Quantitative Aptitude PYQs

Formula Used:

Sum of interior angles of a polygon of n sides = (n - 2) × 180° 

Sum of exterior angles of a polyon of n sides = 360° 

Calculation:

According to the question 

Required ratio = \(\frac{(n - 2) × 180^0}{360^0}\)

⇒ Required ratio = \(\frac{(n - 2)}{2}\)

∴ The ratio of interior and exterior angles of a polygon of n side is  \(\frac{(n - 2)}{2}\).

Shortcut Trick 

If the number is in the form of an + bn and n is an odd number then 

the number an + bn must be divisible by (a + b)

So, the number 4143 + 4343 must be divisible by 41 + 43, that is 84

 

Alternate Method 

Concept Used:

When n is an odd number then

aⁿ + bⁿ = (a + b)(aⁿ⁻¹ b⁰ − aⁿ⁻² b +...........a. bⁿ⁻² + a⁰ bⁿ⁻¹)

Calculation:

We have 4143 + 4343

Since the given number is of the form an + bn

And 43 is an odd number 

So, 4143 + 4343 can be written as

⇒ 4143 + 4343 = (41 + 43) (4143−1 430 − 4143−2 43 +...........41. 4343−2 + 410 4343−1)

⇒ 4143 + 4343 = 84 (4143−1 430 − 4143−2 43 +...........41. 4343−2 + 410 4343−1)

∴ We can say that 4143 + 4343 is completely divisible by 84.

Given:

\(x, = ,7, + ,4√ 3 \)

Formula Used:

a² + b² + 2ab = (a + b)²

Calculation:

We have 

x = 7 + 4√3

⇒ x = 4 + 3 + 2. 2 .√3

⇒ x = (2)² + (√3)² + 2.2.√3

⇒ x = (2 + √3)2

⇒ √x = 2 + √3      -----(i)

So, \(\frac{1}{\sqrt {x}} = \frac{1}{2 + \sqrt 3}\) 

After rationalizing we get 

\(\frac{1}{\sqrt x{}} = 2 - \sqrt 3\)         -----(ii) 

On adding equation (i) and (ii), we get 

\(\sqrt x , + ,\frac{1}{{\sqrt x }}\)=  2 + √3 +  2 - √3   

∴  \(\sqrt x , + ,\frac{1}{{\sqrt x }} = 4\).

Given:

The number is 461 + 462 + 463 + 464

Concept Used:

If n = a × b × c ...

Then n must be divisible by a, b, c, .....

Calculation:

We have 461 + 462 + 463 + 464

Taking 4⁶¹ common from this number, we get

⇒ 4⁶¹ (1 + 4¹ + 4² + 4³)

⇒ 461 (1 + 4 + 16 + 64)

⇒ 461 (85)

⇒ 461 × 5 × 17

So, We can write 

461 + 462 + 463 + 464 = 461 × 5 × 17

So, the number can be divisible by 5, 17 and 4ⁿ {n ≤ 61}

From option we can say that 

∴ 461 + 462 + 463 + 464 is divisible by 17.

Shortcut Trick

Given p : q = 14 : 1 and pq = 1134.

Substitute 14q for p: 14q × q = 1134 ⇒ q² = 81 ⇒ q (HCF) = 9.

Then p (LCM) = 14 × 9 = 126.

Let numbers be 9a and 9b, where a × b = 14 (LCM/HCF). Factors of 14 are (1, 14) or (2, 7).

Differences: 9(14 − 1) = 117 or 9(7 − 2) = 45. Since 45 is in the options, it is the intended answer.

∴ The correct answer is 45.

Alternate Method

Step 1: Find HCF and LCM values

⇒ p = 14q

⇒ (14q) × q = 1134

⇒ 14q² = 1134

⇒ q² = 1134 ÷ 14 = 81

⇒ q = 9 (HCF) and p = 126 (LCM)

Step 2: Find the numbers

⇒ Let the numbers be x and y. x × y = p × q = 1134.

⇒ Since HCF is 9, let x = 9a and y = 9b where a and b are co-prime.

