Important Questions: Fractions and Decimals | Mathematics (Maths) Class 7 PDF Download

Q1: Ritu ate at least (3/5) part of an apple, and then the remaining apple was eaten by ritu’s brother Shyam. How many parts of the apple did Shyam eat? Who had the larger share? And By how much?
Ans:
From the above question, it is given that,
The part of apple eaten by Ritu is equal to (3/5)
And the part of apple eaten by Shyam is = 1 – the part of apple eaten by Ritu is?
= 1 – (3/5)
The LCM of numbers 1 and 5 is = 5
Now, let us change these given fractions into an equivalent fraction by taking ten as the denominator number.
= [(1/1) × (5/5)] – [(3/5) × (1/1)]
= (5/5) – (3/5)
= (5 – 3)/5
= 2/5
Hence the part of apple eaten by Shyam is (2/5)
So, (3/5) is greater than (2/5); hence, Ritu ate the larger apple.
Now, the difference between the shares is = (3/5) – (2/5)
= (3 – 2)/5
= 1/5
Thus, Ritu’s share is greater than the share of Shyam by (1/5)

Q2: Michae finished colouring a picture on (7/12) hour. On the other hand, Vaibhav finished colouring the same picture in (3/4) hour. Who worked longer colouring the picture? By what fraction was it longer?
Ans:
From the above question, it is given that,
Time taken by Michae to colour the picture is = (7/12)
Time taken by Vaibhav to colour the picture is = (3/4)
The LCM of numbers 12, 4 = 12
Now, let us change these given fractions into an equivalent fractions by using 12 as the denominator number.
(7/12) = (7/12) × (1/1) = 7/12
The same method is applied to,
(3/4) = (3/4) × (3/3) = 9/12
As seen, (7/12) is less than (9/12)
Hence, (7/12) < (3/4)
Thus, Vaibhav worked for a longer time as compared.
Now, Vaibhav worked longer time by = (3/4) – (7/12)
= (9/12) – (7/12)
= (9 – 7)/12
= (2/12)
= (1/6) of an hour.

Q3: Vidya and Pratik went on a picnic. Their mother gave them a mineral water bottle that contained 5 litres of water. Vidya consumed around 2/5 of the water, and Pratik consumed the remaining water.
(i) How many litres of water did Vidya drink?
(ii) And What fraction of the total quantity of water did Pratik drink?
Ans:
(i) From the above question, it is given that,
The amount of water in the water bottle is = 5 litres.
The amount of water consumed by Vidya is = 2/5 of 5 litres.
= (2/5) × 5
= 2 litres
Hence, the total amount of water drank by Vidya is 2 litres.
(ii) From the above question, it is given that,
Amount of water present in the water bottle = 5 litres
Then,
The amount of water consumed by Pratik = (1 – water consumed by Vidya)
= (1 – (2/5))
= (5-2)/5
= 3/5
Hence the Total Amount of water consumed by Pratik = is 3/5 of 5 litres.
= (3/5) × 5
= 3 litres
So, the total amount of water consumed by Pratik is 3 litres.

Q4: Which of the following is greater:
(i) (2/7) of (3/4) or the fraction (3/5) of (5/8)
Ans:
We have seen that,
= (2/7) × (3/4) and the (3/5) × (5/8)
Hence, By the rule of Multiplication of the fraction,
The product of fraction is equal to (product of numerator)/ (product of denominator)
Then,
= (2/7) × (3/4)
= (2 × 3)/ (7 × 4)
= (1 × 3)/ (7 × 2)
= (3/14) … [i]
And,
= (3/5) × (5/8)
= (3 × 5)/ (5 × 8)
= (3 × 1)/ (1 × 8)
= (3/8) … [ii]
Now, convert the [i] and [ii] into like fractions,
LCM of 14 and 8 become 56
Now, let us change each of these given fractions into an equivalent fraction that has 56 as its denominator number.
[(3/14) × (4/4)] is equal to (12/56) [(3/8) × (7/7)] = (21/56)
Clearly, it is seen,
(12/56) is less than (21/56)
Hence,
(3/14) is less than (3/8)

(ii) (1/2) of (6/7) or the (2/3) of (3/7)
Ans:
We have seen that,
= (1/2) × (6/7) and the (2/3) × (3/7)
By the rule of Multiplication of the fraction,
Product of fraction = (product of numerator) divided by (product of denominator)
Then,
= (1/2) × (6/7)
= (1 × 6)/ (2 × 7)
= (1 × 3)/ (1 × 7)
= (3/7) … [i]
And,
= (2/3) × (3/7)
= (2 × 3)/ (3 × 7)
= (2 × 1)/ (1 × 7)
= (2/7) … [ii]
By comparing [i] and [ii],
Clearly,
(3/7) > (2/7)

Q5: Saili plants four saplings successively in a row in her garden. The distance between the two adjacent saplings is ¾ m. Now Find the distance between the first and the last planted sapling.
Ans:
From the above question, it is given that,
The distance between the two adjacent saplings is = ¾ m
The number of saplings which are planted by Saili in a row is = 4
Then, the number of the gap in the saplings = ¾ × 4
= 3
Hence The total distance between the first and the last saplings becomes = three × ¾
= (9/4) m
= 2 ¼ m
Thus, the distance between the first and the last saplings is two ¼ m.

