Search Previous Year Questions

Shortcut Trick

Price after 40% discount = 60% of 1200 = ₹720.

Final Sale Price = ₹520.

Additional Discount value = 720 − 520 = ₹200.

Additional Discount % = (200 ÷ 720) × 100 = (5 ÷ 18) × 100 = 250/9% = 27(7)/(9)%.

∴ The correct answer is 27(7)/(9)%.

Alternate Method

Given:

Marked Price (MP) = ₹1,200

First Discount = 40%

Final Sale Price (SP) = ₹520

Formula:

Price after discount = MP × (100 − Discount%) ÷ 100

Calculations:

⇒ Price after 40% discount = 1200 × (1 − 0.40)

⇒ Price after 40% discount = 1200 × 0.60 = ₹720

⇒ Total Additional Discount needed = 720 − 520 = ₹200

⇒ Additional Discount % = (Additional Discount ÷ Mid Price) × 100

⇒ Additional Discount % = (200 ÷ 720) × 100

⇒ Additional Discount % = (5 ÷ 18) × 100 = 500/18%

⇒ Additional Discount % = 250/9% = 27 7/9%

∴ The correct answer is 27(7)/(9)%.

Additional Information

Successive Discounts

When two discounts a% and b% are given, the equivalent single discount is calculated as: [a + b − (ab ÷ 100)]%.

Marked Price (MP)

The price printed on the product. Discounts are always calculated on the Marked Price unless stated otherwise.

Selling Price (SP)

The actual price at which an item is sold after all discounts. SP = MP − Discount.

Given:

\(\sqrt{\sqrt{\sqrt{256} + \sqrt{144} + (2\sqrt{2})^2} + \sqrt{841} \times (\sqrt{2})^2}\)

Formula used:

√x² = x

(a√b)² = a² × b

Calculations:

⇒ \(\sqrt{\sqrt{16 + 12 + 8} + 29 \times 2}\)

⇒ \(\sqrt{\sqrt{36} + 58}\)

⇒ \(\sqrt{6 + 58}\)

⇒ \(\sqrt{64}\)

⇒ 8

Now, checking the options:

A. (√2)² = 2

B. (2√2)² = 4 × 2 = 8

C. (4√2)² = 16 × 2 = 32

D. (3√2)² = 9 × 2 = 18

∴ The correct answer is option (2).

Shortcut Trick

Effective rate of Compound Interest for 2 years at 2% = 2 + 2 + (2 × 2)/100 = 4.04%.

Given Compound Interest (4.04%) = ₹404.

⇒ Principal (y) = (404 ÷ 4.04) × 100 = ₹10,000.

Total Amount = Principal + Interest = 10,000 + 404 = ₹10,404.

∴ The correct answer is ₹10,404.

Alternate Method

Given: Interest (CI) = ₹404, Rate (r) = 2% per annum, Time (n) = 2 years.

Formula Used: Amount (A) = P(1 + r/100)ⁿ and CI = A − P

⇒ CI = y[(1 + 2/100)² − 1]

⇒ 404 = y[(51/50)² − 1]

⇒ 404 = y[2601/2500 − 1]

⇒ 404 = y[101/2500]

⇒ y = (404 × 2500) ÷ 101

⇒ y = 4 × 2500 = 10,000

⇒ Total Amount (A) = Principal (y) + Interest

⇒ Total Amount (A) = 10,000 + 404 = 10,404

∴ The correct answer is ₹10,404.

Additional Information

Effective Percentage Rate

For two successive changes of a% and b%, the effective change is (a + b + ab/100)%.

CI vs SI Difference

The difference between Compound Interest and Simple Interest for 2 years is P(r/100)².

Amount Property

In Compound Interest, the Total Amount of the previous year becomes the Principal for the next year.

Shortcut Trick

A natural number is a positive integer starting from 1 (1, 2, 3, ...).

Evaluate the expression 2x + 1 sequentially:

For x = 1, 2, 3 → 2x + 1 = 3, 5, 7 (all are prime numbers)

For x = 4 → 2(4) + 1 = 9 (composite number, not prime)

∴ The correct answer is 4.

Alternate Method

Given:

An algebraic expression: 2x + 1

We need to find the smallest natural number x such that 2x + 1 is not a prime number.

Definition and Method:

A natural number belongs to the set {1, 2, 3, 4, ...}.

A prime number is a positive integer greater than 1 that has no positive divisors other than 1 and itself.

We substitute sequential natural numbers for x into the expression and check for primality.

Calculations:

⇒ For x = 1:

⇒ 2(1) + 1 = 3 (Prime number)

⇒ For x = 2:

⇒ 2(2) + 1 = 5 (Prime number)

⇒ For x = 3:

⇒ 2(3) + 1 = 7 (Prime number)

⇒ For x = 4:

⇒ 2(4) + 1 = 9 (Composite number, since 9 = 3 × 3)

Since 9 is divisible by 3, it is not a prime number. Hence, the smallest natural number satisfying the given condition is x = 4.

∴ The correct answer is 4.

