Quantitative Aptitude PYQs

Given:

37.98% of 549.99 + √(49.02) ÷ ? = 7.99 × (2.97)³

Rules of Approximation:

1. If a number has digits to the right of the decimal less than 5, then just drop the digits to the right of the decimal. The number so obtained will be the approximated value.
2. If a number has digits to the right of the decimal more than 5, then just drop the digits to the right of the decimal and raise the remaining number by '1'.The number so obtained will be the approximated value.

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

37.98% of 549.99 + √(49.02) ÷ ? = 7.99 × (2.97)³

⇒ 38% of 550 + √49 ÷ ? = 8 × 3³

⇒ (38/100) × 550 + 7 ÷ ? = 8 × 27

⇒ 209 + 7 ÷ ? = 216

⇒ 7 ÷ ? = 7

⇒ ? = 1

∴ 1 should come in place of the question mark (?).

Given:

9.78% of 79.94 + ?% of 9.67 = 10.94 × 2.99 + √100 × 2.98

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

9.78% of 79.94 + ?% of 9.67 = 10.94 × 2.99 + √100 × 2.98

⇒ 10% of 80 + ?% of 10 = 11 × 3 + √100 × 3

⇒ (10/100) × 80 + (?/100) × 10 = 33 + 10 × 3

⇒ 8 + (?/10) = 33 + 30

⇒ 8 + (?/10) = 63

⇒ (?/10) = 63 - 8 

⇒ (?/10) = 55 

⇒ ? = 55 × 10 = 550

∴ The value of ? is 550.

Given:

Profit = 425%

C.P. = Increase is 5%

S.P. =Constant

Discount = 80% on mark up

Formula used :

Profit = S.P. – C.P.

Calculation:

Quantity I :

Assume the cost price = Rs 100

⇒ Selling price = 100 + 425

⇒ Selling price = 525

⇒ Profit from new C.P. increase = 525 – (100 + (0.05 × 100))

⇒ Profit from new C.P. increase = 525 – 105

⇒ Profit from new C.P. increase = 420%

⇒ New profit margin = (420/105) × 100

⇒ New profit margin % = 400%

Quantity II :

Let the marked price be Rs 100

Selling price = 100 – (0.80 × 100)

⇒ Selling price = 20

⇒ S.P. = C.P. = 20

⇒ Percentage markup = ((100 – 20)/20) × 100

⇒ Percentage markup = 400%

∴ Quantity I = Quantity II

GIVEN:

Payment = Rs 15000

(A + B + C)'s time = 6 days

A's time = 12 days

B's time = 15 days

CONCEPT:

Wages are distributed in the ratio of efficiency.

For a unit work done efficiency is inversely proportional to the time taken.

CALCULATION:

Let the work be 1 unit

A's efficiency = 1/12

B's efficiency = 1/15

(A + B + C)'s efficiency = 1/6

C's efficiency = 1/6 - (1/12 + 1/15)

⇒ (10 - 5 - 4)/60 = 1/60

C's wages = [15000/1(/6)] × 1/60 = Rs 1500

∴ Quantity A = Quantity B

ALTERNATE SOLUTION:

Let the work be 60 (L.C.M of 6, 12, 15)

A's efficiency = 60/12 = 5

B's efficiency = 60/15 = 4

(A + B + C)'s efficiency = 60/6 = 10

C's efficiency = 10 - (5 + 4) = 1

C's wages = [15000/10] × 1 = Rs 1500

∴ Quantity A = Quantity B

GIVEN:

Quantity = 63 l

Initial ratio (alcohol : water) = 5 : 4

Final ratio (alcohol : water) = 3 : 2

EXPLANATION:

Total mixture (ratio) = 5 + 4 = 9 ratio

9 ratio = 63 l

→ 1 ratio = 7 l

∴ Alcohol = 5 ratio = 35 l

and, water = 4 ratio = 28 l

Since the amount of water will remain same,

For final ratio, 2 ratio = 28 l

∴ 3 ratio = 42 l

So, alcohol quantity in the new mixture = 42 l

∴ Quantity A < Quantity B

Quantity I:

Let the present ages of Yash and Kush be 'Y' years and 'K' years respectively.

