Decimals Class- 7th RS Aggarwal Exe-3C Goyal Brothers ICSE Math Solution . We provide step by step Solutions of lesson-3 Decimals for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan . Our Solutions contain all type Questions of Exe-3C to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.
Decimals Class- 7th RS Aggarwal Exe-3C Goyal Brothers ICSE Math Solution
| Board | ICSE |
| Publications | Goyal brothers Prakashan |
| Subject | Maths |
| Class | 7th |
| Chapter-3 | Decimals |
| Writer | RS Aggrawal |
| Book Name | Foundation |
| Topics | Solution of Exe-3C |
| Academic Session | 2023 – 2024 |
Exercise – 3C
Decimals Class- 7th RS Aggarwal Exe-3C Goyal Brothers ICSE Math Solution
1. Multiply :
(i) 6.5 × 10
Solution: 6.5 × 10
= 65.0
= 65
(ii) 0.8 × 10
Solution: 0.8 × 10
= 8.0
= 8
(iii) 9.07 × 10
Solution: 9.07 × 10
= 90.7
(iv) 2.345 × 10
Solution: 2.345 × 10
= 23.45
(v) 4.63 × 100
Solution: 4.63 × 100
= 463.00
= 463
(vi) 8.279 × 100
Solution: 8.279 × 100
= 827.9
(vii) 7.8 × 100
Solution: 7.8 × 100
= 780.0
= 780
(viii) 0.09 × 100
Solution: 0.09 × 100
= 09.00
= 9
(ix) 0.283 × 1000
Solution: 0.283 × 1000
= 283.000
= 283
(x) 6.25 × 1000
Solution: 6.25 × 1000
= 6250.00
= 6250
(xi) 5.4 × 1000
Solution: 5.4 × 1000
= 5400.00
= 5400
(xii) 0.3 × 1000
Solution: 0.3 × 1000
= 300.00
= 300
2. Multiply :
(i) 2.4 × 16
Solution: 2.4 × 16
= 38.4
(ii) 3.45 × 17
Solution: 3.45 × 17
= 58.65
(iii) 0.86 × 14
Solution: 0.86 × 14
= 12.04
(iv) 2.68 × 30
Solution: 2.68 × 30
= 80.40
= 80.4
(v) 0.023 × 65
Solution: 0.023 × 65
= 1.495
(vi) 0.0006 × 15
Solution: 0.0006 × 15
= 0.0090
= 0.009
3. Find the product :
(i) 5.4 × 1.4
Solution: 5.4 × 1.4
= 7.84 [Decimal point 1 + 1 = 2 places]
(ii) 2.35 × 7.2
Solution: 2.35 × 7.2
= 16.92 [Decimal point at 2 + 1 = 3 places]
(iii) 0.37 × 0.26
Solution: 0.37 × 0.26
= 0.0962 [Decimal point at 2 + 2 = 4 places]
(iv) 0.74 × 6.7
Solution: 0.74 × 6.7
= 4.958 [Decimal point after 2 + 1 = 3 places]
(v) 5.64 × 0.08
Solution: 5.64 × 0.08
= 0.4512 [Decimal point at 2 + 2 = 4 places]
(vi) 2.75 × 1.7
Solution: 2.75 × 1.7
= 4.675 [Decimal point after 2 + 1 = 3 places]
(vii) 0.04 × 0.36
Solution: 0.04 × 0.36
= 0.014 [Decimal point at 2 + 2 = 4 places]
(viii) 34.2 × 1.86
Solution: 34.2 × 1.86
= 63.612 [Decimal point after 2 + 1 = 3 places]
(ix) 0.028 × 0.9
Solution: 0.028 × 0.9
= 0.0252 [Decimal point after 3 + 1 = 4 places]
4. Find the product :
(i) 2.3 × 0.23 × 0.1
Solution: 2.3 × 0.23 × 0.1
= 23 × 23 × 1
= 529 [Decimal point at 1 + 2 + 1 = 4 places]
= 0.0529
(ii) 1.2 × 3.5 × 0.3
Solution: 1.2 × 3.5 × 0.3
= 12 × 35 × 3
= 126 [Decimal point at 1 + 1 + 1 = 3 places]
= 1.26
(iii) 0.6 × 1.5 × 0.7
Solution: 0.6 × 1.5 × 0.7
= 6 × 15 × 7
= 63 [Decimal point at 1 + 1 + 1 = 3 places]
= 0.63
(iv) 0.2 × 0.2 × 0.02
Solution: 0.2 × 0.2 × 0.02
= 2 × 2 × 2
= 8 [Decimal point at 1 + 1 + 2 = 4 places]
= 0.0008
(v) 1.1 × 0.1 × 0.11
Solution: 1.1 × 0.1 × 0.11
= 11 × 1 × 11
= 121 [Decimal point at 1 + 1 + 2 = 4 places]
= 0.0121
(vi) 0.6 × 0.06 × 0.006
Solution: 0.6 × 0.06 × 0.006
= 6 × 6 × 6
= 216 [Decimal point at 1 + 2 + 3 = 6 places]
= 0.000216
5. Evaluate :
(i) (1.3)²
Solution: (1.3)²
= 1.3 × 1.3
= 1.69 [Decimal places 1 + 1 = 2 places]
(ii) (0.06)²
Solution: (0.06)²
= 0.06 × 0.06
= 0.0036 [Decimal places 2 + 2 = 4 places]
(iii) (0.2)³
Solution: (0.2)³
= 0.2 × 0.2 × 0.2
= 0.008 [Decimal places 1 + 1 + 1 = 3 places]
(iv) (0.8)³
Solution: (0.8)³
= 0.8 × 0.8 × 0.8
= 0.512 [Decimal places 1 + 1 + 1 = 3 places]
6. The cost of one pen is ₹42.25. Find the cost of one dozen such pens.
Solution: Cp of 1 pen = 42.25
Cp of 12 pen = ?
= 42.25 × 12
= 507.00 [Decimal point of 2 places]
= ₹507
7. A car moves at a constant speed of 56.4 km per hour. How much distance does it cover in 3.5 hours?
Solution: Car moves in 1 hours = 56.4 km
It will moves in 3.5 hours
= 56.4 × 3.5
= 197.40 km
= 197.4 km
8. A room is 4.5 m long and 3.8 m broad. Calculate the area of the floor of the room.
Solution: Length of room = 4.5 m
Breadth of room = 3.8
Area of the floor = ?
⇒ Area of its floor = Length × breadth
Area = 4.5 × 3.8 m²
Area = 17.10 m²
Area = 17.1 m²
9. The cost of 1 liter of refined oil is ₹124.75. What is the cost of 6.2 liters of this oil?
Solution: CP of 1 liter refined oil = 124.75
CP of 6.2 liter = ?
= 124.75 × 6.2
= 773.450
= ₹773.45
— : end of Decimals Class- 7th RS Aggarwal Exe-3C Goyal Brothers ICSE Math Solution:–
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