Quantitative Aptitude PYQs

Concept used:

Pythagoras Theorem states,

Hypotenuse² = Base² + Perpendicular²

Calculation:

According to the given data, this diagram is drawn.

Let, v and p are the centers of semi-circle S1 and circle S respectively.

In ΔpvC, ∠ pCv = 90° 

vp = Sum of the radius of S3 and S = (R + r)

pC = Radius of semi-circle S1 - Radius of circle S

⇒ DC - r = (2R - r)     [∵ DC = AC]

So, vp2 = pC2 + Cm2  

⇒ (R + r)2 = (2R - r)2 + R2

⇒ R2 + 2Rr + r2 = 4R2 - 4Rr + r2 + R2

⇒ ⇒ R2 + 2Rr + r2 = 5R2 - 4Rr + r2 

⇒ 5R2 - 4Rr + r2  - R2 - 2Rr - r2 = 0

⇒ 4R2 - 6Rr = 0

⇒ 4R2 = 6Rr 

⇒ 2R = 3r

∴ Option 4 is the correct answer.

Given:

Area of circle S = m

Area of semi-circle S1 = n

Concept used:

Area of a circle = πR²

Area of a semi-circle = (πR2)/2

Where, R = radius, and π = 22/7

Calculation:

Radius of circle S = r

Radius of semi-circle S1 = 2R

Let, v and p are the centers of semi-circle S1 and circle S respectively.

In ΔpvC, ∠ pCv = 90° 

vp = Sum of the radius of S3 and S = (R + r)

pC = Radius of semi-circle S1 - Radius of circle S

⇒ DC - r = (2R - r)     [∵ DC = AC]

So, vp2 = pC2 + Cm2  

⇒ (R + r)2 = (2R - r)2 + R2

⇒ R2 + 2Rr + r2 = 4R2 - 4Rr + r2 + R2

⇒ ⇒ R2 + 2Rr + r2 = 5R2 - 4Rr + r2 

⇒ 4R2 - 6Rr = 0

⇒ 2R = 3r

⇒ r = (2R)/3

Now, \(\frac{m}{n} = \frac{\Pi r^{2}}{\frac{\Pi (2R)^{2}}{2}} \)

⇒ \(\frac{m}{n} = \frac{\Pi (\frac {2R}{3})^{2}}{\frac{\Pi (2R)^{2}}{2}} \) 

⇒ \(\frac{m}{n} = \frac{\frac {4R^{2}}{9}}{\frac{4R^{2}}{2}} \)

⇒ \(m \times {\frac{4R^{2}}{2}} = n \times {\frac {4R^{2}}{9}} \)

⇒ m/2 = n/9

⇒ 9m = 2n

∴ Option 1 is the correct answer.

Concept:

Componendo Dividendo Rule:

If (a + b ) : (a – b) = (c + d) : (c – d), than 

a : b = c : d

Calculation:

According to the question,

\(\rm \frac{(x-a)(x-b)}{(x-m a)(x-m b)}=\frac{(x+a)(x+b)}{(x+m a)(x+m b)}\)

⇒ \((x - a) (x - b) \over (x + a) (x + b) \) = \((x - ma) (x - mb) \over (x + ma) (x + mb)\)

⇒ \(x^2 - (a + b)x + ab \over x^2 + (a + b)x + ab\) = \(x^2 - (ma + mb)x + m^2ab \over x^2 + (ma + mb)x + m^2ab \)

By componendo dividendo method,

⇒ \(x^2 + ab \over (a + b)x \) = \(x^2 + m^2 ab \over (ma + mb)x \)

⇒ \(x^2 + ab \over x^2 + m^2 ab\) = \((a + b) \over (a + b) m\)

⇒ \(x^2 + ab \over x^2 + m^2 ab\) = \(1 \over m\)

∴ The equation \(\rm \frac{x^2+a b}{x^2+m^2 a b}\)  is equal to \(1 \over m\).