Difference between the ages of Yash and Kush is 6 years.

Therefore, either Y - K = 6 or K - Y = 6 ... (a)

Three years later, the sum of their ages will be 40 years.

Therefore, Y + 3 + K + 3 = 40

⇒ Y + K = 34 ...(b)

Case (i) when Y - K = 6 and Y + K = 34

Then, 2Y = 40

⇒ Y = 20

4 years ago Yash was = 20 - 4 = 16 years old  

Case (ii) when K - Y = 6 and Y + K = 34

Then, 2Y = 28

⇒ Y = 14

4 years ago Yash was = 14 - 4 = 10 years old  

∴ Quantity I = 16 years or 10 years

Quantity II:

Let the present age of Viru and Veer be 'x' years and 'y' years respectively.  

Two years later, Viru's age will be two - third of Veer's age at that time.  

Therefore, x + 2 = (2/3) × (y + 2)  

⇒ 3x + 6 = 2y + 4  

⇒ -3x + 2y = 2 ...(a) 

Sum of their present ages = 26 years  

So, x + y = 26 ...(b)  

From (a) and (b)  

-3x + 2 × (26 - x) = 2  

⇒ x = 10  

∴ Quantity II = 10 years

Hence, the correct answer is Quantity - I ≥ Quantity - II.

Shortcut Trick

Let the third booth voters be 100 units.

First booth = 52 units, Second booth = 46 units.

Difference (52 − 46) = 6 units → 7686.

Total units = 52 + 46 + 100 = 198 units.

Total voters = (7686 ÷ 6) × 198 = 1281 × 198 = 2,53,638.

∴ The correct answer is 2,53,638 voters.

Alternate Method

Given:

Voters in Booth 3 = x

Voters in Booth 1 = 52% of x = 0.52x

Voters in Booth 2 = 46% of x = 0.46x

Calculations:

⇒ Difference = 0.52x − 0.46x = 7686

⇒ 0.06x = 7686

⇒ x = 7686 ÷ 0.06 = 1,28,100

⇒ Total voters = Booth 1 + Booth 2 + Booth 3

⇒ Total = 0.52x + 0.46x + x = 1.98x

⇒ Total = 1.98 × 1,28,100

⇒ Total = 2,53,638

∴ The correct answer is 2,53,638 voters.

Additional Information

Percentage Base Change

When multiple values are compared to a single base (the third booth here), they can be added or subtracted directly as long as the base remains constant.

Unitary Method

Value of 1 unit = (Given Difference) ÷ (Difference in units). Total Value = (1 unit value) × (Total sum of units).

Successive Percentage

If a value changes by a% and then b%, the net change is [a + b + (ab/100)]%.

Note:

This rule does not apply when values are independent parts of a whole.

Shortcut Trick

Assume Marked Price (MP) = 100.

Cost Price (CP) = 100 × (100 − 20)% = 80.

Selling Price (SP) = 80 × (100 − 10)% = 72.

If CP = MP = 100, then Loss = 100 − 72 = 28.

Loss% = (28 ÷ 100) × 100 = 28%.

∴ The correct answer is 28%.

Alternate Method

Given → Discount = 20% on MP | Loss = 10% on CP | Target = Loss% if CP = MP.
Formulas: CP = MP × (100 − D%) ÷ 100 | SP = CP × (100 − L%) ÷ 100 | New Loss% = (MP − SP) ÷ MP × 100.

⇒ Let Marked Price = x.
⇒ Cost Price (CP) = x − 0.20x = 0.80x.
⇒ Selling Price (SP) = 0.80x − (10% of 0.80x) = 0.80x − 0.08x = 0.72x.
⇒ If article was bought at Marked Price, New CP = x.
⇒ Loss = New CP − SP = x − 0.72x = 0.28x.
⇒ Loss Percentage = (0.28x ÷ x) × 100 = 28%.
∴ The correct answer is 28%.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price. SP = MP × (100 − Discount%)/100.

