Quantitative Aptitude – Modern Maths – Progressions – If a1 = 1/(2*5), a2 = 1/(5*8)

Quantitative Aptitude – Modern Maths – Progressions

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Modern Maths - Progressions - If a1 = 1(25), a2 = 1(58)
If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is

A) 25/151
B) 1/2
C) 1/4
D) 111/55

Answer

Option (A)

Solution

From CAT 2017 – Quantitative Aptitude – Modern Maths – Progressions, we can see that,
The 100th term will be 1/299*302
The series is:
1/(2*5) + 1/(5*8) + 1/(8*11) + ……………+ 1/(299*302)
It can also be written as
3[1/2 – 1/5 + 1/5 – 1/8 + ………………………+ 1/299 – 1/302]
3(1/2 – 1/302) = 300/(3*2*302)
25/151
Option (A)

Download CAT 2017 Question Paper with answers and detailed solutions in PDF

CAT 2017 Questions from Quantitative Aptitude – Modern Maths

Quantitative Aptitude – Modern Maths – Progressions – Q1: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is
Quantitative Aptitude – Modern Maths – Progressions – Q2: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.
Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?
Quantitative Aptitude – Modern Maths – Progressions – Q4: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?
Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?
Quantitative Aptitude – Modern Maths – P&C – Q3: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?
Quantitative Aptitude – Modern Maths – P&C – Q4: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

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Quantitative Aptitude – Modern Maths – Progressions – If a1 = 1/(2*5), a2 = 1/(5*8)
5 (100%) 51 votes

Quantitative Aptitude – Modern Maths – Progressions – An infinite GP a1, a2, a3,….

Quantitative Aptitude – Modern Maths – Progressions

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Modern Maths - Progressions - An infinite GP a1, a2, a3,....
An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is

A) 1/32
B) 2/32
C) 3/32
D) 4/32

Answer

Option (C)

Solution

From CAT 2017 – Quantitative Aptitude – Modern Maths – Progressions, we can see that,
For any n ≥ 1, an = 3 (a(n+1) + a(n+2) + ……..)
So, a1 = 3 (a2 + a3 + ……) or r = 1/4 and
a1 + a2 + a3 +………… = 4a1/3 = 32
So, a1 = 24
GP = 24, 6, 1.5,…….
a5 = 1.5/16 = 3/32
Option (C)

Download CAT 2017 Question Paper with answers and detailed solutions in PDF

CAT 2017 Questions from Quantitative Aptitude – Modern Maths

Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is
Quantitative Aptitude – Modern Maths – Progressions – Q2: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.
Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?
Quantitative Aptitude – Modern Maths – Progressions – Q4: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?
Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?
Quantitative Aptitude – Modern Maths – P&C – Q3: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?
Quantitative Aptitude – Modern Maths – P&C – Q4: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

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Quantitative Aptitude – Modern Maths – Progressions – An infinite GP a1, a2, a3,….
5 (100%) 52 votes

Quantitative Aptitude – Algebra – Functions – Let f(x) = 2x-5 and g(x) = 7-2x

Quantitative Aptitude – Algebra – Functions

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Algebra - Functions - Let f(x) = 2x-5 and g(x) = 7-2x
Let f(x) = 2x-5 and g(x) = 7-2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if

A) 5/2 < x < 7/2
B) x ≤ 5/2 or x ≥ 7/2
C) x < 5/2 or x ≥ 7/2
D) 5/2 ≤ x ≤ 7/2

Answer

Option (D)

Solution

From CAT 2017 – Quantitative Aptitude – Algebra – Functions, we can see that,
|f(x) + g(x)| = |f(x)| + |g(x)|
Putting value of f(x) and g(x), we get,
|2x-5| + |7-2x| = 2