Concept:

Componendo Dividendo Rule:

If (a + b ) : (a – b) = (c + d) : (c – d), than 

a : b = c : d

Calculation:

According to the question,

\(\rm \frac{(x-a)(x-b)}{(x-m a)(x-m b)}=\frac{(x+a)(x+b)}{(x+m a)(x+m b)}\)

⇒ \((x - a) (x - b) \over (x + a) (x + b) \) = \((x - ma) (x - mb) \over (x + ma) (x + mb)\)

⇒ \(x^2 - (a + b)x + ab \over x^2 + (a + b)x + ab\) = \(x^2 - (ma + mb)x + m^2ab \over x^2 + (ma + mb)x + m^2ab \)

By componendo dividendo method,

⇒ \(x^2 + ab \over (a + b)x \) = \(x^2 + m^2 ab \over (ma + mb)x \)

⇒ \(x^2 + ab \over x^2 + m^2 ab\) = \((a + b) \over (a + b) m\)

⇒ \(x^2 + ab \over x^2 + m^2 ab\) = \(1 \over m\)

⇒ mx² + mab = x² + m²ab

⇒ mx² - x² = m²ab - mab

⇒ x² (m - 1) = mab (m - 1)

⇒ x² = mab

⇒ x = \(\pm \sqrt{\mathrm{mab}}\)

∴ The value of x is equal to \(\pm \sqrt{\mathrm{mab}}\).

Given:

Electricity bill for 8 rooms and 240 units is Rs. 400

Electricity bill for 6 rooms and 200 units is Rs. 320

Concept used:

The concept of linear equations in two variables is used, considering

the charge per room as one variable and the charge per unit as

another variable.

Calculation:

The equations can be represented as:

8x + 240y = 400 ......(i)

6x + 200y = 320 ......(ii)

Solving these equations:

Subtract equation 2 from equation 1,

⇒ (8x - 6x) + (240y - 200y) = (400 - 320)

⇒ 2x + 40y = 80

⇒ x + 20y = 40

⇒ x = 40 - 20y

Substituting x = 40 - 20y in Equation 2,

⇒ 6(40 - 20y) + 200y = 320

⇒ 240 - 120y + 200y = 320

⇒ 240 + 80y = 320

⇒ 80y = 80

⇒ y = 1

Substituting y = 1 in equation 1,

⇒ 8x + 240 = 400

⇒ 8x = 160

⇒ x = 20

∴ The monthly electricity bill for a house with m rooms and consuming n units is 20m + n.

Given:

Electricity bill for 8 rooms and 240 units is Rs. 400

Electricity bill for 6 rooms and 200 units is Rs. 320

Formula used:

Total cost = cost per room × number of rooms + cost per unit

electricity × number of units consumed

Concept used:

Use Of simultaneous equations to find the cost per room (x) and cost

per unit of electricity (y) and find the total cost by adding the cost per

room and cost per unit of electricity.

Calculation:

The equations can be represented as:

8x + 240y = 400 ......(i)

6x + 200y = 320 ......(ii)

Solving these equations:

Subtract equation 2 from equation 1,

⇒ (8x - 6x) + (240y - 200y) = (400 - 320)

⇒ 2x + 40y = 80

⇒ x + 20y = 40

⇒ x = 40 - 20y

Substituting x = 40 - 20y in Equation 2,

⇒ 6(40 - 20y) + 200y = 320

⇒ 240 - 120y + 200y = 320

⇒ 240 + 80y = 320

⇒ 80y = 80

⇒ y = 1

Substituting y = 1 in equation 1,

⇒ 8x + 240 = 400

⇒ 8x = 160

⇒ x = 20

To find the bill for 7 rooms and 300 units.

Using the formula,

⇒ Total cost = 20 × 7 + 1 × 300

⇒ Total cost = 140 + 300

⇒ Rs. 440

∴ The monthly electricity bill for a house with 7 rooms and consuming 300 units is Rs. 440.

Given:

Average weekly wages earned by a worker = Rs. 3,520

Formula used:

Average Mean = \(\frac{\sum fi \times xi}{\sum fi}\)

x(i) = (L₁ + L₂)/2

L₁ = upper limit &  L₂= lower limit

Where f(i) = median of class  & x(i) = Mean of class

Calculation:

 

3520 = \(\frac{(126500 + 3300k)}{(35 + k)}\)

123200 + 3520k = 126500 + 3300k

220k = 3300

k = 15

∴ Correct answer is 15.

Formula used:

Median = \(L + (\frac{\frac{N}{2} - F}{f}) × h\)

L = lower limit of median class

N = total frequency

F = C.F of class just preceding median class

f = frequency of median class

h = width of median class

Calculation:

 

Here, h = 3550 - 3050 = 500, L = 3550

Median = \(3050 + (\frac{\frac{50}{2} - 15}{15}) × 500\)

Median = 3050 + (10/15) × 500

Median = 3050 + 333.33 = 3383.33 ≈ 3383

∴ The correct answer is 3383.

