Quantitative Aptitude PYQs

Shortcut Trick

For the quadratic equation x² − 7x + 1 = 0, the sum of roots (α + β) = 7 and product (αβ) = 1.

Using the property: α² + β² = (α + β)² − 2αβ = 7² − 2(1) = 47.

Then, α⁴ + β⁴ = (α² + β²)² − 2(αβ)² = 47² − 2(1)² = 2209 − 2 = 2207.

∴ The correct answer is 2207.

Alternate Method

Given: The quadratic equation is x² − 7x + 1 = 0.

Formula Used:

1. Sum of roots (α + β) = −b/a
2. Product of roots (αβ) = c/a
3. α² + β² = (α + β)² − 2αβ
4. α⁴ + β⁴ = (α² + β²)² − 2(αβ)²

Calculations:

⇒ Comparing with ax² + bx + c = 0, we have a = 1, b = −7, c = 1.

⇒ α + β = −(−7)/1 = 7

⇒ αβ = 1/1 = 1

⇒ α² + β² = 7² − 2(1) = 49 − 2 = 47

⇒ α⁴ + β⁴ = 47² − 2(1)²

⇒ α⁴ + β⁴ = 2209 − 2 = 2207

∴ The correct answer is 2207.

Additional Information

Roots and Coefficients

For a quadratic equation ax² + bx + c = 0, the sum of roots is −b/a and the product is c/a.

Algebraic Identity for Fourth Power

The sum of fourth powers can be derived iteratively from the sum of squares: x⁴ + y⁴ = (x² + y²)² − 2x²y².

Reciprocal Roots Case

If the product of roots is 1 (c = a), the roots are reciprocals of each other, satisfying the relation x + 1/x = −b/a.

Concept:

For the number "a", which can be written as,

 \(a=p_1^{\alpha_1} \cdot p_2^{\alpha_2}....\) 

The total number of positive factors is given as,

T(a) = (a1+1)(a2+1)

Here, p1 and p2 are the prime numbers.

Sum of factors = \(\frac{P_1^{\alpha_1+1} - 1}{P_1 - 1} × \frac{P_2^{\alpha_2+1} - 1}{P_2 - 1} × \dots\)

Calculation:

360 = 2³ × 3² × 5¹

Here, 2, 3 & 5 are the only prime number.

Total number of factor = (3 + 1) × (2 + 1) × (1 + 1) = 24

Sum of factors = \(\frac{2^{3+1}-1}{2-1}× \frac{3^{2+1}-1}{3-1} × \frac{5^{1+1}-1}{5-1}\)

Sum of factors = (2⁴ - 1) × [(3³ - 1)/2] × [(5² - 1)/4]

Sum of factors = (15) × (13) × 6

Sum of factors = 1170

∴ The number of factors of 360 is 24 and the sum of all factors is 1170.

Calculation:

XYXYXY = 100000X + 10000Y + 1000X + 100Y + 10X + Y

XYXYXY = (100000X + 1000X + 10X) + (10000Y + 100Y + Y)

XYXYXY = 101010X + 10101Y

XYXYXY = 10101 × (10X) + 10101(Y)

XYXYXY = 10101.[10X + Y]

Prime factors of 10101 are  3 x 7 x 13 x 37

XYXYXY =  3 x 7 x 13 x 37 (10X + Y)

∴ XYXYXY is divisible by 3, 7, 13 & 37.

Concept:

HCF: It is the product of the smallest power of the common factors involved in

the prime factorization.

HCF of (a^m - 1) & (aⁿ - 1) is [a^{HCF (m, n)} - 1]

Calculation:

329 - 9 = 3² (3²⁷ - 1)

338 - 9  = 3² (3³⁶ - 1)

By using the above concept

HCF = 3² [3^{HCF (27, 36)} - 1]

Let's find the HCF of (27, 36) first

27 = 3³

36 = 2² × 3²

HCF of (27, 36) = 3²

Now, the required value

HCF = 3² (3⁹ - 1) = 3¹¹ -  9

Mistake Points Do not forget to multiply with 3² as this will also be the common prime factor.

Calculation:

Given that 

x =\(√{4 √{4 √{4 √{4 \ldots}}}}\)

This can be written as 

⇒ x = √4x

⇒ x² = 4x

⇒ x² - 4x = 0

⇒ x(x - 4) = 0

⇒ x = 0 & 4

∴ The required value is 4.