⇒ 9a × 9b = 1134 ⇒ 81ab = 1134 ⇒ ab = 14.

Step 3: Evaluate pairs and difference

⇒ Pair 1: a=1, b=14. Numbers = (9, 126). Difference = 126 − 9 = 117.

⇒ Pair 2: a=2, b=7. Numbers = (18, 63). Difference = 63 − 18 = 45.

⇒ Comparing with given options, 45 is the only available valid difference.

∴ The correct answer is 45.

Additional Information

Product of Numbers Rule

For any two positive integers x and y, x × y = HCF(x, y) × LCM(x, y).

LCM and HCF Relationship

The HCF is always a factor of the LCM. The ratio LCM/HCF gives the product of the co-prime factors of the numbers.

Difference Property

The difference between two numbers x and y, both having HCF 'q', is always a multiple of 'q'. In this case, Difference = 9(b − a).

Shortcut Trick

The given expression follows the algebraic pattern (a³ − b³) ÷ (a² + ab + b²).

This expands and simplifies directly to (a − b).

Here, a = 5.4 and b = 0.4 (since 0.4³ = 0.064).

Required value = 5.4 − 0.4 = 5.

∴ The correct answer is 5.

Alternate Method

Given:

Expression: (5.4³ − 0.064) ÷ (5.4² + 2.16 + 0.16)

Formula Used:

a³ − b³ = (a − b)(a² + ab + b²)

Calculations:

⇒ Let a = 5.4 and b = 0.4

⇒ 0.064 = (0.4)³ = b³

⇒ 2.16 = 5.4 × 0.4 = ab

⇒ 0.16 = (0.4)² = b²

⇒ Now substitute a and b into the expression:

⇒ Expression = (a³ − b³) ÷ (a² + ab + b²)

⇒ Expression = [(a − b)(a² + ab + b²)] ÷ (a² + ab + b²)

⇒ The term (a² + ab + b²) cancels out from the numerator and denominator.

⇒ Expression = a − b

⇒ Substitute the values of a and b back into the result:

⇒ Result = 5.4 − 0.4 = 5

∴ The correct answer is 5.

Additional Information

Sum of Cubes Identity

a³ + b³ = (a + b)(a² − ab + b²). When an expression has a plus sign in the numerator matching this form, it simplifies to just a + b.

Cube of a Binomial

(a + b)³ = a³ + b³ + 3ab(a + b) and (a − b)³ = a³ − b³ − 3ab(a − b).

Difference of Squares

a² − b² = (a − b)(a + b). This foundational identity is highly useful for directly canceling out terms in division problems.

Given:

If the man walks @ 3 km/hr, he reaches 40 min early

If the man walks @ 2 km/hr, he reaches 40 min late

Formula Used:

Total distacne = Speed × Time 

Caluclation:

Let the distance between residence and office be D 

The usual speed is S km/hr and usual time be T hr

Now, According to the question 

D = 3 (T - 40/60)      -----(ii)

And D = 2 (T + 40/60)    ----(iii)

From (ii) and (iii),we get 

3 (T - 40/60) = 2 (T + 40/60)

⇒ 3T - 2 = 2T + 4/3

⇒ T = (2 + 4/3) hr

⇒ T = 10/3 hr

From (ii)

D = 3 [(10/3) - (40/60)] km

⇒ D = 3 [(10/3) - (2/3)] km

⇒ D = 3 (8/3) km

⇒ D = 8 km

∴ The distance between residence and office is 8 km.

