Samacheer Kalvi Class 8 Maths Solutions Chapter 4 Life Mathematics InText Questions

Try These (Text Book Page No. 122)

 

Question 1. What percentage of a day is 10 hours?
Answer: A day has 24 hours in total. To find what percentage 10 hours is of 24 hours, we set up a fraction \( \frac{10}{24} \) and then multiply by 100. This gives us the percentage of the day that 10 hours represents.
In a day, there are 24 hours.
\( \text{Percentage} = \frac{10}{24} \times 100 \)
\( = 41.67\% \)
In simple words: To find what part 10 hours is of a full day (24 hours), we divide 10 by 24 and multiply by 100 to get the percentage.

🎯 Exam Tip: Remember that "a day" in such problems usually means 24 hours, unless specified otherwise. Always convert parts to percentage by multiplying the fraction by 100.

 

Question 2. P, Q and R such that P gets 50% of what Q gets and Q gets 50% of what R gets.
Answer: Let's assume R gets an amount 'x'. Since Q gets 50% of what R gets, Q's share will be half of R's. Then, P gets 50% of what Q gets, which means P's share is half of Q's share.
Let R get \( x \).
Q gets 50% of R's amount.
\( \implies \) Q gets \( \frac{50}{100} \times x = \frac{x}{2} \)
P gets 50% of Q's amount.
\( \implies \) P gets \( \frac{50}{100} \times \frac{x}{2} = \frac{x}{4} \)
If the total amount divided among the three is 350, then we add up their shares:
\( 350 = x + \frac{x}{2} + \frac{x}{4} \)
To add these fractions, find a common denominator, which is 4:
\( 350 = \frac{4x}{4} + \frac{2x}{4} + \frac{x}{4} \)
\( 350 = \frac{4x+2x+x}{4} \)
\( 350 = \frac{7x}{4} \)
Now, solve for \( x \):
\( x = \frac{350 \times 4}{7} \)
\( x = 50 \times 4 \)
\( x = 200 \)
So, R gets 200.
Q gets \( \frac{x}{2} = \frac{200}{2} = 100 \)
P gets \( \frac{x}{4} = \frac{200}{4} = 50 \)
So, P = 50, Q = 100, R = 200.
In simple words: If R gets a certain amount, Q gets half of that, and P gets half of what Q got. If the total is 350, then R gets 200, Q gets 100, and P gets 50.

🎯 Exam Tip: When dealing with percentages of amounts, it's often easiest to define one amount as 'x' and express the others in terms of 'x' before setting up an equation for the total.

Think (Text Book Page No. 124)

Question. With a lot of pride, the traffic police commissioner of a city reported that the accidents had decreased by 200% in one year. He came up with this number by stating that the increase in accidents from 200 to 600 is clearly a 200% rise and now that it had gone down from 600 last year to 200 this year should be a 200% fall. Is this decrease from 600 to 200, the same 200% as reported by him? Justify.
Answer: The commissioner's report is incorrect. A percentage decrease cannot be 200%. The maximum percentage decrease for any value is 100%, which would mean the value dropped to zero. Here's why:
First, let's calculate the percentage increase from 200 to 600:
\( \% \text{ increase } = \frac{\text{Change in value}}{\text{original value}} \times 100 \)
\( = \frac{600-200}{200} \times 100 \)
\( = \frac{400}{200} \times 100 = 200\% \text{ increase} \)
The initial calculation for the increase is correct. The original value for the increase was 200.
Now, let's calculate the percentage decrease from 600 to 200:
\( \% \text{ decrease } = \frac{\text{Change in value}}{\text{original value}} \times 100 \)
Here, the original value for the decrease is 600.
\( \% \text{ decrease } = \frac{600-200}{600} \times 100 \)
\( = \frac{400}{600} \times 100 \)
\( = \frac{2}{3} \times 100 \)
\( = 66.67\% \text{ decrease} \)
Therefore, an increase from 200 to 600 is a 200% increase, but a decrease from 600 to 200 is only a 66.67% decrease. These are not the same percentage values because the base (original value) for calculation is different in each case.
In simple words: The commissioner is wrong. When accidents increased from 200 to 600, it was a 200% rise. But when they fell from 600 to 200, it was only a 66.67% drop. You can't have a 200% decrease because the most something can decrease is 100%, which means it's gone completely.

