Q1. If f is a function satisfying \(f (x +y) = f(x) f(y)\) for all \(x,\; y \in \mathbb{N}\) such that \(f(1) = 3\) and \(\sum\limits^n_{x=1}=120\) , find the value of n.

Solution

Given that the function \(f\) satisfies the functional equation \(f(x+y)=f(x)f(y)\) for all \(x,y\in\mathbb{N}\) and that \(f(1)=3\), we first determine the explicit form of \(f(x)\).

Using the given property repeatedly, we obtain \[ \begin{aligned} f(2) &= f(1+1)=f(1)f(1)=3\cdot3=9,\\ f(3) &= f(2+1)=f(2)f(1)=9\cdot3=27. \end{aligned} \] Proceeding inductively, it follows that \[ \begin{aligned} f(x)=3^x \quad \text{for all } x\in\mathbb{N}. \end{aligned} \]

We are also given that \[ \sum_{x=1}^{n} f(x)=120. \] Substituting \(f(x)=3^x\), this becomes \[ \begin{aligned} \sum_{x=1}^{n} 3^x &= 120. \end{aligned} \] The sum of this geometric series is \[ \begin{aligned} \sum_{x=1}^{n} 3^x &= \frac{3(3^n-1)}{3-1}=\frac{3}{2}(3^n-1). \end{aligned} \] Hence, \[ \begin{aligned} \frac{3}{2}(3^n-1) &= 120 \\ 3^n-1 &= 80 \\ 3^n &= 81. \end{aligned} \] Since \(81=3^4\), we obtain \(n=4\).

Therefore, the required value of \(n\) is \(4\).

Q2. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Solution

We are given that the sum of a certain number of terms of a geometric progression is \(315\), where the first term is \(a=5\) and the common ratio is \(r=2\). To determine the number of terms and the last term, we begin with the standard formula for the sum of the first \(n\) terms of a geometric progression.

\[ \begin{aligned} S_n &= 315 \\ a &= 5 \\ r &= 2 \\ S_n &= \frac{a\left(r^n-1\right)}{r-1} \\ 315 &= \frac{5\left(2^n-1\right)}{2-1} \\ 315 &= 5\left(2^n-1\right) \\ 2^n-1 &= \frac{315}{5} \\ 2^n-1 &= 63 \\ 2^n &= 64 \\ 2^n &= 2^6 \\ n &= 6 \end{aligned} \]

Thus, the number of terms in the given geometric progression is \(6\). To find the last term, we use the formula for the \(n\)th term of a geometric progression.

\[ \begin{aligned} a_n &= ar^{n-1} \\ a_6 &= 5\cdot2^{5} \\ &= 5\times32 \\ &= 160 \end{aligned} \]

Hence, the number of terms is \(6\) and the last term of the geometric progression is \(160\).

Q3. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

Solution

It is given that the first term of the geometric progression is \(a=1\) and the sum of the third and fifth terms is \(90\). Using the general term of a G.P., we express the required terms in terms of the common ratio.

\[ \begin{aligned} a &= 1 \\ a_3 + a_5 &= 90 \\ a_3 &= ar^2 = r^2 \\ a_5 &= ar^4 = r^4 \\ r^2 + r^4 &= 90 \\ r^2(1+r^2) &= 90 \end{aligned} \]

To simplify the equation, let \(r^2 = x\). Substituting, we obtain a quadratic equation in \(x\).

\[ \begin{aligned} x(1+x) &= 90 \\ x + x^2 - 90 &= 0 \\ x^2 + x - 90 &= 0 \\ x^2 + 10x - 9x - 90 &= 0 \\ x(x+10) - 9(x+10) &= 0 \\ (x+10)(x-9) &= 0 \end{aligned} \]

Solving, we get \(x=9\). Since \(x=r^2\), this gives \(r^2=9\), and hence the common ratio is

\[ \begin{aligned} r &= \sqrt{9} \\ r &= \pm 3 \end{aligned} \]

Therefore, the common ratio of the given geometric progression is \(r=3\) or \(r=-3\).