Given:

\(\frac{1}{{5\sqrt 4 , + ,4\sqrt 5 }}, + ,\frac{1}{{6\sqrt 5 , + ,5\sqrt 6 }}, + ,\frac{1}{{7\sqrt 6 , + ,6\sqrt 7 }}, + ,\frac{1}{{8\sqrt 7 , + ,7\sqrt 8 }}, + ,\frac{1}{{9\sqrt 8 , + ,8\sqrt 9 }}\)

Concept used:

(a + b)(a - b) = a2 - b2

Calculation:

Consider the first term only, i.e. \(\frac{1}{{5\sqrt 4 , + ,4\sqrt 5 }},\)

Rationalize the denominator

⇒ \(\frac{1}{{5\sqrt 4 , + ,4\sqrt 5 }},\times \frac{5\sqrt 4 , - ,4\sqrt 5 }{{5\sqrt 4 , - ,4\sqrt 5 }},\)

⇒ \(\frac{5\sqrt 4 , - ,4\sqrt 5}{{(5\sqrt 4)^2 , - ,(4\sqrt 5)^2 }},\)

⇒ \(\frac{5\sqrt 4 , - ,4\sqrt 5}{{100-80 }}, = \frac{5\sqrt 4}{20} - \frac{4\sqrt5}{20}\)

⇒ \(\frac{1}{\sqrt 4} - \frac{1}{\sqrt 5}​​\)

After doing a similar operation in other terms, we will get

\(\frac{1}{\sqrt 4} - \frac{1}{\sqrt 5}+\frac{1}{\sqrt 5} - \frac{1}{\sqrt 6}+\frac{1}{\sqrt 6} - \frac{1}{\sqrt 7}+\frac{1}{\sqrt 7} - \frac{1}{\sqrt 8}+\frac{1}{\sqrt 8} - \frac{1}{\sqrt 9}\)

⇒ \(\frac{1}{\sqrt 4} - \frac{1}{\sqrt 9}​​\)

⇒ \(\frac{1}{2} - \frac{1}{3}​​\)

⇒ \(\frac{3;-;2}{6}\)

⇒ \(\frac{1}{6}\)

∴ Required value of the expression is \(\frac{1}{6}\)

Given:

x = 9999

Formula Used:

a² - b² = (a - b) (a + b)

Calculation:

To find \(\frac{{4{x^3}, - ,x}}{{\left( {2x, + ,1} \right)\left( {6x, - ,3} \right)}}\)

⇒ \(\frac{{x \ (4x^2 - 1)}} {{\left( {2x, + ,1} \right)\left( {6x, - ,3} \right)}}\)

⇒ \(\frac{{x \ [ \ (2x)^2 \ - \ (1)^2 \ ]}} {{\left( {2x, + ,1} \right)\left( {6x, - ,3} \right)}}\)

⇒ \(\frac{{x \ (2x - 1) (2x + 1)}} {{3\left( {2x, + ,1} \right)\left( {2x, - ,1} \right)}}\)

⇒ x/3

⇒ 9999/3 = 3333

∴ The value of the expression at x = 9999 is 3333.

Given:

\(\left( {x, + ,\sqrt {1,, + ,,{x^2}} } \right)\left( {y, + ,\sqrt {1,, + ,,{y^2}} } \right), = ,1\)

Concept used:

A² - B² = (A + B) (A - B)

(A + B)² = A² + B² + 2AB

(A - B)2 = A2 + B2 - 2AB

Shortcut TrickGiven that

\(\left( {x, + ,\sqrt {1,, + ,,{x^2}} } \right)\left( {y, + ,\sqrt {1,, + ,,{y^2}} } \right), = ,1\)

Put x = 0 and y = 0 (Real number)

⇒ \(\left( {0, + ,\sqrt {1,, + ,,{0^2}} } \right)\left( {0, + ,\sqrt {1,, + ,,{0^2}} } \right), = ,1\)

⇒ 1 × 1 = 1  satisfied.

Hence, value (x + y)² at x = 0 and y = 0

⇒ (0 + 0)² = 0

∴ (x + y)² = 0 

Alternate Method

\(\left( {x, + ,\sqrt {1,, + ,,{x^2}} } \right)\left( {y, + ,\sqrt {1,, + ,,{y^2}} } \right), = ,1\)

⇒ \(({x\ +\ \sqrt{1\ +\ x^2}})\) = \(1 \over ({y\ +\ \sqrt{1\ +\ y^2}})\)

⇒ \(({x\ +\ \sqrt{1\ +\ x^2}})\) = \(({\sqrt{1\ +\ y^2}\ -\ y}) \over ({\sqrt{1\ +\ y^2}\ +\ y}) × ({\sqrt{1\ +\ y^2}\ - y})\)

