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Given

Scores in first 5 innings = [50, 56, 42, 38, 78]

Target average = 60

Concept:

The average is calculated by the total sum divided by the number of elements.

Calculation:

⇒ Total score in first 5 innings = Sum of given scores = 50 + 56 + 42 + 38 + 78 = 264

⇒ Target total score for 6 innings = Average × Total innings = 60 × 6 = 360

⇒ Score needed in 6^th innings = Target total score - Total score in first 5 innings = 360 - 264 = 96

Therefore, the batsman needs to score 96 in his 6^th innings.

Shortcut Trick

For any sequence of consecutive even or odd numbers, the numbers form an Arithmetic Progression (AP).

In an AP, the average of all terms is equal to the average of the first and last terms.

⇒ Average = (Smallest Number + Largest Number) ÷ 2

⇒ Smallest + Largest = 2 × Average

⇒ Sum = 2 × 35 = 70

∴ The correct answer is 70.

Alternate Method

Given:

Average of 6 consecutive even numbers = 35

Formula Used:

Average = Sum of observations ÷ Number of observations

30 Smallest

32

34

36

38

40 Largest

Avg = 35

Sum = 30 + 40 = 70

Calculations:

⇒ Let the 6 consecutive even numbers be x, x + 2, x + 4, x + 6, x + 8, and x + 10.

⇒ Sum of these 6 numbers = x + (x + 2) + (x + 4) + (x + 6) + (x + 8) + (x + 10)

⇒ Sum = 6x + 30

⇒ Average = (6x + 30) ÷ 6 = x + 5

⇒ We are given that the average is 35.

⇒ x + 5 = 35

⇒ x = 35 − 5 = 30

⇒ The smallest number = x = 30

⇒ The largest number = x + 10 = 30 + 10 = 40

⇒ Sum of smallest and largest number = 30 + 40 = 70

∴ The correct answer is 70.

Additional Information

Average of an Arithmetic Progression (AP)

For any sequence with a constant difference (like consecutive numbers), the average is simply (First Term + Last Term) ÷ 2.

Consecutive Even/Odd Numbers

Any sequence of consecutive even or odd numbers has a common difference of 2. For n numbers, the difference between the largest and smallest term is exactly 2 × (n − 1).

Sum of 'n' terms in AP

The total sum of terms in an arithmetic sequence can be calculated quickly as: Sum = n × Average.

Shortcut Trick

Net multiplier for 3 years ahead at 10% growth = (1.1)³ = 1.331

Net multiplier for 2 years ago at 10% growth = 1 ÷ (1.1)² = 1 ÷ 1.21

Difference = (1,21,000 × 1.331) − (1,21,000 ÷ 1.21)

Difference = 1,61,051 − 1,00,000 = 61,051

∴ The correct answer is 61,051.

Alternate Method

Given:

Current Population (P) = 1,21,000

Annual Growth Rate (r) = 10%

Formula Used:

Future Population = P × (1 + r/100)ⁿ

Past Population = P ÷ (1 + r/100)ⁿ

Calculations:

⇒ Population 3 years from now = 1,21,000 × (1 + 10/100)³

⇒ Population 3 years from now = 1,21,000 × (1.1)³ = 1,21,000 × 1.331 = 1,61,051

⇒ Population 2 years ago = 1,21,000 ÷ (1 + 10/100)²

⇒ Population 2 years ago = 1,21,000 ÷ (1.1)² = 1,21,000 ÷ 1.21 = 1,00,000

⇒ Required Difference = 1,61,051 − 1,00,000 = 61,051

∴ The correct answer is 61,051.

Additional Information

Population Growth Formula

The population after n years is given by P_n = P₀(1 + R/100)ⁿ, where P₀ is the initial population and R is the annual rate of growth.

Population Decrease/Depreciation

If the population decreases at a rate R, the formula becomes P_n = P₀(1 − R/100)ⁿ.

Successive Growth Rates

If the growth rates for different years are R₁, R₂, etc., then P_final = P₀(1 + R₁/100)(1 + R₂/100)...

Given:

Total amount = ₹17,650

The rate of interest = 8.5%

Formula used:

S.I. = (P × R × T)/100

Where, S.I. = Simple interest; P = Total amount; R = Rate of interest; T = Time (years)

Calculation:

The total amount = Total simple interest + Total invested amount

= [(17650 × 8.5 × 5)/100] + 17650

= 7501.25 + 17650

= ₹25151.25

= ₹25151 (to nearest ₹)

∴ Seema will have ₹25151 after five years.

Given:

Meenu is 38 years old.

Her daughter is 8 years old.

Calculation:

Let, after x years will Meenu be double her daughter's age

As per the question,

⇒ 38 + x = 2 × (8 + x)

⇒  x = 22 years

∴ The correct option is 1

Face value = 480 Rs

Discount = 25% 

So, Selling price = 360 Rs

Loss = 40 Rs

Cost Price = 400 Rs

Loss percent = 40/400 * 100 = 10%

Given:

Speed of one person, S₁ = 18 km/hour

Speed of second person, S₂ = 24 km/hour

T₁ – T₂ = 20 minutes

Formula used:

Speed = Distance/Time

Calculation:

Let the distance be x km.

