### CAT 2019 – Slot 1 – Quantitative Ability –Let x and y be positive real numbers such that log(base 5) (x + y) + log(base 5) (x − y) = 3

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Q. 1: Let x and y be positive real numbers such that log(base 5) (x + y) + log(base 5) (x − y) = 3, and log(base 2)y − log(base 2)x = 1 − log(base 2)3. Then xy equals?
1. 150
2. 100
3. 25
4. 250

Given, log(base5) (x + y) + log(base5) (x − y) = 3

Or log(base5) (x + y)*(x-y) =3

Or x^2 –y^2 = 5^3 = 125————-1)

log(base2) y − log(base2) x = 1 − log(base2) 3= log(base2) 2 – log(base2) 3

log(base2) y/x = log(base2) 2/3

y/x = 2/3

y = 2x/3

from eq 1) x^2 – (2x/3)^2 = 125

x^2 – (4x^2/9) = 125

5x^2 = 125*9 or x^2 = 225

x = 15

y= 2x/3 = 30/3 = 10

xy = 15*10 =150

##### Q. 2: One can use three different transports which move at 10, 20, and 30 kmph, respectively. To reach from A to B, Amal took each mode of transport 1/3 of his total journey time, while Bimal took each mode of transport 1/3 of the total distance. The percentage by which Bimal’s travel time exceeds Amal’s travel time is nearest to1. 192.  213.  204.  22

let total distance = 180x

Total time taken by Bimal = 60x/10 + 60x/20 + 60x/30 = 11x hr

Let time taken by Amal in each mode of transport is same so distance traveled in each mode will be in the ratio of speed that is 10:20:30= 1:2:3

So distance traveled in each mode = (1/6 *180x = 30x ) , (2/6 *180x= 60x ) and (3/6 *180x = 90x)

So total time taken = 3*(30x/10) = 9x

Thus The percentage by which Bimal’s travel time exceeds Amal’s travel time = (11-9)/9 *100 = 22.22 = 22 (approx)

Q. 3: If the rectangular faces of a brick have their diagonals in the ratio 3 : 2√3 : √15, then the ratio of the length of the shortest edge of the brick to that of its longest edge is?
1. 2 : √5
2. 1 : √3
3. √2 : √3
4. √3 : 2

Let the size of bricks are l*b*h such that l > b> h

As we know diagonals = (l^2 + b^2 )^1/2 , (l^2 + h^2 )^1/2 , and (h^2 + b^2 )^1/2

Thus ratio of squares of diagonals = (l^2 + b^2 ) : (l^2 + h^2) : (h^2 + b^2 ) = (√15)^2 : (2√3)^2 : 3^2

Or (l^2 + b^2 ) : (l^2 + h^2) : (h^2 + b^2 ) = 15 : 12 : 9 = ( 9 + 6) : (9 + 3) : (3 + 6)

By comparing we can say l^2 = 9, h^2 = 3 and b^2 = 6

So l = 3 and h = √3

Required ratio of h/l = √3/3 = 1: √3

Q. 4: If the population of a town is p in the beginning of any year then it becomes 3+2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be
1. (997)2^14 + 3
2. (997)^15 – 3
3. (1003)^15 + 6
4. (1003)2^15 – 3

Given as population of the town is p in the beginning of any year then it becomes 3+2p in the beginning of the next year

Population in 2019 = 1000 = (1003)*2^0 -3

So in 2020 it will be = 3 + 2*1000 = 2003 = (1003)*2^1 -3

In 2021 it will be = 3 + 2*2003 = 6009 = (1003)*2^2 -3

In 2022 it will be = 3 + 2*6009 = 12021 = ((1003)*2^3 –3

Thus in the year 2034 the population will be (1003)*2^15 -3

Q. 5: A person invested a total amount of Rs 15 lakh. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is Rs 76000 then the amount (in Rs lakh) invested in the fixed deposit was

96. Final interest at the end of the year = 76000,

Interest rate = 76000/1500000 *100 = 76/15 = 5 1/15

Combine interest rate from amount deposited at 4% and 3% interest rate = (4*2 + 3)/(2+1) =11/3 %

Let the amount ratio of amount deposited in fixed deposit and other deposit be x:y so

x/y = ( 76/15 -11/3 )/(6 – 76/15)

x/y = (76-55)/(90 -76) = 19:14

amount deposited in 6% = 19/33 *15 =8.63 lakh = 9 lakh

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