Quantitative Aptitude PYQs

Given:

Sequence: 1, 3, 6, 10, …

Formula Used:

n-th term of the sequence, \(T_n\) = \(\frac{n(n+1)}{2}\)

Calculation:

Sixth term, \(T_6\) :

\(T_6 = \frac{6(6+1)}{2}\)

\(T_6 = \frac{6 \times 7}{2}\)

\(T_6 = 21\)

Seventh term, \(T_7\) :

\(T_7 = \frac{7(7+1)}{2} \)

\(T_7 = \frac{7 \times 8}{2}\)

\(T_7 = 28\)

Ratio of sixth and seventh terms, \(\frac{T_6}{T_7}\) :

\(\frac{21}{28} = \frac{3}{4}\)

The ratio of the sixth and seventh terms of the sequence is 3 ∶ 4.

Given:

The average of 6 terms = 30

The first term is twice the sum of the remaining terms.

Formula Used:

Average = Sum of terms / Number of terms

Calculation:

Let the first term be x . Let the sum of the remaining 5 terms be S .

Since the average of 6 terms is 30:

Sum of all terms = 30 × 6 = 180

So, x + S = 180

Given that the first term is twice the sum of the remaining terms:

x = 2S

Substitute x = 2S into the sum equation:

2S + S = 180

3S = 180

S = 60

Now, x = 2S

x = 2 × 60

x = 120

The first term is 120.

Given:

Three successive discounts: 20%, 30%, and 10%

Formula used:

Successive Discount Formula:

Net Discount % = A + B + C - (AB + AC + BC)/100 + (ABC)/10000

Where A = 20, B = 30, C = 10

Calculation:

Net Discount % = 20 + 30 + 10 - (20×30 + 20×10 + 30×10)/100 + (20×30×10)/10000

⇒ = 60 - (600 + 200 + 300)/100 + 6000/10000

⇒ = 60 - 11 + 0.6 = 49.6%

∴ The net discount is 49.6%.

Given:

Borrowing Rate = 4% per annum.

Lending Rate = 8% per annum compounded half-yearly.

Gain = ₹166.4 per year.

Formula Used:

Compound Interest Formula for half-yearly = A = \(P \left(1 + \frac{r}{2 \times 100}\right)^{2t} \)

Simple Interest Formula = SI = \(\frac{P \times r \times t}{100}\)

Calculation:

Let the amount of money borrowed be ₹P.

Interest paid at the end of the year = \(\frac{P \times 4 \times 1}{100} = 0.04P\)

Interest received at the end of the year (compounded half-yearly) = \(P \left(1 + \frac{8}{2 \times 100}\right)^{2} - P\)

Interest received = \(P \left(1 + 0.04\right)^{2} - P\)

Interest received = \(P \left(1.04^{2}\right) - P \)

Interest received = \(P \left(1.0816\right) - P\)

Interest received = 1.0816P - P

Interest received = 0.0816P

Gain = Interest received - Interest paid

₹166.4 = 0.0816P - 0.04P

₹166.4 = 0.0416P

⇒ P = \(\frac{166.4}{0.0416}\)

⇒ P = 4000

The amount of money he borrows is ₹4,000.

Given:

Selling Price of 1^st item = ₹500

Profit on 1^st item = 25%

Loss on 2^nd item = 10%

Overall profit/loss = 0 (no profit, no loss)

Formula used:

Cost Price (CP) = \( \frac{SP}{1 + \frac{Profit%}{100}} \) (for profit)

CP = \( \frac{SP}{1 - \frac{Loss%}{100}} \) (for loss)

Total CP (both items) = Total SP (since no profit, no loss)

Calculation:

CP of 1^st item = \( \frac{500}{1 + \frac{25}{100}} = \frac{500}{1.25} = ₹400 \)

Let CP of 2^nd item = x

Loss = 10% ⇒ SP = \( x - \frac{10x}{100} = 0.9x \)

Total SP = ₹500 + 0.9x

Total CP = ₹400 + x

Given: Total SP = Total CP

⇒ 500 + 0.9x = 400 + x

⇒ 100 = 0.1x

⇒ x = ₹1,000

∴ The second item cost him ₹1,000.

