Search Previous Year Questions

Shortcut Trick

Evaluate the square roots directly: √6400 = 80 and √2401 = 49

Subtract the values inside the bracket: 80 − 49 = 31

Multiply the result by 1/4: 31 × 1/4 = 7.75

∴ The correct answer is 7.75.

Alternate Method

Given:

(√6400 − √2401) × (1/4)

Concept Used:

BODMAS Rule (Brackets, Orders, Division/Multiplication, Addition/Subtraction)

Calculation:

⇒ First, evaluate the square roots present inside the bracket.

⇒ √6400 = 80 (since 80 × 80 = 6400)

⇒ √2401 = 49 (since 50² = 2500, so 49² = 2401)

⇒ Substitute these values back into the expression:

⇒ (80 − 49) × (1/4)

⇒ Solve the subtraction inside the bracket first according to BODMAS:

⇒ 31 × (1/4)

⇒ 31 ÷ 4

⇒ 7.75

∴ The correct answer is 7.75.

Additional Information

BODMAS Rule

Mathematical operations must always be solved in the exact sequence: Brackets, Orders (Powers/Roots), Division, Multiplication, Addition, and Subtraction.

Square Root Shortcut Trick

For finding the square root of a number like 2401, check the nearest base square. 50² = 2500. Since 2401 ends in 1 and is slightly less than 2500, the square root logically must be 49.

Division by 4 Shortcut

To divide any number by 4 quickly without long division, simply halve the number twice. Half of 31 is 15.5, and half of 15.5 is 7.75.

Shortcut Trick

Find square roots quickly using unit digits and base squares.

√5776: 70² = 4900, 80² = 6400. Unit digit 6 → ends in 4 or 6. Since 5776 > 5625 (75²), √5776 = 76.

√4761: 60² = 3600, 70² = 4900. Unit digit 1 → ends in 1 or 9. Since 4761 is close to 4900, √4761 = 69.

Expression = (76 − 69) × (1/2) = 7/2 = 3.5.

∴ The correct answer is 3.5.

Alternate Method

Given:

(√5776 − √4761) × (1/2)

Method Used:

Find the exact square roots of the given perfect squares, then apply the BODMAS rule for simplification.

Calculations:

⇒ √5776 = 76

⇒ √4761 = 69

⇒ Substitute these values back into the main expression:

⇒ (76 − 69) × (1/2)

⇒ 7 × (1/2)

⇒ 3.5

∴ The correct answer is 3.5.

Additional Information

BODMAS Rule

Always follow the order: Brackets, Orders (powers/roots), Division, Multiplication, Addition, Subtraction.

Square Root Shortcut

To find √N, identify the base tens (like 60², 70²) it lies between, check the unit digit, and compare N with the square of the middle number (ends in 5).

Squares Ending in 5

To square a number ending in 5 (like N5), multiply N by (N+1) and append 25. Example: 75² → 7 × 8 = 56 → 5625.

Shortcut Trick

Convert all given values into simple fractions: 40% = 2/5, 0.5 = 1/2, and 1/4 remains 1/4.

The given equation becomes: (2/5)A = (1/2)B = (1/4)C.

To find the ratio A : B : C directly, take the reciprocal of their coefficients.

A : B : C = 5/2 : 2/1 : 4/1.

Multiply the entire ratio by 2 to eliminate the denominator: 5 : 4 : 8.

∴ The correct answer is 5 : 4 : 8.

Alternate Method

Given:

40% of A = 0.5 of B = 1/4 of C

Formula Used:

If p × A = q × B = r × C = k, then A = k/p, B = k/q, and C = k/r

A 5 units

B 4 units

C 8 units

Ratio A : B : C = 5 : 4 : 8

Calculations:

⇒ 40% = 40/100 = 2/5

⇒ 0.5 = 5/10 = 1/2

⇒ (2/5) × A = (1/2) × B = (1/4) × C

⇒ Let (2/5) × A = (1/2) × B = (1/4) × C = k

⇒ A = 5k/2

⇒ B = 2k

⇒ C = 4k

⇒ Ratio A : B : C = 5k/2 : 2k : 4k

⇒ Multiply all terms by 2 to clear fractions:

⇒ A : B : C = 5 : 4 : 8

∴ The correct answer is 5 : 4 : 8.

