Quantitative Aptitude PYQs

Given :

x3 + x2 + 16 is exactly divisible by x, where x is a positive integer

Calculations :

Let a number N be equal to x3 + x2 + 16. 

N = x3 + x2 + 16 

Now divide N/x 

N/x = (x3 + x2 + 16)/x 

⇒ x² + x + (16/x)

As N is divisible by x so 16 must be divisible by x 

Values of x that can divide 16 are 1, 2, 4, 8 and 16 

So there are total 5 values of x that can divide N without leaving the remainder.

∴ Option 3 is the correct answer.

Given :

There are 3 number a, b and c where abc = 30.

Calculations :

All divisors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. 

Selecting any 3 numbers from all divisors so that we get multiplication as 30 

1. 1 × 1 × 30 = 30 

Number of ways to arrange 1, 1 and 30 = 3!/2 = 3 

2. 1 × 3 × 10 = 30 

Number of ways to arrange 1, 3 and 10 = 3! = 6 

3. 2 × 3 × 5 = 30 

Number of ways to arrange 2, 3 and 5 = 3! = 6

4. 1 × 5 × 6 = 30 

Number of ways to arrange 1, 5 and 6 = 3! = 6 

5. 1 × 2 × 15 = 30 

Number of ways to arrange 1, 2 and 15 = 3! = 6   

Total solutions = 6 + 6 + 6 + 6 + 3

⇒ 27 ways 

∴ Option 2 will be the correct choice.

Alternate MethodNumber of the integral solution for abc = 30 are:

1 × 3 × 10 ⇒ Permutation = 3 !

15 × 2 × 1  ⇒  Permutation = 3 !

5 × 3 × 2 ⇒ Permutation = 3 !

5 × 6 × 1 ⇒ Permutation = 3 !

30 × 1 × 1 ⇒ Permutation = 3! / 2!

Total solutions = (3! × 4) + 3 = 27

Given:

The quadratic equation x2 - 4x - log10 N = 0 

Concept Used:

The quadratic equation ax² + bx + c = 0,

For minimum value of b² - 4ac ≥ 0

If log_a b = c 

Then b = a^c

Calculation:

The quadratic equation x2 - 4x - log10 N = 0 

Compare the above equation ax² + bx + c = 0 

Then a = 1, b = - 4 and c = - log₁₀ N

Putting the value of b² - 4ac ≥ 0

⇒ (- 4)² - 4⋅1⋅(- log₁₀ N) ≥ 0

⇒ 16 = - 4⋅log₁₀ N

⇒ log₁₀ N = - 4

⇒ N = (10)^-4

⇒ N = \(\rm 1\over 10000\)

∴The minimum value of N = \(\rm 1\over 10000\)

Given : 

x,y and z is a positive integer and x + y + z = 12 

Concept used :

When we have n positive integrals and we have to distribute them to r different things then 

No of ways = ^{n - 1}c_{r - 1} 

Calculations :

x, y and z can be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12 

Total number to select are 3 

Total number of solutions = ^{12 - 1}c_{3 - 1} = ¹¹c₂

⇒ 11!/[(11 - 2)!.2!]

⇒ 11!/(9! × 2!) 

⇒ 55 

∴ Option 3 will be the correct answer.

Given :

I = a2 + b2  + c2, where a and b are consecutive integers and c = ab.

Calculations :

Let a and b be 1 and 2 respectively    (a and b are consecutive numbers)

So, c = 1 × 2 = 2 

Now 

⇒ I = 1² + 2² + 2²

⇒ I = 9 

Now let the values of a and b be 2 and 3 respectively 

So, c = 2 × 3 = 6 

Now, 

⇒ I = 2² + 3² + 6² 

⇒ I = 49

In both cases, it is coming out to be the square of an odd integer.

∴ Option 4 will be the correct choice.

Given :

An eleven digit number 23P62971335.

Concept used :

A number is divisible by 9 when the sum of all its digits is divisible by 9. 

Calculations :

23P62971335 is divisible by 9 (the smallest odd composite number is 9) 

So sum of digits of  23P62971335 must be divisible by 9 

Sum of digits = 2 + 3 + P + 6 + 2 + 9 + 7 + 1 + 3 + 3 + 5 

⇒ 41 + P 

At P = 4 we get 45 which is divisible by 9 

∴ Option 1 will be the correct answer.

