Practicing with Maths Mela Class 5 Solutions Chapter 8 Weight and Capacity Question Answer NCERT Solutions improves a student’s confidence in the subject.
Class 5 Maths Chapter 8 Weight and Capacity Question Answer Solutions
Weight and Capacity Class 5 Maths Solutions
Class 5 Maths Chapter 8 Solutions
Check! Check! (NCERT Pg 104)
Question 1.
Anu has recorded the weights of the items in her house. Check if she has recorded them correctly by putting a tick against them if they look correct.
(i) Iron Almirah – 40 g
(ii) Bed – 60 kg
(iii) Rice Bag – 5 kg
(iv) Sofa -30 g
(v) Bucket – 1 kg 800 g
(vi) Water Bottle -650 g
(vii) Refrigerator – 50 g
Answer:
(i) Incorrect, an iron almirah is much heavier than 40 grams. It should be in kilogram (kg).
(ii) Correct, a bed can reasonably weight around 60 kg.
(iii) Correct, a bag of rice can commonly be 5 kg.
(iv) Incorrect, a sofa is much heavier than 30 grams. It should be in kilograms.
(v) Correct, a bucket, when empty or with a small amount of content, can weight around 1 kg 800 gm .
(vi) Correct, a water bottle can weight around 650 grams when empty or partially filled.
(vii) Incorrect, a refrigerator is significantly heavier than 50 grams. It should be in kg.
Let Us Do (NCERT Pg 104)
Question 2.
Read the scales. Write the correct weight in the space given below.
Answer:
(i) Here, each small divisions represent 100 grams. The pointer exactly on the 6th small division after 0 kg.
So, the reading = 0 kg + 6 × 100 g
= 600 g
(ii) Here, each small divisions represent 100 grams. The pointer exactly on the 8th small division after 1 kg.
So, the reading = 1 kg + 8 × 100 g
= 1 kg 800 gm
(iii) Here, each small division represent 100 grams. The pointer exactly on the 5th small division after 2 kg.
So, the reading = 2 kg + 5 × 100 g
= 2 kg 500 gm
(iv) Here, each small divisions represent 100 grams. The pointer exactly on the 6 th small division after 2 kg.
So, reading = 2 kg + 6 × 100 g = 2 kg 600 g
(v) Here, each big divisions represent 50 g and each small division represent 10 g. The pointer exactly on the 3rd big division after 0 kg.
So, reading = 0 + 3 × 50 g = 150 g
(vi) Here, each big division represent 50 g and each small division represent 10 g. The pointer exactly on the 1st small division after 500 g and 3rd big division.
So, reading = 500 gm + 3 × 50 g + 10 g
= 500 gm + 150 g + 10 g = 660 gm
Different Units but Same Measure (NCERT Pg 105)
Question 3.
I. Match the bags that have the same weights. You can use the double number line given below.
II.
Answer:
I. We know that Tkg = 1000 g
(i) (e) 5 kg = 5 × 1000 g = 5000 g
(ii) (c) 10 kg = 10 × 1000 g = 10000 g
(iii) (a) 3 kg = 3 × 1000 g = 3000 g
(iv) (b) 6 kg = 6 × 1000 g = 6000 g
(v) (f) 25 kg = 25 × 1000 g = 25000 g
(vi) (d) 30 kg = 30 × 1000 g = 30000 g
II. 1000 g × 3 = 3000 g
8 kg = 8 × 1000 g = 8000 g
15000 g = \(\frac{15000}{1000}\) g = 15 kg
20 kg = 20 × 1000 g = 20000 g
25000 g = \(\frac{25000}{1000}\) kg = 25 kg
30 kg = 30 × 1000 g = 30000 g
Let Us Find (NCERT Pg 106)
Question 4.
Shamim and Rehan observed someone buying sugar weighing 5 kg 50 g. They thought of the quantity in grams. How much is it?
Answer:
We have, 5 kg 50 g = 5 × 1000 g + 50 g [∵ 1 kg = 1000 g]
= 5000 g + 50 g
= 5050 g
Therefore, Shamim is right because his calculation of 5050 g is correct, while Rehan’s calculation is 5500 g , which is incorrect.
Question 5.
I. Complete the conversions by filling in the blanks. You can use the double number line given below on which some numbers have been marked.
