Quantitative Aptitude PYQs

Let the selling price of both computers be ‘91x’ each.

Cost price of first computer = 91x × (100/130) = 70x

Cost price of second computer = 91x × (100/70) = 130x

Total cost price = 70x + 130x = 200x

Total selling price = 91x × 2 = 182x

Loss amount = 200x – 182x = 18x

Loss percent = (18x/200x) × 100 = 9%

Shortcut Trick

Saving 1 unit = 600 Rs

∴ Monthly income of A = 600 × 4 = 2400 Rs.

Alternate Method

Formula used: 

Income = Expenditure + Savings

Calculation:

Let, the monthly income of A and B be ‘4x’ and ’3x’ respectively.

Expenditure of A = (4x – 600)

Expenditure of B = (3x – 600)

According to the question,

⇒ \(\frac{4x-600}{3x-600} \) = \(\frac{3}{2}\)

⇒ 8x – 1200 = 9x – 1800

⇒ x = 600

Monthly income of A = 4x = Rs. 2400

∴ Rs. 2400 is the correct option.

Given :

Ratio of train fare and bus fare = 3 : 4

The train fare increases by 20% and bus fare increases by 30%

Calculations :

Let the original train fare and the bus fare be ‘3x’ and ‘4x’ respectively.

Increase train fare = 120% of 3x

⇒ 3.6x

Increase bus fare = 130% of 4x

⇒ 5.2x

Required ratio = 3.6x: 5.2x

⇒ 9 : 13

∴ The ratio between revised train fare and revised bus fare is \(\frac{9}{13}\).

Given:

Quotient = 182

Concept used:

Number = Divisor × quotient + remainder

Calculation:

According to the question,

Remainder = 182 – 175

⇒ Remainder = 7

Now,

N = 182 × 17 + 7

⇒ N = 3094 + 7

⇒ N = 3101

∴ The value of N is 3101

Given:

A stock of food grains is enough for 240 men for 48 days.

Formula used:

Work = Efficiency × Time

Calculation:

Since the stock of food grains is enough for 240 men for 48 days.

Total stock = 240 × 48

Suppose the same stock for 160 men will last for ‘N’ days.

Now,

⇒ 240 × 48 = 160 × N

⇒ N = 72 days

∴ For 72 days the same stock will last for 160 men.

Let the diameter and height of the cylinder are ‘2x’ each.

Volume of cylinder \( = V = {\rm{\pi ;}}{\left( {\frac{{2x}}{2}} \right)^2}.2x = 2{\rm{\pi }}{x^3}\)

Diameter of sphere of maximum possible volume = Diameter of cylinder = 2x

Volume of sphere \(= \frac{4}{3};{\rm{\pi }}{\left( {\frac{{2x}}{2}} \right)^3} = \frac{4}{3}{\rm{\pi ;}}{x^3} = \frac{2}{3}{\rm{}} \times {\rm{}}2{\rm{\pi }}{x^3} = \frac{{2V}}{3}\)

Volume of water remains in the vessel = Volume of the cylinder – Volume of the sphere

= V – 2V/3 = V/3

Now,

\(\frac{{PQ}}{{QR}}; = ;\frac{{LM}}{{MN}}\)

\(\Rightarrow ;\frac{3}{9} = \frac{{LM}}{{10.5}}\)

\(\Rightarrow LM = \frac{{3{\rm{;}} \times {\rm{;}}10.5}}{9}\)

⇒ LM = 3.5 cm

Area of circle = π × 4² = 16π cm²

Let the angle formed by arc be θ.

Angle of sector = 25.6 = \(\frac{{\rm{\theta }}}{{360^\circ }} \times {\rm{\pi }} \times {\rm{;}}{4^2} = \frac{{16{\rm{\pi \theta }}}}{{360^\circ }}\)

\(\Rightarrow {\rm{}}\theta = \frac{{25.6{\rm{;}} \times {\rm{;}}360}}{{16{\rm{;}} \times {\rm{;\pi }}}} = {\frac{{576}}{{\rm{\pi }}}^0}\)

⇒ θ = 3.2 Radian

Concept:

Orthocenter:

Point of intersection of altitudes that are drawn perpendicular from the vertex

to the opposite sides of a triangle. In a triangle, it is that point where all the

three altitudes of a triangle intersect.

• The orthocenter lies inside the triangle for an acute angle triangle.
• The orthocenter lies outside the triangle for an obtuse angle triangle.
• The orthocenter lies on the vertex of the right angle of the right triangle.

 

Explanation:

Using the above concept, we can conclude that, for a right-angled triangle

It's orthocentre lies on the triangle.

Suppose BC = a, AC = b and AB = c

We know that,

AX divides BC is the same ratio equivalent to the ratio of sides AB and AC respectively.

