Quantitative Aptitude PYQs

Formula used:

a3 + b3 + c3 - 3abc = (a + b +c)(a2 + b2 + c2 - ab - bc - ac)

Calculation:

a2 - bc = α  ] × a

a3 - abc = aα        ------(1)

b2 - ac = β  ] × b

b3 - abc = bβ        ------(2)

c2 - ab = γ  ] × c

c3 - abc = cγ        ------(3)

Adding equation (1),(2) and (3)

a3 + b3 + c3 - 3abc = aα + bβ + cγ  

a2 - bc + b2 - ac + c2 - ab = α +  β +  γ

⇒ \(\rm \frac{a \alpha+b \beta+c \gamma}{(a+b+c)(\alpha+\beta+\gamma)}\) = \(\frac{a^3 + b^3 +c^3 -3abc}{(a+b+c)(a^2 +b^2 +c^2 - ab - bc -ca)}\)

∴ \(\frac{a^3 + b^3 +c^3 -3abc}{a^3 + b^3 +c^3 -3abc}\) = 1

Calculation:

(x - 1)³ = x³ - 3x² + 3x - 1

∴ Other factor will be x + 1

Confusion PointsFocus only on the constant terms of the expression (α - 4)x³ + βx² + (γ - 2)x. Try to find a factor to divide it by, such that subtracting it from -1 makes the expression equal to zero

Shortcut Trick

⇒ 11(x + y) = 55

⇒ x + y = 5 -----(1)

⇒ 9(x - y) = 45

⇒ x - y = 5 -----(2)

From equation (1) and (2)

⇒ x × y = 5 × 0 = 0

∴ The correct answer is 0.

Solution:-

Calculation:

Let 1^st digit = x & 2^nd digit = y

2-digit number = 10x + y

Reversing the order of the digits of the number = 10y + x

According to question

The sum of the number and the number obtained by reversing the order of the digits of the number = 55

⇒ 10x + y + 10y + x = 55

⇒ 11x + 11y = 55

⇒ x + y = 5 ---------(1)

The difference of the given number and the number obtained by reversing the order of the digits of the number = 45

⇒ 10x + y - 10y - x = 45

⇒ 9x - 9y = 45

⇒ x - y = 5 -----------(2)

Adding Equation (1) & (2)

⇒ x + y + x - y = 10

⇒ 2x = 10 ⇒ x = 5

Put the value of x in equation (1)

⇒ 5 + y = 5 ⇒ y = 0

The product of the digits = 5 × 0 = 0

∴ The correct answer is 0.

Shortcut Trick

Let Total Work be LCM of (10, 12, 15) = 60 units.

Efficiency of 2(A + B + C) = (60/10) + (60/12) + (60/15) = 6 + 5 + 4 = 15 units/day.

Combined Efficiency of (A + B + C) = 15 ÷ 2 = 7.5 units/day.

Time for Half work (30 units) = 30 ÷ 7.5 = 4 days.

∴ The correct answer is 4 days.

Alternate Method

Given:

Time taken by (A + B) = 10 days

Time taken by (B + C) = 12 days

Time taken by (C + A) = 15 days

Formula Used: Efficiency = Total Work ÷ Time

⇒ Total Efficiency of [ (A + B) + (B + C) + (C + A) ] = 6 + 5 + 4

⇒ 2(A + B + C) = 15 units/day

⇒ Efficiency of (A + B + C) = 15 ÷ 2 = 7.5 units/day

⇒ Half of the total work = 60 ÷ 2 = 30 units

⇒ Time taken by (A + B + C) = 30 ÷ 7.5 = 4 days

∴ The correct answer is 4 days.

Additional Information

Combined Efficiency Property

When individuals work in pairs, the sum of their efficiencies represents twice the combined efficiency of all individuals: 2(A+B+C) = Σ(Pair Efficiencies).

Work and Time Relationship

Total Work = Efficiency × Time. If work is halved, for a constant efficiency, the time required will also be halved.

LCM Method

Taking the Least Common Multiple (LCM) of time durations as total work simplifies fractional calculations into integer efficiencies.

Shortcut Trick 

R = 12/4 = 3% and T = 9/3 = 3 year

Interest = (10000 / 1000000) × 92727 =  927.27

∴ The correct answer is  Rs. 927.27.

