Selina Concise Solutions for ICSE Class 7 Mathematics Chapter 7 Unitary Method Including Time and Work

Unitary Method (Including Time and Work)

The unitary method is a way of solving problems where we first find the value of a single item. Once we know the value of one item, we can easily calculate the value of any number of items.

There are two main types of variations in this method:

• Direct variation
• Inverse variation

Direct variation: In this type, if one quantity goes up, the other quantity also goes up. Similarly, if one quantity goes down, the other goes down as well.

Inverse variation: In this type, when one quantity goes up, the other quantity goes down. If one quantity goes down, the other goes up. This usually happens in questions about speed, time, or work.

Exercise 7(A)

Question 1. Weight of 8 identical articles is 4.8 kg. What is the weight of 11 such articles ?
Answer:
Given that the total weight of 8 identical items is 4.8 kg.
First, we calculate the weight of 1 item by dividing the total weight by the number of items:
Weight of 1 item = \( \frac{4.8}{8} \) kg = 0.6 kg
Next, we calculate the weight of 11 such items by multiplying this single value:
Weight of 11 items = 0.6 \( \times \) 11 = 6.6 kg
So, the weight of 11 items is 6.6 kg.
In simple words: First, find the weight of one single item by dividing. Then, multiply that answer by the number of items you want to find.

Exam Tip: In direct variation, always write down the given values clearly and find the value of one unit first by dividing, before multiplying to get the final answer.

Question 2. 6 books weigh 1 .260 kg. How many books will weigh 3.150 kg ?
Answer:
We are given that a weight of 1.260 kg corresponds to 6 books.
First, we find the number of books that would weigh 1 kg:
Number of books per kg = \( \frac{6}{1.260} \)
Now, we calculate the number of books that will weigh 3.150 kg:
Number of books = \( \frac{6}{1.260} \times 3.150 \)
To make the calculation simpler, we can remove the decimal points since both numbers have three decimal places:
Number of books = \( \frac{6 \times 3150}{1260} \)
Dividing 1260 by 6 gives 210, so we have:
Number of books = \( \frac{3150}{210} \) = 15 books
So, 15 books are needed to weigh 3.150 kg.
In simple words: Find out how many books make up 1 kg first. Then, multiply that amount by the total weight you want to find.

Exam Tip: Pay close attention to decimal places. Removing decimals by multiplying both numerator and denominator by 1000 can prevent calculation errors.

Question 3. 8 men complete a work in 6 hours. In how many hours will 12 men complete the same work ?
Answer:
This is a case of inverse variation because more workers will take less time to complete the same task.
Time taken by 8 men to finish the work = 6 hours
If only 1 man works, he will need more time, which is the product of the number of men and hours:
Time taken by 1 man = 6 \( \times \) 8 = 48 hours
Now, if 12 men are working, the time required will be:
Time taken by 12 men = \( \frac{6 \times 8}{12} \) hours
Time taken = \( \frac{48}{12} \) = 4 hours
So, 12 men will complete the work in 4 hours.
In simple words: If you have more people working, the job gets done faster. First, find how long it takes for one person, then divide that time by the new number of people.

Exam Tip: Remember that "men and hours" is an inverse proportion. Never divide for 1 man; instead, multiply because 1 man will take more time than many men.

Question 4. If a 25 cm long candle burns for 45 minutes, how long will another candle of the same material and same thickness but 5 cm longer than the previous one, burn ?
Answer:
A candle of length 25 cm burns for 45 minutes.
First, we find the burning time for 1 cm of the candle:
Burning time for 1 cm = \( \frac{45}{25} \) minutes
The new candle is 5 cm longer, so its total length is:
Length = 25 + 5 = 30 cm
Now, we calculate the burning time for this 30 cm candle:
Burning time for 30 cm = \( \frac{45}{25} \times 30 \) minutes
Burning time = \( \frac{45 \times 30}{25} \) = \( \frac{1350}{25} \) = 54 minutes
Thus, the new candle will burn for 54 minutes.
In simple words: Find the burning time for just 1 cm of the candle by dividing. Then, add 5 cm to find the new candle's total length (30 cm), and multiply to get the final time.

Exam Tip: Be careful to add the extra 5 cm to the original length of 25 cm first to get the correct new length of 30 cm before doing the calculations.

Question 5. A typist takes 80 minutes to type 24 pages. How long will he take to type 87 pages ?
Answer:
The time needed to type 24 pages is 80 minutes.
First, we calculate the time required to type 1 page:
Time for 1 page = \( \frac{80}{24} \) minutes
Now, we calculate the time required to type 87 pages:
Time for 87 pages = \( \frac{80 \times 87}{24} \) minutes
Simplifying the fraction by dividing both 80 and 24 by 8, we get \( \frac{10}{3} \):
Time for 87 pages = \( \frac{10 \times 87}{3} \) = 10 \( \times \) 29 = 290 minutes
So, the typist will take 290 minutes.
In simple words: Find the time it takes to type one single page first by dividing. Then, multiply that time by 87 to find the total time.

Exam Tip: Simplify the fraction first (like dividing by 8) to make the multiplication much easier and avoid mistakes with large numbers.

