JKSSB Aptitude MCQs

Explanation

1. Define the variables:

• Let the number of brown pairs of socks in the original order = $x$

• The number of black pairs of socks in the original order = $4$

• Let the price of a brown pair of socks = $P$

• Since the price of a black pair is double that of a brown pair, the price of a black pair = $2P$

2. Calculate the original bill:

$$\text{Original Bill} = (4 \times 2P) + (x \times P) = 8P + xP = P(8 + x)$$

3. Calculate the mistaken bill:

The clerk interchanged the number of pairs, meaning there are now $x$ pairs of black socks and $4$ pairs of brown socks.

$$\text{Mistaken Bill} = (x \times 2P) + (4 \times P) = 2xP + 4P = P(2x + 4)$$

4. Set up the equation:

We are told that the mistake increased the bill by $50\%$. This means the mistaken bill is $150\%$ (or $1.5$ times) of the original bill:

$$\text{Mistaken Bill} = 1.5 \times \text{Original Bill}$$

$$P(2x + 4) = 1.5 \times P(8 + x)$$

Since the price $P$ cannot be zero, we can divide both sides by $P$:

$$2x + 4 = 1.5(8 + x)$$

$$2x + 4 = 12 + 1.5x$$

Now, solve for $x$:

$$2x - 1.5x = 12 - 4$$

$$0.5x = 8$$

$$x = 16$$

So, the original order contained $16$ pairs of brown socks.

5. Find the ratio:

The ratio of the number of black pairs to brown pairs in the original order is:

$$\text{Black} : \text{Brown} = 4 : 16 = 1 : 4$$

Explanation

16% P.a.

Explanation

• Set up the equation:

  According to the problem, "17 times the cost price is equal to 8 times the sum of the cost price and the selling price":

  $$17 \times CP = 8 \times (CP + SP)$$

• Simplify the equation:

  Distribute the $8$ on the right side:

  $$17 \times CP = 8 \times CP + 8 \times SP$$

  Bring all $CP$ terms to one side:

  $$17 \times CP - 8 \times CP = 8 \times SP$$

  $$9 \times CP = 8 \times SP$$

• Find the ratio of Selling Price to Cost Price:

  $$\frac{SP}{CP} = \frac{9}{8}$$

  This means if the Cost Price ($CP$) is $8$ units, the Selling Price ($SP$) is $9$ units. Since the selling price is greater than the cost price, it is a gain (profit).

• Calculate the Gain Percentage:

  $$\text{Gain} = SP - CP = 9 - 8 = 1 \text{ unit}$$

  $$\text{Gain Percentage} = \left( \frac{\text{Gain}}{CP} \right) \times 100\%$$

  $$\text{Gain Percentage} = \frac{1}{8} \times 100\% = 12.5\%$$

Explanation

Let the Cost Price (CP) of the T.V. be $100\%$.

• Initial Scenario (8% Gain): The T.V. is sold at $108\%$ of the CP.

• Alternative Scenario (15% Loss): The T.V. is sold at $85\%$ of the CP ($100\% - 15\%$).

The transition from an $8\%$ profit to a $15\%$ loss represents a total percentage drop of:

$$\text{Total Percentage Change} = 8\% - (-15\%) = 23\%$$

This $23\%$ drop in value corresponds exactly to the price reduction of Rs. 2553.

Calculation:

1. Find $1\%$ of the Cost Price:

   $$23\% \text{ of CP} = 2553$$

   $$1\% \text{ of CP} = \frac{2553}{23} = 111$$

2. Calculate the Target Selling Price (for an 18% Gain):

   To make an $18\%$ profit, the new selling price must be $118\%$ of the CP ($100\% + 18\%$).

   $$\text{Target Selling Price} = 118\% \text{ of CP}$$

   $$\text{Target Selling Price} = 118 \times 111 = \mathbf{13098}$$

Explanation

Step 1: Determine the total work and individual efficiencies

Find the Least Common Multiple (LCM) of the time taken by A, B, and C:

• Time: A = 10 days, B = 12 days, C = 15 days.

• Total Work (LCM of 10, 12, 15): 60 units.

Now, calculate how many units each person completes per day (Efficiency):

• Efficiency of A ($E_A$): $60 / 10 = \mathbf{6\text{ units/day}}$

• Efficiency of B ($E_B$): $60 / 12 = \mathbf{5\text{ units/day}}$

• Efficiency of C ($E_C$): $60 / 15 = \mathbf{4\text{ units/day}}$

Step 2: Calculate work done in phases

Phase 1: First 2 days (A, B, and C work together)

• Combined Efficiency: $6 + 5 + 4 = 15\text{ units/day}$.

