The amount paid in the from of penalty is in the from of following AP: Rs 200, Rs 250, Rs 300, Rs 350, ... Here, a = 200, d = 250 - 200 = 50 and n = 30. The sum of n terms of an AP is given by S_n = n/2[2a + (n - 1)d] ⇒ S_30 = 30/2[2(200) + (30 - 1)(50)] = 15[400 + 1450] = 15 (1850) = 27750 Hence, tRead more
The amount paid in the from of penalty is in the from of following AP:
Rs 200, Rs 250, Rs 300, Rs 350, …
Here, a = 200, d = 250 – 200 = 50 and n = 30.
The sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d]
⇒ S_30 = 30/2[2(200) + (30 – 1)(50)]
= 15[400 + 1450]
= 15 (1850) = 27750
Hence, the contractor has to pay Rs 27750 as penalty for the delay of 30 days.
Let the amount for first prize = Rs x number of prizes = 7 Total prize amount = Rs 700 Therefore, the series of 7 prizes are as follows: (x) + (x - 20) + (x - 40) + (x - 60) + (x - 80) + (x - 100) + (x - 120) = 700 ⇒ 7x - 420 = 700 ⇒ 7x = 1120 ⇒ x = 1120/7 = 160 Hence, the seven prizes are Rs 160, RRead more
Let the amount for first prize = Rs x
number of prizes = 7
Total prize amount = Rs 700
Therefore, the series of 7 prizes are as follows:
(x) + (x – 20) + (x – 40) + (x – 60) + (x – 80) + (x – 100) + (x – 120) = 700
⇒ 7x – 420 = 700
⇒ 7x = 1120
⇒ x = 1120/7 = 160
Hence, the seven prizes are Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60 and Rs 40.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: rupay 200 for the first day, rupay 250 for the second day, rupay 300 for the third day, etc., the penalty for each succeeding day being rupay 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
The amount paid in the from of penalty is in the from of following AP: Rs 200, Rs 250, Rs 300, Rs 350, ... Here, a = 200, d = 250 - 200 = 50 and n = 30. The sum of n terms of an AP is given by S_n = n/2[2a + (n - 1)d] ⇒ S_30 = 30/2[2(200) + (30 - 1)(50)] = 15[400 + 1450] = 15 (1850) = 27750 Hence, tRead more
The amount paid in the from of penalty is in the from of following AP:
See lessRs 200, Rs 250, Rs 300, Rs 350, …
Here, a = 200, d = 250 – 200 = 50 and n = 30.
The sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d]
⇒ S_30 = 30/2[2(200) + (30 – 1)(50)]
= 15[400 + 1450]
= 15 (1850) = 27750
Hence, the contractor has to pay Rs 27750 as penalty for the delay of 30 days.
A sum of rupay 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is rupay 20 less than its preceding prize, find the value of each of the prizes.
Let the amount for first prize = Rs x number of prizes = 7 Total prize amount = Rs 700 Therefore, the series of 7 prizes are as follows: (x) + (x - 20) + (x - 40) + (x - 60) + (x - 80) + (x - 100) + (x - 120) = 700 ⇒ 7x - 420 = 700 ⇒ 7x = 1120 ⇒ x = 1120/7 = 160 Hence, the seven prizes are Rs 160, RRead more
Let the amount for first prize = Rs x
See lessnumber of prizes = 7
Total prize amount = Rs 700
Therefore, the series of 7 prizes are as follows:
(x) + (x – 20) + (x – 40) + (x – 60) + (x – 80) + (x – 100) + (x – 120) = 700
⇒ 7x – 420 = 700
⇒ 7x = 1120
⇒ x = 1120/7 = 160
Hence, the seven prizes are Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60 and Rs 40.