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Shortcut Trick

Use the direct formula for percentage less: [r ÷ (100 + r)] × 100

Given r = 20%

⇒ Required % = [20 ÷ (100 + 20)] × 100

⇒ Required % = [20 ÷ 120] × 100 = 100 ÷ 6 = 50 ÷ 3

∴ The correct answer is 50/3 %.

Alternate Method

Given:

Let the salary of Y be 100.

Salary of X is 20% more than Y.

⇒ Salary of X = 100 + 20% of 100 = 120

Calculations:

Difference between salaries = 120 − 100 = 20

Percentage by which Y is less than X = (Difference ÷ Salary of X) × 100

⇒ Required % = (20 ÷ 120) × 100

⇒ Required % = (1 ÷ 6) × 100

⇒ Required % = 50 ÷ 3

∴ The correct answer is 50/3 %.

Additional Information

Percentage Increase/Decrease Formula

If A is r% more than B, then B is less than A by [r ÷ (100 + r)] × 100%.

Inverse Case

If A is r% less than B, then B is more than A by [r ÷ (100 − r)] × 100%.

Fraction Conversion

20% = 1/5. If X is 1/5 more than Y, then Y is 1/(5+1) = 1/6 less than X. 1/6 = 16.66% or 50/3%.

Given:

A student has to secure 40% of the marks to pass an exam. 

He gets only 45 marks and fails by 5 marks. 

Calculation:

A student gets 45 marks and failed by 5 marks.

Passing marks = (45 + 5) = 50 marks.

Now, according to the question

⇒ 40% = 50

⇒ 100% = [(50 × 100)/40]

⇒ 125 marks.

∴ The maximum marks are 125.

Given:

The number which has to be added to each term of the ratio 49 ∶ 68

Calculation:

Let the number be x.

According to the question

⇒ \(\frac{49+x}{{68+x}}\) = \(\frac{3}{{4}}\)

⇒ 196 + 4x = 204 + 3x

⇒ x = 8

∴ The required number is 8.

Shortcut Trick

Sum of ages 5 years ago = 75 − (5 × 2) = 65 years.

Product of ages 5 years ago = 750.

Find two factors of 750 that sum to 65: 50 × 15 = 750 and 50 + 15 = 65.

Father's age 5 years ago = 50 years.

Father's present age = 50 + 5 = 55 years.

∴ The correct answer is 55.

Alternate Method

Given:

Present sum of ages = 75 years

Product of ages 5 years ago = 750

Formula Used:

(x − a) × (y − a) = Product; where x + y = Sum

5 Years Ago Present

Prod: 750

Sum: 75

• 5 Years

Let Father's present age be F years.

Then, Son's present age = (75 − F) years.

5 years ago, Father's age = (F − 5) years.

5 years ago, Son's age = (75 − F − 5) = (70 − F) years.

According to the question:

⇒ (F − 5) × (70 − F) = 750

⇒ 70F − F² − 350 + 5F = 750

⇒ −F² + 75F − 350 − 750 = 0

⇒ F² − 75F + 1100 = 0

⇒ F² − 55F − 20F + 1100 = 0

⇒ F(F − 55) − 20(F − 55) = 0

⇒ (F − 55)(F − 20) = 0

⇒ F = 55 or F = 20

Since a father must be older than his son, Father's age = 55.

∴ The correct answer is 55.

Additional Information

Constant Age Difference

The difference between the ages of two people remains constant throughout their lives, regardless of the time elapsed.

Average Age Property

If the average age of n people is X, then after k years, their average age will be X + k.

Quadratic Roots in Ages

In problems involving product of ages, the larger root of the quadratic equation typically represents the older person's age.

Shortcut Trick

To find the fourth proportional to three numbers a, b, and c, use the direct formula: (b × c) ÷ a.

Given values are a = 7, b = 11, and c = 14.

Fourth Proportional = (11 × 14) ÷ 7 = 11 × 2 = 22.

This result matches the value given in option (4).

∴ The correct answer is 22.

Alternate Method

Given:

First term (a) = 7

Second term (b) = 11

Third term (c) = 14

Formula Used:

If a, b, c, and d are in proportion, then a : b = c : d, which means Product of Extremes = Product of Means (a × d = b × c).

Calculations:

⇒ Let the fourth proportional be x.

⇒ 7 : 11 :: 14 : x

⇒ 7 / 11 = 14 / x

⇒ 7 × x = 11 × 14

⇒ x = (11 × 14) ÷ 7

⇒ x = 11 × 2

⇒ x = 22

Therefore, the value 22 confirms that option (4) is the correct choice.