Question 6. Rs. 750 support a family for 15 days. For how many days will Rs. 2,500 support the same family ?
Answer:
A budget of Rs. 750 can sustain a family for 15 days.
First, we find out how long Re. 1 would last for the family:
Support for Re. 1 = \( \frac{15}{750} \) days
Now, we calculate how many days Rs. 2,500 will support them:
Support for Rs. 2,500 = \( \frac{15}{750} \times 2500 \) days
Simplifying \( \frac{15}{750} \) gives \( \frac{1}{50} \):
Support for Rs. 2,500 = \( \frac{2500}{50} \) = 50 days
Therefore, Rs. 2,500 will support the family for 50 days.
In simple words: Find out how many days just 1 Rupee will last by dividing. Then, multiply that small value by 2,500 to find the total days.

Exam Tip: Simplify the fraction \( \frac{15}{750} \) to \( \frac{1}{50} \) right away to make multiplying by 2500 very straightforward.

Question 7. 400 men have provisions for 23 weeks. They are joined by 60 men. How long will the provisions last ?
Answer:
This is an inverse proportion problem because an increase in the number of men means the food will last for fewer weeks.
Provisions are sufficient for 400 men for 23 weeks.
If there were only 1 man, the food would last him much longer:
Provisions for 1 man = 23 \( \times \) 400 weeks
Since 60 more men join, the new total number of men is:
Total men = 400 + 60 = 460 men
Now, we calculate how long the provisions will last for 460 men:
Provisions for 460 men = \( \frac{23 \times 400}{460} \) weeks
Provisions = \( \frac{9200}{460} \) = 20 weeks
So, the food provisions will last for 20 weeks.
In simple words: More people eating means the food runs out sooner. Multiply the original men and weeks to find how long food lasts for one person, then divide by the new total number of people.

Exam Tip: Always calculate the new total number of people first (400 + 60 = 460) before finding the final number of weeks.

Question 8. 200 men have provisions for 30 days. If 50 men left, the same provisions would last for the remaining men, in how many days?
Answer:
This is an inverse variation problem since fewer men will take longer to finish the same food provisions.
Provisions are sufficient for 200 men for 30 days.
For 1 man, the same food would last much longer:
Provisions for 1 man = 30 \( \times \) 200 days
The number of remaining men after 50 men leave is:
Remaining men = 200 - 50 = 150 men
Now, we calculate how long the provisions will last for these 150 men:
Provisions for 150 men = \( \frac{30 \times 200}{150} \) days
Provisions = \( \frac{6000}{150} \) = 40 days
Hence, the provisions will last for 40 days.
In simple words: Since some people left, there is more food for the remaining people, so it will last longer. Find the total days for one person, then divide by the remaining number of people.

Exam Tip: Be careful to subtract the men who left (200 - 50 = 150) instead of adding them, as the group size has decreased.

Question 9. 8 men can finish a certain amount of provisions in 40 days. If 2 more men join with them, find for how many days the same amount of provisions be sufficient ?
Answer:
This is an inverse variation problem since adding more people reduces the number of days the provisions will last.
Provisions are enough for 8 men for 40 days.
If only 1 man was eating, the food would last:
Provisions for 1 man = 40 \( \times \) 8 = 320 days
When 2 more men join, the new group size is:
Total men = 8 + 2 = 10 men
Now, we calculate the number of days the provisions will last for 10 men:
Provisions for 10 men = \( \frac{320}{10} \) = 32 days
So, the provisions will be sufficient for 32 days.
In simple words: When more people join, the food gets eaten faster. Find out how many days the food lasts for one person by multiplying, then divide that by the new total number of people.

Exam Tip: In inverse proportion, always multiply the initial values first (8 \( \times \) 40) to find the single unit value before dividing by the new total.

Question 10. If interest on Rs. 200 be Rs. 25 in a certain time, what will be the interest on Rs 750 for the same time ?
Answer:
This is a case of direct proportion because more principal money will yield more interest.
Interest earned on Rs. 200 = Rs. 25
First, we find the interest earned on Re. 1:
Interest on Re. 1 = Rs. \( \frac{25}{200} \)
Now, we calculate the interest for Rs. 750:
Interest on Rs. 750 = Rs. \( \frac{25}{200} \times 750 \)
We can simplify \( \frac{25}{200} \) to \( \frac{1}{8} \):
Interest on Rs. 750 = Rs. \( \frac{750}{8} \) = Rs. 93.75
So, the interest on Rs. 750 will be Rs. 93.75.
In simple words: Find the interest for just 1 Rupee first by dividing. Then, multiply that amount by 750 to get the total interest.

Exam Tip: Simplify the fraction \( \frac{25}{200} \) to \( \frac{1}{8} \) before multiplying by 750 to keep the division simple.

