Quantitative Aptitude PYQs

Area of the first face of cuboid = x

Length of the first side = √x

Length of the second side = √y

Length of the third side = √z

Volume of cuboid = V = √x × √y × √z

⇒ V = √(xyz)

⇒ V² = xyz

Length of altitude of an equilateral triangle =\(l = \frac{{\sqrt 3 }}{2}a\)

⇒ a = 2l/√3

Area of equilateral triangle \(= \frac{{\sqrt 3 }}{4}{a^2} = \frac{{\sqrt 3 }}{4}{\rm{}} \times {\rm{}}{\left( {\frac{{2l}}{{\sqrt 3 }}} \right)^2} = \frac{{\sqrt 3 {l^2}}}{3}\)

Formula used:

The perimeter of the circle = 2πr

The perimeter of the sector = \(2r+\frac{θ}{360^{\circ}}\)

Where r is the radius of the circle and

θ is the subtended angle at the centre of the circle

Calculation:

Using the above formula, 

Length of arc = \(2{\rm{π }} \times {\rm{}}20{\rm{}} \times {\rm{}}\frac{{150}}{{360}} = \frac{{50{\rm{π }}}}{3}\) cm

Let the length of the circle = R

The perimeter of circle = Length of arc

⇒  50π/3 = 2πR

⇒ R = 25/3 cm

∴ The radius of the circle will be 25/3 cm.

Since P, Q and R are mid-points of sides YZ, XY and XZ respectively.

QR = YZ/2

⇒ YZ = 2QR

Area of ΔXYZ : Area of ΔPQR = YZ² : QR²

⇒ 128 : Area of ΔPQR = (2QR)² : QR² = 4 : 1

⇒ Area of ΔPQR = 128/4 = 32 cm²

Similarly,

BC = QR/2

⇒ QR = 2BC

Area of ΔPQR : Area of ΔABC = QR² : BC²

⇒ 32 : Area of ΔABC = (2BC)² : BC² = 4 : 1

⇒ Area of ΔABC = 32/4 = 8 cm²

Given:

AC : AD = 4 : 9 , p ll q

Calculation: 

 Shortcut TrickΔBAC ∼ ΔEAD (By AA similarity)

⇒ Area of ΔBAC: Area of ΔEAD = AC2 : AD2= 16 : 81

 Alternate MethodConcept:

If two triangles are similar, the ratio of the area of triangles is equal to their corresponding ratios.

Calculation:

We have,p II q  and

∴ ∠ CBE = ∠ BED & ∠ BCD = ∠ CDE (Alternate angles)

By AA Similarity ΔBAC ∼ ΔEAD 

∴ \(\frac{\Delta BAC}{\Delta EAD}=\frac{BA^{2}}{EA^{2}}=\frac{AC^{2}}{AD^{2}}=\frac{BC^{2}}{ED^{2}}\)

⇒ \(\frac{\Delta BAC}{\Delta EAD}=\frac{AC^{2}}{AD^{2}}\)

⇒  \(\frac{\Delta BAC}{\Delta EAD}=\frac{4^{2}}{9^{2}}\) = \(\frac{16}{81}\)

∴ 16 : 81 is the correct option.

Since an equilateral triangle and a square are constructed using metallic wires of equal length which means that the perimeter of the triangle and the square is same.

Let the side of the triangle and the square is ‘a’ and ‘b’ respectively.

Now,

3a = 4b

⇒ a/b = 4/3

Suppose a = 4x and b = 3x

Area of equilateral triangle = \(\frac{{\sqrt 3 }}{4}{a^2} = 4\sqrt 3 {x^2}\)

Area of square = b² = (3x)² = 9x²

∴ Required ratio = (4√3 x²) / (9x²) = 4√3 : 9

Since all the four sides of a parallelogram are of equal length, then it is a rhombus.

Area of rhombus = 1/2 × First diagonal × Second diagonal

Sum of the length of diagonals = 12 cm

First diagonal = 12 × (1/3) = 4 cm

Second diagonal = 12 × (2/3) = 8 cm

Area of parallelogram = Area of rhombus = (1/2) × 4 × 8 = 16 cm²

AB = 5 cm, BC = 10 cm

AC = \(\sqrt {{5^2} + {{10}^2}} = 5\sqrt 5\) cm

BM is perpendicular to AC.

Area of ∆ABC = 1/2 × AB × BC = 1/2 × AC × BM

⇒ 5 × 10 = 5√5 × BM

⇒ BM = 2√5 cm

Let the height of the cylinders are ‘2h’ and ‘3h’ respectively.

Let the radii of the cylinders are ‘x’ and ‘y’ respectively.

Now,

Volume of the first cylinder = Volume of the second cylinder

πx² × 2h = πy² × 3h

⇒ x²/y² = 3/2

⇒ x : y = √3 : √2

Let the length and breadth of the rectangle is ‘x’ and ‘y’ respectively.