Shortcut Trick If y =  \(\sqrt{p \sqrt{p \sqrt{p \sqrt{p \ldots}}}}\) 

than always answer is p

∴  x = \(\sqrt{4 \sqrt{4 \sqrt{4 \sqrt{4 \ldots}}}}\) 

∵ x  = 4

Shortcut Trick

We are given the expression 33m + 22n, which can be factored as 11(3m + 2n).

For this to be divisible by 121 (which is 11 × 11), the term (3m + 2n) must be a multiple of 11.

To minimize the sum (m + n), we evaluate the smallest positive multiple, setting 3m + 2n = 11.

Testing natural numbers: if m = 3 and n = 1, then 3(3) + 2(1) = 11, giving m + n = 4.

Alternatively, if m = 1 and n = 4, then 3(1) + 2(4) = 11, giving m + n = 5.

∴ The correct answer is 4.

Alternate Method

Given:

Expression: 33m + 22n

Divisor: 121

m and n are natural numbers (m, n ≥ 1).

Formula Used:

If a number N is divisible by D, then N = k × D for some positive integer k.

Calculations:

⇒ 33m + 22n = 121k

⇒ 11(3m + 2n) = 11 × 11k

⇒ 3m + 2n = 11k

To minimize (m + n), we should take the smallest possible positive integer value for k. Let k = 1.

⇒ 3m + 2n = 11

⇒ 2n = 11 − 3m

Since n must be a natural number, 2n is an even positive integer. Therefore, (11 − 3m) must be an even positive integer.

Let's check possible natural number values for m:

⇒ If m = 1: 2n = 11 − 3(1) = 8 ⇒ n = 4. The sum is m + n = 1 + 4 = 5.

⇒ If m = 2: 2n = 11 − 3(2) = 5 (Not possible, since n would not be an integer).

⇒ If m = 3: 2n = 11 − 3(3) = 2 ⇒ n = 1. The sum is m + n = 3 + 1 = 4.

⇒ If m ≥ 4: 11 − 3m will be negative or zero, which cannot yield a natural number for n.

Thus, the possible pairs of (m, n) for k = 1 are (1, 4) and (3, 1).

The sums are 5 and 4 respectively. The minimum value of (m + n) is 4.

∴ The correct answer is 4.

Additional Information

Divisibility Rule of 11

A number is divisible by 11 if the difference between the sum of its digits at odd positions and the sum of its digits at even positions is either 0 or a multiple of 11.

Linear Diophantine Equations

An equation of the form ax + by = c has integer solutions if and only if the greatest common divisor of a and b perfectly divides c.

Natural Numbers

Natural numbers are the set of positive integers starting from 1 (i.e., 1, 2, 3, 4, ...). Zero, fractions, decimals, and negative numbers are not considered natural numbers.

Shortcut Trick

Using the identity: (Sum)² = Sum of squares + 2 × Product

⇒ (Sum)² = 4896 + 2 × 2160 = 4896 + 4320 = 9216

⇒ Sum of the two numbers = √9216 = 96

⇒ Mean = Sum ÷ 2 = 96 ÷ 2 = 48

*(

Note:

The given HCF = 12 is redundant and not needed for this method)*

∴ The correct answer is 48.

Alternate Method

Given:

Product of two numbers = 2160

Sum of their squares = 4896

HCF of the numbers = 12

Sum of two numbers = 96
Mean = 96 ÷ 2 = 48
Mean = 48
Mean = 48

Calculations:

⇒ Let the two numbers be 12x and 12y (where x and y are co-prime).

⇒ Product = 12x × 12y = 144xy = 2160

⇒ xy = 2160 ÷ 144 = 15

⇒ Sum of squares = (12x)² + (12y)² = 4896

⇒ 144x² + 144y² = 4896

⇒ x² + y² = 4896 ÷ 144 = 34

⇒ Using the algebraic identity: (x + y)² = x² + y² + 2xy

⇒ (x + y)² = 34 + 2 × 15 = 34 + 30 = 64

⇒ x + y = √64 = 8

⇒ Sum of the numbers = 12x + 12y = 12(x + y) = 12 × 8 = 96

⇒ Mean of the numbers = Sum ÷ 2 = 96 ÷ 2 = 48

∴ The correct answer is 48.

