Question 1: A man saved Rs. 16500 in ten years. In each year after the first he saved Rs. 100 more than he did in the receding year. How much did he save in the first year ?

Answer:

Let the amount in Rs saved by the man in the first year = x

\therefore x + ( x+100) + ( x+200) + \ldots + (x + 900) = 16500

\Rightarrow 10x + [ 100+ 200 + \ldots  + 900] = 16500

\Rightarrow 10x + \frac{9}{2} \Big[ 2(100) + ( 9-1) (100) \Big] = 16500

\Rightarrow 10x + 4500 = 16500

\Rightarrow 10x = 12000

\Rightarrow x = 1200

Therefore the man saved 1200 Rs. in the first year.

\\

Question 2: A man saves Rs. 32 during the first year, Rs. 36 in the second year and in this way he increases his savings by Rs. 4 every year. Find in what time his saving will be Rs. 200 .

Answer:

Given series is 32, 36, 40, \ldots

\therefore a = 32, \ \ \ \ \ d = 36-32=4, \ \ \ \ \ \ S_n = 200

We know, S_n = \frac{n}{2} [ 2(32) + ( n-1)(4)  ] 

\Rightarrow 200 = n [ 32 + 2n-2]

\Rightarrow 2n^2 + 30 n - 200 = 0

\Rightarrow n^2 + 15n - 100 = 0

\Rightarrow (n-5)(n+20) = 0

\Rightarrow n = 5 \ or \ n = -20 (this is not possible)

Therefore man will save 200 Rs. in 5 year.

\\

Question 3: A man arranges to pay off a debt of Rs. 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of ,the first instalment.

Answer:

Let the first instalment be a  and the common difference be d  . Therefore

a + ( a+d) + ( a+2d) + \ldots + (a+39d) = 3600 

\Rightarrow 3600 = \frac{40}{2} [ 2(a) + ( 40-1) d] 

\Rightarrow 3600 = 20 [ 2a +39 d] 

\Rightarrow 2a + 39d = 180       … … … … … i)

In the first 30  instalments, the man has paid 2400  Rs. Therefore

2400 = \frac{30}{2} [ 2(a) + ( 30-1) d ] 

\Rightarrow 2400 = 15 [ 2a + 29 d] 

\Rightarrow 2a + 29 d = 160       … … … … … ii)

Solving i) and ii) we get

180 - 39d = 160 - 29 d      \Rightarrow 20 = 10 d       \Rightarrow d = 2 

Substituting in i) we get

\therefore 2a = 180 - 39 ( 2)  

\Rightarrow a = 51 

\\

Question 4: A manufacturer of radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find (i) the production in the first year (ii) the total product in 7 years and (iii) the product in the 10^{th} year.

Answer:

Given a_3 = 600  \ \ \ \ \ a_7 = 700

We know a_n = a + ( n-1) d

\Rightarrow 600 = a+(3-1) d

\Rightarrow a + 2d = 600      … … … … … i)

Also 700 = a + (7-1) d

\Rightarrow a + 6d = 700      … … … … … ii)

Solving i) and ii)  we get

600-2d = 700 - 6d   \Rightarrow 4d = 100   \Rightarrow d = 25

\therefore a = 700-6(25) = 700 - 150 = 550

i) Production in first year = 550 units

ii) S_7 = \frac{7}{2} [ 2 ( 550) + ( 7-1)(25) ] = \frac{7}{2} [ 1250] = 4375

iii) a_{10} = 550 + ( 10-1)(25) = 550 + 225 = 775

\\

Question 5:  There are 25 trees at equal distances of 5 meters in a line with a well, the distance of the well from the nearest tree being 10 meters. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.

Answer:

The series will be 20, 30, 40, \ldots

\therefore a = 20 \ \ \ \ \ d = 30-20 = 10 \ \ \ \ \ n = 25

We know S_n = \frac{n}{2} [ 2a+(n-1)d]

\therefore S_{25}= \frac{25}{2} [ 2(20) + ( 25-1)(10) ] = \frac{25}{2} [ 40 + 240] = 3560

The total distance the gardener will cover in order to water all the trees = 3560 m.

\\

Question 6: A man is employed to count Rs. 10710 . He counts at the rate of Rs. 180 per minute for half an hour. After this he counts at the rate of Rs. 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount

Answer:

Amount of Rs. counted in first 30 minutes = 180 \times 30 = 5400 Rs.

Amount of money left for counting = 10710 - 5400 = 5310 Rs.

