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Given:

The difference between the two numbers is 1140.

If 6% of one number is equal to 10% of the other number.

Calculation:

Let one number = x and other number = y

According to question

6x% = 10y%

3x = 5y 

x = 5y/3

Now, x - y = 1140

⇒ 5y/3 - y = 1140 

⇒ 2y/3 = 1140 

⇒ y/3 = 570 

⇒ y = 570 × 3 = 1710

Now, x = 5 × (1710/3)

⇒ x = 5 × 570 = 2850

∴ The greater number is 2850.

Given:

If the cost price of 15 articles is equal to the selling price of 10 articles

And the discount on 12 articles is equal to the profit on 4 articles. 

Formula used:

Profit % = $\frac{{\Pr ofit}}{{CP}} × 100$

Mark Price = Selling Price + Discount 

Discount % = $\frac{{Discount}}{{Markprice}} × 100$

Calculation:

Cost price of 15 articles = Selling price of 10 articles

⇒ CP × 15 = SP × 10

⇒ (CP/SP) = 2/3

Profit % = (1/2) × 100 = 50 %

Discount on 12 articles is equal to profit on 4 articles.

⇒ 12 M.P - 12 S.P = 4 S.P - 4 C.P

⇒ 3 M.P - 3 S.P = S.P - 2/3 S.P

⇒ 3 M.P = 10/3 S.P

⇒ M.P/S.P = 10/9

Discount % = [(10 - 9)/10] × 100 = 10 %

∴ Difference of Profit and Discount percent is (50 - 10) = 40 %

Shortcut Trick

Identify perfect squares between 20 and 70: 25, 36, 49, 64.

Count of numbers (n) = 4.

Sum = 25 + 36 + 49 + 64 = 174.

Average = 174 ÷ 4 = 43.5.

∴ The correct answer is 43.5.

Alternate Method

Given: Range is between 20 and 70.

Formula Used: Average = Sum of observations ÷ Number of observations

⇒ Perfect squares are numbers of the form n².

⇒ 5² = 25 (lies between 20 and 70)

⇒ 6² = 36 (lies between 20 and 70)

⇒ 7² = 49 (lies between 20 and 70)

⇒ 8² = 64 (lies between 20 and 70)

⇒ 9² = 81 (outside the range)

⇒ Sum = 25 + 36 + 49 + 64 = 174

⇒ Average = 174 ÷ 4 = 43.5

∴ The correct answer is 43.5.

Additional Information

Sum of first n square numbers

The sum of squares of first n natural numbers is given by [n(n + 1)(2n + 1)] ÷ 6.

Average of first n square numbers

The average of squares of first n natural numbers is [(n + 1)(2n + 1)] ÷ 6.

Perfect Square Property

A perfect square never ends with the digits 2, 3, 7, or 8 at the unit place.

Given:

Wall Dimensions = 50m × 30m × 40m

Brick Dimensions = 3m × 2m × 1m

Ordered Bricks = 6500

Formula used:

Volume of cuboid = length × breath × height

Calculations: 

Volume of Wall = 50m × 30m × 40m = 60000 m³

Volume of Single Brick = 3m × 2m × 1m = 6 m³

Total Bricks Required = Volume of Wall/Volume of Single Brick = 60000/6 = 10000 bricks

Additional Bricks Required = Total Bricks Required - Ordered Bricks = 10000 - 6500

Hence, the contractor would require 3500 more bricks.

Given:

Total mixture = 90 L

Initial ratio of milk and water = 4 : 5

Final ratio of milk and water = 5 : 4

Calculation:

Let, the added quantity of the milk = a 

Initial quantity of milk

⇒ 90 × 4/(4 + 5) = 90 × 4/9 = 40 L

Initial quantity of water

⇒ 90 × 5/(4 + 5) = 90 × 5/9 = 50 L

According to the question,

(40 + a) : 50 = 5 : 4

⇒ (40 + a)/50 = 5/4

⇒ (40 + a) × 4 = 5 × 50

⇒ 160 + 4a = 250

⇒ 4a = 250 - 160

⇒ a = 90/4 = 22.5 L

∴ The quantity of milk to be further added is 22.5 L.

Given:

Profit percentage increases from 15% to 40%.

SP is increased by 40%.

Concept used:

New = Initial × (100 + Increase%)/100

CP = SP × 100/(100 + Profit%)

Calculation:

Let, Initial CP of the article = 100 units

Initial SP = 100 × 115/100 = 115 units

After increment in SP,

New SP = 115 × 140/100 = 161 units

Now, New CP = 161 × 100/140 = 115 units

So, Increase in CP = (New CP - Initial CP)/Initial CP × 100

⇒ (115 - 100)/100 × 100 = 15%

∴ The percentage increase in the cost price is 15%.

Given:

Jai can complete a work in 20 days.

Naresh can complete the work in 30 days.

