Quantitative Aptitude PYQs

Concept:

• Congruence triangle: If two or more triangles are congruent, then all of their corresponding angles and sides are as well congruent.
• 'Side-Angle-Side' triangle congruence theorem: If the two sides of a triangle are equal to the two sides of another triangle, and the angle formed by these sides in the two triangles are equal, then these two triangles are congruent.
• Concentric Circles: Concentric circles are circles with a common center.
• The perpendicular from the center of a circle to a chord bisects the chord.

 

Calculation:

Statement: 1

Given that,  AB and AC are two equal chords of the circle.

In △APB and △APC,

⇒ AB = AC       [Given]

⇒ ∠BAP = ∠CAP    [Given]

⇒ AP = AP       [Common]

⇒ △APB ≅ △APC   [by SAS congruence criterion]

⇒ So, by property of congruence triangle we have

⇒ BP = CP and ∠APB = ∠APC

⇒ ∠APB + ∠APC = 180° [Linear pair]

⇒ 2∠APB = 180° [∠APB = ∠APC]

⇒ ∠APB = 90°

Now, BP = CP and ∠APB = 90°

Therefore, AP is the perpendicular bisector of chord BC.

Hence, AP passes through the center O of the circle.

Statement: 2

Draw a perpendicular from the center of the circle OM to the line AD.

We can see that BC and AD are the chords of the smaller and bigger circle respectively.

We know that a perpendicular drawn from the center of the circle bisects the chord.

⇒ BM = MC             ------(1)

⇒ AM = MD             ------(2)

Subtracting (2) from (1), we get

AM − BM = DM − CM

⇒ AB = CD

⇒ AB + BC = CD + bC

⇒ AC = BD

∴ Both 1 and 2 are correct.

Given:

Radius = 25 cm

Length of ​chord = 48 cm

Concept used:

The perpendicular from the center of a circle to a chord bisects the chord.

Pythagoras Theorem: Hypotenuse2 = Base2 + Perpendicular²

Calculation:

OA = OB = 25 cm  (Radius of circle)

AB = 48 cm

PB = AB/2 = 48/2

Using the above concept

PA = PB = 24 cm

Applying Pythagoras theorem in ΔOPB,

OB² = OP² + PB²

⇒ 25² = OP² + 24²

⇒ OP² = 625 - 576

⇒ OP2 = 49

⇒ OP = 7 cm

∴ The length of perpendicular drawn from the centre of the circle to the chord is 7 cm.

Given :

The surface area of a cube = surface area of a sphere

Length of cube = x

The radius of the sphere = y 

Formula used :

1. Surface area of cube = 6 × (side)² 
2. Surface area of sphere = 4πr²
3. Volume of cube = side³
4. Volume of sphere = (4/3)π r³

 

Calculations :

The surface area of a cube = surface area of a sphere

Using equations (1) and (2), we get

⇒ 6x² = 4πy²

⇒ x²/y² = 4π/6

⇒ x2/y² = (4/6) × (22/7)

⇒ x2/y² = 44/21

⇒ \(\frac{x}{y} = \frac{√{44}}{√{21}}\)

If 2x > 3y then, 4x² > 9y²

⇒ 4(√44)² > 9(√21)²

⇒ 4 × 44 > 9 × 21

⇒ 176 < 189 ≠ 2x > 3y

Statement I is incorrect

Using equations (3) and (4), we get

⇒ The volume of cube = (√44)³ 

⇒ 44√44 unit³ = 88√11 unit³

⇒ The volume of sphere = (4/3) × (22/7) × (√21)³

⇒ 88√21 unit³ 

The volume of a sphere is greater than the volume of a cube

∴ Neither statement 1 nor statement 2 is correct.

Shortcut Trick

For a right circular cone, Volume ∝ r²h.

Given the ratio r : h = 3 : 7, let r = 3x and h = 7x.

Volume = (1/3) × (22/7) × (3x)² × (7x) = 66x³.

We are given the volume as 528, so 66x³ = 528 ⇒ x³ = 8 ⇒ x = 2.

The height of the cone is 7x = 7 × 2 = 14 cm.

∴ The correct answer is 14 cm.

Alternate Method

Given:

The ratio of the radius to the height of the cone = 3 : 7

Volume of the right circular cone = 528 cm³

Formula Used:

Volume of a right circular cone = (1/3) × π × r² × h

Calculations:

Let the radius (r) be 3x and the height (h) be 7x.

⇒ Volume = (1/3) × π × (3x)² × 7x = 528

⇒ (1/3) × (22/7) × 9x² × 7x = 528

⇒ 22 × 3x³ = 528

⇒ 66x³ = 528

⇒ x³ = 528 ÷ 66

⇒ x³ = 8

⇒ x = 2

Now, substitute the value of x into the expression for height.

