(i) 0.2 x 6 = 1.2 (ii) 8 x 4.6 = 36.8 (iii) 2.71 x 5 = 13.55 (iv) 20.1 x 4 = 80.4 (v) 0.05 x 7 = 0.35 (vi) 211.02 x 4 = 844.08 (vii) 2 x 0.86 = 1.72 Class 7 Maths Chapter 2 Exercise 2.6 Question 1, 2, 3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
(i) 0.2 x 6 = 1.2
(ii) 8 x 4.6 = 36.8
(iii) 2.71 x 5 = 13.55
(iv) 20.1 x 4 = 80.4
(v) 0.05 x 7 = 0.35
(vi) 211.02 x 4 = 844.08
(vii) 2 x 0.86 = 1.72
We have to find the difference of 42.6 km and 28 km. Difference = 42.6 – 28.0 = 14.6 km Therefore 14.6 km less is 28 km than 42.6 km. Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 9
We have to find the difference of 42.6 km and 28 km.
Difference = 42.6 – 28.0 = 14.6 km
Therefore 14.6 km less is 28 km than 42.6 km.
Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 9
Total weight of fruits bought by Shyam = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g Total weight of fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g On comparing the quantity of fruits, 8 kg 550 g < 8 kg 950 g Therefore, Sarala bought more fruits. Class 7 Maths Chapter 2 Exercise 2.5 QuesRead more
Total weight of fruits bought by Shyam = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g
Total weight of fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g
On comparing the quantity of fruits, 8 kg 550 g < 8 kg 950 g
Therefore, Sarala bought more fruits.
Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 9
Distance travelled by Dinesh when he went from place A to place B = 7.5 km and from place B to C = 12.7 km. Total distance covered by Dinesh = AB + BC = 7.5 + 12.7 = 20.2 km Total distance covered by Ayub = AD + DC = 9.3 + 11.8 = 21.1 km On comparing the total distance of Ayub and Dinesh, 21.1 kmRead more
Distance travelled by Dinesh when he went from place A to place B = 7.5 km and from
place B to C = 12.7 km.
Total distance covered by Dinesh = AB + BC
= 7.5 + 12.7 = 20.2 km
Total distance covered by Ayub = AD + DC
= 9.3 + 11.8 = 21.1 km
On comparing the total distance of Ayub and Dinesh,
21.1 km > 20.2 km
Therefore, Ayub covered more distance by 21.1 – 20.2 = 0.9 km = 900 m
(i) Place value of 2 in 2.56 = 2 x 1 = 2 ones (ii) Place value of 2 in 21.37 = 2 x 10 = 2 tens (iii) Place value of 2 in 10.25 = 2x1/10 = 2 tenths (iv) Place value of 2 in 9.42 = 2x1/100 = 2 hundredth (v) Place value of 2 in 63.352 = 2x1/1000 = 2 thousandth Class 7 Maths Chapter 2 Exercise 2.5 QuestRead more
(i) Place value of 2 in 2.56 = 2 x 1 = 2 ones
(ii) Place value of 2 in 21.37 = 2 x 10 = 2 tens
(iii) Place value of 2 in 10.25 = 2×1/10 = 2 tenths
(iv) Place value of 2 in 9.42 = 2×1/100 = 2 hundredth
(v) Place value of 2 in 63.352 = 2×1/1000 = 2 thousandth
20.03 = 2 x 10 + 0 x 1 + 0 x 1/10+3×1/100 Class 7 Maths Chapter 2 Exercise 2.5 Question 5, 6, 7 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Let us consider, 1000 g = 1 kg ⇒ 1g= 1/1000 kg 200 g = (200x1/1000) kg =0.2kg Class 7 Maths Chapter 2 Exercise 2.5 Question 3, 4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Let us consider,
1000 g = 1 kg
⇒ 1g= 1/1000 kg
200 g = (200×1/1000) kg =0.2kg
Class 7 Maths Chapter 2 Exercise 2.5 Question 3, 4
(i) Express 5 cm in meter and kilometer. ∵ 100 cm = 1 meter ∴ 1 cm = 1/100 meter ⇒ 5 cm = 5/100 = 0.05 meter. Now, ∵ 1000 meters = 1 kilometers ∴ 1 meter = 1/1000 kilometers 0.05 meter = 0.05/1000 = 0.00005 kilometer (ii) Express 35 mm in cm, m and km. ∵ 10 mm = 1 cm ∴ 1 mm = 1/10 cm ⇒ 3.5 cm = 3.5/Read more
(i) Express 5 cm in meter and kilometer.
∵ 100 cm = 1 meter
∴ 1 cm = 1/100 meter
⇒ 5 cm = 5/100 = 0.05 meter.