Formula used:

ax² + bx + c = 0

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Calculation:

x = \( {(2a\ +\ 2b)\ \pm\ \sqrt{4(a\ +\ b)^2\ -\ 4[(a\ +\ b)^2\ -\ c^2]} \over 2(a\ +\ b\ +\ c)}\) 

Where, [(a +b + c)(a +b - c) = (a + b)² - c²]

⇒ x =  \( {(2a\ +\ 2b)\ \pm\ \sqrt{4(a\ +\ b)^2\ -\ 4(a\ +\ b)^2\ +\ 4c^2} \over 2(a\ +\ b\ +\ c)}\)

⇒ x =  \( {(2a\ +\ 2b)\ \pm\ \sqrt{4c^2} \over 2(a\ +\ b\ +\ c)}\)

⇒ x = \( {(2a\ +\ 2b)\ \pm\ {2c} \over 2(a\ +\ b\ +\ c)}\)

⇒ x = \( {a\ +\ b\ +\ {c} \over a\ +\ b\ +\ c}\) = 1, and

x = \( {a\ +\ b\ -\ {c} \over a\ +\ b\ +\ c}\) 

∴ x has two roots 1, \( {a\ +\ b\ -\ {c} \over a\ +\ b\ +\ c}\)

Given:

(a + b + c)x² - (2a + 2b)x + (a + b - c) = 0

Concept:

For quadratic equation Ax² + Bx + C = 0

Sum of roots (α + β) = -B/A , 

Product of root (αβ) = C/A

Calculation:

Statement: 1

If we take a = b = c = 1

⇒ (1 + 1 + 1)x2 - (2 × 1 + 2 × 1)x + (1 + 1 - 1) = 0

3x2 - 4x + 1 = 0

3x2 - x - 3x + 1 = 0

(3x - 1)(x - 1) = 0

x = 1 & 1/3

Hence, statement 1 is correct.

Statement:2

If we take a = b = c = -1

⇒ (-1 -1 -1)x2 - [2 × (-1) + 2 × (-1)]x + [-1 - 1 - (-1)] = 0

⇒ -3x2 + 4x - 1 = 0

⇒ -3x2 + 3x + x - 1 = 0

⇒ -3x(x - 1) + 1(x - 1) = 0

⇒ (-3x + 1)(x - 1) = 0

⇒ x = 1/3 & x = 1

Hence, statement 2 is incorrect.

∴ Only statement 1 is correct

Given:

Edges of cuboid are 8 cm, 12 cm and 24 cm

The sphere passes through the corners of the cuboid

Concept Used:

If the sphere passes through the corners of the cuboid, the diameter of the sphere is the body diagonal of the cuboid.

Formula Used:

Diagonal of Cuboid = \(\sqrt{l^2 +b^2 + h^2} \)

Calculation:

Diagonal of Cuboid = \(\sqrt{8^2 +12^2 + 24^2} \)

Diagonal of Cuboid = \(\sqrt{784} \) = 28 cm

⇒ Diagonal of the Sphere = 28 cm

⇒ Radius of the Sphere = 28 ÷ 2 = 14 cm

∴ The radius of the sphere is 14 cm.

Given:

Circumferences of top and bottom are 30 cm & 40 cm respectively.

The perpendicular distance between the ends is 6 cm.

Calculation:

Here, Ratio of circumference = ST : PQ = 3 : 4 

We know that, Circumference ∝ Radius

Therefore, \(\triangle\)OST ∼ \(\triangle\)OPQ​

⇒ OS : OP = 3 : 4

We know that, SP = 1 unit = 6 cm

⇒ OS = 3 unit = 6 × 3 = 18 cm

∴ The distance of its vertex from the top is 18 cm.

Given:

Rate = 20% per annum

log10 2 = 0.301

log10 3 = 0.477

Formula Used:

1. A = P(1 + R/100)ⁿ

2. log mⁿ  = n log m

3. log mn = log m + log n

Where, 

A = Amount, P = Principal,

R = Rate, n = No. of years

Calculation:

According to the question condition,

100P = P(1 + 20/100)ⁿ = P(1 + 2/10)n

⇒ 10² = (12/10)n 

Taking log on both sides

⇒ 2 log 10 = n log (12/10)

⇒ 2 log 10 = n log (4 × 3) - log 10     

⇒ 2 = n [log 4 + log 3] - 1       [∵ log 10 = 1]

⇒ 2 = n [log 2² + log 3] - 1

⇒ 2 = n [2 log 2 + log 3 - 1]

⇒ 2 = n (2 × 0.301 + 0.477 - 1)          (given)

⇒ 2 = n (0.602 + 0.477 - 1) = 0.079 n

⇒ n = 2/0.079 = 25.32 

To ensure the amount reaches 100 times its initial value, the minimum number of years needed is 26.

