Quantitative Aptitude – Algebra – Quadratic Equations – The minimum possible value

Quantitative Aptitude – Algebra – Quadratic Equations

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Algebra - Quadratic Equations - The minimum possible value
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

Answer

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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CAT 2017 Questions from Quantitative Aptitude – Algebra

Quantitative Aptitude – Algebra – Quadratic Equations – Ques: If x + 1 = x^2 and x > 0, then 2x^4 is
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Quantitative Aptitude – Algebra – Functions – Let f(x) = x^2 and g(x) = 2^x

Quantitative Aptitude – Algebra – Functions

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Algebra - Functions - Let f(x) = x^2 and g(x) = 2^x
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

Answer

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)

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: 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
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Quantitative Aptitude – Algebra – Logarithms – If x is a real number

Quantitative Aptitude – Algebra – Logarithms

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Algebra - Logarithms - If x is a real number
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

Answer

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.

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

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
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 – Number Systems – If the product of three consecutive

Quantitative Aptitude – Number Systems

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Number Systems - If the product of three consecutive
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

Answer

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

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

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)?
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Quantitative Aptitude – Geometry – Triangles – Let P be an interior point

Quantitative Aptitude – Geometry – Triangles

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Geometry - Triangles - Let P be an interior point
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

Answer

16

Solution

From CAT 2017 – Quantitative Aptitude – Geometry – Triangles, we can see that,
Quantitative Aptitude - Geometry - Triangles - Let P be an interior point
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
Answer: 16

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

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.
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Quantitative Aptitude – Algebra – Maxima Minima – If three sides of a rectangular

Quantitative Aptitude – Algebra – Maxima Minima

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Algebra - Maxima Minima - If three sides of a rectangular
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

Answer

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.
Answer: 200

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

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
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Quantitative Aptitude – Geometry – Circles – ABCD is a quadrilateral inscribed

Quantitative Aptitude – Geometry – Circles

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Geometry - Circles - ABCD is a quadrilateral inscribed
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

Answer

90

Solution

From CAT 2017 – Quantitative Aptitude – Geometry – Circles, we can see that,
Quantitative Aptitude - Geometry - Circles - ABCD is a quadrilateral inscribed
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
Answer: 90 degrees

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

CAT 2017 Questions from Quantitative Aptitude – Geometry

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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
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Quantitative Aptitude – Geometry – Coordinate – The points (2, 5) and (6, 3)

Quantitative Aptitude – Geometry – Coordinate

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Geometry - Coordinate - The points (2, 5) and (6, 3)
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

Answer

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)

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

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
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Quantitative Aptitude – Geometry – Mensuration – The base of a vertical pillar

Quantitative Aptitude – Geometry – Mensuration

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Geometry - Mensuration - The base of a vertical pillar
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

Answer

Option (C)

Solution

From CAT 2017 – Quantitative Aptitude – Geometry – Mensuration, we can see that,
Quantitative Aptitude - Geometry - Mensuration - The base of a vertical pillar
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)

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

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
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Quantitative Aptitude – Geometry – Polygons – Let ABCDEF be a regular hexagon

Quantitative Aptitude – Geometry – Polygons

Question

CAT 2017 - Afternoon slot - Quantitative Aptitude - Geometry - Poygons - Let ABCDEF be a regular hexagon
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

Answer

Option (B)

Solution

From CAT 2017 – Quantitative Aptitude – Geometry – Polygons, we can see that,
Quantitative Aptitude - Geometry - Poygons - Let ABCDEF be a regular hexagon
Angle (ABC) = 120 deg
CAT-2017-Quantitative-Aptitude-Geometry-Poygons-Let-ABCDEF-be-a-regular-hexagon-
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.

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

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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