Math, asked by MERCURY1234, 1 year ago

PLEASE ANSWER THIS QUESTIONS

BRAINLIEST QUESTION

If a man increases his speed by 25%, he would take 10 minutes less to reach his destination. If he increases his speed by 100/9%, find the number of minutes he saves to reach his destination.

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Answers

Answered by gaurav2013c
10
Let the distance between starting position to the final position be a m

And the original speed be b m/s

Time taken = a / b

New speed = b + 25 % of b = b + b/4 = 5b/4

Time taken when speed is increased by 25% = [ a / (5b/4)] = 4a/5b

____________________________

When speed is increased by 100/9 %

Speed = b + 100/9 % of b

= b + ( 100 / 9) b/ 100

= b + b/9

= 10b / 9

Time taken when speed is increased by 100/9 % = [ a / (10b /9)] = 9a/10b

___________________________

According to question,


a/b - 4a/5b = 10 /60

=> a / 5b = 1/6

=> a/b = 5 /6 -----(1)

Time saved to reach the destination when speed is increased by 100/9 % = a/b - 9a/10b hours

= ( 10a - 9a) / 10b hours

= a/b × 1/10 hours

= 5/6 × 1/10 hours

= 1 / 12 hours

= 1 × 60 / 12 minutes

= 5 minutes

MERCURY1234: thankyou so much
MERCURY1234: you are really genius
Answered by ɪᴛᴢᴛʀᴀɢɪᴄɢɪʀʟ
0

☃️ required answer ::-

Let the distance between starting position to the final position be a m

And the original speed be b m/s

Time taken = a / b

New speed = b + 25 % of b = b + b/4 = 5b/4

Time taken when speed is increased by 25% = [ a / (5b/4)] = 4a/5b

____________________________

When speed is increased by 100/9 %

Speed = b + 100/9 % of b

= b + ( 100 / 9) b/ 100

= b + b/9

= 10b / 9

Time taken when speed is increased by 100/9 % = [ a / (10b /9)] = 9a/10b

___________________________

According to question,

a/b - 4a/5b = 10 /60

=> a / 5b = 1/6

=> a/b = 5 /6 -----(1)

Time saved to reach the destination when speed is increased by 100/9 % = a/b - 9a/10b hours

= ( 10a - 9a) / 10b hours

= a/b × 1/10 hours

= 5/6 × 1/10 hours

= 1 / 12 hours

= 1 × 60 / 12 minutes

= 5 minutes

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