## Slot -2 – Quantitative Aptitude – Algebra – Functions – Let f(x)=max{5x, 52-2x^2}

Let f(x)=max{5x, 52-2x^2}, where x is any positive real number. Then the minimum possible value of f(x) is?

Solution:
For f(x) to be minimum , 5𝑥 = 52 − 2𝑥^2
2𝑥^2+5𝑥 −52 =0
𝑥−42𝑥+13=0
𝑆𝑜 𝑥 = 4
Thus minimum value of f(x) = 5x = 5*4 = 20

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## Slot -2 – Quantitative Aptitude – Algebra – Inequalities – If N and x are positive integers

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?

Solution: 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.

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## Slot -2 – Quantitative Aptitude – Algebra – Logarithms – If p^3 = q^4 = r^5 = s^6

If p^3 = q^4 = r^5 = s^6, then the value of log_s⁡pqr is equal to?

a) 24/5

b) 16/5

c) 47/10

d) 1

Solution:

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

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## Slot -2 – Quantitative Aptitude – Algebra – Logarithms – 1/log(base2⁡)100 -1/log(base4⁡)100 +1/log(base5⁡)100

1/log(base2⁡)100 -1/log(base4)⁡100 +1/log(base5)⁡100 -1/log(base10)⁡100 +1/log(base20⁡)100 -1/log(base25)⁡100 +1/log(base50)⁡100 —?

a) ½

b) 10

c) -4

d) 0

Solution:
Using log(basea)⁡b = 1/log(baseb⁡)a

1/log(base2⁡)100 -1/log(base4⁡)100 +1/log(base5⁡)100 -1/log(base10)⁡100 +1/log(base20)⁡100 -1/log(base25)⁡100 +1/log(base50)⁡100

= log(base100)⁡2- log(base100)⁡4+ log(base100⁡)5-log(base100⁡)10+ log(base100⁡)20-log(base100⁡)25+log(base100⁡)50

=log(base100)⁡ (2×5×20×50)/(4×10×25)

=log(base(10^2 ))⁡10=1/2

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## Slot -2 – Quantitative Aptitude – Algebra – Quadratic Equations – If a and b are integers such that 2x^2

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?

Solution: 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

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## Slot -2 – Quantitative Aptitude – Algebra – Inequalities – The smallest integer n such that n^3

The smallest integer n such that n^3 – 11n^2 + 32n – 28 > 0 is?

Solution:
Given, n^3 – 11n^2 + 32n – 28 > 0
(n-7) (n-2)^2>0

Therefore n must be greater than 7.
So smallest integral value of n = 8

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## Slot -1 – Quantitative Aptitude – Algebra – Logarithms – If x is a positive quantity such that 2^x

If x is a positive quantity such that 2^x = 3^log(base5)^⁡2 , then x is equal to?

a) log(base5⁡)^9

b) 1 + log(base5)⁡ 3/5

c) log(base5⁡)^8

d) 1 + log(base3) ⁡5/3

Answer: b) 1 + log(base5)⁡ 3/5

Solution:
Given , 2^x = 3^log(base5)⁡2

taking log of both sides ,

x log⁡2 = log(base5) 2 log⁡ 3 = (log⁡ 2 log⁡ 3 )/log⁡5
Or x = log 3/log⁡ 5 = log(base5)⁡ 3 = 1+ log(base5)⁡ 3/5

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## Slot -1 – Quantitative Aptitude – Algebra – Polynomials – If u^2 + (u-2v-1)^2

If u^2 + (u-2v-1)^2 = -4v(u + v), then what is the value of u + 3v?

a) -1/4

b) ½

c) 0

d) ¼

Solution:
Given, u^2 + (u-2v-1)^2 = -4v(u + v)

Or u^2+4vu+4 v^2 +(u-2v-1)^2 = 0

(u+2v)^2+ (u-2v-1)^2= 0

This will be zero only if u = -2v = 2v + 1

Or v = -1/4 & u = ½

So u + 3v = -1/4
Option a) -1/4 is correct.

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## Slot -1 – Quantitative Aptitude – Algebra – Polynomials – Given that x^2018 y^2017

Given that x^2018 y^2017 =1/2 and x^2016 y^2019= 8,the value of x^2 + y^3 is?a) 33/4

b) 35/4

c) 31/4

d) 37/4

Solution:
Given , x^2018 y^2017 =1/2——————1)

x^2016 y^2019= 8——————-2)

By dividing eq 1) with eq 2)

x^2/y^2 =1/16——–3)

By multiplying eq 1) with eq 2)
x^4034 y^4036=(xy)^4034 y^2= 4——–4)

From eq 3) & eq ) 4 we can say x = ½ and y = 2
So x^2 + y^3=1/4+8=33/4

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## Slot -1 – Quantitative Aptitude – Algebra – Functions – If f(x + 2) = f(x) + f(x + 1)

If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals?

Solution:

Given , f(x + 2) = f(x) + f(x + 1)

f(15) = f(13) + f(14)

f(13) + f(14) = 617 —————1)

f(12) + f(13) = f(14) ————-2)

f(11) + f(12) = f(13)————-3)

from eq 1) , 2) & 3)

2f(11) + 3f(12) = 617

Putting f(11) = 91

f(12) = (617 – 2*91)/3 = 145

f(12) = f(11) + f(10)

so f(10) = f(12) – f(11) = 145 – 91 =54