CAT 2019 Question Paper Slot 2

It is given that starting from 10:01 am, every minute a motorcycle leaves A and moves towards B.

Forty-five such motorcycles reach B by 11 am.

It means that the forty-fifth motorcycle starts at 10:45 AM at A and reaches B by 11:00 AM i.e 15 minutes.

Since the speed of all the motorcycles is the same, all the motorcycles will take the same duration i.e 15 minutes.

If the cyclist doubles the speed, then he will reach B by 10:30 AM. (Since if the speed is doubled, time is reduced by half)

Since each motorcycle takes 15 minutes to reach B, 15 motorcycles would have reached B by the time the cyclist reaches B

It is given that the volume of all the cylinders is the same, so the volume of each cylinder = HCF of (405, 783, 351)

=27

The number of iron cylinders = $$\ \frac{\ 405}{27}$$ = 15

The number of aluminium cylinders = $$\ \frac{\ 783}{27}$$ = 29

The number of copper cylinders = $$\ \frac{351\ }{27}$$ = 13

15*$$\pi\ r^2h=$$ 405  

15*$$\pi\ 9*h=$$ 405

$$\pi\ h=3$$

Now we have to calculate the total surface area of all the cylinders 

Total number of cylinders = 15+29+13 = 57

Total surface area of the cylinder = 57*($$2\pi\ rh+2\pi\ r^2$$)

=57(2*3*3 + 2*9*$$\pi\ $$)

=$$1026(1 + \pi)$$

Let $$2^{3x}$$ = v

$$2^{6x} + 2^{3x + 2} - 21 = 0$$

= $$v^2+4v-21=0$$

=(v+7)(v-3)=0

v=3, -7

$$2^{3x}$$ = 3 or $$2^{3x}$$ = -7(This can be negated)

3x=$$\log_23$$

x=$$\log_23$$/3

$$2^4 \times 3^5 \times 10^4$$

=$$2^4 \times 3^5 \times 2^4*5^4$$

=$$2^8 \times 3^5 \times 5^4$$

For the factor to be a perfect square, the factor should be even power of the number.

In $$2^8$$, the factors which are perfect squares are $$2^0, 2^2, 2^4, 2^6, 2^8$$ = 5

Similarly, in $$3^5$$, the factors which are perfect squares are $$3^0, 3^2, 3^4$$ = 3

In $$5^4$$, the factors which are perfect squares are $$5^0, 5^2, 5^4$$ = 3

Number of perfect squares greater than 1 = 5*3*3-1

=44

Let the six-digit number be ABCDEF

F = A+B+C, E= A+B, C=A, B= 2A, D= E+F.

Therefore D = 2A+2B+C = 2A + 4A + A= 7A.

A cannot be 0 as the number is a 6 digit number.

A cannot be 2 as D would become 2 digit number.

Therefore A is 1 and D is 7.

Speed of John = 6kmph 

Speed of Mary = 7.5 kmph 

Lengths of tracks A and B = 325 m

Let the length of track A be a, then the length of track B = 325-a

9 rounds of John on track A = 5 rounds of Mary on track B

$$\ \frac{\ 9\times\ a}{6\ \times\ \ \frac{\ 5}{18}}\ =\ \ \frac{\ 5\cdot\left(325-a\right)}{7.5\times\ \ \frac{\ 5}{18}}$$

On solving we get , 13a=1300

a=100

The length of track A = 100m, track B = 225m

Mary makes one round of track A = $$\ \frac{\ 100}{7.5\times\ \ \frac{\ 5}{18}}$$

= 48 sec

Let the number of fiction and non-fiction books in 2010 = 100a, 100b respectively

It is given that the total number of books in 2010 = 11500

100a+100b = 11500            -------Eq 1

The number of fiction and non-fiction books in 2015 = 110a, 112b respectively

110a+112b = 12760           -------Eq 2

On solving both the equations we get, b=55, a= 60

The number of fiction books in 2015 = 110*60=6600

It is given that John works altogether 172 hours i.e including regular and overtime hours.

Let a be the regular hours, 172-a will be the overtime hours

John's income from regular hours = 57*a 

John's income for working overtime hours = (172-a)*114

It is given that his income from overtime hours is 15% of his income from regular hours

a*57*0.15 = (172-a)*114

a=160

The number of hours for which he worked overtime = 172-160=12 hrs

Let us first find the number of terms 

47=1+(n-1)2

n=24

24*2n+1+3+5+....47=5280

48n+576=5280

48n=4704

n=98

Sum of first 98 terms = 98*99/2

=4851

CP of the table at which the shopkeeper procured each table = p

It is given that shopkeeper sold the tables to Amal and Asim at a profit of 20% and at a loss of 20%, respectively

The selling price of the tables = 1.2p and 0.8p to Amal and Asim respectively.

Amal sells his table to Bimal at a profit of 30%

So, CP of the table by Bimal (x)= 1.2p*1.3 = 1.56p

Asim sells his table to Barun at a loss of 30%

So, CP of the table by Barun (y)= 0.7*0.8p = 0.56p

(x-y)/p = (1.56p-0.56p)/p = p/p=1