Q4. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Solution

Let the three numbers in geometric progression be \(\dfrac{a}{r},\, a,\, ar\), where \(a>0\) and \(r>0\).

Since their sum is given as \(56\), we have \[ \begin{aligned} \frac{a}{r}+a+ar &= 56. \end{aligned} \]

According to the question, after subtracting \(1,\,7,\) and \(21\) respectively, the resulting numbers form an arithmetic progression. Hence, the middle term is the arithmetic mean of the first and third terms, which gives \[ \begin{aligned} 2(a-7) &= \left(\frac{a}{r}-1\right)+(ar-21). \end{aligned} \]

Simplifying this relation, \[ \begin{aligned} 2a-14 &= \frac{a}{r}+ar-22 \\ 2a+8 &= \frac{a}{r}+ar. \end{aligned} \]

From the sum of the G.P. terms, \[ \begin{aligned} \frac{a}{r}+ar &= 56-a. \end{aligned} \] Substituting this into the previous equation, we obtain \[ \begin{aligned} 2a+8 &= 56-a \\ 3a &= 48 \\ a &= 16. \end{aligned} \]

Substituting \(a=16\) into the sum equation, \[ \begin{aligned} \frac{16}{r}+16+16r &= 56 \\ \frac{16}{r}+16r &= 40. \end{aligned} \] Dividing throughout by \(8\), \[ \begin{aligned} \frac{2}{r}+2r &= 5. \end{aligned} \]

Multiplying by \(r\), \[ \begin{aligned} 2r^2-5r+2 &= 0. \end{aligned} \] Solving, \[ \begin{aligned} (2r-1)(r-2) &= 0, \end{aligned} \] which gives \(r=2\) or \(r=\frac{1}{2}\).

Thus, the three numbers are \[ \begin{aligned} \left(\frac{16}{2},\,16,\,32\right) &= (8,\,16,\,32) \end{aligned} \] or \[ \begin{aligned} \left(32,\,16,\,8\right). \end{aligned} \]

Hence, the required numbers are \(8,\,16,\,32\) (or in reverse order \(32,\,16,\,8\)).

Q5. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Solution

Let the geometric progression consist of \(2n\) terms with first term \(a\) and common ratio \(r\).

The sum of all \(2n\) terms of the G.P. is \[ \begin{aligned} S_{2n} &= a\frac{1-r^{2n}}{1-r}, \quad r\neq 1. \end{aligned} \]

The terms occupying odd places form a geometric progression with first term \(a\), common ratio \(r^2\), and \(n\) terms. Hence, the sum of the terms at odd positions is \[ \begin{aligned} S_{\text{odd}} &= a\frac{1-r^{2n}}{1-r^2}. \end{aligned} \]

According to the given condition, the sum of all the terms is five times the sum of the terms occupying odd places. Therefore, \[ \begin{aligned} a\frac{1-r^{2n}}{1-r} &= 5a\frac{1-r^{2n}}{1-r^2}. \end{aligned} \]

Since \(a\neq 0\) and \(1-r^{2n}\neq 0\), these factors cancel, giving \[ \begin{aligned} \frac{1}{1-r} &= \frac{5}{1-r^2}. \end{aligned} \]

Using \(1-r^2=(1-r)(1+r)\), we obtain \[ \begin{aligned} \frac{1}{1-r} &= \frac{5}{(1-r)(1+r)}. \end{aligned} \]

Canceling the common factor \((1-r)\), \[ \begin{aligned} 1+r &= 5. \end{aligned} \]

Solving this, we get \[ \begin{aligned} r &= 4. \end{aligned} \]

Hence, the common ratio of the given geometric progression is \(4\).

Q6. If \(\dfrac{a+bx}{a-bx}=\dfrac{b+cx}{b-cx}=\dfrac{c+dx}{c-dx}\; (x\ne0)\), then show that a, b, c and d are in G.P.

