Quantitative Aptitude PYQs

Total production by India = 120 + 150 + 180 + 210 + 240 = 900 lakh

Total production by China = 180 + 210 + 240 + 270 + 330 = 1230 lakh

Required Percentage = (1230 – 900)/1230 x 100 = 26.83% ≈ 26.8%

China for Company C = 420 – 180 = 240 lakh

Total production (India + China) for Company B = 360 lakh

Required ratio = 240 : 360 = 2 : 3

Premium phones manufactured by China for Company E = 30% of 330 = 99 lakh

Premium phones manufactured by India for Company D = 20% of 210 = 42 lakh

Required Percentage = (99 – 42)/42 x 100 = 135.71% ≈ 135.7% approx.

China's production for Companies A and D = 180 + 270 = 450 lakh

India's production for Companies B and C = 150 + 180 = 330 lakh

Required Percentage = [(450 – 330)/330] x 100 = 36.36% = 36.4% approx.

China's production for Company B = (India + China) – India = 360 – 150 = 210 lakh

Average total production (India + China) for Companies C, D and E = (420 + 480 + 570)/3 = 490 lakh

Required Percentage = (210/490) x 100 = 42.857% = 43% approx.

Shortcut Trick

Let the total salary be 100 units.

Remaining after house rent (15%) = 85 units.

Remaining after groceries (40% of 85) = 85 − 34 = 51 units.

Other expenses = 1/3 of 51 = 17 units. Final remaining = 51 − 17 = 34 units.

Since 34 units = ₹8500, then 1 unit = ₹250.

Other expenses (17 units) = 17 × 250 = ₹4250.

∴ The correct answer is ₹4250.

Alternate Method

Given: Rent = 15%, Groceries = 40% of remaining, Other = 1/3 of next remaining, Final balance = ₹8500.

Calculations:

⇒ Let total salary be S.

⇒ After rent: S × (1 − 0.15) = 0.85S

⇒ After groceries: 0.85S × (1 − 0.40) = 0.85S × 0.6 = 0.51S

⇒ Final balance after other expenses (1/3^rd spent, so 2/3^rd left): 0.51S × (2/3) = 0.34S

⇒ 0.34S = 8500

⇒ S = 8500 ÷ 0.34 = ₹25000

⇒ Amount before other expenses = 0.51 × 25000 = ₹12750

⇒ Other expenses = 1/3 × 12750 = ₹4250

∴ The correct answer is ₹4250.

Additional Information

Successive Percentage Change

When values are spent from the remaining amount, use the multiplying factor: Final = Initial × (1 − r₁) × (1 − r₂)...

Fraction to Percentage Conversion

Spending 1/3 is equivalent to spending 33.33%. The remaining fraction is always (1 − spent fraction).

Shortcut Trick

CI for 2 years at 10% = 21% of Principal.

Scheme II Interest = 21% of 6000 = ₹1260.

Scheme I Interest = 1260 + 900 = ₹2160.

Rate (x) = (Interest × 100) ÷ (Principal × Time) = (2160 × 100) ÷ (6000 × 3) = 12%.

∴ The correct answer is 12.

Alternate Method

Given: Principal (P) = ₹6000, Time for SI = 3 years, Rate for SI = x%, Time for CI = 2 years, Rate for CI = 10%.

Formula Used: SI = (P × R × T) ÷ 100; CI = P[(1 + R/100)ⁿ − 1]

⇒ Interest from Scheme II (CI) = 6000 × [(1 + 10/100)² − 1]

⇒ CI = 6000 × [1.21 − 1] = 6000 × 0.21 = ₹1260

⇒ Interest from Scheme I (SI) = 1260 + 900 = ₹2160

⇒ SI = (6000 × x × 3) ÷ 100 = 180x

⇒ 180x = 2160

⇒ x = 2160 ÷ 180 = 12

∴ The correct answer is 12.

Additional Information

Effective Rate of Compound Interest

For 2 years at R%, effective rate = (2R + R²/100)%. For 10%, it is (20 + 1) = 21%.

SI vs CI Difference for 2 Years

Difference = P × (R/100)². This is used when the principal and rate are the same for both.

Simple Interest Property

Simple interest remains constant every year if the principal and rate remain unchanged.

