Quantitative Aptitude – Algebra - Quadratic Equations – If a and b are integers such that 2x^2
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?
Answer: 36
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
(
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Quantitative Aptitude – Algebra - Logarithms – 1/log(base2)100 -1/log(base4)100 +1/log(base5)100
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
Answer: a) ½
Solution:
Using log(basea)b = 1/log(baseb)a
1/log(base2)100 -1/log(base4)100 +1/log(base
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Quantitative Aptitude – Algebra - Logarithms – If p^3 = q^4 = r^5 = s^6
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_spqr is equal to?
a) 24/5
b) 16/5
c) 47/10
d) 1
Answer: c) 47/10
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
Other p
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Quantitative Aptitude – Algebra - Inequalities – If N and x are positive integers
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?
Answer: 10
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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Quantitative Aptitude – Algebra - Functions – Let f(x)=max{5x, 52-2x^2}
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?
Answer: 20
Solution:
For f(x) to be minimum , 5 = 52 − 2^2
2^2+5 −52 =0
−42+13=0
= 4
Thus minimum value of f(x) = 5x = 5*4 = 20
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Permutation and Combination – Fundamental Principle
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Quantitative Aptitude – Algebra - Inequalities – The smallest integer n such that n^3
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?
Answer: 8
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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Permutation and Combination – Fundamental Principle of
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Quantitative Aptitude – Algebra - Logarithms – If x is a positive quantity such that 2^x
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 log2 = log(base5) 2 log 3 = (log 2
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Quantitative Aptitude – Algebra - Polynomials – If u^2 + (u-2v-1)^2
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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Quantitative Aptitude – Algebra - Polynomials – Given that x^2018 y^2017
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
Answer: a) 33/4
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------
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Quantitative Aptitude – Algebra - Functions – If f(x + 2) = f(x) + f(x + 1)
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?
Answer: 54
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) =
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