Therefore the series would be (180-3), (180-6), (180-9) , \ldots or 177, 174, 171, \ldots

\therefore a = 177 \ \ \ \ \ d= 174-177 = -3 \ \ \ \ \ S_n = 5310

Let the time taken to count the rest of the money be n minutes

We know S_n = \frac{n}{2} [2a+(n-1)d]

\therefore 5310 = \frac{n}{2} [ 2 ( 177) + ( n-1) (-3) ]

\Rightarrow 5310 = \frac{n}{2} [ 354 - 3n + 3]

\Rightarrow 10620 = n [ 357 - 3n]

\therefore 3n^2 - 357n + 10620 = 0

(n-60)(n-59) = 0

Therefore the time taken  to count is 59 minutes.

Hence the total time taken to count the money = 30 + 59 = 89 minutes.

\\

Question 7: A piece of equipment cost a certain factory Rs. 600,000 . If it depreciates in value, 15\% the first, 13.5\% the next year, 12\% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost ?

Answer:

The percentage of depreciation of 1^{st}, 2^{nd}, 3^{rd} \ldots years is 15, 13.5, 12, \ldots

\therefore a = 15 \ \ \ \ d = 13.5 - 15 = - 1.5 \ \ \ \ \ n = 10

Therefore total depreciation percentage is S_{10}

S_{10} = \frac{10}{2} [ 2(15) + ( 10 - 1) ( -1.5) ] = 5[30-13.5] = 82.5

Therefore the total depreciation = \frac{82.5}{100} \times 600000 = 495000 Rs.

Therefore the value at the end of 10^{th} year = 600000-495000 = 105000 Rs.

\\

Question 8: A farmer buys a used tractor for Rs. 12000 . He pays Rs. 6000 cash and agrees to pay the balance in annual instalments of Rs. 500 plus 12\% interest on the unpaid amount. How much the tractor cost him?

Answer:

Cost of the tractor = 12000 Rs.

Unpaid amount = 12000 - 6000 = 6000 Rs.

The farmer pays 500 Rs. per month. This means that the farmer will pay the remaining principal in 12 instalments.

But he also has to pay the interest on the remaining principal at 12\% .

Therefore the interest paid would be represented by a series i.e.

6000 \times \frac{12}{100}, 5500 \times \frac{12}{100}, 5000 \times \frac{12}{100}, \ldots    or 720, 660, 600, \ldots

\therefore a = 720 \  \ \ \ \ d = 660-720 =-60 \ \ \ \ \ n = 12

\therefore S_{12} = \frac{12}{2} [ 2(720) + ( 12-1) ( -60) ] = 6[ 1440 - 660] = 4680 Rs.

Therefore the total cost of the tractor = 12000 + 4680 = 16680 Rs.

\\

Question 9: Shamshad Ati buys a scooter for Rs. 22000 . He pays Rs. 4000 cash and agrees to pay the balance in annual instalments of Rs. 1000 plus 10\% interest on the unpaid amount. How much the scooter will cost him.

Answer:

Cost of the scooter = 22000 Rs.

Unpaid amount = 22000 - 4000 = 18000 Rs.

The farmer pays 1000 Rs. per month. This means that the farmer will pay the remaining principal in 18 instalments.

But he also has to pay the interest on the remaining principal at 10\% .

Therefore the interest paid would be represented by a series i.e.

18000 \times \frac{10}{100}, 17000 \times \frac{10}{100}, 16000 \times \frac{10}{100}, \ldots    or 1800, 1700, 1600, \ldots

\therefore a = 1800 \  \ \ \ \ d = 1700-1800 =-100 \ \ \ \ \ n = 18

\therefore S_{18} = \frac{18}{2} [ 2(1800) + ( 18-1) ( -100) ] = 9[ 3600 - 1700] = 17100 Rs.

Therefore the total cost of the scooter = 22000 + 17100 = 39100 Rs.

\\

Question 10: The income of a person is Rs. 300,000 in the first year and he receives an increase of Rs. 10000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.

Answer:

Given a = 300000 \ \ \ \ \ d = 10000 \ \ \ \ \ n = 20

\therefore S_{20} = \frac{20}{2} [ 2 ( 300000) + ( 20 -1) ( 10000) ] = 10 [ 600000 + 190000 ] = 7900000 Rs.

Therefore total amount received = 7900000 Rs.

\\

Question 11: A man starts repaying a loan as the first installment of  Rs. 100 . If he increases the instalments by Rs. 5 every month , what amount he will pay in the 30^{th} instalment?