• Work done in 2 days: $15 \times 2 = \mathbf{30\text{ units}}$.

Phase 2: Next 2 days (A has left, only B and C work)

• Combined Efficiency (B + C): $5 + 4 = 9\text{ units/day}$.

• Work done in 2 days: $9 \times 2 = \mathbf{18\text{ units}}$.

Total work completed so far:

• $30\text{ units} + 18\text{ units} = \mathbf{48\text{ units}}$.

Step 3: Calculate the remaining work and time

• Remaining Work: $60\text{ units} - 48\text{ units} = \mathbf{12\text{ units}}$.

• After the 4th day, C also leaves, leaving only B to finish the work.

Time taken by B to finish the remaining work:

$$\text{Time} = \frac{\text{Remaining Work}}{\text{Efficiency of B}} = \frac{12}{5} = \mathbf{2.4\text{ days}}$$

Answer:

The remaining work will be completed in 2.4 more days.

(The total time for the entire project would be $2 + 2 + 2.4 = 6.4\text{ days}$).

Explanation

Step 1: Determine Total Work and Efficiencies

• Total Work = LCM(15, 25) = 75 units

• A's efficiency = $75 \div 15 = \mathbf{5 \text{ units/day}}$

• B's efficiency = $75 \div 25 = \mathbf{3 \text{ units/day}}$

Step 2: Calculate Work Done at the End

B leaves 7 days before the work is finished. This means for the final 7 days, only A worked.

• Work done by A in last 7 days = $7 \text{ days} \times 5 \text{ units/day} = \mathbf{35 \text{ units}}$

Step 3: Calculate Time for Initial Phase

• Remaining work (done by A and B together) = $75 - 35 = \mathbf{40 \text{ units}}$

• Combined efficiency $(A + B) = 5 + 3 = \mathbf{8 \text{ units/day}}$

• Time spent together = $40 \div 8 = \mathbf{5 \text{ days}}$

Step 4: Total Time

• Total days = (Days together) + (Days A worked alone)

• Total days = $5 + 7 = \mathbf{12 \text{ days}}$

Explanation

The Compound Ratio of two or more ratios is found by multiplying the corresponding terms together (antecedent with antecedent, and consequent with consequent).

• Ratio 1: $a : b$ (or $\frac{a}{b}$)

• Ratio 2: $c : d$ (or $\frac{c}{d}$)

To compound them, multiply the first terms ($a$ and $c$) and the second terms ($b$ and $d$):

$$(a \times c) : (b \times d) = \mathbf{ac : bd}$$

Explanation

1. Initial State:

   The ratio of wine to water is 3:1.

   In terms of concentration, wine is $\frac{3}{3+1} = \frac{3}{4}$ of the total volume.

2. Target State:

   The final ratio is 1:1.

   In terms of concentration, wine is now $\frac{1}{1+1} = \frac{1}{2}$ of the total volume.

3. The Process:

   When you draw off a fraction of the mixture and replace it with water, you are adding a substance that contains 0% wine.

Calculation (Alligation Method)

• Initial Wine Concentration: $\frac{3}{4}$

• Replacement (Water) Wine Concentration: $0$

• Mean (Target) Wine Concentration: $\frac{1}{2}$

Subtract across the diagonals:

• $(\frac{3}{4} - \frac{1}{2}) = \frac{1}{4}$ (This represents the amount of original mixture left)

• $(\frac{1}{2} - 0) = \frac{1}{2}$ (This represents the amount of water added)

The ratio of Remaining Mixture to Water Added is:

$$\frac{1}{2} : \frac{1}{4}$$

Multiply by 4 to simplify:

2 : 1

Conclusion

Out of a total of 3 parts (2 parts old mixture + 1 part water), 1 part was replaced.

Fraction to be drawn off = $\frac{1}{3}$

Explanation

Step 1: Find the Cost Price

Let the Marked Price (MP) be 100.

• Case 1: A 10% discount is given.

  • $SP = 100 - 10 = 90$

• At this price, the publisher gains 20%.

  • $SP = CP \times (1 + \text{gain}\%)$

  • $90 = CP \times 1.20$

  • $CP = \frac{90}{1.20} = \mathbf{75}$

Step 2: Apply the New Discount

• Case 2: The discount is increased to 15%.