Q6: Lipika always reads a book for one ¾ hour every day. She reads an entire book in 6 days. How many hours in total were required by her to finish the book?
Ans:
From the above question, it is clearly given that,
Lipika reads her book for = one ¾ hours every day, which is equal to 7/4 hours for six days.
The number of days she took to finish the entire book = was six days
Hence, the Total number of hours required by her to fully complete the book = (7/4) × 6
= (7/2) × 3
= 21/2
= 10 ½ hours
Thus, the total number of hours required by her to complete the book is 10 ½ hours.

Q7: A car runs around 16 km using 1 litre of petrol. How much distance will the car cover use two ¾ litres of petrol?
Ans:
From the above question, it is given that,
The total distance travelled by car in 1 litre of petrol is = 16 km.
Then,
The Total quantity of petrol becomes = two ¾ litre = 11/4 litres
The total distance travelled by car in 11/4 litres of petrol is = (11/4) × 16
= 11 × 4
= 44 km
Hence, the total distance travelled by car in 11/4 litres of petrol is 44 km.

Q8: Find the reciprocal of each of these following fractions. Also, Classify these reciprocals as proper fractions, improper fractions and whole numbers.
(i) 3/7
Ans:
Reciprocal of (3/7) is (7/3) [which is ((3/7) × (7/3)) = 1]
So, it is an improper fraction.
An improper fraction is defined as a fraction in which the numerator is always greater than its denominator.

(ii) 5/8
Ans:
Reciprocal of (5/8) is (8/5) [which is ((5/8) × (8/5)) = 1]
So, it is also an improper fraction.
An improper fraction is defined as a fraction in which the numerator is greater than its denominator.

(iii) 9/7
Ans:
Reciprocal of (9/7) is (7/9) [which is ((9/7) × (7/9)) = 1]
Hence, it is a proper fraction.
A proper fraction is defined as that fraction in which the denominator is greater than the numerator of their fraction.

(iv) 6/5
Ans:
Reciprocal of (6/5) is (5/6) [which is ((6/5) × (5/6)) = 1]
Hence, it is a proper fraction.
A proper fraction is defined as a fraction in which the denominator is greater than the numerator of any fraction.

(v) 12/7
Ans:
Reciprocal of (12/7) is (7/12) [which is ((12/7) × (7/12)) = 1]
Hence, it is a proper fraction.
A proper fraction is defined as a fraction in which the denominator is greater than the numerator of the given fraction.

(vi) 1/8
Ans:
Reciprocal of the fraction (1/8) is (8/1) or 8 as [∵ ((1/8) × (8/1)) = 1]
Hence, it is a whole number.
Whole numbers are the total collection of all positive integers, including the number 0.

(vii) 1/11
Ans:
Reciprocal of the fraction (1/11) is (11/1) or 11 which is [∵ ((1/11) × (11/1)) = 1]
So, it is a whole number.
Whole numbers are the total collection of all positive integers, including 0.

Q9: Find:
(i) (7/3) ÷ 2
Ans:
We have,
= (7/3) × reciprocal of 2
= (7/3) × (1/2)
= (7 × 1) / (3 × 2)
= 7/6

(ii) (4/9) ÷ 5
Ans:
We have,
= (4/9) × reciprocal of 5
= (4/9) × (1/5)
= (4 × 1) / (9 × 5)
= 4/45

(iii) (6/13) ÷ 7
Ans:
We have,
= (6/13) × reciprocal of 7
= (6/13) × (1/7)
= (6 × 1) / (13 × 7)
= 6/91

Q10: Which of the following is greater?
(i) 0.5 or 0.05
Ans:
By comparing the whole number, we get 0 = 0
By comparing the tenths place digit, we get 5 > 0
∴ 0.5 > 0.05

(ii) 0.7 or 0.5
Ans:
By comparing the whole number, 0 = 0
By comparing the tenths place digit we receive, 7 > 5
∴ 0.7 > 0.5