Additional Information

Algebraic Identity for Sum of Squares

The standard formula (a + b)² = a² + b² + 2ab directly connects the sum, sum of squares, and product of two variables without finding the individual numbers.

Properties of HCF

Any two numbers can always be expressed as (HCF × x) and (HCF × y), where x and y are co-prime integers relative to each other.

Arithmetic Mean

The arithmetic mean or average of exactly two numbers a and b is given by (a + b) ÷ 2, representing their exact central balanced value.

Calculation:

Given that

Q = P + 3

Q = 2S

We can equate these and solve for P and S.

P + 3 = 2S

⇒ P = 2S - 3            ----(1)

We also know that

R = 2P and R - S = 30,

We can equate these and solve for P and S.

2P = S + 30

⇒ P = (S + 30)/2     ------(2)

2S - 3 = (S + 30) / 2

⇒ 4S - 6 = S + 30

⇒ 3S = 36

⇒ S = 12

⇒ P =  2 × 12 - 3            [From equation (1)

⇒  P = 24 - 3 = 21

⇒ Q = P + 3 = 21 + 3 = 24

Now, the sum of the ages of P and Q

P + Q = 21 + 24 = 45 years.

Concept:

If A > B then √A > √B 

Triangle inequality theorem:

For the triangle with sides a, b & c

a + b > c, b + c > a, c + a > b

Explanation:

By using the triangle property

a + b > c

a + b - c > 0

Since, a > √a, b > √b, c > √c

∴ √a + √b - √c > 0

Concept:

LCM: LCM is the product of the greatest power of each prime factor, involved in the numbers.

Calculation:

LCM of 15, 25, 35, and 45.

15 = 3 × 5

25 = 5 × 5

35 = 5 × 7

45 = 3 × 3 × 5

LCM = 3² × 5² × 7 = 1575 min = 26.25 hr

It menas, bell ring at every 26.25 hr

Since, 72/26.25 = 2.74 

Therefore, In 72 hr, bell rings 2 times.

Concept:

For the sum of four numbers to be even, you either need to have

• All four numbers be even
• All four numbers be odd
• Two of them be even and two of them be odd

 

Explanation:

a + b + c + d = 200   (even)

As discussed above

Case:1 All four numbers be even

S = 1 + 1 + 1 + 1 = 4

Case:2 All four numbers be odd

S  = - 1 - 1 - 1 - 1 = - 4

Case:3 Two of them be even and two of them be odd

S = 1 + 1 - 1 - 1 = 0

∴ There will be three possible values of S.

Concept:

aⁿ - bⁿ, (where n is even number) is always divisible by (a - b) & (a + b)

Calculation:

We have 97³⁰ - 14³⁰

By using the above concept, 

9730 - 1430 is divisible by

(97 - 14) = 83

And

(97 + 14) = 111 = 37 × 3

∴ 9730 - 1430 is divisible by both 37 and 83.

Concept:

• For any base b which is a transcendental number the logarithm log_bn of every integer n ≥ 2 is an irrational number.
• A rational number is a number, which can be written as p/q where p & q are integers and q ≠ 0.

Formula used:

• log mn = log m + log n
• log_aa = 1
• log_aⁿb = (1/n) log_ab

 

Calculation:

Statement:1 log₁₀50 is a rational number.

log1050 = log1050 =  log10(5 × 10)

log1050 = log105 + log1010   

log1050 = log105 + 1

As discussed, log105 is an irrational number, log1050 will be also irrational.

Statement:2  log100 10 is an irrational number.

log₁₀₀10 = log_10²10 = (1/2)log₁₀10 = 1/2

Which is a rational number.

∴ Both statements are incorrect.

Calculation:

Let's say the capacity of a man is m and the capacity of a woman is w. Then,

we can set up the following equations:

5(17w + 24m) = 6(12w + 23m)

85w + 120m = 72w + 138m

13w = 18m

∴ Efficiency of 13 women = Efficiency of 18 men

Shortcut Trick

Total capacity of the tank = LCM(6, 12) = 12 units.

Efficiency of (A+B+C) = 12 ÷ 6 = 2 units/hr; Efficiency of C = 12 ÷ 12 = 1 unit/hr.

Efficiency of (A+B) = 2 − 1 = 1 unit/hr.

Work done by (A+B) in the last 8 hours = 1 × 8 = 8 units.

