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Shortcut Trick

In a parallelogram, opposite angles are equal and adjacent angles are supplementary (sum is 180°).

Given angle = 70°

Opposite angle = 70°

Adjacent angle = 180° − 70° = 110°

Angle opposite to the adjacent angle = 110°

The other three angles are 70°, 110°, and 110°.

∴ The correct answer is 70°, 110°, 110°.

Alternate Method

Given:

One angle of the parallelogram = 70°

Formula used:

1. Sum of adjacent angles of a parallelogram = 180°
2. Opposite angles of a parallelogram are equal.

Let the parallelogram be ABCD where ∠A = 70°.

⇒ ∠A + ∠B = 180° (Adjacent angles are supplementary)

⇒ 70° + ∠B = 180°

⇒ ∠B = 180° − 70°

⇒ ∠B = 110°

Since opposite angles are equal:

⇒ ∠C = ∠A = 70°

⇒ ∠D = ∠B = 110°

∴ The other three angles are 110°, 70°, and 110° (which can be written as 70°, 110°, 110°).

∴ The correct answer is 70°, 110°, 110°.

Additional Information

Sum of Interior Angles

For any quadrilateral, including a parallelogram, the sum of all interior angles is always 360°.

Properties of Diagonals

The diagonals of a parallelogram bisect each other, although they are not necessarily equal in length unless it is a rectangle.

Consecutive Angles

Any pair of consecutive angles in a parallelogram will always sum to 180° because they are interior angles on the same side of a transversal cutting parallel lines.

Shortcut Trick

The Effective Rate of interest for 2 years at 8% compounded annually = 8 + 8 + (8 × 8)/100 = 16.64%.

Total Compound Interest = 16.64% of ₹5,000.

⇒ CI = 5000 × (16.64 / 100) = 50 × 16.64 = ₹832.

∴ The correct answer is ₹832.

Alternate Method

Given:

Principal (P) = ₹5,000

Rate (R) = 8% per annum

Time (T) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)^T

Compound Interest (CI) = A − P

Calculations:

⇒ A = 5000 × (1 + 8/100)²

⇒ A = 5000 × (108/100) × (108/100)

⇒ A = 5000 × 1.08 × 1.08

⇒ A = 5000 × 1.1664

⇒ A = ₹5,832

⇒ CI = 5832 − 5000

⇒ CI = ₹832

∴ The correct answer is ₹832.

Additional Information

Difference between CI and SI

For a period of 2 years, the difference between Compound Interest and Simple Interest is given by D = P × (R/100)².

Effective Percentage Method

The net change for two successive percentage increases of a% and b% is calculated as [a + b + (ab/100)]%.

Compounding Periods

If interest is compounded half-yearly, the rate is halved (R/2) and the time is doubled (2n) in the standard formula.

Shortcut Trick

Net Multiplier = (1 + 20%) × (1 + 25%)

⇒ Multiplier = 1.20 × 1.25 = 1.5

Shopkeeper Cost Price = B pays ÷ Net Multiplier

⇒ CP = 3,000 ÷ 1.5 = 2,000

∴ The correct answer is ₹2,000.

Alternate Method

Given:

Amount paid by B = ₹3,000

Profit made by A = 25%

Profit made by shopkeeper = 20%

Calculations:

Let the cost price for the shopkeeper be x.

The shopkeeper sells it to A at 20% profit:

⇒ Cost price for A = x × (120 ÷ 100) = 1.2x

A sells it to B at 25% profit:

⇒ Cost price for B = (1.2x) × (125 ÷ 100)

⇒ 3,000 = 1.2x × 1.25

⇒ 3,000 = 1.5x

⇒ x = 3,000 ÷ 1.5

⇒ x = 2,000

∴ The correct answer is ₹2,000.

Additional Information

Successive Profit Percentage

When there are two successive profits of a% and b%, the effective profit percentage is given by [a + b + (a × b ÷ 100)]%.

