CAT 2018 Question Paper Slot 2

It is given that there were a total of 3 rooms and seven candidates. Ganeshan said that he was one among the two candidates allotted to Room 102 whereas Erina said that she is the only person in the room she was allotted to. Therefore, we can say that there were 1 and 4 candidates in either 101 or 103 room. But it is given that Balaram was the third person to enter in the Room 101 therefore we can say that there were 4 candidates in Room 101 and only 1 candidate in room 103. 

Fatima said that three people including Akhil were already in the room that I was allotted to when I entered it. Hence, we can say that Fatima was the last person to enter in 101 and Akhil is the first person who entered in room 101. 

Chitra said that she was the last person to enter the room she was allotted to. Hence, we can say that Chitra was allotted room no 102 and she entered after Ganeshan.  

Erina was the only person in room no 103. 

Balaram said he was third to enter room no 101. Hence, we can say that Divya was second person who entered in room 101. 

Since Chitra and Fatima were already in by 7:40 AM we can say that the candidate who entered at 7:45 am is Erina. 

From the table we can see that Erina reached the venue at 7:45 am. Hence, option A is the correct answer. 

It is given that there were a total of 3 rooms and seven candidates. Ganeshan said that he was one among the two candidates allotted to Room 102 whereas Erina said that she is the only person in the room she was allotted to. Therefore, we can say that there were 1 and 4 candidates in either 101 or 103 room. But it is given that Balaram was the third person to enter in the Room 101 therefore we can say that there were 4 candidates in Room 101 and only 1 candidate in room 103. 

Fatima said that three people including Akhil were already in the room that I was allotted to when I entered it. Hence, we can say that Fatima was the last person to enter in 101 and Akhil is the first person who entered in room 101. 

Chitra said that she was the last person to enter the room she was allotted to. Hence, we can say that Chitra was allotted room no 102 and she entered after Ganeshan.  

Erina was the only person in room no 103. 

Balaram said he was third to enter room no 101. Hence, we can say that Divya was second person who entered in room 101. 

Since Chitra and Fatima were already in by 7:40 AM we can say that the candidate who entered at 7:45 am is Erina. 

In the question it is given that Ganeshan entered the venue before Divya. Therefore, we can say that Ganesh must have entered with Akhil at 7:10 am. In that case, Divya and Balaram must have entered at 7:15 am and 7:25 am respectively. Hence, option A is the correct answer. 

Let '100x' be the number of smartphones sold in year 2016. 

Total revenue generated by Azra = 40x*15000 = Rs. 600000x

Total revenue generated by Bysi = 25x*20000 = Rs. 500000x

Total revenue generated by Cxqi = 15x*30000 = Rs. 450000x

Total revenue generated by Dipq = 20x*25000 = Rs. 500000x

We can see that revenue generated by Azra is the highest among all four brands. Hence, option C is the correct answer. 

Let '100x' be the number of smartphones sold in year 2016. 

Total revenue generated by Azra = 40x*15000 = Rs. 600000x

Profitability is defined as the profit as a percentage of the revenue. Therefore, profit generated by Azra = $$\dfrac{10}{100}$$*600000x = Rs. 60000x

Total revenue generated by Bysi = 25x*20000 = Rs. 500000x

Profit generated by Bysi  = $$\dfrac{30}{100}$$*500000x = Rs. 150000x

Total revenue generated by Cxqi = 15x*30000 = Rs. 450000x

Profit generated by Cxqi  = $$\dfrac{40}{100}$$*450000x = Rs. 180000x

Total revenue generated by Dipq = 20x*25000 = Rs. 500000x

Profit generated by Dipq  = $$\dfrac{30}{100}$$*500000x = Rs. 150000x

We can see that profit generated by Cxqi is the highest among all four brands. Hence, option C is the correct answer. 

Let '100x' be the number of smartphones sold in year 2016. Then the number of smartphones sold in 2017 = 1.4*100x = 140x

It is given that Cxqi offered a 40% discount on its unit selling price in 2017 i.e. selling price in 2017 = 0.6*30000 = Rs. 18000

Also Cxqi's merket share increased by 15% whereas the other three brands lost 5% market share. 

Amount of profit generated by Azra = $$\dfrac{10}{100}$$*15000*49x = 73500x

Amount of profit generated by Bysi = $$\dfrac{30}{100}$$*20000*28x = 168000x

Amount of profit generated by Cxqi = $$\dfrac{20}{100}$$*18000*42x = 151200x

Amount of profit generated by Dipq = $$\dfrac{30}{100}$$*25000*21x = 157500x

We can see that brand Bysi generated maximum profit in year 2017. Hence, option A is the correct answer.

