Quantitative Aptitude PYQs

Given: 

log x = 1.25 and y = x^{log x}

Concept Used:

log ab = b log a

Calculation:

y = xlog x

Taking logarithm on both sides,

⇒ log y = log (x^{log x} )     

⇒ log y = (log x) log x       [∵ log a^b = b log a]

But, log x = 1.2500

⇒ log y = 1.25 × 1.25

∴ Value of log y is 1.5625.

Given:

f(x) is divided by (x – α)(x – β)

Formula Used:

According to the remainder theorem, when any function f(x) is divided by (x – a), then the remainder is equal to f(a)

Dividend = Divisor × Quotient + Remainder      ----(1)

⇒ f(x) = (x – a) × q(x) + f(a)

Calculation:

When f(x) is divided by (x – α), remainder = f(α)

⇒ f(x) = (x – α) × q₁(x) + f(α)

⇒ f(x)/(x – α) = q₁(x) + f(α)/(x – α)      ----(2)

Similarly,

When f(x) is divided by (x – β), remainder = f(β)

⇒ f(x) = (x – β) × q₂(x) + f(β)

⇒ f(x)/(x – β) = q₂(x) + f(β)/(x – β)      ----(3)

Subtracting (3) from (2), we get,

\( \Rightarrow \frac{{f\left( x \right)}}{{x - \alpha }} - \frac{{f\left( x \right)}}{{x - \beta }} = {q_1}\left( x \right) - {q_2}\left( x \right) + \frac{{f\left( \alpha \right)}}{{x - \alpha }} - \frac{{f\left( \beta \right)}}{{x - \beta }};\)

\( \Rightarrow \frac{{\left( {x - \beta - x + \alpha } \right)f\left( x \right)}}{{\left( {x - \alpha } \right)\left( {x - ;\beta } \right)}} = {q_1}\left( x \right) - {q_2}\left( x \right) + \frac{{\left( {x - \beta } \right)f\left( \alpha \right) - \left( {x - \alpha } \right)f\left( \beta \right)}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}}\)

\( \Rightarrow \frac{{\left( {\alpha - ;\beta } \right)f\left( x \right)}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}} = {q_1}\left( x \right) - {q_2}\left( x \right) + \frac{{(x - \beta f\left( \alpha \right) - \left( {x - ;\alpha } \right)f\left( \beta \right)}}{{\left( {x - \alpha } \right)\left( {x - ;\beta } \right)}}\)

\( \Rightarrow f\left( x \right) = \frac{{\left[ {{q_1}\left( x \right) - {q_2}\left( x \right)} \right]}}{{\left( {\alpha - \beta } \right)}}\left( {{\rm{x;}}-{\rm{;\alpha }}} \right)\left( {{\rm{x;}}-{\rm{;\beta }}} \right) + \frac{{\left( {{\rm{x}} - {\rm{\beta }}} \right)f\left( \alpha \right) - \left( {x - \alpha } \right)f\left( \beta \right)}}{{\left( {{\rm{\alpha }} - {\rm{;\beta }}} \right)}}\)

Comparing with eq. (1), we can write,

When f(x) is divided by (x – α)(x – β),

∴ Reminder is\(\rm \frac{{\left( {{\rm{x}} - {\rm{\beta }}} \right){\rm{f}}\left( {\rm{\alpha }} \right) - \left( {{\rm{x}} - {\rm{\alpha }}} \right){\rm{f}}\left( {\rm{\beta }} \right)}}{{\left( {{\rm{\alpha }} - {\rm{\beta }}} \right)}}\)

Given: 

Area of square = 2401x⁴ + 196x² + 4

Formula Used: 

Area of square = (Side)²

Calculation:

Area of square = 2401x4 + 196x2 + 4

⇒ Area of square = (49x²)² + (2 × 49x² × 2) + (2)²

∵ (a² + 2ab + b²) = (a + b)²

⇒ Area of square = (49x² + 2)²

⇒ Area of square = (Side)² = (49x² + 2)²

∴ Side of square is 49x² + 2

Given:

x ∝ yz

Calculation:

x ∝ yz

⇒ \(\frac{x}{z}\) ∝ y

⇒ \(\frac{z}{x}\) ∝ \(\frac{1}{y}\)