Answer:
3 kg = 3 × 1000 g = 3000 g [∵ 1 kg = 1000 g]
4000 g = 4 × 1000 g = 4 kg [∵ 1000 g = 1 kg]
kg = 7 × 1000 g = 7000 g [∵ 1 kg = 1000 g]
8 kg = 8 × 1000 g = 8000 g [∵ 1 kg = 1000 g]
10000 g = 10 × 1000 g = 10 kg [∵ 1000 g = 1 kg]
12000 g = 12 × 1000 g = 12 kg [∵ 1000 g = 1 kg]
II. (i) 7 kg 67 g = ……….. g
(ii) 3 kg 300 g = ……….. g
(iii) 8 kg 69 g = ……….. g
(iv) 10,760 g = ……….. kg ……….. g
(v) 4,080 g = ……….. kg ……….. g
(vi) 12,042 g = ……….. kg ……….. g
Answer:
(i) We have, 7 kg 67 g = 7 × 1000 g + 67 g [∵ 1 kg = 1000 g]
= 7000 g + 67 g = 7067 g
(ii) We have, 3 kg 300 gm
= 3 × 1000 g + 300 g [∵ 1 kg = 1000 g]
= 3000 g + 300 g = 3300 g
(iii) We have, 8 kg 69 g = 8 × 1000 g + 69 g [∵ 1 kg = 1000 g]
= 8000 g + 69 g = 8069 g
(iv) We have, 10760 g = 10000 g + 760 g
= 10 × 1000 g + 760 g
= 10 kg 760 g [∵ 1000 g = 1 kg]
(v) We have, 4080 g = 4000 g + 80 g
= 4 × 1000 g + 80 g
= 4 kg 80 g [∵ 1000 g = 1 kg]
(vi) We have, 12042 g = 12000 g + 42 g
= 12 × 1000 g + 42 g
= 12 kg 42 g [∵ 1000 g = 1 kg]
Comparison between Different Weights (NCERT Pg 107)
Question 6.
Harpreet’s family planned a picnic over the weekend. Her mother and father packed different food items to take along. The following is the list of fruits they carried.
Among the fruits they carried, which one has the
(i) highest weight? ………..
(ii) least weight? ………..
(ii) arrange the items in descending order of their weight. ………..
Answer:
Given, the weight of Watermelon
= 3 kg
= 3 × 1000 g [∵ 1 kg = 1000 g]
= 3000 g
Weight of Pineapple = 1 kg 750 g
= 1 × 1000 g + 750 g [∵ 1 kg = 1000 g]
= 1750 g
Weight of Apples = 1 kg 250 g
= 1 × 1000 g + 250 g [∵ 1000 g = 1 kg]
= 1250 g
and the weight of Mangoes = 2 kg
= 2 × 1000 g [∵ 1000 g = 1 kg]
= 2000 g
(i) Clearly, the highest weight is 3000 g or 3 kg, which corresponds to Watermelon.
(ii) Clearly, the least weight is 1250 g or 1 kg 250 g, which corresponds to Apples.
(iii) The arrangement of the items in descending order is
Watermelon (3 kg) > Mangoes (2 kg) > Pineapple (1 kg 750 g) > Apples (1 kg 250 g)
Question 7.
Compare the weights using <, = ,> signs.
Answer:
(i) We have, 1 kg 600 g = 1 × 1000 g + 600 g [∵ 1 kg = 1000 g]
= 1000 g + 600 g
= 1600 g
∵ 1600 g < 1705 g
∴ 1 kg 600 g < 1700 g.
(ii) We have, 1 kg 600 g
= 1 × 1000 g + 600 g [∵ 1 kg = 1000 g]
and 1600 g 1 kg 60 g
= 1 × 1000 g + 60 g [∵ 1 kg = 1000 g]
= 1060 g ∵ 1600 g > 1060 g
∴ 1 kg 600 g > 1 kg 60 g.
(iii) We have, 10 kg 35 g = 10 × 1000 g + 35 g [∵ 1 kg = 1000 g]
= 10000 g + 35 g
= 10035 g
∴ 10 kg 35 g = 100035 g.
(iv) We have, 1 kg 600 g = 1 × 1000 g + 600 g [∵ 1 kg = 1000 g]
= 1000 g + 600 g
= 1600 g
and 2 kg 500 g = 2 × 1000 g + 500 g [∵ 1 kg = 1000 g]
= 2000 g + 500 g = 2500 g
∵ 1600 g < 2500 g
∴ 1 kg 600 g < 2 kg 500 g.
