Advertisements
Advertisements
Question
A man completed a trip of 136 km in 8 hours. Some part of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.
Advertisements
Solution
Total distance of the trip = 136 km
Let part of the trip covered at 18 km/hr = x km
∴ Distance of the trip covered at 15 km/hr = (136 - x)km
Time taken by the man to cover x km = `"Distance"/"speed" = "x"/18` hours
Time taken by the man to cover (136 - x) km = `(136 - "x")/15` hours
Time taken by the man to cover a trip of 136 km = 8 hours
`therefore "x"/18 + (136 - "x")/15 = 8`
`=> "x"/18 xx 90 + (136 - "x")/15 xx 90 = 8 xx 90`
....[Multiplying each term by 90 becuase L.C.M. of denominatorsn = 90]
⇒ 5x + 6 (136 - x) = 720
⇒ 5x + 816 - 6x = 720
⇒ 5x - 6x = 720 - 816
⇒ -x = -96
⇒ x = 96
∴ Part of the trip covered at 18 km/hr = 96 km
APPEARS IN
RELATED QUESTIONS
Solve the following equation:
15 + x = 5x + 3
Solve the following equation:
`(3"x" + 2)/("x" - 6) = -7`
Solve: `(4-3"x")/5 + (7 - "x")/3 + 4 1/3 = 0` Hence, find the value of 'p', if 3p - 2x + 1 = 0
Five less than 3 times a number is -20. Find the number.
A man’s age is three times that of his son, and in twelve years he will be twice as old as his son would be. What are their present ages.
Solve: `(2"x")/3 - (3"x")/8 = 7/12`
Solve: `(3"x" - 2)/3 + (2"x" + 3)/2 = "x" + 7/6`
Solve: `"x" - ("x" - 1)/2 = 1 - ("x" - 2)/3`
The ages of A and B are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.
The cost of one pen is ₹ 8 and it is available in a sealed pack of 10 pens. If Swetha has only ₹ 500, how many packs of pens can she buy at the maximum?