∴ y varies inversely as \(\frac{z}{x}\)

Given:

The points P and Q represents real number \(0.7\bar 3\) and \(0.5\bar 6\) on the number line

Calculation:

Let, x = \(0.7\bar 3\)= 0.733333…

⇒ 100x = 73.3333…

⇒ 10x = 7.3333…

⇒ 100x – 10x = 73.3333… – 7.3333…

⇒ 90x = 66

⇒ x = 66/90 = 11/15

Similarly,

Let, y = \(0.5\bar 6\) = 0.566666…

⇒ 100y = 56.6666…

⇒ 10y = 5.6666…

⇒ 100y – 10y = 56.6666… – 5.6666…

⇒ 90y = 51

⇒ y = 51/90 = 17/30

⇒ x – y = (11/15) – (17/30) = 1/6

∴ Distance between point P and Q is 1/6

Given:

5x + 2y = 7xy

10x + 3y = 8xy

Calculation:

5x + 2y = 7xy        ----(1)

10x + 3y = 8xy       ----(2)

Multiply equation 1 by 2 and subtract equation 2 from it, we get

⇒ y = 6xy

⇒ x = 1/6

Now, from equation 1:

⇒ 5/6 + 2y = 7y/6

⇒ 5y/6 = -5/6

⇒ y = -1

∴ \(\left( {\frac{1}{6},; - 1} \right)\) is the required point.

Shortcut Trick

Calculate successive differences step-by-step using multiplication factors 1.2 and 1.1.

Diff 2011 = 5000 × 1.2 − 2000 × 1.1 = 6000 − 2200 = 3800

Diff 2012 = 6000 × 1.2 − 2200 × 1.1 = 7200 − 2420 = 4780

Diff 2013 = 7200 × 1.2 − 2420 × 1.1 = 8640 − 2662 = 5978

In 2013, the price difference 5978 clearly exceeds 5000.

∴ The correct answer is 2013.

Alternate Method

Given:

Initial Price of X in 2010 = ₹5000, rate of increase = 20%

Initial Price of Y in 2010 = ₹2000, rate of increase = 10%

Formula Used:

Price after n years = Initial Price × (1 + Rate/100)ⁿ

Price Difference Progression

Year
Price X
Price Y
Difference

2010
₹5000
₹2000
₹3000

2011
₹6000
₹2200
₹3800

2012
₹7200
₹2420
₹4780

2013
₹8640
₹2662
₹5978

Calculations:

⇒ In 2010: Difference = 5000 − 2000 = ₹3000

⇒ In 2011: X = 5000 × 1.2 = ₹6000, Y = 2000 × 1.1 = ₹2200

⇒ Difference in 2011 = 6000 − 2200 = ₹3800

⇒ In 2012: X = 6000 × 1.2 = ₹7200, Y = 2200 × 1.1 = ₹2420

⇒ Difference in 2012 = 7200 − 2420 = ₹4780

⇒ In 2013: X = 7200 × 1.2 = ₹8640, Y = 2420 × 1.1 = ₹2662

⇒ Difference in 2013 = 8640 − 2662 = ₹5978

⇒ As ₹5978 > ₹5000, the difference exceeds ₹5000 for the first time in 2013.

∴ The correct answer is 2013.

Additional Information

Successive Percentage Increase

When a value increases continuously over years, the formula is: Final Value = Initial Value × (1 + R/100)ⁿ, which uses the exact same principles as the Compound Interest formula.

Multiplying Factor Method

To increase a given number by R%, multiply it directly by the factor (1 + R/100). For example, a 20% continuous yearly increase means multiplying by 1.2 repeatedly each year.

Percentage Change Calculation

The overall percentage change between two values is calculated as: % Change = (Final Value − Initial Value) ÷ Initial Value × 100.

