Question

One thief run at the speed of 100m/min .After 1 min a policeman run behind him to catch him with the speed of 100m/min and increases his speed 10m/min in every succeeding min .
find how much time will be taken by policeman to catch the thief.



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Answer 1

ANSWER

This question is of A. P(Arithmetic progression)

Now,

let the time taken to catch the thief by police be n.

Let the time taken by thief being caught by police be (n+1)

(Since the thief ran 1 minute before the police)

Hence, distance travelled by thief in n+1 seconds is 100(n+1) m

Now we know that =>

Speed of police in 1st second =100m/s

Similaly speed of police in 2nd minute =110m/s

Since the speed got increased by 10m/s we got the AP as >

100,110,120,130,.........(A)n(last term)

So the total distance travelled by police in n minutes will be=>

$\dfrac{n}{2}${2a +(n-1)d}

=>$\dfrac{n}{2}${2×100 +(n-1)×10} - - - - - (i)

Also we know that

Distance travelled by police before catching the thief =100(n+1) - - - (ii)

Now from (i ) and (ii) we get =>

100(n+1) = $\dfrac{n}{2}${2×100 +(n-1)×10}

=>100n+100 = 100n +5n(n-1)

On canceling 100n from both side we get=>

100 = 5n(n-1)

=>5$n^2$ - 5n - 100 =0

=>5($n^2$ - n - 20) =0

=>($n^2$ - n - 20) =0

Since this is an quadratic equation we will have two values for n.

Solving by middle term split methord we get =>

($n^2$ - 5n + 4n - 20)=0

=>n(n-5) +4(n-5) = 0

=>(n-5)(n+4) =0

So,

either

n = 5

or

n = -4

(neglected as time can't be negative)

Hence the policeman takes 5 minutes to catch the thief.

Remember

1)Sn= $\dfrac{n}{2}${2×100 +(n-1)×10}

-for finding sum till nth terms of an AP.

2)An = a +(n-1)d

-for finding nth term of an AP.

Answer 2

$\sf\huge{\underline{Question}}$

▶One thief run at the speed of 100m/min .After 1 min a policeman run behind him to catch him with the speed of 100m/min and increases his speed 10m/min in every succeeding min .

find how much time will be taken by policeman to catch the thief.

$\sf\huge{\boxed{\underline{\mathbb{Answer}}}}$➡

Let the policeman catch theif in 'n' minutes

According to the question,

theif has a uniform speed of 100m/min

Policeman with the same speed run back after him after the interval of 1 min!.

hence!

➡ (n+1) minutes = 100(n+1) minutes==(1)

since it is mentioned that, speed increases by 10m/min per each min, so!

Speed of policeman in 1s , 2s , 3s ...ns

is 100, 110,120,....(100+n)s

from the above series , it is clear that, the values are in ap!

i.e, 100, 110,120,...(100+n)

➡ a2-a1 = 110-100= 10

i.e, common difference is d=10

we have!

sum of n terms,,

S(n)=$\frac{n(2a+(n-1)d}{2}$

according to above equation! (1)

it is clear that distance travelled by thief and police is same!

hence..

100(n+1) = $\frac{n(2a+(n-1)d}{2}$

we have, a= 100

substituting!

100n + 100 = $\frac{n(2\times 100+(n-1)d}{2}$

200n + 200 = $n{2\times 100+(n-1)d}$

we have, d = 10

200n + 200 = $n{2\times 100+(n-1)10}$

200n + 200 = $n{2\times 100+(10n-10}$

200n + 200 = $n{200+10n-10}$

200n + 200 = $n{190+10n}$

200n + 200 = 190n + $10\times n^{2}$

190n + $10\times n^{2}$-200n-200=0

$10\times n^{2}$ - 10n - 200= 0

dividing whole equation by 10

$n^{2}$ - n - 20 = 0

$n^{2}$- 5n+4n - 20= 0

n(n-5)+4(n-5)=0

(n-5)(n+4)=0

n-5=0

n = 5

n+4=0

n = -4

since n can't be negative! therfore!

n = 5 minutes

Policeman takes 5 minutes to catch the theif!