(v) We have, 5 kg 50 g = 5 × 1000 g + 50 g [∵ 1 kg = 1000 g]
= 5000 g + 50 g = 5050 g and 4 kg 500 g
= 4 × 1000 g + 500 g [∵ 1 kg = 1000 g]
= 4000 g + 500 g = 4500 g
∵ 5050 g > 4500 g
∴ 5 kg 50 g > 4 kg 500 g.
(vi) We have, 900 g + 7000 g = 7900 g
And 7 kg + 900 g = 7 × 1000 g + 900 g
[∵ 1 kg = 1000 g]
= 7000 g + 900 g = 7900 g
∴ 900 g + 7000 g = 7 kg + 900 g.
Let Us Find (NCERT Pg 108-109)
Question 8.
If a sugar sachet weighs 5 g , how much will it be in milligrams?
Answer:
Given, the weight of a sugar sachet = 5 g So, the weight of the sugar sachet in
milligrams = 5 × 1000 mg [∵ 1 g = 1000 mg]
= 5000 mg
Question 9.
Complete the double number line below appropriately.
Answer:
Question 10.
A goldsmith has made an ornament weighing 10 g 500 mg. What will its weight be in milligrams? ………..
Answer:
Given, the weight of an ornament = 10 g 500 mg
So, the weight of the ornament in milligrams
= 10 × 1000 mg + 500 mg [∵ 1 g = 1000 mg]
= 10000 mg + 500 mg
= 10500 mg
Question 11.
Compare the weights using <, =, > signs.
(i) 20 g ……….. 200 mg
(ii) 16 g 50 mg ……….. 50 g 16 mg
(iii) 2,010 mg ……….. 2 g 100 mg
(iv) 9,000 mg ……….. 90 g
(v) 5,000 g ……….. 7,500 g
(vi) 800 mg + 88 mg ……….. 880 mg + 8 mg
Answer:
(i) We have, 20 g = 20 × 1000 mg = 20000 mg [∵ 1 g = 1000 mg]
∵ 20000 mg > 200 mg
∴ 20 g > 200 mg.
(ii) We have, 16 g 50 mg = 16 × 1000 mg + 50 mg [∵ 1 g = 1000 mg]
= 16000 mg + 50 mg
= 16050 mg
and 50 g 16 mg = 50 × 1000 mg + 16 mg
[∵ 1 g = 1000 mg]
= 50000 mg + 16 mg
= 50016 mg
∵ 16050 mg < 50016 mg
∴ 16 g 50 mg < 50 g 16 mg.
(iii) We have, 2010 mg
and 2 g 100 mg = 2 × 1000 mg + 100 mg [1 g = 1000 mg]
= 2000 mg + 100 mg
= 2100 mg
∵ 2010 mg < 2100 mg
∴ 2010 mg < 2 g 100 mg.
(iv) We have, 9000 mg
and 90 g = 90 × 1000 mg = 90000 mg
∵ 9000 mg < 90000 mg
∴ 9000 mg < 90 g.
(v) We have, 5000 g and 7500 g. Clearly, 5000 g < 7500 g.
(vi) We have, 800 mg + 88 mg = 888 mg and 880 mg + 8 mg = 888 mg.
∴ 800 mg + 88 mg = 880 mg + 8 mg.
Question 12.
Observe the pictures given below and fill in the blanks.
Answer:
Given, the weight of elephant = 5000 kg
So, weight of whale
= Weight of elephant × 40
= 5000 kg × 40
= 200000 kg
Question 13.
Answer the following questions.
(i) 5,000 kg = ……….. quintals = ……….. tonnes
(ii) 9,000 kg = ……….. quintals
(iii) ……….. kg = 8 tonnes
Answer:
King’s Weight (NCERT Pg 109)
Question 14.
In a kingdom, the king donates wheat grains equal to 10 times his weight on his birthday.
(i) If he donates 800 kg of wheat grain this birthday, what is his current weight? ……….. kg.
(ii) If he had donated 780 kg of wheat grain on his last birthday, what was his weight last year? ……….. kg.
(iii) How much weight did he gain in a year until this birthday? ……….. kg.
Answer:
Since, the king denotes wheat grains equal to 10 times his weight.
(i) Given, the quantity of wheat grain denoted by him = 800 kg.
So, his current weight = \(\frac{800}{10}\) = 80 kg.
(ii) Given, the quantity of wheat grain denoted by him last year = 780 kg.
So, his weight last year = \(\frac{780}{10}\) = 78 kg.
(iii) We have, the current weight of king = 80 kg and the weight of king last year = 78 kg [from part (i) and (ii)]
So, weight gained = 80 kg – 78 kg = 2 kg.