Additional Information

Prime Numbers

A prime number is a positive integer greater than 1 that has exactly two distinct positive divisors: 1 and itself. Examples include 2, 3, 5, 7, 11, etc.

Composite Numbers

A composite number is a positive integer greater than 1 that has more than two distinct positive divisors. For example, 4, 6, 8, 9, 10 are composite numbers.

Natural Numbers

Natural numbers are the set of positive integers starting from 1, i.e., {1, 2, 3, 4, ...}. Note that 0 is not considered a natural number in this context.

Primality of 2x + 1

For any integer x, the expression 2x + 1 always represents an odd number. Thus, we are looking for the smallest odd composite number greater than 1, which is 9.

Shortcut Trick

When the sum (S) and difference (D) of two ages are known:

Son's age (Smaller value) = (Sum − Difference) ÷ 2

Son's age = (56 − 24) ÷ 2

Son's age = 32 ÷ 2 = 16 years

∴ The correct answer is 16 years.

Alternate Method

Given:

⇒ Sum of Father and Son's ages = 56 years

⇒ Difference between Father and Son's ages = 24 years

⇒ Let the age of the son be x years.

⇒ According to the question, the father is 24 years older than the son.

⇒ Father's age = (x + 24) years

⇒ Sum of their ages: x + (x + 24) = 56

⇒ 2x + 24 = 56

⇒ 2x = 56 − 24

⇒ 2x = 32

⇒ x = 32 ÷ 2

⇒ x = 16

∴ The correct answer is 16 years.

Additional Information

Linear Equations in Ages

Age problems are typically solved by forming a linear equation where the unknown age is represented by a variable like 'x'.

Constant Age Difference

The difference between the ages of two individuals never changes over time. If the father is 24 years older now, he will always be 24 years older.

Sum and Difference Rule

If (A + B) = S and (A − B) = D, then the larger value A = (S + D) ÷ 2 and the smaller value B = (S − D) ÷ 2.

Shortcut Trick

Error in total sum = 37 − 22 = 15 marks.

Given average marks increase = 5/4 = 1.25.

Number of students = Total Increase in Sum ÷ Increase in Average.

⇒ 15 ÷ 1.25 = (15 × 4) ÷ 5 = 12.

∴ The correct answer is 12.

Alternate Method

Given:

Incorrect entry = 37

Correct entry = 22

Average increase = 5/4

Formula used:

Average = Total Sum of Marks ÷ Number of Students

Total Increase in Sum = Number of Students × Increase in Average

Calculations:

⇒ Net effect on total sum = 37 − 22 = 15.

⇒ Let the total number of students be n.

⇒ (New Sum − Old Sum) ÷ n = 5/4

⇒ 15 ÷ n = 5/4

⇒ 5 × n = 15 × 4

⇒ 5 × n = 60

⇒ n = 60 ÷ 5 = 12.

∴ The correct answer is 12.

Additional Information

Average Property (Constant Change)

If every value in a data set is increased or decreased by a constant k, the average also increases or decreases by k.

Relationship between Sum and Average

The change in total sum is always equal to the number of terms × the change in average (i.e., ΔSum = n × ΔAvg).

Correction of Errors

Correct Average = [Incorrect Sum + (Correct Value − Incorrect Value)] ÷ Total Number of Items.

Shortcut Trick

Ratio of numbers = 19 : 7 : 13

Since 19, 7, and 13 are all prime numbers, they are co-prime to each other.

For co-prime ratios, LCM = (Product of Ratios) × HCF

8645 = (19 × 7 × 13) × HCF ⇒ 8645 = 1729 × HCF

HCF = 8645 ÷ 1729 = 5

∴ The correct answer is 5.

Alternate Method

Given:

Ratio of three numbers = 19 : 7 : 13

Least Common Multiple (LCM) = 8645

Formula Used:

Let the HCF be x. Then the numbers are ax, bx, and cx.

LCM of (ax, bx, cx) = x × LCM(a, b, c)

Calculations:

⇒ Let the three numbers be 19x, 7x, and 13x, where x is the HCF.

⇒ Since 19, 7, and 13 are prime numbers, their LCM is the product of these numbers and their common factor x.

⇒ LCM = 19 × 7 × 13 × x

⇒ 8645 = 1729 × x

⇒ x = 8645 ÷ 1729

⇒ x = 5

Therefore, the Highest Common Factor (HCF) of the numbers is 5, which corresponds to Option 4.

∴ The correct answer is 5.

Additional Information

Relationship between HCF and LCM

For any two numbers a and b, the product of the numbers is equal to the product of their HCF and LCM: a × b = HCF(a, b) × LCM(a, b).

HCF of Co-prime Numbers

The HCF of two or more co-prime numbers (numbers that have no common factor other than 1) is always 1.

LCM and HCF of Fractions

LCM of fractions = (LCM of Numerators) ÷ (HCF of Denominators). HCF of fractions = (HCF of Numerators) ÷ (LCM of Denominators).

Shortcut Trick

Use Option Verification to identify the number satisfying both conditions.

