CAT 2019 Question Paper Slot 1

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