(iii) 7 or 0.7
Ans:
By comparing these whole numbers, 7 > 0
∴ 7 > 0.7

(iv) 1.37 or 1.49
Ans:
By comparing these whole numbers, 1 = 1
By comparing the tenths place digit, we get, 3 < 4
∴ 1.37 < 1.49

(v) 2.03 or 2.30
Ans:
By comparing the whole number, 2 = 2
By comparing the tenths place digit, we get, 0 < 3
∴ 2.03 < 2.30

(vi) 0.8 or 0.88
Ans:
By comparing these whole numbers, 0 = 0
By comparing the tenths place digit, we get, 8 = 8
Also, by comparing the hundredths place digit, 0 < 8
∴ 0.8 < 0.88

Q11: Express as rupees as decimals:
(i) 7 paise
Ans:
We know that,
= Rs. 1 = 100 paise
= 1 paise = Rs. (1/100)
∴ 7 paise = Rs. (7/100)
= Rs. 0.07

(ii) 7 rupees 7 paise
Ans:
We know that,
= Rs. 1 = 100 paise
= 1 paise = Rs. (1/100)
∴ 7 rupees 7 paise = Rs. 7 + Rs. (7/100)
= Rs. 7 + Rs. 0.07
= Rs. 7.07

(iii) 77 rupees 77 paise
Ans:
We know that,
= Rs. 1 = 100 paise
= 1 paise = Rs. (1/100)
∴ 77 rupees 77 paise = Rs. 77 + Rs. (77/100)
= Rs. 77 + Rs. 0.77
= Rs. 77.77

(iv) 50 paise
Ans:
We know that,
= Rs. 1 = 100 paise
= 1 paise = Rs. (1/100)
∴ 50 paise = Rs. (50/100)
= Rs. 0.50

(v) 235 paise
Ans:
We know that,
= Rs. 1 = 100 paise
= 1 paise = Rs. (1/100)
∴ 235 paise = Rs. (235/100)
= Rs. 2.35

Q12: (i) Express 5 centimeters in meter and kilometre
Ans:
We all know that,
= 1 meter = 100 centimeter
Then,
= 1 cm = (1/100) meter
= 5 cm = (5/100)
= 0.05 m
Now,
= 1 km = 1000 meter
Then,
= 1 m = (1/1000) kilometer
= 0.05 m = (0.05/1000)
= 0. 00005 kilometer

(i) Express 35 mm in cm, m and km
Ans:
We know that,
= 1 cm = 10 mm
Then,
= 1 mm = (1/10) cm
= 35 mm = (35/10) cm
= 3.5 cm
And,
= 1 meter = 100 cm
Then,
= 1 cm = (1/100) m
= 3.5 cm = (3.5/100) m
= (35/1000) m
= 0.035 m
Now,
= 1 km = 1000 m
Then,
= 1 m = (1/1000) km
= 0.035 m = (0.035/1000)
= 0. 000035 km

Q13: Express in kilogram:
(i) 200 gram
Ans:
We know that,
= 1 kg = 1000 gram
Then,
= 1 g = (1/1000) kg
= 200 g = (200/1000) kilogram
= (2/10)
= 0.2 kg

(ii) 3470 gram
Ans:
We know that,
= 1 kg = 1000 gram
Then,
= 1 g = (1/1000) kilogram
= 3470 g = (3470/1000) kilogram
= (3470/100)
= 3.470 kg

(ii) 4 kg 8 g
Ans:
We know that,
= 1 kg = 1000 gram
Then,
= 1 g = (1/1000) kilogram
= 4 kg 8 g is equal to 4 kg + (8/1000) kilogram
= 4 kg + 0.008
= 4.008 kilogram

Q14: Write the following in decimal numbers in their expanded form:
(i) 20.03
Ans:
We have that,
20.03 = (2 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(ii) 2.03
Ans:
We have that,
2.03 is equal to (2 × 1) + (0 × (1/10)) + (3 × (1/100))

(iii) 200.03
Ans:
We have,
200.03 is (2 × 100) + (0 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(iv) 2.034
Ans:
We have,
2.034 gets equal to (2 × 1) + (0 × (1/10)) + (3 × (1/100)) + (4 × (1/1000)).

Q15: Shyama bought around 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala then bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Ans:
From the above question, it is given that,
Fruits bought by Shyama is = 5 kg 300 g
= 5 kg + (300/1000) kg
= 5 kg + 0.3 kg
= 5.3 kg
Fruits bought by Sarala is = 4 kg 800 g + 4 kg 150 g
= (4 + (800/1000)) + (4 + (150/1000))
= (4 + 0.8) kg + (4 + .150) kg
= 4.8 kg + 4.150kg
= 8.950 kg
So, Sarala bought more fruits.