Let Us Do (NCERT Pg 111)
Question 15.
A restaurant owner uses 5 kg 200 g, 8 kg 900 g and 12 kg 600 g of onions over 3 days. What is the total weight of onions used by the restaurant owner in 3 days?
Answer:
Given, the weight of onion used by restaurant owner on day 1
= 5 kg 200 g = 5 × 1000 g + 200 g [∵ 1 kg = 1000 g]
= 5200 g
On day 2 = 8 kg 900 g = 8 × 1000 g + 900 g [∵ 1 kg = 1000 g]
= 8900 g
On day 3 = 12 kg 600 g = 12 × 1000 g + 600 g [∵ 1 kg = 1000 g]
= 12600 g
So, the total weight of onions used by the restaurant owner in 3 days
= 5200 g + 8900 g + 12600 g
= 26700 g = 26000 g + 700 g
= 26 × 1000 g + 700 g
= 26 kg 700 g [∵ 1000 g = 1 kg]
Question 16.
Aarav is helping his grandfather at the fruit stall. He lifts two baskets of apples, weighing 2 kg 100 g and 3 kg 950 g. What is the total weight of apples he lifted?
Answer:
Given, the weight of 1 st apple basket
= 2 kg 100 g = 2 × 1000 g + 100 g
= 2100 g [∵ 1 kg = 1000 g]
and the weight of 2 nd apple basket
= 3 kg 950 g = 3 × 1000 g + 950 g
= 3950 g [∵ 1 kg = 1000 g]
So, the total weight of apples, he lifted
= 2100 g + 3950 g = 6050 g
= 6000 g + 50 g
= 6 × 1000 g + 50 g
= 6 kg 50 g [∵ 1000 g = 1 kg]
Question 17.
4 kg 500 g of sand is used from a sack weighing 10 kg. How much sand is left in the sack?
Answer:
Given, the total weight of sack
= 10 kg
= 10 × 1000 g [∵ 1 kg = 1000 g]
= 10000 g
and the weight of used sand from sack
= 4 kg 500 g
= 4 × 1000 g + 500 g
= 4500 g
So, the weight of send is left in the sack
= 10000 g – 4500 g
= 5500 g
= 5000 g + 500 g
= 5 × 1000 g + 500 g
= 5 kg 500 g [∵ 1000 g = 1 kg]
Question 18
A rice sack weighs 9 kg 750 g. After some rice is used, it weighs 3 kg 700 g. How much rice was used?
Answer:
Given, the total weight of rice sack
= 9 kg 750 g
= 9 × 1000 g + 750 g [∵ 1 kg = 1000 g]
= 9750 g
and the weight of sack after used some rice
= 3 kg 700 g
= 3 × 1000 g + 700 g [∵ 1 kg = 1000 g]
= 3700 g
So, the weight of used rice
= 9750 g – 3700 g
= 6050 g
= 6000 g + 50 g
= 6 × 1000 g + 50 g
= 6 kg 50 g [∵ 1000 g = 1 kg]
Question 19.
A delivery truck delivered 17 kg 900 g of supplies in the morning and 12 kg 700 g in the afternoon. How much total supplies did it deliver?
Answer:
Given, the weight of supplies delivered by delivery truck in the morning
= 17 kg 900 g
= 17 × 1000 g + 900 g
[∵ 1 kg = 1000 g]
= 17900 g
and in the evening = 12 kg 700 g
= 12 × 1000 g + 700 g
[∵ 1 kg = 1000 g]
= 12700 g
So, the weight of total supplies delivered by delivery truck
= 17900 g + 12700 g
= 30600 g
= 30000 g + 600 g
= 30 × 1000 g + 600 g
= 30 kg 600 g [∵ 1000 g = 1 kg]
Question 20
A box of books weighs 14 kg 750 g . After removing some books, the weight of the box is 10 kg 500 g. What is the weight of the books removed?
Answer:
Given, the total weight of box
= 14 kg 750 g = 14 × 1000 g + 750 g
= 14750 g [∵ 1 kg = 1000 g]
and the weight of box after removing some books
= 10 kg 500 g = 10 × 1000 g + 500 g
= 10500 g [∵ 1 kg = 1000 g]
So, the weight of the removed books
= 14750 g – 10500 g = 4250 g
= 4000 g + 250 g
= 4 × 1000 g + 250 g
[∵ 1000 g = 1 kg]
= 4 kg 250 g
Question 21.