This is because rounding off to 25 years would result in a compounded amount that is less than 100 times the initial value.

⇒ n = 26

∴ The value of n is 26.

Given:

Equilateral triangular plate cut to hexagonal plate

Formula used:

Area of equilateral triangle (T) = √3/4 × a²

Area of regular hexagon (H) = (3√3)/2 × a²

Where a =  length of side

Calculation:

Let the side of the equilateral triangle = 3a

       

 

Area of equilateral triangle (T) = √3/4 × (3a)² = (9√3)/4 × a²

Area of regular hexagon (H) = (3√3)/2 × a²

⇒  \(\frac{T}{H} = \frac{\frac{9√3}{4} \ ×\ a^2}{\frac{3√3}{2}\ \times \ a^2}\) \(= \frac{9√3}{4} \times \frac{2}{3\sqrt3}\) = 3/2

Hence, The required ratio is 3 : 2

Shortcut Trick

For two equal arcs of different circles, the radius is inversely proportional to the angle subtended at the center.

Formula: r₁θ₁ = r₂θ₂

⇒ r₁ × 60° = r₂ × 75°

⇒ r₁ / r₂ = 75 / 60

⇒ r₁ : r₂ = 5 : 4

∴ The correct answer is 5 : 4.

Alternate Method

Given:

Two equal arcs are taken from two different circles C1 and C2.

Angle subtended by the arc of C1 at the centre = 60°

Angle subtended by the arc of C2 at the centre = 75°

Let the radii of C1 and C2 be r1 and r2, respectively.

Formula or Concept Used:

Arc length = r × θ

Since the two arcs are equal in length:

⇒ r1 × θ1 = r2 × θ2

Calculation:

⇒ r1 × 60° = r2 × 75°

⇒ r1 ÷ r2 = 75 ÷ 60

⇒ r1 ÷ r2 = 5 ÷ 4

⇒ r1 : r2 = 5 : 4

∴ The required ratio of the radius of C1 to the radius of C2 is 5 : 4.

Additional Information

Arc Length in Radians

When the angle is measured in radians, the formula for arc length simplifies to l = r × θ, where θ is in radians.

Area of a Sector

The area of a sector subtending an angle θ is given by A = (θ / 360) × πr² or A = (1/2) × l × r, where l is the arc length.

Degree to Radian Conversion

To convert degrees to radians, multiply by π / 180. For example, 180° = π radians and 90° = π / 2 radians.

Shortcut Trick

The area of a triangle formed by joining the mid-points of the sides of a triangle is always 1/4^th of the area of the original triangle.

Area of ΔDEF = 1/4 × Area of ΔABC

Semi-perimeter (s) of ΔABC = (41 + 28 + 15) ÷ 2 = 42 cm

Area of ΔABC = √[42 × (42 − 41) × (42 − 28) × (42 − 15)] = √[42 × 1 × 14 × 27] = 126 cm²

Area of ΔDEF = 126 ÷ 4 = 31.5 cm²

∴ The correct answer is 31.5 square cm.

Alternate Method

Given: Sides of ΔABC are a = 28 cm, b = 15 cm, and c = 41 cm. D, E, F are mid-points.

Formula Used: Heron's Formula: Area = √[s(s − a)(s − b)(s − c)], where s = (a + b + c)/2.

⇒ s = (28 + 15 + 41) ÷ 2

⇒ s = 84 ÷ 2 = 42 cm

⇒ Area of ΔABC = √[42 × (42 − 28) × (42 − 15) × (42 − 41)]

⇒ Area of ΔABC = √[42 × 14 × 27 × 1]

⇒ Area of ΔABC = √[(14 × 3) × 14 × (3 × 9)] = √[14² × 3² × 3²]

⇒ Area of ΔABC = 14 × 3 × 3 = 126 cm²

⇒ Area of ΔDEF = (1/4) × Area of ΔABC

⇒ Area of ΔDEF = 126 ÷ 4 = 31.5 cm²

∴ The correct answer is 31.5 square cm.