Given:

Ratio of the quantity of rice and wheat = 5:3

1/4^th of rice and half of wheat are mixed, and the difference between them is 10 kg

30% of the initial quantity of rice and y% of the initial quantity of wheat is 180 kg

Calculation:

Difference = |(1/4 × Rice) - (1/2 × Wheat)|

Total quantity equation: 30% of Rice + y% of Wheat = 180 kg

Let the quantity of rice be 5x and the quantity of wheat be 3x

1/4^th of rice = 1/4 × 5x = 5x/4

1/2 of wheat = 1/2 × 3x = 3x/2

The difference between the mixed quantities is 10 kg

⇒ |5x/4 - 3x/2| = 10

Since 3x/2 > 5x/4:

⇒ 3x/2 - 5x/4 = 10

⇒ 6x/4 - 5x/4 = 10

⇒ x/4 = 10

⇒ x = 40

Quantity of rice = 5x = 5 × 40 = 200 kg

Quantity of wheat = 3x = 3 × 40 = 120 kg

30% of rice + y% of wheat = 180 kg

⇒ 0.3 × 200 + (y/100) × 120 = 180

⇒ 60 + 1.2y = 180

⇒ 1.2y = 120

⇒ y = 120 / 1.2

⇒ y = 100

∴ The value of y is 100.

Given 

Ace Spade is lost.

Concept used

P(E) = Favourable outcomes / Total outcomes

Calculation

As a Spade has lost, 

Total number of cards = 52 -1 = 51.

Total number of black cards = 26 - 1 = 25.

P(Black Card)  = 25/51

⇒ Probability of getting a Black Card = 25/51

∴ Required answer is 25/51.

Solution:

Let the efficiency of B be 100

Then, the efficiency of C = {(100 – 25)/100} ×  100 = (75/100) ×  100 = 75

Ratio of B’s efficiency to C’s efficiency = 100: 75 = 4: 3

Ratio of time taken by B to that of C = 3: 4

Time taken by B to complete the work = (3/4) ×  12 = 9 days

x + 5 = 9

x = 4 days

So, total work = LCM of 4, 9 and 12 = 36 units

A’s efficiency = 36/4 = 9 units/day

B’s efficiency = 36/9 = 4 units/day

And, C’s efficiency = 36/12 = 3 units/day

In one day, work done = 9 + 4 + 3 = 16 units

Now, required time = 36/16 = 9/4 = 2 (1/4) days

Common solution:

Since, the sum of the percentages in a pie chart = 100%

10 + x + (x + 2) + (x + 4) + (x + 8) = 100

4x + 24 = 100

4x = 76

So, x = 19

 

Calculation:

Applications received by Companies A and B together = 4,800 + 9,120 = 13,920

Applications received by Company E = 12,960

Required Difference = 13,920 − 12,960 = 960

Solution:

Investment of A = Rs. 3500

Investment of B = Rs. 4000

Investment of C = Rs. 5000

To find the value of x

3500 ×  x : 4000 ×  (x + 2): 5000 ×  (12 - (x + 2)) = 14 : 24 : 30

3500  × x /(5000  × (10 - x)) = 14/30

(35  × x)/(50  × (10 -x)) = 7/15

525x = (3500 - 350x)

875x = 3500

x = 4

Given:

Speed of train A: 15 m/s

Speed of train B: 20 m/s

Time taken by train A to cross the standing man is 4 seconds less than the time taken by train B

Length of train B is 180 meters more than the length of train A

Concept Used:

Distance = Speed × Time.

Solution:

Let L_A be the length of train A.

The time taken by train A to cross the man is t_A = L_A / 15

Let L_B be the length of train B.

The time taken by train B to cross the man is t_B = L_B / 20

Given that L_B = L_A + 180 meters.

The time taken by train A is 4 seconds less than the time taken by train B:

t_A = t_B - 4

Substituting the expressions for t_A and t_B:

L_A / 15 = (L_B / 20) - 4

Substitute L_B = L_A + 180 into the equation, we get

L_A / 15 = (L_A + 180) / 20 - 4

Multiply through by 60 to clear the denominators:

4L_A = 3(L_A + 180) - 240

4L_A = 3L_A + 540 - 240

4L_A = 3L_A + 300

L_A = 300

L_B = L_A + 180 = 300 + 180 = 480

Sum of lengths = L_A + L_B = 300 + 480 = 780

The sum of the lengths of train A and train B is 780 meters.

Shortcut Trick

Net discount % for Case 1 = 10 + 20 − (10 × 20) ÷ 100 = 28%.

Net discount % for Case 2 = 15 + 15 − (15 × 15) ÷ 100 = 27.75%.

Difference in discount % = 28% − 27.75% = 0.25%.