Question 11. If 3 dozen eggs cost Rs. 90, find the cost of 3 scores of eggs. (1 score = 20)
Answer:
First, we convert both measurements into single units:
Number of eggs in 3 dozen = 3 \( \times \) 12 = 36 eggs
Number of eggs in 3 scores = 3 \( \times \) 20 = 60 eggs
We are given that 36 eggs cost Rs. 90.
First, we find the cost of 1 egg:
Cost of 1 egg = Rs. \( \frac{90}{36} \)
Now, we calculate the cost of 60 eggs:
Cost of 60 eggs = Rs. \( \frac{90}{36} \times 60 \)
Simplifying \( \frac{90}{36} \) by dividing by 18 gives \( \frac{5}{2} \):
Cost of 60 eggs = Rs. \( \frac{5 \times 60}{2} \) = Rs. 5 \( \times \) 30 = Rs. 150
So, 3 scores of eggs will cost Rs. 150.
In simple words: Change both dozen and scores into the actual number of eggs. Find the price of one egg by dividing, and then multiply by the total number of eggs you want to buy.

Exam Tip: Remember to write down the conversions clearly: 1 dozen is 12 items and 1 score is 20 items. This is a very common starting point for such questions.

Question 12. If the fare for 48 km is Rs. 288, what will be the fare for 36 km ?
Answer:
The total fare for a distance of 48 km is Rs. 288.
First, we find the fare for a distance of 1 km:
Fare for 1 km = Rs. \( \frac{288}{48} \) = Rs. 6
Now, we calculate the fare for a distance of 36 km:
Fare for 36 km = Rs. 6 \( \times \) 36 = Rs. 216
Therefore, the fare for 36 km is Rs. 216.
In simple words: Divide the total fare by the total distance to find the cost of traveling 1 km. Then, multiply that rate by 36 km.

Exam Tip: First divide Rs. 288 by 48 to find the rate per kilometer (Rs. 6), which makes the next step of multiplying by 36 very simple.

Question 13. What will be the cost of 3.20 kg of an item, if 3 kg of it costs Rs. 360 ?
Answer:
The price for 3 kg of an item is Rs. 360.
First, we determine the cost of 1 kg of that item:
Cost of 1 kg = Rs. \( \frac{360}{3} \) = Rs. 120
Now, we calculate the cost for 3.20 kg:
Cost of 3.20 kg = Rs. 120 \( \times \) 3.20 = Rs. 384
Thus, the total cost for 3.20 kg of the item is Rs. 384.
In simple words: Find the price for 1 kg of the item first by dividing 360 by 3. Then, multiply that price by 3.20 kg.

Exam Tip: Be comfortable with decimal multiplication. Since Rs. 120 \( \times \) 3.20 is the same as 12 \( \times \) 32, the answer is easily found to be Rs. 384.

Question 14. If 9 lines of a print, in a column of a book contains 36 words. How many words will a column of 51 lines cqntain ?
Answer:
We are given that there are 36 words in 9 lines of print.
First, we find the number of words in a single line:
Words in 1 line = \( \frac{36}{9} \) = 4 words
Now, we calculate the number of words in 51 lines of print:
Words in 51 lines = 4 \( \times \) 51 = 204 words
Thus, there will be 204 words in 51 lines.
In simple words: Find the number of words in one single line by dividing. Then, multiply that number by 51.

Exam Tip: Make sure to do the division step first (\( \frac{36}{9} = 4 \)) so that you only have to multiply 4 by 51, which is quick and easy.

Question 15. 125 pupil have food sufficient for 18 days. If 25 more pupil join them, how long will the food last now ? What assumption have you made to come to your answer ?
Answer:
This is an inverse variation problem because more pupils eating means the food will run out faster.
The initial number of pupils = 125
With 25 more pupils joining, the new total is:
Total pupils = 125 + 25 = 150 pupils
For 125 pupils, the food provisions last for 18 days.
For 1 pupil, the food would last for:
Duration for 1 pupil = 18 \( \times \) 125 days
Now, we calculate how long the food will last for 150 pupils:
Duration for 150 pupils = \( \frac{18 \times 125}{150} \) days
We simplify the fraction by dividing both 125 and 150 by 25:
Duration = \( \frac{18 \times 5}{6} \) days
Duration = 3 \( \times \) 5 = 15 days
Assumption: We assume that each pupil consumes the same constant amount of food every day.
In simple words: When more pupils join, the food finishes quicker. Find how long the food lasts for just one pupil, and then divide that by the new total of 150 pupils.

Exam Tip: Don't forget to write down the assumption clearly when asked, as this is worth separate marks in the marking scheme.

Question 16. A carpenter prepares a new chair in 3 days, working 8 hours a day. Atleast how many hours per day must he work in order to make the same chair in 4 days ?
Answer:
This is an inverse variation problem because if the carpenter has more days to complete the work, he can work fewer hours each day.
Working 8 hours per day, the carpenter finishes the chair in 3 days.
If he had to complete it in 1 day, he would need to work:
Total working hours = 8 \( \times \) 3 = 24 hours
Since he now has 4 days to complete the chair, the hours required per day are:
Hours per day = \( \frac{24}{4} \) = 6 hours
So, he must work 6 hours per day.
In simple words: Find the total number of hours needed to make the chair (24 hours). Then, divide this total by 4 days to find how many hours he needs to work each day.

Exam Tip: Since "days and hours per day" are in inverse proportion, multiply the initial values (3 \( \times \) 8) to find the total work in man-hours before dividing by the new number of days.