Alternate Method 

Given:

Amount = Rs. 10,000

Rate,R = 12% 

Time ,n  = 9 months

Concept used:

For quarterly, Rate = R/4 & Time = n × 4

Calculation:

R = 12%/4 = 3% & n = 9 × 4 = 36 months ⇒  3 years

The interest of 1^st year = 10000 × 3% = 300

The interest of 2^nd year = 300 + 300 × 3% = 309

The interest of 3^rd year = 309 + 309 × 3% = 318.27

Total interest after 3 years = 300 + 309 + 318.27 = Rs. 927.27 

∴ The correct answer is  Rs. 927.27

Given:

Sinθ = (a + b)/2√ab

Concept:

-1 ≤ sinθ ≤ +1

Calculation:

-1 ≤ (a + b)/2√ab ≤ +1

We solve one of them

(a + b)/2√ab ≤ +1

⇒ a + b ≤ 2√ab

⇒ a + b - 2√ab ≤ 0

⇒ (√a - √b)²  ≤ 0

⇒ √a = √b = a = b

∴ The correct answer is a = b

Shortcut Trick We can solve this question by option to satisfy equation

put a = b

then Sinθ = (a + a)/ 2√(a × a)

Sinθ = 2a/2a = 1 (sinθ ≤ 1)

Given:

A, B, and C together can do a piece of work in 36 days

A and B together can do five times as much work as C alone

B and C together can do as much work as A alone

Calculation:

Ratios of Efficiency 

(A + B) : C = 5 : 1

⇒ A + B = 5a, C = a --(1)

⇒ A + B + C = 6a

(B + C) : A = 1 : 1

⇒ (B + C) = a, A = a

⇒ A + B + C = 2a 

Efficiency should be equal so 

 A + B + C = 2a × 3 = 6a  now 

 (B + C) = 3a, A = 3a --(2)

By solving (1) and (2)

A = 3a , C = a and B = 3a - a = 2a

Efficiency of (A + C) = n(B)

(3a + a) = n × 2a

n = 4a/2a = 2

∴ The correct answer is n = 2

Given:

2a/3 = 4b/5 =  3c/4

Calculation:

LCM of numerator value (2, 4, 3) is 12

\((\frac{2 a}{3}=\frac{4 b}{5}=\frac{3 c}{4})\) × \(\frac{1}{12} \)

a/18 = b/15 = c/16

a : b : c = 18 : 15 : 16

a = 18u, b = 15u, c = 16u

Put the value in \(\frac{18}{a} \sqrt{a^2+c^2-b^2}\)

\(\frac{18}{18 u} \sqrt{(18u)^2+(16u)^2-(15u)^2}\)

⇒ √355

∴ The correct answer is √355.

Given:

Sum of deviations of n numbers from 10 and 20 are a, b respectively. 

​​\(\frac{b}{a}\) = - 4

Formula used:

Mean = A + \(\frac{∑ d}{n}\)

Where A is assumed to mean, 

∑d is sum of deviation, n = total terms

Calculation:

Mean = 10 + a/n       ----(1)

Mean = 20 + b/n       ----(2)

The mean is equal in equation (1) & (2). 

⇒ 10 + a/n = 20 + b/n

⇒ a/n - b/n = 10

⇒ a/n - (-4a)/n = 10       (∵ b/a = -4)

⇒ 5a/n = 10

⇒ a/n = 2

Put value of a/n in mean of equation (1)

Mean = 10 + 2 = 12

∴ The correct answer is 12.

Concept:

Median (M): The median is the middle number in a set to find the median, arrange all of the numbers (n) in the data set in ascending order, and then find the middle number. this is called median.

Formula used:

if n is odd then M = {(n + 1)/2}^th term

if n is even then M = {(n/2)th term + (n/2 + 1)th term} ÷ 2

Calculation:

Arrange the numbers in ascending order

1, 8, 12, 17, 21, 28, 35, 45, 54, 61

Total n = 10 then

We use M = {(n/2)th term + (n/2 + 1)th term} ÷ 2

(21 + 28)/2 = 24.5

2M + 5 = 2 × 24.5 + 5 = 54

∴ The correct answer is 54.

Calculation:

√ (x + 9) = x - 3

⇒ x + 9 = (x - 3)2

⇒ x + 9 = x2 - 6x + 9

⇒ x2 - 7x = 0

⇒ x(x - 7) = 0

For x = 0, the given equation doesn't satisfy.

For x = 7, the given equation satisfies.

∴ No. of real roots is one.

