CAT 2019 Question Paper Quant Slot 1

Let the speed of cars be a and b and the distance =d

Minimum time taken by 1st car = 6 hours,

For maximum difference in time taken by both of them, car 1 has to start at 10:00 AM and car 2 has to start at 11:00 AM. 

Hence, car 2 will take 5 hours. 

Hence a= $$\ \frac{\ d}{6}$$ and b = $$\ \frac{\ d}{5}$$

Hence the speed of car 2 will exceed the speed of car 1 by $$\ \dfrac{\ \ \frac{\ d}{5}-\ \frac{\ d}{6}}{\ \frac{\ d}{6}}\times\ 100$$ = $$\ \dfrac{\ \ \frac{\ d}{30}}{\ \frac{\ d}{6}}\times\ 100$$ = 20

We have, $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$$

Now, $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}}$$ = $$\frac{\sqrt{a_2} - \sqrt{a_1}}{(\sqrt{a_2} + \sqrt{a_1})(\sqrt{a_2} - \sqrt{a_1})}$$   (Multiplying numerator and denominator by $$\sqrt{a_2} - \sqrt{a_1}$$)

= $$\frac{\sqrt{a_2} - \sqrt{a_1}}{({a_2} - {a_1}}$$

=$$\frac{\sqrt{a_2} - \sqrt{a_1}}{d}$$   (where d is the common difference)

Similarly, $$ \frac{1}{\sqrt{a_2} + \sqrt{a_3}}$$ = $$\frac{\sqrt{a_3} - \sqrt{a_2}}{d}$$ and so on.

Then the expression $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$$

can be written as $$\ \frac{\ 1}{d}(\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_3}+..........................\sqrt{a_{n+1}} - \sqrt{a_{n}}$$

= $$\ \frac{\ n}{nd}(\sqrt{a_{n+1}}-\sqrt{a_1})$$ (Multiplying both numerator and denominator by n)

= $$\ \frac{n(\sqrt{a_{n+1}}-\sqrt{a_1})}{{a_{n+1}} - {a_1}}$$     $$(a_{n+1} - {a_1} =nd)$$

= $$\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$$

Since AB is a diameter,  AQB and APB will right angles.

In right triangle APB, AP = $$\sqrt{10^2-6^2}=8$$

Now, 2AQ=AP  => AQ= 8/2=4

In right triangle AQB, AP = $$\sqrt{10^2-4^2}=9.165$$ =9.1 (Approx)

We have, $$(5.55)^x = (0.555)^y = 1000$$

Taking log in base 10 on both sides,

x($$\log_{10}555$$-2) = y($$\log_{10}555$$-3) = 3

Then, x($$\log_{10}555$$-2) = 3.....(1)

y($$\log_{10}555$$-3) = 3 .....(2)

From (1) and (2)

=> $$\log_{10}555$$=$$\ \frac{\ 3}{x}$$+2=$$\ \frac{\ 3}{y}+3$$

=> $$\frac{1}{x} - \frac{1}{y}$$ = $$\frac{1}{3}$$

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 $$\ \ \ \frac{\ \ \frac{\ 11t}{9}-t}{t}\times\ 100 = 22.22%$$

Assuming the maximum marks =100a, then Meena got 40a

After increasing her score by 50%, she will get 40a(1+50/100)=60a

Passing score = 60a+35

Post review score after 20% increase = 60a*1.2=72a

=>Hence, 60a+35+7=72a

=>12a=42   =>a=3.5

=> maximum marks = 350 and passing marks = 210+35=245

=> Passing percentage = 245*100/350 = 70

Assuming the amount invested in the ratio 2:1 was 200x and 100x, then the fixed deposit investment = 1500000-300x

Hence, the interest = 200x*4/100 = 8x and 100x*3/100=3x

Interest from the fixed deposit = (1500000-300x)*6/100 = 90000-18x

Hence the total interest  = 90000-18x+8x+3x=90000-7x =76000

=> 7x=14000   => x=2000

Hence, the fixed deposit investment = 1500000-300*2000 = 900000 = 9 lakhs

Assume the number of members who can play exactly 1 game = I

The number of members who can play exactly 1 game = II

The number of members who can play exactly 1 game = III

I+2II+3III=144+123+132=399....(1)

I+II+III=256......(2)

=> II+2III=143.....(3)

Also, II+3III=58+25+63=146 ......(4)

