CAT 2018 Question Paper Slot 2

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)

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

$$\Rightarrow$$ $$103 = 4k+7$$

$$\Rightarrow$$ $$k = 24$$