⇒ \(({x\ +\ \sqrt{1\ +\ x^2}})\) = \(({\sqrt{1\ +\ y^2}\ -\ y}) \over {(\sqrt{1\ +\ y^2})^2\ -\ y^2}\)

⇒ \(({x\ +\ \sqrt{1\ +\ x^2}})\) = \(({\sqrt{1\ +\ y^2}\ -\ y}) \over {(1\ +\ y^2)\ -\ y^2}\)

⇒ \(({x\ +\ \sqrt{1\ +\ x^2}})\) = \(({\sqrt{1\ +\ y^2}\ -\ y}) \over 1\)

⇒ \(({x\ +\ \sqrt{1\ +\ x^2}})\) = \(({\sqrt{1\ +\ y^2}\ -\ y})\)

⇒ \(x\ +\ y\) = \({\sqrt{1\ +\ y^2}}\ -\ \sqrt{1\ +\ x^2}\)

⇒ \((x\ +\ y)^2\) = \(({\sqrt{1\ +\ y^2}}\ -\ \sqrt{1\ +\ x^2})^2\) (Squaring both side)

⇒ \({x^2\ +\ y^2\ + 2xy}\) = \({1\ + y^2\ +\ 1\ +\ x^2\ - 2 {\sqrt{1\ +\ y^2}\ {\sqrt{1\ +\ x^2}}}}\)

⇒ \({xy}\) = \({1\ -\ {\sqrt{1\ +\ y^2}\ {\sqrt{1\ +\ x^2}}}}\)

⇒ \({\sqrt{1\ +\ y^2}\ {\sqrt{1\ +\ x^2}}}\) = \({1\ -\ xy}\)

⇒ \(({\sqrt{1\ +\ y^2}\ {\sqrt{1\ +\ x^2}}})^2\) = \(({1\ -\ xy})^2\) (Squaring both side)

⇒ \((1\ +\ y^2)\ (1\ +\ x^2)\) = \({1\ +\ x^2y^2\ -2xy}\)

⇒ \({x^2y^2\ +\ x^2\ +\ y^2\ +\ 1}\) = \({1\ +\ x^2y^2\ -2xy}\)

⇒ \({x^2\ +\ y^2\ +\ 2xy}\) = \(0\)

⇒ \((x\ +\ y)^2\) = 0

∴ The value of (x + y)2 is 0.

Concept:

Using trigonometric ratios of complementary angles:-

• sin(90° - θ) = cosθ
• cos(90° - θ) = sinθ
• tan(90° - θ) = cotθ
• cot(90° - θ) = tanθ
• cotθ = \(\frac{1}{tanθ}\)

 

Calculation:

 \(\frac{{2\sin ,68^\circ }}{{\cos 22^\circ }}, - ,\frac{{2\cot ,15^\circ }}{{5\tan 75^\circ }} - \frac{{3\tan 20^\circ \tan 40^\circ \tan 45^\circ \tan 50^\circ \tan 70^\circ }}{5}\)

⇒  \(\frac{2sin68°}{cos(90° - 68°)} - \frac{2cot15°}{5tan(90° - 15°)} - \frac{3tan(90° - 70°) tan(90° - 50°) tan45° tan50° tan70°}{5}\)

⇒ \(\frac{2sin68°}{sin68°} - \frac{2cot15°}{5cot15°} - \frac{3 tan45° cot70° cot50° tan50° tan70°}{5}\)

⇒ \(\frac{2sin68°}{sin68°} - \frac{2cot15°}{5cot15°} - \frac{3 tan45°. \frac{1}{tan70°}. \frac{1}{tan50°}. tan50°. tan70°}{5}\)

⇒ \(\frac{2sin68°}{sin68°} - \frac{2cot15°}{5cot15°} - \frac{3 tan45°}{5}\)

⇒ \(2 - \frac{2}{5} - \frac{3\times 1}{5}\)    [∵ tan 45° = 1]

⇒ \(2 - \frac{5}{5}\)

⇒ 2 - 1 = 1

∴ The value of \(\frac{{2\sin ,68^\circ }}{{\cos 22^\circ }}, - ,\frac{{2\cot ,15^\circ }}{{5\tan 75^\circ }} - \frac{{3\tan 20^\circ \tan 40^\circ \tan 45^\circ \tan 50^\circ \tan 70^\circ }}{5}\)  is 1.