Original area of rectangle = xy

Increased length of rectangle = 120% of x = 1.2x

Increased breadth of rectangle = 110% of y = 1.1y

Increased area of rectangle = 1.2x × 1.1y = 1.32xy

Increment in the area = 1.32xy – xy = 0.32xy

Percent increment = (0.32xy/xy) × 100 = 32%

Hypotenuse is given as 10 cm.

Since it is a right-angle triangle, its maximum area will be possible only if the triangle is isosceles.

Other 2 sides when hypotenuse is 10 cm = √(10²/2) = 5√2 cm

Hence, area = 1/2 × 5√2 × 5√2 = 25 cm²

From the figure,

Diagonal of circle = diameter = 2r

Side of the square = 2r/√2 = r√2

Now,

Area of circle = πr²

Area of square = side² = (r√2)² = 2r²

Hence,

Ratio of area of circle to that of square = πr² : 2r² = 22 : 14 = 11 : 7

Given:

Diameter of 20 cm then, radius = 10 cm

The curved surface area is 1000 cm².

Formula used:

The surface area of the cylinder = 2πrh

Volume of cylinder = πr²h

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

Calculation:

Using the above formula

⇒ 2πrh = 1000

⇒ 2 × π × 10 × h = 1000

⇒ h = 50/π

Now,

Volume of cylinder = πr²h

⇒ π × 10 × 10 × 50/π = 5000 cm³

∴ Volume of the cylinder is 5000 cm3

Length of arc = 33 cm

Radius of arc = 14 cm

Then,

Angle subtended = l/r × 180/π

⇒ 33/14 × 180° × 7/22

⇒ 135°

In the semicircle:

a² + a²/4 = r²

⇒ 5a² = 4r²

⇒ a = 2r/√5

In the circle with radius r:

Side of square = r√2

∴ Required ratio = (2r/√5)² / (r√2)²

⇒ 2 : 5

On dividing (x⁴ - x² + 7x + 5) by (x + 2):

Quotient = x³ - 2x² + 3x + 1

Comparing the coefficients with ax³ + bx² + cx + d:

⇒ a = 1, b = -2, c = 3, d = 1

Sides: a = 30 cm, b = 28 cm and c = 16 cm

s = (a + b + c)/2 = (30 + 28 + 16)/2 = 74/2 = 37

Area of the triangle = \(\sqrt {\left[ {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \right]} \)

⇒ \(\sqrt {\left[ {37\left( {37 - 30} \right)\left( {37 - 28} \right)\left( {37 - 16} \right)} \right]} \)

⇒ \(\sqrt {\left[ {37 \times 7 \times 9 \times 21} \right]} \)

To get the value of this expression, we need the log values of 37, 21, 9, 7.

Suppose X = (0.001995)^1/8

Taking log:

⇒ log₁₀X = 1/8 log₁₀0.001995

⇒ log₁₀X = 1/8 log₁₀ (1995 × 10^-6)

⇒ log₁₀X = 1/8 [log₁₀ 1995 + log₁₀ 10^-6]

⇒ log₁₀X = 1/8 [3.3 - 6]

⇒ log₁₀X = -2.7/8 = -0.3375

⇒ X = \(\frac{1}{{{{10}^{0.3375}}}}\)

(x - a) (x - b) (x - c)

Expanding the expression:

⇒ (x² - ax - bx + ab) (x - c)

⇒ (x³ - ax² - bx² + abx - cx² + acx + bcx - abc)

⇒ (x³ - ax² - bx² + abx - cx² + acx + bcx - abc)

⇒ x³ - (a + b + c)x² + (bc + ca + ab)x - abc

Side of equilateral triangle = 7 cm

A = √3/4 × 7² = √3/4 × 49  \(=\sqrt3:\times \frac{7}{2} \times \frac{7}{2}\) = 3.5 × 3.5 × √3

Now,

log₁₀A⁴ = 2 log₁₀A²

⇒ 2 log₁₀ (3.5 × 3.5 × √3)²

⇒ 2 log₁₀ [(3.5 × 3.5)² × 3]

⇒ 2 log₁₀ [(3.5)³ × 3.5 × 3]

⇒ 2 [log₁₀ 3.5³ + log₁₀10.5]

⇒ 2 [3log₁₀ 3.5 + log₁₀1050/100]

⇒ 2 [3log₁₀ 35/10 + log₁₀1050/100]

⇒ 2 [3(log₁₀ 35 - log₁₀ 10) + (log₁₀1050 - log₁₀ 100)]

Putting log₁₀1050 = 3.0212 and log₁₀35 = 1.5441

⇒ 2 [3 × (1.5441 - 1) + (3.0212 - 2)]

⇒ 2 [1.6323 + 1.0212]

⇒ 5.3070