This page 10th CBSE math solution for exercise 5.3 part 9 is going to provide you solution for every problems that you find in the exercise no 5.3 part 9

10th CBSE math solution for exercise 5.3 part 9

(14) Find the sum of odd numbers between 0 and 50.

Solution:

1 + 3 + 5 + 7 + .............. + 49

a = 1 d = 3 - 1 = 2 a n = 49

a n = a + (n - 1) d

49 = 1 + (n - 1) 2

49 -1 = (n-1) 2

48/2 = n - 1

24 = n - 1

n = 24 + 1

= 25

Sum of 25 terms

Sn= (n/2) [ 2 a + (n - 1) d]

S₂₅
= (25/2) [ 2 (1) + (25 - 1) (2)]

= (25/2) [ 2 + 24 (2)]

= (25/2) [ 50]

= 625

(15)
A contract on construction job specifies a penalty for delay for
completion beyond a certain due date as follows. Rs 200 for the first
day,Rs 250 for the second day, Rs 300 for the third day etc., the
penalty for each succeeding day being Rs 50more than for the preceding
day. How much money the contractor has to pay as penalty,if he has
delayed the work be 30 days.

Solution:

For the first day of delay he has to pay Rs 200

for the second day he has to pay Rs 250

200 + 250 + 300 + .............

To find the money that has to pay for 30 days delay,we have to find the sum of 30 terms

a = 200 d = 250 - 200 n = 30

= 50

200 + 250 + 300 + ...........

Sn= (n/2) [ 2 a + (n - 1) d]

S
₃₀= (30/2) [2(200) + (30-1) (50)]

= 15 [400 + 29(50)]

= 15 [400 + 1450]

= 15 [1850]

= 27750

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In the page 10th CBSE math solution for exercise 5.3 part 9 next we are going to see the solution of next problem.

(16)
A sum of Rs 700 is to be used to seven cash prizes to students of a
school for their overall academic performance. If each prize is Rs 20
less than the preceding prize, find the value of each prizes.