⇒ Height (h) = 7x = 7 × 2 = 14 cm

∴ The correct answer is 14 cm.

Additional Information

Volume of a Right Circular Cone

V = (1/3) × π × r² × h, where r is the base radius and h is the vertical height of the cone.

Curved Surface Area (CSA)

CSA = π × r × l, where l = √(r² + h²) is the slant height of the cone.

Total Surface Area (TSA)

TSA = π × r × (r + l), which is the total sum of the curved surface area and the circular base area.

Given:

Inner diameter = 14 cm

Water flows rate = 154 litres per minute

Concept used:

Volume of a cylinder = πr²h

Here, r = radius

h = height or length

Calculation:

The Volume of water that flows in one hour = (154 × 60) liters. [∵ 1 hour = 60 mins]

Volume in cm³ = (154 × 60 × 1000) cm³ [∵ 1 litre = 1000 cm³]

Inner radius of the pipe = 14/2 = 7 cm.

Let the length of column of water that flows in 1 hour be H cm.

According to the question,

πr²H = (154 × 60 × 1000)

⇒ (22/7) × 7² × H = (154 × 60 × 1000)

⇒ 22 × 7 × H = (154 × 60 × 1000)

⇒ 154H = (154 × 60 × 1000)

⇒ H = 60000 cm or 0.6 km

∴ The speed of water in km/hr is 0.6

Given:

Diameter of the spherical ball of lead = 6 cm

The diameters of the balls are 2 cm, 4 cm and x cm

Concept used:

The volume of a sphere = (4πr³)/3

Calculation:

Radius of the ball = 6/2 = 3 cm

So, volume of the ball = 4/3 × (22/7) × 3³

According to the question,

4/3 × (22/7) × 33 = 4/3 × (22/7) × 13 + 4/3 × (22/7) × 23 + 4/3 × (22/7) × (x/2)3

⇒ 4/3 × (22/7) × 33 = 4/3 × (22/7)[13 + 23 + (x/2)3]

⇒ 27 = [1 + 8 + x³/8]

⇒ 27 = [(72 + x3)/8]

⇒ 216 = (72 + x3)

⇒ 144 = x3

So, 5³ < 144 < 6³

Here we will take 5 as the base value.

Now from the options we will check cube of 5.2

So, 5.2³ = 140.608 which is less than 144

Now 5.4³ = 157.464 which is more than 144

So, from this we can conclude that x is in between 5.2 and 5.4

∴ Required answer is 5.2 cm < x cm < 5.4 cm

Given :

The length of a rectangle is 10 units and 8 units.

This rectangle is split into two squares and two rectangles

Area of square = x square unit

Area of rectangle = y square unit

Formula used :

Area of square = side²    

Area of rectangle = Length × breadth   

Calculations :

The rectangle of the length of 10 units and breadth of 8 units can be split as

y is the area of the rectangle and x is the area of square

Using the above formula

⇒ Area of square  = (5)²

⇒ x = 25 unit²

⇒ Area of rectangle = 5 × 3

⇒ y = 15 unit²  

Therefore, y < x

∴ Only statement 2 is correct.

Formula used:

Area of square = (side)²

Diagonal = √2 × (side)

Calculation:

Let a and A  be the side of n identical square and large square.

Diagonal of square = A√2 

⇒ A√2 = m      (given)

⇒ A = m/√2         

Area of large square = sum of the area of small square

⇒ A² = n × a²

⇒ m²/2 = n × a²      [∵A = m/√2]

⇒ a = m/√(2n)

∴ The side of the smaller square sheet is m/√(2n).

Given:

Length of rectangular sheet = x, Breadth of rectangular sheet = y

Volume of the cylinder formed by rolling the sheet along with its breadth = p

Volume of the cylinder formed by rolling the sheet along with its length = q

Formula Used:

Volume of Cylinder = πr²h 

Circumference = 2πr

Where r and h are the radius and height of the cylinder respectively.

Calculation:

Let, x and y are length and breadth of cylinder

 

According to the question, q = 2p

⇒ \(π\frac{x^2}{4π^2}y = 2π\frac{y^2}{4π^2}x\)

⇒ x²y = 2y²x

∴ x = 2y

Given:

Radius of the base the cylindrical container = 18 cm

Radius of metal balls = 0.9 cm

Water level raised by 3 cm

Concept used:

Volume of cylinder = πr²h

Volume of sphere = (4πr³)/3

Calculation:

Volume of the increased water level = π × 18² × 3

Let n number of metal balls be required

According to the question,

[(4 × π × 0.9³)/3} × n = π × 182 × 3

⇒ n = (324 × 3 × 3 × 10³)/(4 × 93) [∵ 0.9 = 9/10]

⇒ n = (324 × 103)/(4 × 81)