Additional Information

Mid-point Theorem

The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.

Medians and Area

A median divides a triangle into two triangles of equal area. The four triangles formed by joining mid-points are congruent and have equal areas.

Perimeter Relation

The perimeter of the triangle formed by joining the mid-points is exactly half the perimeter of the original triangle.

Given:

Sides of a triangle = 13 cm, 14 cm, and 15 cm

Formula:

For triangle with 3 sides a, b & c,

Semi-perimeter, S = (a + b + c)/2

Area = \( \sqrt{S(S - a) (S - b)(S - c) } \)

Area = 1/2 × Base × Perpendicular

Concept:

The length of  a triangle's smallest altitude is always opposite its longest side.

Calculation:

Semi-perimeter of the triangle, S = (13 + 14 + 15)/2 = 21 cm

Area of the triangle = \(\sqrt{21(21 - 13) (21 - 14) (21 - 15)}\)

⇒ Area of triangle =\(​​​​\sqrt{21 × 8 × 7 × 6}\) = 84 cm²

Let, the smallest altitude (AD) = h cm

According to the question,

⇒ 1/2 × 15 × h = 84

⇒ h = 84/15 × 2 = 11.2 cm

∴ The smallest altitude of the triangle is 11.2 cm.

Shortcut Trick

Difference between Circumference and Diameter = 16.8 cm.

Using the ratio method: (π − 1) × d = 16.8

⇒ (22/7 − 1) × d = 16.8

⇒ (15/7) × d = 16.8

⇒ d = (16.8 × 7) ÷ 15 = 1.12 × 7 = 7.84 cm.

∴ The correct answer is 7.84 cm.

Alternate Method

Given:

Circumference (C) − Diameter (d) = 16.8 cm

Formula Used:

Circumference of a circle = πd

Calculations:

⇒ πd − d = 16.8

⇒ d(π − 1) = 16.8

⇒ d(\(\frac{22}{7}\) − 1) = 16.8

⇒ d(\(\frac{22 - 7}{7}\)) = 16.8

⇒ d × \(\frac{15}{7}\) = 16.8

⇒ d = \(\frac{16.8 \times 7}{15}\)

⇒ d = 1.12 × 7

⇒ d = 7.84 cm

∴ The correct answer is 7.84 cm.

Additional Information

Area of Circle

The space occupied by the circle is given by Area = πr² or (π/4)d².

Relation between Radius and Diameter

The diameter is always twice the radius of the circle, d = 2r.

Semicircle Perimeter

The perimeter of a semicircle includes the arc and the diameter: P = πr + 2r = r(π + 2).

Given:

Right angle triangle ABC, hypotenuse 3√5

If AB is doubled and BC is tripled then hypotenuse is 15

Formula used:

(hypotenuse)² = (height)² + (base)²

Calculation:

(3√5)² = (AB)² + (BC)²

⇒ 45 = (AB)2 + (BC)2  --(1)

If AB is doubled and BC is tripled then

⇒ (15)² = (2AB)² + (3BC)²   

⇒ 225 = 4AB² + 9BC² --(2)

By subtracting eq (2) - {4 × eq(1)}

225 - 180 = 4AB2 + 9BC2 - 4AB2 - 4BC²

⇒ 45 = 5BC2

⇒ BC² = 9

⇒ BC = 3 (side is always positive)

Put BC in eq (1)

45 - 9 =  (AB)2

⇒ AB = 6

Now AB + BC = 6 + 3 = 9 cm

∴ The correct answer is 9 cm

Shortcut TrickOriginal triangle (BC, AB, 3√5)

If AB is doubled and BC is tripled then (2AB, 3BC, 15)

(3BC, 2AB, 15) by observing this it is triplet of (9, 12, 15)

2AB = 12, 3BC = 9

AB = 6 cm, BC = 3 cm

AB + BC = 6 + 3 = 9 cm

Calculation:

Let the radius of the bigger circle = R 

The radius of smaller circle = r

AB = 14 cm.

AM = 14/2 = 7 cm

In triangle OAM 

OA2 = OM2 + AM2

R2 = r2 + 72

R2 - r2 = 49

Area of between two triangle = π(R2 - r2)

⇒ (22/7) × 49 = 154 square cm

∴ Correct answer is 154 square cm