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
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
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