In a community kitchen of a Gurdwara, 65 kg of flour was purchased on one day. Out of this, 42 kg 275 g flour was used for preparing langar. The next day, an additional 52 kg 500 g of flour was bought. What is the total quantity of flour now available in the kitchen store?
Answer:
Given, the initial weight of flour
= 65 kg
= 65 × 1000 g [∵ 1 kg = 1000 g]
= 65000 g
and the weight of used flour
= 42 kg 275 g
= 42 × 1000 g + 275 g
[∵ 1 kg = 1000 g]
= 42275 g
So, the weight remaining flour
= 65000 g – 42275 g = 22725 g
Now, the weight of addition flour bought
= 52 kg 500 g
= 52 × 1000 g + 500 g
= 52500 g [∵ 1 kg = 1000 g]
Hence, the total weight of flour available in the kitchen store
= 22725 g + 52500 g = 75225 g
= 75000 g + 225 g
= 75 × 1000 g + 225 g
= 75 kg 225 g [∵ 1000 g = 1 kg]
More Operations on Weight (NCERT Pg 112)
Question 22.
A box of nuts weighing 4 kg 800 g is equally distributed into 4 smaller boxes. What is the weight of each small box in grams?
4 kg ÷ 4 = 1 kg and 800 g ÷ 4 = 200 g
So, 4 kg 800 g + 4 = 1 kg 200 g
We can also convert the quantity into grams and divide 4800 g + 4 = ?
Answer:
Given, the weight of box in kilograms
= 4 kg 800 g = 4 × 1000 g + 800 g
[∵ 1 kg = 1000 g]
= 4800 g.
Since, the total weight is equally distributed into 4 smaller boxes.
So, the weight of small box = \(\frac{4800 g}{4}\)
= 1200 g.
Let Us Do (NCERT Pg 112-113)
Question 23
The cost of some grocery items is given in the following table.
Find the total cost of each item.
Answer:
(i) Given, the cost of 1 kg rice = ₹ 60
So, the cost of 12 kg 500 g rice
= ₹ 60 × 12 kg + ₹ 60 × \(\frac{1}{2}\) kg
= ₹ 720 + ₹ 30 = ₹ 750
(ii) Given, the cost of 1 kg rice = ₹ 40
So, the cost of 7 kg 250 g rice
(iii) Given, the cost of 1 kg sugar = ₹ 45
So, the cost of 5 kg sugar = 5 × ₹ 45 = ₹ 225
(iv) Given, the cost of 1 kg Chana daL = ₹ 70
So, the cost of 3 kg 600 g Chana dal
(v) Given, the cost of 1 kg Besan = ₹ 60
So, the cost of 4 kg Beson = ₹ 60 × 4 = ₹ 240
(vi) Given, the cost of 1 kg Jaggery = ₹ 50
So, the cost of 1 kg 400 g
Question 24.
4 people need 500 g rice for a meal. How much rice will be needed for 8 people if they eat similar quantity of rice?
Answer:
Given, the quantity of rice needed for 4 people = 500 g
So, the quantity of rice needed by 8 people
= 2 × 500 g = 1000 g
[Since, the number of people is doubled. So, the amount of rice needed will also doubled.]
Question 25.
5 kg of tomatoes cost ₹ 73. How much will 10 kg of tomatoes cost?
Answer:
Given, the cost of 5 kg tomatoes = ₹ 73
So, the cost of 10 kg tomatoes = 2 × ₹ 73
= ₹ 146
[Since, the quantity of tomatoes is doubled.
So, the cost of tomatoes will also doubled]
Question 26.
Nitesh is a scrap dealer. How much would he have paid for
(i) 16 kg of old newspaper, if he paid ₹ 8 for every 1 kg of newspaper?
(ii) 20 kg iron, if he paid ₹ 200 for every 10 kg of iron?
(iii) 10 kg plastic,-if. he 7 paid ₹ 30 for 5 kg of plastic?
(iv) Make double number lines for answering (ii) and (iii).
Answer:
(i) Given, the cost of 1 kg newspaper = ₹ 8
So, the cost of 16 kg newspaper = 16 × ₹ 8
= ₹ 128
(ii) Given, the cost of 10 kg iron = ₹ 200
So, the cost of 20 kg iron = 2 × ₹ 200
= ₹ 400
[Since, the quantity of iron is doubled.
So, the cost of iron will also doubled.]
(iii) Given, the cost of 5 kg plastic = ₹ 30
So, the cost of 10 kg plastic = 2 × ₹ 30
= ₹ 60
[Since, the quantity of plastic is doubled.