Shortcut Trick 

{1⁵ + (2²)² × 2 + (3²)² × 3 + 4⁵ + (5²)² × 5} / 4

{1 + 0 + (1 × 3) + 0 + (1 × 1)} / 4 = 5/4 = 1

∴ Option 2 will be the correct answer.

Alternate Method 

Concept Used:

Divisibility rule of 4:

A number is divisible by 4 if its last two digits are divisible by 4

Calculations :

Remainder when 1⁵  divided by 4 = 1

Remainder when 2⁵ or 32 divided by 4 = 0 

Remainder when 3⁵ or 243 divided by 4 = 3 

Remainder when 4⁵ or 1024 divided by 4 = 0 

Remainder when 5⁵ or 3125 divided by 4 = 1 

Remainder = 1 + 3 + 1 = 5 

Remainder when 5 divided by 4 = 1 

∴ Option 2 will be the correct answer.

Given:

3⁹⁹

Concept Used:

Unit digit of 3ⁿ

3¹ unit digit 3

3² unit digit 9

3³ unit digit 7

3⁴ unit digit 1

3⁵ unit digit 3... 

After 4 power unit digits repeat again.

Calculation:

⇒ 3⁹⁹ = (3⁴)²⁴ × (3³)

⇒ 3⁹⁹ = (1)²⁴ × 7 = (1) × 7

Unit place = 1 × 7 = 7

∴ The place of 3⁹⁹ is 7

Shortcut Trick 

X + 28X = 1740

⇒ x = 60

⇒ 60 × (28 × 60) = 240 × y

⇒ y = 420

∴ The other number is 420.

Alternate Method 

Given:

LCM of two number = 28 × HCF

First number = 240

HCF + LCM = 1740

Formula used:

LCM × HCF = first number × second number

Calculation:

Let HCF be x and the second number be y

HCF = x

LCM = 28x

According to the question,

The Sum of the HCF and the LCM is 1740

⇒ HCF + LCM = 1740

⇒ x + 28x = 1740

⇒ 29x = 1740

⇒ x = 60

∴ HCF = 60, LCM = 28 × 60 = 1680

By using formula,

LCM × HCF = first number × second number

⇒ 1680 × 60 = 240 × y

⇒ y = 420

∴ The other number is 420.

Calculations :

(xn - an) is divisible by (x-a) where n can be any natural number. 

∴ Option 1 will be the correct answer.

Additional InformationNatural Number:

The numbers that we use when we are counting or ordering {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 …}.

Odd Number:

A whole number that is not able to be divided by two into two equal whole numbers The numbers 1, 3, 5, and 7 are odd numbers.

Even Number:

A whole number that is able to be divided by two into two equal whole numbers The numbers 0, 2, 4, 6, and 8 are even numbers.

Prime Number:

Prime numbers are numbers that have only 2 factors: 1 and themselves. For example, the first 5 prime numbers are 2, 3, 5, 7, and 11.

Shortcut Trick

172020 is divided by 18 then

\(\rm = {17^{2020}} = {{(18\ -\ 1)^{2020}}\over 18} = (-\ 1)^{2020} = 1\)

Alternate Method

Concept used:

Divisibility rule of 18:

A number that is divisible by 18 must be divisible by both 2 and 9. 

Formula used:

(xⁿ - aⁿ) → divisible by (x + a) if n is even number.

Calculation:

By using the formula, we get

(xn - an) → divisible by (x + a) if n is even number.

Where,

x = 17, n = 2020, a = 1

⇒ (17²⁰²⁰ - 1²⁰²⁰)/(17 + 1) = (-1)²⁰²⁰/18

When 17²⁰²⁰ is divided by 18 getting the remainder 1

∴ The required answer is 1.

Additional InformationDivisibility rule of 2:

If a number ends in a 0, 2, 4, 6 or 8, it is divisible by 2. 

Divisibility rule of 9:

If the sum of digits of the number is divisible by 9, then the number itself is divisible by 9.

