Question

32. A thief, after committing a theft, runs ata uniform speed of 50 m/minute. After 2 minutes
a policeman runs to catch him. He runs 60 m in the first minute and increases his speed
by 5 m/minute every succeeding minute. After how many minutes, the policeman will
catch the thief?
(CBSE 2016​

Answer 1

Question:

A thief, after committing a theft, runs at a uniform speed of 50 m/minute. After 2 minutes a policeman runs to catch him. He runs 60 m in the first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?

Given:

• Uniform speed of the thief= 50meter/minute.
• After 2 minutes
• a policeman runs to catch him.
• Policeman runs in the first minute=60m
• And he (policeman) increases his speed by 5 meter/minute every succeeding minute.

To Find:

• When The policeman would catch the thief.

Solution:

We know, that the uniform speed of thief is 50 meters per minute and the policeman runs to catch him after 2 minutes.

$\\ \therefore\mathsf{Distance\: covered\: by\: thief\: after\: 2\: minutes=50\times 2=100m}$

Hence, the distance between policeman and the thief after 2 minutes is 100 meters.

Now, Lets assume that the policeman would catch the thief after $\bold{x}$ minutes.

Also, the thief is running at an uniform speed.

$\therefore\mathsf{Distance\: covered\: by\: thief=50(x+2)}$

$\\ \hookrightarrow\mathsf{Distance\: covered\: by\: thief=50x+100}$

Now, The policeman runs 60 meters in first minute and increases his speed by 5m/minute. Thus, the policeman runs in an sequence that is in form of Arithmetic progression. That is,

$\mathsf{60, 65, 70, 75, 80…}$

Here,

$\rightarrow\mathsf{First\: term,a=60}$

$\rightarrow\mathsf{Common\: Difference,d=5}$

Now, Distance covered by policeman in $\bold{x}$ minutes equals to Sum of $\bold{x}$ terms. And the sum of Arithmetic progression is given by,

$\sf{{S}_{n}=\large\frac{n}{2}[2a+(n-1)d]}$

Putting all the values in the formula we get,

$\rightarrow\mathsf{{S}_{n}=\large\frac{n}{2}[2a+(n-1)d]}$

$\rightarrow\mathsf{{S}_{n}=\large\frac{x}{2}[2\times 60+(x-1)5]}$

$\rightarrow\mathsf{{S}_{n}=\large\frac{x}{2}[120+(5x-5)]}$

$\rightarrow\mathsf{{S}_{n}=\large\frac{x}{2}[115+5x]}$

When the policeman would catch the thief. We have,

$\dashrightarrow\mathsf{\large\frac{x}{2}[115+5x]=50x+100}$

$\dashrightarrow\mathsf{115x+{5x}^{2}=(50x+100)\times 2}$

$\dashrightarrow\mathsf{115x+{5x}^{2}=100x+200}$

$\dashrightarrow\mathsf{115x+{5x}^{2}-100x=200}$

$\dashrightarrow\mathsf{15x+{5x}^{2}-200=0}$

$\dashrightarrow\mathsf{5(3x+{x}^{2}-40)=0}$

$\dashrightarrow\mathsf{3x+{x}^{2}-40=\large\frac{0}{5}}$

$\dashrightarrow\mathsf{3x+{x}^{2}-40=0}$

$\dashrightarrow\mathsf{{x}^{2}+3x-40=0}$

$\dashrightarrow\mathsf{{x}^{2}+8x-5x-40=0}$

$\dashrightarrow\mathsf{(x+8)(x-5)=0}$

$\dashrightarrow\mathsf{x+8=0\: or\: x-5=0}$

$\dashrightarrow\mathsf{x=-8\: or\: x=5}$

As $\bold{x}$ cannot be negative. Therefore,

$\bold{x=5}$

Hence, the policeman catches the thief after 5 minutes.

_________________________________________________________