So, the cost of plastic will also doubled.]
(iv) For answer (ii),
For answer (iii),
Measuring Capacity (NCERT Pg 113)
Question 27.
(i) You must have seen tea being prepared at your home. How much water and milk do we need to make 2 cups of tea?
(ii) Do we need 1 L of water to make 2 cups of tea?
(iii) Is 500 ml of water enough for 2 cups of tea?
Answer:
(i) Do yourself.
(ii) No, 1 L(1000 ml) of water is significantly more than needed for just 2 cups of tea.
(iii) Yes, 500 ml of water would be more than enough for 2 cups of tea.
Question 28.
A bucket can hold a maximum of 20 ml of water. Is this statement correct? Which unit should be used in such a situation?
Answer:
No, this statement is incorrect. 20 ml is a very small quantity, equivalent to about four teaspoons of water.
A bucket typically holds a much larger quantity of water.
For expressing the capacity of bucket, we should use the larger unit like litre (L).
Big to Small, Small to Big (NCERT Pg 113-114)
Question 29.
Ramiz brings a 500 ml water bottle to school. He drinks two bottles at school. How much water does he drink at school?
(i) Ramiz drinks ……….. ml + ……….. ml = ……….. ml.
(ii) Ramiz drinks ……….. L of water in a day.
Answer:
Given, the capacity of water bottle = 500 ml. Since, Ramiz drinks 2 water bottles at school.
(i) Ramiz in a day drinks water
= 500 ml + 500 ml = 1000 ml
(ii) Ramiz drinks water is a day in litres
= \(\frac{1000}{1000}\) L [∵ 1 ml = \(\frac{1}{1000}\) L]
= 1 L
Question 30.
Muskaan drinks 3 L of water in a day.
(i) How many times would she need to refill a 500 ml water-bottle? ………..
(ii) Muskaan drinks ……….. ml of water in a day.
Answer:
The quantity of water drunk by Muskaan in a day = 3 L
= 3 × 1000 ml [∵ 1 L = 1000 ml]
= 3000 ml
(i) The number of times would she need to refill a 500 ml water bottle
= \(\frac{3000 ml}{500 ml}\)
= 6 times.
(ii) Muskaan drinks 3000 ml of water in a day.
Question 31.
Write the total-eapacity of the followin containers ín each blank.
Answer:
(i) The containers show marking 1 L, 500 ml and 100 ml .
So, the total capacity = 1 L + 500 ml + 100 ml
= 1 L + 600 ml
= 1 L 600 ml
(ii) The containers shows marking 1 L and 500 ml.
So, the total capacity = 1 L + 500 ml
= 1 L 500 ml
(iii) The containers shows marking 100 ml, 100 ml and 500 ml.
So, the total capacity
= 100 ml + 100 ml + 500 ml
= 700 ml
(iv) The containers shows marking 1 L, 1 L and 100 ml.
So, the total capacity = 1 L + 1 L + 100 ml
= 2 L + 100 ml
= 2 L 100 ml
Different Units but Same Measure (NCERT Pg 114)
Question 32.
The Milkman’s Delivery
Khayal chacha delivers fresh cow milk to homes. Bhalerao’s family orders 2 L of milk everyday.
This family has a vessel marked in ml only. What mark will you see in the vessel corresponding to 2 L ?
Answer:
From the figure, the mark we will see in the vessel corresponding to 2 L is 2000 ml.
Question 33.
Khayal chacha đelivers the following amounts of milk each week to different families.
Answer:
(i) Given, the quantity of milk delivered to Arora’s family in a week in L = 8 L
So, the quantity of milk in ml = 8 × 1000 ml
[∵ 1 L = 1000 ml]
= 8000 ml
(ii) Given, the quantity of milk delivered to Nair’s family in a week in L = 14 L
So, the quantity of milk in ml = 14 × 1000 ml
[∵ 1 L = 1000 ml]
= 14000 ml
(iii) Given, the quantity of milk delivered to Shrivastava’s family in ml = 12000 ml
So, the quantity of milk in L = 12 × 1000 ml
= 12 L [∵ 1000 ml = 2 L]
(iv) Given, the quantity of milk delivered to Das’s family in ml = 20000 ml
So, the quantity of milk in L = 20 × 1000 ml
= 20 L [∵ 1000 ml = 1 L]
(v) Given, the quantity of milk delivered to Rao’s family in ml = 25000 ml
So, the quantity of milk in L = 25 × 1000 ml = 25 L [∵ 1000 ml = 1 L]
Let Us Think (NCERT Pg 115)
Question 34.
