrac{450}{b}$$

=> $$\dfrac{150}{b}=\dfrac{75}{60}$$

=> b = 120 km/hr

=> $$\dfrac{450}{a}=\dfrac{45}{60}+\dfrac{450}{120}$$

=> $$\dfrac{450}{a}=\dfrac{3}{4}+\dfrac{15}{4}=\dfrac{18}{4}$$

Thus, a = 100 km/hr

One Trader calculates the percentage of profit on the buying price and another calculates on the selling price.

Let the Selling Price (same for both) = X

For First Trader, we will calculate it on CP.

If CP = C

20% profit means

X = C + 0.2C = 1.2C

Therefore, C = $$\dfrac{X}{1.2}$$

Thus, the actual profit for the first trader

Actual profit = $$X-\dfrac{X}{1.2}$$ = $$\dfrac{X}{6}$$

Similarly, we calculate for the second trader, but on SP.

20% profit on SP means:

Profit = 0.2X

If CP = K

X−K=0.2X 

K = 0.8X

$$0.2X-\dfrac{X}{6}$$ = $$\dfrac{0.2X}{6}=85$$

0.2X = 85*6 = 510

Therefore, X = 510*5 = 2550

Thus, the selling price is Rs 2550.

Let the cost price of one egg be Rs. A

Therefore, in Rs. 24, the number of eggs that can be purchased = $$\dfrac{24}{A}$$

Due to an increase of 20% in the price of eggs, the new price of one egg = Rs. 1.2 A

Now, with this amount the number of eggs that can be purchased = $$\dfrac{24}{1.2A}$$

It is given that their difference is 2.

Thus, $$\dfrac{24}{A}$$ when subtracted from $$\dfrac{24}{1.2A}$$, we will get 2

$$\dfrac{24}{A}-\dfrac{24}{1.2A}=2$$

=> $$\dfrac{24}{A}-\dfrac{20}{A}=2$$

=> $$\dfrac{4}{A}=2$$

Thus, A = Rs 2 per egg

We need to find the present rate of eggs per dozen is

Present rate of one egg = 2*1.2 = 2.4

Therefore, the cost of 1 dozen eggs = 12*2.4 = Rs.28.8/dozen

$$\sin\left(\dfrac{13\pi}{6}\right)\ =?$$

We know the property that $$\sin\left(2\pi+\theta\ \right)=\sin\theta\ $$

Similarly, $$\sin\left(\dfrac{13\pi}{6}\right)=\sin\left(2\pi+\dfrac{\pi}{6}\right)=\sin\left(\dfrac{\pi}{6}\right)$$

Which is equal to 1/2 or 0.5.

Option A is the correct answer.

It is given 9 boys ate $$\dfrac{3}{5}th$$ of them in 2 hours.

or, 9 boys ate $$\dfrac{3}{10}th$$ of them in 1 hour.

or, 1 boy ate $$\dfrac{3}{10\times\ 9}=\dfrac{1}{30}th$$ of them in 1 hour.

Now, $$\dfrac{3}{5}th$$ of fruits have been eaten

So, amount of fruits remaining are $$\dfrac{2}{5}th$$.

Also, number of persons that will eat the remaining fruits = $$9-6=3$$

So, required time to eat = $$\dfrac{\dfrac{2}{5}}{\dfrac{1}{30}\times\ 3}=\dfrac{\dfrac{2}{5}}{\dfrac{1}{10}}=\dfrac{2}{5}\times\ 10=4$$ hours

The minimum/maximum value of quadratic $$ax^2+bx+c=0$$ occurs at point $$x=-\frac{b}{2a}$$

So, for the quadratic, $$f(x) = x^{2} + 2bx + 2c^{2}$$, minimum value will occur at $$x=-\dfrac{2b}{2}=-b$$

Similarly, for the quadratic, $$g(x) = - x^{2} - 2cx + b^{2}$$, minimum value will occur at $$x=-\dfrac{-2c}{-2}=-c$$

Now, according to question,

$$\left(-b\right)^2+2b\left(-b\right)+2c^2>-\left(-c\right)^2-2c\left(-c\right)+b^2$$

or, $$b^2-2b^2+2c^2>-c^2+2c^2+b^2$$

or, $$-2b^2>-c^2$$

or, $$c^2>2b^2$$

or, $$\left|c\right|>\sqrt{\ 2}\left|b\right|$$ 

Case-1 $$(x>-2)$$:

So, $$x+2>0$$ and so $$\left|x+2\right|=x+2$$

or, $$x^2-\left(x+2\right)+x>0$$

or, $$x^2-x-2+x>0$$

or, $$x^2-2>0$$

or, $$x>\sqrt{\ 2}$$ and $$x<-\sqrt{\ 2}$$

But here, $$x>-2$$

so, $$-2<x<-\sqrt{\ 2}$$ and $$x>\sqrt{\ 2}$$

Case-2 $$(x<-2)$$:

So, $$x<-2$$ and $$x+2<0$$

or, $$\left|x+2\right|=-\left(x+2\right)$$

or, $$x^2-\left(-\left(x+2\right)\right)+x>0$$

or, $$x^2+x+2+x>0$$

or, $$x^2+2x+2>0$$

or, $$\left(x+1\right)^2+1>0$$ and this is always true that $$\left(x+1\right)^2+1>0$$ for any value of $$x$$

So, for $$x<-2$$, the relation always holds true

So, combining case-1 and case-2, $$x$$ belongs to $$(-\infty, -\surd{2}) \cup (\surd{2}, \infty)$$

46080, 3840, 384, 48, 8, 2, ?

If we divide the consecutive terms.

$$\dfrac{46080}{3840}=12$$

$$\dfrac{3840}{384}=10$$

$$\dfrac{384}{48}=8$$

$$\dfrac{48}{8}=6$$

$$\dfrac{8}{2}=4$$

A series is formed 12,10,8,6,4,... so next term should be 2

Therefore, it should be $$\dfrac{2}{x}=2$$

Thus, x = 1

The floor is 15x6 , meaning the length is 15 and the breadth is 6. So the area of the base (ie floor) is 15x6=90 sq cm.

Assuming that the room is a cuboid, we can say that the roof will also be 90 sq cm.

If h is the height of the room then , area of the 4 walls will be hx15, hx6, hx15 and hx6 respectively. (imagine a cuboid which has opposite faces as equal.

SO the total area of the walls is 42h , which is equal to (90+90)=180.

So h=180/42 =4.29 cm

The sum of third and ninth term of an A.P is 8. Find the sum of the first 11 terms of the progression.

Let the first term of AP = a

Common difference = d

Third term  = a+2d

Ninth term = a+8d

Third term + Ninth term  = 8

2a+10d = 8

We know sum of an AP = $$\dfrac{n}{2}\left[2a+\left(n-1\right)d\right]$$

The sum of the first 11 terms = $$\dfrac{11}{2}\left[2a+10d\right]$$

We know that 2a+10d = 8

$$\dfrac{11\times8}{2}=44$$