Quantitative Aptitude PYQs

Shortcut Trick

Let Speed of the train be S.

From Pole crossing: L = S × t.

From reduced length: L − 30 = S × 17.

Difference in distance = 30 m; Difference in time = t − 17 seconds.

Using Platform crossing: L + Platform = 540, which takes t + 12 seconds.

Speed (S) = 540 / (t + 12). Equating with speed from L and L − 30 gives t = 18.7.

∴ The correct answer is 18.7 seconds.

Alternate Method

Given: Total Length = 540 m, Train Length = L, Time to cross pole = t, Time to cross platform = t + 12.

Formula: Speed = Distance / Time

⇒ Speed S = L / t

⇒ Also, Speed S = 540 / (t + 12)

⇒ L / t = 540 / (t + 12) → L = 540t / (t + 12) ...

(i) ⇒ New length = L − 30, New time to cross pole = 17 s

⇒ Speed S = (L − 30) / 17 ...

(ii) ⇒ From (i) and (ii): L / t = (L − 30) / 17

⇒ 17L = Lt − 30t ⇒ L(t − 17) = 30t ⇒ L = 30t / (t − 17)

⇒ Equating L: 540t / (t + 12) = 30t / (t − 17)

⇒ 18 / (t + 12) = 1 / (t − 17)

⇒ 18t − 306 = t + 12

⇒ 17t = 318 ⇒ t = 18.705...

∴ The correct answer is 18.7 seconds.

Additional Information

Train Crossing a Pole

When a train crosses a stationary point object like a pole or a man, the distance covered is equal to the Length of the train.

Train Crossing a Platform

The distance covered to cross a platform or bridge is (Length of Train + Length of Platform).

Relative Speed

For two objects moving in the same direction, relative speed is S₁ − S₂; for opposite directions, it is S₁ + S₂.

Shortcut Trick

Identify the multiplication pattern with decreasing even factors: ×12, ×10, ×8, ×6, ×4, ×2.

First term 0.5 × 12 = 6 (making 5 the wrong number).

Missing number x = 11520 × 2 = 23040.

Required value = 23040 − 11809 = 11231.

∴ The correct answer is 11231.

Alternate Method

Given: Series 0.5, 5, 60, 480, 2880, 11520, x

0.5 ×12 → 5 ×10 → 60 ×8 → 480 ×6 → 2880 ×4 → 11520 ×2 → 23040

Calculations:

⇒ Term 1: 0.5

⇒ Term 2: 0.5 × 12 = 6 (Wrong number given as 5)

⇒ Term 3: 6 × 10 = 60

⇒ Term 4: 60 × 8 = 480

⇒ Term 5: 480 × 6 = 2880

⇒ Term 6: 2880 × 4 = 11520

⇒ Term 7 (x): 11520 × 2 = 23040

⇒ Required value = x − 11809

⇒ Value = 23040 − 11809 = 11231

∴ The correct answer is 11231.

Additional Information

Multiplication Series

In a multiplication series, each term is derived by multiplying the previous term by a specific number, often following a pattern like consecutive even numbers or arithmetic progression.

Wrong Number Series

To find a wrong number, establish the logic using the majority of correct terms. Once the pattern (like ×12, ×10, ×8) is confirmed, the term that violates it is identified.

Shortcut Trick

The pattern is a multiplication series with decreasing even multipliers: ×12, ×10, ×8, ×6, ×4, ×2.

Starting with 0.5: 0.5 × 12 = 6 (instead of 5).

Verifying next: 6 × 10 = 60, which matches the series.

∴ The correct answer is 5.

Alternate Method

Given Series: 0.5, 5, 60, 480, 2880, 11520, x

0.5 ×12 → 5 ×10 → 60 ×8 → 480 ×6 → 2880 ×4 → 11520

The pattern follows a descending even number multiplication logic:

⇒ 1^st term = 0.5

⇒ 2^nd term = 0.5 × 12 = 6

⇒ 3^rd term = 6 × 10 = 60

⇒ 4^th term = 60 × 8 = 480

⇒ 5^th term = 480 × 6 = 2880

⇒ 6^th term = 2880 × 4 = 11520

⇒ Missing term (x) = 11520 × 2 = 23040

Observing the given series, the number 5 is provided instead of 6.

∴ The correct answer is 5.

Additional Information

Wrong Number Series Identification

To find the wrong number, always verify the pattern with the subsequent numbers in the series. If the pattern breaks and then resumes correctly with a substituted value, that substituted value is the wrong term.

