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The distance of the School and house of Suresh is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed speed of Suresh?
- a)28 kmph
- b)25 kmph
- c)20 kmph
- d)24 kmph
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The distance of the School and house of Suresh is 80km. One day he was...
80/x – 80/(x+4) = 1
x(x+20) – 16(x+20) = 0
x = 16kmph
Increased speed = 20 kmph
x(x+20) – 16(x+20) = 0
x = 16kmph
Increased speed = 20 kmph
Most Upvoted Answer
The distance of the School and house of Suresh is 80km. One day he was...
80/x – 80/(x+4) = 1
x(x+20) – 16(x+20) = 0
x = 16kmph
Increased speed = 20 kmph
x(x+20) – 16(x+20) = 0
x = 16kmph
Increased speed = 20 kmph
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Community Answer
The distance of the School and house of Suresh is 80km. One day he was...
Given:
Distance between School and house of Suresh = 80 km
Let, the normal speed of Suresh be x km/h
He was late by 1 hour, so he increased his speed by 4 km/h and reached college at normal time.
To find: The changed speed of Suresh
Concept Used:
Time = Distance/Speed
Calculation:
Let's assume that Suresh normally takes T hours to reach college.
So, according to the given data:
Time taken at normal speed = T hours
Time taken at increased speed = T - 1 hours
Using the concept of Time = Distance/Speed:
Time taken at normal speed = Distance/Normal speed
T = 80/x
Time taken at increased speed = Distance/Increased speed
T - 1 = 80/(x+4)
Now, we can solve these two equations to find the value of x.
80/x = 80/(x+4) + 1
80(x+4) = 80x + x(x+4)
80x + 320 = x^2 + 4x
x^2 - 76x - 320 = 0
(x-20)(x-56) = 0
As speed cannot be negative, we will take x = 20 km/h.
Now, to find the changed speed of Suresh, we need to add 4 km/h to his normal speed.
Therefore, the changed speed of Suresh = 20 + 4 = 24 km/h.
Hence, the correct answer is option (d) 24 kmph.
Distance between School and house of Suresh = 80 km
Let, the normal speed of Suresh be x km/h
He was late by 1 hour, so he increased his speed by 4 km/h and reached college at normal time.
To find: The changed speed of Suresh
Concept Used:
Time = Distance/Speed
Calculation:
Let's assume that Suresh normally takes T hours to reach college.
So, according to the given data:
Time taken at normal speed = T hours
Time taken at increased speed = T - 1 hours
Using the concept of Time = Distance/Speed:
Time taken at normal speed = Distance/Normal speed
T = 80/x
Time taken at increased speed = Distance/Increased speed
T - 1 = 80/(x+4)
Now, we can solve these two equations to find the value of x.
80/x = 80/(x+4) + 1
80(x+4) = 80x + x(x+4)
80x + 320 = x^2 + 4x
x^2 - 76x - 320 = 0
(x-20)(x-56) = 0
As speed cannot be negative, we will take x = 20 km/h.
Now, to find the changed speed of Suresh, we need to add 4 km/h to his normal speed.
Therefore, the changed speed of Suresh = 20 + 4 = 24 km/h.
Hence, the correct answer is option (d) 24 kmph.
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The distance of the School and house of Suresh is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed speed of Suresh?a)28kmphb)25 kmphc)20kmphd)24kmphCorrect answer is option 'C'. Can you explain this answer?
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