S Chand ICSE Mathematics Class 6 Solution Sixth Chapter Decimal Fractions Exercise 6C (O.P. Malhotra, S.K. Gupta, Anubhuti Gangal)

S Chand ICSE Mathematics Class 6 Solution Sixth Chapter Decimal Fractions Exercise 6C

(1) (i) 8 P = 1 hundredth of a rupee = 8/100 x 1 = 0.08

(ii) 19 P = 19 hundredths of a rupee = 19/100 x 1 = 0.19

(iii) 75 P = 75 hundredths of a rupee = 75/100 x 1 = 0.75

(iv) 10 P = 0.10

(v) 7.43

(vi) 28.06

(vii) 509.95

(viii) 127.80

(2) (i) 6 mm = 6/10 cm = 0.6 cm

(ii) 9 mm = 9/10 cm = 0.9 cm

(iii) 42 mm = 42 / 10 = 4.2 cm

(iv) 375 mm = 375 / 10 = 37.5 cm

(3) (i) 7 cm = 7 / 100 m = 0.07 m

(ii) 18 cm – 18/100 m = 0.18 m

(iii) 195 cm = 195 / 100 m = 19.5 m

(iv) 1328 cm = 1328 / 100 m = 13.28 m

(4) (i) 8 m = 8 / 1000 km = 0.008 km

(ii) 39 m = 39 / 1000 km = 0.039 km

(iii) 457 m = 457 / 1000 km = 0.457 km

(iv) 3824 m = 3824 / 1000 km = 3.824 km

(5) (i) 6.15

(ii) 65 m 9 cm = 65 + 9/100 = 65 + 0.09 = 65.09 m

(iii) 12 km + 200 / 1000 = 12 km + 0.200 = 12.200 km

(iv) 105 km + 25/1000 = 105 km + 0.025 km = 105.025 km

(v) 829 cm = 829 / 100000 km = 0.00829 km

(6) (i) 18 mg = 18 / 1000 g = 0.018 g

(ii) 125 mg = 125 / 1000 g = 0.125 g

(iii) 6 mg = 6 / 1000 g = 0.006 g

(iv) 18 g 12 mg = 18 g + 12 / 1000 = 18 g + 0.012 g = 18.012 g

(v) 9 g 77 mg = 9 g + 77 / 1000 g = 9 g + 0.077 = 9.077g

(7)(i) 690 g = 690 / 1000 kg= 0.690 kg

(ii) 97 g = 97 / 1000 kg = 0.097 kg

(iii) 2790 g = 2790 / 1000 kg = 2.790 kg

(iv) 1700 g = 1700 / 1000 kg = 1.700 kg

(v) 4 g = 4 / 1000 kg = 0.004 kg

(vi) 4 kg 25 g = 45 kg + 25 /1000 kg = 45 kg + 0.025 kg = 45.025 kg

(vii) 23 kg 179 g

= 23 kg + 179 / 1000 kg

= 23 kg + 0.179 kg

= 23.179 kg

(viii) 110 kg 8 g

= 110 kg + 8 /1000 kg

= 110 kg + 0.008 kg

= 110.008 kg

(8) (i) 5 ml

= 5 / 1000 l

= 0.005 l

(ii) 732

= 732 / 1000

= 0.732 l

(iii) 19 l

= 19 / 1000 kl

= 0.019 kl

(iv) 15 l 25 ml

= 15 l + 25 / 1000 l

= 15 l + 0.025

= 15.025 l

(v) 4 kl 289 l

= 4 kl + 289 / 1000  kl

= 4 kl + 0.289 kl

= 4.289 kl


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  1. The distance between Dhruv’s office and home is 7 km 5 m. He walks 1 km 125 m to the metro station A The distance travelled by metro to metro station B is 4 km 65 m. What distance does he have to wa from the metro station B to the office? Give your answer in kilometres.

    1. 7 km 5 m

      7000 m + 5 m = 7005 m

      1 km 125 m
      4 km 65 m

      5 km 190m = 5000 + 190 = 5190 m

      distance does he have to wa from the metro station B to the office 7005 m – 5190 m = 1815 m = 1 km 815 m

    2. You have not done next exercise 6-D

  2. You have not done the net 6D exercise

  3. You have not done next exercise 6-D

  4. very nice solutionsof every exercise

  5. now i dont have to worry about anything THANK YOU MAM

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