CAT 2018 Slot 2 Question Paper - QA

Let the efficiency of type A pipe be 'a' and the efficiency of type B be 'b'.
In the first case, 10 type A and 45 type B pipes fill the tank in 30 mins.
So, the capacity of the tank  = $$\dfrac{1}{2}$$(10a + 45b)........(i)
In the second case, 8 type A  and 18 type B pipes fill the tank in 1 hour.
So, the capacity of the tank = (8a + 18b)..........(ii)
Equating (i) and (ii), we get
10a + 45b = 16a + 36b
=>6a = 9b
From (ii), capacity of the tank = (8a + 18b) = (8a + 12a) = 20a
In the third case, 7 type A and 27 type B pipes fill the tank.
Net efficiency = (7a + 27b) = (7a + 18a) = 25a
Time taken = $$\dfrac{20\text{a}}{25\text{a}}$$ hour = 48 minutes.
Hence, 48 is the correct answer.

The minimum value of the function will occur when the expressions inside the function are equal.
So, 5$$x$$ = $$52 - 2x^2$$
or, $$2x^2 + 5x - 52$$ = 0
On solving, we get $$x$$ = 4 or $$-\dfrac{13}{2}$$
But, it is given that $$x$$ is a positive number.
So, $$x$$ = 4
And the minimum value = 5*4 = 20
Hence, 20 is the correct answer.

Car 3 meets car 1 at Q, which is 200 km from A.
Therefore, at the time of their meeting car 1 must have travelled 200 km and car 3 must have travelled 100 km.
As the time is same, ratio of speed of car 1 to speed of car 3 = 2 : 1.

Car 3 meets car 2 at P, which is 100 km from A.
Therefore, at the time of their meeting car 2 must have travelled 100 km and car 3 must have travelled 200 km.
As the time is same, ratio of speed of car 2 to speed of car 3 = 1 : 2.

Speed of car 1 : speed of car 3 = 2 : 1
And speed of car 2 : speed of car 3 = 1 : 2
So, speed of car 1 : speed of car 2 : speed of car 3 = 4 : 1 : 2

Hence, option D is the correct answer. 

We can see that at n = 2, $$n^3-11n^2+32n-28=0$$ i.e. (n-2) is a factor of $$n^3-11n^2+32n-28$$

$$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$$

We can further factorize n^2-9n+14 as (n-2)(n-7).

$$n^3-11n^2+32n-28=(n-2)^2(n-7)$$

$$\Rightarrow$$ $$n^3-11n^2+32n-28>0$$

$$\Rightarrow$$ $$(n-2)^2(n-7)>0$$

Therefore, we can say that n-7>0

Hence, n$$_{min}$$ = 8

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

Let 't' pm be the time when the tank is emptied everyday. Let 'a' and 'b' be the liters/hr filled by pump A and pump B respectively.

On Monday, A alone completed filling the tank at 8 pm. Therefore, we can say that pump A worked for (8 - t) hours. Hence, the volume of the tank = a*(8 - t) liters. 

Similarly, on Tuesday, B alone completed filling the tank at 6 pm. Therefore, we can say that pump B worked for (6 - t) hours. Hence, the volume of the tank = b*(6 - t) liters. 

On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. Therefore, we can say that pump A worked for (5 - t) hours  and pump B worked for 2 hours. Hence, the volume of the tank = a*(5 - t)+2b liters. 

We can say that  a*(8 - t) =  b*(6 - t) =  a*(5 - t) + 2b

a*(8 - t) = a*(5 - t) + 2b

$$\Rightarrow$$ 3a = 2b   ... (1)

a*(8 - t) =  b*(6 - t)

Using equation (1), we can say that 

$$a*(8-t)=\dfrac{3a}{2}*(6-t)$$

$$t = 2$$

Therefore, we can say that the tank gets emptied at 2 pm daily. We can see that A takes 6 hours and pump B takes 4 hours alone. 