Mary and Daisy filled their bottle with 1 L 400 ml of water. They wondered about the capacity of the bottle in ml . How much is it?
Who do you think is correct and why?
Answer:
Given, the capacity of bottle
= 1 L 400 ml
= 1 × 1000 ml + 400 ml [∵ 1 L = 1000 ml]
= 1400 ml
Hence, Mary is right because her calculation of 1400 ml is correct, while Daisy’s calculation is 1040 ml , which is incorrect.
Question 35.
Convert and fill in the blanks appropriately. You can use the double number line given earlier,
(i) 3 L 8 ml = ……….. ml
(ii) 9 L 90 ml = ……….. ml
(iii) 14,075 ml = ……….. L ……….. ml
(iv) 8 L 86 ml = ……….. ml
(v) 12,200 ml = ……….. L ……….. ml
(vi) 18,350 ml = ……….. L ……….. ml
Answer:
(i) We have, 3 L 8 ml = 3 × 1000 ml + 8 ml
[∵ 1 L = 1000 ml]
= 3000 ml + 8 ml
= 3008 ml
(ii) We have, 9 L 90 ml = 9 × 1000 ml + 90 ml [∵ 1 L = 1000 ml]
= 9000 ml + 90 ml
= 9090 ml
(iii) We have, 14075 ml = 14000 ml + 75 ml
= 14 × 1000 ml + 75 ml
= 14 L 75 ml [∵ 1000 ml = 1 L]
(iv) We have, 8 L 86 ml = 8 × 1000 ml + 86 ml [∵ 1 L = 1000 ml]
= 8000 ml + 86 ml
= 8086 ml
(v) We have, 12200 ml = 12000 ml + 200 ml
= 12 × 1000 ml + 200 ml
= 12 L + 200 ml [∵ 1000 ml = 1 L]
= 12 L 200 ml
(vi) We have, 18350 ml = 18000 ml + 350 ml
= 18 × 1000 ml + 350 ml
= 18 L 350 ml
[∵ 1000 ml = 1 L]
Let Us Compare (NCERT Pg 115-116)
Question 36.
Kiran owns a petrol pump. She records the details of the sales of petrol in a day.
(i)
(ii) How much more fuel is bought for buses than for trucks?
(iii) What is the total quantity of fuel filled from the petrol pump on that day?
Answer:
(i) (a) Given, the quantity of fuel for one truck = 500 L
So, the total quantity of fuel for 3 trucks
= 3 × 500 L
= 1500 L
(b) Given, the quantity of fuel for one bus
= 300 L
So, the total quantity of fuel for 6 buses
= 6 × 300 L
= 1800 L
(c) Given, the quantity of fuel for one car
= 50 L
So, the total quantity of fuel for 10 cars
= 10 × 50 L
= 500 L
(d) Given, the quantity of fuel for one auto rickshaw = 8 L
So, the total quantity of fuel for 12 auto rickshaw = 12 × 8 L = 96 L
(e) Given, the quantity of fuel for one two-wheeler = 5 L
So, the total quantity of fuel for 25 two-wheeler = 25 × 5 L = 125 L
(ii) We have, the quantity of fuel for trucks = 1500 L
and the quantity of fuel for buses = 1800 L
So, difference = Total fuel for buses – Total fuel for trucks
= 1800 L – 1500 L
= 300 L
(iii) Total quantity of fuel filled from the petrol pump on that day
= 1500 L + 1800 L + 500 L + 96 L + 125 L
= 4021 L
Question 37.
Compare the following quantities using the signs <, = ,>.
Answer:
(i) We have, 5 L 600 ml = 5 × 1000 ml + 600 ml
[∵ 1 L = 1000 ml]
= 5600 ml
∵ 5600 ml > 5400 ml
∴ 5 L 600 ml > 5400 ml
(ii) We have, 10 L 100 ml = 10 × 1000 ml + 100 ml
[∵ 1 L = 1000 ml]
= 10000 ml + 100 ml
= 10100 ml
and 1 L 600 ml = 1 × 1000 ml + 600 ml
= 1600 ml
∵ 10100 ml > 1600 ml
∴ 10 L 100 ml > 1 L 600 ml
(iii) We have, 190 ml + 800 ml = 990 ml and 800 ml + 109 ml = 909 ml
∵ 990 ml > 909 ml
∴ 190 ml + 800 ml2 > 800 ml + 109 ml
(iv) We have, 3 L 600 ml = 3 × 1000 ml + 600 ml
[∵ 1 L = 1000 ml]
= 3000 ml + 600 ml
= 3600 ml
∴ 3 L 600 ml = 3600 ml
(v) Clearly, 4 L 50 ml < 4 L 500 ml
Let Us Solve (NCERT Pg 119)
Question 38
Riya is filling water bottles for a picnic. She fills one 2 L bottle and four 500 ml bottles. Her friend, Aarav fills three 750 ml bottles. Who filled more water, Riya or Aarav? How much more?