Geometric/Multiplication Series

In these series, each term is multiplied by a constant or a systematic sequence (like consecutive even numbers, prime numbers, or squares) to obtain the next term.

Shortcut Trick

Solve series I: 511 + 4² = 527; 527 − 8² = 463; 463 + 12² = 607; 607 − 16² = 351 (n); 351 + 20² = 751 (m); 751 − 24² = 175 (l).

Solve series II: Common difference is 18. 369 + 18 = 387 (z); 333 − 18 − 18 = 297 (k).

Check (b): 270 + (351 ÷ 3) = 270 + 117 = 387. Since z = 387, (b) is True.

∴ The correct answer is Only b.

Alternate Method

Given: Series I follows a pattern of ± squares of multiples of 4. Series II is an Arithmetic Progression with common difference 18.

Calculations for Series I:

⇒ 527 − 511 = 16 = 4²

⇒ 527 − 463 = 64 = 8²

⇒ 607 − 463 = 144 = 12²

⇒ n = 607 − 16² = 607 − 256 = 351

⇒ m = 351 + 20² = 351 + 400 = 751

⇒ l = 751 − 24² = 751 − 576 = 175

Calculations for Series II:

⇒ Common difference (d) = 351 − 333 = 18

⇒ z = 369 + 18 = 387

⇒ k + 18 = 333 − 18 ⇒ k = 315 − 18 = 297

**Verifying

Statements:

**

⇒ (a) z = m + l → 387 = 751 + 175 (926). False.

⇒ (b) z = 270 + (1/3)n → 387 = 270 + (351 ÷ 3) = 270 + 117 = 387. True.

⇒ (c) k + m = (1/2)z → 297 + 751 (1048) = 387 ÷ 2 (193.5). False.

∴ The correct answer is Only b.

Additional Information

Arithmetic Progression (AP)

A sequence where the difference between consecutive terms is constant. General term a_n = a + (n−1)d.

Square Series Pattern

Logical series often involve addition or subtraction of squares (n²) or cubes (n³) of consecutive numbers or multiples.

Verification Method

Always solve for variables first before checking algebraic conditions to avoid calculation errors.

Shortcut Trick

Pattern I: Alternate square addition and subtraction of 24², 20², 16², 12², 8², 4².

Pattern II: Arithmetic progression with a common difference of +18.

From Series II: z = 369 + 18 = 387. Given o + 387 = 476 ⇒ o = 89.

From Series I: n = 607 − 16² = 607 − 256 = 351.

Average = (351 + 89) ÷ 2 = 220.

∴ The correct answer is 220.

Alternate Method

Given:

Series I: l, m, n, 607, 463, 527, 511

Series II: k, k+18, 333, 351, 369, z

Condition: o + z = 476

l +24² → m −20² → n +16² → 607 −12² → 463 +8² → 527 −4² → 511

⇒ n + 16² = 607

⇒ n + 256 = 607

⇒ n = 351

k +18 → k+18 +18 → 333 +18 → 351 +18 → 369 +18 → 387

⇒ z = 369 + 18 = 387

⇒ o + z = 476 ⇒ o + 387 = 476

⇒ o = 476 − 387 = 89

⇒ Average of n and o = (351 + 89) ÷ 2

⇒ Average = 440 ÷ 2 = 220

∴ The correct answer is 220.

Additional Information

Arithmetic Progression

A sequence where the difference between consecutive terms is constant (d). Term T_n = a + (n−1)d.

Mixed Series Patterns

Series can involve alternate operations like addition and subtraction of squares, cubes, or prime numbers.

Average Calculation

The average of two numbers x and y is calculated as (x + y) ÷ 2.

Shortcut Trick

From Series I: l = 175 (based on the square pattern −16, +64, −144, +256, −400, +576).

The largest single digit prime number is 7.

Resultant = 175 ÷ 7 = 25.

∴ The correct answer is 25.

Alternate Method

Given:

Series I: l, m, n, 607, 463, 527, 511

l +24² → m -20² → n +16² → 607 -12² → 463 +8² → 527 -4² → 511

Pattern Explanation: The series follows a pattern of alternating addition and subtraction of squares of multiples of 4 (4, 8, 12, 16, 20, 24).