Question 17. A man earns Rs. 5,800 in 10 days. How much will he earn in the month of February of a leap year?
Answer:
A leap year contains 29 days in the month of February.
The man earns Rs. 5,800 in 10 days.
First, we calculate how much he earns in 1 day:
Earnings for 1 day = Rs. \( \frac{5800}{10} \) = Rs. 580
Now, we calculate his total earnings for the 29 days of February:
Earnings for February = Rs. 580 \( \times \) 29 = Rs. 16,820
So, he will earn Rs. 16,820 during that month.
In simple words: First, find how much money the man makes in one single day by dividing. Then, multiply that daily rate by 29 (since February has 29 days in a leap year).

Exam Tip: Remember that February has 29 days in a leap year (and 28 days in a non-leap year). Mentioning this fact explicitly shows the examiner you understand the context.

Question 18. A machine is used for making rubber balls and makes 500 balls in 30 minutes. How many balls will it make in \( 3\frac{1}{2} \) hours?
Answer:
First, we convert the time into hours for easier calculation:
30 minutes = \( \frac{30}{60} \) hour = \( \frac{1}{2} \) hour
The machine makes 500 balls in \( \frac{1}{2} \) hour.
First, we find how many balls it makes in 1 hour:
Balls made in 1 hour = 500 \( \times \) 2 = 1000 balls
The given time is \( 3\frac{1}{2} \) hours, which is \( \frac{7}{2} \) hours when converted to an improper fraction.
Now, we calculate the balls made in \( \frac{7}{2} \) hours:
Balls made = 1000 \( \times \frac{7}{2} \) = 500 \( \times \) 7 = 3500 balls
Therefore, the machine will make 3500 balls.
In simple words: Since 30 minutes is half an hour, the machine makes 1000 balls in one full hour. Multiply 1000 by 3.5 hours to get the final answer.

Exam Tip: Convert mixed numbers like \( 3\frac{1}{2} \) to improper fractions (\( \frac{7}{2} \)) right away. This makes working with fractions in multiplication much simpler.

Question 19. In a school’s hostel mess, 20 children consume a certain quantity of ration in 6 days. However, 5 children did not return to the hostel after holidays. How long will the same amount of ration last now?
Answer:
This is an inverse variation problem because a decrease in the number of children means the food provisions will last for more days.
The initial number of children in the hostel is 20.
The provisions last for 20 children for 6 days.
If there were only 1 child, the food would last:
Duration for 1 child = 6 \( \times \) 20 = 120 days
Since 5 children did not return after the holidays, the number of children is:
Remaining children = 20 - 5 = 15 children
Now, we calculate how long the provisions will last for 15 children:
Duration for 15 children = \( \frac{120}{15} \) = 8 days
Thus, the provisions will last for 8 days.
In simple words: Since there are fewer children eating, the food will last longer. First, find how long the food lasts for one child, and then divide that by the 15 remaining children.

Exam Tip: Be sure to write down the subtraction (20 - 5 = 15) to show the new number of children before doing the division.

Exercise 7(B)

Question 1. The cost of \( \frac{3}{5} \) kg of ghee is Rs. 96 ; find the cost of : (i) one kg ghee (ii) \( \frac{5}{8} \) kg ghee.
Answer:
(i) We are given that the price of \( \frac{3}{5} \) kg of ghee is Rs. 96.
To find the cost of 1 kg, we divide Rs. 96 by \( \frac{3}{5} \), which means multiplying by the reciprocal \( \frac{5}{3} \):
Cost of 1 kg of ghee = Rs. 96 \( \times \frac{5}{3} \) = Rs. 32 \( \times \) 5 = Rs. 160
(ii) Now, we calculate the price for \( \frac{5}{8} \) kg of ghee by multiplying the price of 1 kg by this fraction:
Cost of \( \frac{5}{8} \) kg of ghee = Rs. 160 \( \times \frac{5}{8} \) = Rs. 20 \( \times \) 5 = Rs. 100
In simple words: Find the price of 1 kg of ghee first by multiplying 96 by the flipped fraction \( \frac{5}{3} \). Then, multiply that rate by \( \frac{5}{8} \) to find the cost of the smaller amount.

Exam Tip: Remember that dividing by a fraction is the same as multiplying by its reciprocal. Showing this step clearly prevents arithmetic errors.

Question 2. \( 3\frac{1}{2} \) m of cloth costs Rs. 168 ; find the cost of \( 4\frac{1}{3} \) m of the same cloth.
Answer:
First, we convert the mixed numbers into improper fractions:
Length of first cloth = \( 3\frac{1}{2} \) m = \( \frac{7}{2} \) m
Length of second cloth = \( 4\frac{1}{3} \) m = \( \frac{13}{3} \) m
We are given that \( \frac{7}{2} \) m of cloth costs Rs. 168.
First, we find the price of 1 m of cloth by multiplying Rs. 168 by the reciprocal of \( \frac{7}{2} \):
Cost of 1 m of cloth = Rs. 168 \( \times \frac{2}{7} \) = Rs. 24 \( \times \) 2 = Rs. 48
Now, we calculate the cost of \( \frac{13}{3} \) m of cloth:
Cost of \( \frac{13}{3} \) m of cloth = Rs. 48 \( \times \frac{13}{3} \) = Rs. 16 \( \times \) 13 = Rs. 208
Thus, the cost of the cloth is Rs. 208.
In simple words: Turn the mixed fractions into improper fractions first. Find the cost of 1 meter of cloth, and then multiply that cost by the second fraction to get your final answer.