Now,
∵ 1000 meters = 1 kilometers
∴ 1 meter = 1/1000 kilometers
0.05 meter = 0.05/1000 = 0.00005 kilometer
(ii) Express 35 mm in cm, m and km.
∵ 10 mm = 1 cm
∴ 1 mm = 1/10 cm
⇒ 3.5 cm = 3.5/100 = 0.035 meter
Again,
∵ 1000 meters = 1 kilometers
∴ 1 meter = 1/1000 kilometer
⇒ 0.035 meter = 0.035/1000 = 0.000035 kilometer
Class 7 Maths Chapter 2 Exercise 2.5 Question 3, 4
Find: (i) 0.2 x 6 (ii) 8 x 4.6 (iii) 2.71 x 5 (iv) 20.1 x 4 (v) 0.05 x 7 (vi) 211.02 x 4 (vii) 2 x 0.86
(i) 0.2 x 6 = 1.2 (ii) 8 x 4.6 = 36.8 (iii) 2.71 x 5 = 13.55 (iv) 20.1 x 4 = 80.4 (v) 0.05 x 7 = 0.35 (vi) 211.02 x 4 = 844.08 (vii) 2 x 0.86 = 1.72 Class 7 Maths Chapter 2 Exercise 2.6 Question 1, 2, 3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
(i) 0.2 x 6 = 1.2
(ii) 8 x 4.6 = 36.8
(iii) 2.71 x 5 = 13.55
(iv) 20.1 x 4 = 80.4
(v) 0.05 x 7 = 0.35
(vi) 211.02 x 4 = 844.08
(vii) 2 x 0.86 = 1.72
Class 7 Maths Chapter 2 Exercise 2.6 Question 1, 2, 3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
How much less is 28 km than 42.6 km?
We have to find the difference of 42.6 km and 28 km. Difference = 42.6 – 28.0 = 14.6 km Therefore 14.6 km less is 28 km than 42.6 km. Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 9
We have to find the difference of 42.6 km and 28 km.
Difference = 42.6 – 28.0 = 14.6 km
Therefore 14.6 km less is 28 km than 42.6 km.
Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 9
See lessShyam bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Total weight of fruits bought by Shyam = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g Total weight of fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g On comparing the quantity of fruits, 8 kg 550 g < 8 kg 950 g Therefore, Sarala bought more fruits. Class 7 Maths Chapter 2 Exercise 2.5 QuesRead more
Total weight of fruits bought by Shyam = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g
Total weight of fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g
On comparing the quantity of fruits, 8 kg 550 g < 8 kg 950 g
Therefore, Sarala bought more fruits.
Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 9
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?
Distance travelled by Dinesh when he went from place A to place B = 7.5 km and from place B to C = 12.7 km. Total distance covered by Dinesh = AB + BC = 7.5 + 12.7 = 20.2 km Total distance covered by Ayub = AD + DC = 9.3 + 11.8 = 21.1 km On comparing the total distance of Ayub and Dinesh, 21.1 kmRead more
Distance travelled by Dinesh when he went from place A to place B = 7.5 km and from
place B to C = 12.7 km.
Total distance covered by Dinesh = AB + BC
= 7.5 + 12.7 = 20.2 km
Total distance covered by Ayub = AD + DC
= 9.3 + 11.8 = 21.1 km
On comparing the total distance of Ayub and Dinesh,
21.1 km > 20.2 km
Therefore, Ayub covered more distance by 21.1 – 20.2 = 0.9 km = 900 m
Class 7 Maths Chapter 2 Exercise 2.5 Question 5, 6, 7
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Write the place value of 2 in the following decimal numbers: (i) 2.56 (ii) 21.37 (iii) 10.25 (iv) 9.42 (v) 63.352
(i) Place value of 2 in 2.56 = 2 x 1 = 2 ones (ii) Place value of 2 in 21.37 = 2 x 10 = 2 tens (iii) Place value of 2 in 10.25 = 2x1/10 = 2 tenths (iv) Place value of 2 in 9.42 = 2x1/100 = 2 hundredth (v) Place value of 2 in 63.352 = 2x1/1000 = 2 thousandth Class 7 Maths Chapter 2 Exercise 2.5 QuestRead more
(i) Place value of 2 in 2.56 = 2 x 1 = 2 ones
(ii) Place value of 2 in 21.37 = 2 x 10 = 2 tens
(iii) Place value of 2 in 10.25 = 2×1/10 = 2 tenths
(iv) Place value of 2 in 9.42 = 2×1/100 = 2 hundredth
(v) Place value of 2 in 63.352 = 2×1/1000 = 2 thousandth
Class 7 Maths Chapter 2 Exercise 2.5 Question 5, 6, 7
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Write the following decimal numbers in the expanded form: 20.03
20.03 = 2 x 10 + 0 x 1 + 0 x 1/10+3×1/100 Class 7 Maths Chapter 2 Exercise 2.5 Question 5, 6, 7 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
20.03 = 2 x 10 + 0 x 1 + 0 x 1/10+3×1/100
Class 7 Maths Chapter 2 Exercise 2.5 Question 5, 6, 7
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Express in kg.: 200 g
Let us consider, 1000 g = 1 kg ⇒ 1g= 1/1000 kg 200 g = (200x1/1000) kg =0.2kg Class 7 Maths Chapter 2 Exercise 2.5 Question 3, 4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Let us consider,
1000 g = 1 kg
⇒ 1g= 1/1000 kg
200 g = (200×1/1000) kg =0.2kg
Class 7 Maths Chapter 2 Exercise 2.5 Question 3, 4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
(i) Express 5 cm in metre and kilometer. (ii) Express 35 mm in cm, m and km.