Remaining work done by (A+B+C) in t hours = 12 − 8 = 4 units.

t = Remaining work ÷ Efficiency(A+B+C) = 4 ÷ 2 = 2.

∴ The correct answer is 2.

Alternate Method

Given:

Time taken by (A+B+C) = 6 hr

Time taken by C = 12 hr

Time for which (A+B) worked alone after C closed = 8 hr

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ Let the total work be 12 units (LCM of 6 and 12).

⇒ Efficiency of (A+B+C) = 12 ÷ 6 = 2 units/hr.

⇒ Efficiency of C = 12 ÷ 12 = 1 unit/hr.

⇒ Efficiency of (A+B) = Efficiency of (A+B+C) − Efficiency of C

⇒ Efficiency of (A+B) = 2 − 1 = 1 unit/hr.

⇒ Work done by all three taps in t hours = 2 × t.

⇒ Work done by (A+B) in the remaining 8 hours = 1 × 8 = 8 units.

⇒ Total Work = (Work in first t hr) + (Work in last 8 hr)

⇒ 12 = 2t + 8

⇒ 2t = 12 − 8

⇒ 2t = 4

⇒ t = 2

∴ The correct answer is 2.

Additional Information

Pipes and Cisterns Efficiency

The efficiency of multiple taps working together is the algebraic sum of their individual efficiencies. If a pipe fills the tank, its efficiency is positive; if it empties, its efficiency is negative.

Total Work Concept

Total work can be assumed as the LCM of the individual times taken to fill or empty the tank, which simplifies the calculation of efficiencies as integers.

Work Equality

Total Work = (Efficiency₁ × Time₁) + (Efficiency₂ × Time₂) + ... for different phases of the task.

Shortcut Trick

Ratio of Time taken by A : B : C = x : 1.5x : 2x = 1 : 1.5 : 2 = 2 : 3 : 4.

When persons work together for the same duration, the Payment Ratio is equal to the Efficiency Ratio.

Efficiency Ratio is the inverse of the Time Ratio = 1/2 : 1/3 : 1/4.

To simplify, multiply by the LCM of denominators (12): (12 ÷ 2) : (12 ÷ 3) : (12 ÷ 4) = 6 : 4 : 3.

∴ The correct answer is 6 : 4 : 3.

Alternate Method

Given:

Time taken by A = x days

Time taken by B = 1.5x days

Time taken by C = 2x days

Formula Used:

Efficiency = Total Work ÷ Time Taken

Wage Ratio = Efficiency Ratio (when time worked is identical for all)

⇒ The Efficiency Ratio of A, B, and C is calculated as 6 : 4 : 3.

⇒ Since they work together to complete the work, they work for the same number of days.

⇒ Ratio of Payment = Efficiency of A : Efficiency of B : Efficiency of C = 6 : 4 : 3.

∴ The correct answer is 6 : 4 : 3.

Additional Information

Relationship between Wages and Efficiency

When multiple people work for the same duration, their wages are directly proportional to their efficiencies. If they work for different durations, wages are proportional to the total work done (Efficiency × Time).

Inverse Variation

For a constant amount of work, Efficiency is inversely proportional to Time (E ∝ 1/T). If time ratio is a : b, efficiency ratio is b : a.

Total Work Calculation

Total Work is usually assumed as the LCM of the individual times taken to simplify the calculation of efficiencies as whole numbers.

Calculation:

Statement:1  n3 - n is divisible by 6.

n³ - n = n(n² - 1) = n(n - 1)(n + 1)

This product is the product of three consecutive integers which is always divisible by 6

Statement:2  n5 - n is divisible by 5

n⁵ - n = n(n⁴ - 1)

We know that (a² - b²) = (a - b)(a + b)

n5 - n = n (n² - 1)(n² + 1)

n5 - n = n (n2 - 1)[(n2 - 4) + 5)]

n5 - n = n (n2 - 1)(n2 - 4) + 5 n (n2 - 1)

n5 - n = n(n - 1)(n + 1)(n - 2)(n + 2) + 5 n (n - 1)(n + 1)

n5 - n = [(n - 2)(n - 1) n (n + 1)(n + 2)] + 5 n (n - 1)(n + 1)

Since the first part is the product of five consecutive numbers which will always be divisible by 5. Therefore, 

(n² - n) is divisible by 5.