=> III = 3 (From 3 and 4)

=> II =137

=> I = 116

The members who play only tennis = 123-58-25+3 = 43

11th term of series = $$a_{11}$$ = Sum of 11 terms - Sum of 10 terms = $$3(2^{11 + 1} - 2)$$-3$$(2^{10 + 1} - 2)$$ 

= 3$$(2^{12} - 2-2^{11} +2)$$=3$$(2^{11})(2-1)$$= 3*$$2^{11}$$ = 6144

$$2 \cos (x(x + 1)) = 2^x + 2^{-x}$$

The maximum value of LHS is 2 when $$\cos (x(x + 1))$$ is 1 and the minimum value of RHS is 2 using AM $$\geq$$ GM 

Hence LHS and RHS can only be equal when both sides are 2. For LHS, cosx(x+1)=1   => x(x+1)=0   => x=0,-1

For RHS minimum value, x=0

Hence only one solution x=0

Assuming A completes a units of work in a day and B completes B units of work in a day and the total work = 1 unit

Hence, 12(a+b)=1.........(1)

Also, 9($$\ \frac{\ a}{2}$$+3b)=1  .........(2)

Using both equations, we get, 12(a+b)= 9($$\ \frac{\ a}{2}$$+3b)

=> 4a+4b=$$\ \frac{\ 3a}{2}$$+9b

=> $$\ \frac{\ 5a}{2}$$=5b

=> a=2b

Substituting the value of b in equation (1),

12($$\ \frac{\ 3a}{2}$$)=1

=> a=$$\ \frac{\ 1}{18}$$

Hence, the number of days required = 1/($$\ \frac{\ 1}{18}$$)=18

Assuming the number of students =100x

Hence, the number of girls = 60x and the number of boys = 40x

We have, 60x-40x=30  => x=1.5

The number of girls = 60*1.5=90

Number of girls that pass = 68x-30=68*1.5-30 = 102-30=72

The number of girls who do not pass = 90-72=18

Hence the percentage of girls who do not pass = 1800/90=20

In figure AE*BE=CE*DE    (The   intersecting chords theorem)

=> 7*15= x(20.5-x)                  (Assuming AE=x)

=> 210=x(41-2x)

=> $$2x^2$$-41x+210=0

=> x=10 or x=10.5   => AE=10 or AE=10.5      Hence BE = 20.5-10=10.5  or BE = 20.5-10.5=10

Required difference= 10.5-10=0.5

Assuming the cost price of pen = 100p and the cost price of book = 100b

So, on selling a pen at 5% loss and a book at 15% gain, net gain =  -5p+15b = 7 ....1

On selling the pen at 5% gain and the book at 10% gain, net gain = 5p+10b = 13  .....2

Adding 1 and 2 we get, 25b=20

Hence 100b= 20*4=80, 

C is the answer. 

The weight/volume(g/L) for liquid 1 = 1000

The weight/volume(g/L) for liquid 2 = 800

The weight/volume(g/L) of the mixture = 480/(1/2) = 960

Using alligation the ratio of liquid 1 and liquid 2 in the mixture = (960-800)/(1000-960) = 160/40 = 4:1

Hence the percentage of liquid 1 in the mixture = 4*100/(4+1)=80

Assume the average of 21 students other than Ramesh = a

Sum of the scores of 21 students other than Ramesh = 21a

Hence the average of 22 students = a+1

Sum of the scores of all 22 students = 22(a+1)

The score of Ramesh = Sum of scores of all 22 students - Sum of the scores of 21 students other than Ramesh = 22(a+1)-21a=a+22  = 82.5   (Given)

=> a = 60.5

Hence, sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353

Now the sum of the scores of students other than Gautam = 21*62 = 1302

Hence the score of Gautam = 1353-1302=51

We have, $$(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$$

Converting both sides in powers of 2 and 3, we get

$$2^{\ \frac{19\ }{2}}3^42^43^{2m}2^{3n}$$ = $$3^n2^{4m}2^{\frac{\ 6}{4}}$$

Comparing the power of 2 we get, $$\ \frac{\ 19}{2}+4+3n\ =4m+\frac{\ 6}{4}\ $$

=> 4m=3n+12 .....(1)

Comparing the power of 3 we get, $$4+2m=n$$

Substituting the value of n in (1), we get

4m=3(4+2m)+12

=> m=-12

Consider the work done by a man in a day = a and that by a machine = b

Since, three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job, hence the efficiency will be double.