Additional Information

Equivalent Ratio

Multiplying or dividing each term of a ratio by the same non-zero number does not change the value of the ratio. For example, a : b = 2a : 2b.

Compound Ratio

The ratio compounded of two ratios a : b and c : d is the ratio of the products of the antecedents and consequents, i.e., ac : bd.

Invertendo Property

If two ratios are equal, i.e., a : b = c : d, then their inverse ratios are also equal, making b : a = d : c.

Shortcut Trick

Sum of ratio parts in container 1 = 2 + 3 = 5 units.

Sum of ratio parts in container 2 = 3 + 2 = 5 units.

Since capacities are equal and sum of parts is the same, we can directly add the respective parts.

Total Milk = 2 + 3 = 5 units and Total Water = 3 + 2 = 5 units.

Ratio of Milk to Water in the final mixture = 5 : 5 = 1 : 1.

∴ The correct answer is 1:1.

Alternate Method

Given:

Ratio of milk and water in 1^st container = 2 : 3

Ratio of milk and water in 2^nd container = 3 : 2

Both containers have equal capacities.

 Container 1 Milk = 2x Water = 3x +     Container 2 Milk = 3x Water = 2x =     Final Mixture Milk = 5x Water = 5x  Resulting Ratio = 5x : 5x ⇒ 1 : 1

Calculations:

⇒ Let the uniform capacity of each container be V units.

⇒ Quantity of milk in the 1^st container = 2V ÷ 5

⇒ Quantity of water in the 1^st container = 3V ÷ 5

⇒ Quantity of milk in the 2^nd container = 3V ÷ 5

⇒ Quantity of water in the 2^nd container = 2V ÷ 5

⇒ Total quantity of milk when combined = (2V ÷ 5) + (3V ÷ 5) = 5V ÷ 5 = V

⇒ Total quantity of water when combined = (3V ÷ 5) + (2V ÷ 5) = 5V ÷ 5 = V

⇒ Resulting ratio of milk to water = V : V = 1 : 1

∴ The correct answer is 1:1.

Additional Information

Unequal Capacities

If the containers are of different volumes, multiply the ratios by their respective overall capacities to find the actual proportional quantity of each component before combining them.

Replacement Concept

When a specific volume is removed from a mixture, the removed portion and the remaining mixture both retain the exact same ratio of ingredients as the original mixture.

Rule of Alligation

When two mixtures with varying concentrations are blended to form a third required mixture, the necessary proportion of each initial mixture can be quickly determined using the cross-difference of their concentration fractions.

Shortcut Trick

Selling Price after 1^st discount = 800 × (100 − 10)% = ₹720

Second discount amount = 720 − 612 = ₹108

Second discount % = (108 ÷ 720) × 100 = 15%

∴ The correct answer is 15%.

Alternate Method

Given:

Marked Price (MP) = ₹800

Final Selling Price (SP) = ₹612

First discount (d₁) = 10%

MP = ₹800

−10%

₹720

−d%

SP = ₹612 Discount = ₹720 − ₹612 = ₹108

Formula Used:

SP = MP × (1 − d₁/100) × (1 − d₂/100)

Calculations:

⇒ Price after first discount = 800 − (10% of 800)

⇒ Price after first discount = 800 − 80 = ₹720

⇒ Second discount amount = 720 − 612 = ₹108

⇒ Second discount % = (Discount ÷ Price after 1^st discount) × 100

⇒ d₂ = (108 ÷ 720) × 100

⇒ d₂ = 15%

∴ The correct answer is 15%.

Additional Information

Single Equivalent Discount

For two successive discounts of a% and b%, the single equivalent discount is given by (a + b − ab/100)%.

Marked Price vs Selling Price

Discount is always calculated on the Marked Price. Hence, Selling Price = Marked Price − Discount.

Successive Discounts Formula

Final SP = MP × (1 − d₁/100) × (1 − d₂/100) × ... × (1 − d_n/100).

Shortcut Trick

Let the Cost Price be 100 units.

Marked Price = 160 units (60% markup).

Selling Price = 160 × 3/4 × 9/10 = 108 units.

Profit = 108 − 100 = 8 units.

8 units = ₹160 ⇒ 1 unit = ₹20.

