Quantitative Aptitude PYQs

Shortcut Trick
The dimensions of the open box formed will be (24 − 2x), (18 − 2x), and x.
Volume of the box = Length × Breadth × Height = x × (24 − 2x) × (18 − 2x)
By checking the given options, we can substitute x = 4 directly into the volume formula:
Volume = 4 × (24 − 2 × 4) × (18 − 2 × 4)
Volume = 4 × 16 × 10 = 640 cm3.
Since this perfectly matches the given volume, we do not need to check further.
∴ The correct answer is 4 cm.

Alternate Method
Given:
Length of the rectangular sheet = 24 cm
Breadth of the rectangular sheet = 18 cm
Side of the square cut from each corner = x cm
Volume of the open box = 640 cm3

Formula Used:
Volume of a cuboid = Length × Breadth × Height
Calculation:
⇒ When squares of side x are cut from the 4 corners, the remaining sheet is folded up to form an open box.
⇒ Length of the box (L) = 24 − 2x
⇒ Breadth of the box (B) = 18 − 2x
⇒ Height of the box (H) = x
⇒ Volume = (24 − 2x) × (18 − 2x) × x = 640
⇒ Taking 2 common from the first two terms:
⇒ 2 × (12 − x) × 2 × (9 − x) × x = 640
⇒ 4x × (12 − x) × (9 − x) = 640
⇒ x(12 − x)(9 − x) = 160
⇒ Since x must be less than 9 (otherwise breadth becomes negative or zero), we test the options.
⇒ If x = 4: 4 × (12 − 4) × (9 − 4) = 4 × 8 × 5 = 160
⇒ This satisfies the algebraic equation perfectly.
∴ The correct answer is 4 cm.

Additional Information
Valid Range for 'x'
Since dimensions cannot be negative, the cut side x must be strictly less than half of the smaller side of the rectangle. Here, 18 − 2x > 0 implies x < 9 cm.
Total Surface Area of Open Box
The outer surface area of the resulting open box is simply the original area of the rectangular sheet minus the area of the 4 cut-out corner squares: Area = (L × B) − 4x2.
Maximized Volume Concept
To find the specific value of x that maximizes the volume of such a box, we use calculus by setting the first derivative equal to zero (dV/dx = 0).

Given:

Volume of cuboid = 400 cm³

Total surface area = 340 cm²

Formula used:

Volume of cuboid = lbh

Total surface area of cuboid = 2(lb + bh + hl)

Where, l = length, b = breadth and h = height

Calculation:

Volume of cuboid

xyz = 400 cm³        ----(1)

The total surface area of a cuboid

A = 2(xy + yz + zx) = 340 cm²

⇒ xy + yz + zx = 170    ----(2)

Now, the required value

⇒ x-1 + y-1 + z-1 = \(\frac{1}{x}\) + \(\frac{1}{y}\) + \(\frac{1}{z}\)

Using equation (1) & (2)

⇒ \(\frac{xy + yz + zx}{xyz}\) = \(\frac{170}{400}\)

⇒ \(\frac{xy + yz + zx}{xyz}\) = \(\frac{17}{40}\)

⇒  x-1 + y-1 + z-1  = \(\frac{17}{40}\)

∴ The required value is \(\frac{17}{40}\).

Formula used:

Area of trapezium = \(\frac{1}{2}\) × sum of parallel sides × height

Calculation:

In ΔAED, by Pythagoras theorem,

⇒ AD² = AE² + DE²

⇒ (8.5)² = (4)² + DE²

⇒ 72.25 = 16 + DE²

⇒ DE² = 56.25

⇒ DE = 7.5 cm

Area of trapezium = \(\frac{1}{2}\) × sum of parallel sides × height

⇒ A = \(\frac{1}{2}\)(29 + 21) × 7.5 cm² 

⇒ A = \(\frac{1}{2}\) × 50 × 7.5 cm²

⇒ A = 187.5 cm²

∴ The required area is 187.5 square cm.