⇒ n = (324 × 103)/324

⇒ n = 103 or 1000

∴ Required number of balls required to raise the water level are 1000

Given:

Area of each smaller equilateral triangle = 9√3 cm2

Concept Used:

Area of equilateral triangle = √3a 2/4

Height of equilateral triangle = √3a/2 

Where, a = side of the equilateral triangle 

Calculation:

Let the side length of each smaller triangle be a cm 

Since the area of each smaller equilateral triangle = 9√3 cm 2

So, √3a2/4 = 9√3

⇒ a = 6 cm 

So, the side of bigger equilateral triangle ABC = 6 × 3 = 18 cm 

So, Height AD of the bigger equilateral triangle = (√3 × 18)/2 = 9√3 cm

∴ The required height of the bigger equilateral triangle is 9√3 cm.

Given:

The perimeter of the rhombus is 100 m

One of the diagonals is 40 m

Formula Used:

Area of a triangle having sides a, b, c is 

A = √s(s - a) (s - b) (s - c)

Where, S (semi perimeter) = (a + b + c)/2

Concept Used:

The sides of the rhombus are equal 

Calculation:

According to the question 

Sides of rhombus = 100/4 = 25 m

So, AB = 25m, AD = 25m, and BD = 40m (given)

S (semi perimeter) = (25 + 25 + 40)/2 = 45 m

⇒  Area of 

 = √s(s - a) (s - b) (s - c)

⇒ Area of ΔABD = √45(45 - 25) (45 - 25) (45 - 40)

⇒ Area of ΔABD = √45 × 20 × 20 × 5

⇒ Area of ΔABD = 300 m²

So, The area of whole rhombus = 300 × 2 = 600 m²

∴ The area of each part is 600 ÷ 4 = 150 m2.

Given:

AB || EF || DC

EF = 10cm

AB - DC = 4cm

Formula Used:

Length of midsegment = \(\frac{1}{2}\)(Sum of parallel sides)

Calculation:

We have

AB - DC = 4cm         ------(1)

Using the  above formula, we can write

EF = \(\frac{1}{2}\)(AB + DC)

⇒  EF = \(\frac{1}{2}\)(4 + DC + DC)      [Fomm equation (1)]

⇒ 20 = 4 + 2DC

⇒ 2DC = 16

⇒ DC = 8cm

Put this value in equation (1), 

⇒ AB - 8 = 4

⇒ AB = 12cm

Hence, the required value 

⇒ AB × DC = 12 × 8

⇒ AB × DC = 96 square cm

∴ AB × DC is 96 sq.cm

Given:

ABCD is a parallelogram 

AB = 15 cm and AD = 8 cm

∠BAD = θ 

Formula Used:

If a and b are adjacent sides and θ is the angle between them

Then the area of parallelogram = ab sinθ 

Calculation:

According to the formula used

Area of parallelogramm ABCD is AD × AB × sinθ 

⇒ Area of ABCD = 8 × 15 × sinθ 

⇒ Area of ABCD = 120 sinθ 

∴ The area of parallelogram ABCD is 120 sinθ cm².

Given:

The perimeter of an isosceles right triangle is \(4\left( {2, + ,√ 2 } \right),{\rm{cm}} \)

Formula Used:

Area of right-angled isosceles triangle = a²/2

Where a = equal sides of the isosceles triangle

Calculation:

Given, AB = BC = x (let)

In ΔABC, ∠B = 90° 

AC² = AB² + BC²

⇒ AC2 = x2 + x2

⇒ AC2 = 2x2

⇒ AC = √2x

So, Perimeter of ABC = AB + BC + CA

⇒ x + x + √2x = 4 (2 + √2)

⇒ 2x + √2x = 4 (2 + √2)

⇒ x (2 + √2) = 4 (2 + √2)

⇒ x = 4

Area of triangle = x²/2

⇒ Area of triangle = 42/2 = 8

∴ The area of triangle is 8 square cm.

Given:

The diagonal of a square is \(12√ 2 ,{\rm{cm}}\) 

The area of an equilateral triangle (AT) is \(64√ 3 \) cm²     ----(i) 

Formula Used:

Perimeter of square = 4x

Area of square = x²

Diagonal of square = √2. x

Where, x = side length of square 

Perimeter of equilateral Δ = 3a 

Area of equilateral Δ = √3a²/4

Calculation:

Let teh length of side of square be x 

And the length of side of equilateral Δ be a

According to the question 

√2. x = 12 √2

⇒ x = 12 

So, Perimeter of square = 4 × 12 = 48 cm   ----(ii) 

Area of square (AS) = 12² = 144 cm²    ----(iii) 

Again according to the question 

√3a2/4 = 64√3

⇒ a2 = 256 cm²

⇒ a = 16 cm

So, Perimeter of equilateral Δ = 3 × 16 = 48 cm     ----(iv) 

From (ii) and (iv), we get 

Perimeter of square = Perimeter of equilateral Δ 

Statemetn 1 is correct

And from (i) and (iii), we get 

\(\frac{A_S }{A_T} = \frac{144} {64√3}\)

⇒ \(\frac{A_S }{A_T} = \frac{3√3} {4}\)

⇒ 4 A_S = 3√3 A_T 

Statement 2 is also correct.