Effective Loss/Gain

When multiple changes occur, the result depends on the sequence of successive percentages applied to the base value.

Calculations:

Let Ramiya’s present age be x.

∴ her age at the time of her marriage is (x - 8) years.

⇒ x = 6/5(x - 8)

⇒ 5x = 6x – 48

⇒ x = 48 years

∴ Ramiya’s brothers age at the time of her marriage = (x – 8) – 10

⇒ 48 – 18

⇒ 30 years

∴ required ratio = 48 : 30

⇒ 8 : 5

The ratio of Ramiya’s present age and her brother’s age at the time of her marriage is 8 : 5.

Shortcut Trick

When two boats move towards each other, one downstream and one upstream, the speed of the stream cancels out.

Relative Speed = Speed of X + Speed of Y = 8 + 7 = 15 km/hr.

Time = Total Distance ÷ Relative Speed = 300 ÷ 15 = 20 hr.

∴ The correct answer is 20 hr.

Alternate Method

Given: Distance = 300 km, Speed of X (u_X) = 8 km/hr, Speed of Y (u_Y) = 7 km/hr.

Formula Used: Time = Distance ÷ Relative Speed; Downstream Speed = u + v; Upstream Speed = u − v.

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

⇒ Speed of Boat X (downstream) = 8 + v km/hr.

⇒ Speed of Boat Y (upstream) = 7 − v km/hr.

⇒ Relative Speed = (8 + v) + (7 − v) = 15 km/hr.

⇒ Time taken to meet = Distance ÷ Relative Speed

⇒ Time = 300 ÷ 15 = 20 hr.

∴ The correct answer is 20 hr.

Additional Information

Relative Speed in Streams

When two boats move towards each other (one up and one down), the stream velocity always cancels out in the relative speed calculation.

Downstream and Upstream

Downstream Speed = u + v and Upstream Speed = u − v, where u is speed in still water and v is stream speed.

Time of Meeting

Time taken to meet = (Initial Distance) ÷ (Relative Speed in opposite direction).

Shortcut Trick

Ratio of Profit Share of A and B is (47000 − 22000) : 22000 = 25 : 22.

Profit Ratio = Ratio of (Investment × Time).

Investment of A = 15000 × 12 months = 180000 units.

Investment of B = (15000 + Y) × 4 + 0.6(15000 + Y) × 8 = 8.8 × (15000 + Y) units.

Ratio 180000 ÷ [8.8 × (15000 + Y)] = 25 ÷ 22.

7200 ÷ (15000 + Y) = 25 ÷ 62.5 ⇒ Y = 3000.

∴ The correct answer is 6000.

Alternate Method

Given → A's investment = Rs. 15000, B's investment = Rs. (15000 + Y), Time = 1 year, Total Profit = Rs. 47000, B's Profit = Rs. 22000.
Formulas: Profit Ratio = (Capital₁ × Time₁) : (Capital₂ × Time₂).

⇒ Profit of A = Total Profit − B's Profit = 47000 − 22000 = Rs. 25000.
⇒ Profit Ratio A : B = 25000 : 22000 = 25 : 22.
⇒ Equivalent Capital of A = 15000 × 12 = 180000.
⇒ Equivalent Capital of B = (15000 + Y) × 4 + (15000 + Y) × (1 − 0.40) × 8.
⇒ B's Capital = 4(15000 + Y) + 4.8(15000 + Y) = 8.8(15000 + Y).
⇒ A/B = 180000 ÷ [8.8(15000 + Y)] = 25 ÷ 22.
⇒ 180000 × 22 = 25 × 8.8 × (15000 + Y).
⇒ 3960000 = 220 × (15000 + Y) ⇒ 18000 = 15000 + Y ⇒ Y = 3000.
⇒ Value of 2Y = 2 × 3000 = 6000.

∴ The correct answer is 6000.

Additional Information

Partnership Principles

When investments vary over time, the profit is distributed according to the weighted average of the investment relative to the time period.

Withdrawal Calculation

Removing 40% means the remaining investment is 60% (or 0.6) of the original amount for the rest of the duration.