Given:

The side of the square plate = 16 cm

Diameter = 14 cm

Weight of copper plate = 128 gm

Formula used:

Area of square = (side)²

Area of circle = \(\frac{\pi}{4}d^2\) 

Calculation:

Using the above concept

Area of square

A₁ = (side)²

A₁ = (16)² = 256 cm²

Now, the area of the circular disc

A₂ = \(\frac{\pi}{4}(14)^2 = \frac{22}{4\times7}\times 14^2\) cm²

⇒ A₂ = 154 cm²

The area of the remaining part of the plate = A₁ - A₂

⇒ A = (256 - 154) cm² = 102 cm²

The weight of 256 cm2 plate = 128 gm

The weight of 102 cm² of the plate = \(\frac{128\times 102}{256}\) = 51 gm

∴ The weight of the remaining part is 51 gm.

Shortcut Trick

Area = (d₁ × d₂) ÷ 2 ⇒ 96 = (12 × d₂) ÷ 2 ⇒ d₂ = 16 cm.

Side (a) = √[(d₁/2)² + (d₂/2)²] = √[6² + 8²] = 10 cm.

Perimeter = 4 × side = 4 × 10 = 40 cm.

∴ The correct answer is 40 cm.

Alternate Method

Given: Area of rhombus = 96 cm², Diagonal 1 (d₁) = 12 cm.

Formula Used: Area = 1/2 × d₁ × d₂; Side (a) = 1/2 × √(d₁² + d₂²); Perimeter = 4a.

⇒ Area = 1/2 × d₁ × d₂

⇒ 96 = 1/2 × 12 × d₂

⇒ 96 = 6 × d₂ ⇒ d₂ = 16 cm

⇒ Side (a) = √[(d₁/2)² + (d₂/2)²]

⇒ a = √[6² + 8²] = √[36 + 64] = √100 = 10 cm

⇒ Perimeter = 4 × a = 4 × 10 = 40 cm

∴ The correct answer is 40 cm.

Additional Information

Properties of Rhombus

All sides are equal and diagonals bisect each other at 90°. The area can also be calculated as base × height.

Relation between Side and Diagonals

4a² = d₁² + d₂², where 'a' is the side and d₁, d₂ are the diagonals.

Pythagorean Triplets

Common triplets like (6, 8, 10) or (3, 4, 5) help in finding the side quickly when half-diagonals are known.

Shortcut Trick

Theorem:

if a right triangle is inscribed in a circle then the hypotenuse will be

the diameter of the circle.

\(\frac{p}{1} = \frac{ab}{c}\)

Calculation:

We know that 

Radius of circle OC = OA = OB = 10 cm

Hypotenuse (AC) = (2 × 10) cm = 20 cm

Now, using the above concept

⇒ \(\frac{8}{1}= \frac{xy}{20}\)

⇒ xy = 160

∴ The required value is 160.

Alternate Method

Area of a triangle with base BC & height AB

A = \(\frac{1}{2}\) xy        -----(1)

Altitude to the hypotenuse BM = 8 cm

From the concept discussed above,

Radius of circle OC = OA = OB = 10 cm

Hypotenuse (AC) = (2 × 10) cm = 20 cm

Since, Area of triangle = 1/2 × base × height

Area of a triangle with base AC & height BM

⇒ A = \(\frac{1}{2}\) × 8 × 20 = 80

From equation (1)

⇒ \(\frac{1}{2}\) xy = 80

⇒ xy = 160

∴ The required value is 160.

Given:

Sum of areas of two circles touching each other externally

= 89π cm2

Distance between the centers = 13 cm

Formula used:

Area of a circle = πx2

Where x = radius of the circle

Calculation:

Let the radii of two circles be R and r

According to the question:

R + r = 13 cm    ----(i)

Also, πR2 + πr2 = 89π

⇒ R2 + r2 = 89

We know that 

(R + r)2 = R2 + r2 + 2Rr

⇒ 169 = 89 + 2Rr

⇒ Rr = 40    ----(ii)

From (i) and (ii), we get

⇒ R(13 - R) = 40

⇒ R2 - 13R + 40 = 0

⇒ (R - 8)(R - 5) = 0

⇒ R = 8, 5; r = 5, 8

∴ Difference between their radii = 3 cm

Shortcut Trick

Let the rectangle's length and breadth be (x + d) and (x − d) cm.