  • New $SP = 100 - 15 = \mathbf{85}$

Step 3: Calculate the New Gain

The cost price remains 75.

• Profit = New $SP - CP$

  • $\text{Profit} = 85 - 75 = 10$

• Gain Percentage:

  • $\text{Gain}\% = \left( \frac{\text{Profit}}{CP} \right) \times 100$

  • $\text{Gain}\% = \left( \frac{10}{75} \right) \times 100 = \frac{2}{15} \times 100$

  • $\text{Gain}\% = \mathbf{13\frac{1}{3}\%}$ (or approximately $13.33\%$)

Explanation

Algebraic Method

1. Sum of $x$ numbers: $15 \times x = 15x$

2. Sum of $y$ numbers: $20 \times y = 20y$

3. Total sum of $(x+y)$ numbers: $18(x+y)$

Set up the equation:

$$15x + 20y = 18(x + y)$$

$$15x + 20y = 18x + 18y$$

$$20y - 18y = 18x - 15x$$

$$2y = 3x$$

$$\frac{x}{y} = \frac{2}{3}$$

Explanation

The expression "$x\%$ of $y = y\%$ of $x$" is written as:

$$\frac{x}{100} \times y = \frac{y}{100} \times x$$

This simplifies to:

$$\frac{xy}{100} = \frac{yx}{100}$$

Logic

Since multiplication is commutative ($a \times b = b \times a$), the statement $\frac{xy}{100} = \frac{yx}{100}$ is always true for any values of $x$ and $y$.

For example:

• $10\%$ of $50 = 5$

• $50\%$ of $10 = 5$

Conclusion

Because the equation is a mathematical identity (it works for any numbers), the relationship between $x$ and $y$ (whether one is larger, smaller, or equal) does not affect the truth of the statement.

The correct option is: D) Cannot be determined

Explanation

1. Assume a starting value: Let the original price be 100.

2. Apply the reduction: A $10\%$ reduction ($100 - 10$) makes the new price 90.

3. Identify the required gain: To return from 90 back to 100, you must increase the price by 10.

4. Calculate the percentage: The percentage increase is always calculated based on the current (new) price:

   $$\text{Percentage Increase} = \frac{\text{Increase needed}}{\text{New price}} \times 100$$

   $$\text{Percentage Increase} = \frac{10}{90} \times 100 = \frac{1}{9} \times 100$$

Final Answer

The new price must be increased by $11\frac{1}{9}\%$ to restore the original price.

Explanation

To find the number of trailing zeros in $x = 2^{10} \times 5^6$:

1. Identify 10s: A trailing zero is formed by the product $2 \times 5 = 10$.

2. Compare Exponents: The number of zeros is determined by the lower exponent of the prime factors 2 and 5.

   • Exponent of 2 is 10.

   • Exponent of 5 is 6.

3. Determine Zeros: Since 6 is the smaller power, you can form exactly six pairs of $(2 \times 5)$.

$$x = 2^4 \times (2 \times 5)^6 = 16 \times 10^6$$

Total zeros: 6

Explanation

To find the number of possible values for $x$, we use the divisibility rule for 3, which states that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.

Step 1: Calculate the sum of the known digits

Sum the digits given in the number $45678x9231$:

$$4 + 5 + 6 + 7 + 8 + 9 + 2 + 3 + 1 = 45$$

Step 2: Set up the divisibility condition

The total sum of the digits is $45 + x$. For the number to be divisible by 3, the expression $(45 + x)$ must be a multiple of 3.

Since 45 is already divisible by 3 ($45 \div 3 = 15$), $x$ must also be a digit that is divisible by 3 to keep the total sum divisible by 3.

Step 3: Identify possible values for $x$

The variable $x$ represents a single digit, so it can be any integer from 0 to 9. Let’s check which digits satisfy the condition:

• If $x = 0$: $45 + 0 = 45$ (Divisible by 3) ✅

• If $x = 1$: $45 + 1 = 46$ (Not divisible)

• If $x = 2$: $45 + 2 = 47$ (Not divisible)

• If $x = 3$: $45 + 3 = 48$ (Divisible by 3) ✅

• If $x = 4$: $45 + 4 = 49$ (Not divisible)

• If $x = 5$: $45 + 5 = 50$ (Not divisible)

• If $x = 6$: $45 + 6 = 51$ (Divisible by 3) ✅

• If $x = 7$: $45 + 7 = 52$ (Not divisible)

• If $x = 8$: $45 + 8 = 53$ (Not divisible)

• If $x = 9$: $45 + 9 = 54$ (Divisible by 3) ✅

The possible values for $x$ are 0, 3, 6, and 9.