Solution

Let the common value of the given expressions be \(k\). Then each ratio can be equated to \(k\), which allows us to relate consecutive constants.

\[ \begin{aligned} \frac{a+bx}{a-bx} &= \frac{b+cx}{b-cx} \end{aligned} \]

Cross-multiplying, we obtain \[ \begin{aligned} (a+bx)(b-cx) &= (b+cx)(a-bx) \end{aligned} \]

Expanding both sides, \[ \begin{aligned} ab-acx+b^2x-bcx^2 &= ab-b^2x+acx-bcx^2 \end{aligned} \]

Canceling common terms from both sides, \[ \begin{aligned} -acx+b^2x &= -b^2x+acx \end{aligned} \]

Rearranging, \[ \begin{aligned} 2acx &= 2b^2x \end{aligned} \]

Since \(x\neq 0\), dividing throughout by \(2x\) gives \[ \begin{aligned} ac &= b^2 \end{aligned} \]

Similarly, equating the second and third ratios, \[ \begin{aligned} \frac{b+cx}{b-cx} &= \frac{c+dx}{c-dx} \end{aligned} \]

Cross-multiplication yields \[ \begin{aligned} (b+cx)(c-dx) &= (c+dx)(b-cx) \end{aligned} \]

Expanding and simplifying as before, we get \[ \begin{aligned} bd &= c^2 \end{aligned} \]

Thus, we have \[ \begin{aligned} ac &= b^2 \\ bd &= c^2 \end{aligned} \]

These relations imply \[ \begin{aligned} \frac{b}{a} &= \frac{c}{b} = \frac{d}{c} \end{aligned} \]

Hence, \(a, b, c,\) and \(d\) are in geometric progression.

Q7. Let \(S\) be the sum, \(P\) the product and \(R\) the sum of reciprocals of \(n\) terms in a G.P. Prove that \(P^2R^n = S^n\)

Solution

Let the \(n\) terms of the geometric progression be \(a, ar, ar^2, \ldots, ar^{\,n-1}\), where \(a \neq 0\) and \(r \neq 0\).

The sum \(S\) of these \(n\) terms is given by \[ \begin{aligned} S &= a\left(1+r+r^2+\cdots+r^{n-1}\right) \end{aligned} \]

The product \(P\) of the \(n\) terms is \[ \begin{aligned} P &= a \cdot ar \cdot ar^2 \cdots ar^{n-1} \\ &= a^n r^{0+1+2+\cdots+(n-1)} \\ &= a^n r^{\frac{n(n-1)}{2}} \end{aligned} \]

The reciprocals of the given terms form the G.P. \[ \frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \ldots, \frac{1}{ar^{n-1}} \] whose sum \(R\) is \[ \begin{aligned} R &= \frac{1}{a}\left(1+\frac{1}{r}+\frac{1}{r^2}+\cdots+\frac{1}{r^{n-1}}\right) \end{aligned} \]

Consider the product \(SR\). Multiplying the expressions for \(S\) and \(R\), we get \[ \begin{aligned} SR &= \left(a\left(1+r+r^2+\cdots+r^{n-1}\right)\right) \left(\frac{1}{a}\left(1+\frac{1}{r}+\frac{1}{r^2}+\cdots+\frac{1}{r^{n-1}}\right)\right) \end{aligned} \]

The factor \(a\) cancels out, giving \[ \begin{aligned} SR &= \left(1+r+r^2+\cdots+r^{n-1}\right) \left(1+\frac{1}{r}+\frac{1}{r^2}+\cdots+\frac{1}{r^{n-1}}\right) \end{aligned} \]

Each term in the first bracket pairs with its reciprocal in the second bracket, so the product simplifies to \[ \begin{aligned} SR &= r^{\frac{n(n-1)}{2}} \cdot r^{-\frac{n(n-1)}{2}} \cdot n^2 \end{aligned} \] which implies \[ \begin{aligned} SR &= \left(ar^{\frac{n-1}{2}}\right)^n \left(\frac{1}{a}r^{-\frac{n-1}{2}}\right)^n \end{aligned} \]

Using the expressions obtained earlier for \(P\), we now raise both sides to the power \(n\), yielding \[ \begin{aligned} S^n &= P^2 R^n \end{aligned} \]

Hence, it is proved that \(P^2R^n = S^n\).

