Quantitative Aptitude PYQs

Total number of articles sold by X, Y and Z in 2018 = 430 

Total number of articles sold in all three years by Y = 390

Total number of articles sold in all three years by Z = 158

Let, articles sold by Z in 2017 = 6x and Z in 2018 = 11x

Let, articles sold by X in 2016 = 7y and Z in 2016 = 8y

So, 8y = 6x + 20 ...............

(1) Y in 2018 = 7y × 5 = 35y

(articles sold by Y in 2016) / 8y = 25 / 28

So, articles sold by Y in 2016 = 200y/28 

X in 2017 = 14y

Now, 11x + 6x + 8y = 158

17x + 8y = 158 ................... (2) After solving equation 1 and 2, we get-

x = 6 and y = 7

Required average = (266 + 390) / 2 = 656 / 2 = 328

Hence, correct answer is option 5.

Total number of articles sold by X, Y and Z in 2018 = 430 

Total number of articles sold in all three years by Y = 390

Total number of articles sold in all three years by Z = 158

Let, articles sold by Z in 2017 = 6x and Z in 2018 = 11x

Let, articles sold by X in 2016 = 7y and Z in 2016 = 8y

So, 8y = 6x + 20 ...............

(1) Y in 2018 = 7y × 5 = 35y

(articles sold by Y in 2016) / 8y = 25 / 28

So, articles sold by Y in 2016 = 200y/28 

X in 2017 = 14y

Now, 11x + 6x + 8y = 158

17x + 8y = 158 ................... (2) After solving equation 1 and 2, we get-

x = 6 and y = 7

Total number of articles sold by Z in 2019 = 56 × 150/100 = 84

Required sum = 84 + 98 = 182

Hence, correct answer is option 3.

Total number of articles sold by X, Y and Z in 2018 = 430 

Total number of articles sold in all three years by Y = 390

Total number of articles sold in all three years by Z = 158

Let, articles sold by Z in 2017 = 6x and Z in 2018 = 11x

Let, articles sold by X in 2016 = 7y and Z in 2016 = 8y

So, 8y = 6x + 20 ...............

(1) Y in 2018 = 7y × 5 = 35y

(articles sold by Y in 2016) / 8y = 25 / 28

So, articles sold by Y in 2016 = 200y/28 

X in 2017 = 14y

Now, 11x + 6x + 8y = 158

17x + 8y = 158 ................... (2) After solving equation 1 and 2, we get-

x = 6 and y = 7

Total number of articles sold by P in 2016 = 245 × 3/5 = 147

Total number of articles sold by P in 2017 = 49 × 9/7 = 63

So, Total number of articles sold by P in 2018 = 332 - 147 - 63 = 122

Hence, correct answer is option 1.

Shortcut Trick

From Mixture C: Milk % = 100% − (168 ÷ 240 × 100)% = 100% − 70% = 30%.

Given (x + 5)% = 30% ⇒ x = 25. (Statement 1 is true).

In Mixture A: Water is 25% more than milk ⇒ Ratio M : W = 100 : 125 = 4 : 5. Milk = 4/9 × 360 = 160 ml. (Statement 2 is false).

In Mixture B: Water is 1.6 × 25 = 40% more than milk ⇒ Ratio M : W = 100 : 140 = 5 : 7. Milk = 5/12 × 480 = 200 ml. (Statement 3 is false).

∴ The correct answer is Only 1.

Alternate Method

Given: Total quantity of Mixture C = 240 ml, Water = 168 ml. Mixture A = 360 ml. Mixture B = 480 ml.

Step 1: Solve for x using Mixture C

⇒ Quantity of milk in Mixture C = Total − Water = 240 − 168 = 72 ml

⇒ Percentage of milk in C = (72 ÷ 240) × 100 = 30%

⇒ x + 5 = 30 ⇒ x = 25 (Statement 1 is True)

Step 2: Check Statement 2 (Mixture A)

⇒ Water is x% = 25% more than milk. Let Milk = m, then Water = 1.25m

⇒ m + 1.25m = 360 ⇒ 2.25m = 360

⇒ Milk in A = 360 ÷ 2.25 = 160 ml (Statement 2 is False)

Step 3: Check Statement 3 (Mixture B)

⇒ Water is 1.6x% = 1.6 × 25% = 40% more than milk. Let Milk = n, then Water = 1.4n

⇒ n + 1.4n = 480 ⇒ 2.4n = 480

⇒ Milk in B = 480 ÷ 2.4 = 200 ml (Statement 3 is False)

∴ The correct answer is Only 1.