Conclusion

There are 4 possible values for $x$.

Explanation

• Property: When two numbers are in the ratio $a/b$ (in simplest form) and their HCF is $H$, the numbers are $aH$ and $bH$.

• Step 1: Find the actual numbers.

  • Ratio = 7/11

  • HCF ($H$) = 5

  • First number = $7 \times 5 = 35$

  • Second number = $11 \times 5 = 55$

• Step 2: Find the LCM.

  • Method 1 (Formula): $\text{LCM} = \text{Ratio}_1 \times \text{Ratio}_2 \times \text{HCF}$

    $$\text{LCM} = 7 \times 11 \times 5 = 385$$

  • Method 2 (Product Rule): $\text{Product of numbers} = \text{HCF} \times \text{LCM}$

    $$35 \times 55 = 5 \times \text{LCM}$$

    $$1925 = 5 \times \text{LCM}$$

    $$\text{LCM} = \frac{1925}{5} = 385$$

Explanation

Finding the HCF (Highest Common Factor):

• Factors of 20: 1, 2, 4, 5, 10, 20

• Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30

• The highest common factor is 10.

2. Finding the LCM (Least Common Multiple):

• Multiples of 20: 20, 40, 60, 80, ...

• Multiples of 30: 30, 60, 90, 120, ...

• The smallest common multiple is 60.

Alternative Method (Prime Factorization):

• $20 = 2 \times 2 \times 5 = 2^2 \times 5^1$

• $30 = 2 \times 3 \times 5 = 2^1 \times 3^1 \times 5^1$

• HCF (lowest powers of common factors): $2^1 \times 5^1 = 10$

• LCM (highest powers of all factors): $2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$

Explanation

• Step 1: Let the two numbers be $3x$ and $4x$ (based on the ratio 3:4).

• Step 2: Find the LCM of $3x$ and $4x$ in terms of $x$.

  • The LCM of 3 and 4 is 12.

  • Therefore, the LCM of $3x$ and $4x$ is $12x$.

• Step 3: According to the problem, the LCM is 48.

  $$12x = 48$$

• Step 4: Solve for $x$:

  $$x = \frac{48}{12} = 4$$

• Step 5: Calculate the smaller number ($3x$):

  $$\text{Smaller number} = 3 \times 4 = 12$$

Explanation

• Fundamental Rule: For any two numbers, the product of the numbers is equal to the product of their HCF and LCM.

  $$\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$$

• Step 1: Plug the given values into the formula.

  • $\text{HCF} = 5$

  • $\text{LCM} = 120$

  • $\text{First number} = 20$

  • Let the second number be $x$.

• Step 2: Form the equation:

  $$5 \times 120 = 20 \times x$$

  $$600 = 20x$$

• Step 3: Solve for $x$:

  $$x = \frac{600}{20} = 30$$

Explanation

• Step 1: Let the required number of years be $x$.

• Step 2: Express their ages after $x$ years:

  • Father's age = $40 + x$

  • Son's age = $10 + x$

• Step 3: According to the problem, after $x$ years, the father will be 3 times as old as the son:

  $$40 + x = 3(10 + x)$$

• Step 4: Solve for $x$:

  $$40 + x = 30 + 3x$$

  $$40 - 30 = 3x - x$$

  $$10 = 2x$$

  $$x = 5 \text{ years}$$

Explanation

• Step 1: Let the present age of the Son be $x$ and the Father be $50 - x$ (since their sum is 50).

• Step 2: Express their ages 5 years ago:

  • Son's age = $x - 5$

  • Father's age = $(50 - x) - 5 = 45 - x$

• Step 3: According to the problem, 5 years ago, the father was 4 times as old as the son:

  $$45 - x = 4(x - 5)$$

  $$45 - x = 4x - 20$$

• Step 4: Solve for $x$:

  $$45 + 20 = 4x + x$$

  $$65 = 5x$$

  $$x = \frac{65}{5} = 13$$

Explanation

• Step 1: Let the present ages of A and B be $7x$ and $9x$ (based on the ratio 7:9).