Given:

\(\frac{1}{{1 + √ 2 }} + \frac{1}{{√ 2 + √ 3 }} + \frac{1}{{√ 3 + √ 4 }} + \ldots + \frac{1}{{√ {99} + √ {100} }}\)

Formula used:

(a² - b²) = (a + b) (a - b)

Calculation:

Rationalizing the denominator, we get

⇒ \(\frac{1}{{\left( {1 + √ 2 } \right)}} \times \frac{{\left( {√ 2 - 1} \right)}}{{\left( {√ 2 - 1} \right)}} + \frac{1}{{\left( {√ 2 + √ 3 } \right)}} \times \frac{{(√ 3 - √ {2)} }}{{\left( {√ 3 - √ 2 } \right)}} + \dots\frac{1}{{(√ {99} + √ {100)} }} \times \frac{{\left( {√ {100} - √ {99} } \right)}}{{\left( {100 - √ {99} } \right)}}\)

On canceling the negative and positive term, we get

⇒ \(\frac{{\left( {√ 2 - 1} \right)}}{1} + \frac{{\left( {√ 3 - √ 2 } \right)}}{1} + \frac{{\left( {√ 4 - √ 3 } \right)}}{1} + \ldots \frac{{\left( {√ {100} - √ {99} } \right)}}{1}\)

⇒ \(√ 2 - 1 + √ 3 - √ 2 + √ 4 - \surd 3 \ldots + √ {100} - √ {99} ;\)

⇒ √100 - 1 = 9

∴ The required value is 9

Given: 

\(\mathbf{x}^{\mathbf{m}}=\sqrt[14]{\mathbf{x} \sqrt{\mathbf{x} \sqrt{\mathbf{x}}}}\)

Concept:

x^a⋅x^b = x^{(a + b)}

(x^a)^b = x^a⋅b

Calculation:

\(\mathbf{x}^{\mathbf{m}}=\sqrt[14]{\mathbf{x} \sqrt{\mathbf{x} \sqrt{\mathbf{x}}}}\)

⇒ \(\rm \mathbf{x}^{\mathbf{m}}=[{\mathbf{x} \sqrt{\mathbf{x} ({\mathbf{x})^{1\over 2}}}}]^{1\over 14}\)

⇒ \(\rm \mathbf{x}^{\mathbf{m}}=[{\mathbf{x} {\mathbf({x})^{1\over 2} ({\mathbf{x})^{1\over 4}}}}]^{1\over 14}\)

⇒ \(\rm x^m = (x^{1 + {1\over 2} +{1\over 4}})^{1\over 14}\)

⇒ \(\rm x^m = [x^{(4 + 2 + 1)\over 4}]^{1\over 14}\)

⇒ \(\rm x^m = (x^{7\over 4})^{1\over 14}\)

⇒ \(\rm x^m = x ^{1\over 8}\)

∴ The value of m = \(\rm 1\over 8\)

Given: 

1, 2, 3, 4, 5, - 1, - 2, - 3, - 4 and - 5

Formula used:

(a + b)² = a² + 2ab +b²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca) 

(a + b + c + d + e + ....................)2 = (a2 + b2 + c2 + d2 + e2 +......) + 2(ab + bc + ca +......) 

1² + 2² + 3² + ...... + n² = \(\rm {{n\cdot(n\ +\ 1)\cdot(2n\ +\ 1)}\over 6}\)

Calculation:

Let, sum of all possible products taken two at a time = X

(1 + 2 + 3 + 4 + 5 - 1 - 2 - 3 - 4 - 5)² = (1)² + (2)² + (3)² + (4)² + (5)² + (- 1)² + (- 2)²  (- 3)² + (- 4)² + ( - 5)² + 2(X)

⇒ 0 = 2[(1)2 + (2)2 + (3)2 + (4)2 + (5)2]] + 2X

⇒ 2X = - 2[(1)2 + (2)2 + (3)2 + (4)2 + (5)2]]

⇒ X = \(\rm -\ {{5\ \times\ 6\ \times\ 11}\over{6}}\)

⇒ X = - 55

∴ The sum of all possible products = - 55

Given:

Length of train = 110 m

Speed of train = 132 km/hr

Length of bridge = 165 m

Formula used:

Distance = Speed × Time

1 km/hr = (5/18) m/sec

Calculation:

Length of train = 110 m

Speed of train = 132 km/hr = 132 × (5/18) = 110/3 m/sec

Length of bridge = 165 m 

Total distance = 110 + 165 = 275 m

Time = Distance/Speed

Time = 275/(110/3)

⇒ Time = 15/2 = 7.5 seconds

∴ The required time is 7.5 seconds.

Shortcut Trick

When the simple interest is a fraction of the principal and Rate = Time:

Time = √((SI ÷ P) × 100)

Given SI ÷ P = 1 ÷ 4

⇒ Time = √((1 ÷ 4) × 100) = √25 = 5 years

∴ The correct answer is 5 years.