Given:

The ratio of speeds of X and Y is 5 : 6.

Y allows X a start of 70 m in a 1.2 km race

Formula:

Distance = Speed × Time

Calculation:

Suppose the speeds of X and Y are 50 m/s and 60 m/s respectively.

Y allows X a start of 70 m in a 1.2 km race;

That means X covers 1130 m and Y covers 1200 m.

Time in which X covers 1130 m = 1130/50 = 22.6 sec

And

Time in which Y covers 1200 m = 1200/60 = 20 sec

It is evident that Y wins the race and X takes 2.6 sec more than Y to finish the race.

∴ Distance by which Y wins the race = 2.6 × 50 = 130 m

Given:

A train takes two hours less for a journey of 300 km

Formula Used:

Distance = Speed × Time

Calculation:

Suppose the usual speed of the train = x km/hr

According to the question:

300/x – 300/(x + 5) = 2

⇒ 300x + 1500 – 300x = 2x² + 10x

⇒ 2x² + 10x – 1500 = 0

⇒ 2x² + 60x – 50x – 1500 = 0

⇒ 2x (x + 30) – 50(x + 30) = 0

⇒ (x + 30) (2x – 50) = 0

⇒ x = 25      (As negative value of x is not possible)

∴ Usual speed of train is 25 km/hr

Given:

6 men and 8 women can do a piece of work in 10 days

13 men and 24 women can do the same work in 4 day

Calculation:

Suppose M and W are the efficiencies (daily work done) of 1 man and 1 woman respectively.

According to the question,

(6M + 8W) × 10 = (13M + 24W) × 4

⇒ 60M + 80W = 52M + 96W

⇒ 8M = 16W

⇒ M = 2W

⇒ M/W = 2/1

∴ M : W is 2 : 1

Given:

The number of students is increased by 2 in each row, then the number of rows decrease by 3

The number of rows decreases by 3. If the number of students is increased by 4 in each row

Concept Used:

By method of elimination of 2 variable

Calculation:

Suppose total number of rows is A and number of students in each row is B;

∴ Total students = A × B

According to the question:

(A – 3) × (B + 2) = A × B

⇒ 2A – 3B = 6      ---- (1)

And

(A – 5) × (B + 4) = A × B

⇒ 4A – 5B = 20      ---- (2)

Multiply equation 1 by 2 and subtract equation 2 from it:

⇒ B = 8

Now from equation 1:

⇒ 2A = 6 + 24

⇒ A = 15

Then  A × B = 120

∴ Total number of students in the class is120

Given:

Time = 2 year

Difference of 1% in interest rates, SI increases by Rs. 24

Formula used:

Simple interest = (P × R × T)/100

Calculation:

Suppose the principal is Rs. P

Time = 2 years

From the question, we can say that due to a difference of 1% in interest rates, SI increases by Rs. 24. So,

24 = (P × 2 × 1)/100

⇒ P = 1200

∴ The principal amount is 1200

Given:

The populations of two villages are 1525 and 2600 respectively

The ratio of male to female population in the first village is 27 : 34

The ratio of male to female population in the second village is 6 : 7

Concept Used:

If the ratio of male and female is a : b

Then male = a/(a + b)

Calculation:

Population of 1^st village = 1525

Ratio of male to female population in the first village is 27 : 34

So,

Number of males in first village = 1525 × (27/61) = 25 × 27 = 675

Number of females in first village = 1525 – 675 = 850

And

Population of 2^nd village = 2600

Ratio of male to female population in the 2^nd village is 6 : 7

So,

Number of males in 2^nd village = 2600 × 6/13 = 200 × 6 = 1200

Number of females in 2^nd village = 2600 – 1200 = 1400

Hence,

Ratio of male to female population of these two villages taken together

= (675 + 1200) : (850 + 1400)

= 1875 : 2250

= 5 : 6

∴ The ratio of male to female population of these two villages 5 : 6 taken together

Given:

The ratio of number of girls to that of boys is 3 : 4. The average height of students in the class is 4.6 feet.