Concept Used:

In a right-angled triangle ABC, If a perpendicular BD is drawn on the hypotenuse AC from the right-angled vertex then 

BD² = AD × CD 

Calculation:

According to the question 

AD = 9 units and 

CD = 16 units 

And we have to find the value of BD 

Now, According to the concept used

BD2 = AD × CD 

⇒ BD2 = 9 × 16 = 144

⇒ BD2 = 12²

⇒ BD = 12 units 

∴ The required length of the perpendicular is 12 units.

Given:

43x × 47y = (2021)2, x ≠ 0, y ≠ 0

To find \(\frac{{4xy, + ,x, + ,y}}{{2xy, - ,x, - ,y}}\)

Calculation:

We have 43x × 47y = (2021)2

On factorization of 2021 we get 

2021 = 43 × 47

So, we can write the equation (i) as

43x × 47y = (43 × 47)2

⇒ 43x × 47y = 432 × 472 

On comparing we get 

x = 2 and y = 2

Now we have to find the value of \(\frac{{4xy, + ,x, + ,y}}{{2xy, - ,x, - ,y}}\)

⇒ \(\frac{4 \times 2 \times 2 + 2 + 2}{2 \times 2 \times 2- 2 - 2}\) = \(\frac{20}{4} = 5\)  

∴ The required value is 5.

Given: 

\(\frac{1}{{{\rm{a}},{\rm{ + }},\frac{1}{{{\rm{b}},{\rm{ + }}\frac{1}{{{\rm{c}},{\rm{ + }},\frac{1}{{\rm{d}}}}}}}}} = ,,\frac{{17}}{{60}}\)

Calculation:

We can write the RHS \(\frac{17}{60} \) as 

⇒ \(\frac{1}{\frac{60}{17}} \) = \(\frac{1}{3+ \frac{9}{17}} \) = \(\frac{1}{3+ \frac{1}{\frac{17}{9}}} \)

⇒ \(\frac{1}{3+ \frac{1}{1 + \frac{8}{9}}} \) = \(\frac{1}{3+ \frac{1}{1 + \frac{1}{\frac{9}{8}}}} \) = \(\frac{1}{3+ \frac{1}{1 + \frac{1}{1 + \frac{1}{8}}}} \)

Now, On comparing 

\(\frac{1}{{{\rm{a}},{\rm{ + }},\frac{1}{{{\rm{b}},{\rm{ + }}\frac{1}{{{\rm{c}},{\rm{ + }},\frac{1}{{\rm{d}}}}}}}}} \) by \(\frac{1}{3+ \frac{1}{1 + \frac{1}{1 + \frac{1}{8}}}} \)

We get a = 3, b = 1, c = 1 and d = 8

So, abcd = 3 × 1 × 1 × 8 = 24

∴ The product of a, b, c, d is 24.

Given:

x2 = 17x + y and y2 = x + 17y, x ≠ y

Formula Used:

a² - b² = (a - b) (a + b)

Calculation:

We have 

x2 = 17x + y      -----(i)

y2 = x + 17y       -----(ii)

On applying (i) - (ii), we get 

x2 - y2 = 16x - 16y

⇒ (x - y) (x + y)  = 16(x - y)

⇒ (x + y) = 16      ----(iii)

Now, On applying (i) + (ii), we get 

x2 + y2 = 18x + 18y

⇒ x2 + y2 = 18(x + y)

⇒ x 2 + y2 = 18 × 16      [from (i)]

⇒ x 2 + y2 = 288    ----(iv)

Now, we have to find the value of \(\sqrt {{x^2}, + ,{y^2}, + ,1} \)

So, from equation (iv), we get 

⇒ \(\sqrt {288 + 1} = \sqrt{289} = 17\)

∴ The required value of \(\sqrt {{x^2}, + ,{y^2}, + ,1} \) is 17.