∴ The correct answer is ₹2000.

Alternate Method

CP = 100x

MP = 160x

SP = 108x

+60%

-25%, -10%

Profit = 8x = ₹160

Given:

Markup % = 60%

Successive discounts = 25% and 10%

Profit = ₹160

Formula Used:

Selling Price (SP) = Marked Price × (100 − D₁)/100 × (100 − D₂)/100

Profit = SP − CP

Calculations:

Let the Cost Price (CP) be 100x.

⇒ Marked Price (MP) = 100x + 60% of 100x = 160x

⇒ SP = 160x × (100 − 25)/100 × (100 − 10)/100

⇒ SP = 160x × 75/100 × 90/100

⇒ SP = 160x × 3/4 × 9/10 = 108x

⇒ Profit = SP − CP = 108x − 100x = 8x

Given that the profit earned is ₹160:

⇒ 8x = 160

⇒ x = 20

⇒ Cost Price (CP) = 100x = 100 × 20 = 2000

∴ The correct answer is ₹2000.

Additional Information

Successive Discounts

When two successive discounts of a% and b% are given, the equivalent single discount percentage is calculated as (a + b − a×b/100)%.

Marked Price and Markup Percentage

The Marked Price (MP) is always marked up above the Cost Price (CP). Markup % = (MP − CP) ÷ CP × 100.

Selling Price Relation

Selling Price (SP) acts as the bridge connecting cost and markings. It can be found from CP as SP = CP × (100 + Profit%)/100, and from MP as SP = MP × (100 − Discount%)/100.

Shortcut Trick

If x + a/x = k, then x³ + a³/x³ = k³ − 3ak.

Here, a = 2 and k = 7.

x³ + 8/x³ = 7³ − 3 × 2 × 7

x³ + 8/x³ = 343 − 42 = 301

∴ The correct answer is 301.

Alternate Method

Given:

x + 2/x = 7

Formula Used:

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculations:

⇒ Cube both sides of the given equation:

⇒ (x + 2/x)³ = 7³

⇒ x³ + (2/x)³ + 3 × x × (2/x) × (x + 2/x) = 343

⇒ x³ + 8/x³ + 6(x + 2/x) = 343

⇒ Substitute the value of (x + 2/x) = 7:

⇒ x³ + 8/x³ + 6 × 7 = 343

⇒ x³ + 8/x³ + 42 = 343

⇒ x³ + 8/x³ = 343 − 42

⇒ x³ + 8/x³ = 301

∴ The correct answer is 301.

Additional Information

Cube of sum with reciprocal

If x + 1/x = k, then x³ + 1/x³ = k³ − 3k.

Cube of difference with reciprocal

If x − 1/x = k, then x³ − 1/x³ = k³ + 3k.

Square with reciprocal

If x + 1/x = k, then x² + 1/x² = k² − 2.

Shortcut Trick

Total annual interest = 9% of ₹6,00,000 = ₹54,000.

Sum of the first and third scholarships = ₹18,000 + ₹22,000 = ₹40,000.

Second scholarship = Total interest − Sum of first and third scholarships.

Second scholarship = 54,000 − 40,000 = ₹14,000.

∴ The correct answer is ₹14,000.

Alternate Method

Given:

Educational Fund Principal (P) = ₹6,00,000

Rate of Interest (R) = 9% per annum

Time (T) = 1 year

First scholarship = ₹18,000

Third scholarship = ₹22,000

Formula Used:

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

Fund = ₹6,00,000

9% Interest = ₹54,000

1^st: ₹18k 2^nd: ? 3^rd: ₹22k

Second = 54k - (18k + 22k) = ₹14,000

Calculation:

⇒ Total annual interest = (6,00,000 × 9 × 1) ÷ 100

⇒ Total annual interest = 54,000

⇒ The sum of all three scholarships equals the total annual interest earned.

⇒ First scholarship + Second scholarship + Third scholarship = ₹54,000

⇒ 18,000 + Second scholarship + 22,000 = 54,000

⇒ Second scholarship + 40,000 = 54,000

⇒ Second scholarship = 54,000 − 40,000

⇒ Second scholarship = ₹14,000

∴ The correct answer is ₹14,000.