1st Case: When x<=5/2
-2x + 5 +7 – 2x = 2
=> x=5/2

2nd Case: 5/2 < x < 7/2
On solving, we get, 2=2, which satisfies the condition

3rd Case: x ≥ 7/2
2x-5 – 7+2x = 2
x=7/2
So, the answer should be 5/2<= x <= 7/2
Option (D)

Download CAT 2017 Question Paper with answers and detailed solutions in PDF

CAT 2017 Questions from Quantitative Aptitude – Algebra – Functions

Quantitative Aptitude – Algebra – Functions – Q1: If f(ab) = f(a)f(b) for all positive integers a and b, then the largest possible value of f(1) is
Quantitative Aptitude – Algebra – Functions – Q2: Let f(x) = x^2 and g(x) = 2^x, for all real x. Then the value of f(f(g(x)) + g(f(x))) at x = 1 is
Quantitative Aptitude – Algebra – Functions – Q3: If f(x) = (5x+2)/(3x-5) and g(x) = x^2 – 2x – 1, then the value of g(f(f(3))) is
Quantitative Aptitude – Algebra – Functions – Q4: If f1(x) = x^2 + 11x + n and f2(x) = x, then the largest positive integer n for which the equation f1(x) = f2(x) has two distinct real roots, is
Quantitative Aptitude – Algebra – Functions – Q5: The area of the closed region bounded by the equation | x | + | y | = 2 in the two-dimensional plane is
Quantitative Aptitude – Algebra – Logarithms
Quantitative Aptitude – Algebra – Quadratic Equations
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Quantitative Aptitude – Algebra – Functions – Let f(x) = 2x-5 and g(x) = 7-2x
5 (100%) 56 votes

Quantitative Aptitude – Algebra – Functions – If f(ab) = f(a)f(b) for all positive

Quantitative Aptitude – Algebra – Functions

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Algebra - Functions - If f(ab) = f(a)f(b) for all positive
If f(ab) = f(a)f(b) for all positive integers a and b, then the largest possible value of f(1) is

Answer

1

Solution

From CAT 2017 – Quantitative Aptitude – Algebra – Functions, we can see that,
Let us take the case when a=b=1
So, f(1) = f(1) f(1)
f(1) = [f(1)]^2
f(1)[f(1)-1] = 0
f(1) = 1
So, the maximum value of f(1) = 1
Answer: 1

Download CAT 2017 Question Paper with answers and detailed solutions in PDF

CAT 2017 Questions from Quantitative Aptitude – Algebra – Functions

Quantitative Aptitude – Algebra – Functions – Q1: Let f(x) = 2x-5 and g(x) = 7-2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if
Quantitative Aptitude – Algebra – Functions – Q2: Let f(x) = x^2 and g(x) = 2^x, for all real x. Then the value of f(f(g(x)) + g(f(x))) at x = 1 is
Quantitative Aptitude – Algebra – Functions – Q3: If f(x) = (5x+2)/(3x-5) and g(x) = x^2 – 2x – 1, then the value of g(f(f(3))) is
Quantitative Aptitude – Algebra – Functions – Q4: If f1(x) = x^2 + 11x + n and f2(x) = x, then the largest positive integer n for which the equation f1(x) = f2(x) has two distinct real roots, is
Quantitative Aptitude – Algebra – Functions – Q5: The area of the closed region bounded by the equation | x | + | y | = 2 in the two-dimensional plane is
Quantitative Aptitude – Algebra – Logarithms
Quantitative Aptitude – Algebra – Quadratic Equations
Quantitative Aptitude – Algebra – Maxima Minima
Quantitative Aptitude – Algebra – Inequalities
Quantitative Aptitude – Algebra – Polynomials
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Quantitative Aptitude – Algebra – Functions – If f(ab) = f(a)f(b) for all positive
5 (100%) 52 votes

Quantitative Aptitude – Modern Maths – P&C – How many four digit numbers

Quantitative Aptitude – Modern Maths – Permutation and Combination

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Modern Maths - P&C - How many four digit numbers
How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