Statement:

3  n5 - 5n3 + 4n is divisible by 120.

n5 - 5n3 + 4n

n(n⁴ - 5n² + 4)

n(n² - 4)(n² - 1)

n(n - 2)(n + 2)(n - 1)(n + 1)

(n - 2)(n - 1) n (n + 1)(n + 2)

Which is always divisible by 120.

∴ All three statements are correct.

Concept:

The last digit of a number that is being raised to a power often follows a cycle.

For example, the last digit of 9ⁿ repeats every 2 numbers (9, 1, 9, 1, ...).

Similarly, 7ⁿ repeats every 4 numbers (7, 9, 3, 1, 7, 9, 3, 1, ...).

Calculation:

For 9ⁿ, the last digit is 9 if n is odd and 1 if n is even.

In this case, 9²⁷ has a last digit of 9 because 27 is odd.

For 27⁹, we only need to consider the last digit of 27, which is 7.

To find the last digit of 27⁹, we need to find the remainder when 9 is divided

by 4 (because the cycle length is 4). That remainder is 1.

 So the last digit of 27⁹ is the first digit in the cycle, which is 7.

Now

9 (from 9²⁷) + 7 (from 27⁹) = 16.

∴ The last digit of the sum S = 9²⁷ + 27⁹ is 6.

Formula used:

1. (a³ - b³) = (a - b)³ + 3ab(a -b)
2. (a - b)² = a² - 2ab + b²
3. (a + b)2 = a2 + 2ab +b2
4. (a - b)(a + b) = a2 - b2

 

Calculation:

Given that

x = \(\frac{√{3}+1}{√{3}-1}\) and y = \(\frac{√{3}-1}{√{3}+1}\)

⇒ xy = \(\frac{√{3}+1}{√{3}-1}\) × \(\frac{√{3}-1}{√{3}+1}\) = 1       -----(1)

⇒ x - y = \(\frac{√{3}+1}{√{3}-1}\) - \(\frac{√{3}-1}{√{3}+1}\)

⇒ x - y = \(\frac{(√{3}+1)^2 - (√{3}-1)^2}{(√{3}-1)(√{3}+1)}\)

By using the  formula (2), (3) & (4)

⇒ x - y = \(\frac{({3}+1+2√3) - ({3}+1-2√3)}{({3}-1)}=\frac{4√ 3}{2}\)

⇒ x - y = 2√3                     -----(3)

We know that 

(x3 - y3) = (x - y)3 + 3xy(x -y)

⇒ (x3 - y3) = (2√3)³ + 3 × 1 (2√3) 

⇒ (x3 - y3) = (2√3)3 + 3 × 1 (2√3) 

⇒ (x3 - y3) = 8 × 3√3 + 6√3 

∴  (x3 - y3) = 30√3

Shortcut Trick

Since time is constant, the ratio of speeds is equal to the ratio of distances covered.

Downstream Speed (S_d) : Upstream Speed (S_u) = 42 : 28 = 3 : 2

Ratio of Still Water Speed (u) to Stream Speed (v) = (3 + 2) : (3 − 2) = 5 : 1

Given 5 units = 15 km/hr ⇒ 1 unit = 3 km/hr.

∴ The correct answer is 3 km/hr.

Alternate Method

Given:

Speed of boat in still water (u) = 15 km/hr

Let the speed of the stream be v km/hr.

Downstream speed = (15 + v) km/hr

Upstream speed = (15 − v) km/hr

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Time downstream = Time upstream

⇒ 42 ÷ (15 + v) = 28 ÷ (15 − v)

⇒ 3 ÷ (15 + v) = 2 ÷ (15 − v) [Dividing both sides by 14]

⇒ 3 × (15 − v) = 2 × (15 + v)

⇒ 45 − 3v = 30 + 2v

⇒ 45 − 30 = 2v + 3v

⇒ 15 = 5v

⇒ v = 15 ÷ 5

⇒ v = 3 km/hr

∴ The correct answer is 3 km/hr.

Additional Information

Speed in Still Water

It is the average of downstream and upstream speeds: u = (S_d + S_u) ÷ 2.

Speed of Stream

It is half the difference between downstream and upstream speeds: v = (S_d − S_u) ÷ 2.

Time, Distance, and Speed Relationship

For a constant time, Distance₁ ÷ Distance₂ = Speed₁ ÷ Speed₂.