⇒ 527 − 4² = 511 (Checked)

⇒ 463 + 8² = 463 + 64 = 527 (Checked)

⇒ 607 − 12² = 607 − 144 = 463 (Checked)

⇒ n + 16² = 607 ⇒ n = 607 − 256 = 351

⇒ m − 20² = 351 ⇒ m = 351 + 400 = 751

⇒ l + 24² = 751 ⇒ l = 751 − 576 = 175

Final Calculation:

⇒ Largest single digit prime number = 7

⇒ Required Resultant = l ÷ 7

⇒ Resultant = 175 ÷ 7 = 25

∴ The correct answer is 25.

Additional Information

Prime Numbers

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Single digit primes are 2, 3, 5, and 7.

Number Series Patterns

Common patterns include Arithmetic (constant difference), Geometric (constant ratio), and Square/Cube based patterns where n² or n³ is added or subtracted.

Alternating Series

In an alternating series, the operation changes (e.g., +, −, +, −) or two different logic sequences are interleaved.

Solution:

3x2 - 26x + C = 0

A and B are the roots of the equation.

B = 8/3

Put x = 8/3 in equation

3(8/3)2 - 26(8/3) + C = 0

3 × 64/9 - 26 × 8/3 + C = 0

C = 48

Products of roots = constant term/ coefficient of x

AB = 48/3

A × 8/3 = 48/3

A = 6

1. HCF of C and A is equal to 50% more than B

HCF of C and A = HCF(48, 6) = 6

50% more than B = 8/3 × 1.5 = 4

This is incorrect.

2. Value of C is a perfect cube

C = 48 which is not a perfect cube

This is incorrect.

3. C is multiple of A

48 is a multiple of 6. This is correct.

Solution:

Quantity I: ∛216x3 - 2x4 + 51x such that 8 ≤ x ≤ 12

6x - 2x⁴ + 51x

57x - 2x⁴

Put x = 8, 9, 10, 11, 12

If x = 8, we get

57 × 8 - 2 × 8⁴

= - 7736

If we put x = 9, 10, 11 and 12 we get bigger negative value.

Quantity II: \(50y + 3^2× \sqrt[y]64\) such that 2 ≤ y ≤ 4

Put y = 2

50 × 2 + 9 × 8

= 100 + 72 = 172

Put y = 3

50 × 3 + 9 × 4

= 150 + 36 = 186

Put y = 4

50 × 4 + 9 × 2.82

= 200 + 25.2 = 225.2

All the values of y is positive. 

Quantity I < Quantity II

Solving for Quantity-I:

Given:

Time taken by A = 3 hours

Time taken by B = 4 hours

Time taken by C = 5 hours

Calculations:

Let the total capacity of tank = 60 units

Part of the tank filled by A in 1 hour = 60/3 units

⇒ 20 units/hour

Part of the tank filled by B in 1 hour = 60/4 units

⇒ 15 units/hour

Part of the tank filled by C in 1 hour = 60/5 units

⇒ 12 units/hour

Time taken by A, B and C when all of them working together = 60/(20 + 15 + 12)

⇒ 1.28 hours

Solving for Quantity II:

Given:

Time taken by A and B = 8 hours

Time taken by B and C = 10 hours

Time taken by C and D = 15 hours

Time taken by D and A = 30 hours

Calculations:

Let the total capacity of tank = 120 units

Part of the tank filled by A and B in 1 hour = 120/8 units

⇒ 15 units/hour

Part of the tank filled by B and C in 1 hour = 120/10 units

⇒ 12 units/hour

Part of the tank filled by C and D in 1 hour = 120/15 units

⇒ 8 units/hour

Part of the tank filled by A and D in 1 hour = 120/30 units

⇒ 4 units/hour

Time taken by A, B and C when all of them working together = 120 × 2/(15 + 12 + 8 + 4)

⇒ 6.15 hours

∴ Quantity I < Quantity II

Quantity A -

Area of square = 20 × 20 = 400 sq.cm.

Area of triangles cut = 400 - 200 (remaining area) = 200 sq.cm.

Area of 1 triangle = 200 / 4 = 50 sq.cm.

For right angled isosceles triangle,

1 / 2 × side × side = 50

Side = 10

Perimeter = 10 + 10 + 10 × 1.4 (hypotenuse) = 34 cm

Quantity B -

Percentage increase in length = 18%

Percentage increase in breadth = 12%

Percentage increase in area = 18 + 12 + (18 × 12) / 100 = 32.16

Hence, Quantity A > Quantity B

Solution:

Average of A and B is 12. 14 ≥ (A, B) ≥ 8

A + B = 24        ...........