Hence, working together both can fill the tank in = \dfrac{6*4}{6+4} = 2.4 hours or 2 hours and 24 minutes. 

The pumps started filling the tank at 2:00 pm. Hence, the tank will be filled by 4:24 pm.  

Given that the arithmetic mean of x, y and z is 80.

$$\Rightarrow$$ $$\dfrac{x+y+z}{3} = 80$$

$$\Rightarrow$$ $$x+y+z = 240$$  ... (1)

Also,  $$\dfrac{x+y+z+v+u}{5} = 75$$

$$\Rightarrow$$ $$\dfrac{x+y+z+v+u}{5} = 75$$

$$\Rightarrow$$ $$x+y+z+v+u = 375$$

Substituting values from equation (1),

$$\Rightarrow$$ $$v+u = 135$$

It is given that u=(x+y)/2 and v=(y+z)/2.

$$\Rightarrow$$ $$(x+y)/2+(y+z)/2 = 135$$

$$\Rightarrow$$ $$x+2y+z = 270$$

$$\Rightarrow$$ $$y = 30$$   (Since $$x+y+z = 240$$) 

Therefore, we can say that $$x+z = 240 - y = 210$$. We are also given that x ≥ z, 

Hence, $$x_{min}$$ = 210/2 = 105. 

Time taken to cover first 50 km at 100 km/hr = $$\frac{1}{2}$$ hr.
Time taken to cover second 50 km at 50 km/hr = 1 hr.
Time taken to cover last 50 km at 25 km/hr = 2 hr.
When car 2 starts, car 1 has already covered 20 km.
So, time taken by car 1 to reach B after car 1 starts = total time - time required to travel first 20 km
= 3 hr 30 min - 12 min = 3 hr 18 min
Distance travelled by car 2 = (50 + 50 + 45) = 145 km
Distance from B = (150 - 145) km = 5 km

Hence, 5 is the correct answer.

Let 'a' and 'b' are those two numbers. 

$$\Rightarrow$$ $$a^2+b^2 = 97$$

$$\Rightarrow$$ $$a^2+b^2-2ab = 97-2ab$$

$$\Rightarrow$$ $$(a-b)^2 = 97-2ab$$

We know that $$(a-b)^2$$ $$\geq$$ 0

$$\Rightarrow$$ 97-2ab $$\geq$$ 0

$$\Rightarrow$$ ab $$\leq$$ 48.5

Hence, ab $$\neq$$ 64. Therefore, option D is the correct answer.

Let 'a' and 'b' be the length of sides of the rectangle. (a > b)

Area of the rectangle = a*b

Perimeter of the rectangle = 2*(a+b)

$$\Rightarrow$$ $$\dfrac{a*b}{(2*(a+b))^2}=\dfrac{1}{25}$$

$$\Rightarrow$$ $$25ab=4(a+b)^2$$

$$\Rightarrow$$ $$4a^2-17ab+4b^2=0$$

$$\Rightarrow$$ $$(4a-b)(a-4b)=0$$

$$\Rightarrow$$ $$a = 4b$$ or $$\dfrac{b}{4}$$

We initially assumed that a > b, therefore a $$\neq$$ $$\dfrac{b}{4}$$.

Hence, a = 4b

$$\Rightarrow$$ b : a = 1 : 4

Let us draw the diagram according to the available information.

We can see that triangle BPC and BQC are inscribed inside a semicircle. Hence, we can say that

$$\angle$$ BPC = $$\angle$$ BQC = 90°

Therefore, we can say that BQ  $$\perp$$ AC and CP $$\perp$$ AB.

In triangle ABC,

Area of triangle = (1/2)*Base*Height = (1/2)*AB*CP = (1/2)*AC*BQ

$$\Rightarrow$$ BQ = $$\dfrac{AB*CP}{AC}$$ = $$\dfrac{30*20}{25}$$ = 24 cm.