🎯 Exam Tip: Always remember that percentage change (increase or decrease) is calculated with respect to the *original* value. An increase can be over 100%, but a decrease can never be more than 100%.

Try These (Text Book Page No. 126)

 

Question 1. If the selling price of an article is less than the cost price of the article, then there is a ______.
Answer: If the selling price of an article is less than its cost price, then there is a **loss**. This happens when you sell something for less than you bought or made it for.
In simple words: When you sell something for less money than you paid for it, you get a loss.

🎯 Exam Tip: Understand the basic definitions: Profit = Selling Price - Cost Price (when SP > CP), and Loss = Cost Price - Selling Price (when CP > SP).

 

Question 2. An article costing Rs. 5000 is sold for Rs. 4850. Is there a profit or loss? What percentage is it?
Answer:First, compare the cost price (CP) and selling price (SP) to see if there's a profit or loss.
Cost Price (CP) = Rs. 5000
Selling Price (SP) = Rs. 4850
Since SP (Rs. 4850) is less than CP (Rs. 5000), there is a **Loss**.
To calculate the loss amount:
Loss = CP - SP
Loss = Rs. 5000 - Rs. 4850
Loss = Rs. 150
Now, calculate the percentage of this loss:
\( \text{Percentage of Loss} = \frac{\text{Loss}}{\text{CP}} \times 100 \)
\( = \frac{150}{5000} \times 100 \)
\( = \frac{150}{50} \) (since \( \frac{100}{5000} = \frac{1}{50} \))
\( = 3\% \)
So, there is a loss of 3%.
In simple words: If an item bought for Rs. 5000 is sold for Rs. 4850, it's a loss because the selling price is lower. The loss amount is Rs. 150, which is 3% of the original cost.

🎯 Exam Tip: Always calculate profit or loss percentage based on the Cost Price (CP) unless specifically asked to calculate it on the Selling Price (SP).

 

Question 3. If the ratio of cost price and the selling price of an article is 5:7, then the profit / gain is ______ %.
Answer:Given the ratio of cost price (CP) to selling price (SP) is 5:7.
Let CP = \( 5x \)
Let SP = \( 7x \)
Since SP \( (7x) \) is greater than CP \( (5x) \), there is a profit.
Profit = SP - CP
Profit = \( 7x - 5x = 2x \)
Now, calculate the profit percentage:
\( \text{Profit percentage } = \frac{\text{Profit}}{\text{CP}} \times 100 \)
\( = \frac{2x}{5x} \times 100 \)
\( = \frac{2}{5} \times 100 \)
\( = 2 \times 20 \)
\( = 40\% \)
So, the profit is 40%.
In simple words: If an item's cost is 5 parts and its selling price is 7 parts, then the profit is 2 parts. This means the profit is 40% of the cost.

🎯 Exam Tip: When given ratios for CP and SP, use a common variable (like 'x') to represent the actual prices, which simplifies calculations for profit or loss percentage.

Think (Text Book Page No. 129)