Shortcut Trick

Let speed of boat in still water be u km/h and current v = 4 km/h.

Total time: 20/(u − 4) + 15/(u + 4) = 29/12 hours.

By Option/Value verification, if u = 16: 20/12 + 15/20 = 5/3 + 3/4 = (20 + 9)/12 = 29/12 (Matches).

Upstream speed = 16 − 4 = 12 km/h.

Time for 64 km upstream = 64 ÷ 12 = 16/3 = 5(1/3) hours.

∴ The correct answer is 5(1/3) hours.

Alternate Method

Given: Upstream distance = 20 km, Downstream distance = 15 km, Current speed (v) = 4 km/h, Total time = 29/12 hours.

Formula: Time = Distance ÷ Speed; Upstream speed = u − v; Downstream speed = u + v.

⇒ 20/(u − 4) + 15/(u + 4) = 29/12

⇒ [20(u + 4) + 15(u − 4)] / (u² − 16) = 29/12

⇒ [20u + 80 + 15u − 60] / (u² − 16) = 29/12

⇒ 12(35u + 20) = 29(u² − 16)

⇒ 420u + 240 = 29u² − 464

⇒ 29u² − 420u − 704 = 0

⇒ Solving the quadratic equation, we get u = 16 km/h.

⇒ Upstream speed = 16 − 4 = 12 km/h.

⇒ Required time = 64 ÷ 12 = 16/3 = 5(1/3) hours.

∴ The correct answer is 5(1/3) hours.

Additional Information

Relative Speed in Water

Downstream speed (D) = u + v and Upstream speed (U) = u − v, where u is boat speed and v is stream speed.

Calculations for u and v

Speed of boat in still water u = (D + U) ÷ 2 and speed of stream v = (D − U) ÷ 2.

Average Speed

If distances are same, average speed = (D × U) ÷ u = (u² − v²) ÷ u.

Shortcut Trick

Days for P + Q + R = 180/11 days and P + Q = 20 days.

Total Work (LCM of 180 and 20) = 180 units.

Efficiency of R = Efficiency(P+Q+R) − Efficiency(P+Q) = 11 − 9 = 2 units/day.

60% of Total Work = 0.6 × 180 = 108 units.

Time for R = 108 ÷ 2 = 54 days.

∴ The correct answer is 54 days.

Alternate Method

Given: Time taken by P, Q, and R = 16(4/11) days = 180/11 days. Time taken by P and Q = 20 days.

⇒ Efficiency of R = (Efficiency of P+Q+R) − (Efficiency of P+Q)

⇒ Efficiency of R = 11 − 9 = 2 units/day

⇒ 60% of Total Work = (60/100) × 180 = 108 units

⇒ Time taken by R alone = 108 ÷ 2 = 54 days

∴ The correct answer is 54 days.

Additional Information

Total Work Formula

Total Work = Efficiency × Time. For fractional days, use the LCM of numerators for Total Work.

Efficiency Calculation

Efficiency of a specific person is found by subtracting the combined efficiency of others from the total group efficiency.

Percentage of Work

Time required for partial work = (Percentage of Total Work) ÷ (Efficiency of the person).

Shortcut Trick

Apply the Weighted Average deviation method.

Deviation of Boys = 90 − 92 = −2 per boy.

Total negative deviation for 3 boys = 3 × (−2) = −6.

To balance, 2 girls must have a total positive deviation of +6.

Deviation per Girl = +6 ÷ 2 = +3.

Average marks of girls = 92 + 3 = 95.

∴ The correct answer is 95 marks.

Alternate Method

Given: Average marks of Boys = 90, Average of Class = 92, Ratio of Boys : Girls = 3 : 2.

Formula Used: Average of Class = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂)

Let the average marks of Girls be x.

⇒ (3 × 90 + 2 × x) ÷ (3 + 2) = 92

⇒ (270 + 2x) ÷ 5 = 92

⇒ 270 + 2x = 92 × 5

⇒ 270 + 2x = 460

⇒ 2x = 460 − 270

⇒ 2x = 190

⇒ x = 95

∴ The correct answer is 95 marks.