We know that area of the triangle = 32 sq. units, BC = 8 units

Therefore, the height of the perpendicular drawn from point A to BC = 2*32/8 = 8 units.

Let us draw a possible diagram of the given triangle. 

We can see that if A coincide with (-4, 0) then the distance between A and (0, 0) = 4 units.

If we move the triangle up or down keeping the base BC on x = 4, then point A will move away from origin as vertical distance will come into factor whereas horizontal distance will remain as 4 units.

Hence, we can say that minimum distance between A and origin (0, 0) = 4 units. 

Area of the semicircle with AB as a diameter = $$\dfrac{1}{2}*\pi*(\dfrac{AB^2}{4})$$

$$\Rightarrow$$ $$\dfrac{1}{2}*\pi*(\dfrac{AB^2}{4})$$ = $$72*\pi$$

$$\Rightarrow$$ $$AB = 24 cm$$

Given that area of the rectangle ABCD = 768 sq.cm

$$\Rightarrow$$ AB*BC = 768

$$\Rightarrow$$ BC = 32 cm

We can see that the perimeter of the remaining shape = AD + DC + BC + Arc(AB)

$$\Rightarrow$$ 32+24+32+$$\pi*24/2$$

$$\Rightarrow$$ $$88+12\pi$$

If we carefully observe set A, then we find that $$6^{2n} -35n - 1$$ is divisible by 35. So, set A contains multiples of 35. However, not all the multiples of 35 are there in set A, for different values of $$n$$.
For $$n = 1$$, the value is 0, for $$n = 2$$, the value is 1225 which is the 35th multiple of 3.
If we observe set B, it consists of all the multiples of 35 including 0.
So, we can say that every member of set A will be in B while every member of set B will not necessarily be in set A.
Hence, option A is the correct answer.

$$4^{n} > 17^{19}$$

$$\Rightarrow$$ $$16^{n/2} > 17^{19}$$

Therefore, we can say that n/2 > 19

n > 38

Hence, option D is the correct answer. 

Final quantity of alcohol in the mixture = $$\dfrac{700}{700+175}*(\dfrac{90}{100})^2*[700+175]$$ = 567 ml

Therefore, final quantity of water in the mixture = 875 - 567 = 308 ml

Hence, we can say that the percentage of water in the mixture = $$\dfrac{308}{875}\times 100$$ = 35.2 %

Let f(x) = $$2x^2−ax+2$$. We can see that f(x) is a quadratic function. 

For, f(x) > 0, Discriminant (D) < 0

$$\Rightarrow$$ $$(-a)^2-4*2*2<0$$

$$\Rightarrow$$ (a-4)(a+4)<0

$$\Rightarrow$$ a $$\epsilon$$ (-4, 4)

Therefore, integer values that 'a' can take = {-3, -2, -1, 0, 1, 2, 3}

Let g(x) = $$x^2−bx+8$$. We can see that g(x) is also a quadratic function. 

For, g(x)≥0, Discriminant (D) $$\leq$$ 0

$$\Rightarrow$$ $$(-b)^2-4*8*1<0$$

$$\Rightarrow$$ $$(b-\sqrt{32})(b+\sqrt{32})<0$$

$$\Rightarrow$$ b $$\epsilon$$ (-$$\sqrt{32}$$, $$\sqrt{32}$$)

Therefore, integer values that 'b' can take = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

We have to find out the largest possible value of $$2a−6b$$. The largest possible value will occur when 'a' is maximum and 'b' is minimum. 

a$$_{max}$$ = 3, b$$_{min}$$ = -5

Therefore, the largest possible value of $$2a−6b$$ = 2*3 - 6*(-5) = 36. 