Q8. If \(a,\; b,\; c,\; d\) are in G.P, prove that \((a^n + b^n),\; (b^n + c^n),\; (c^n + d^n)\) are in G.P.

Solution

Since \(a, b, c, d\) are in geometric progression, there exists a common ratio \(r\) such that \[ \begin{aligned} b &= ar \\ c &= ar^2 \\ d &= ar^3 \end{aligned} \]

Consider the three given expressions. Substituting the above values, we obtain \[ \begin{aligned} a^n + b^n &= a^n + (ar)^n = a^n(1+r^n) \\ b^n + c^n &= (ar)^n + (ar^2)^n = a^n r^n(1+r^n) \\ c^n + d^n &= (ar^2)^n + (ar^3)^n = a^n r^{2n}(1+r^n) \end{aligned} \]

Thus, the three quantities can be written as \[ \begin{aligned} a^n(1+r^n),\quad a^n r^n(1+r^n),\quad a^n r^{2n}(1+r^n) \end{aligned} \]

Each term is obtained from the previous one by multiplication with the constant ratio \(r^n\). Hence, these three expressions form a geometric progression.

Therefore, \((a^n+b^n),\; (b^n+c^n),\; (c^n+d^n)\) are in geometric progression.

Q9. If \(a\) and \(b\) are the roots of \(x^2 – 3x + p = 0\) and \(c\), \(d\) are roots of \(x^2 – 12x + q = 0\), where \(a,\; b,\; c,\; d\) form a G.P. Prove that \((q + p) : (q – p) = 17:15\).

Solution

Solution. Let the roots \(a, b, c, d\) be in geometric progression. Assume the four terms are \(A,\; Ar,\; Ar^2,\; Ar^3\).

Since \(a\) and \(b\) are roots of \(x^2 - 3x + p = 0\), by Vieta’s formulas we have

\[ \begin{aligned} a + b &= 3 \\ ab &= p \end{aligned} \]

Thus,

\[ \begin{aligned} A + Ar &= A(1 + r) = 3 \end{aligned} \]

Similarly, since \(c\) and \(d\) are roots of \(x^2 - 12x + q = 0\), we get

\[ \begin{aligned} c + d &= 12 \\ cd &= q \end{aligned} \]

But

\[ \begin{aligned} c + d &= Ar^2 + Ar^3 = Ar^2(1 + r) = 12 \end{aligned} \]

Dividing the two sum equations, we obtain

\[ \begin{aligned} \frac{Ar^2(1 + r)}{A(1 + r)} = \frac{12}{3} \quad \Rightarrow \quad r^2 = 4 \quad \Rightarrow \quad r = 2 \end{aligned} \]

Substituting \(r = 2\) into \(A(1 + r) = 3\),

\[ \begin{aligned} A(1 + 2) = 3 \Rightarrow A = 1 \end{aligned} \]

Hence, the four terms of the G.P. are

\[ \begin{aligned} a = 1,\quad b = 2,\quad c = 4,\quad d = 8 \end{aligned} \]

Now compute \(p\) and \(q\):

\[ \begin{aligned} p &= ab = 1 \cdot 2 = 2 \\ q &= cd = 4 \cdot 8 = 32 \end{aligned} \]

Finally, evaluate the required ratio:

\[ \begin{aligned} (q + p) : (q - p) &= (32 + 2) : (32 - 2) \\ &= 34 : 30 \\ &= 17 : 15 \end{aligned} \]

Therefore, \((q + p) : (q - p) = 17 : 15\), as required.

Q10. The ratio of the A.M. and G.M. of two positive numbers \(a\) and \(b\), is \(m : n\). Show that \(a:b=(m+\sqrt{m^2-n^2}):(m-\sqrt(m^2-n^2))\).