Additional Information

Weighted Average Formula

The overall average of two groups combined is given by A_avg = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Alligation Rule

Ratio of quantities = (Higher Value − Mean) : (Mean − Lower Value). This is very effective for mixture and average problems.

Deviation Method

Net sum of deviations from the mean is always zero. Use this to find missing components quickly without large calculations.

Shortcut Trick

Let the total quantity taken out from vessel A be x litres.

Since equal quantities of milk and water are taken out, milk taken out = x/2 litres, and water taken out = x/2 litres.

Mixture P consists of 4 litres of honey, x/2 litres of milk, and x/2 litres of water.

Since water is 40% of mixture P, we have: x/2 = 40% of (4 + x) ⇒ x/2 = 0.4 × (4 + x) ⇒ x/2 = 1.6 + 0.4x ⇒ 0.5x − 0.4x = 1.6 ⇒ 0.1x = 1.6 ⇒ x = 16 litres.

Percentage of mixture taken out from vessel A = (16 ÷ 45) × 100% ≈ 35.56%, which lies between 32% to 36%.

∴ The correct answer is Between 32% to 36%.

Alternate Method

Given:

Total volume of mixture in vessel A = 45 litres.

Quantity of honey in vessel B = 4 litres.

Equal quantities of milk and water are taken out from vessel A and added to vessel B to form mixture P.

Percentage of water in mixture P = 40%.

Calculations:

Let the quantity of milk taken out be y litres.

Therefore, the quantity of water taken out is also y litres.

⇒ Total quantity of mixture taken out from vessel A = 2y

⇒ Total quantity of mixture P in vessel B = Honey + Milk + Water = 4 + y + y = 4 + 2y

⇒ Water percentage in mixture P = (Water ÷ Total Mixture P) × 100

⇒ 40 = [y ÷ (4 + 2y)] × 100

⇒ 40 ÷ 100 = y ÷ (4 + 2y)

⇒ 0.4 = y ÷ (4 + 2y)

⇒ 0.4 × (4 + 2y) = y

⇒ 1.6 + 0.8y = y

⇒ 1.6 = y − 0.8y

⇒ 1.6 = 0.2y

⇒ y = 1.6 ÷ 0.2 = 8 litres

⇒ Total quantity taken out from vessel A = 2y = 2 × 8 = 16 litres

⇒ Required Percentage = (Total quantity taken out ÷ Total mixture in vessel A) × 100

⇒ Required Percentage = (16 ÷ 45) × 100

⇒ Required Percentage = 320 ÷ 9 = 35.56%

⇒ The percentage value 35.56% lies in the interval between 32% to 36%.

∴ The correct answer is Between 32% to 36%.

Additional Information

Mixture Concentration Formula

The concentration of any ingredient in a mixture is calculated as (Quantity of the ingredient ÷ Total volume of the mixture) × 100.

Equal Quantity Replacement Concept

When equal volumes of different components are removed from one container and combined in another, their ratio in the new container is 1:1, making the algebraic expression linear and easy to solve.

Shortcut Trick

Total CP = 400 × 2 = 800. Let CP of A be x and B be (800 − x).

Total SP = 1.1x + 1.2(800 − x) = 928.

1.1x + 960 − 1.2x = 928 ⇒ 0.1x = 32 ⇒ x = 320.

CP of B = 800 − 320 = 480.

∴ The correct answer is Rs. 480.

Alternate Method

Given: Average CP = Rs. 400, Profit A = 10%, Profit B = 20%, Total SP = Rs. 928. Formulas: Total CP = Average CP × 2 | SP = CP × (100 + P%)/100

⇒ Total Cost Price (CP) of A and B = 400 × 2 = Rs. 800. ⇒ Total Profit earned = Total SP − Total CP = 928 − 800 = Rs. 128.

⇒ Let CP of shirt B be y. Then CP of shirt A = (800 − y).

⇒ Profit from A + Profit from B = Total Profit.

⇒ (800 − y) × 10/100 + y × 20/100 = 128.

⇒ 80 − 0.1y + 0.2y = 128 ⇒ 0.1y = 48 ⇒ y = 480.