We know that $$\dfrac{1}{log_{a}{b}}$$ = $$\dfrac{log_{x}{a}}{log_{x}{b}}$$

Therefore, we can say that $$\dfrac{1}{log_{2}{100}}$$ = $$\dfrac{log_{10}{2}}{log_{10}{100}}$$

$$\Rightarrow$$ $$\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$$

$$\Rightarrow$$ $$\dfrac{log_{10}{2}}{log_{10}{100}}$$-$$\dfrac{log_{10}{4}}{log_{10}{100}}$$+$$\dfrac{log_{10}{5}}{log_{10}{100}}$$-$$\dfrac{log_{10}{10}}{log_{10}{100}}$$+$$\dfrac{log_{10}{20}}{log_{10}{100}}$$-$$\dfrac{log_{10}{25}}{log_{10}{100}}$$+$$\dfrac{log_{10}{50}}{log_{10}{100}}$$

We know that $$log_{10}{100}=2$$

$$\Rightarrow$$ $$\dfrac{1}{2}*[log_{10}{2}-log_{10}{4}+log_{10}{5}-log_{10}{10}+log_{10}{20}-log_{10}{25}+log_{10}{50}]$$

$$\Rightarrow$$ $$\dfrac{1}{2}*[log_{10}{\dfrac{2*5*20*50}{4*10*25}}]$$

$$\Rightarrow$$ $$\dfrac{1}{2}*[log_{10}10]$$

$$\Rightarrow$$ $$\dfrac{1}{2}$$

Given that, p$$^{3}$$ = q$$^{4}$$ = r$$^{5}$$ = s$$^{6}$$

p$$^{3}$$=s$$^{6}$$

p = s$$^{\frac{6}{3}}$$ = s$$^{2}$$   ...(1)

Similarly, q = s$$^{\frac{6}{4}}$$ = s$$^{\frac{3}{2}}$$   ...(2)

Similarly, r = s$$^{\frac{6}{5}}$$   ...(3)

$$\Rightarrow$$ $$log_{s}{(pqr)}$$ 

By substituting value of p, q, and r from equation (1), (2) and (3) 

$$\Rightarrow$$ $$log_{s}{(s^{2}*s^{\frac{3}{2}}*s^{\frac{6}{5}})}$$ 

$$\Rightarrow$$ $$log_{s}(s^{\frac{47}{10}})$$ 

$$\Rightarrow$$ $$\dfrac{47}{10}$$ 

Hence, option A is the correct answer. 

It is given that in drum 1, A and B are in the ratio 18 : 7. 

Let us assume that in drum 2, A and B are in the ratio x : 1.

It is given that drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7.

By equating concentration of A 

$$\Rightarrow$$ $$\dfrac{3*\dfrac{18}{18+7}+4*\dfrac{x}{x+1}}{3+4} = \dfrac{13}{13+7}$$

$$\Rightarrow$$ $$\dfrac{54}{25}+\dfrac{4x}{x+1} = \dfrac{91}{20}$$

$$\Rightarrow$$ $$\dfrac{4x}{x+1} = \dfrac{239}{100}$$

$$\Rightarrow$$ $$x = \dfrac{239}{161}$$

Therefore, we can say that in drum 2, A and B are in the ratio $$\dfrac{239}{161}$$ : 1 or 239 : 161. 

Let 'R' and 'G' be the amount of work that Ramesh and Ganesh can complete in a day. 

It is given that they can together complete a work in 16 days. Hence, total amount of work = 16(R+G) ... (1)

For first 7 days both of them worked together. From 8th day, Ramesh worked at 70% of his original efficiency whereas Ganesh worked at his original efficiency. It took them 17 days to finish the same work. i.e. Ramesh worked at 70% of his original efficiency for 10 days.

$$\Rightarrow$$ 16(R+G) = 7(R+G)+10(0.7R+G)

$$\Rightarrow$$ 16(R+G) = 14R+17G

$$\Rightarrow$$ R = 0.5G ... (2)

Total amount of work left when Ramesh got sick = 16(R+G) - 7(R+G) = 9(R+G) = 9(0.5+G) = 13.5G

Therefore, time taken by Ganesh to complete the remaining work = $$\dfrac{13.5G}{G}$$ = 13.5 days.