Cost Price and Selling Price Relation

Selling Price (SP) = CP × (100 + Profit%) ÷ 100. Conversely, Cost Price (CP) = SP × 100 ÷ (100 + Profit%).

Chain Rule in Profit and Loss

In a chain of transactions, the final price = Initial CP × (1 + r₁) × (1 + r₂) × ... × (1 + r_n), where r is the profit/loss fraction.

Shortcut Trick

To find the least beneficial scheme, we calculate the effective discount percentage for each case when buying 9 toys.

Scheme A: 27% + 39% − (27 × 39 / 100) ≈ 55.47%

Scheme B: (39% + 7% − 2.73%) then with 40% → effective ≈ 65.96%

Scheme C: Flat 26% for the first 9 toys → 26%

Scheme D: 7 free on 9 items → (7 / 16) × 100 ≈ 43.75%

Comparing all, Scheme C provides the lowest discount (26%).

∴ The correct answer is Scheme C.

Alternate Method

Given: Number of toys to be bought = 9. We evaluate the discount % for each scheme.

Formula Used:

1. Successive Discounts (x, y): d_eff = x + y − (xy/100)
2. Buy X Get Y Free: d_eff = [Y / (X + Y)] × 100

Calculations:

Scheme A: Successive 27% and 39%

⇒ d = 27 + 39 − (27 × 39 / 100) = 66 − 10.53 = 55.47%

Scheme B: Successive 39%, 7%, and 40%

⇒ Remainder factor = (1 − 0.39) × (1 − 0.07) × (1 − 0.40)

⇒ 0.61 × 0.93 × 0.60 = 0.34038

⇒ Effective Discount = (1 − 0.34038) × 100 = 65.96%

Scheme C: 26% discount on first 9 toys

⇒ Since customer buys exactly 9 toys, the discount is a flat 26%.

Scheme D: 7 toys free on buying 9 toys

⇒ Total items = 9 + 7 = 16 | Free items = 7

⇒ Discount % = (7 / 16) × 100 = 43.75%

The least beneficial scheme for the customer is the one with the minimum discount percentage.

⇒ 26% (C) < 43.75% (D) < 55.47% (A) < 65.96% (B)

∴ The correct answer is Scheme C.

Additional Information

Successive Discount Rule

For multiple discounts, the final price is P × (1−d₁/100) × (1−d₂/100) ... Total discount = 100 − [100 × (Product of remainders)].

Buy X Get Y Free

Effective Discount % = [Free Items / (Purchased Items + Free Items)] × 100.

Comparison of Schemes

The scheme most beneficial to the customer is the one with the highest discount, while the scheme most beneficial to the shopkeeper is the one with the lowest discount.

Shortcut Trick

Total Distance = 200 + 300 = 500 m

Relative Speed (Opposite) = 500 ÷ 20 = 25 m/s

Relative Speed (Same) = 500 ÷ 48 = 10.417 m/s

Faster Speed = (25 + 10.417) ÷ 2 = 17.708 m/s = 17.708 × 18/5 km/h

Faster Speed = 63.75 km/h

∴ The correct answer is 63.75 km/h.

Alternate Method

Given:

⇒ Length of Train P (L₁) = 200 m

⇒ Length of Train Q (L₂) = 300 m

⇒ Time taken in same direction (T_same) = 48 sec

⇒ Time taken in opposite direction (T_opp) = 20 sec

Formula Used:

⇒ Relative Speed (Opposite) = (L₁ + L₂) ÷ T_opp = S_f + S_s

⇒ Relative Speed (Same) = (L₁ + L₂) ÷ T_same = S_f − S_s

Calculations:

⇒ Total distance to be covered in both cases = 200 + 300 = 500 m

⇒ Let Faster Train Speed = S_f and Slower Train Speed = S_s

⇒ S_f + S_s = 500 ÷ 20 = 25 m/s −−− (1)