Perimeter = 2(x + d + x − d) = 4x. For the same perimeter, the square's side is x.

Area difference = Area of Square − Area of Rectangle = x² − (x + d)(x − d) = d².

Given d² = 1 ⇒ d = 1 cm.

Difference between length and breadth = (x + d) − (x − d) = 2d = 2(1) = 2 cm.

∴ The correct answer is 2 cm.

Alternate Method

Given:

Perimeter of square = Perimeter of rectangle

Difference in Areas = 1 cm²

Formula Used:

Area of square = a², Perimeter of square = 4a

Area of rectangle = l × b, Perimeter of rectangle = 2(l + b)

Algebraic identity: (l + b)² − 4lb = (l − b)²

Calculations:

⇒ Perimeter of square = Perimeter of rectangle

⇒ 4a = 2(l + b)

⇒ a = (l + b) / 2

For a given perimeter, a square encloses a greater area than a rectangle. So, Area of Square − Area of Rectangle = 1

⇒ a² − (l × b) = 1

⇒ ((l + b) / 2)² − lb = 1

⇒ (l² + b² + 2lb) / 4 − lb = 1

⇒ (l² + b² + 2lb − 4lb) / 4 = 1

⇒ (l² + b² − 2lb) / 4 = 1

⇒ (l − b)² / 4 = 1

⇒ (l − b)² = 4

⇒ l − b = √4 = 2 cm

∴ The correct answer is 2 cm.

Additional Information

Maximum Area for Given Perimeter

For a fixed perimeter, a regular polygon (like a square) always encloses the maximum area compared to any non-regular polygon (like a rectangle) of the same perimeter.

Minimum Perimeter for Given Area

Conversely, for a fixed area, a square has the minimum possible perimeter among all valid rectangles.

Direct Area Difference Identity

The difference between the area of a square and a rectangle sharing the same perimeter can be calculated directly using: Area_sq − Area_rect = ((l − b) / 2)².

Given:

Two rectangles are of the same area equal = 480 cm²

They differ in length by 6 cm and breadth by 4 cm

Formula used:

Area of rectangle = l × b

Perimeter of rectangle = 2(l + b)

Where, l = length and b = breadth

Shortcut Trick

For the same figure,

The difference in perimeter = Difference in the sum of sides

⇒ P₁ - P₂ = 2(l + b) - 2(l + 6 + b - 4)

⇒ P1 - P2 = 2(6 - 4) = 4

Alternate Method

Let the length and breadth of two rectangles be l₁b₁ and l₂b₂ respectively.

According to the question

⇒ l₁b₁ = 480         ------(1)

⇒ l₂b₂ = 480         ------(2)

They differ in length by 6 cm and breadth by 4 cm

Then, length of second rectangle (l₂) = (l₁ + 6) cm

Then, breadth of second rectangle (b₂) = (b₁ – 4) cm

Perimeter of first rectangle = 2(l₁ + b₁) 

Perimeter of second rectangle = 2(l₁ + 6 + b₁ – 4)

⇒ 2(l₁ + b₁) + 4 

The difference in their perimeters is

⇒ 2(l1 + b1) – 2(l1 + b1) + 4

⇒ 4 cm

∴ The required difference in their perimeters is 4 cm.

Mistake PointsIf the length of the first rectangle increases then to make the area

the same, the breadth will decrease.

Given:

The external length, breadth & height are 12 cm, 10 cm, & 8 cm

The inner surface area of the box = 376 cm²

Concept used:

For a cuboid of external length, breadth and height l, b & h 

The external surface area = 2(lb + bh + lh)

If t is the thickness of the cuboid then internal dimensions

Length = l - 2t, breadth = b - 2t, height = h - 2t

Shortcut TrickLet thickens of wood = 1 cm

 

Area = 2( 10 × 8 + 8 × 6 + 6 × 10) 

Area  = 2(80 + 48 + 60) = 376 = given area

∴ The thickness of the box is 1 cm. 

Alternate Method

Let the thickness of the box be t cm.