Amount of interest paid by Ishan to Gopal if the borrowed amount is Rs. (X+Y) = $$\dfrac{10}{100}*(X+Y)$$ = 0.1(X+Y)

Gopal also borrowed Rs. X from Ankit at 8% per annum. Therefore, he has to return Ankit Rs. 0.08X as the interest amount on borrowed sum. 

Hence, the interest retained by gopal = 0.1(X+Y) - 0.08X = 0.02X + 0.1Y   ... (1)

It is given that the net interest retained by Gopal is the same as that accrued to Ankit.

Therefore, 0.08X = 0.02X + 0.1Y 

$$\Rightarrow$$ X = (5/3)Y   ... (2)

Amount of interest paid by Ishan to Gopal if the borrowed amount is Rs. (X+2Y) = $$\dfrac{10}{100}*(X+2Y)$$ = 0.1X+0.2Y

In this case the amount of interest retained by Gopal = 0.1X+0.2Y - 0.08X = 0.02X + 0.2Y   ... (3)

It is given that the interest retained by Gopal increased by Rs. 150 in the second case. 

$$\Rightarrow$$ (0.02X + 0.2Y) - (0.02X + 0.1Y) = 150

$$\Rightarrow$$ Y = Rs. 1500

By substituting value of Y in equation (2), we can say that X = Rs. 2500

Therefore, (X+Y) = Rs. 4000.

Let 'a', 'b' and 'c' be the concentration of salt in solutions A, B and C respectively. 

It is given that three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.

$$\Rightarrow$$ $$\dfrac{a+2b+3c}{1+2+3} = 20$$

$$\Rightarrow$$ $$a+2b+3c = 120$$ ... (1)

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.

$$\Rightarrow$$ $$\dfrac{3a+2b+c}{1+2+3} = 30$$

$$\Rightarrow$$ $$3a+2b+c = 180$$ ... (2)

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

$$\Rightarrow$$ $$b+2c = 45$$

$$\Rightarrow$$ $$b = 45 - 2c$$

Also, on subtracting (1) from (2), we get

$$a - c = 30$$

$$\Rightarrow$$ $$a = 30 + c$$

In solution D, B and C are mixed in the ratio 2 : 7

So, the concentration of salt in D = $$\dfrac{2b + 7c}{9}$$ = $$\dfrac{90 - 4c + 7c}{9}$$ = $$\dfrac{90 + 3c}{9}$$

Required ratio = $$\dfrac{90 + 3c}{9a}$$ = $$\dfrac{90 + 3c}{9 (30 + c)}$$ = $$1 : 3$$

Hence, option B is the correct answer.

Let 'a' and 'b' be the speed (in km/hr) of cars starting from both A and B respectively. 

If they both move in east direction, then B will catch A if and only if b > a. 

Relative speed of both the cars when they move in east direction = (b - a) km/hr

It takes them 7 hours to meet. i.e. they travel 350 km in 7 hours with a relative speed of (b-a) km/hr. 

Hence, (b - a) = $$\dfrac{350}{7}$$ = 50 km/hr.

It is given that $$t_{1}+t_{2}+…+t_{n} = 2n^{2}+9n+13$$, for every positive integer $$n \geq 2$$. 

We can say that $$t_{1}+t_{2}+…+t_{k} = 2k^{2}+9k+13$$   ... (1) 

Replacing k by (k-1) we can say that 

 $$t_{1}+t_{2}+…+t_{k-1} = 2(k-1)^{2}+9(k-1)+13$$   ... (2)

On subtracting equation (2) from equation (1)

$$\Rightarrow$$ $$t_{k} = 2k^{2}+9k+13 - 2(k-1)^{2}+9(k-1)+13$$

$$\Rightarrow$$ $$103 = 4k+7$$

$$\Rightarrow$$ $$k = 24$$