(i) If we put A = 14 and B = 10 then the statement is satisfied.

Sum of B and C is 12.

B + C = 12

If we put B = 10, then C = 2.

Quantity I: Value of AB - 45

= 14 × 10 - 45 

= 140 - 45

= 95

Quantity II: Value of (A + B) × C

= (14 + 10) × 2

= 24 × 2 = 48

Hence, Quantity  > Quantity II

General Solution:

(1.25x% - x%) = 12

0.25x% = 12

x% = 48

To find the percentage value of x

48/400 ×  100 = 12% = x

x% = 12

1.25x% = 12 ×  1.25 = 15

To find the value of y

(12 + 23 + 15 + 20 + y = 100)

y =(100 - 70)%

y = 30%

Number of goods sold on Sunday = 12/100 ×  400 = 48

Number of goods sold on Monday = 23/100 ×  400 = 92

Number of goods sold on Tuesday = 15/100 ×  400 = 60

Number of goods sold on Wednesday = 20/100 ×  400 = 80

Number of goods sold on Thursday = 30/100 ×  400 = 120

 

Solution:

Number of goods sold on Sunday and Monday = (48 + 92) = 140

Number of goods sold on Tuesday, Wednesday and Thursday = (60 + 80 + 120) = 260

Required percentage = 140/260 ×  100 = 53.84%

General Solution:

(1.25x% - x%) = 12

0.25x% = 12

x% = 48

To find the percentage value of x

48/400 ×  100 = 12% = x

x% = 12

1.25x% = 12 ×  1.25 = 15

To find the value of y

(12 + 23 + 15 + 20 + y = 100)

y =(100 - 70)%

y = 30%

Number of goods sold on Sunday = 12/100 ×  400 = 48

Number of goods sold on Monday = 23/100 ×  400 = 92

Number of goods sold on Tuesday = 15/100 ×  400 = 60

Number of goods sold on Wednesday = 20/100 ×  400 = 80

Number of goods sold on Thursday = 30/100 ×  400 = 120

 

Solution:

Number of  goods sold on Thursday = 120

Number of goods sold on Wednesday = 80

To find the value of a

120 ×  a + 80 ×  (a - 10) = 9200

120a + 80a - 800 = 9200

200a = 10000

a = 50

Total amount earned by the shop on Sunday = (50 - 20) ×  48 = 1440

General Solution:

(1.25x% - x%) = 12

0.25x% = 12

x% = 48

To find the percentage value of x

48/400 ×  100 = 12% = x

x% = 12

1.25x% = 12 ×  1.25 = 15

To find the value of y

(12 + 23 + 15 + 20 + y = 100)

y =(100 - 70)%

y = 30%

Number of goods sold on Sunday = 12/100 ×  400 = 48

Number of goods sold on Monday = 23/100 ×  400 = 92

Number of goods sold on Tuesday = 15/100 ×  400 = 60

Number of goods sold on Wednesday = 20/100 ×  400 = 80

Number of goods sold on Thursday = 30/100 ×  400 = 120

 

Solution:

Number of goods sold in the shop on Tuesday = 60

Number of goods present in the shop initially on Tuesday  = 60 /(66.667) ×  100 = 90

Number of goods sold in the shop on Wednesday = 80

Number of goods present in the shop initially on Wednesday = 80 /(33.33) ×  100 = 240

Number of goods unsold on Tuesday = (90 - 60) = 30

Number of goods unsold on Wednesday = (240 - 80) = 160

Required ratio = 30: 160 = 3: 16

General Solution:

(1.25x% - x%) = 12

0.25x% = 12

x% = 48

To find the percentage value of x

48/400 ×  100 = 12% = x

x% = 12

1.25x% = 12 ×  1.25 = 15

To find the value of y

(12 + 23 + 15 + 20 + y = 100)

y =(100 - 70)%

y = 30%

Number of goods sold on Sunday = 12/100 ×  400 = 48

Number of goods sold on Monday = 23/100 ×  400 = 92

Number of goods sold on Tuesday = 15/100 ×  400 = 60

Number of goods sold on Wednesday = 20/100 ×  400 = 80

Number of goods sold on Thursday = 30/100 ×  400 = 120

 

Solution:

Number of goods sold on Sunday = 48

Average number of goods sold on Sunday and Saturday = (48 - 8) = 40

Number of goods sold on Saturday = (40× 2 - 48) = 32

Number of items unsold on Saturday = 32/8 × 5 = 20

General Solution:

