### CAT 2018 – Slot 2 – Quantitative Aptitude – If a and b are integers such that 2x^2 −ax + 2 > 0

Q. 1: If a and b are integers such that 2x^2 −ax + 2 > 0 and x^2 −bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a−6b is?

Given,

2x^2 −ax + 2 > 0

2{ (x-a/4)^2 – a^2/16+1} > 0 ∀ x ∈R

-a^2/16+1 > 0

a ∈{ -3,-2,-1,0,1,2,3}

x^2 −bx + 8 ≥ 0

(x-b/2)^2 – b^2 /4 + 8>0 ∀ x ∈R -b^2 /4 + 8>0

b ∈{-5,-4,-3,-2,-1,0,1,2,3,4,5 }

So largest possible value of 2a – 6b = 3*2 – 6(-5) = 36

##### Q. 2: A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?

Suppose inlets of type A fill A liters per minute and type B fills B liters per minute.

So, capacity of tank = 30(10A + 45B) = 60(8A + 18B)

⇒10A + 45B = 16A + 36B

⇒6A = 9B

⇒A = 1.5B

⇒Capacity of Tank = 30(15B + 45B) = 30*60B = 1800 B

Time taken to fill the tank with 7A and 27B, which is 10.5B and 27B, which is 37.5B = 1800 B / 37.5 B = 48 minutes

Q. 3: If N and x are positive integers such that N^N = 2^160 and N^2 + 2^N is an integral multiple of 2^x, then the largest possible x is?

N^N = (2^5)^32

N^N = 32^32

N=32

32^2 + 2^32 = (2^5)^2 + 2^32

32^2 + 2^32 = 2^10 + 2^32

32^2 + 2^32 = 2^10(1 + 2^22)

Hence, Largest possible value of x is 10.

Q. 4: Let t1, t2,… be real numbers such that t1+t2+…+tn = 2n2+9n+13, for every positive integer n ≥ 2. If tk=103, then k equals?

Given, t(base1)+t(base2)+…+t(base n) = 2n^2+9n+13

So t(base1) + t(base2) = 2×2^2+9×2+13=39

t(base1) + t(base2) + t(base3) = 2×3^2+9×3+13=58 means t(base 3) = 58 – 39 = 19

t(base1) + t(base2) + t(base3)+t(base4) = 2×4^2+9×4+13=81 means t(base4) = 81 – 58 = 23

t(base1) + t(base2) + t(base3)+t(base4) + t(base5) = 2×5^2+9×5+13=108 means t(base4) = 108 – 81 = 27

t(base1) + t(base2) + t(base3)+t(base4) + t(base5) + t(base6) = 2×6^2+9×6+13=139 means t(base4) = 139 – 108 = 31

So we can see that t(base3), t(base4) , t(base5) , t(base6) form an A.P. 19, 23, 27, 31

t(basek) = 4(k+1) + 3

Given , t(basek) = 103

4(k+1) + 3 = 103

k = 24

Q. 5: If p^3 = q^4 = r^5 = s^6, then the value of log_s⁡pqr is equal to?
1. 24/5
2. 16/5
3. 47/10
4. 1

Let p^3 = q^4 = r^5 = s^6=k

So p=k^(1/3), q=k^(1/4), r=k^(1/5) and s=k^(1/6)

Thus log(base s)⁡pqr = log(base(k^(1/6) ))⁡ k^(1/3+1/4+1/5) =6 log(base k)⁡ k^((20+15+12)/60)=6×47/60 log(base k)⁡ k=47/10

##### Checkout Other Questions of CAT 2018 Slot 2 Paper:

Verbal Ability :              |   Q.01- Q.05  |  Q.06- Q.9  |  Q.10- Q.14  |  Q.15- Q.19  |  Q.20- Q.24  |  Q.25- Q.29  |  Q.30 – Q.34  |

Logical Reasoning :    |   Q.29 – Q.32  |

Quantitative Aptitude: |