Question. A shopkeeper marks the price of a marker board 15% above the cost price and then allows a discount of 15% on the marked price. Does he gain or lose in the transaction?
Answer: To find out if the shopkeeper gains or loses, let's assume an original cost price for the marker board.
Let the Cost Price (CP) of the marker board be Rs. 100.
The shopkeeper marks the price 15% above the CP.
Marked Price (MP) = CP + 15% of CP
\( = 100 + \frac{15}{100} \times 100 \)
\( = 100 + 15 \)
\( = 115 \)
So, the Marked Price is Rs. 115.
Next, the shopkeeper allows a discount of 15% on the Marked Price.
Discount amount = 15% of MP
\( = \frac{15}{100} \times 115 \)
\( = 0.15 \times 115 \)
\( = 17.25 \)
Now, calculate the Selling Price (SP) after the discount:
Selling Price (SP) = MP - Discount
\( = 115 - 17.25 \)
\( = 97.75 \)
Compare the Selling Price with the original Cost Price:
CP = Rs. 100
SP = Rs. 97.75
Since SP \( (97.75) \) is less than CP \( (100) \), the shopkeeper experiences a **loss**.
Loss = CP - SP
Loss = \( 100 - 97.75 = 2.25 \)
The shopkeeper loses Rs. 2.25 in this transaction.
In simple words: If you increase a price by 15% and then decrease that new price by 15%, you don't end up with the original price. The shopkeeper actually loses a small amount because the discount is calculated on a higher marked price.

🎯 Exam Tip: When a price is increased by a certain percentage and then decreased by the *same* percentage, there is always a net loss. This is because the percentage decrease is applied to a larger base (the marked-up price).

Try These (Text Book Page No. 129)

 

Question 1. The formula to find the simple interest for a given principal is ______.
Answer: The formula to find the simple interest for a given principal, rate, and time is:
\( I = \frac{\text{PNR}}{100} \)
Where:
I = Simple Interest
P = Principal amount (the initial money)
N = Number of years
R = Rate of interest per annum (per year)
In simple words: To calculate simple interest, you multiply the main amount (P), the number of years (N), and the interest rate (R), then divide by 100.

🎯 Exam Tip: Ensure the rate 'R' is a percentage (e.g., 8% means R=8 in the formula) and the time 'N' is in years. If time is given in months or days, convert it to years before applying the formula.

 

Question 2. Find the simple interest on Rs. 900 for 73 days at 8% p.a.
Answer: We need to find the simple interest using the formula \( I = \frac{\text{PNR}}{100} \).
Given values:
Principal (P) = Rs. 900
Time (N) = 73 days
Rate (R) = 8% p.a.
First, convert the time from days to years. There are 365 days in a year.
\( N = \frac{73}{365} \text{ years} = \frac{1}{5} \text{ years} \)
Now, apply the formula:
\( I = \frac{P \times N \times R}{100} \)
\( = \frac{900 \times \frac{1}{5} \times 8}{100} \)
\( = \frac{900 \times 1 \times 8}{5 \times 100} \)
\( = \frac{7200}{500} \)
\( = \frac{72}{5} \)
\( = 14.4 \)
So, the simple interest is Rs. 14.4.
In simple words: To find the interest for 73 days, which is one-fifth of a year, on Rs. 900 at 8% interest, you multiply 900 by one-fifth, then by 8, and divide by 100. The interest comes out to be Rs. 14.4.

🎯 Exam Tip: Always convert time into years when calculating simple interest, by dividing the number of days by 365 or months by 12, as the annual rate is provided "per annum".

 

Question 3. In how many years will Rs. 2000 become Rs. 3600 at 10% p.a simple interest?
Answer: We need to find the time (N) in years.
Given:
Principal (P) = Rs. 2000
Amount (A) = Rs. 3600
Rate (R) = 10% p.a.
First, calculate the Simple Interest (I):
Interest (I) = Amount - Principal
\( I = 3600 - 2000 = 1600 \)
Now, use the simple interest formula \( I = \frac{\text{PNR}}{100} \) and rearrange it to solve for N:
\( 1600 = \frac{2000 \times N \times 10}{100} \)
\( 1600 = \frac{20000 \times N}{100} \)
\( 1600 = 200 \times N \)
\( N = \frac{1600}{200} \)
\( N = 8 \)
So, it will take 8 years.
In simple words: The money earned Rs. 1600 in interest. To find how many years it took, we divide the interest (1600) by the yearly interest earned from Rs. 2000 at 10% (which is Rs. 200). This gives us 8 years.

🎯 Exam Tip: When the amount is given instead of interest, first calculate the interest by subtracting the principal from the amount, then use the simple interest formula to find the missing variable.

