10th CBSE math solution for exercise 5.2 part 7

This page 10th CBSE math solution for exercise 5.2 part 7 is going to provide you solution for every problems that you find in the exercise no 5.2 part 6

10th CBSE math solution for exercise 5.2 part 7

(13) How many three digit numbers are divisible by 7?

Solution:

Generally the first three digit number is 100 and the last three digit number is 999

Now we need to find how many two digit numbers are divisible by 7.


105 is the first three digit number which is divisible by 7 and 994 is the last three digit number which is divisible by 7.

 105 + 112 + ........... + 994

  a n = a + (n - 1) d

  a = 105      d = 112 - 105

                      = 7

  994 = 105 + (n - 1) 7

  994 - 105 = (n - 1) 7

    889 = (n- 1) 7

   889/7 = n - 1

      127 = n - 1

      n = 127 + 1

      n = 128

Therefore 128 three digit numbers are divisible by 7.


(14) How many multiples of 4 lie between 10 and 250?

Solution:


12,16,20,.......................,248

 a n = a + (n - 1) d

a = 12      d = 16 - 12       a n = 248

                  = 4                 

  248 = 12 + (n - 1) 4

  248 - 12 = (n - 1) 4

   236 = (n - 1) 4

   236/4 = n - 1

     59 = n - 1

    n = 59 + 1

   n = 60


In this topic 10th CBSE math solution for exercise 5.2 part 7 we are going to see solution of 15th question

(15) For what value of n, are the nth terms of two APs 63,65,67,....... and 3,10,17,........ equal?

Solution:

Let a n and b n are two nth terms of the given sequence respectively

a n = a + (n - 1) d

 a = 63   d = 65 - 63

                 = 2

 a n = 63 + (n - 1) 2

b n = b + (n - 1) d

 b = 3     d = 10 - 3

                = 7

  b n = 3 + (n - 1) 7

 a n = b n

  63 + (n - 1) 2 = 3 + (n - 1) 7

  63 + 2 n - 2 = 3 + 7n - 7

    61 + 4 = 7 n - 2 n

      65 = 5 n

      n = 65/5

      n = 13




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