IPMAT Rohtak 2019 Question Paper

Based on the question, the following figure is obtained. The shaded area denotes the area that we need to find. 

Here, area of the equilateral triangle is $$A=\frac{\sqrt{\ 3}}{4}\left(side\right)^2=\frac{\sqrt{\ 3}}{4}\times\ 24^2=144\sqrt{\ 3}sq\ units$$

The inradius of the circle will be given by $$r=\frac{A}{s}$$ where A is the area of the triangle and s is the semi perimeter which is found to be $$\frac{24+24+24}{2}=36$$

So, $$r=\frac{144\sqrt{\ 3}}{36}=4\sqrt{\ 3}units$$

Hence, area of the circle is $$\pi\ \times\ \left(4\sqrt{\ 3}\right)^2=48\pi\ \ sq\ units$$

Finally, the area of the remaining area will be teh difference, which is $$144\surd{3} - 48 \pi cm^{2}$$ sq units. Hence , option A is the correct choice. 

A clock strikes 4, taking 9 seconds.

There are three intervals between the four strikes.

Time taken = 9 seconds

Therefore, the rate of interval taken = 9/3 = 3 seconds per interval.

To strike for 12, there will be 11 intervals.

Time per interval = 3 seconds

Total time taken = 11*3 = 33 seconds.

Let, John's present age be "x" years.

So, age of father two years ago = 4x

So, current age of father = (4x+2) years

It is given, Raman's 12th birthday was celebrated 2 years ago, so Raman's current age = 12+2 = 14 years

So, Raman's age after 10 years = 14+10 = 24 years

So, John's father's age after 10 years = 2*24 = 48 years

So, present age of John's father = (48-10) years = 38 years

So, 4x+2=38

or, 4x=36

or, x=9

So, John's present age is 9 years.

Raj invested Rs 76000 in a business.

After a few months, Monty joined him and invested Rs 57000.

At the end of the year, both of them share the profits at a ratio of 2:1

We know that the ratio of profits will equal the ratio of investment.

We know Raj invested for 12 months; let us assume Monty invested for N months.

Thus, $$\dfrac{76000\times12}{57000\times N}=\dfrac{2}{1}$$

=> $$\dfrac{4\times12}{3\times N}=\dfrac{2}{1}$$

Thus, N = 8

Therefore, Monty invested for 8 months.

So, Monty invested after 4 months of Raju's investment.

It is given that work done by Q in one day is thrice the work done by R in one day.

Let us assume R completes K units of work per day.

Thus, Q will complete 3K units of work per day.

It is also given that the work done by P in one day is double the work done by Q in one day.

Therefore, P will complete 6K units of work per day.

Together, P, Q and R complete the total work in one day = 6K+3K+K = 10K units per day

They complete the work in 30 days.

So, total work done by them in 30 days = 30*10K = 300K units

If P works alone, he will complete 6K units daily.

So, the number of days taken by P alone to complete the work = 300K/6K = 50 days.

Let us assume that the speed of boat = B and speed of stream = S

Now, it is given that A boatman goes 2 km against the stream's current in 1 hour.

Net speed of boat against the stream (upstream) = B-S

(B-S)*1 = 2 

B-S = 2   -----equation (1)

Now, it is also given that the boat goes 1 km along the current in 10 minutes.

The net speed of the boat in the direction of the stream (downstream) = B+S

(B+S)*1/6 = 1 (As 10 minutes = 1/6 hour)

B+S = 6   ------equation (2)

Adding both, we get 2B = 8 or B = 4 km/hr

Thus, S = 2 km/hr

We need to find a boat's time to go 5 km in stationary water.

Distance/Speed = Time

5 km/4 km/hr = 1 hour 15 minutes

$$25^{(2.7)}\times 5^{(4.2)} \div 5^{(5.4)} = ?$$

$$25^{2.7}=\left(5\right)^{2\times2.7}=5^{5.4}$$

So, $$5^{(5.4)}\times5^{(4.2)}\div5^{(5.4)}=5^{\left(4.2\right)}$$

As $$5^{(5.4)}$$ will get cancelled.

18800 / 470 / 20

=> $$\dfrac{18800}{470}=40$$

=> $$\dfrac{40}{20}=2$$

The HCF of $$\dfrac{2}{3}, \dfrac{4}{6}, \dfrac{8}{27}$$

$$\dfrac{2}{3}, \dfrac{2}{3}, \dfrac{8}{27}$$

HCF of fractions = HCF of Numerator/ LCM of Denominator

HCF of $$\dfrac{2}{3}, \dfrac{2}{3}, \dfrac{8}{27}$$ = HCF of (2,2,8) / LCM of (3,3,27) = $$\dfrac{2}{27}$$

If $$\log 2, \log(2^{x}-1)$$ and $$\log(2^{x}+3)$$ are in A.P

We know when A, B and C are in AP.

2B = A+C

Thus, $$2\log(2^x-1)\ =\log2+\log\left(2^x+3\right)$$

Assuming $$2^x=p$$

$$2\log(p-1)\ =\log2+\log\left(p+3\right)$$

Now, applying the properties of logarithms

$$\log_{a}{xy} = \log_{a}{x}+\log_{a}{y}$$

$$(p-1)^2\ =2\left(p+3\right)$$

$$\ p^2-2p+1=2p+6$$

$$\ p^2-4p-5=0$$

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

Thus, p = 5 or -1

$$2^x$$ = 5 as it cannot be negative.

Thus, x = $$\log_25$$