Try These (Text Book Page No. 141)

 

Question 1. Classify the given examples as direct or inverse proportion:
(i) Weight of pulses to their cost.
(ii) Distance travelled by bus to the price of ticket.
(iii) Speed of the athlete to cover a certain distance.
(iv) Number of workers employed to complete a construction in a specified time.
(v) Area of a circle to its radius.
Answer:
(i) **Weight of pulses to their cost.**
As the weight of pulses increases, their cost also increases. This is a **direct proportion**.
(ii) **Distance travelled by bus to the price of ticket.**
As the distance travelled by bus increases, the price of the ticket also increases. This is a **direct proportion**.
(iii) **Speed of the athlete to cover a certain distance.**
As the speed of the athlete increases, the time taken to cover the same distance decreases. This is an **inverse proportion**.
(iv) **Number of workers employed to complete a construction in a specified time.**
As the number of workers increases, the amount of time taken to complete the same work decreases. This is an **inverse proportion**.
(v) **Area of a circle to its radius.**
If the radius of a circle increases, its area also increases (Area \( = \pi r^2 \)). This is a **direct proportion**.
In simple words: Direct proportion means if one thing goes up, the other goes up too. Inverse proportion means if one thing goes up, the other goes down. Things like weight and cost are direct, while speed and time are inverse.

🎯 Exam Tip: To determine if a relationship is direct or inverse, imagine increasing one quantity and observe how the other quantity changes. If it increases, it's direct; if it decreases, it's inverse.

 

Question 2. A student can type 21 pages in 15 minutes. At the same rate, how long will it take a student to type 84 pages?
Answer: This is a direct proportion problem because as the number of pages increases, the time taken to type them will also increase.
Let the number of pages be \( P_1 = 21 \) and \( P_2 = 84 \).
Let the time taken be \( T_1 = 15 \) minutes and \( T_2 = x \) minutes.
For direct proportion, the ratio of pages to time is constant:
\( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)
\( \frac{21}{15} = \frac{84}{x} \)
Now, solve for \( x \):
\( 21 \times x = 84 \times 15 \)
\( x = \frac{84 \times 15}{21} \)
\( x = \frac{1260}{21} \)
\( x = 60 \)
So, it will take the student 60 minutes (or 1 hour) to type 84 pages.
In simple words: If typing more pages takes more time, it's a direct relationship. Since 84 pages are 4 times more than 21 pages, it will take 4 times more time, so 4 times 15 minutes, which is 60 minutes.

🎯 Exam Tip: In direct proportion, the ratio of the two quantities remains constant. When setting up the equation, ensure that corresponding values are placed in the same positions (e.g., pages on top, time on bottom for both ratios).

 

Question 3. If 35 women can do a piece of work in 16 days, In how many days will 28 women do the same work?
Answer: This is an inverse proportion problem because as the number of women working decreases, the number of days taken to complete the same work will increase.
Let the number of women be \( W_1 = 35 \) and \( W_2 = 28 \).
Let the number of days be \( D_1 = 16 \) days and \( D_2 = x \) days.
For inverse proportion, the product of the quantities is constant:
\( W_1 \times D_1 = W_2 \times D_2 \)
\( 35 \times 16 = 28 \times x \)
Now, solve for \( x \):
\( x = \frac{35 \times 16}{28} \)
\( x = \frac{5 \times 16}{4} \) (by dividing 35 and 28 by 7)
\( x = 5 \times 4 \) (by dividing 16 and 4 by 4)
\( x = 20 \)
So, 28 women will take 20 days to complete the same work.
In simple words: If fewer women are doing the same job, it will take them longer to finish. We multiply the number of women by days for the first group and set it equal to the number of women by days for the second group to find the missing days.

🎯 Exam Tip: In inverse proportion, if one quantity increases, the other decreases proportionally. Remember the formula \( X_1 Y_1 = X_2 Y_2 \) for inverse variation.