Sunil can complete the same work in 60 days.

Jai started the work alone on the first day.

Every alternate day, Jai is assisted by Naresh and Sunil.

Concept used:

Efficiency = 1-day work

Required time = Total work/Efficiency

Calculation:

Total work = LCM of (20, 30, 60) = 60 units

1-day work of (Jai + Naresh + Sunil) = 3 + 2 + 1 = 6 units

Back to back 2 days of work

⇒ 1-day work of Jai + 1-day work of (Jai + Naresh + Sunil)

⇒ 3 + (3 + 2 + 1) = 3 + 6 = 9 units

54 is the last number divisible by 9, which is less than 60.

So, 12 days of work = 6 pairs of days work  = 6 × 9 = 54 units

Remaining work = 60 - 54 = 6 units

Jai does 3 units of work on (12 + 1) = 13^th day

Now, last remaining work = 6 - 3 = 3 units

Three of them completed the last remaining work in

⇒ 3/6 = 1/2 days

So, Total no. of days = 12 + 1 + 1/2 = $13\frac{1}{2}$ days

∴ In $13\frac{1}{2}$ days Jai will complete the work, as per the given condition.

Given:

Expression, $\rm x+\frac{1}{x}=6$ and $\rm y=\frac{2^{18}+2^{20}+2^{22}}{2^{20}+2^{22}+2^{24}}$

Concept used:

a^m + a^m+n + a^m+p = a^m(1 + aⁿ + a^p)

a^m/aⁿ = a^m-n

a^-m = 1/a^m

Calculation:

We have $\rm x+\frac{1}{x}=6$

Squaring both sides

⇒ x² + 1 = 6x     -----(1)

Now, $\rm y=\frac{2^{18}+2^{20}+2^{22}}{2^{20}+2^{22}+2^{24}}$

⇒ $\rm y=\frac{2^{18}(1+2^{2}+2^{4})}{2^{20}(1+2^{2}+2^{4})}$

⇒ $\rm y=\frac{2^{18}}{2^{20}}$

⇒ y = 2^18-20

⇒ y = 2^-2 = 1/(2²)

⇒ y = 1/4 = 0.25     -----(2)

Now the expression $\rm \frac{3x}{2x^2+2-5x}:y$ can be written as

⇒ $\rm \frac{3x}{2(x^2+1)-5x}:y$

Put the values from equation(1) and equation(2)

⇒ $\rm \frac{3x}{2(6x)-5x}:0.25$

⇒ 3x/7x : 0.25

⇒ 3/7 : 0.25 = 12 : 7

Hence, the required ratio is 12 : 7.

Given:

LCM of two prime numbers = 667

Prime number x > y

Concept used:

If the LCM of two prime numbers x and y is N

Then, x and y must be distinct prime factors of N.

Calculation:

The prime factorization of 667 = 23 × 29

We have x > y

Therefore, x = 29 and y = 23

Put the values in the expression 5y - 3x

⇒ (5 × 23) - (3 × 29)

⇒ 115 - 87

⇒ 28

Hence, the value of 5y - 3x is 28.

Shortcut Trick

Total Work (LCM of 15, 20) = 60 units.

Efficiency: A = 4 units/day, B = 3 units/day.

Work done in 2 days (A + B) = 4 + 3 = 7 units.

In 8 days (4 cycles of 2 days), work done = 4 × 7 = 28 units.

Unfinished part = (60 − 28) ÷ 60 = 32/60 = 8/15.

∴ The correct answer is 8/15.

Alternate Method

Given:

Time taken by A = 15 days

Time taken by B = 20 days

Total duration of work = 8 days (alternate starts with A)

Calculations:

⇒ Total work = LCM(15, 20) = 60 units.

⇒ Efficiency of A = 60 ÷ 15 = 4 units/day.

⇒ Efficiency of B = 60 ÷ 20 = 3 units/day.

⇒ Work on Day 1 (A) = 4 units.

⇒ Work on Day 2 (B) = 3 units.

⇒ Work in first 2-day cycle = 4 + 3 = 7 units.

⇒ Total work done in 8 days (4 cycles) = 4 × 7 = 28 units.

⇒ Remaining work = Total Work − Completed Work = 60 − 28 = 32 units.

⇒ Part of work unfinished = Remaining work ÷ Total work = 32 ÷ 60.

⇒ Unfinished part = 8/15.

∴ The correct answer is 8/15.

Additional Information

Alternate Day Work

In alternate day problems, calculate the work done in one full cycle (e.g., 2 days for 2 people) and then multiply by the number of complete cycles within the given time.

Work-Efficiency Relation

Total Work = Efficiency × Time. Efficiency is inversely proportional to the time taken to complete the work (Efficiency ∝ 1/Time).

Leftover Work Fraction

Fraction of work left = (Total Work − Work Done) ÷ Total Work. To find the part completed, use (Work Done ÷ Total Work).