Concept used:

(a + b)² = a² + 2ab + b²

Calculation:

x = 97 + 56√3     ...(i)

Comparing the equation (i) with the given concept,

⇒ x = (a + b)2

⇒ 97 + 56√3 = a2 + 2ab + b2

So, 2ab = 56√3 = 2 × 7 × 4√3

Now, x = (7 + 4√3)²

⇒ √x = 7 + 4√3     ...(ii)

Comparing the equation (ii) with the given concept,

⇒ √x = (a + b)2

⇒ 7 + 4√3 = a2 + 2ab + b2

Here, 2ab = 2 × 2 × √3

Now, √x = (2 + √3)2

⇒ \(​​\sqrt [4] {x}\)  = 2 + √ 3

Similarly, \(1\over ​​\sqrt [4] {x}\) = 2 - √3

Now,

⇒ \(\sqrt [4] {x} \) + \(1 \over \sqrt[4]{x}\) = 2 + √ 3 + 2 - √3 = 4

∴ The answer is 4.

Given:

L : H = 3 : 2

Sum of two numbers = 45

Concept used:

Numbers are multiples of HCF.

Product of numbers = LCM × HCF

HCF (Highest Common Factor) is the largest number that can divide the given set of two or more numbers.

Calculation:

Let, the common factor of the ratio of L and H be m**.**

Now, LCM = 3m and HCF = 2m

According to the concept of HCF,

The numbers considered are 2ma and 2mb.

According to the question,

⇒ 2ma × 2mb = 3m × 2m

⇒ ab = 3/2     ---(i)

According to the question,

⇒ 2ma + 2mb = 45

⇒ 2m(a + b) = 45     ---(ii)

Now, using equations (i) and (ii), we can't find the value of all three variables a, b, and m.

∴ The answer cannot be determined due to insufficient data.

Given:

Distance between Home to Office = D

With S₁ = 3 km/hr, reaches 40 minutes early

With S₂ = 2 km/hr, reaches 40 minutes late

Formula Used:

Distance = Speed × Time

Calculation:

Let, original time to reach the office = T minutes

According to the question,

⇒ 3 × (T - 40) = 2 × (T + 40)

⇒ 3T - 120 = 2T + 80

⇒ 3T - 2T = 80 + 120

⇒ T = 200

Now, D = 3 × (T - 40)/60

⇒ D = 3 × (200 - 40)/60

⇒ D = 3 × 160/60

⇒ D = 3 × 8/3 = 8 km

∴ The distance is 8 km. 

Shortcut Trick 

According to the question,

⇒ 1 unit = 80 min

⇒ 2 unit = 80 × 2 = 160 min. = 8/3 hour

⇒ Distance = 3 × (8 / 3) = 8 km

∴ The distance is 8 km.

Calculation:

3x - 1 + 33 - x = 6

⇒ 3x - 1 + 33 - x = 3¹ + 3¹

Comparing the power of 3 in both side of the equation,

x - 1 = 1

⇒ x = 2

Similarly, 3 - x = 1

⇒ x = 2

So, x = 2

Now, 

⇒ 2x - 1 + 23 - x

⇒ 22 - 1 + 23 - 2 = 21 + 21 = 2 + 2 = 4

∴ The answer is 4. 

Shortcut Trick We know that 3 + 3 = 6

so put x = 2

3² - 1 + 33 - 2 = 6

than 

22 - 1 + 23 - 2 = 4

Given:

\(x\left(a-b+\frac{a b}{a-b}\right)\) = \(y\left(a+b-\frac{a b}{a+b}\right)\) and x + y = 2a³

Formula used:

a³ + b³ = (a + b) (a² + b² - ab)

(a + b)² = a² + b² + 2ab

Calculation:

We have

\(\left(a-b+\frac{a b}{a-b}\right)\) = \(\frac{(a\ -\ b )^2 \ +\ ab}{a\ -\ b}\) 

⇒ \(\left(a-b+\frac{a b}{a-b}\right)\) = \(\frac{a^2\ +\ b^2\ -\ ab }{a\ -\ b}\) 

Put the value of (a2 + b2 - ab) = (a3 + b3)/(a + b)

⇒ \(\frac{a^2\ +\ b^2\ -\ ab }{a\ -\ b}\) = \(\frac{a^3\ +\ b^3}{(a\ +\ b)(a\ - \ b)}\)