Let '100x' be the number of smartphones sold in year 2016. 

Total revenue generated by Azra = 40x*15000 = Rs. 600000x

Profitability is defined as the profit as a percentage of the revenue. Therefore, profit generated by Azra = $$\dfrac{10}{100}$$*600000x = Rs. 60000x

Total revenue generated by Bysi = 25x*20000 = Rs. 500000x

Profit generated by Bysi  = $$\dfrac{30}{100}$$*500000x = Rs. 150000x

Total revenue generated by Cxqi = 15x*30000 = Rs. 450000x

Profit generated by Cxqi  = $$\dfrac{40}{100}$$*450000x = Rs. 180000x

Total revenue generated by Dipq = 20x*25000 = Rs. 500000x

Profit generated by Dipq  = $$\dfrac{30}{100}$$*500000x = Rs. 150000x

It is given that the market sales increased by 40%. Therefore, the number of smartphones sold in 2017 = 1.4*100x = 140x

It is given that Cxqi offered a 40% discount on its unit selling price in 2017 i.e. selling price in 2017 = 0.6*30000 = Rs. 18000

Also Cxqi's merket share increased by 15% whereas the other three brands lost 5% market share. 

Amount of profit generated by Azra = $$\dfrac{10}{100}$$*15000*49x = 73500x

Amount of profit generated by Bysi = $$\dfrac{30}{100}$$*20000*28x = 168000x

Amount of profit generated by Cxqi = $$\dfrac{20}{100}$$*18000*42x = 151200x

Amount of profit generated by Dipq = $$\dfrac{30}{100}$$*25000*21x = 157500x

We can see that profit of brands Azra, Bysi and Dipq increased in the year 2017 as compared to 2016. Hence, option A is the correct answer. 

Let the efficiency of type A pipe be 'a' and the efficiency of type B be 'b'.
In the first case, 10 type A and 45 type B pipes fill the tank in 30 mins.
So, the capacity of the tank  = $$\dfrac{1}{2}$$(10a + 45b)........(i)
In the second case, 8 type A  and 18 type B pipes fill the tank in 1 hour.
So, the capacity of the tank = (8a + 18b)..........(ii)
Equating (i) and (ii), we get
10a + 45b = 16a + 36b
=>6a = 9b
From (ii), capacity of the tank = (8a + 18b) = (8a + 12a) = 20a
In the third case, 7 type A and 27 type B pipes fill the tank.
Net efficiency = (7a + 27b) = (7a + 18a) = 25a
Time taken = $$\dfrac{20\text{a}}{25\text{a}}$$ hour = 48 minutes.
Hence, 48 is the correct answer.

The minimum value of the function will occur when the expressions inside the function are equal.
So, 5$$x$$ = $$52 - 2x^2$$
or, $$2x^2 + 5x - 52$$ = 0
On solving, we get $$x$$ = 4 or $$-\dfrac{13}{2}$$
But, it is given that $$x$$ is a positive number.
So, $$x$$ = 4
And the minimum value = 5*4 = 20
Hence, 20 is the correct answer.

Car 3 meets car 1 at Q, which is 200 km from A.
Therefore, at the time of their meeting car 1 must have travelled 200 km and car 3 must have travelled 100 km.
As the time is same, ratio of speed of car 1 to speed of car 3 = 2 : 1.

Car 3 meets car 2 at P, which is 100 km from A.
Therefore, at the time of their meeting car 2 must have travelled 100 km and car 3 must have travelled 200 km.
As the time is same, ratio of speed of car 2 to speed of car 3 = 1 : 2.

Speed of car 1 : speed of car 3 = 2 : 1
And speed of car 2 : speed of car 3 = 1 : 2
So, speed of car 1 : speed of car 2 : speed of car 3 = 4 : 1 : 2

Hence, option D is the correct answer. 

We can see that at n = 2, $$n^3-11n^2+32n-28=0$$ i.e. (n-2) is a factor of $$n^3-11n^2+32n-28$$

$$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$$

We can further factorize n^2-9n+14 as (n-2)(n-7).

$$n^3-11n^2+32n-28=(n-2)^2(n-7)$$

$$\Rightarrow$$ $$n^3-11n^2+32n-28>0$$

$$\Rightarrow$$ $$(n-2)^2(n-7)>0$$

Therefore, we can say that n-7>0

Hence, n$$_{min}$$ = 8