Given:

The numbers are 275 and 146.

When divided by the required number, the remainders are 5 and 6 respectively.

Formula Used:

Required number = HCF of (Number1 - Remainder1, Number2 - Remainder2)

Calculation:

Number1 - Remainder1 = 275 - 5 = 270

Number2 - Remainder2 = 146 - 6 = 140

We need to find the HCF of 270 and 140.

Prime factorization of 270:

270 = 2 × 3³ × 5

Prime factorization of 140:

140 = 2 × 2 × 5 × 7 = 2² × 5 × 7

Common factors: 2 and 5

HCF = 2 × 5 = 10

The correct answer is option 2.

Given:

Marked selling price for 12% profit = ₹152

New selling price = ₹125

Formula Used:

Cost Price (CP) = Marked Selling Price / (1 + Profit Percentage)

Profit or Loss Percentage = ((New Selling Price - Cost Price) / Cost Price) × 100

Calculation:

To find the Cost Price (CP):

Profit Percentage = 12%

Marked Selling Price (SP) = ₹152

CP = SP / (1 + Profit Percentage)

CP = 152 / (1 + 0.12)

CP = 152 / 1.12

CP = 135.71

To find the Profit or Loss Percentage:

New Selling Price = ₹125

Profit or Loss Percentage = ((125 - 135.71) / 135.71) × 100

Profit or Loss Percentage = (-10.71 / 135.71) × 100

Profit or Loss Percentage = -7.89%

The correct answer is option 2.

Shortcut Trick

When Profit% at SP₁ equals Loss% at SP₂, the Cost Price (CP) is the simple average of the two selling prices.

CP = (1800 + 1160) ÷ 2 = 2960 ÷ 2 = ₹ 1,480.

To earn 20% profit, New SP = CP × 1.2.

New SP = 1480 × 1.2 = ₹ 1,776.

∴ The correct answer is ₹ 1,776.

Alternate Method

Given: SP₁ = ₹ 1,800, SP₂ = ₹ 1,160, Profit% = Loss%, Target Profit = 20%.

Formulas: Profit = SP − CP | Loss = CP − SP | SP = CP × (100 + P%) ÷ 100

⇒ Let CP be x. Since Profit% = Loss%, then (1800 − x) ÷ x × 100 = (x − 1160) ÷ x × 100

⇒ 1800 − x = x − 1160

⇒ 2x = 2960 ⇒ x = 1480

⇒ Required SP = 1480 × (120 ÷ 100) = 148 × 12

⇒ Required SP = ₹ 1,776

∴ The correct answer is ₹ 1,776.

Additional Information

Equal Profit and Loss Percentage

If the numerical value of profit percentage and loss percentage is the same for two different selling prices, the Cost Price is always the Arithmetic Mean of the two selling prices.

Profit Calculation

To find the final selling price for a specific profit percentage, use the multiplier method: SP = CP × (1 + P/100).

Given:

Numbers: 78, 104, 338 and 136, 357

Formula Used:

HCF (Highest Common Factor)

Calculation:

To find HCF of 78, 104, 338:

Prime Factorization:

78 = 2 × 3 × 13

104 = 2³ × 13

338 = 2 × 13 × 13

Common factors are 2 and 13

HCF = 2 × 13 = 26

To find HCF of 136, 357:

Prime Factorization:

136 = 2³ × 17

357 = 3 × 7 × 17

Common factor is 17

HCF = 17

A + B = 26 + 17

⇒ A + B = 43

The correct answer is option 1.

Given:

The price of tea has decreased by 15%.

Formula Used:

Percentage increase in consumption = \(\frac{\text{Percentage decrease in price}}{100 - \text{Percentage decrease in price}} \times 100\)

Calculation:

Percentage decrease in price = 15%

Remaining price = 100 - 15 = 85%

Percentage increase in consumption = \(\frac{15}{85} \times 100\)

⇒ \(\frac{15}{85} \times 100\)

⇒ \(\frac{1500}{85}\)

⇒ 17.647%

⇒ \(17 \frac{11}{17} \) %

The correct answer is option 2.

