IPMAT Rohtak 2019 Question Paper

$$B = (2^{64} + 2^{63} + 2^{62} + ... + 2^{0})$$ is a GP with first term $$2^{64}$$ and common ratio 2 and the number of terms 65. 

So its sum can we written as $$B=(2^{64}+2^{63}+2^{62}+...+2^0=\ \frac{2^{65}-1}{2-1}=2^{65}-1=A-1$$

So we get B= A -1 , which means A is larger than B by 1. So option D is the correct choice. 

A line from the centre to the circumference of a circle is called a radius.

A line passing from one point on the circumference to another point on the circumference through the centre is called the diameter.

We are given 4 blue balls 5 white balls and 6 green balls

Total balls = 4 + 5 + 6 = 15 balls

We are drawing 2 balls at random, and we need to find the probability that exactly one of them is white.

So, total ways  $$ = \binom{15}{2} = \frac{15 \times 14}{2} = 105 $$

We want one white ball and one non-white ball (i.e., either blue or green).

Number of ways to choose 1 white ball = $$ \binom{5}{1} = 5$$

Number of ways to choose 1 non-white ball = $$ \binom{10}{1} = 10 $$ (since 4 blue + 6 green = 10)

Favorable outcomes= $$ 5 \times 10 = 50 $$

So, Probability $$= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{50}{105} = \frac{10}{21} $$

Let, the volume of the tank be 1500 liters

So, efficiency of pipe P = $$\dfrac{1500}{20}=75$$ liters/ hour.

Efficiency of pipe Q = $$\dfrac{1500}{25}=60$$ liters/ hour.

and, efficiency of pipe R = $$-\dfrac{1500}{30}=-50$$ liters/ hour.

(Since, pipe R is emptying)

So, when all the pipes work together, volume of the tank filled by them = $$\left(75+60-50\right)\times\ 10$$ liters = $$850$$ liters

Now, volume of the tank, that is required to be filled = $$1500-850\ =\ 650$$ liters

And, this has to be filled by pipe P and Q

So, time required to fill the remaining part = $$\dfrac{650}{75+60}=\dfrac{650}{135}=\dfrac{130}{27}$$ hours

So, total time = $$\dfrac{130}{27}+10=\dfrac{400}{27}$$ hours

$$\dfrac{a}{b}=\dfrac{2}{3}$$  ----(1)

$$\dfrac{b}{c}=\dfrac{5}{2}$$   ---(2)

$$\dfrac{c}{d}=\dfrac{1}{4}$$ -------(3)

Multiplying (1) and (2), we get $$\dfrac{a}{b}\times\dfrac{b}{c}=\dfrac{2}{3}\times\dfrac{5}{2}$$

$$\dfrac{a}{c}=\dfrac{5}{3}$$

If we assume a = 10, then c = 6

Similarly, from $$\dfrac{a}{b}=\dfrac{2}{3}$$

We will get b = 15

Therefore, a:b:c = 10:15:6

It is given that there are Rs 495 in a bag.

One-rupee, 50 paisa and 25 paisa coins are in the ratio 1:8:16.

Let us assume the one-rupee coins are N,

50 paisa coins will be 8N and 25 paisa coins will be 16N.

Let us convert all amounts to paisa

One-rupee amount = 1*100*N = 100N Paisa

50 Paisa amount = 50*8N = 400N Paisa

25 Paisa amount = 25*16N = 400N Paisa

Total amount in Paisa = 49500 (Rs.495 = 49500 Paisa)

100N+400N+400N = 49500

900N = 49500 paisa

N = 55

Therefore, 50 paisa coins are 8N = 8*55 = 440 

1 hour = 60 minutes

1 minute = 60 seconds

Thus, 1 hour = 60*60 = 3600 seconds

Therefore, 1 second = $$\dfrac{1}{3600}$$ hour

1 second = .00027 hour

$$0.0056$$ = $$\dfrac{56}{10000}$$

=> $$\dfrac{28}{5000}=\dfrac{14}{2500}=\dfrac{7}{1250}$$

Let us assume the first number is A and the second number is B.

One-fifth of a number is equal to $$\dfrac{5}{8}th$$ of another number.

$$\dfrac{A}{5}=\dfrac{5B}{8}$$

=> $$A=\dfrac{25B}{8}$$   -----(1)

Also, if 35 is added to the first number, it becomes four times the second number.

$$A+35=4B$$   .......(2)

Substituting the value of (1) in (2)

$$\dfrac{25B}{8}+35=4B$$

$$\dfrac{7B}{8}=35$$

B = 40

The given equation is 43 * 561 + 500 - 100 / 10 = ?

We need to change the symbols according to If * = +, / = -, + = *, - = /

43 + 561* 500 / 100 - 10 = ?

Now applying the BODMAS rule, we get the equation as

$$43+561\times\left(\dfrac{500}{100}\right)-10$$

=> $$43+\left(561\times5\right)-10$$

=> $$43+2805-10\ =\ 2838$$