Alternate Method

Sum = P

R = T

SI = P/4

SI = (P × R × T) ÷ 100

Simple Interest Analysis

Given:

Simple Interest (SI) = P ÷ 4

Rate of Interest (R) = Time in years (T)

Formula Used:

SI = (P × R × T) ÷ 100

Calculations:

⇒ P ÷ 4 = (P × T × T) ÷ 100

⇒ 1 ÷ 4 = T² ÷ 100

⇒ T² = 100 ÷ 4

⇒ T² = 25

⇒ T = √25 = 5

∴ The correct answer is 5 years.

Additional Information

Direct Formula for Equal Rate and Time

When Simple Interest is 1/n of the principal and the rate percentage equals the time in years, directly apply T = √(100 ÷ n).

Simple Interest vs Compound Interest

For the first year, both SI and CI are equal for a given principal and rate of interest. After the first year, CI calculates interest on the accumulated sum.

Rule of 72

A mental math shortcut to estimate the doubling time of an investment under compound interest: Time ≈ 72 ÷ Annual Interest Rate.

Given :

A 60-page book has n lines per page, when the number of lines were reduced by 3 on each page, then the number of pages will increase by 10. 

Calculations :

Total number of lines = Page × Lines per page 

⇒ 60 × n = 60n 

Now the number of lines when 3 lines are reduced = n - 3 

Number of pages when lines are reduced = 70 

Total number of lines = 70 × (n - 3) = 70n - 210 

So, 

60n = 70n - 210 

⇒ 10n = 210 

⇒ n = 21  

∴ Option 2 is the correct answer.

Given : 

x men working x hours per day can do x units of work in x days

Shortcut Trick

Formula Used:

M1D1H1/W₁ = M2D2H2/W2

Where, M = men, D = days, H = hours, w = works

Calculations:

x.x.x/x = y.y.y/k

⇒ x² = y³/k

⇒ k =  y3/x² = y³x^-2

∴ The correct answer is option (4).

Alternate Method

Concept used :

Total work = Number of persons × Total time × Each person's work per unit time 

Calculations :

Let the work done by each person per hour be 'w' units 

According to the questions 

x = x × x × x × w         (total time - total hours × total days)

⇒ w = 1/x²

Now work done by y men in y hours and y days 

Total work(k) = y × y × y × (1/x²) 

⇒ y³/x²

⇒ y³x^-2 

∴ Option 4 will be the correct choice.

Given :

d(n) is the number of positive divisors of a positive number n. 

Calculations :

d(5) = 2       (positive divisors of 5 are 1 and 5) 

d(11) = 2     (positive divisors of 11 are 1 and 11) 

d(16) = 5     (positive divisors of 16 are 1, 2, 4, 8 and 16) 

d(55) = 4     (positive divisors of 55 are 1, 5, 11 and 55)

Checking Choices 

1. d(5) = d(11)

⇒ 2 = 2 

2. d(5).d(11) = d(55) 

⇒ 2 × 2 = 4 

⇒ 4 = 4 

3. d(5) + d(11) = d(16) 

⇒ 2 + 2 = 5 

We can conclude that choice 1 and 2 are correct. 

∴ Option 2 is the correct answer.

Given: 

An = Pn + 1, where \(P_n\) is the product of the first n prime numbers

Concept:

Odd number + odd number = always an even number 

Odd number + even number = always an odd number 

Even number + enev number = always an even number

Calculation:

Here P_n = 2, 3, 5, 7, 11, 13,............

We  know that when we add one in the product of first prime numbers it becomes also a prime number

So, A_n is a prime number, not a composite number

All the prime numbers (except 2) are an odd number 

∵ Odd number + even number = always an odd number 

So, A_n + 2 is always an odd number. 

 ∵ Odd number + odd number = always even number    

So, A_n +  1 is always an even number.       

∴ Option (4) is correct.

Alternate Method

Prime  numbers are 2, 3, 5, 7, 11, ......

Let n = 1

⇒ P1 = 2 (i.e. product of 1 prime number)

⇒ A1 = P1 + 1 = 2 + 1 = 3  (i.e. odd prime number)

Hence, A1 is a prime number

A1 + 2 = 3 + 2 = 5  (i.e. odd number)

Hence, statement 2 is correct.

A1 + 2 = 3 + 1 = 4  (i.e. even number)

Hence, statement 3 is also correct.

∴ Option (4) is correct.