Exam Tip: Be precise when simplifying \( 168 \times \frac{2}{7} \). Dividing 168 by 7 gives 24, which makes the multiplication much simpler.

Question 3. A wrist watch loses 10 sec in every 8 hours; in how much time will it lose 15 sec. ?
Answer:
This is a direct variation problem since losing more seconds will require a longer duration of time.
Time taken to lose 10 seconds = 8 hours
First, we find the time taken to lose 1 second:
Time to lose 1 second = \( \frac{8}{10} \) hours
Now, we calculate the time taken to lose 15 seconds:
Time to lose 15 seconds = \( \frac{8}{10} \times 15 \) hours
Simplifying the fraction:
Time = \( \frac{120}{10} \) = 12 hours
Hence, the watch will lose 15 seconds in 12 hours.
In simple words: Find out how many hours it takes to lose just 1 second by dividing. Then, multiply that number by 15 to get the total hours.

Exam Tip: Keep the fraction as \( \frac{8}{10} \) instead of converting to a decimal, as multiplying by 15 makes it very easy to simplify to a whole number of hours.

Question 4. In 2 days and 20 hours, a watch gains 20 sec ; find how much time will the watch take to gain 35 sec. ?
Answer:
First, we convert the time into hours:
Total hours = (2 days \( \times \) 24 hours/day) + 20 hours = 48 + 20 = 68 hours
We know that the watch gains 20 seconds in 68 hours.
First, we find the time needed to gain 1 second:
Time to gain 1 second = \( \frac{68}{20} \) hours
Now, we calculate the time required to gain 35 seconds:
Time to gain 35 seconds = \( \frac{68}{20} \times 35 \) hours
Simplifying by dividing 35 and 20 by 5, we get \( \frac{7}{4} \):
Time to gain 35 seconds = \( \frac{68 \times 7}{4} \) = 17 \( \times \) 7 = 119 hours
To convert 119 hours back into days, we divide by 24:
119 hours = 4 days and 23 hours
Therefore, the watch will take 4 days and 23 hours to gain 35 seconds.
In simple words: Turn the days into hours first. Find how many hours it takes to gain 1 second, multiply by 35, and then convert that total back into days and hours.

Exam Tip: Be sure to convert the final answer back to days and hours by dividing by 24, as leaving the answer only in hours might lose you marks.

Question 5. 50 men mow 32 hectares of land in 3 days. How many days will 15 men take to mow it?
Answer:
This is an inverse variation problem since the area of land (32 hectares) is constant, and fewer men will take more days to complete the work.
Time taken by 50 men = 3 days
If only 1 man were to mow the land, it would take him much longer:
Time taken by 1 man = 3 \( \times \) 50 = 150 days
Now, we calculate the time required for 15 men to complete the same work:
Time taken by 15 men = \( \frac{3 \times 50}{15} \) days
Time taken = \( \frac{150}{15} \) = 10 days
Hence, 15 men will take 10 days to mow the land.
In simple words: Since the amount of land is the same, fewer workers means the job takes longer. Find the total days needed for one person, and then divide that by 15.

Exam Tip: Since the area of land (32 hectares) is constant, it does not enter into the calculation. Focus only on the inverse relationship between the number of men and days.

Question 6. The wages of 10 workers for a six days week are Rs, 1,200. What are the one day wages: (i) of one worker ? (ii) of 4 workers?
Answer:
(i) We are given that 10 workers earn Rs. 1,200 in 6 days.
First, we find the amount earned by 10 workers in 1 day:
Earnings of 10 workers in 1 day = Rs. \( \frac{1200}{6} \) = Rs. 200
Now, we find the daily earning of 1 worker:
Earnings of 1 worker in 1 day = Rs. \( \frac{200}{10} \) = Rs. 20
(ii) Now, we calculate the total daily earnings for 4 workers:
Earnings of 4 workers in 1 day = Rs. 20 \( \times \) 4 = Rs. 80
In simple words: First, find how much 10 workers make in a single day by dividing by 6. Then, find what one worker makes by dividing by 10. Finally, multiply that single rate by 4 to get the wages of 4 workers.

Exam Tip: Work in steps: first divide the total wages by the number of days to get the daily rate for the group, and then divide by the group size to get the individual daily rate.