If the larger number is 9, then the smaller number is 9 − 8 = 1.

Check the second condition: (2 × 9) − (5 × 1) = 18 − 5 = 13.

The result matches the condition given in the problem.

∴ The correct answer is 9.

Alternate Method

Given:

Difference between numbers = 8

Result of (2 × Larger − 5 × Smaller) = 13

Formula Used:

Method of Linear Equations using Substitution.

Calculations:

⇒ Let the larger number be x and the smaller number be y.

⇒ From the first condition: x − y = 8 → y = x − 8

⇒ From the second condition: 2x − 5y = 13

⇒ Substitute the value of y in the second equation:

⇒ 2x − 5(x − 8) = 13

⇒ 2x − 5x + 40 = 13

⇒ −3x = 13 − 40

⇒ −3x = −27

⇒ x = −27 ÷ (−3) = 9

∴ The correct answer is 9.

Additional Information

Linear Equations in One Variable

Equations where the highest power of the variable is 1, generally expressed as ax + b = 0.

Substitution Method

A technique to solve simultaneous equations by expressing one variable in terms of another and substituting it into the other equation.

Verification Technique

Fastest method for objective exams where you plug option values into the given constraints to see which one satisfies all conditions.

Shortcut Trick

Equate the total work done in both cases: (15M + 19W) × 8 days = (9M + 4W) × 14 days.

Divide both sides by 2: (15M + 19W) × 4 = (9M + 4W) × 7.

Expand and group terms: 60M + 76W = 63M + 28W ⇒ 3M = 48W.

Thus, 1 man is equivalent to 16 women.

∴ The correct answer is 16.

Alternate Method

Given:

Case 1: (15 men + 19 women) can do the work in 8 days.

Case 2: (9 men + 4 women) can do the same work in 14 days.

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ 15M + 19W = 7 ... (i)

⇒ 9M + 4W = 4 ... (ii)

⇒ To find the ratio 1M/1W, use cross-multiplication or elimination.

⇒ From (ii), 4W = 4 − 9M.

⇒ Substitute into (i) by multiplying (i) by 4: 60M + 19(4W) = 28.

⇒ 60M + 19(4 − 9M) = 28.

⇒ 60M + 76 − 171M = 28.

⇒ 111M = 48 ⇒ M = 48/111 = 16/37.

⇒ Substituting M in 4W = 4 − 9M ⇒ 4W = 4 − 9(16/37) = (148 − 144)/37 = 4/37.

⇒ W = 1/37.

⇒ Ratio M/W = (16/37) ÷ (1/37) = 16.

⇒ 1 man = 16 women.

∴ The correct answer is 16.

Additional Information

Work-Efficiency Relation

Efficiency (E) is inversely proportional to Time (T). For the same work, E₁ × T₁ = E₂ × T₂.

Man-Day Concept

Total work is often measured in man-days, calculated as (Number of persons × Number of days taken).

Composite Efficiency

When different types of workers (men, women, children) work together, their individual efficiencies are summed to find the group's total efficiency.

Shortcut Trick

Net Profit Ratio = (True Weight ÷ Faulty Weight) × (Markup Factor) × (Discount Factor)

Net Ratio = (1000 ÷ 960) × (120 ÷ 100) × (90 ÷ 100)

Net Ratio = (25/24) × (6/5) × (9/10) = 45/40 = 9/8

Profit % = [(9 − 8) ÷ 8] × 100 = 1/8 × 100 = 12.5%

∴ The correct answer is 12.5% profit.

Alternate Method

Given:

Cost Price (CP) per kg = ₹48

Markup percentage = 20%

Discount percentage = 10%

Faulty weight used = 960 g instead of 1000 g

Calculations:

⇒ Marked Price (MP) for 1000 g = 48 + (20 ÷ 100) × 48 = 48 + 9.6 = ₹57.6

⇒ Selling Price (SP) for 1000 g = 57.6 − (10 ÷ 100) × 57.6 = 57.6 − 5.76 = ₹51.84

⇒ The shopkeeper delivers only 960 g, so his actual cost is for 960 g.

⇒ Actual CP for 960 g = (48 ÷ 1000) × 960 = 0.048 × 960 = ₹46.08

⇒ Total Profit = SP for 1000 g (received from customer) − Actual CP for 960 g

⇒ Total Profit = 51.84 − 46.08 = ₹5.76

⇒ Profit % = (Total Profit ÷ Actual CP) × 100

⇒ Profit % = (5.76 ÷ 46.08) × 100 = 1/8 × 100 = 12.5%

∴ The correct answer is 12.5% profit.

Additional Information

Dishonest Dealer Formula

Profit % = [ (True Weight − False Weight) ÷ False Weight ] × 100 when selling at cost price.

Successive Discounts/Markups

Effective change of two percentages x and y is given by the formula: [ x + y + (xy ÷ 100) ].

General Profit Relation

Selling Price (SP) = CP × (100 + Profit%) ÷ 100 = MP × (100 − Discount%) ÷ 100.