(i) Express 5 cm in meter and kilometer. ∵ 100 cm = 1 meter ∴ 1 cm = 1/100 meter ⇒ 5 cm = 5/100 = 0.05 meter. Now, ∵ 1000 meters = 1 kilometers ∴ 1 meter = 1/1000 kilometers 0.05 meter = 0.05/1000 = 0.00005 kilometer (ii) Express 35 mm in cm, m and km. ∵ 10 mm = 1 cm ∴ 1 mm = 1/10 cm ⇒ 3.5 cm = 3.5/Read more
(i) Express 5 cm in meter and kilometer.
∵ 100 cm = 1 meter
∴ 1 cm = 1/100 meter
⇒ 5 cm = 5/100 = 0.05 meter.
Now,
∵ 1000 meters = 1 kilometers
∴ 1 meter = 1/1000 kilometers
0.05 meter = 0.05/1000 = 0.00005 kilometer
(ii) Express 35 mm in cm, m and km.
∵ 10 mm = 1 cm
∴ 1 mm = 1/10 cm
⇒ 3.5 cm = 3.5/100 = 0.035 meter
Again,
∵ 1000 meters = 1 kilometers
∴ 1 meter = 1/1000 kilometer
⇒ 0.035 meter = 0.035/1000 = 0.000035 kilometer
Class 7 Maths Chapter 2 Exercise 2.5 Question 3, 4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Express as rupees using decimals: (i) 7 paise (ii) 7 rupees 7 paise (iii) 77 rupees 77 paise (iv) 50 paise (v) 235 paise
∵ 100 paise = ₹1 ∴ 1 paisa = ₹ 1/100 (i) 7 paise = ₹ 7/100 ₹ 0.07 (ii) 7 rupees 7 paise = ₹ 7 + ₹ 7/100 = ₹ 7 + ₹ 0.07 = ₹ 7.07 (iii) 77 rupees 77 paise = ₹ 77 + ₹ 77/100 = ₹ 77 + ₹ 0.77 = ₹ 77.77 (iv) 50 paise = ₹ 50/100 = ₹ 0.50 (v) 235 paise = ₹ 235/100 = ₹ 2.35 Class 7 Maths Chapter 2 Exercise 2Read more
∵ 100 paise = ₹1
∴ 1 paisa = ₹ 1/100
(i) 7 paise = ₹ 7/100 ₹ 0.07
(ii) 7 rupees 7 paise = ₹ 7 + ₹ 7/100 = ₹ 7 + ₹ 0.07 = ₹ 7.07
(iii) 77 rupees 77 paise = ₹ 77 + ₹ 77/100 = ₹ 77 + ₹ 0.77 = ₹ 77.77
(iv) 50 paise = ₹ 50/100 = ₹ 0.50
(v) 235 paise = ₹ 235/100 = ₹ 2.35
Class 7 Maths Chapter 2 Exercise 2.5 Question 1, 2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Which is greater: (i) 0.5 or 0.05 (ii) 0.7 or 0.5 (iii) 7 or 0.7 (iv) 1.37 or 1.49 (v) 2.03 or 2.30 (vi) 0.8 or 0.88
(i) 0.5 > 0.05 (ii) 0.7 > 0.5 (iii) 7 > 0.7 (iv) 1.37 < 1.49 (v) 2.03 < 2.30 (vi) 0.8 < 0.88 Class 7 Maths Chapter 2 Exercise 2.5 Question 1, 2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
(i) 0.5 > 0.05 (ii) 0.7 > 0.5 (iii) 7 > 0.7
(iv) 1.37 < 1.49 (v) 2.03 < 2.30 (vi) 0.8 < 0.88
Class 7 Maths Chapter 2 Exercise 2.5 Question 1, 2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/