Using the above concept, internal surface area

A_i = 2[(12 - 2t)(10 - 2t) + (10 - 2t)(8 - 2t) + (12 - 2t)(8 - 2t)]

According to the question, A_i = 376

⇒ 2[(12 - 2t)(10 - 2t) + (10 - 2t)(8 - 2t) + (12 - 2t)(8 - 2t)] = 376

⇒ 8[(t² - 11t + 30) + (t² - 10t + 24) + (t² - 9t + 20)] = 376

⇒ 3t² - 30t + 74 = 47

⇒ 3t² - 30t + 27 = 0

⇒ t² - 10t + 9 = 0

⇒ t² - 9t - t + 9 = 0

⇒ t(t - 9) - 1(t - 9) = 0

⇒ (t - 9)(t - 1) = 0

⇒ t = 9 or 1

Therefore, from the given options

∴ The thickness of the box is 1 cm.

Given: 

X, Y and Z are three equilateral triangles.

The side of X = 6 cm

The side of Y = 8 cm

Concept used:

Area of equilateral triangle = \(\frac{√3}{{4}}\)a²

Calculation:

Area of equilateral triangle = \(\frac{√3}{{4}}\)a²

According to the question

A_X + A_Y = A_Z

⇒ (\(\frac{√3}{{4}}\) × 6²) + (\(\frac{√3}{{4}}\) × 8²) = (\(\frac{\sqrt3}{{4}}\) × x²)

⇒ 36 + 64 = x²

⇒ 100 = x²

⇒ x = 10

∴ The side length of Z is 10 cm.

Given:

Diameter of a circular park = 210 m

Speed of the person = 6 km/hr

Concept:

Convert km/h into m/min multiply by 50/3

Formula Used:

Distance = Speed × Time

Circumference of circle = 2πr

Calculation:

We have,

Diameter of a circular park = 210 m

Radius of a circular park = 105 m

Circumference of circle = \(2 \times \frac{22}{7} \times 105\)

Circumference = 660 m

Distance = 660 m

Speed = \(6km/h \times \frac{50}{3}\) = 100m/min

⇒ Time = Distance/Speed

⇒ T = \(\frac{660}{100}\)

⇒ T = 6.6 min

∴ Time taken by a person to cover one round of circular park is 6.6 min.

Given:

Side of an equilateral triangle = \(2\sqrt3 \)cm

A circle is inscribed in a triangle.

Formula Used:

 

Radius of inscribed circle(r) = \(\frac{side}{2\sqrt3}\)

Area of equilateral triangle = \(\frac{\sqrt3}{4}(side)^2\)

Area of circle = πr²

Calculation:

We have,

Radius of circle = \(\frac{2\sqrt3}{2\sqrt3}\) = 1 cm

Area of shaded region = Area of an equilateral triangle - Area of a circle

Area of shaded region = \(\frac{\sqrt3}{4}(2\sqrt3)^2 - \pi (1)^2 \)

Area of shaded region = \(\frac{\sqrt3}{4} \times 4 \times 3 - \pi \)

Area of shaded region = (\(3\sqrt3 - \pi \)) cm²

∴ Area of the remaining portion of the triangle is (\(3\sqrt3 - \pi \)) cm2

Given:

A triangle has side lengths x cm, x + 13 cm and x + 26 cm

Area of triangle = 126 square cm

Formula Used:

Heron's formula:- Δ = \(\sqrt{s(s - a)(s - b)(s-c)} \)

Perimeter:- s = \(\frac{a + b + c}{2}\)

Calculation:

We have,

⇒ Δ = \(\sqrt{s(s - a)(s - b)(s-c)} \)

⇒ a = x, b = x + 13, c = x + 26

By option selection method, 

Put x = 15 

⇒ a = 15, b = 28, c = 41

⇒ s = \(\frac{15 \ + \ 28 \ + \ 41}{2}\) = 84/2 = 42 

⇒ Δ = \(\sqrt{42(42 - 15)(42 - 28)(42 - 41)}\)

⇒ Δ = \(\sqrt{42 × 27 × 14 × 1}\)

⇒ Δ = \(\sqrt{14 × 81 × 14}\)

⇒ Δ = 14 × 9

⇒ Δ = 126 cm²

∴ The value of x is 15 cm.

Given:

ABCD is a parallelogram.

A circle through A, B and C intersects CD (produced) at E. 