=> 3a+8b = 2(3b+8a)

=> 13a=2b

Hence work done by 13 men in a day = work done by 2 machines in a day.

=> If two machines can finish the job in 13 days, then same work will be done 13 men in 13 days. 

Hence the required number of men = 13

The given figure can be divided into 9 regions or equilateral triangles of equal areas as shown below,

Now the hexagon consists of 6 regions and the triangle consists of 9 regions.

Hence the ratio of areas = 6/9 =2:3

We have, $$\log_{5}{(x + y)} + \log_{5}{(x - y)} = 3$$

=> $$x^2-y^2=125$$......(1)

$$\log_{2}{y} - \log_{2}{x} = 1 - \log_{2}{3}$$

=>$$\ \frac{\ y}{x}$$ = $$\ \frac{\ 2}{3}$$

=> 2x=3y   => x=$$\ \frac{\ 3y}{2}$$

On substituting the value of x in 1, we get

$$\ \frac{\ 5x^2}{4}$$=125

=>y=10, x=15

Hence xy=150

Sum of the area of region I and II is the required area.

Now, required area =  $$\ 4\times\frac{\ 1}{2}\times\ 1\times\ 1$$ = 2

The number of paths from (1, 1) to (8, 10) via (4, 6) = The number of paths from (1,1) to (4,6) * The number of paths from (4,6) to (8,10)

To calculate the number of paths from (1,1) to (4,6), 4-1 =3 steps in x-directions and 6-1=5 steps in y direction

Hence the number of paths from (1,1) to (4,6) = $$^{(3+5)}C_3$$ = 56

To calculate the number of paths from (4,6) to (8,10), 8-4 =4 steps in x-directions and 10-6=4 steps in y direction

Hence the number of paths from (4,6) to (8,10) = $$^{(4+4)}C_4$$ = 70

The number of paths from (1, 1) to (8, 10) via (4, 6) = 56*70=3920

Assuming the dimensions of the brick are a, b and c and the diagonals are 3, 2 $$\surd3$$ and $$\surd{15}$$

Hence, $$a^{2\ }+\ b^2$$ = $$3^2$$   ......(1)

$$b^{2\ }+\ c^2$$ = $$(2\sqrt{3})^2$$ ......(2)

$$c^{2\ }+\ a^2$$ = $$(\sqrt{15})^2$$ ......(3)

Adding the three equations, 2($$a^2+b^2+c^2$$) = 9+12+15=36

=>$$a^2+b^2+c^2$$ = 18......(4)

Subtracting (1) from (4), we get $$c^2$$ = 9    =>c=3

Subtracting (2) from (4), we get $$a^2$$ = 6    =>a=$$\sqrt{6}$$

Subtracting (3) from (4), we get $$b^2$$ = 3    =>b=$$\sqrt{3}$$

The ratio of the length of the shortest edge of the brick to that of its longest edge is = $$\ \frac{\ \sqrt{3}}{3}$$ = $$1 : \sqrt{3}$$

For x <0, -x($$6x^2+1$$) = $$5x^2$$

=> ($$6x^2+1$$) = -5x

=> ($$6x^2 + 5x+ 1$$) = 0 

=>($$6x^2 + 3x+2x+ 1$$) = 0

=> (3x+1)(2x+1)=0    =>x=$$\ -\frac{\ 1}{3}$$  or x=$$\ -\frac{\ 1}{2}$$

For x=0, LHS=RHS=0    (Hence, 1 solution)

For x >0, x($$6x^2+1$$) = $$5x^2$$

=> ($$6x^2 - 5x+ 1$$) = 0 

=>(3x-1)(2x-1)=0    =>x=$$\ \frac{\ 1}{3}$$   or   x=$$\ \frac{\ 1}{2}$$

Hence, the total number of solutions = 5

In any right triangle, the circumradius is half of the hypotenuse. Here,L=$$\ \frac{\ 1}{2}\ $$* the length of the hypotenuse = $$\ \frac{\ 1}{2}$$($$\sqrt{\ 15^2+9^2}$$) = $$\ \frac{\ 1}{2}\ $$*$$\sqrt{\ 306}$$ = $$\ \frac{\ 1}{\ 2}\times\ $$17.49 = 8.74

Hence, the integer close to L = 9