Selina solutions

Selina solutions

Grade 6

Decimal Fractions | Exercise 15(E)

Question 10

Q10) Three strings of lengths 50 m 75 cm; 68 m 58 cm and 121 m 3 cm, respectively, are joined together to get a single string of greatest length, And the length of the single string obtained.

If this single string is then divided into 12 equal pieces ; find the length of each piece.

Solution:

1st string 50m 75cm = 50.75 m

2nd string 68m 58cm = 68.58 m

3rd string 121m 3cm = 121.03 m

On joining three total length = 240.36 m

Now , one string = 240.36 m

Dividing 12 parts = \frac{240.36}{12}=20.3m

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Question 10

Q10) Three strings of lengths 50 m 75 cm; 68 m 58 cm and 121 m 3 cm, respectively, are joined together to get a single string of greatest length, And the length of the single string obtained.

If this single string is then divided into 12 equal pieces ; find the length of each piece.

Solution:

1st string 50m 75cm = 50.75 m

2nd string 68m 58cm = 68.58 m

3rd string 121m 3cm = 121.03 m

On joining three total length = 240.36 m

Now , one string = 240.36 m

Dividing 12 parts = \frac{240.36}{12}=20.3m

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

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Book a free class now
subject-cta
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