Question 7. If 32 apples weigh 2 kg 800 g. How many apples will there be in a box, containing 35 kg of apples ?
Answer:
First, we convert the weight into a uniform unit of kilograms:
Weight of 32 apples = 2 kg 800 g = 2.8 kg
We need to find the number of apples in a box weighing 35 kg.
First, we calculate how many apples weigh 1 kg:
Number of apples per kg = \( \frac{32}{2.8} \)
Now, we calculate the number of apples in 35 kg:
Number of apples = \( \frac{32}{2.8} \times 35 \)
We can simplify this by multiplying the numerator and denominator by 10 to clear the decimal:
Number of apples = \( \frac{32 \times 35 \times 10}{28} \)
Simplifying the fraction by dividing both 35 and 28 by 7 gives \( \frac{5}{4} \):
Number of apples = \( \frac{32 \times 10 \times 5}{4} \) = 8 \( \times \) 10 \( \times \) 5 = 400 apples
So, there are 400 apples in the box.
In simple words: First, convert 2 kg 800 g to 2.8 kg. Find out how many apples make 1 kg by dividing. Then, multiply that rate by 35 kg to find the total number of apples.

Exam Tip: Convert mixed units (like kg and g) into decimals of a single unit (like kg) before you start. This prevents confusion and keeps the calculations simple.

Question 8. A truck uses 20 litres of diesel for 240 km. How many litres will be needed for 1200 km?
Answer: To cover a distance of 240 km, the amount of diesel required is 20 litres.
Thus, to travel 1 km, the truck requires: \[ \frac{20}{240} \text{ litres} \]
Therefore, for a journey of 1200 km, the total fuel needed is: \[ \frac{20}{240} \times 1200 \text{ litres} = 100 \text{ litres} \]
In simple words: Find how much fuel the truck uses for just 1 km. Then multiply that by the total distance of 1200 km to find the total fuel.

Exam Tip: Identify the relationship between distance and fuel first - since they increase together, you can set it up as a direct proportion.

Question 9. A garrison of 1200 men has provisions for 15 days. How long will the provisions last if the garrison be increased by 600 men ?
Answer: Food supplies for 1200 men are enough to last for 15 days.
If only 1 man were to consume these supplies, they would last for: \[ 15 \times 1200 \text{ days} \]
When the number of men is increased by 600, the new total is: \[ 1200 + 600 = 1800 \text{ men} \]
So, for 1800 men, the supplies will last for: \[ \frac{15 \times 1200}{1800} \text{ days} \] \[ = 10 \text{ days} \]
In simple words: If there are more men to feed, the food will finish much faster. We find how long the food lasts for 1 man first, and then divide that by the new total number of men.

Exam Tip: This is an inverse proportion problem because more men means fewer days of food. Be careful not to use direct multiplication.

Question 10. A camp has provisions for 60 pupil for 18 days. In how many days, the same provisions will finish off if the strength of the camp is increased to 72 pupil ?
Answer: For 60 students, the food supply lasts for 18 days.
For just 1 student, the same supply would last for: \[ 18 \times 60 \text{ days} \]
If the total count of students rises to 72, the food supply will last for: \[ \frac{18 \times 60}{72} \text{ days} \] \[ = 15 \text{ days} \]
In simple words: Since there are more pupils in the camp, the food will finish sooner. We calculate how many days the food lasts for just 1 pupil, then divide it by 72 pupils.

Exam Tip: Always calculate the total food units first (pupils multiplied by days) to easily find the answer for any new number of pupils.

Exercise 7(C)

Question 1. A can do a piece of work in 6 days and B can do it in 8 days. How long will they take to complete it together ?
Answer: Time taken by A to finish the work is 6 days.
So, the fraction of work completed by A in 1 day is: \[ \frac{1}{6} \]
Time taken by B to finish the same task is 8 days.
Thus, the fraction of work completed by B in 1 day is: \[ \frac{1}{8} \]
Combined work done by both A and B in a single day: \[ = \frac{1}{6} + \frac{1}{8} \] \[ = \frac{4 + 3}{24} = \frac{7}{24} \]
Therefore, the total time required for both to finish the task together is: \[ \frac{24}{7} \text{ days} = 3\frac{3}{7} \text{ days} \]
In simple words: Find out what part of the work each person does in one single day. Add those parts together, and flip the final fraction to find the total days they need together.

Exam Tip: Do not simply average the days (like (6+8)/2 = 7 days). Instead, always add their one-day work rates as fractions.

Question 2. A and B working together can do a piece of work in 10 days B alone can do the same work in 15 days. How long will A alone take to do the same work ?
Answer: Working together, A and B complete the task in 10 days, while B alone finishes it in 15 days.
This means the work done by A and B together in 1 day is: \[ \frac{1}{10} \]
The work done by B alone in 1 day is: \[ \frac{1}{15} \]
To find the work done by A in 1 day, we subtract B's daily progress from their combined daily progress: \[ = \frac{1}{10} - \frac{1}{15} \] \[ = \frac{3 - 2}{30} = \frac{1}{30} \]
Hence, A alone will complete the task in: \[ 30 \text{ days} \]
In simple words: Subtract B's daily work fraction from the combined daily work fraction of both. The result is A's daily work fraction, which we flip to get the total days.

Exam Tip: Keep track of your fractions carefully when subtracting, and remember to find a common denominator.