Answer

50

Solution

From CAT 2017 – Quantitative Aptitude – Modern Maths – Permutation and Combination, we can see that,
For a number to be divisible by 6, it should have both 2 and 3 as factors.
For 3: The sum of digits should be divisible by 3
For 2: The last digit of the number should be even
There can be three such cases, (0,2,3,4), (0,2,4,6) and (2,3,4,6)

1st case: (0,2,3,4)
When the last digit is ‘0’
The number of combinations can be 3! = 6
When the last digit is 2/4
The number of combinations will be 2*2*1*2 = 8
Total = 14

2nd case: (0,2,4,6)
When the last digit is ‘0’
The number of combinations = 3! = 6
When the last digit is 2/4/6
The number of combinations = 2*2*1*3 = 12
Total = 18

3rd case: (2,3,4,6)
Last digit has to be 2/4/6
So, the number of combinations = 3*2*1*3 = 18
Total number of combinations = 14+18+18 = 50
Answer: 50

Download CAT 2017 Question Paper with answers and detailed solutions in PDF

CAT 2017 Questions from Quantitative Aptitude – Modern Maths

Quantitative Aptitude – Modern Maths – P&C – Q1: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?
Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?
Quantitative Aptitude – Modern Maths – P&C – Q3: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?
Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is
Quantitative Aptitude – Modern Maths – Progressions – Q2: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.
Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?
Quantitative Aptitude – Modern Maths – Progressions – Q4: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Quantitative Aptitude – Modern Maths – Progressions – Q5: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is

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Quantitative Aptitude – Modern Maths – P&C – How many four digit numbers
5 (100%) 51 votes

Quantitative Aptitude – Modern Maths – P&C – In how many ways can 8 identical

Quantitative Aptitude – Modern Maths – Permutation and Combination

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Modern Maths - P&C - In how many ways can 8 identical
In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?

Answer

6

Solution

From CAT 2017 – Quantitative Aptitude – Modern Maths – Permutation and Combination, we can see that,
Number of pens that Amal gets = a+1
Number of pens that Bimal gets = b+2
Number of pens that Kamal gets = k+3
So, (a+1) + (b+2) + (k+3) = 8
a + b + k = 2
So, we get (2 + 3 – 1)C(3-1) = 6 [Based on the formula, (n+r-1)C(r-1)]
Answer: 6

Download CAT 2017 Question Paper with answers and detailed solutions in PDF

CAT 2017 Questions from Quantitative Aptitude – Modern Maths

Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?
Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?
Quantitative Aptitude – Modern Maths – P&C – Q3: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?
Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is
Quantitative Aptitude – Modern Maths – Progressions – Q2: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.
Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?
Quantitative Aptitude – Modern Maths – Progressions – Q4: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Quantitative Aptitude – Modern Maths – Progressions – Q5: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is

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Quantitative Aptitude – Modern Maths – P&C – In how many ways can 8 identical
5 (100%) 53 votes

Quantitative Aptitude – Algebra – Number of integer solutions – 1/a + 1/b = 1/9

Quantitative Aptitude – Algebra – Number of integer solutions

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Algebra - Number of integer solutions - 1a + 1b = 19
How many different pairs (a, b) of positive integers are there such that a ≤ b and
1/a + 1/b = 1/9

Answer

3

Solution

From CAT 2017 – Quantitative Aptitude – Algebra – Number of integer solutions, we can see that,
9(a + b) = ab
ab – 9a – 9b + 81 = 81
(a – 9) (b – 9) = 81 = 34
As a, b > 0 and a ≤ b, there are only 3 ordered pairs, given by a – 9 = 1, 3 or 9 and correspondingly b – 9 = 81, 27, 9.
We have to make sure that we satisfy the condition, a≤b

These are the following pairs of (a,b) that satisfy the condition
(a,b) = (10,90), (12, 36), (18,18)
Answer: 3

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Quantitative Aptitude – Algebra – Number of integer solutions – 1/a + 1/b = 1/9
5 (100%) 52 votes

Quantitative Aptitude – Modern Maths – Progressions – Let a1, a2, a3, a4, a5

Quantitative Aptitude – Modern Maths – Progressions

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Modern Maths - Progressions - Let a1, a2, a3, a4, a5
Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.

