A thief runs with a uniform speed of 100 m/minute.

Question 11:- A thief runs with a uniform speed of 100 m/minute. After one minute, a policeman ran after to catch him. He goes with a speed of 100 m/minute in the first minute and increase his speed by 10 m/minute every succeeding minute. After how many minute the policeman will catch the thief ?                 [CBSE   2016]

Solution:-  Let the total time taken to catch the thief be ‘n’ minutes

Since, speed of thief = 10 m/minute

∴ Total distance covered by the thief  = 100n

Now the speed of the policeman in the first minute is 100m/min, in the second minute is 110 m/min, in the third minute is 120 m/min and so on.

Then, the speed forms an A. P. with a constant increasing speed 10 m/min. Thus, the series is :

100, 110, 120, 130  .   .   .

AS the policeman start after a minute, so time taken by the policeman to catch the thief is (n – 1) minutes.

∴ Total distance covered by the policeman

100, 110, 120, 130  .   .   . (n – 1) terms

∴

⇒ 200n = (n-1)[200 + 10n – 20]

⇒ 200n = (n-10)[180 + 10n]

⇒ 200n = 180n – 180 + 10n² – 10n

⇒ 10n² – 30n – 180 = 0

⇒ n² – 3n – 18 =0

⇒ n² – 6n + 3n – 18 = 0

⇒ (n – 6)(n – 3) = 0

n = 6, – 3

Since n = -3 is not possible

Hence, the policeman takes (n – 10 = 5 minutes to catch the theif.

Some other question

Question 1: Ramkali would require Rs 5000 for getting her daughter admitted in a school after a year. She saved Rs 150 in the first month and increased her monthly by Rs 50 every month. Find, if she will be able to arrange the required money after 12 months. Which value is reflected in her efforts ?

Solution: For solution click here

Question 2: The first term of an A.P. is -5 and the last term is 45. If the sum of  the terms of the A.P. is 120, then find the number of terms and the common difference.

Solution: For solution Click Here

Question 3: If the sum of the first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms.

[CBSE 2016, 13, 12]

Solution: For solution click here

Question 4: The digits of a positive number of three digits are in A.P.  and  their sum is 15. The number obtained by reversing the digits  is 594 less than the original number. Find the number.

Solution : For solution click here

Question 5: If the sum of first m terms of an A.P. is some as the sum of its first n terms, show that the sum of its first (m + n) terms is zero.                          [CBSE  2019]

Solution: for solution click here

Question 6: If the sum of first ‘p’ terms of an A.P. is ‘q’ and the sum of first ‘q’ terms is ‘p’ ; then show that the sum of the first (p + q) terms is {-(p + q)}.

Solution: For solution click here

Question 7: Show that the sum of all terms of an A.P. whose first term is a, the second term is b and the last term is c is equal to $\dfrac{(a + c)(b + c -2a)}{2(b-a)}$.                                                    [CBSE      2020]

Solution:  For solution click here

Question 9: A child puts one five-rupees coin of her savings in the piggy bank on the first day. She increases her saving by one five-rupee coin daily. If the piggy bank can hold 190 coins of five rupees in all, find the number of days she can continue to put the five -rupee coins into it and find the total money she saved.                           [CBSE   2017]

Solution: For solution click here

Question 10: A sum of ₹ 4,250 is to be used to give 10 cash prizes to students of a school for their overall academic performance. If each prize is ₹ 50 less than its preceding prize, find the value of each of the prizes.

Solution: For solution click here

Question 12:- Jaspal Singh repays his total loan Rs 1,18,000 by paying every month starting with the Ist instalment of Rs 1000, If he increases the instalment by Rs 100 every month, what amount will be paid by him in the 30th instalment ? What amount of loan will he still have to pay after the 30th instalment ?      [CBSE   2012]

Solution:- See full solution