Concept used:

Theorem:

1. In Parallelogram ABCD, opposite angles are always equal.
2. In a cyclic quadrilateral, the sum of a pair of opposite angles is 180°
3. The sum of angles of a linear pair is always equal to 180°.
4. The angles opposite to the equal sides are equal in measure.

 

Calculation:

∠ ABC = ∠ ADC = θ         (Theorem 1)

∠ABC + ∠AED = 180°      (Theorem 2)

⇒ ∠ AED = 180° - θ        ------(1)

∠ADE + θ = 180°             (Theorem 3) 

⇒ ∠ADE = 180° - θ        ------(2)

From equation (1) &  (2)

∠AED = ∠ADE

Statement: 1  

We know that, ∠AED = ∠ADE

From theorem 4, 

AE = AD

Hence, statement 1 is correct.

Statement: 2

CD = DE is only possible if D become the centre of the circle and CD, DE radius. but we take D from the centre, parallelogram ABCD will not be possible such that A, B & C lies on the circle and AB = CD & BC = AD

Hence,  statement 2 is incorrect.

∴ Only statement 1 is correct.

Concept:

Centroid:- The point in which the three medians of the triangle intersect.

Centroid of a triangle is the point located at 2/3^rd of the distance from the vertex along the median.

Centre of circumcircle:- The point where the perpendicular bisectors of the sides of that particular triangle intersect.

Cyclic quadrilateral:- A quadrilateral which has all its four vertices lying on a circle.

• Property of cyclic quadrilateral:- The sum of a pair of opposite angles is 180°.

Rectangle:- It is a four-sided polygon, having all the internal angles equal to 90°.

Property of parallelogram:- Opposite angles of a parallelogram are equal.

Property of triangle:- Sum of all angles of a triangle is 180°.

Calculation:

Statement I: In an equilateral triangle, the centroid and center of the circumcircle coincide.

Let G be the centroid of △ABC.

In △BEC and △BFC,

⇒ ∠B = ∠C = 60°

⇒ BC = BC 

⇒ BF = CE   ---- (1)[Since AB = AC and E and F are the mid points]

⇒ △BEC ≅ △BFC

Similarly, △CAF ≅ △CAD

⇒ CF = AD    ------ (2)

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

⇒ AD = BE = CF

⇒ \(\frac{2}{3}AD = \frac{2}{3}BE =\frac{2}{3}CF\)

⇒ GA = GB = GC

G is equidistant from the vertices.

G is the circumcentre of △ABC.

Hence, the centroid and circumcentre coincide.

So, the statement I is correct.

Statement II: Angle bisectors of a cyclic quadrilateral form another cyclic quadrilateral.

We have, 

ABCD is a quadrilateral and PQRS is a quadrilateral formed by the angle bisectors of ∠A, ∠B, ∠C, and ∠D.

In △APD

⇒ ∠PAD + ∠ADP + ∠APD = 180°     ------ (1)

Similarly, In △BQC, 

⇒ ∠QBC + ∠BCQ + ∠BQC = 180°     ------ (2)

Adding (1) and (2), we get

⇒ ∠PAD + ∠ADP + ∠APD + ∠QBC + ∠BCQ + ∠BQC = 180° + 180°

⇒ ∠PAD + ∠ADP + ∠APD + ∠QBC + ∠BCQ + ∠BQC = 360°

But ∠PAD + ∠ADP + ∠QBC + ∠BCQ = \(\frac{1}{2}\)(∠A + ∠B + ∠C + ∠D)

⇒ ∠PAD + ∠ADP + ∠QBC + ∠BCQ = \(\frac{1}{2} \times 360° \) = 180°

⇒ ∠APD + ∠BQC = 360° - 180° = 180°

So, these are the sum of opposite angles of quadrilateral PQRS.

So, PQRS is a cyclic quadrilateral.

Hence, Statement II is correct.

Statement III: Every cyclic parallelogram is a rectangle.

In a cyclic parallelogram ABCD,

⇒ ∠A + ∠C = 180°    ------ (1) [by property of cyclic quad.]