• Step 2: After 4 years, their ages will be $7x + 4$ and $9x + 4$. According to the problem, this new ratio is 9:11.

  $$\frac{7x + 4}{9x + 4} = \frac{9}{11}$$

• Step 3: Cross-multiply to solve for $x$:

  $$11(7x + 4) = 9(9x + 4)$$

  $$77x + 44 = 81x + 36$$

  $$81x - 77x = 44 - 36$$

  $$4x = 8 \implies x = 2$$

• Step 4: Calculate A's present age:

  $$\text{A's present age} = 7x = 7 \times 2 = 14 \text{ years}$$

• (Note: B's present age would be $9 \times 2 = 18$ years.)

• Shortcut Method:

  • Initial Ratio ($A:B$) = $7:9$ (Difference = 2 units)

  • Ratio after 4 years = $9:11$ (Difference = 2 units)

  • Since the difference between the parts is the same (9 - 7 = 2 and 11 - 9 = 2), we can say that an increase of 2 units in the ratio corresponds to 4 years.

  • $2 \text{ units} = 4 \text{ years} \implies 1 \text{ unit} = 2 \text{ years}$.

  • A's present age = $7 \text{ units} = 7 \times 2 = 14 \text{ years}$.

Explanation

• Step 1: Let the initial number of boys be $5x$ and the initial number of girls be $6x$ (based on the ratio 5:6).

• Step 2: According to the problem, 5 more boys join the class, but the number of girls remains the same. The new ratio is 10:11.

  $$\frac{5x + 5}{6x} = \frac{10}{11}$$

• Step 3: Cross-multiply to solve for $x$:

  $$11(5x + 5) = 10(6x)$$

  $$55x + 55 = 60x$$

  $$60x - 55x = 55$$

  $$5x = 55 \implies x = 11$$

• Step 4: Calculate the number of girls:

  $$\text{Number of girls} = 6x = 6 \times 11 = 66$$

Explanation

• Step 1: To find the ratio of $a/c$ when $a/b$ and $b/c$ are given, you can multiply the two ratios together:

  $$\frac{a}{c} = \frac{a}{b} \times \frac{b}{c}$$

• Step 2: Substitute the given values ($4/7$ and $14/15$):

  $$\frac{a}{c} = \frac{4}{7} \times \frac{14}{15}$$

• Step 3: Simplify the fraction. Since 14 is divisible by 7:

  $$\frac{a}{c} = \frac{4 \times 2}{1 \times 15} = \frac{8}{15}$$

• Alternative Method (Making 'b' equal):

  • Ratio $a:b = 4:7$

  • Ratio $b:c = 14:15$

  • To combine them, make the value of '$b$' the same in both. Multiply the first ratio by 2:

    $$a:b = (4 \times 2) : (7 \times 2) = 8:14$$

  • Now, $a:b = 8:14$ and $b:c = 14:15$.

  • Therefore, $a:b:c = 8:14:15$, which gives $a:c = 8:15$.

Explanation

• Step 1: Let the two numbers be $7x$ and $11x$ (based on the ratio 7:11).

• Step 2: According to the problem, their product is 693.

  $$(7x) \times (11x) = 693$$

  $$77x^2 = 693$$

• Step 3: Solve for $x$.

  $$x^2 = \frac{693}{77} = 9$$

  $$x = \sqrt{9} = 3$$

• Step 4: Find the difference between the numbers.

  $$\text{Difference} = 11x - 7x = 4x$$

  $$\text{Difference} = 4 \times 3 = 12$$

• (Alternatively: The numbers are $7 \times 3 = 21$ and $11 \times 3 = 33$. Difference: $33 - 21 = 12$.)

Explanation

• Step 1: Calculate the daily work rates.

  • Work rate of A = $1/15$ work per day.

  • Work rate of B = $1/20$ work per day.

• Step 2: Calculate the combined work rate (A + B).

  $$\text{Combined Rate} = \frac{1}{15} + \frac{1}{20}$$

  $$\text{Using 60 as the common denominator:}$$

  $$\text{Combined Rate} = \frac{4}{60} + \frac{3}{60} = \frac{7}{60} \text{ work per day}$$

• Step 3: Calculate the total work done in 4 days.

  $$\text{Work Done} = \text{Rate} \times \text{Time} = \frac{7}{60} \times 4 = \frac{7}{15}$$

• Step 4: Calculate the remaining fraction of work.

  $$\text{Work Left} = 1 - \text{Work Done}$$

  $$\text{Work Left} = 1 - \frac{7}{15} = \frac{8}{15}$$

• LCM Method:

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

  • Efficiency of A = $60 / 15$ = 4 units/day.