Additional Information

Percentage Comparison

If A is k% more than B, then A = B × (1 + k/100). The ratio A : B = (100 + k) : 100.

Mixture Concentration

Component Quantity = Total Quantity × (Component % / 100). Sum of all component percentages always equals 100%.

Ratio Division

To divide a quantity Q in ratio a:b, the first part is [a / (a + b)] × Q and the second part is [b / (a + b)] × Q.

Shortcut Trick

Ratio of 3^rd year CI to 2^nd year CI = (1 + r/100) = 11/10.

Rate of interest R% = (1/10) × 100 = 10%.

For 10% rate, Effective CI for 2 years = 10 + 10 + (10×10)/100 = 21%.

Given 21% of (X + 1500) = 1890 ⇒ (X + 1500) = 1890/0.21 = 9000.

X = 9000 − 1500 = 7500.

∴ The correct answer is Rs. 7500.

Alternate Method

Given: Ratio of CI₃ / CI₂ = 11/10. Time = 2 years, CI = Rs. 1890 on Principal = (X + 1500).

Formula Used: CI for n^th year = P(1 + R/100)ⁿ⁻¹ × (R/100). Total CI = P[(1 + R/100)ⁿ − 1].

⇒ CI₃ / CI₂ = [P(1 + R/100)² × (R/100)] / [P(1 + R/100)¹ × (R/100)] = 11/10

⇒ 1 + R/100 = 1.1 ⇒ R/100 = 0.1 ⇒ R = 10%

⇒ For 2 years at 10%, CI = (X + 1500) × [(1.1)² − 1] = 1890

⇒ (X + 1500) × 0.21 = 1890

⇒ X + 1500 = 1890 / 0.21 = 9000

⇒ X = 9000 − 1500 = 7500

∴ The correct answer is Rs. 7500.

Additional Information

Successive Year CI Ratio

The ratio of Compound Interest earned in any two consecutive years (n+1) and n is always (1 + R/100).

Effective Interest Rate

For 2 years, the effective CI rate can be calculated using x + y + (xy/100). For 10%, it is 10 + 10 + 1 = 21%.

CI vs SI Difference

The difference between CI and SI for 2 years is P(R/100)² and for 3 years is P(R/100)² × (3 + R/100).

Concept used:

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved second, and following BODMAS rule in the bracket

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

[{(1.2 × 2.5) + (0.50 × 18)} – (0.500 ÷ 0.25)] ÷ 10 = ?

⇒ [{3 + 9} – 2] ÷ 10 = ?

⇒  [12 – 2] ÷ 10 = ?

⇒  10 ÷ 10 = ?

⇒ ? = 1

Hence, the correct answer is 1.

Solution :

Concept used :

Follow the BODMAS rule to solve this question, as per the question is given below:

Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated.

⇒ 29 x 3 + ? = 117 + 81 x 3 - 173

⇒ 87 + ? = 360 - 173 

⇒ ? = 360 - 173 - 87

⇒ ? = 360 - 260 

⇒ ? = 100

Hence, the correct answer is option 3.

Given:

74.99% of 499.97 + 359.99 ÷ 59.98 = ? + 15.05 × 3.98 + 14.99 × 5.98

Concept used:

Follow the 'BODMAS' Rule

B ⇒ Bracket, O ⇒ of, D ⇒ Division, M ⇒ Multiplication, A ⇒ Addition, S ⇒ Subtraction

Calculation:

⇒ 74.99% of 499.97 + 359.99 ÷ 59.98 = ? + 15.05 × 3.98 + 14.99 × 5.98

⇒ 75/100 × 500 + 360/60 = ? + 15 × 4 + 15 × 6

⇒ 375 + 6 = ? + 60 + 90

⇒ ? = 381 - 150

⇒ ? = 231

∴ The value of '?' is 231.

Given:

338.99 + 266.98 ÷ 3 + 15.982 + 16.08 = ? × 77.77% of 89.98 - 19.99

Concept used:

Follow the 'BODMAS' Rule

B ⇒ Bracket, O ⇒ of, D ⇒ Division, M ⇒ Multiplication, A ⇒ Addition, S ⇒ Subtraction

Calculation:

⇒ 338.99 + 266.98 ÷ 3 + 15.982 + 16.08 = ? × 77.77% of 89.98 - 19.99

⇒ 339 + 267 ÷ 3 + 162 + 16 = ? × 7/9 × 90 - 20

⇒ 339 + 89 + 256 + 16 = ? × 70 - 20

⇒ ? × 70 =  700 + 20 

⇒ ? = 720/70

⇒? = 10.28 ≈ 10

∴ The value of '?' is 10.

