CAT 2018 Question Paper Slot 2

Let the score of Amal and Bimal be 11k and 14k
Let the scores be increased by x
So, after increment, Amal's score =  11k + x and Bimal's score = 14k + x
According to the question,
$$\dfrac{\text{11k + x}}{\text{14k + x}}$$ = $$\dfrac{47}{56}$$
On solving, we get x = $$\dfrac{42}{9}$$k
Ratio of Bimal's new score to his original score

= $$\dfrac{\text{14k + x}}{\text{14k}}$$

=$$\dfrac{14k +\frac{42k}{9}}{14k}$$

=$$\dfrac{\text{168k}}{\text{14*9k}}$$

=$$\dfrac{4}{3}$$

Hence, option A is the correct answer.

Let 'ab' be the two digit number. Where b $$\neq$$ 0.

We will get number 'ba' after interchanging its digit. 

It is given that 10a+b > 3*(10b + a)

7a > 29b

If b = 1, then a = {5, 6, 7, 8, 9}

If b = 2, then a = {9}

If b = 3, then no value of 'a' is possible. Hence, we can say that there are a total of 6 such numbers.

 P = {1,2,3,4} and  Q = {2,3,5,6,}
PΔQ = {1, 4, 5, 6}
R = {1,3,7,8,9} and S = {2,4,9,10}
RΔS = {1, 2, 3, 4, 7, 8, 10}
(PΔQ)Δ(RΔS) = {2, 3, 5, 6, 7, 8, 10}
Thus, there are 7 elements in (PΔQ)Δ(RΔS) .
hence, 7 is the correct answer.

We can see that area of parallelogram ABCD = 2*Area of triangle ACD 

48 = 2*Area of triangle ACD 

Area of triangle ACD = 24

$$(1/2)*CD*DA*sinADC=24$$

$$AD*sinADC=6$$

We know that $$sin\theta$$ $$\leq$$ 1, Hence, we can say that AD $$\geq$$ 6

$$\Rightarrow$$ s $$\geq$$ 6

Let the volume of the first and the second solution be 100 and 300.
When they are mixed, quantity of ethanol in the mixture 
= (20 + 300S)
Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.
So, the quantity of ethanol in the final solution 
= (20 + 300S + 80) = (300S + 100)
It is given that, 31.25% of 800 = (300S + 100)
or, 300S + 100 = 250
or S = $$\frac{1}{2}$$ = 50%
Hence, 50 is the correct answer.

In a tournament, there are 43 junior level and 51 senior level participants.

Let 'n' be the number of girls on junior level. It is given that the number of girl versus girl matches in junior level is 153.

$$\Rightarrow$$ nC2 = 153

$$\Rightarrow$$ n(n-1)/2 = 153

$$\Rightarrow$$ n(n-1) = 306

=> n$$^{2}$$-n-306 = 0

=> (n+17)(n-18)=0

=> n=18  (rejecting n=-17)

Therefore, number of boys on junior level = 43 - 18 = 25. 

Let 'm' be the number of boys on senior level. It is given that the number of boy versus boy matches in senior level is 276.

$$\Rightarrow$$ mC2 = 276

$$\Rightarrow$$ m = 24

Therefore, number of girls on senior level = 51 - 24 = 27. 

Hence, the number of matches a boy plays against a girl  = 18*25+24*27 = 1098

We are given that AB = 5 cm and $$\angle$$ AOB = 60°

Let us draw OM such that OM $$\perp$$ AB. 

In right angle triangle AMO,

$$sin 30° = \dfrac{AM}{AO}$$

$$\Rightarrow$$ AO = 2*AM = 2*2.5 = 5 cm. Therefore, we can say that the radius of the circle = 5 cm. 

In right angle triangle PNO,

$$sin 60° = \dfrac{PN}{PO}$$

$$\Rightarrow$$ PN = $$\dfrac{\sqrt{3}}{2}$$*PO = $$\dfrac{5\sqrt{3}}{2}$$

Therefore, PQ = 2*PN = $$5\sqrt{3}$$ cm

Let 'x' be the average of all 52 positive integers $$\ a_{1},a_{2}...a_{52}\ $$.

$$a_{1}+a_{2}+a_{3}+...+a_{52}$$ = 52x ... (1)

Therefore, average of $$a_{2}$$, $$a_{3}$$, ....$$a_{52}$$ = x+1

$$a_{2}+a_{3}+a_{4}+...+a_{52}$$ = 51(x+1) ... (2)

From equation (1) and (2), we can say that

$$a_{1}+51(x+1)$$ = 52x

$$a_{1}$$ = x - 51. 

We have to find out the largest possible value of $$a_{1}$$. $$a_{1}$$ will be maximum when 'x' is maximum. 

(x+1) is the average of terms $$a_{2}$$, $$a_{3}$$, ....$$a_{52}$$. We know that $$a_{2}$$ < $$a_{3}$$ < ... < $$a_{52}\ $$ and $$a_{52}$$ = 100.

Therefore, (x+1) will be maximum when each term is maximum possible. If $$a_{52}$$ = 100, then $$a_{52}$$ = 99, $$a_{50}$$ = 98 ends so on.

$$a_{2}$$ = 100 + (51-1)*(-1) = 50.

Hence,  $$a_{2}+a_{3}+a_{4}+...+a_{52}$$ = 50+51+...+99+100 = 51(x+1)

$$\Rightarrow$$ $$\dfrac{51*(50+100)}{2} = 51(x+1)$$

$$\Rightarrow$$ $$x = 74$$

Therefore, the largest possible value of $$a_{1}$$ = x - 51 = 74 - 51 = 23. 

S = 7 x 11 + 11 x 15 + 15 x 19 + ...+ 95 x 99

Nth term of the series can be written as $$T_{n} = (4n+3)*(4n+7)$$

Last term, (4n+3) = 95 i.e. n = 23

$$\sum_{n=1}^{n=23} (4n+3)*(4n+7)$$

$$\Rightarrow$$ $$\sum_{n=1}^{n=23}16n^2+40n+21$$

$$\Rightarrow$$ $$16*\dfrac{23*24*47}{6}+40*\dfrac{23*24}{2}+21*23$$

$$\Rightarrow$$ $$80707$$

It is given that $$N^{N}$$ = $$2^{160}$$

We can rewrite the equation as $$N^{N}$$ = $$(2^5)^{160/5}$$ = $$32^{32}$$

$$\Rightarrow$$ N = 32

$$N{^2} + 2^{N}$$ = $$32^2+2^{32}=2^{10}+2^{32}=2^{10}*(1+2^{22})$$

Hence, we can say that $$N{^2} + 2^{N}$$ can be divided by $$2^{10}$$

Therefore, x$$_{max}$$ = 10