∴ The correct answer is Rs. 480.

Additional Information

Alligation Method

Overall Profit % = (128/800) × 100 = 16%. Alligating 10% and 20% with mean 16% gives CP ratio of (20−16):(16−10) = 4:6 = 2:3.

Cost Price Calculation

CP of B = Total CP × (Ratio share) = 800 × (3/5) = 480.

Shortcut Trick

Total Interest % in Scheme A = 15% × 2 = 30%

Total Interest % in Scheme B = 12% × 2 = 24%

Equation: 30% of x + 24% of (x + 500) = 4224

54% of x + 120 = 4224 ⇒ 54% of x = 4104 ⇒ x = 7600

∴ The correct answer is 7600.

Alternate Method

Given:

Principal in A = x, Rate = 15%, Time = 2 years

Principal in B = x + 500, Rate = 12%, Time = 2 years

Formula Used:

Simple Interest (SI) = (P × R × T) ÷ 100

Calculations:

⇒ SI_A = (x × 15 × 2) ÷ 100 = 0.30x

⇒ SI_B = ((x + 500) × 12 × 2) ÷ 100 = 0.24(x + 500)

⇒ SI_A + SI_B = 4224

⇒ 0.30x + 0.24x + 120 = 4224

⇒ 0.54x = 4224 − 120

⇒ 0.54x = 4104

⇒ x = 4104 ÷ 0.54 = 7600

∴ The correct answer is 7600.

Additional Information

Simple Interest Calculation

Interest remains constant every year if the principal is the same. Total SI % = Rate × Time.

Effective Rate Method

To find the total value, sum the products of each principal and its effective interest rate percentage.

Shortcut Trick

Let original length be 3x and breadth be 2x.

New Area = (3x + 5) × 2x = 2600 ⇒ 6x² + 10x − 2600 = 0 ⇒ 3x² + 5x − 1300 = 0.

Solving for x: x = 20 (as x must be positive).

Original Perimeter = 2(3x + 2x) = 10x = 10 × 20 = 200 m.

∴ The correct answer is 200 m.

Alternate Method

Given: Ratio of Length (L) to Breadth (B) = 3 : 2. Let L = 3x and B = 2x. New Length = 3x + 5. New Area = 2600 m².

Formula Used: Area of Rectangle = Length × Breadth; Perimeter = 2(Length + Breadth).

⇒ (3x + 5) × 2x = 2600

⇒ 6x² + 10x = 2600

⇒ 6x² + 10x − 2600 = 0

⇒ 3x² + 5x − 1300 = 0

⇒ 3x² + 65x − 60x − 1300 = 0

⇒ x(3x + 65) − 20(3x + 65) = 0

⇒ (x − 20)(3x + 65) = 0

⇒ x = 20 (Dimensions cannot be negative)

⇒ Original Length = 3 × 20 = 60 m; Original Breadth = 2 × 20 = 40 m

⇒ Original Perimeter = 2(60 + 40) = 2 × 100 = 200 m

∴ The correct answer is 200 m.

Additional Information

Area and Perimeter of Rectangle

Area = Length × Breadth and Perimeter = 2 × (Length + Breadth). The diagonal of a rectangle is √(L² + B²).

Quadratic Roots Method

For a quadratic equation ax² + bx + c = 0, roots can be found using factorisation or the formula x = [−b ± √(b² − 4ac)] / 2a.

Ratio and Scaling

When dimensions are in ratio a:b, always use a common multiplier x to represent actual values as ax and bx for algebraic consistency.

300 * 0.5 - 1 = 149

149 * 1 -1 = 148

148* 1.5 - 1 = 221

221 * 2 - 1 = 441 

441 * 2.5 - 1 = 1101.5

∴ The correct answer is 149.

100 - 25 = 75,

256 - 100 = 156,

476 - 256 = 220,

745 - 476 = 269,

1050 - 745 = 305

Again taking the difference 

156 - 75 = 81(9²)

220 - 156 = 64(82​)

269 - 220 = 49(72​)

305 - 269 = 36(6²)

Hence 100 is the correct answer