\(BX; = ;BC{\rm{;}} \times {\rm{;}}\frac{{AB}}{{AB; + ;AC}}; = ;AB{\rm{;}} \times {\rm{;}}\frac{a}{{c; + ;b}}\)

Since, we know that,

Sum of any two sides of a triangle is always greater than the third largest side.

c + b > a

Hence,

BX < AB

AB > BX

log₁₀(cos θ) + log₁₀(sin θ) + log₁₀(tan θ) + log₁₀(cot θ) + log₁₀(sec θ) + log₁₀(cosesc θ)

⇒ log₁₀(cos θ × sin θ × tan θ × cot θ × sec θ × cosesc θ)

\(\Rightarrow {\rm{lo}}{{\rm{g}}_{10}}\left( {\cos {\rm{\theta }} \times {\rm{}}\sin {\rm{\theta }} \times {\rm{}}\tan {\rm{\theta ;}} \times {\rm{;}}\frac{1}{{{\rm{tan;\theta }}}}{\rm{}} \times {\rm{}}\frac{1}{{{\rm{cos;\theta }}}}{\rm{}} \times {\rm{}}\frac{1}{{{\rm{sin\theta }}}}} \right)\)

⇒ log₁₀1 = 0

cos²x + cosx = 1

cosx = 1 – cos²x

cosx = sin²x

sin¹²x + 3sin¹⁰x + 3sin⁸x + sin⁶x

⇒ (sin⁴x)³ + 3 × sin⁴x × sin²x × (sin⁴x + sin²x) + (sin²x)³

⇒ (sin⁴x + sin²x)³

⇒ [sin²x(sin²x + 1)]³

⇒ [cosx(cosx + 1)]³

⇒ (cos²x + cosx)³

⇒ 1³ = 1

sinθ = 3/5 = Perpendicular/Hypotenuse

Base = \(\sqrt {{5^2} - {3^2}} = 4\)

cotθ = Base/Perpendicular = 4/3

x = cotθ = 4/3

Now,

1 + 3x + 9x² + 27x³ + 81x⁴ + 243x⁵

\(\Rightarrow {\rm{}}1{\rm{}} + {\rm{}}3{\rm{}} \times {\rm{}}\frac{4}{3}{\rm{}} + {\rm{}}9{\rm{}} \times {\rm{}}{\left( {\frac{4}{3}} \right)^2}{\rm{;}} + {\rm{;}}27{\rm{}} \times {\rm{}}{\left( {\frac{4}{3}} \right)^3}{\rm{}} + {\rm{}}81{\rm{}} \times {\rm{}}{\left( {\frac{4}{3}} \right)^4}{\rm{}} + {\rm{}}243{\rm{}} \times {\rm{}}{\left( {\frac{4}{3}} \right)^5}\)

⇒ 1 + 4 + 16 + 64 + 256 + 1024 = 1365

tanθ = h/x

⇒ tan(90° – θ) = 2h/y

cotθ = 2h/y

tanθ = y/2h

Now,

⇒ h/x = y/2h

⇒ 2h² = xy

\(\frac{{\sin 19^\circ }}{{\cos 71^\circ }} + \frac{{\cos 73^\circ }}{{\sin 17^\circ }}\)

\(\Rightarrow {\rm{}}\frac{{\sin \left( {90^\circ - 71^\circ } \right)}}{{\cos 71^\circ }} + \frac{{\cos (90^\circ - 17^\circ )}}{{\sin 17^\circ }}\)

\(\Rightarrow {\rm{}}\frac{{cos71^\circ }}{{\cos 71^\circ }} + \frac{{\sin 17^\circ }}{{\sin 17^\circ }}\)

= 1 + 1 = 2

Suppose AB = a, BC = b and AC = c

Since, ABPC is parallelogram, which means; AB = PC = a

ABCQ is parallelogram, which means; AB = CQ = a

Similarly, AQ = AR = BC = b

And, BR = BP = AC = c

Perimeter of ABC = a + b + c = 22

Perimeter of PQR = 2a + 2b + 2c = 2(a + b + c) = 2 × 22 = 44 cm

In a trapezium-

We know that: S₁ = S₃

S₁ = S₃ = 3 cm²

S₂ + S₃ = 8 cm²

S₂ = 5 cm²

We also know that:

\({S_1} = \sqrt {{S_2}{S_4}}\)

\(\Rightarrow {\rm{}}3 = \sqrt {5{\rm{}} \times {\rm{}}{S_4}}\)

⇒ 9 = 5S­₄

⇒ S₄ = 9/5 = 1.8 cm²

Let point P on the circle which means triangle will be in the semi-circle and we know that triangle in a semi-circle is always a right-angle triangle.

A = x, BP = y and AB = 2 × 6.5 = 13 cm

AP² + BP² = AB²

⇒ x² + y² = 13²

Options (2), (3) and (4) will satisfy the above relation.

So, option (1) will be our correct answer.

Since ΔABC and ΔPQR are similar.

Perimeter of ΔABC: Perimeter of ΔPQR = One side of ΔABC: Corresponding side of ΔPQR

⇒ 75 : 50 = x : 20

⇒ x = (75 × 20)/50

⇒ x = 30 cm

Concept:

When the side of an equilateral triangle is 'a'.

Its area, \(A=\frac{\sqrt 3}{4}a^2\) 

And area of the normal triangle =  \(\frac{1}{2}\times Base \times Height\)

Since QRS is an equilateral triangle of side 10 cm. That means PQRS is a rhombus and diagonal of a rhombus are always perpendicular to each other.

Area of ΔQRS = \(\frac{{\sqrt 3 }}{4}{a^2} = \frac{1}{2}{\rm{;}} \times {\rm{;}}QS{\rm{;}} \times {\rm{;}}OR\)

\(\frac{{\sqrt 3 }}{4}{\rm{}} \times {\rm{}}{10^2} = \frac{1}{2}{\rm{}} \times {\rm{}}10{\rm{}} \times {\rm{}}OR\)

⇒ OR = 5√3 cm

⇒ PR = 2OR = 10√3 cm