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 ?

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.

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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. $200$.

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.

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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.

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

$\therefore 2a = 180 - 39 ( 2)$

$\Rightarrow a = 51$

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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.

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

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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.

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.

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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

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.

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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 ?

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.

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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?

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.

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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.

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.

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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.

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.

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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?

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

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

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

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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?

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.

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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.

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 $= 3420^{\circ}$

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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?

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.

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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?

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.

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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?

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.

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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?

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.