Question 3. A can do a piece of work in 4 days and B can do the same work in 5 days. Find, how much work can be done by them working together in :
(i) one day
(ii) 2 days.
What part of work will be left, after they have worked together for 2 days ?
Answer: A finishes the task in 4 days, which means A does \( \frac{1}{4} \) of the work in 1 day.
B completes the same task in 5 days, so B does \( \frac{1}{5} \) of the work in 1 day.
(i) Together, the portion of work they complete in 1 day is: \[ \frac{1}{4} + \frac{1}{5} = \frac{5 + 4}{20} = \frac{9}{20} \]
(ii) In 2 days, the portion of work they complete is: \[ \frac{9}{20} \times 2 = \frac{9}{10} \]
The fraction of work remaining after these 2 days is: \[ 1 - \frac{9}{10} = \frac{10 - 9}{10} = \frac{1}{10} \]
In simple words: Add their daily fractions to find what they do in 1 day. Multiply that by 2 to get what they do in 2 days. Then subtract this from 1 to find what is left.

Exam Tip: Remember that "whole work" is always represented by the number 1. Subtract the completed fraction from 1 to find the remaining work.

Question 4. A and B take 6 hours and 9 hours respectively to complete a work. A works for 1 hour and then B works for two hours.
(i) How much work is done in these 3 hours ?
(ii) How much work is still left ?
Answer: A requires 6 hours to complete the job, meaning A's work in 1 hour is \( \frac{1}{6} \).
B requires 9 hours to complete the job, meaning B's work in 1 hour is \( \frac{1}{9} \).
Therefore, in 2 hours, B completes: \[ \frac{1}{9} \times 2 = \frac{2}{9} \text{ of the work} \]
(i) The total work finished during these 3 hours (1 hour by A and 2 hours by B) is: \[ \frac{1}{6} + \frac{2}{9} = \frac{3 + 4}{18} = \frac{7}{18} \]
(ii) The remaining portion of the work is: \[ 1 - \frac{7}{18} = \frac{18 - 7}{18} = \frac{11}{18} \]
In simple words: Calculate how much work A does in 1 hour and how much B does in 2 hours. Add these two parts together, and then subtract the sum from 1 to see how much is left.

Exam Tip: Watch out for the different working times for each person - do not assume they worked together for the entire duration.

Question 5. A, B and C can do a piece of work in 12, 15 and 20 days respectively. How long will they take to do it working together ?
Answer: The individual times taken by A, B, and C to complete the work are 12 days, 15 days, and 20 days respectively.
This gives their 1-day work as:
- A's 1-day progress = \( \frac{1}{12} \)
- B's 1-day progress = \( \frac{1}{15} \)
- C's 1-day progress = \( \frac{1}{20} \)
Combined 1-day progress when working together: \[ = \frac{1}{12} + \frac{1}{15} + \frac{1}{20} \] \[ = \frac{5 + 4 + 3}{60} = \frac{12}{60} = \frac{1}{5} \]
Thus, working together, they will finish the task in: \[ 5 \text{ days} \]
In simple words: Find the fraction of work each person does in 1 day. Add all three fractions together, and then turn the final fraction upside down to get the total days.

Exam Tip: Double-check your LCM calculations when adding three fractions to avoid simple arithmetic mistakes.

Question 6. Two taps can fill a cistern in 10 hours and 8 hours respectively. A third tap can empty it in 15 hours. How long will it take to fill the empty cistern, if all of them are opened together ?
Answer: The first tap takes 10 hours to fill the tank, so it fills \( \frac{1}{10} \) of the tank in 1 hour.
The second tap takes 8 hours to fill the tank, so it fills \( \frac{1}{8} \) of the tank in 1 hour.
The third tap can completely empty the tank in 15 hours, so it drains \( \frac{1}{15} \) of the tank in 1 hour.
When all three taps operate simultaneously, the net change in 1 hour is: \[ = \frac{1}{10} + \frac{1}{8} - \frac{1}{15} \] \[ = \frac{12 + 15 - 8}{120} = \frac{19}{120} \]
Consequently, the time required to completely fill the empty tank is: \[ \frac{120}{19} \text{ hours} = 6\frac{6}{19} \text{ hours} \]
In simple words: The first two taps add water to the tank, while the third tap takes it away. Add the first two fractions and subtract the third one, then flip the result.

Exam Tip: Remember to use a minus sign for the tap that empties the tank, as it performs negative work.

Question 7. Mohit can complete a work in 50 days, whereas Anuj can complete the same work in 40 days.
Find:
(i) work done by Mohit in 20 days.
(ii) work left after Mohit has worked on it for 20 days.
(iii) time taken by Anuj to complete the remaining work.
Answer: Mohit can complete the whole task in 50 days, while Anuj can do it in 40 days.
This means Mohit does \( \frac{1}{50} \) of the work in a single day, and Anuj does \( \frac{1}{40} \) in a single day.
(i) Portion of work completed by Mohit in 20 days: \[ = \frac{1}{50} \times 20 = \frac{2}{5} \]
(ii) Remaining portion of the work: \[ = 1 - \frac{2}{5} = \frac{3}{5} \]
(iii) Time required for Anuj to finish this remaining \( \frac{3}{5} \) of the work: \[ = 40 \times \frac{3}{5} \text{ days} = 24 \text{ days} \]
In simple words: Mohit works for 20 days, so we find his progress and subtract it from 1. Then we calculate how many days Anuj needs to finish this leftover part.