Given:

Length of a diagonal = 30 m

Length of the perpendiculars from two vertices to the diagonal = 15 m and 25 m

Formula Used:

Area of a quadrilateral = 12×Diagonal×(Sum of lengths of perpendiculars)12×Diagonal×(Sum of lengths of perpendiculars)

Calculation:

Diagonal (d) = 30 m

Sum of lengths of perpendiculars = 15 m + 25 m

Sum of lengths of perpendiculars = 40 m

Area of the quadrilateral

⇒ Area = 12×30×4012×30×40

⇒ Area = 15 × 40

⇒ Area = 600 m²

The area of the quadrilateral is 600 m².

Given:

a ∶ b = 9 ∶ 14

b ∶ c = 16 ∶ 7

Formula Used:

To find a ∶ b ∶ c, we need to find a common value for b in both ratios and then combine them.

Calculation:

Given a ∶ b = 9 ∶ 14

Given b ∶ c = 16 ∶ 7

To make the 'b' term common in both ratios, we need to find the least common multiple (LCM) of 14 and 16.

LCM of 14 and 16 = 112

Convert the ratios to have the same 'b' value:

a ∶ b = 9 ∶ 14

Multiply both terms by 8:

a ∶ b = 9 × 8 ∶ 14 × 8

a ∶ b = 72 ∶ 112

b ∶ c = 16 ∶ 7

Multiply both terms by 7:

b ∶ c = 16 × 7 ∶ 7 × 7

b ∶ c = 112 ∶ 49

Now, combine the ratios:

a ∶ b ∶ c = 72 ∶ 112 ∶ 49

The correct answer is option 1.

Given:

Radius (r) : Height (h) = 5 : 18

Volume of the cone (V) = 1,61,700 cm³

\(\pi = \frac{22}{7}\)

Formula Used:

Volume of the cone, V = \( \frac{1}{3} \pi r^2 h\)

Slant height of the cone, l = \(\sqrt{r^2 + h^2}\)

Calculation:

Let the radius be 5x and the height be 18x.

Volume of the cone, V = \(\frac{1}{3} \pi r^2 h\)

1,61,700 = \(\frac{1}{3} × \frac{22}{7} × (5x)^2 × 18x\)

1,61,700 = \(\frac{1}{3} × \frac{22}{7} × 25x^2 × 18x\)

1,61,700 = \(\frac{1}{3} × \frac{22}{7} × 450x^3\)

\(1,61,700 × 3 = \frac{22 × 450x^3}{7}\)

4,85,100 ×  7 = 22 × 450x³

33,95,700 = 9900x3

x3 = \(\frac{33,95,700}{9900}\)

x3 = 343

x = \(\sqrt[3]{343}\)

x = 7

So, radius (r) = 5x = 5 × 7 = 35 cm

Height (h) = 18x = 18 × 7 = 126 cm

Slant height (l), \(l = \sqrt{r^2 + h^2}\)

l = \(\sqrt{35^2 + 126^2} \)

l = \(\sqrt{1225 + 15876}\)

l = \(\sqrt{17101} \)

The correct answer is option 3.

Given:

Principal (P) = ₹800

Amount (A) = ₹926.10

Rate (r) = 10% per annum compounded half-yearly ⇒ 5% per half-year

Formula used:

A = P\((1+\frac{r}{100})^n\)

Where, n = number of half-years

Calculation:

926.10 = 800\((1+\frac{5}{100})^n\)

⇒ \(\frac{926.10}{800}= (1+\frac{5}{100})^n\)

⇒ \(1.157625 = (1.05)^n\)

Check powers of 1.05:

1.05¹ = 1.05

1.05² = 1.1025

1.05³ = 1.157625

⇒ n = 3 half-years

Time (t) = \(\frac{3}{2} \) = \(1 \frac{1}{2}\) years

∴ The correct answer is \(1 \frac{1}{2}\) years

Given:

Nitish marks up his goods by 20% over the cost price but gives a 15% discount on the marked price. He sells the item for ₹1,020.

Formula Used:

Selling Price after discount = Marked Price × (1 - Discount%)

Marked Price = Cost Price × (1 + Markup%)

Calculation:

Let the Cost Price be ₹C.