Solution :

 Concept used :

 Follow the BODMAS rule to solve this question, as per the question is given below:

 Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

 Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

 Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

 Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated

 Given :

⇒ 320.001÷8.041 - 14.99×14.01 + 27.0223^1/3 = ?

Approximate the number to its nearest value 

⇒ 320÷8 -15×14 +27^1/3 = ?

⇒ 40 - 210 + (3)^3×1/3 = ?

⇒ 40 - 210 + 3 = ?

⇒ -170 + 3 = ?

⇒ ? = -167

Hence - 167 is the correct answer**.**

Sum of roots = 1/5 + 2/3 = 13/15

Product of roots = 1/5 × 2/3 = 2/15

ax² + bx + c = 0

Comparing the options with this equation, we get the right answer from option 2.

15x² - 13x + 2 = 0

sum of roots = -b/a = - (-13/15) = 13/15

Products of roots = c/a = 2/15

So, 15x2 - 13x + 2 = 0 is the correct equation.

Hence, correct answer is option 2.

Sum of roots = 1/5 + 2/3 = 13/15

Product of roots = 1/5 × 2/3 = 2/15

ax2 + bx + c = 0

So, equation will be = 15x2 - 13x + 2 = 0

After diving by 2, equation is = (15x2 - 13x + 2) / 2 = 0 / 2

7.5x2 - 6.5x + 1 = 0

Now,

7.5x2 - 6.5x + 1 +  2.5x2 - 0.5x = 0

10x2 - 7x + 1 = 0

Now, comparing with equation- ax2 + bx + c = 0

Product = c/a = 1/10

Hence, correct answer is option 1.

Sum of roots = 1/5 + 2/3 = 13/15

Product of roots = 1/5 × 2/3 = 2/15

Required difference = 13/15 - 2/15 = 11/15

Hence, correct answer is option 3.

Shortcut Trick

When Cost Price (CP) of two articles is the same, net profit or loss percent is the arithmetic mean of individual profit/loss percentages.

Net Gain/Loss% = (P₁% − L₂%) ÷ 2

Net Change = (12% − 8%) ÷ 2 = 4% ÷ 2 = 2% profit.

Quantity A = 2% profit and Quantity B = 2% profit.

∴ The correct answer is Quantity A = Quantity B.

Alternate Method

Given: CP of each bag = Rs 800, Profit₁ = 12%, Loss₂ = 8%.

Formulas: Profit/Loss Value = CP × %; Net % = (Total Profit/Loss ÷ Total CP) × 100.

⇒ Profit on first bag = 12% of 800 = (12 ÷ 100) × 800 = Rs 96.
⇒ Loss on second bag = 8% of 800 = (8 ÷ 100) × 800 = Rs 64.
⇒ Overall profit = 96 − 64 = Rs 32.
⇒ Total Cost Price = 800 + 800 = Rs 1600.
⇒ Gain % = (32 ÷ 1600) × 100 = 2%.
⇒ Comparing Quantity A (2%) and Quantity B (2%): Quantity A = Quantity B.

∴ The correct answer is Quantity A = Quantity B or No relation.

Additional Information

Equal CP Case

If two items are bought at the same price, the net gain/loss depends only on the numerical sum of individual percentages.

Equal SP Case

If two items are sold at the same price with x% gain and x% loss, there is always a net loss = x² ÷ 100 %.

Given:

Ghee : Dalda = 5 : a

Calculation:

Let the quantity of ghee = 5x litres

⇒ the quantity of Dalda = ax litres

According to the question, ax = 50% of (5x + ax + 5)

⇒ 2ax = 5x + ax + 5

⇒ ax = 5x + 5 ….(i)

⇒ 5x = 40% of (5x + ax + 2)

⇒ 25x = 10x + 2ax + 4

⇒ ax = (15x – 4)/2 ….(ii)

From the equation (i) and (ii)

⇒ 5x + 5 = (15x – 4)/2

⇒ 5x = 14

⇒ x = 14/5

Put the value of x in the equation (i)

⇒ a = 95/14

The quantity of ghee in the original mixture = 5x = 14 litres

⇒ the quantity of dalda = ax

⇒ quantity of dalda = 19 litres

Quantity I : when, 10 litres of ghee were added in the original mixture

ghee = (14 + 10) / (14 + 19 + 10)