Time taken by first person

⇒ Time = Distance/Speed

⇒ x/18 hour

Time taken by second person

⇒ x/24 hour

T₁ – T₂ = 20 minutes

⇒ x/18 – x/24 = 20/60

⇒ (4x – 3x)/72 = 1/3

⇒ x = 72/3

⇒ x = 24 km

∴ The distance travelled is 24 km.

 

Required distance = {(S₁ × S₂)/(S₂ – S₁)} × (T₁ – T₂)

⇒ {(18 × 24)/(24 – 18)} × (20/60)

⇒ (18 × 24)/6) × (1/3)

⇒ 24 km

∴ The distance travelled is 24 km

Shortcut Trick

Stoppage time per hour = [(Faster Speed − Slower Speed) ÷ Faster Speed] × 60 min

Stoppage time per hour = [(90 − 75) ÷ 90] × 60

Stoppage time per hour = (15 ÷ 90) × 60

Stoppage time per hour = (1 ÷ 6) × 60 = 10 minutes

∴ The correct answer is 10 minutes.

Alternate Method

Given:

Speed without stoppages = 90 kmph

Speed with stoppages = 75 kmph

Without Stoppage (90 kmph)

Distance covered in 1 hr = 90 km With Stoppage (75 kmph)

Distance covered in 1 hr = 75 km

15 km

10 min stop

Calculations:

⇒ Distance covered in 1 hr without stop = 90 km

⇒ Distance covered in 1 hr with stop = 75 km

⇒ Decrease in distance due to stoppage = 90 − 75 = 15 km

⇒ Stoppage time = Time taken to cover the lost 15 km at normal speed of 90 kmph

⇒ Stoppage time (in hr) = Distance ÷ Speed = 15 ÷ 90 = 1 ÷ 6 hr

⇒ Stoppage time (in min) = (1 ÷ 6) × 60 = 10 minutes

∴ The correct answer is 10 minutes.

Additional Information

Direct Stoppage Time Formula

Stoppage time per hour = (Difference in Speed ÷ Speed without Stoppages) × 60 minutes.

Average Speed

Average Speed = Total Distance ÷ Total Time. It represents the uniform speed needed to cover the same distance in the same total time.

Unit Conversions

To convert kmph to m/s, multiply by 5 ÷ 18. To convert m/s to kmph, multiply by 18 ÷ 5.

Shortcut Trick

Initial salary ratio of A and B = 5 : 6 (gap between terms is 1).

New salary ratio after ₹2000 increment = 11 : 13 (gap between terms is 2).

To make the gap equal, multiply the initial ratio by 2 → 10 : 12.

Now, the increment in ratio parts for both A and B = 11 − 10 = 13 − 12 = 1 part.

Since 1 ratio part = ₹2000, the new salary of A (11 parts) = 11 × 2000 = ₹22000.

∴ The correct answer is ₹22000.

Alternate Method

Given:

Initial salary ratio of A and B = 5 : 6

Increment for both = ₹2000

New salary ratio of A and B = 11 : 13

Calculations:

Let the initial salaries of A and B be 5x and 6x respectively.

After an increment of ₹2000, their salaries become (5x + 2000) and (6x + 2000).

⇒ (5x + 2000) ÷ (6x + 2000) = 11 ÷ 13

⇒ 13 × (5x + 2000) = 11 × (6x + 2000)

⇒ 65x + 26000 = 66x + 22000

⇒ 66x − 65x = 26000 − 22000

⇒ x = 4000

We need to find the new monthly salary of A.

⇒ New salary of A = 5x + 2000

⇒ New salary of A = 5 × 4000 + 2000

⇒ New salary of A = 20000 + 2000 = ₹22000

∴ The correct answer is ₹22000.

Additional Information

Compound Ratio

For two given ratios a : b and c : d, the compound ratio is calculated by multiplying the antecedents together and the consequents together, giving ac : bd.

Duplicate & Triplicate Ratios

The duplicate ratio of a : b is defined as a² : b². Similarly, the triplicate ratio is a³ : b³.

Invertendo & Alternendo

If four variables are in proportion a : b = c : d, then by the property of Invertendo b : a = d : c, and by Alternendo a : c = b : d.

Given:

Total coin = 220

Total money = Rs. 160

There are thrice as many 1 Rupee coins as there are 25 paise coins.

Concept used:

Ratio method is used.

Calculation:

Let the 25 paise coins be 'x'

So, one rupees coins = 3x

50 paise coins = 220 – x – (3x) = 220 – (4x)

According to the questions,

3x + [(220 – 4x)/2] + x/4 =160

⇒ (12x + 440 – 8x + x)/4 = 160

⇒  5x + 440 = 640

⇒ 5x = 200

⇒ x = 40

So, 50 paise coins = 220 – (4x) = 220 – (4 × 40) = 60

∴ The number of 50 paise coin is 60.