Change in Selling Price = 0.25% of Marked Price = 0.0025 × 10000 = 25.

∴ The correct answer is Rs. 25.

Alternate Method

Given → Marked Price (MP) = Rs. 10000, Case 1 Discounts = 10% and 20%, Case 2 Discounts = 15% and 15%.

Formulas: SP = MP × (100 − D₁)% × (100 − D₂)% | Net Discount = a + b − ab ÷ 100

⇒ Selling Price 1 (SP₁) = 10000 × (90 ÷ 100) × (80 ÷ 100) = 10000 × 0.9 × 0.8 = 7200.

⇒ Selling Price 2 (SP₂) = 10000 × (85 ÷ 100) × (85 ÷ 100) = 10000 × 0.85 × 0.85 = 7225.

⇒ Change in SP = |SP₂ − SP₁| = |7225 − 7200| = 25.

∴ The correct answer is Rs. 25.

Additional Information

Successive Discounts

Applying multiple discounts one after another is called successive discounting. The total discount is always less than the sum of individual percentages because each subsequent discount is applied to the reduced price.

Constant Sum Property

When the sum of two successive discounts is constant (e.g., 10+20=30 and 15+15=30), the pair with the larger difference provides a higher total discount and a lower SP.

Shortcut Trick

Volume ratio V_cyl : V_cone = (πr₁²h₁) : (1/3πr₂²h₂) = 3(r₁/r₂)² × (h₁/h₂).

From Statement I: r₁ : r₂ = 1 : 2 is known.

From Statement II: h₁ = 21 cm and h₂ = 21 − 7 = 14 cm. Thus, h₁ : h₂ is known.

Since both radius ratio and height ratio are required, I and II together are sufficient.

∴ The correct answer is Both statements I and II are required to answer the question.

Alternate Method

Given: Ratio of volumes of Cylinder and Cone.

Formula Used: Volume of Cylinder = πr²h; Volume of Cone = (1/3)πR²H

⇒ Volume Ratio = (πr₁²h₁) / (1/3πr₂²h₂) = 3 × (r₁/r₂)² × (h₁/h₂)

⇒ From Statement I: r₁/r₂ = 1/2. (Ratio of radii is 1 : 2)

⇒ From Statement II: h₁ = 21 cm, h₂ = 21 − 7 = 14 cm. So, h₁/h₂ = 21/14 = 3/2.

⇒ Required Ratio = 3 × (1/2)² × (3/2) = 3 × (1/4) × (3/2) = 9/8.

⇒ Statement III alone or with I/II without the other height info is insufficient to find specific heights.

∴ The correct answer is Both statements I and II are required to answer the question.

Additional Information

Volume of Solids

The volume of a Cylinder is πr²h and a Cone is 1/3πr²h. If they have same radius and height, the cylinder volume is 3 times the cone volume.

Data Sufficiency for Ratios

To find a ratio of volumes, we only need the ratio of radii and the ratio of heights, not necessarily their absolute values.

Surface Area Formulas

Curved Surface Area (CSA) of cylinder = 2πrh; CSA of cone = πrl, where l is the slant height √(r²+h²).

Shortcut Trick

To find the speed of the boat in still water (x), we need the upstream speed (u) and downstream speed (d).

Statement I gives one linear equation: 14/u + 35/d = 4.

Statement II gives another linear equation: 29/u + 24/d = 5.

Two distinct linear equations with two variables (1/u and 1/d) can be solved to find unique values for u and d, and thus x = (d + u) ÷ 2.

∴ The correct answer is both the statements together are needed.

Alternate Method

Let the speed of boat in still water be x kmph and speed of stream be y kmph.

Upstream speed (u) = x − y and Downstream speed (d) = x + y.

From Statement I: 14/u + 35/d = 4 → (Eq 1)

From Statement II: 29/u + 24/d = 5 → (Eq 2)

⇒ Let 1/u = A and 1/d = B.

⇒ 14A + 35B = 4 and 29A + 24B = 5.

⇒ Solving these two equations gives specific values for A and B, which provides u and d.

⇒ Finally, x = (u + d) ÷ 2.

∴ The correct answer is both the statements together are needed.

Additional Information

Boat and Stream Formulas

Downstream speed (d) = x + y; Upstream speed (u) = x − y. Speed of boat x = (d + u) ÷ 2; Speed of stream y = (d − u) ÷ 2.

Data Sufficiency Tip

For two unknown variables, you generally require two independent equations to find unique values. Always check if equations are redundant or inconsistent.