Exam Tip: To find the time taken for the remaining work, multiply the leftover fraction by the total days the second person takes for the full job.

Question 8. Joseph and Peter can complete a work in 20 hours and 25 hours respectively.
Find :
(i) work done by both together in 4 hrs.
(ii) work left after both worked together for 4 hrs.
(iii) time taken by Peter to complete the remaining work.
Answer: Joseph takes 20 hours to finish the task, while Peter needs 25 hours.
Thus, Joseph completes \( \frac{1}{20} \) of the task in 1 hour, and Peter completes \( \frac{1}{25} \) of the task in 1 hour.
Their combined progress in 1 hour: \[ = \frac{1}{20} + \frac{1}{25} = \frac{5 + 4}{100} = \frac{9}{100} \]
(i) Total work completed by both together in 4 hours: \[ = \frac{9}{100} \times 4 = \frac{9}{25} \]
(ii) Portion of work still remaining: \[ = 1 - \frac{9}{25} = \frac{16}{25} \]
(iii) Time taken by Peter to complete the leftover \( \frac{16}{25} \) of the work: \[ = 25 \times \frac{16}{25} \text{ hours} = 16 \text{ hours} \]
In simple words: Find out how much they both do in 1 hour and multiply it by 4. Subtract this from 1 to see the leftover part, then find how many hours Peter needs to finish it.

Exam Tip: Always make sure to calculate the combined rate first before finding the work done in a specific number of hours.

Question 9. A is able to complete 1/3 of a certain work in 10 hrs and B is able to complete 2/5 of the same work in 12 hrs.
Find:
(i) how much work can A do in 1 hour ?
(ii) how much work can B do in 1 hour ?
(iii) in how much time will the work be completed, if both work together.
Answer: A finishes \( \frac{1}{3} \) of the work in 10 hours.
Thus, the total time A requires to complete the entire job is: \[ = 10 \times 3 = 30 \text{ hours} \]
B finishes \( \frac{2}{5} \) of the work in 12 hours.
Thus, the total time B requires to complete the entire job is: \[ = \frac{12 \times 5}{2} = 30 \text{ hours} \]
(i) The fraction of work A can complete in 1 hour: \[ = \frac{1}{30} \]
(ii) The fraction of work B can complete in 1 hour: \[ = \frac{1}{30} \]
(iii) Combined work done by both A and B in 1 hour: \[ = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \]
Hence, if both work together, they will finish the task in 15 hours.
In simple words: First, find how long each person takes to complete the entire job alone. Then find their 1-hour work fractions, add them together, and flip the fraction.

Exam Tip: Ensure you find the time for "full work" first. For example, if 1/3 work takes 10 hours, full work takes 30 hours.

Question 10. Shaheed can prepare one wooden chair in 3 days and Shaif can prepare the same chair in 4 days. If they work together, in how many days will they prepare :
(i) one chair ?
(ii)14 chairs of the same kind?
Answer: Shaheed requires 3 days to build a chair, so Shaheed's 1-day progress is \( \frac{1}{3} \).
Shaif requires 4 days to build a chair, so Shaif's 1-day progress is \( \frac{1}{4} \).
Their combined progress in a single day: \[ = \frac{1}{3} + \frac{1}{4} = \frac{4 + 3}{12} = \frac{7}{12} \]
(i) Number of days they take to prepare 1 chair together: \[ = \frac{12}{7} \text{ days} = 1\frac{5}{7} \text{ days} \]
(ii) Number of days they take to build 14 such chairs: \[ = \frac{12}{7} \times 14 \text{ days} = 24 \text{ days} \]
In simple words: First, find how long they take to make 1 chair together by adding their daily work fractions. Then, multiply that time by 14 to find the total days for 14 chairs.

Exam Tip: Keep your fractions in improper form (like 12/7) during calculation, as it makes multiplying by the number of chairs much easier.

Question 11. A, B and C together finish a work in 4 days. If A alone can finish the same work in 8 days and B in 12 days, find how long will C take to finish the work.
Answer: Working together, A, B, and C complete the job in 4 days, so their combined progress in 1 day is \( \frac{1}{4} \).
A alone completes the job in 8 days, so A's 1-day progress is \( \frac{1}{8} \).
B alone completes the job in 12 days, so B's 1-day progress is \( \frac{1}{12} \).
To find C's 1-day progress, we subtract the sum of A's and B's individual progress from the collective daily progress: \[ = \frac{1}{4} - \left(\frac{1}{8} + \frac{1}{12}\right) \] \[ = \frac{1}{4} - \left(\frac{3 + 2}{24}\right) \] \[ = \frac{1}{4} - \frac{5}{24} \] \[ = \frac{6 - 5}{24} = \frac{1}{24} \]
Hence, C alone will take 24 days to finish the work.
In simple words: Add the daily work fractions of A and B. Subtract this from their combined daily rate with C to find C's daily fraction, then flip it.

Exam Tip: Group the known individual rates in parentheses first when subtracting them from the total combined rate to maintain a clear path of calculation.