Decimals Class- 7th RS Aggarwal Exe-3C Goyal Brothers ICSE Math Solution

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

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:–

Return to- ICSE Class -7 RS Aggarwal Goyal Brothers Math Solutions

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