⇒ ghee = 24/43

Quantity II: when 3 litres of dalda were added in the original mixture

⇒ dalda = (19 + 3) / (14 + 19 + 3)

⇒ dalda = 11/18

∴ Quantity I < Quantity II

Using statement I

Let the price of Masoor dal and Arhar dal be x and y respectively

Selling price of dal = 36.66/kg 

A profit of 20% is earned

∴ Cost price of dal = 36.66 × (100/120) = 30.56

⇒ (12x + 15y)/27 = 30.56

⇒12x + 15y = 30.56 × 27 = 825.12 ≈ 825

Statement I alone is not sufficient to answer the question

Using statement II

The ratio of cost price of Masoor dal and Arhar dal is 5 ∶ 7

⇒ Cost price of Massor dal and Arhar dal be 5x and 7x respectively 

Statement II alone is not sufficient to answer the question

Using statement I and statement II together 

⇒ (12 × 5x + 15 × 7x) = 825

⇒ 60x + 105x = 825

⇒ x = 825/165 = 5

∴ Cost price of Masoor dal = 5 × 5 = Rs.25/kg

Cost price of 15 kg Masoor dal

⇒ 25 × 15 = Rs.375

∴ The data in both statements I and II together are needed to answer the question.

Shortcut Trick

Ratio of profit sharing: John : (Johnny + Janardan) = 90% : 10% = 9 :

1. Increase in John's income = 25% − 22% = 3% of his investment.

3% of John's investment = Rs. 540 ⇒ 1% = Rs. 180 ⇒ 100% = Rs. 18000.

Since profit is shared in the ratio of investment: 9 units = Rs. 18000.

1 unit (Johnny + Janardan) = Rs. 18000 ÷ 9 = Rs. 2000. ∴ The correct answer is Rs. 18000, Rs. 2000.

Alternate Method

Given → John's Profit Share = 90%, Johnny's = 5%, Janardan's = 5%. Profit rate increase = 3% (25% − 22%).
Formulas: Profit = Investment × Rate | Investment Ratio = Profit Sharing Ratio.

⇒ Let John's investment be J₁. 3% of J₁ = 540.
⇒ (3 ÷ 100) × J₁ = 540 ⇒ J₁ = (540 × 100) ÷ 3 = Rs. 18000.
⇒ Ratio of John to Others = 90 : 10 = 9 :

1. ⇒ Investment of Johnny + Janardan = (1 ÷ 9) × 18000 = Rs. 2000.

∴ The correct answer is Rs. 18000, Rs. 2000.

Additional Information

Partnership Principle

When the time period is constant, the Profit Sharing Ratio is directly proportional to the Investment Ratio.

Income Change

Change in Income = Investment × Change in Profit Rate.

Shortcut Trick

Ratio of Boat speed (u) to Stream speed (v) = 900 : 100 = 9 :

1. Downstream speed (D) = 9 + 1 = 10 units; Upstream speed (U) = 9 − 1 = 8 units.

Time Ratio: T_down = 150% of T_up ⇒ T_down / T_up = 3/2.

Distance Ratio: [2(M + 15) ÷ 10] / [(M + 31) ÷ 8] = 3/2 ⇒ (M + 15) / 5 × 8 / (M + 31) = 3/2.

Solving for M: 16(M + 15) = 15(M + 31) ⇒ 16M + 240 = 15M + 465 ⇒ M = 225.

Value required: 225% of 500 = 2.25 × 500 = 1125.

∴ The correct answer is 1125.

Alternate Method

Given: Still water speed (u) = 900% of stream (v) ⇒ u = 9v.

Downstream speed = u + v = 10v; Upstream speed = u − v = 8v.

Time Downstream (T_D) = 2(M + 15) ÷ 10v.

Time Upstream (T_U) = (M + 31) ÷ 8v.

According to question: T_D = 1.5 × T_U

⇒ 2(M + 15) / 10v = 1.5 × (M + 31) / 8v

⇒ (M + 15) / 5 = 3/2 × (M + 31) / 8

⇒ (M + 15) / 5 = 3(M + 31) / 16

⇒ 16(M + 15) = 15(M + 31)

⇒ 16M + 240 = 15M + 465

⇒ M = 225

⇒ Required value = 225% of 500 = (225/100) × 500 = 1125.

∴ The correct answer is 1125.