Similarly,

\(\left(a+b-\frac{a b}{a+b}\right)\) = \(\frac{a^3\ -\ b^3}{(a\ +\ b)(a\ - \ b)}\)

⇒ \(\left(a+b-\frac{a b}{a+b}\right)\) = \(\frac{a^3\ -\ b^3}{(a\ +\ b)(a\ - \ b)}\)

Now,

\(\frac{x}{y}\ =\ \frac{a^3\ -\ b^3}{(a\ +\ b)(a\ - \ b)}\ \times \frac{(a\ +\ b)(a\ - \ b)}{a^3\ +\ b^3}\)

\(⇒ \frac{x}{y} = \frac{a^3\ -\ b^3}{a^3\ +\ b^3}\)

Here, x = a3 - b³ and y = a3 + b3 

⇒ x - y = (a3 - b³) - (a3 + b³) = -2b³

Hence the value of x - y is -2b³.

Given:

Expression 3√3x3 + 2√2y3 - 18xy + 6√6

Formula used:

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Calculation:

The Expression 3√3x3 + 2√2y3 - 18xy + 6√6 can be written as

⇒ (√3x)³ + (√2y)³ + (√6)³ - 3 × √3x × √2y × √6

Here a = √3x, b = √2y and c = √6

Using a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)

The factors can be written as,

⇒ (√3x + √2y + √6) (3x² + 2y² + 6 - √6xy - √12y - √18x)

Hence the required factor is 3x2 + 2y2 + 6 - √6xy - √12y - √18x.

Given:

The number = 125¹⁰⁰

log₁₀2 = 0.301

Formula Used:

log mⁿ = n log m

log (m/n) = log m - log n

Calculation:

Let x = 125¹⁰⁰

Taking log on both sides

⇒ log₁₀x = 100 log₁₀125

⇒ log10x = 100 log₁₀53

⇒ log₁₀x = 100 × 3 log₁₀5

⇒ log10x = 300log₁₀(10/2)

⇒ log₁₀x = 300 × (log₁₀10 - log₁₀2)

⇒ log10x = 300 × (1 - 0.301)

⇒ log₁₀x = 209.69

The number of digits = 209 + 1 = 210

Hence, the number of digits in the expansion of  125100 is 210.

Calculation:

Expression 1:

acx3 + bcx2 +  adx2 + acdx + bdx + bcd

(acx3 + bcx2) + (adx2 + bdx) + (acdx + bcd)

cx²(ax + b) + dx (ax + b) + cd(ax + b)

(ax + b)(cx² + dx + cd)

Expression 2:

dx3 + acx2 + bdx2 + bcx + acdx + bcd

(adx3 + bdx2) + (acx2 + bcx) + (acdx + bcd)

dx²(ax + b) + cx(ax + b) + cd(ax + b) 

(ax + b)(dx² + cx + cd)

We know that, HCF is the highest power of each common prime factor.

∴ HCF = ax + b

Given:

xn - pyn + qzn is divisible by x2 + abyz - bzx - axy

Calculation:

We have 

x2 + abyz - bzx - axy

This can be written as

⇒ x2 - bzx - axy + abyz

⇒ x(x - bz) - ay(x - bz)

⇒ x² + (bz - ay)x + abyz

⇒ (x - bz) (x - ay)

Therefore, bz & ay will be factor of (x2 + abyz - bzx - axy)

As, x2 + abyz - bzx - axy is factor of xn - pyn + qzn

We can write 

xn - pyn + qzⁿ = (x - bz) (x - ay) × k    -----(1)

Put x = bz

⇒ bⁿzⁿ - pyⁿ + qzⁿ = 0     -----(2)

Put x = ay

⇒ aⁿyⁿ - pyⁿ + qzⁿ = 0     -----(3)

Equation (3) - Equation (1) 

⇒ bⁿzⁿ - aⁿyⁿ = 0

⇒ bnzn = anyn

From equation (3)

⇒ bnzn - pyn + qzn = 0          (∵ bnzn = anyn)

⇒ 1 - (pyn/bnzn) + q/bn = 0

⇒ 1 - (pyⁿ/aⁿyⁿ) + q/bⁿ = 0    (∵ bnzn = anyn)

⇒ \(\rm \frac{p}{a^n}-\frac{q}{b^n}\) = 1

Hence, the value of \(\rm \frac{p}{a^n}-\frac{q}{b^n}\) = 1.