Problem 1 :
A construction company will be penalized each day of delay in construction for bridge. The penalty will be $4000 for the first day and will increase by $10000 for each following day. Based on its budget, the company can afford to pay a maximum of $ 165000 toward penalty. Find the maximum number of days by which the completion of work can be delayed
Solution :
Let us write the penalty amount paid by the construction company from the first day as sequence
4000,5000,6000,..............
The company can pay 165000 as penalty for this delay at maximum.
So, we have to write this amount as series
4000 + 5000 + 6000 +.....
and the sum of the penalty amount is 165000.
Sn = 165000
(n/2)[2a +(n - 1)d] = 165000
Substitute a = 4000 and d = 1000.
(n/2)[2(4000) + (n -1)1000] = 165000
(n/2)[8000 + (n - 1)1000] = 165000
(n/2)[8000 + 1000n - 1000] = 165000
(n/2)[7000 + 1000n] = 165000
n[7000 + 1000n] = 165000 x 2
7000n + 1000n^{2} = 330000
Divide each side by 1000.
7n + n^{2 }= 330
n^{2} + 7n - 330 = 0
(n - 15)(n + 22) = 0
n = 15, -22
Here 'n' represents number of days delayed.
So it must be positive.
Therefore the correct answer is 15.
Problem 2 :
The sum of $1000 is deposited every year at 8% simple interest. Calculate the interest at the end of each year. Do these interest amounts form an A.P?. If so,find the total interest at the end of 30 years.
Solution :
First let us find the interest using simple interest formula.
Simple Interest = (PNR) / 100
Substitute P = 1000, N = 1 and R = 8.
Simple Interest = (1000 x 1 x 8) / 100
Simple Interest = 80
In the first year, amount deposited is $1000.
Then, interest earned at the end of the first year is
= $80.
In the second year, amount deposited is $1000.
Total deposit for the second year is
= 1000 + 1000
= $2000
Then, interest earned at the end of the first year is
= 80 + 80
= $160
So, the interest amounts from the first year are
80, 160, 240..........
This sequence is an arithmetic progression.
Therefore interest amounts form an arithmetic progression.
To find the total interest for 30 years, we have to find the sum of 30 terms in the above arithmetic progression.
Formula to find sum of 'n' terms in an arithmetic progression is
Sn = (n/2) [2a + (n - 1)d]
Substitute a = 80, d = 80 and n = 30.
= (30/2) [2(80) + (30 - 1)80]
= 15[160 + 29(80)]
= 15[160 + 2320]
= 15[2480]
= 37200
So, the total interest earned at the and of 30 years is $37200.
Problem 3 :
If a clock strikes once at 1'o clock,twice at 2'o clock and so on. How many times will it strike a day?
Solution :
The clock strikes once at 1'o clock, twice at 2'o clock and so on.
1, 2, 3, ..............................12
1, 2, 3, ..............................12
The above sequences are arithmetic sequences.
Find the sum of terms in both the sequences.
= 2[1 + 2 + 3 + ............ + 12]
Because the sequences are arithmetic progressions, we can use the formula to find sum of 'n' terms of an arithmetic series.
= 2 x (n/2)[a + l]
Substitute n = 12, a = 1 and l = 12.
= 2 x (12/2)[1 + 12]
= 12[13]
= 156
Therefore the clock will strike 156 times in a day.
Problem 4 :
A gardener plans to construct a trapezoidal shaped structure in his garden. The longer side of trapezoid needs to start with a row of 97 bricks. Each row must be decreased by 2 bricks on each end and the construction should stop at 25^{th} row. How many bricks does he need to buy ?
Solution :
The longer side of trapezoid shaped garden is containing 97 and each row must be decreased by 2.
The construction has to be stopped when it reaches the 25^{th} row.
If we write the number of bricks in each row as a sequence, we get
97, 93, 89,............
The above sequence is an arithmetic progression.
We can use the formula to find the sum of 'n' terms of an arithmetic series.
Sn = (n/2)[2a + (n - 1)d]
Substitute a = 97, d = -4 and n = 25.
= (25/2)[2(97) + (25 - 1)(-4)]
= (25/2)[194 + (24)(-4)]
= (25/2)[194 - 96]
= (25/2)(98)
= 25 x 49
= 1225
So, the gardener needs 1225 bricks.
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