Answer:

Given a = 100 \ \ \ \ \ d = 5 \ \ \ \ \ n = 30

\therefore a_{30} = 100 + ( 30 -1)(5) = 245

Therefore the 30^{th} instalment = 245 Rs.

\\

Question 12: A carpenter was hired to build 192 window frames. The first day he made five frames and each day thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?

Answer:

Given a = 5, \ \ \ \ d= 2 \ \ \ \ \ \ S_n = 192

S_n = \frac{n}{2} [ 2a + ( n-1) d]

\Rightarrow 192 = \frac{n}{2} [ 2(5) + ( n-1) (2) ]

\Rightarrow 192 = \frac{n}{2} [ 10 + 2n - 2]

\Rightarrow 192 = n ( 4 + n)

\Rightarrow n^2 + 4n - 192 = 0

\Rightarrow (n-12)(n+16) = 0

\Rightarrow n = 12 \ or \ n = -16 (this is not possible)

\Rightarrow n = 12

Therefore the carpenter will take 12 days to complete the task.

\\

Question 13: We know that the sum of the interior angles of a triangle is 180^{\circ} . Show that the sums of the interior angles of polygons with 3, 4, 5, 6, \ldots   sides form an arithmetic progression. Find the sum of the interior angles for a 21- sided polygon.

Answer:

We know that:

The sum of interior angles of a polygon with 3 sides (a_1) = 180^{\circ}

The sum of interior angles of a polygon with 4 sides (a_2) = 360^{\circ}

The sum of interior angles of a polygon with 5 sides (a_3) = 540^{\circ}

\therefore a = 180^{\circ} \ \ \ \ d = 360^{\circ}-180^{\circ}= 180^{\circ}

Since the sum of the interior angles with 3 sides = a , then the sum of 21 sided polygon would be a_{19}

\therefore a_{19} = 180^{\circ} + ( 19-1)(180^{\circ}) = 180^{\circ} + 3240^{\circ} = 3420^{\circ}

The sum of the interior angles for a 21- sided polygon = 3420^{\circ}

\\

Question 14: In a potato race 20 potatoes are placed in a line at intervals of 4 meters with, the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Answer:

Given a = 2 \times 24 = 48 \ \ \ \ d = 2 \times 4 = 8 \ \ \ \ n =20

\therefore S_{20} = \frac{20}{2} [ 2 ( 48) + ( 20-1) ( 4) ] = 10 [ 96 + 152] = 2480

Therefore the contestant would run 2480 meters to bring back all the potatoes.

\\

Question 15: A man accepts a position with an initial salary of  Rs. 5200   per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month
ii)  What is his total earnings during the first year?

Answer:

Given a = 5200  \ \ \ \ d = 320

i) a_{10} = 5200 + ( 10-1) ( 320 ) = 5200 + 2880 = 8080

Therefore 10^{th} month salary = 8080 Rs.

ii) S_{12} = \frac{12}{2} [ 2 ( 5200) + ( 12 -1) ( 320) ] = 6 [ 10400 + 3520 ] = 83520

Therefore the total earnings in the first year = 83520 $ Rs.

\\

Question 16: A man saved Rs. 66000 in 20 years. In each succeeding year after the, first year he saved Rs. 200   more than what he saved in the previous year. How much did he save in the first year?

Answer:

Let the savings for the first year = a

Given d = 200, \ \ \ n = 20 \ \ \  S_{20} = 66000

We know S_n = \frac{n}{2} [ 2(a) + ( n-1) (d) ]

66000 = \frac{20}{2} [ 2(a)+(20-1)(200) ]

\Rightarrow 6600 = 2a + 3800

\Rightarrow 2a = 2800

\Rightarrow a = 1400

Therefore the man saved 1400 Rs. in the first year.

\\

Question 17: In a cricket team tournament 16 teams participated. A sum of Rs. 8000 is to be awarded among themselves as prize money.  If the last placed team is awarded Rs. 275   in prize money and the award increases by the same amount  for successive winning places , how much amount would the first team receive?

Answer:

Given: a = 275 \ \ \ n = 16  \ \ \ S_{16} = 8000

We know S_n = \frac{n}{2} [ 2(a) + ( n-1) (d) ]

\therefore 8000 = \frac{16}{2} [ 2(275) + ( 16-1) ( d) ]

\Rightarrow 1000 = 550 + 15 d

\Rightarrow 450 = 15 d

\Rightarrow d = 30

\therefore a_{16} = 275 + ( 16-1) ( 30) = 725

Therefore the winner will get 725 Rs.