⇒ S_f − S_s = 500 ÷ 48 = 125/12 m/s −−− (2)

⇒ Adding (1) and (2): 2 × S_f = 25 + 125/12

⇒ 2 × S_f = (300 + 125) ÷ 12 = 425 ÷ 12

⇒ S_f = 425 ÷ 24 m/s

⇒ To convert to km/h: S_f = (425 ÷ 24) × (18 ÷ 5)

⇒ S_f = 85 × (3 ÷ 4) = 255 ÷ 4 = 63.75 km/h

∴ The correct answer is 63.75 km/h.

Additional Information

Relative Speed Concept

When two objects move in the opposite direction, their relative speed is the sum (u + v). When they move in the same direction, it is the difference (u − v).

Unit Conversion

To convert speed from meters per second (m/s) to kilometers per hour (km/h), multiply by 18/5 or 3.6.

Crossing Distance

The total distance covered when two trains cross each other is always the sum of their lengths (L₁ + L₂), regardless of their direction.

Shortcut Trick

The given sequence 3, 7, 11, ..., 51 is an Arithmetic Progression (AP).

For any AP, the Arithmetic Mean is simply the average of the first term and the last term.

First term (a) = 3, Last term (l) = 51

Arithmetic Mean = (a + l) ÷ 2 = (3 + 51) ÷ 2 = 54 ÷ 2 = 27.

∴ The correct answer is 27.

Alternate Method

Given: First term (a) = 3, Common difference (d) = 7 − 3 = 4, Last term (l) = 51

Formula Used: n = [(l − a) ÷ d] + 1; Sum (S_n) = (n ÷ 2) × (a + l); Mean = Sum ÷ n

⇒ Number of terms (n): 51 = 3 + (n − 1) × 4

⇒ 48 = (n − 1) × 4 ⇒ 12 = n − 1 ⇒ n = 13

⇒ Sum (S_n): (13 ÷ 2) × (3 + 51) = 6.5 × 54 = 351

⇒ Arithmetic Mean: 351 ÷ 13 = 27

∴ The correct answer is 27.

Additional Information

Properties of Arithmetic Progression (AP)

In an AP, the arithmetic mean is always equal to the median of the series. For a symmetric distribution or linear sequence, Mean = Median = (First + Last) ÷ 2.

General Term of an AP

The n^th term is calculated as a_n = a + (n − 1)d, where 'a' is the first term and 'd' is the common difference.

Sum of n terms

The sum of the first n terms of an AP can also be expressed as S_n = (n ÷ 2) × [2a + (n − 1)d].

Shortcut Trick

LCM of 54, 81, and 108 = 324 units (Total Capacity).

Efficiency of C = 6, D = 4, and Net (C+D−E) = 3.

Efficiency of Drain E = (6 + 4) − 3 = 7 units/min.

Time for E = 324 ÷ 7 = 46.2857... ≈ 46.29 min.

∴ The correct answer is 46.29 min.

Alternate Method

Given: Filling time for Tap C = 54 min, Tap D = 81 min. Combined filling time with Drain E = 108 min.

Formula: Net Efficiency = Efficiency of Filling Taps − Efficiency of Emptying Taps

⇒ Efficiency of Drain E = Efficiency(C) + Efficiency(D) − Efficiency(C + D − E)

⇒ Efficiency of Drain E = 6 + 4 − 3 = 7 units/min.

⇒ Time required by Drain E alone = Total Work ÷ Efficiency of E

⇒ Time = 324 ÷ 7

⇒ Time ≈ 46.29 min.

∴ The correct answer is 46.29 min.

Additional Information

Pipes and Cisterns Principle

If a pipe fills a tank in 'x' hours and another empties it in 'y' hours, the net work in 1 hour is (1/x − 1/y).

Efficiency and Time

Efficiency is inversely proportional to the time taken to complete the work (E ∝ 1/T).