Additional Information

Simple Interest (SI) Formula

Simple Interest is strictly calculated on the initial principal amount throughout the entire period. Formula: I = (P × R × T) ÷ 100.

Concept of Compound Interest (CI)

Compound Interest is calculated on the principal as well as the accumulated interest from previous periods. Formula for amount: A = P(1 + R/100)^T.

SI vs CI Difference

For the first year, both Simple Interest and Compound Interest remain the same. The difference for 2 years can be directly found using: Difference = P(R ÷ 100)².

Shortcut Trick

Since the rate (R = 5%) and time (T = 2 years) are constant, the ratio of Interest to Principal remains the same in both investments.

Interest ÷ Principal = (R × T) ÷ 100 = (5 × 2) ÷ 100 = 1 ÷ 10

For Case 1: Y ÷ X = 1 ÷ 10

For Case 2: Z ÷ Y = 1 ÷ 10

Equating both ratios: Y ÷ X = Z ÷ Y ⇒ Y² = XZ

∴ The correct answer is Y² = XZ.

Alternate Method

Given:

Principal 1 = ₹X, Interest 1 = ₹Y

Principal 2 = ₹Y, Interest 2 = ₹Z

Rate (R) = 5%, Time (T) = 2 years

Formula Used:

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

₹ X Principal 1

₹ Y Interest 1 / Prin. 2

₹ Z Interest 2

R=5%, T=2

R=5%, T=2 Interest to Principal Ratio is Constant Y ÷ X = Z ÷ Y

Calculations:

⇒ For the first investment, substituting values in the formula:

⇒ Y = (X × 5 × 2) ÷ 100

⇒ Y = 10X ÷ 100 = X ÷ 10

⇒ X = 10Y

⇒ For the second investment:

⇒ Z = (Y × 5 × 2) ÷ 100

⇒ Z = 10Y ÷ 100 = Y ÷ 10

⇒ Y = 10Z

⇒ Let's find the ratio of Interest to Principal in both cases:

⇒ Y ÷ X = 1 ÷ 10

⇒ Z ÷ Y = 1 ÷ 10

⇒ Equating both ratios: Y ÷ X = Z ÷ Y

⇒ Cross multiplying, we get: Y² = XZ

∴ The correct answer is Y² = XZ.

Additional Information

Simple Interest (SI)

It is calculated only on the principal amount, keeping the interest constant every year. Formula: SI = (P × R × T) ÷ 100.

Amount in Simple Interest

The total amount (A) accumulated at the end of the time period is the sum of Principal and Simple Interest. Formula: A = P + [(P × R × T) ÷ 100].

Effective Interest Rate

For simple interest, the total effective percentage of interest accumulated over T years is directly given by (R × T)%. In this given question, it is 5% × 2 = 10%.

Shortcut Trick

For an object at height h, if angles of depression to two points on opposite sides are α and β:

Total distance = h × (cot α + cot β)

Here, h = 300 m, α = 45° and β = 30°

Distance = 300 × (cot 45° + cot 30°)

Distance = 300 × (1 + √3) = 300(√3 + 1) m

∴ The correct answer is 300(√3 + 1) m.

Alternate Method

A B C D

300 m Boat 1 Boat 2

45° 30° 45° 30°

Total Distance = ?

Given:

⇒ Height of airplane (AD) = 300 m

⇒ Angle of depression of Boat 1 (∠B) = 45°

⇒ Angle of depression of Boat 2 (∠C) = 30°

Formula Used:

⇒ tan θ = Perpendicular / Base

Calculations:

⇒ In right-angled ΔADB, tan 45° = AD / BD

⇒ 1 = 300 / BD

⇒ BD = 300 m

⇒ In right-angled ΔADC, tan 30° = AD / DC

⇒ 1 / √3 = 300 / DC

⇒ DC = 300√3 m

⇒ Total distance between the boats (BC) = BD + DC

⇒ BC = 300 + 300√3 = 300(√3 + 1) m

∴ The correct answer is 300(√3 + 1) m.

Additional Information

Same Side vs Opposite Sides

If two objects are located on the same side of the observer, the distance between them is h × |cot α − cot β|.

Angle of Elevation & Depression

The angle of elevation from an object to an observer is always mathematically equal to the angle of depression from the observer to the object due to alternate interior angles.