Try These (Text Book Page No. 145)

 

Question 1. If x and y vary directly, find k when x = y = 5.
Answer: If x and y vary directly, it means their ratio is a constant value, 'k'.
The relationship is expressed as:
\( \frac{x}{y} = k \)
Given that \( x = 5 \) and \( y = 5 \).
Substitute these values into the direct variation equation:
\( k = \frac{5}{5} \)
\( k = 1 \)
So, the constant of proportionality \( k \) is 1.
In simple words: When two things change directly together, their division always equals the same number. If x is 5 and y is 5, then that number (k) is 1.

🎯 Exam Tip: For direct variation, the constant \( k \) is found by dividing \( y \) by \( x \) (or \( x \) by \( y \), as long as it's consistent). Make sure to show your steps clearly.

 

Question 2. If x and y vary inversely, find the constant of proportionality when x = 64 and y = 0.75.
Answer: If x and y vary inversely, it means their product is a constant value, 'k'.
The relationship is expressed as:
\( xy = k \)
Given that \( x = 64 \) and \( y = 0.75 \).
Substitute these values into the inverse variation equation:
\( k = 64 \times 0.75 \)
To calculate \( 64 \times 0.75 \):
\( 0.75 = \frac{3}{4} \)
\( k = 64 \times \frac{3}{4} \)
\( k = 16 \times 3 \)
\( k = 48 \)
So, the constant of proportionality (or constant of variation) \( k \) is 48.
In simple words: When two things change inversely, multiplying them always gives the same number. If x is 64 and y is 0.75, their product is 48, which is the constant.

🎯 Exam Tip: For inverse variation, the constant \( k \) is found by multiplying \( x \) and \( y \). Convert decimals to fractions if it makes the multiplication easier.

Activity (Text Book Page No. 145)

Question. Draw a circle of a given radius. Then, draw its radii in such a way that the angles between any two consecutive pair of radii are equal. Start drawing 3 radii and end with drawing 12 radii in the circle. List and prepare a table for the number of radii to the angle between a pair of consecutive radii and check whether they are in inverse proportion. What is the proportionality constant?
Answer:When you draw radii in a circle such that angles between consecutive radii are equal, the total angle of the circle (360°) is divided equally among them. As the number of radii increases, the angle between any consecutive pair of radii decreases. This indicates an inverse proportion.
The relationship is: Number of radii \( \times \) Angle between radii \( = 360^\circ \) (constant).
The table is given below:

When we check the products:
\( 3 \times 120^\circ = 360^\circ \)
\( 4 \times 90^\circ = 360^\circ \)
\( 5 \times 72^\circ = 360^\circ \)
...and so on for all pairs, the product is consistently \( 360^\circ \).
This confirms that the number of radii and the angle between consecutive radii are in inverse proportion.
The proportionality constant \( k \) is \( 360^\circ \).
In simple words: When you slice a circle into more parts (radii), the angle of each slice gets smaller. This is an inverse relationship because as the number of slices goes up, the angle size goes down, but their product always stays 360 degrees (the whole circle). So, 360 degrees is the constant.

🎯 Exam Tip: Remember that the total angle in a circle is always \( 360^\circ \). When this total is divided into equal parts, the number of parts and the size of each part are inversely proportional, with \( 360^\circ \) as the constant.

Try These (Text Book Page No. 147)

 

Question 1. 24 men can make 48 articles in 12 days. Then, 6 men can make _______ articles in 6 days.
Answer: This is a combined variation problem involving men, days, and articles. We can use the formula:
\( \frac{P_1 D_1}{W_1} = \frac{P_2 D_2}{W_2} \)
Where P = Men, D = Days, W = Work (Articles)
Given values:
\( P_1 = 24 \) men, \( D_1 = 12 \) days, \( W_1 = 48 \) articles
\( P_2 = 6 \) men, \( D_2 = 6 \) days, \( W_2 = x \) articles (what we need to find)
Substitute the values into the formula:
\( \frac{24 \times 12}{48} = \frac{6 \times 6}{x} \)
Simplify the left side:
\( \frac{288}{48} = \frac{36}{x} \)
\( 6 = \frac{36}{x} \)
Now, solve for \( x \):
\( 6x = 36 \)
\( x = \frac{36}{6} \)
\( x = 6 \)
So, 6 men can make 6 articles in 6 days.
In simple words: We compare how much work a group of men does over some days to how much work another group of men does over different days. Using a special formula that links men, days, and articles, we find that 6 men will make 6 articles in 6 days.

