CAT 2019 Question Paper Slot 1

Assuming the income of Bimla = 100a, then the income of Amala will be 120a.

And the income of Kamala will be 120a*100/80=150a

If Kamala's income goes down by 4%, then new income of Kamala = 150a-150a(4/100) = 150a-6a=144a

If Bimla's income goes up by 10 percent, her new income will be 100a+100a(10/100)=110a

=> Hence the Kamala income will exceed Bimla income by (144a-110a)*100/110a=31

Distance covered by A in 1 revolution = 2$$\pi\ $$*30 = 60$$\pi\ $$

Distance covered by B in 1 revolution = 2$$\pi\ $$*40 = 80$$\pi\ $$

Now, (5000+n)60$$\pi\ $$ = 80$$\pi\ $$n

=> 15000= 4n-3n   =>n=15000

Then distance travelled by B = 15000*80$$\pi\ $$ cm = 12$$\pi\ $$ km

Hence, the speed = $$\ \frac{\ 12\pi\times\ 60\ }{45}$$ = 16$$\pi\ $$

We have, $$\mid x^2 - x - 6 \mid = x + 2$$

=> |(x-3)(x+2)|=x+2

For x<-2, (3-x)(-x-2)=x+2

=> x-3=1   =>x=4 (Rejected as x<-2)

For -2$$\le\ $$x<3, (3-x)(x+2)=x+2    =>x=2,-2

For x$$\ge\ $$3, (x-3)(x+2)=x+2   =>x=4

Hence the product =4*-2*2=-16

Assuming the length of race course = x and the speed of three horses be a,b and c respectively.

Hence, $$\ \frac{\ x}{a}=\ \frac{\ x-11}{b}$$......(1)

and $$\ \frac{\ x}{a}=\ \frac{\ x-90}{c}$$......(2)

Also, $$\ \frac{\ x}{b}=\ \frac{\ x-80}{c}$$......(3)

From 1 and 2, we get, $$\ \frac{\ x-11}{b}=\ \frac{\ x-90}{c}$$ .....(4)

Dividing (3) by (4), we get, $$\ \frac{\ x-11}{x}=\ \frac{\ x-90}{x-80}$$  

=> (x-11)(x-80)=x(x-90)

=> 91x-90x=880  => x=880

The population of town at the beginning of 1st year = p

The population of town at the beginning of 2nd year = 3+2p

The population of town at the beginning of 3rd year = 2(3+2p)+3 = 2*2p+2*3+3 =4p+3(1+2)

The population of town at the beginning of 4th year = 2(2*2p+2*3+3)+3 = 8p+3(1+2+4)

Similarly population at the beginning of the nth year = $$2^{n-1}$$p+3($$2^{n-1}-1$$) = $$2^{n-1}\left(p+3\right)$$-3 

The population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be $$(2^{2034-2019})\left(1000+3\right)$$-3 = $$2^{15}\left(1003\right)$$-3

f (x + y) = f (x) f (y)

Hence, f(2)=f(1+1)=f(1)*f(1)=2*2=4

f(3)=f(2+1)=f(2)*f(1)=4*2=8

f(4)=f(3+1)=f(3)*f(1)=8*2=16

.......=> f(x)=$$2^x$$

Now, f(a + 1) +f (a + 2) + ... + f(a + n) = 16 (2$$^n$$ - 1)

On putting n=1 in the equation we get, f(a+1)=16   => f(a)*f(1)=16  (It is given that f (x + y) = f (x) f (y))

=> $$2^a$$*2=16

=> a=3

Assuming the investment of Amala, Bina, and Gouri be 300x, 400x and 500x, hence the interest incomes will be 300x*6/100=18x, 400x*5/100=20x and 500x*4/100 = 20x

Given, Bina's interest income exceeds Amala by 20x-18x=2x=250  => x=125

Now, total interest income = 18x+20x+20x=58x = 58*125 = 7250

Assuming m is even, then 8f(m+1)-f(m)=2

m+1 will be odd

So, 8(m+1+3)-m(m+1)=2

=> 8m+32-$$m^2-m$$=2

=> $$m^2-7m-30=0$$

=> m=10,-3 

Rejecting the negative value, we get m=10

Assuming m is odd, m+1 will be even.

then, 8(m+1)(m+2)-m-3=2

=> 8($$m^2+3m+2$$)-m-3=2

=> $$8m^2+23m+11=0$$

Solving this, m = -2.26 and -0.60

Hence, the value of m is not integral. Hence this case will be rejected.

Assume the numbers are a and b, then ab=616

We have, $$\ \ \frac{\ a^3-b^3}{\left(a-b\right)^3}$$ = $$\ \frac{\ 157}{3}$$

=> $$\ 3\left(a^3-b^3\right)\ =\ 157\left(a^3-b^3+3ab\left(b-a\right)\right)$$

=> $$154\left(a^3-b^3\right)+3*157*ab\left(b-a\right)$$ = 0

=> $$154\left(a^3-b^3\right)+3*616*157\left(b-a\right)$$ = 0        (ab=616)

=>$$a^3-b^3+\left(3\times\ 4\times\ 157\left(b-a\right)\right)$$    (154*4=616)

=> $$\left(a-b\right)\left(a^2+b^2+ab\right)\ =\ 3\times\ 4\times\ 157\left(a-b\right)$$

=> $$a^2+b^2+ab\ =\ 3\times\ 4\times\ 157$$

Adding ab=616 on both sides, we get

$$a^2+b^2+ab\ +ab=\ 3\times\ 4\times\ 157+616$$

=> $$\left(a+b\right)^2=\ 3\times\ 4\times\ 157+616$$ = 2500

=> a+b=50

Assume the total distance between A and B as d and time taken by Amal = t

Since Amal travelled $$\frac{1}{3}^{rd}$$ of his total journey time in different speeds
d = $$\ \frac{\ t}{3}\times\ 10+\ \frac{\ t}{3}\times\ 20+\frac{\ t}{3}\times\ 30\ \ =\ 20t$$

$$\text{Total time taken by Bimal} = \ \frac{d_1}{s_1}+\frac{d_2}{s_2}+\frac{d_3}{s_3}$$

$$=\ \frac{20t}{3}\times\ \frac{1}{10}+\frac{20t}{3}\times\ \frac{1}{20}+\frac{20t}{3}\times\ \frac{1}{30}\ \ =\frac{20t\left(6+3+2\right)}{3\ \times30}\ =\frac{11}{9}t$$

Hence, the ratio of time taken by Bimal to time taken by Amal = $$\frac{\frac{11t}{9}}{t}=\frac{11}{9}$$
Therefore, Bimal will exceed Amal's time by $$\ \ \ \frac{\ \ \frac{\ 11t}{9}-t}{t}\times\ 100 = 22.22%$$