Question

The minimum possible value of the sum of the squares of the roots of the equation x^2 + (a + 3)x – (a + 5) = 0 is

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

Option (C)

Solution

From CAT 2017 – Quantitative Aptitude – Algebra – Quadratic Equations, we can see that,
b and c can be the roots of the given equation.
We have to find, b^2 + c^2 = (b+c)^2 – 2bc
b+c = -(a+3) and bc = -(a+5)
b^2 + c^2 = (a+3)^2 + 2(a+5) = a^2 + 8a + 19
Min value of a quadratic equation = -Discriminant (D)/4*First term
D = b^2 – 4ac = 64 – 76 = -12
Min value = 12/4 = 3
Option (C)

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Question

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

A) 16
B) 18
C) 36
D) 40

Option (C)

Solution

From CAT 2017 – Quantitative Aptitude – Algebra – Functions, we can see that,
f(g(x)) = 2^(2x)
g(f(x)) = 2^((x)^2)
f(f(g(x)) + g(f(x)) = (2^(2x) + 2^(x^2))^2
at x = 1, we get 36
Option (C)

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: 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 – 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
Quantitative Aptitude – Algebra – Simple Equations

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Question

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

A) 0 < x < 3
B) 23 < x < 30
C) x > 30
D) 3 < x < 23

Option (D)

Solution

From CAT 2017 – Quantitative Aptitude – Algebra – Logarithms, we can see that,
Log(base 3)5 lies between 1 and 2 because Log(base 3)3 = 1 and Log(base 3)9 = 2
1 < Log(base 3)5 < 2
So, log(base 5)(2+x) should also lie between 1 and 2
1 < log(base 5)(2+x) < 2
5^1 < 2+x < 5^2
5 < 2+x < 25
3 < x < 23
Option D is the right answer.

Logarithm Concepts Questions and Answers for CAT 2018 Quant Preparation

Q1: 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

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

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

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Question

If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

A) 1777
B) 1785
C) 1875
D) 1877

Option (D)

Solution

From CAT 2017 – Quantitative Aptitude – Number Systems, we can see that,
a (a+1) (a+2) = 15600
The consecutive numbers will be 24, 25 and 26
We have to find,
a^2 + (a+1)^2 + (a+2)^2 = (24)^2 + (25)^2 + (26)^2 = 1877

CAT 2017 Questions from Quantitative Aptitude – Number Systems

Quantitative Aptitude – Number Systems – Q1: If a, b, c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio 2:1, then which one of the following is a possible value of (a + b + c)?
Quantitative Aptitude – Number Systems – Q2: The numbers 1, 2,…,9 are arranged in a 3 X 3 square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.
Quantitative Aptitude – Algebra
Quantitative Aptitude – Arithmetic
Quantitative Aptitude – Modern Maths
Quantitative Aptitude – Geometry

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Question

Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB. If the perpendicular distance of P from each of AB, BC, and CA is 4 (√2 – l) cm, then the area, in sq cm, of the triangle ABC is

16

Solution

From CAT 2017 – Quantitative Aptitude – Geometry – Triangles, we can see that,

PQ = PR = PS = 4(√2-1)
CS = PR
(PC)^2 = (PS)^2 + (CS)^2
On solving, we get, PC = 4√2(√2-1)
So, QC = PC + PQ = 4
Area of a right angled triangle = ½ * Base * Height
So, ½ * AC * BC = ½ * QC * AB
On solving, we get a = 4√2
Area of triangle = ½ * a^2 = 16

CAT 2017 Questions from Quantitative Aptitude – Geometry

Quantitative Aptitude – Geometry – Triangles – Q1: Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively.
Quantitative Aptitude – Geometry – Triangles – Q2: From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC.
Quantitative Aptitude – Geometry – Circles – Q1: ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is
Quantitative Aptitude – Geometry – Circles – Q2: Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC.
Quantitative Aptitude – Geometry – Coordinate – Q1: The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is
Quantitative Aptitude – Geometry – Coordinate – Q2: The shortest distance of the point (½, 1) from the curve y = |x -1| + |x + 1| is
Quantitative Aptitude – Geometry – Mensuration
Quantitative Aptitude – Geometry – Polygons – Ques: Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

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Question

If three sides of a rectangular park have a total length 400 ft, then the area of the park is maximum when the length (in ft) of its longer side is

200

Solution

From CAT 2017 – Quantitative Aptitude – Algebra – Maxima Minima, we can see that,
Let a and b be the two sides of a rectangle.
a + 2b = 400
Area of rectangle = ab (We have to maximize it)
b = (400-a)/2
Put the value of b in area of rectangle.

a(400-a)/2 = (400a-a^2)/2
On differentiating the above equation, we get
400-2a = 0 => a=200
b = 100
Area will be max when length of longer side = 200.