🎯 Exam Tip: For problems involving multiple variables (like men, days, work), use the combined variation formula \( \frac{P_1 D_1}{W_1} = \frac{P_2 D_2}{W_2} \). Ensure that the 'work' or 'output' (like articles made) is in the denominator.

 

Question 2. 15 workers can lay a road of length 4 km in 4 hours. Then, _______ workers can lay a road of length 8 km in 8 hours.
Answer: This is a combined variation problem involving workers, road length (work), and hours.
We can use the formula:
\( \frac{W_1 H_1}{L_1} = \frac{W_2 H_2}{L_2} \) (where W = Workers, H = Hours, L = Length of road / Work)
Given values:
\( W_1 = 15 \) workers, \( H_1 = 4 \) hours, \( L_1 = 4 \) km
\( W_2 = x \) workers (what we need to find), \( H_2 = 8 \) hours, \( L_2 = 8 \) km
Substitute the values into the formula:
\( \frac{15 \times 4}{4} = \frac{x \times 8}{8} \)
Simplify both sides:
\( 15 = x \)
So, 15 workers can lay a road of length 8 km in 8 hours.
In simple words: If 15 workers can build a 4 km road in 4 hours, and we need to build an 8 km road in 8 hours, it's like both the work and the time have doubled. These changes balance each other out, meaning the same number of workers (15) will be needed.

🎯 Exam Tip: Be careful to identify which quantities are "work" (output) and which are "efforts" (input). Work done always goes in the denominator of the \( \frac{P_1 D_1}{W_1} \) or similar formula.

 

Question 3. 25 women working 12 hours a day can complete a work in 36 days. Then, 20 women working _______ hours a day to complete the same work in 30 days.
Answer: This is a combined variation problem involving women, hours per day, and total days to complete a fixed amount of work.
We can use the formula:
\( P_1 D_1 H_1 = P_2 D_2 H_2 \) (where P = Persons/Women, D = Days, H = Hours per day)
Given values:
\( P_1 = 25 \) women, \( D_1 = 36 \) days, \( H_1 = 12 \) hours/day
\( P_2 = 20 \) women, \( D_2 = 30 \) days, \( H_2 = x \) hours/day (what we need to find)
Substitute the values into the formula:
\( 25 \times 36 \times 12 = 20 \times 30 \times x \)
Multiply the known values:
\( 10800 = 600 \times x \)
Now, solve for \( x \):
\( x = \frac{10800}{600} \)
\( x = 18 \)
So, 20 women will need to work 18 hours a day to complete the same work in 30 days.
In simple words: We have a group of women working a certain number of hours each day for a certain number of days to finish a task. If we change the number of women and days, we need to find out how many hours per day the new group must work to finish the same task. The total effort (women x days x hours) must be equal.

🎯 Exam Tip: For questions where the amount of work is constant ("the same work"), you can use the formula \( P_1 D_1 H_1 = P_2 D_2 H_2 \). Always double-check your calculations to avoid errors.

 