⇒ ∠A = ∠C     ------ (2)

From equation (1) and (2)

⇒ ∠A + ∠A = 180°

⇒ 2∠A = 180°

⇒ ∠A = 90°

⇒ ∠A = ∠C = 90°

Similarly, ∠ A = ∠B = ∠C = ∠D = 90°

Hence, cyclic parallelogram is a rectangle.

So, statement III is correct.

∴ All the statements are correct.

Given:

ABCD is a trapezium in which AB is parallel to DC. 

The vertices A, B, C and D pass through a circle. 

Concept used:

In cyclic quadrilateral, the sum of a pair of opposite angles is 180°

If two lines are parallel, Then the angles on the same side of the transversal are 180° .

Calculation:

Let ∠BCD = 110°

Using the above concept

∠BCD + ∠DAF = 180° 

⇒ 110° + ∠DAF = 180°

⇒ ∠DAF = 70° 

Draw ⊥ CE &  DF on AB

In ΔDAF, ∠AFD = 90° 

⇒ 70° + 90° + ∠ADF = 180° 

⇒ ∠ADF = 20° 

Simiarly,  ∠BCF = 70°  

Statement 1.

⇒ AD = BC

⇒ ∠F = ∠E  [90°]

⇒ DF = CE

So, ΔADF ~ ΔBCE

⇒  AD = BC

So, statement 1 is true.

Statement 2.

As discussed above in cyclic quadrilateral ABCD, the sum of a pair

of opposite angles is 180°

⇒ ∠A + ∠C = 180°

So, statement 2 is true.

Statement 2.

⇒ ∠A + ∠D = (70° + 110°)

⇒ ∠A + ∠D = 180° 

So, statement 3 is also true.

∴ The required answer is 1, 2 and 3.

Concept:

Similar Triangle:

• Both have the same shape but sizes may be different
• Each pair of corresponding angles are equal.
• The ratio of corresponding sides is the same.

 

Cyclic Quadrilateral: A cyclic quadrilateral has opposite angles that sum 180 degrees, the total sum of the angles equals 360 degrees.

Calculation:

Statement I: ΔEBC is similar to ΔEAD.

In △EBC and △EAD, 

⇒ ∠E = ∠E    [common]

From the figure

 ∠EBC = α, ∠EAD = θ   

So, ∠EBC ≠ ∠EAD   

We know that Each pair of corresponding angles of a similar triangle should

be equal.

Therefore,

But △EBC ~ △EAD is not correct   

Statement II: 

⇒ ∠CBE + ∠DAE = α + θ 

We know that 

α + θ  = 180° will be possible if ABCD is square. In this case, line AB & CD

will not intersect at E.

α + θ  ≠ 180°

Hence, Statement II is incorrect.

∴ Both statement I & II are incorrect.

Concept:

Area of similar triangle Theorem:

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Calculation:

We have,

⇒ DE || BC

⇒ ar(△ADE) = ar(△trap.DEBC)

⇒ ar(△ADE) = ar(△ABC) - ar(DBCE )

⇒ 2ar(△ADE) = ar(△ABC)      ------ (1)

In △ADE and △ABC

⇒ ∠A = ∠A    [common]

⇒ ∠ADE = ∠ABC    [corresponding angles]

⇒ △ADE ~ △ABC    [by AA similarity]

By area of similar triangle theorem

⇒ \(\frac{ar(△ADE)}{ar(△ABC)} = \frac{AD^2}{AB^2}\)

⇒ \(\frac{ar(△ADE)}{2ar(△ADE)} = \frac{AD^2}{AB^2}\)      by using (1)

⇒ \(\frac{1}{2} = \frac{AD^2}{AB^2}\)

Taking square root

⇒ \(\frac{1}{\sqrt2} = \frac{AD}{AB}\)

⇒ \(\frac{1}{\sqrt2} = \frac{AB \ - \ BD}{AB}\)

⇒ \(\frac{1}{\sqrt2} = 1 - \frac{BD}{AB}\)

⇒ \(\frac{BD}{AB} = 1 - \frac{1}{\sqrt2}\)

⇒ \(\frac{BD}{AB} = \frac{\sqrt2 \ - \ 1}{\sqrt2}\)

∴ The ratio of BD/AB is \(\frac{\sqrt2 \ - \ 1}{\sqrt2}\)