Answer:
Given, Riya fills one 2 L bottle and four 500 ml bottles.
So, the total water filled by Riya
= 2 L + (4 × 500 ml)
= 2 L + 2000 ml
= 2 × 1000 ml + 2000 ml
[∵ 1 L = 1000 ml]
= 4000 ml
Also given, Aarav fills three 750 ml bottles.
So, the total water filled by Aarav
= 3 × 750 ml
= 2250 ml
Cleraly, Riya filled more water.
Now, the required difference
= 4000 ml-2250 ml
= 1750 ml
= 1000 ml + 750 ml
= 1 L 750 ml [∵ 1 L = 1000 ml]
Question 39.
A bottle of milk is poured equally into 8 glasses, leaving 120 ml of milk in the bottle.
(i) If each glass has a capacity of 360 ml, what is the total capacity of 8 glasses?
(ii) How much milk was there in the bottle initially?
(iii) If 1 L of milk costs ₹ 40, how much will 3 L milk cost?
Answer:
(i) Given, the capacity of one glass
= 360 ml
So, the total capacity of 8 glasses
= 360 ml × 8
= 2880 ml
= 2000 ml + 880 ml
= 2 L 880 ml [∵ 1000 ml = 1 L]
(ii) We have, the quantity of milk poured into 8 glasses = 2880 ml
and the quantity of milk remaining in the bottle = 120 ml
So, the total milk initially in the bottle
= Milk poured + Milk remaining
= 2880 ml + 120 ml
= 3000 ml
= 3 L [∵ 1000 ml = 1 L]
(iii) Given, the cost of 1 L of milk = ₹ 40
So, the cost of 3 L milk = 3 × ₹ 40 = ₹ 120
Question 40
A juice vendor has a 5 L container of orange juice. Each glass has a capacity 250 ml.
(i) How many full glasses can he serve before the container becomes empty?
(ii) If he has already served 10 glasses, how much juice is left?
(iii) If 250 ml of juice is sold at ₹ 25, how much will he earn by selling 5 L juice?
Answer:
(i) Given, the total quantity of orange juice
= 5 L
= 5 × 1000 ml[∵ 1 L = 1000 ml]
= 5000 ml
and the capacity of one glass = 250 ml
So, the number of glasses he can serve
= \(\frac{5000 ml}{250 ml}\) = 20 glasses
(ii) The amount of juice served
= 10 glasses × 250 ml
= 2500 ml
∵ Amount of juice is left
= 5000 ml – 2500 ml
= 2500 ml
= 2000 ml + 5000 ml
= 2 × 1000 ml + 500 ml
= 2 L 500 ml [∵ 1000 ml = 1 L]
(iii) Given, the cost of 250 ml juice = ₹ 25
So, the cost of 1 ml juice = ₹ \(\frac{25}{250}\) ml
= ₹ \(\frac{1}{10}\)
Now, the cost of 5 L or 5000 ml juice
= 5000 ml × ₹ \(\frac{1}{10}\)
= ₹ 500
Question 41.
In a factory, 8 L 400 ml of oil needs to be equally poured into 7 containers for storage. How much oil will each container hold?
Answer:
Given, the total quantity of oil
= 8 L 400 ml
= 8 × 1000 ml + 400 ml
[∵ 1 L = 1000 ml]
= 8400 ml
Since, the oil is equally poured into 7 containers.
So, the quantity of oil hold by each container
= \(\frac{8400 ml}{7}\) = 1200 ml
= 1000 ml + 200 ml
= 1 L 200 ml [∵ 1000 ml = 1 L]
Question 42.
If one container can hold 1L 75 ml of buttermilk, how much buttermilk will be there in 8 such containers?
Answer:
Given, the quality of buttermilk hold by one container
= 1 L 75 ml
So, the quantity of buttermilk hold by 8 such containers
= 8 × (1 L 75 ml)
= 8 × 1 L + 8 × 75 ml
= 8 L + 600 ml
= 8 L 600 ml