Question 4. In a camp there are 420 kg of rice sufficient for 98 persons for 45 days. The number of days that 60 kg of rice will last for 42 persons is _______.
Answer: This is a combined variation problem involving the amount of rice, number of persons, and days.
We can use the formula:
\( \frac{R_1}{P_1 D_1} = \frac{R_2}{P_2 D_2} \) or \( \frac{P_1 D_1}{R_1} = \frac{P_2 D_2}{R_2} \) (where R = Rice, P = Persons, D = Days)
Let's use the second form: \( \frac{P_1 D_1}{R_1} = \frac{P_2 D_2}{R_2} \)
Given values:
\( R_1 = 420 \) kg, \( P_1 = 98 \) persons, \( D_1 = 45 \) days
\( R_2 = 60 \) kg, \( P_2 = 42 \) persons, \( D_2 = x \) days (what we need to find)
Substitute the values into the formula:
\( \frac{98 \times 45}{420} = \frac{42 \times x}{60} \)
Simplify the left side:
\( \frac{4410}{420} = \frac{42x}{60} \)
\( 10.5 = \frac{42x}{60} \)
Now, solve for \( x \):
\( 10.5 \times 60 = 42x \)
\( 630 = 42x \)
\( x = \frac{630}{42} \)
\( x = 15 \)
So, 60 kg of rice will last 42 persons for 15 days.
In simple words: We are comparing how long a certain amount of rice lasts for a group of people. If the amount of rice and the number of people change, we need to find how many days the new amount of rice will last for the new group of people. The relationship means less rice or more people will mean fewer days the rice lasts.

🎯 Exam Tip: When using combined variation, clearly identify which quantities are directly proportional (like rice and days for a fixed number of people) and which are inversely proportional (like people and days for a fixed amount of rice) to correctly set up your equation.

Try These (Text Book Page No. 150)

 

Question 1. Vikram can do one-third of work in p days. He can do three-fourths of work In _______ days.
Answer: This is a direct proportion problem between the amount of work done and the number of days taken.
Given:
\( \frac{1}{3} \) of the work is done in \( p \) days.
To find the time to complete the full work:
If \( \frac{1}{3} \) work takes \( p \) days, then 1 (full) work takes \( 3 \times p = 3p \) days.
Now, we need to find the time to do \( \frac{3}{4} \) of the work.
Time for \( \frac{3}{4} \) of the work = \( \frac{3}{4} \times (\text{time for full work}) \)
\( = \frac{3}{4} \times 3p \)
\( = \frac{9}{4}p \)
\( = 2\frac{1}{4}p \) days.
So, Vikram can do three-fourths of the work in \( 2\frac{1}{4}p \) days.
In simple words: If Vikram does one-third of a job in 'p' days, it means the whole job would take him 3 times 'p' days. To find how long three-fourths of the job would take, we simply multiply the total job time by three-fourths.

🎯 Exam Tip: In work-and-time problems, always consider the fraction of work and relate it to the total time. If you know the time for a fraction, you can easily find the time for the whole work and then any other fraction.

 

Question 2. If m persons can complete a work in n days, then 4m persons can complete the same work in _______ days and m/4 persons can complete the same work in _______ days.
Answer: This is an inverse proportion problem between the number of persons and the number of days to complete the same work. The total "man-days" (persons × days) for a given work is constant.
Given:
\( m \) persons complete a work in \( n \) days.
Total work = \( mn \) person-days.
**Part 1: 4m persons**
Let \( D_1 \) be the number of days for \( 4m \) persons.
\( m \times n = 4m \times D_1 \)
\( D_1 = \frac{m \times n}{4m} \)
\( D_1 = \frac{n}{4} \) days.
So, \( 4m \) persons can complete the same work in \( \frac{n}{4} \) days.
**Part 2: m/4 persons**
Let \( D_2 \) be the number of days for \( \frac{m}{4} \) persons.
\( m \times n = \frac{m}{4} \times D_2 \)
\( D_2 = \frac{m \times n}{\frac{m}{4}} \)
\( D_2 = m \times n \times \frac{4}{m} \)
\( D_2 = 4n \) days.
So, \( \frac{m}{4} \) persons can complete the same work in \( 4n \) days.
In simple words: If more people do the same work, it takes less time. If 4 times more people work, it takes 1/4 of the time. If 1/4 the number of people work, it takes 4 times longer.

🎯 Exam Tip: Remember the fundamental concept of work and time: Work = Number of workers \( \times \) Time taken. If the work is constant, then \( \text{Person}_1 \times \text{Days}_1 = \text{Person}_2 \times \text{Days}_2 \).