A thief, after committing a theft runs at a uniform speed of 50 m/min. After 2 min, a policeman runs to catch him. He goes 60 m in the first min and increases his speed by 5 m/min every succeeding min. After how many minutes, the policeman will catch the thief?
Answers
Answer:
after 7 mins of committing a theft , thief was caught
Policeman caught him in 5 mins after policeman started
Step-by-step explanation:
Thief speed = 50m/min
Let say he was caught after n mins
then Distance covered by thief = 50n m
Policeman runs after 2 mins so he will catch him in n-2 mins
Distance covered by Policeman
in first minute = 60 m
in second minute = 60 + 5 = 65 m
in third minute = 65 + 5 = 70 m
this forms an AP where
a = first term = 60 m
d = common difference = 5 m
Sum of n terms = (n/2)(a + a + (n-1)d)
he catches in n-2 minutes
Distance covered in n-2 minutes
= ((n-2)/2)(a + a + (n-3)d)
=((n-2)/2)(2a + (n-3)d)
= ((n-2)/2)(2×60 +(n-3)5)
= ((n-2)/2)(120 +5n - 15)
= ((n-2)/2)(105 +5n)
Distance covered by thief = 50n
((n-2)/2)(105 +5n) = 50 n
105n + 5n² - 210 - 10n = 100n
=> 5n² - 5n - 210 = 0
=> n² - n - 42 = 0
=> n² - 7n + 6n - 42 = 0
=> n(n-7) + 6(n-7) = 0
=> (n+6)(n-7) = 0
=> n = 7
after 7 mins of committing a theft , thief was caught
Policeman caught him in 7-2 = 5 mins
Answer:
According to the question,
Distance between thief and police after 2 minutes = 100 m
Police starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.
Let the police chase the thief for t minutes.
Implies, thief covers total 100 + 50t m
Police runs with following pattern 60,65,70...
This is an Arithmetic Progression with a = 60, d = 5 & number of terms = t
t/2(2×60+(t−1)×5)=50t+100
t×(120+5t−5)=100t+200
5t
2
+15t−200=0
t
2
+3t−40=0
t={−3±
(9+160)}/2
=(−3±13)/2=+5
Implies, police will catch thief in 5 minutes
or the thief will be caught in 7 minutes