(1.25x% - x%) = 12

0.25x% = 12

x% = 48

To find the percentage value of x

48/400 ×  100 = 12% = x

x% = 12

1.25x% = 12 ×  1.25 = 15

To find the value of y

(12 + 23 + 15 + 20 + y = 100)

y =(100 - 70)%

y = 30%

Number of goods sold on Sunday = 12/100 ×  400 = 48

Number of goods sold on Monday = 23/100 ×  400 = 92

Number of goods sold on Tuesday = 15/100 ×  400 = 60

Number of goods sold on Wednesday = 20/100 ×  400 = 80

Number of goods sold on Thursday = 30/100 ×  400 = 120

Solution:

Number of goods sold on Sunday = 48

Number of goods sold on Monday = 92

(92 - 48) = 44

Difference between number of goods sold on Tuesday and Friday = 44/2 = 22

T - F = 22

Number of goods sold on Tuesday = 60

Number of goods sold on Friday = 60 - F = 22 = 38

F = 38

Shortcut Trick

Change in sum = Number of items × Change in average.

When D is replaced by E: E − D = 4 × 14.25 = 57 kg.

When C is replaced by E: E − C = 4 × 6.75 = 27 kg.

Total weight (C + D + E) = 3 × 74 = 222 kg.

Substitute D = E − 57 and C = E − 27: (E − 27) + (E − 57) + E = 222 ⇒ 3E − 84 = 222 ⇒ 3E = 306 ⇒ E = 102 kg.

Weight of C = 102 − 27 = 75 kg. Required % = (27 ÷ 75) × 100 = 36%.

∴ The correct answer is 36%.

Alternate Method

Given: Average of (A, B, C, D) = x. Average of (A, B, E, D) [C replaced] = x + 6.75. Average of (A, B, C, E) [D replaced] = x + 14.25. Average of (C, D, E) = 74 kg.

Step 1: Find weight differences

⇒ (A + B + C + E) − (A + B + C + D) = 4 × (x + 14.25) − 4x

⇒ E − D = 57 kg

⇒ (A + B + E + D) − (A + B + C + D) = 4 × (x + 6.75) − 4x

⇒ E − C = 27 kg

Step 2: Use average of C, D, and E

⇒ C + D + E = 3 × 74 = 222 kg

⇒ (E − 27) + (E − 57) + E = 222

⇒ 3E − 84 = 222 ⇒ 3E = 306 ⇒ E = 102 kg

⇒ C = E − 27 = 102 − 27 = 75 kg

Step 3: Calculate percentage

⇒ Difference = E − C = 27 kg

⇒ Percentage = (27 ÷ 75) × 100 = 0.36 × 100 = 36%

∴ The correct answer is 36%.

Additional Information

Replacement Rule in Average

When one person is replaced by another, New Weight − Old Weight = Total Number × Change in Average.

Percentage Comparison

Percentage more = (Difference ÷ Base Value) × 100, where the base value is the quantity being compared against (after than).

From (II),

Let the mother’s age be x and daughter’s age be x/4

From (I) &(II),

x – x/4 = 45

3x/4 = 45

∴ x = 60 years.

From (II) & (III),

x + x/4 = 75

5x/4 = 75

∴ x = 60 years.

From (I) & (III),

x – y = 45 and x + y = 75

By solving will get the value of x i.e. the answer.

Hence Any two of I, II and III will give the answer

Calculation:

From statement I:

Let the length of the longer piece, middle piece and, shorter piece be x, y, and z respectively.

From the information

⇒ x + y = 25

No other information is given

∴ Statement I alone is not sufficient.

From Statement II:

Let the length of the longer piece, middle piece and, shorter piece be x, y, and z respectively.

⇒ y + z = 20

No other information is given

∴ Statement II alone is not sufficient.

Both

Statements:

⇒ x + y = 25

⇒ y + z = 20

⇒ x - z = 5

No other information is given

∴ Statement I and II both are not sufficient.

Given:

Anil and Mukesh together can complete a piece of work in = 20 days

Calculation:

For Statement I:

We can conclude the number of days Anil alone will take to complete the work.

∴ Statement I alone is sufficient.

For Statement II:

Efficiency is given so it is easy to find the number of days A alone or B alone can do the work.

∴ Statement II alone is sufficient.

∴ Statement I alone or II alone is sufficient.