Marked Price = C × (1 + 20/100)

⇒ Marked Price = C × 1.2

Selling Price = Marked Price × (1 - 15/100)

⇒ Selling Price = C × 1.2 × 0.85

Given that Selling Price = ₹1,020,

⇒ 1,020 = C × 1.2 × 0.85

⇒ 1,020 = C × 1.02

⇒ C = 1,020 / 1.02

⇒ C = 1,000

The Cost Price is ₹1,000.

Given:

Average age of grandchildren = 12 years

Average age of parents = 40 years

Average age of grandparents = 68 years

Formula Used:

Family average age = (Sum of all ages) / (Total number of members)

Calculation:

Let the number of grandchildren be 2.

Total age of grandchildren = 2 × 12 = 24 years

Let the number of parents be 2.

Total age of parents = 2 × 40 = 80 years

Let the number of grandparents be 2.

Total age of grandparents = 2 × 68 = 136 years

Total age of all family members = 24 + 80 + 136

⇒ Total age of all family members = 240 years

Total number of family members = 2 (grandchildren) + 2 (parents) + 2 (grandparents)

⇒ Total number of family members = 6

Family average age = 240 / 6

⇒ Family average age = 40 years

The family average age is 40 years.

Given:

Speed₁ = 25 km/h ⇒ reaches 9 min late

Speed₂ = 30 km/h ⇒ reaches 9 min early

Time difference = 9 + 9 = 18 min = \(\frac{18}{60} = \frac{3}{10}\) hours

Formula used:

Time = Distance / Speed

Let distance = x km

Calculation:

\(\frac{x}{25} - \frac{x}{30} = \frac{3}{10}\)

⇒ \(\frac{6x - 5x}{150} = \frac{3}{10}\)

⇒ \(\frac{x}{150} = \frac{3}{10}\)

⇒ x = \(\frac{3}{10} × 150 = 45\)

∴ The correct answer is 45 km

Given:

Krisha alone takes 4 hrs more than together

Krupa alone takes 9 hrs more than together

Let time taken together = x hours

Then, Krisha alone takes = x + 4 hrs

Krupa alone takes = x + 9 hrs

Formula used:

Work done = 1 (complete work)

Work per hour = 1 / time

Calculation:

Work done in 1 hour by both = \( \frac{1}{x+4} + \frac{1}{x+9} = \frac{1}{x} \)

Take LCM:

⇒ \( \frac{(x+9)+(x+4)}{(x+4)(x+9)} = \frac{1}{x} \)

⇒ \( \frac{2x + 13}{(x+4)(x+9)} = \frac{1}{x} \)

Cross multiply:

⇒ \( x(2x + 13) = (x+4)(x+9) \)

⇒ \( 2x^2 + 13x = x^2 + 13x + 36 \)

⇒ \( 2x^2 + 13x - x^2 - 13x = 36 \)

⇒ \( x^2 = 36 \)

⇒ x = 6

∴ The correct answer is 6 hours

Given:

Simple Interest (SI) = ₹1260

Rate₁ = 5% per annum

Time₁ = 9 years

Time₂ = 5 years

Formula used:

SI = \(\frac{P × R × T}{100}\)

Calculation:

1260 = \(\frac{P × 5 × 9}{100}\)

⇒ 1260 = \(\frac{45P}{100}\)

⇒ P = \(\frac{1260 × 100}{45} = 2800\)

Now use P = ₹2800, SI = ₹1260, T = 5 years

1260 = \(\frac{2800 × R × 5}{100}\)

⇒ 1260 = \(\frac{14000R}{100}\)

⇒ 1260 = 140R

⇒ R = \(\frac{1260}{140} = 9\)

∴ The correct answer is 9% per annum

Given:

Number to find prime factors of: 2145

Formula Used:

The sum of all prime factors of a number.

Calculation:

Prime factorize 2145:

2145 ÷ 5 = 429

429 ÷ 3 = 143

143 ÷ 11 = 13

143 ÷ 13 = 11

Prime factors of 2145 are: 5, 3, 11, 13

Sum of prime factors = 5 + 3 + 11 + 13

Sum of prime factors = 32

The number greater than 32 from the options is 41.

The correct answer is option 1.