  • Efficiency of B = $60 / 20$ = 3 units/day.

  • Combined Efficiency = $4 + 3$ = 7 units/day.

  • Work done in 4 days = $7 \times 4$ = 28 units.

  • Remaining Work = $60 - 28$ = 32 units.

  • Fraction Left = $32 / 60 = 8/15$.

Explanation

• Step 1: Determine the time taken by B.

  • A takes 18 days.

  • B takes half the time as A = $18 / 2$ = 9 days.

• Step 2: Calculate individual work rates.

  • Work rate of A = $1/18$ work per day.

  • Work rate of B = $1/9$ work per day.

• Step 3: Calculate the combined work rate (A + B).

  $$\text{Combined Rate} = \frac{1}{18} + \frac{1}{9}$$

  $$\text{Using 18 as the common denominator:}$$

  $$\text{Combined Rate} = \frac{1}{18} + \frac{2}{18} = \frac{3}{18} = \frac{1}{6} \text{ work per day}$$

• Step 4: Find the total days taken together.

  $$\text{Total Time} = \frac{1}{\text{Combined Rate}} = 6 \text{ days}$$

• Shortcut Formula:

  $$\text{Total Time} = \frac{xy}{x + y} = \frac{18 \times 9}{18 + 9} = \frac{162}{27} = 6 \text{ days}$$

Explanation

• Step 1: Calculate the work rates (work done per day).

  • Combined rate (A + B) = $1/10$ work per day.

  • Individual rate of A = $1/30$ work per day.

• Step 2: Find B's work rate by subtracting A's rate from the combined rate.

  $$\text{B's rate} = \frac{1}{10} - \frac{1}{30}$$

  $$\text{Using 30 as the common denominator:}$$

  $$\text{B's rate} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \text{ work per day}$$

• Step 3: Convert the rate back into time.

  $$\text{Time taken by B alone} = \frac{1}{\text{Rate}} = 15 \text{ days}$$

• Alternative LCM Method:

  • Let Total Work = LCM of 10 and 30 = 30 units.

  • Efficiency of (A + B) = $30 / 10$ = 3 units/day.

  • Efficiency of A = $30 / 30$ = 1 unit/day.

  • Efficiency of B = $(A + B) - A = 3 - 1$ = 2 units/day.

  • Time for B = $\text{Total Work} / \text{Efficiency} = 30 / 2$ = 15 days.

Explanation

• Step 1: Calculate the combined work rate of the man and woman.

  $$\text{Work rate (Man + Woman)} = \frac{1}{8} \text{ work per day}$$

• Step 2: Calculate the work rate of the man alone.

  $$\text{Work rate (Man)} = \frac{1}{10} \text{ work per day}$$

• Step 3: Subtract the man's rate from the combined rate to find the woman's work rate.

  $$\text{Work rate (Woman)} = \frac{1}{8} - \frac{1}{10}$$

  $$\text{To subtract, find a common denominator (40):}$$

  $$\text{Work rate (Woman)} = \frac{5}{40} - \frac{4}{40} = \frac{1}{40} \text{ work per day}$$

• Step 4: Convert the rate back into time.

  $$\text{Time taken by woman alone} = \frac{1}{\text{Work rate}} = 40 \text{ days}$$

Explanation

• The sum of the first $n$ natural numbers is given by the formula:

  $$\text{Sum} = \frac{n(n + 1)}{2}$$

• The average is defined as the $\frac{\text{Sum}}{\text{Total Count}}$:

  $$\text{Average} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2}$$

Explanation

• Method 1 (Algebraic):

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

  The average is the sum divided by the count:

  $$\frac{x + (x+2) + (x+4) + (x+6) + (x+8) + (x+10)}{6} = 25$$

  $$\frac{6x + 30}{6} = 25$$

  $$x + 5 = 25 \implies x = 20$$

  • Smallest number ($x$) = 20

  • Largest number ($x+10$) = $20 + 10$ = 30

• Method 2 (Logic):

  For a set of consecutive numbers, the average is always the middle value. For 6 numbers, the average lies exactly between the $3^{rd}$ and $4^{th}$ numbers.

  Middle value = 25.

  The two even numbers surrounding 25 are 24 ($3^{rd}$) and 26 ($4^{th}$).

  The sequence is: $20, 22, 24, \text{[25]}, 26, 28, 30$.