CAT 2019 – Slot 1 – Quantitative Ability – Let x, y, z be three positive real numbers in a geometric progression such

Q. 1: Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is
1. 3/6
2. 3/2
3. 5/2
4. 1/6

Given x,y, z are in GP . let common ratio of this GP is r so y=xr & z=xr^2

Now 5x, 16y and 12z are in A.P. So

12z-16y=16y-5x

32y=12z+5x

32xr=12xr^2+5x

12r^2-32r+5=0

(6r-1)(2r-5)=0

So r=1/6 or 5/2

As given x < y < z so r >1

Thus r = 5/2

 

Q. 2: A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes areon? 

Let the rate of each filling pipes be ‘x lts/hr’ similarly, the rate of each draining pipes be ‘y lts/hr’. As per the first condition,

Capacity of tank = (6x-5y)×6. (i)

Similarly, from the second condition, Capacity of tank = (5x – 6y)×60 (ii)

On equating (i) and (ii), we get (6x – 5y) × 6 = (5x – 6y)×60

or, 6x – 5y = 50x – 60y or, 44x = 55y

or, 4x = 5y or, x = 1.25y

Therefore, the capacity of the tank = (6x – 5y) × 6 = (7.5y – 5y) × 6 = 15y lts Effective rate of 2 filling pipes and 1 draining pipe = (2x – y) = (2.5y – y) = 1.5y Hence, the required time = 15y/1.5y=10 hours.


Q. 3: Given that x^2018 y^2017 =1/2 and x^2016 y^2019= 8,the value of x^2 + y^3 is?
1.  33/4
2.
 35/4
3. 
31/4
4.
 37/4

Given, x^2018 y^2017 =1/2——————1)

x^2016 y^2019= 8——————-2)

By dividing eq 1) with eq 2)

x^2/y^2 =1/16——–3)

By multiplying eq 1) with eq 2)

x^4034 y^4036=(xy)^4034 y^2= 4——–4)

From eq 3) & eq ) 4 we can say x = ½ and y = 2

So x^2 + y^3=1/4+8=33/4

Q. 4: Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is

Ratio of time taken by car 1 and car 2 = (x/2v) : (3x/v) = 1 : 6 = t : 6t

Given 6t – t = 60 minutes

Or 5t = 60

Thus t = 12 minutes

Answer: 12

Q. 5: If log2(5 + log3a) = 3 and log5(4a + 12 + log2 b) = 3, then a + b is equal to?
1.
 59
2.
 40
4.
 67
5. 32

log2(5 + log3 a) = 3

5 + log3 a = 2^3 = 8

log3 a = 3

so a =  3^3 = 27

Now log(base5⁡) (4a+12+log(base2)⁡ b)=3

Or 4a + 12 + log(base2) b) = 125

log(base2)⁡ b = 125-12-4×27=5

So b = 2^5 = 32

Thus a+b = 59

Inspiring Education… Assuring Success!!

Ghatkopar | Borivali | Andheri | Online

Head Office

Site Links

Competitive Exams

Free Material Area

Ⓒ 2016 - All Rights Are Reserved