∴ Both the statements are correct.

Concept:

S.A.S. Congruence:- If two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the triangles are congruent.

Median:- A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side.

Calculation:

Statement 1:

In ∆ADB and ∆EDC

AD = DE (by construction)

BD = DC (given) and

∠ADB = ∠EDC (vertically opposite angles)

⇒ ∆ADB = ∆EDC (by SAS congruency property)

⇒ AB = CE (by the congruency of a triangle)

Now in ∆ACE

AC + CE > AE

⇒ AC + AB > AE (∵ AB = CE)

⇒ AC + AB > 2AD 

Hecne, statement 1 is incorrect.

Statement 2:

In the △ABC, D, E, and F are the midpoints of sides BC, CA, and AB respectively.

We know that the sum of two sides of a triangle is greater than twice the median bisecting the third side.

Hence, In △ABD, AD is a median,

⇒ AB + AC > 2AD    -----(1)

Similarly, we get

⇒ BC + AC > 2CF    -----(2)

⇒ BC + AB > 2BE    -----(3)

On adding the above equations, we get

⇒ (AB + AC) + (BC + AC) + (BC + AB)  >  2AD + 2CF + 2BE

⇒ 2(AB + BC + AC) > 2(AD + BE + CF)

⇒  AB + BC + AC  >  AD + BE + CF

∴ Only statement 2 is correct.

Concept:

Mid Point Theorem:- The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.

S.S.S. Congruence:- If all the three sides of one triangle are equivalent to the corresponding three sides of the second triangle, then the two triangles are said to be congruent.

Calculation:

According to the midpoint theorem,

⇒ DE = \(\frac{1}{2}\)AB and DE || AB

⇒ DE = AF = BF           ------(1)

Similarly,

⇒ FD = \(\frac{1}{2}\)AC and FD || AC

⇒ FD = AE = EC            -------(2)

⇒ EF = \(\frac{1}{2}\)BC and EF || BC

⇒ EF = BD = DC           -------(3)

⇒ In △ AEF and △ DEF

⇒ AF = DE         [from(1)]

⇒ AE = DF         [from(2)]

⇒ EF = EF         [common]

⇒ △AEF ≅ △DEF   [by SSS congruency]

⇒ In △FBD and △DEF

⇒ DF = DF        [common]

⇒ BF = DE        [from(1)]

⇒ BD = EF         [from(3)]

⇒ △FBD ≅  △DEF [by SSS congruency]

⇒ In △EDC and △DEF

⇒ DE = DE       [common]

⇒ CE = DF       [from(2)]

⇒ CD = EF       [from(3)]

⇒ △EDC ≅ △DEF [by SSS congruency]

Hence, △DEF is congruent to △AFE, △FBD, and △EDC.

∴ The correct answer is option (4).

Given:

In a triangle ABC, AB = AC

∠ACD = x

Formula Used:

The sum of the linear pair of angles = 180° 

∠A + ∠B + ∠C = 180° 

Calculation:

Since, ΔABC is an isosceles triangle 

So, ∠ABC = ∠ACB   [∵ AB = AC]     ----(i)

According to the figure

∠ACB + ∠ACD = 180° 

⇒ ∠ACB + x = 180° 

⇒ ∠ACB = (180° - x)     ----(ii)

From (i) and (ii), we have 

⇒ ∠ABC = (180° - x)       ----(ii)

Now, In Δ ABC

∠ABC + ∠ACB + ∠BAC = 180° 

⇒ (180° - x) + (180° - x) + ∠BAC = 180° 

⇒ ∠BAC = 180° - 360° + 2x

⇒ ∠BAC = 2x - 180° 

∴ The value of ∠BAC is 2x - 180°.

Given:

ABC is a right-angled triangle, ∠B = 90° 

AC = 2BC, ∠A = x

Formula Used:

The Sum of the angle of a triangle is 180° 

sin30° = 1/2

Calculation:

Let ∠C = y and BC = a 

So, AC = 2a

In ΔABC

sinx = BC/AC

⇒ sinx = a/2a = 1/2 

⇒ sinx = sin30° 

⇒ x = 30°     -----(i)

We know that 

∠A + ∠B + ∠C = 180° 

⇒ 30° + 90° + ∠C = 180° 

⇒ ∠C = 60° 

⇒ y = 60° 

⇒ y = 2 × 30°

From (i)

⇒ y = 2 × x = 2x

∴ The value of ∠C is 2x.