Negative Work

An outlet pipe (drain) performs negative work because it removes water from the cistern.

Shortcut Trick

Net price after 20% discount = 24,000 × 80% = ₹19,200

Final Selling Price after additional ₹800 reduction = 19,200 − 800 = ₹18,400

∴ The correct answer is ₹18,400.

Alternate Method

Given:

Marked Price (MP) of the washing machine = ₹24,000

Festival Discount percentage = 20%

Additional cash discount = ₹800

Formula Used:

Selling Price (SP) = [MP × (100 − Discount%)/100] − Additional Discount

Calculations:

⇒ Price after 20% discount = 24,000 − (20% of 24,000)

⇒ Price after 20% discount = 24,000 − (20/100 × 24,000)

⇒ Price after 20% discount = 24,000 − 4,800 = ₹19,200

⇒ Final Selling Price = 19,200 − 800

⇒ Final Selling Price = ₹18,400

∴ The correct answer is ₹18,400.

Additional Information

Marked Price (MP)

The price printed on the label of an item before any discount is applied. It is also known as List Price or Tag Price.

Net Discount Formula

For two successive percentage discounts a% and b%, the single equivalent discount = [a + b − (a × b)/100]%.

Selling Price (SP) Calculation

SP = MP − Discount. If profit is given, SP = CP + Profit. If loss is given, SP = CP − Loss.

Shortcut Trick

Use the Algebraic Identity: (Difference)² = (Sum)² − 4 × (Product)

⇒ (Difference)² = (44)² − 4 × 228

⇒ (Difference)² = 1936 − 912 = 1024

⇒ Difference = √1024 = 32

∴ The correct answer is 32.

Alternate Method

Given:

Sum of two positive numbers (x + y) = 44

Product of two positive numbers (xy) = 228

Formula Used:

(x − y)² = (x + y)² − 4xy

Calculations:

⇒ (x − y)² = (44)² − 4 × 228

⇒ (x − y)² = 1936 − 912

⇒ (x − y)² = 1024

⇒ x − y = √1024

⇒ x − y = 32

∴ The correct answer is 32.

Additional Information

Fundamental Algebraic Identities

1. (x + y)² = x² + y² + 2xy

2. (x − y)² = x² + y² − 2xy

Relationship between Sum, Difference, and Product

(x + y)² − (x − y)² = 4xy. This identity is extremely useful for finding the Difference when the Sum and Product are known.

Quadratic Roots Property

If x and y are roots of a quadratic equation t² − (Sum)t + (Product) = 0, then the Difference between the roots is √D, where D = (Sum)² − 4(Product).

Shortcut Trick

Total Height (Building + Tower) = Distance × tan(Angle to Top)

Total Height = 70 × tan(60°) = 70√3 m

Tower Height = Total Height − Building Height

Tower Height = 70√3 − 30 m

∴ The correct answer is 70√3 − 30 m.

Alternate Method

Given:

Height of the building (BT) = 30 m

Distance from the point (P) to the building (B) = 70 m

Angle of elevation to top of tower (C) = 60°

Angle of elevation to bottom of tower (T) = 30°

Formula Used:

In a right-angled triangle, tan θ = Perpendicular / Base

Calculations:

Let the height of the tower be h.

In right-angled triangle ΔPBC (Base to Top of Tower):

⇒ tan 60° = BC / PB

⇒ √3 = (BT + TC) / 70

⇒ √3 = (30 + h) / 70

⇒ 30 + h = 70 × √3

⇒ h = 70√3 − 30

∴ The correct answer is 70√3 − 30 m.

Additional Information

Angle of Elevation

The angle formed between the horizontal line and the line of sight when an observer looks at an object above the horizontal level.

Trigonometric Ratios

Standard values used in height and distance problems: tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3.

Total Height Relation

For a structure placed on a base, the total height from the ground is the sum of the base height and the structure height (H = h_base + h_object).