If the sum of the numbers in the new sequence is 450, then a5 is

Answer

51

Solution

From CAT 2017 – Quantitative Aptitude – Modern Maths – Progressions, we can see that,
5 consecutive odd numbers are a1 , a2 , a3 , a4 , a5.
5 consecutive even numbers are 2a3 – 8, 2a3 – 6, 2a3 – 4, 2a3 – 2, 2a3
Sum of these 5 numbers = 10a3 – 20 = 450
a3 = 47 and a5 = 51.

Download CAT 2017 Question Paper with answers and detailed solutions in PDF

CAT 2017 Questions from Quantitative Aptitude – Modern Maths

Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is
Quantitative Aptitude – Modern Maths – Progressions – Q2: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is
Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?
Quantitative Aptitude – Modern Maths – Progressions – Q4: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?
Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?
Quantitative Aptitude – Modern Maths – P&C – Q3: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?
Quantitative Aptitude – Modern Maths – P&C – Q4: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

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Quantitative Aptitude – Modern Maths – Progressions – Let a1, a2, a3, a4, a5
4.9 (98.71%) 62 votes

Quantitative Aptitude – Algebra – Logarithms – If log (2^a × 3^b × 5^c)

Quantitative Aptitude – Algebra – Logarithms

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Algebra - Logarithms - If log (2^a × 3^b × 5^c)
If log (2^a × 3^b × 5^c) is the arithmetic mean of log (2^2 × 3^3 × 5), log (2^6 × 3 × 5^7), and log(2 × 3^2 × 5^4), then a equals

Answer

3

Solution

From CAT 2017 – Quantitative Aptitude – Algebra – Logarithms, we can see that,
log (2^a. 3^b. 5^c) = [log (2^2.3^3.5) + log (2^6.3.5^7) + log (2.3^2.5^4)]/3
3 * log (2^a. 3^b. 5^c) = log (2^9.3^6.5^12)
log (2^a. 3^b. 5^c)^3 = log (2^9.3^6.5^12)
log (2^3a. 3^3b. 5^3c) = log (2^9.3^6.5^12)
3a = 9
a=3
Answer: 3

Download CAT 2017 Question Paper with answers and detailed solutions in PDF

Logarithm Concepts Questions and Answers for CAT 2018 Quant Preparation

Q1: If x is a real number such that log(base 3)5 = log(base 5)(2 + x), then which of the following is true?
Check answer of logarithm Q1

Q2: The value of log (base 0.008) √5 + log (base√3) 81 – 7 is equal to
Check answer of logarithm Q2

Q3: Suppose, log(base3)x = log(base12)y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log(base6)G is equal to
Check answer of logarithm Q3

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Quantitative Aptitude – Algebra – Logarithms – If log (2^a × 3^b × 5^c)
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Quantitative Aptitude – Algebra – If 9^(x-1/2) – 2^(2x-2) = 4^x

Quantitative Aptitude – Algebra

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Algebra - If 9^(x-12) – 2^(2x-2) = 4^x
If 9^(x-1/2) – 2^(2x-2) = 4^x – 3^(2x-3) , then x is

A) 3/2
B) 2/5
C) 3/4
D) 4/9

Answer

Option (A)

Solution

From CAT 2017 – Quantitative Aptitude – Algebra, we can see that,
You can solve the question easily by putting in values from the options given.
When we put the value of x as 3/2, it satisfies the equation. So, 3/2 is the correct answer.
Option (A)

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Quantitative Aptitude – Algebra – If 9^(x-1/2) – 2^(2x-2) = 4^x
5 (100%) 54 votes