Given :

194480 + n = m4

Concept used :

When we square a number, on squaring the unit digit can only be 0, 1, 4, 5, 6, 9

Calculations :

We know that,

On squaring,  the unit digit can only be 0, 1, 4, 5, 6, 9

Thus, a number of squares never have 2, 3, 7, and 8 as unit digit

Equation (1) can be written as 

194480 + n = (m²)²       ----- (1)

When we again square the number  0, 1, 4, 5, 6, 9, we get unit digit as 0, 1, 6, 5 only.

Putting n = 1 in equation (1), we get

⇒ 194480 + 1 = (m2)² 

⇒ (√194481) = m²

⇒ √ 441 = m

⇒ m = 21

∴ The least value of n is 1.

Given:

\(\frac{{x - y}}{{x√ y , + ,y√ x }}, = ,\frac{1}{{√ x }};\left( {x, > ,0,y, > ,0} \right)\)

Formula Used:

a² - b² = (a + b) (a - b)

Calculation:

We have\(\frac{{x - y}}{{x√ y , + ,y√ x }}, = ,\frac{1}{{√ x }}\)

⇒ \(\frac{{(√ x)^2 - (√ y)^2}}{{√ x √ x√ y , + , √ y √ y√ x }}, = ,\frac{1}{{√ x }}\)

⇒ \(\frac{(√ x + √ y) (√ x - √ y)}{{√ x .√ y \ (√ x , + , √ y) }}, = ,\frac{1}{{√ x }}\)

⇒ \(\frac{√ x - √ y}{√ x . √ y} = \frac{1}{√ x}\)

⇒ √x - √y = √y 

⇒ √x = 2√y 

On squaring both side, we get 

⇒ x = 4y 

⇒ x/y = 4

∴ The required value of x/y is 4.

Given:

Three equal circles of radius 4 cm touch each other 

Formula Used:

Area of circle = πr2

Area of equilateral triangle = √3a²/4

Where, r = radius of circle and 

a = side of an equilateral triangle 

Calculation:

According to the question 

We can draw figure like this 

Now, From figure, we can say that 

AB = BC = CA = 8 cm

So, ΔABC is an equilateral triangle 

And we know that angles of equilateral triangle is 60° 

Now, Area of shaded region = Area of ΔABC - Area of sectors formed in three circles 

⇒ Area of shaded region = \( \frac{\sqrt3}{4} (8)^2 - 3 \times \frac{60^0}{360^0} \pi (4)^2\)

⇒ \(16{\sqrt3}- 8 \pi\)  square cm

∴ The area of shaded region is \(16{\sqrt3}- 8 \pi\)  square cm.

Given:

A car travels from A to B at a speed of 40 km/hr

Travels back from B to A at a speed of 30 km/hr

And again goes from A to B at a speed of 60 km/hr

Formula Used:

Speed = Distance/Time

Average speed = Total distance/Total time 

Calculation:

Let the distance between A and B is d km 

According to the question 

Total distance = d + d + d = 3d 

Total time = \(\frac{d}{40} + \frac{d}{30} + \frac{d}{60}\)

Now, Average speed = \(\frac{3d}{\frac{d}{40} + \frac{d}{30} + \frac{d}{60}}\)

⇒ \(\frac{3d}{\frac{3d + 4d + 2d}{120}}\) = \(\frac{3d \times 120}{9d}\) = 40 km/hr

∴ The average speed of car in the whole journey is 40 km/hr.

Given:

Given number is 9410

Concept used:

Long division method to find the square root

Calculation:

So, the number should be 9409 which is 97²

The smallest natural number which must be subtracted is 9410 - 9409 = 1

∴ The required answer is 1

Alternate Method ⇒ 95² < 9410 < 100²

⇒ 97² = 9409

∴ The perfect square = 9410 - 1 = 9409