∴ The correct answer is 22.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is √(a × b).

Third Proportional

If a, b, and c are in continued proportion (a:b = b:c), then the third proportional c = b² ÷ a.

Invertendo and Alternendo

If a : b = c : d, then b : a = d : c (Invertendo) and a : c = b : d (Alternendo).

Compound Ratio

The compound ratio of (a : b) and (c : d) is (a × c) : (b × d).

Given:

The average weight of a class of 15 boys and 10 girls = 38.4 kg

Concept used:

Average weight = (Total weight of girls + Total weight of boys)/(No. of boys + No. of girls)

Calculation:

Let the average weight of girls be x

Total weight of boys = (40 × 15) = 600 kg

Now, according to the above concept,

⇒ 38.4 = \(\frac{600+10 \times x}{{15 +10}}\)

⇒ 38.4 = \(\frac{600+10x}{{25}}\)

⇒ 960 = 600 + 10x

⇒ 10x = 360

⇒ x = 36 kg

∴ The average weight of the girls is 36 kg.

Shortcut Trick

Ratio of Boys to Girls in the class = 70 : 30 = 7 : 3.

Deviation of Boys' average (75) from Total average (72) = +3.

Total positive deviation for 7 parts of boys = 7 × 3 = +21.

This must be balanced by a −21 deviation for 3 parts of girls.

Average deviation per girl = −21 ÷ 3 = −7.

Average marks of Girls = 72 − 7 = 65.

∴ The correct answer is 65 marks.

Alternate Method

Given:

Total students in class = 100

Number of Boys = 70

Average marks of Boys = 75

Average marks of Class = 72

Formula Used:

Total marks = Number of students × Average marks

Calculations:

⇒ Total marks of the complete class = 100 × 72 = 7200

⇒ Total marks of the boys = 70 × 75 = 5250

⇒ Number of girls = 100 − 70 = 30

⇒ Total marks of the girls = 7200 − 5250 = 1950

⇒ Average marks of the girls = 1950 ÷ 30

⇒ Average marks of girls = 65

∴ The correct answer is 65 marks.

Additional Information

Weighted Average Formula

The average of two groups combined is A_avg = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂), where n is the count and A is the average.

Alligation Rule

The ratio of the quantities of two components is inversely proportional to the difference between their averages and the mixture average: n₁ ÷ n₂ = (A₂ − A_avg) ÷ (A_avg − A₁).

Deviation Method

The sum of the deviations of all observations from the mean is always zero. This is used to find missing values when the group mean is known.

Given:

The sum of two numbers = 7

The sum of their squares = 25

Formula used:

(a + b)² = a² + b² + 2ab

Calculation:

Let the two numbers be x and y

x + y = 7           -----(1)

x² + y² = 25      -----(2)

We know that

(a + b)2 = a2 + b2 + 2ab

⇒ (x + y)² = x² + y² + 2xy

⇒ xy = [(x + y)² - (x² + y²)]/2       ----(3)

From equation (1) & (2), we get

⇒ xy = [(7)² - 25)/2]

⇒ xy = [(49 - 25)/2]

⇒ xy = (24/2) = 12

∴ The product of the two numbers is 12.

Given:

Difference between CI and SI for 2 years = Rs.2000

Rate = 4% 

Formula used:

Difference between CI and SI for 2 years = P × (R/100)²

Where, P = Principal, R = Rate 

Calculation:

⇒ P × R²/10000 = 2000

⇒ P × 16/10000 = 2000

⇒ P = 2000 × 10000/16

⇒ P = 1250000

∴ The required Sum is Rs.1250000.

Given:

An article is sold at a 20% discount, the selling price is Rs. 24

Calculation:

According to the question

⇒ 80% = 24

Then, 70% = (\(\frac{70 × 24}{{80}}\))

⇒ Rs. 21

∴ The required selling price is Rs. 21

Alternate Method

Formula used:

SP = {MP × (100 - D)}/100

Calculation:

According to the above concept,

SP = {MP × (100 - D)}/100

⇒ 24 = {MP × (100 - 20)}/100

⇒ 24 = (MP × 80/100)

⇒ MP = Rs. 30

Again using the same formula and considering D = 30%

SP = {MP × (100 - D)}/100

⇒ SP = {30 × (100 - 30)}/100

⇒ SP = (30 × 70/100)

⇒ SP = Rs. 21

∴ The required selling price is Rs. 21