Additional Information

Relative Speed in Water

Downstream speed = Speed of boat in still water + Speed of stream. Upstream speed = Speed of boat in still water − Speed of stream.

Boat Speed Calculation

Speed in still water (u) = (Downstream speed + Upstream speed) ÷ 2. Stream speed (v) = (Downstream speed − Upstream speed) ÷ 2.

Percentage Conversion

To find x% of y, use the formula (x/100) × y. For example, 225% of 500 is 2.25 × 500 = 1125.

Shortcut Trick

Speed of both trains is 117 km/hr ⇒ 117 × 5/18 = 32.5 m/s.

Since train P takes 10 seconds to cross a pole: Length of train P = Speed × Time = 32.5 × 10 = 325 m.

Given, Length of P = x + 10 ⇒ x = 315 m (which is the length of the platform).

Time taken by train P to cross the platform = (Length of P + Length of platform) ÷ Speed = (325 + 315) ÷ 32.5 = 640 ÷ 32.5 ≈ 19.69 seconds (approximated to 19.6 seconds as per option constraints).

∴ The correct answer is 19.6 sec.

Alternate Method

Given:

Length of train P = (x + 10) m

Length of train Q = (x + 40) m

Speed of both trains = 117 km/hr

Time taken by train P to cross a pole = 10 seconds

Formula Used:

Speed in m/s = Speed in km/hr × 5/18

Distance = Speed × Time

Calculations:

→ Convert Speed of trains into m/s:

⇒ Speed = 117 × 5/18 = 32.5 m/s

→ Find the length of train P using its pole-crossing time:

⇒ Length of train P = Speed × Time

⇒ x + 10 = 32.5 × 10

⇒ x + 10 = 325

⇒ x = 315 m

→ Find the total distance to cross the platform:

⇒ Total Distance = Length of train P + Length of platform

⇒ Total Distance = 325 + 315 = 640 m

→ Time taken by train P to cross the platform:

⇒ Time = Total Distance ÷ Speed

⇒ Time = 640 ÷ 32.5 ≈ 19.6 sec

∴ The correct answer is 19.6 sec.

Additional Information

Unit Conversion of Speed

To convert km/hr to m/s, multiply the speed by 5/18. To convert m/s to km/hr, multiply the speed by 18/5.

Train Crossing Stationary Objects

When a train crosses a pole or a standing man, the distance covered equals the length of the train. When it crosses a platform or bridge, the distance covered equals the sum of the lengths of the train and the platform.

Relative Speed of Trains

When two trains of lengths L₁ and L₂ travel in opposite directions with speeds S₁ and S₂, the relative speed is (S₁ + S₂). If they travel in the same direction, the relative speed is |S₁ − S₂|.

Shortcut Trick

Interest for 4 years (13 − 9) = ₹31725 − ₹25425 = ₹6300.

Interest per year = 6300 ÷ 4 = ₹1575.

Principal (P) = 25425 − (9 × 1575) = 25425 − 14175 = ₹11250.

Original rate r = (1575 ÷ 11250) × 100 = 14%.

New rate = 14 + 1 = 15%. New Amount = 11250 + (11250 × 15% × 12) = 11250 + 20250 = ₹31500.

∴ The correct answer is ₹31500.

Alternate Method

Given: Amount after 9 years (A₉) = ₹25425, Amount after 13 years (A₁₃) = ₹31725.

Formula: SI = (P × R × T) ÷ 100 | Amount = P + SI

⇒ SI for 4 years = A₁₃ − A₉ = 31725 − 25425 = ₹6300

⇒ SI for 1 year = 6300 ÷ 4 = ₹1575

⇒ SI for 9 years = 1575 × 9 = ₹14175

⇒ Principal (P) = 25425 − 14175 = ₹11250

⇒ Rate (r) = (1575 × 100) ÷ 11250 = 14%

⇒ New Rate = r + 1 = 15%

⇒ SI for 12 years at 15% = (11250 × 15 × 12) ÷ 100 = ₹20250

⇒ Final Amount = 11250 + 20250 = ₹31500

∴ The correct answer is ₹31500.

Additional Information

Simple Interest Property

In Simple Interest, the interest earned remains constant every year because it is always calculated on the original Principal.

Effective Rate of Interest

The total percentage interest earned over T years is (R × T)%. Total Amount as a percentage of Principal is (100 + RT)%.

Difference between Amounts

The difference between the amounts of two different time periods (for the same Principal and Rate) is equal to the Simple Interest for the time difference.