CAT 2017 Questions from Quantitative Aptitude – Algebra

Quantitative Aptitude – Algebra – Maxima Minima – Q1: If a, b, c, and d are integers such that a + b + c + d = 30, then the minimum possible value of (a – b)^2 + (a – c)^2 + (a – d)^2 is
Quantitative Aptitude – Algebra – Maxima Minima – Q2: An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg. What is the maximum possible number of people in the group?
Quantitative Aptitude – Algebra – Functions
Quantitative Aptitude – Algebra – Logarithms
Quantitative Aptitude – Algebra – Quadratic Equations
Quantitative Aptitude – Algebra – Inequalities
Quantitative Aptitude – Algebra – Polynomials
Quantitative Aptitude – Algebra – Simple Equations

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Question

ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is

90

Solution

From CAT 2017 – Quantitative Aptitude – Geometry – Circles, we can see that,

OD = OC (Radius of circle)
So, angle (ODC) = angle (OCD) = 30 deg
Angle (DOA) = 60 degrees
Angle (BAC) = 30 degrees (Given)
OA = OD (radius of circle)
Angle (ODA) = angle (OAD) = 60 deg
Sum of Opposite angles in a cyclic quad are 180 deg
Angle (BAD) + angle (BCD) = 180
So, angle (BCD) = 90 deg

CAT 2017 Questions from Quantitative Aptitude – Geometry

Quantitative Aptitude – Geometry – Circles – Ques: Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC.
Quantitative Aptitude – Geometry – Triangles – Q1: Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB.
Quantitative Aptitude – Geometry – Triangles – Q2: Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively.
Quantitative Aptitude – Geometry – Triangles – Q3: From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is
Quantitative Aptitude – Geometry – Coordinate – Q1: The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is
Quantitative Aptitude – Geometry – Coordinate – Q2: The shortest distance of the point (½, 1) from the curve y = |x -1| + |x + 1| is
Quantitative Aptitude – Geometry – Mensuration
Quantitative Aptitude – Geometry – Polygons – Ques: Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

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Question

The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is

A) -5
B) -6
C) -7
D) -8

Option (D)

Solution

From CAT 2017 – Quantitative Aptitude – Geometry – Coordinate, we can see that,
The diagonals of a rectangle bisect each other. Mid points of the diagonal are (4,4)
These points fall on the line with equation y = 3x + c
Putting the coordinates (4,4) in the equation, we get
c= -8
Option (D)

CAT 2017 Questions from Quantitative Aptitude – Geometry

Quantitative Aptitude – Geometry – Coordinate – Ques: The shortest distance of the point (½, 1) from the curve y = |x -1| + |x + 1| is
Quantitative Aptitude – Geometry – Circles – Q1: ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is
Quantitative Aptitude – Geometry – Circles – Q2: Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC.
Quantitative Aptitude – Geometry – Triangles – Q1: Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB.
Quantitative Aptitude – Geometry – Triangles – Q2: Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively.
Quantitative Aptitude – Geometry – Triangles – Q3: From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is
Quantitative Aptitude – Geometry – Mensuration
Quantitative Aptitude – Geometry – Polygons – Ques: Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

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Question

The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length. The perpendicular distance between the parallel sides of the trapezium is 12 cm. If the height of the pillar is 20 cm, then the total area, in sq cm, of all six surfaces of the pillar is

A) 1300
B) 1340
C) 1480
D) 1520

Option (C)

Solution

From CAT 2017 – Quantitative Aptitude – Geometry – Mensuration, we can see that,

Given, the non-parallel sides are equal. Let the non-parallel sides be x cm each
x= √(12^2 + 5^2) = 13
So, we have 6 faces, out of which 2 are trapezoid faces and 4 are rectangular faces.
Area of trapezium = 1/2(sum of two parallel sides)(height)
Area of 2 trapeziums
= 2[(1/2)(12)(10+20)] = 360 cm^2
Area of rectangle = base*height
Area of 4 rectangles
= 2[13 × 20] + 20(20) + 10(20) = 1120 cm^2
Total area = 1120 + 360 = 1480 cm^2
Option (C)

CAT 2017 Questions from Quantitative Aptitude – Geometry

Quantitative Aptitude – Geometry – Mensuration – Q1: A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9 π cm^3 .
Quantitative Aptitude – Geometry – Mensuration – Q2: A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8: 27: 27.
Quantitative Aptitude – Geometry – Coordinate – Q1: The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is
Quantitative Aptitude – Geometry – Coordinate – Q2: The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length.
Quantitative Aptitude – Geometry – Circles – Q1: ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is
Quantitative Aptitude – Geometry – Circles – Q2: Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC.
Quantitative Aptitude – Geometry – Triangles – Q1: Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB.
Quantitative Aptitude – Geometry – Triangles – Q2: Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively.
Quantitative Aptitude – Geometry – Triangles – Q3: From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is
Quantitative Aptitude – Geometry – Polygons – Ques: Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

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Question

Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

A) 3√2
B) 3
C) 4
D) √3

Option (B)

Solution

From CAT 2017 – Quantitative Aptitude – Geometry – Polygons, we can see that,

Angle (ABC) = 120 deg

According to the formula,
Cos (theta) = (b^2 + c^2 – a^2)/2bc
Cos (120) = [(AB)^2 + (BC)^2 – (AC)^2]/2*AB*BC
-1/2 = [1+1 – (AC)^2]/2
=> On solving, we get, AC = √3
Area of square = (side)^2 = (√3)^2 = 3
Option B is the right answer.

CAT 2017 Questions from Quantitative Aptitude – Geometry

Quantitative Aptitude – Geometry – Mensuration – Q1: The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length.
Quantitative Aptitude – Geometry – Mensuration – Q2: A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9 π cm^3 .
Quantitative Aptitude – Geometry – Mensuration – Q3: A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8: 27: 27.
Quantitative Aptitude – Geometry – Coordinate – Q1: The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is
Quantitative Aptitude – Geometry – Coordinate – Q2: The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length.
Quantitative Aptitude – Geometry – Circles – Q1: ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is
Quantitative Aptitude – Geometry – Circles – Q2: Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC.
Quantitative Aptitude – Geometry – Triangles – Q1: Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB.
Quantitative Aptitude – Geometry – Triangles – Q2: Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively.
Quantitative Aptitude – Geometry – Triangles – Q3: From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is

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