The average height of the boys in the class is 4.8 fee

Concept Used:`

The ratio of a and b is a₁ : b₁, total average is P, and average of a and b is c and d respectively.

Total average (P) =  (a₁ × c + b₁ × d)/(a₁ + b₁)

Calculation:

Suppose the number of girls & boys in the class are 3a and 4a;

The average height of all students is 4.6 feet and average height of the boys is 4.8 feet.

Suppose average height of the girls is X feet;

So,

4a × 4.8 + 3a × X = 4.6 × (3a + 4a)

19.2 + 3X = 32.2

3X = 13

X = 4.33

∴ Average height of the girls in the class is more than 4.3 feet but less than 4.4 feet.

Given:

The data 3, 5, 9, 4, 6, 11, 18

Formula Used:

Median = [(n + 1)/2]^th terms

Calculation:

Data in ascending order = 3, 4, 5, 6, 9, 11, 18

Total numbers (n) = 7 (Odd)

Median = [(n + 1)/2]^th terms

Median = [(7 + 1)/2]^th terms

Median = 4^th term

And 4^th term is 6

∴ Median of 4^th term is 6

Given:

The ratio of the angles of the sectors is 1 : 2 : 3

Concept Used:

The sum of angles of in a pie chart = 360°

Calculation:

The ratio of the angles of the sectors is 1 : 2 : 3

We know the sum of angles of in a pie chart = 360°

So, largest angle = [3/(1 + 2 + 3)] × 360° = 360°/2 = 180°

∴ The largest angle is 180°.

Given:

Maximum marks are converted from 250 to 50.

 The highest marks scored were 170 and the lowest marks were 70

Calculation:

Maximum marks are converted from 250 to 50.

Highest marks 170 will be converted to: 170 × (50/250) = 34

Lowest marks 70 will be converted to: 70 × (50/250) = 14

Difference = 34 – 14 = 20

∴ The difference is 20.

Given:

Formula used:

If a, b and c is the numbers of student in class

% of a = [a/(a + b + c)] × 100

Calculation:

Percentage of trained convicts:

In jail A: [45/(86 + 45 + 25)] × 100 = [45/156] × 100 = 28.8%

In jail B: [903/(1305 + 903 + 461)] × 100 = [903/2669] × 100 = 33.8%

In jail C: [940/(2019 + 940 + 474)] × 100 = [940/3433] × 100 = 27.3%

In jail D: [869/(1166 + 869 + 416)] × 100 = [869/2451] × 100 = 35.4%

In jail E: [544/(954 + 544 + 254)] × 100 = [544/1752] × 100 = 31.05%

In jail F: [464/(1198 + 464 + 174)] × 100 = [464/1836] × 100 = 25.2%

∴ Jails with highest and smallest percentage of trained convicts are D and F respectively.

Given:

Formula used:

If a, b and c is the numbers of student in class

% of a = [a/(a + b + c)] × 100

Calculation:

Placement rate of trained convicts:

In jail A: [25/45] × 100 = 55.55%

In jail B: [461/903] × 100 = 51.05%

In jail C: [(474/940) × 100] = 50.42%

In jail D: [416/869] × 100 = 47.87%

In jail E: [(254/544) × 100] = 46.7%

In jail F: [174/464] × 100 = 37.5%

∴ Jail with highest placement rate of trained convicts is jail A.

Given:

Formula used:

If a, b and c is the numbers of student in clss

% of a = [a/(a + b + c)] × 100

Calculation:

Placement rate of trained convicts:

In jail A: [25/45] × 100 = 55.55%

In jail B: [461/903] × 100 = 51.05%

In jail C: [474/940] × 100 = 50.42%

In jail D: [416/869] × 100 = 47.87%

In jail E: [254/544] × 100 = 46.69%

In jail F: [174/464] × 100 = 37.50%

∴ From Jails A, B and C, more than half of the trained convicts are offered jobs.