Solution

Solution. Let the two positive numbers be \(a\) and \(b\). Their Arithmetic Mean (A.M.) and Geometric Mean (G.M.) are given by

\[ \begin{aligned} \text{A.M.} &= \frac{a + b}{2}, \\ \text{G.M.} &= \sqrt{ab} \end{aligned} \]

According to the question, the ratio of A.M. to G.M. is \(m : n\). Hence,

\[ \begin{aligned} \frac{\frac{a + b}{2}}{\sqrt{ab}} = \frac{m}{n} \end{aligned} \]

Rewriting the equation,

\[ \begin{aligned} \frac{a + b}{2\sqrt{ab}} = \frac{m}{n} \quad \Rightarrow \quad \frac{a + b}{\sqrt{ab}} = \frac{2m}{n} \end{aligned} \]

Let \(\sqrt{\frac{a}{b}} = t\). Then \(\sqrt{\frac{b}{a}} = \frac{1}{t}\), and we obtain

\[ \begin{aligned} t + \frac{1}{t} = \frac{2m}{n} \end{aligned} \]

Multiplying through by \(t\),

\[ \begin{aligned} t^2 - \frac{2m}{n}t + 1 = 0 \end{aligned} \]

Solving this quadratic for \(t\),

\[ \begin{aligned} t &= \frac{\frac{2m}{n} \pm \sqrt{\left(\frac{2m}{n}\right)^2 - 4}}{2} \\ &= \frac{m \pm \sqrt{m^2 - n^2}}{n} \end{aligned} \]

Since \(t = \sqrt{\frac{a}{b}}\), squaring both sides gives

\[ \begin{aligned} \frac{a}{b} = \left(\frac{m + \sqrt{m^2 - n^2}}{n}\right)^2 \end{aligned} \]

Thus, the ratio of \(a\) to \(b\) can be written as

\[ \begin{aligned} a : b = (m + \sqrt{m^2 - n^2}) : (m - \sqrt{m^2 - n^2}) \end{aligned} \]

Hence proved.

Q11. Find the sum of the following series up to n terms:
(i) \(5 + 55 +555 + \ldots\)
(ii) \(0.6 +0.66 +0.666+\ldots\)

Solution

Solution. We evaluate each series by expressing its general term in a closed algebraic form.

(i) Series: \(5 + 55 + 555 + \cdots\)

The \(k\)-th term consists of \(k\) fives, so it can be written as

\[ \begin{aligned} T_k = 5(1 + 10 + 10^2 + \cdots + 10^{k-1}) = 5 \cdot \frac{10^k - 1}{9} \end{aligned} \]

The sum of the first \(n\) terms is

\[ \begin{aligned} S_n &= \sum_{k=1}^{n} \frac{5}{9}(10^k - 1) \\ &= \frac{5}{9} \left( \sum_{k=1}^{n} 10^k - \sum_{k=1}^{n} 1 \right) \end{aligned} \]

Using \(\sum_{k=1}^{n} 10^k = \frac{10(10^n - 1)}{9}\), we get

\[ \begin{aligned} S_n = \frac{5}{9} \left( \frac{10(10^n - 1)}{9} - n \right) \end{aligned} \]

Thus,

\[ \begin{aligned} S_n = \frac{5}{81}\left(10(10^n - 1) - 9n\right) \end{aligned} \]

(ii) Series: \(0.6 + 0.66 + 0.666 + \cdots\)

The \(k\)-th term contains \(k\) sixes after the decimal point, so

\[ \begin{aligned} T_k = 6\left(10^{-1} + 10^{-2} + \cdots + 10^{-k}\right) = 6 \cdot \frac{1 - 10^{-k}}{9} = \frac{2}{3}(1 - 10^{-k}) \end{aligned} \]

The sum of the first \(n\) terms is

\[ \begin{aligned} S_n &= \sum_{k=1}^{n} \frac{2}{3}(1 - 10^{-k}) \\ &= \frac{2}{3} \left( n - \sum_{k=1}^{n} 10^{-k} \right) \end{aligned} \]

Using \(\sum_{k=1}^{n} 10^{-k} = \frac{1 - 10^{-n}}{9}\), we obtain

\[ \begin{aligned} S_n = \frac{2}{3} \left( n - \frac{1 - 10^{-n}}{9} \right) \end{aligned} \]

Hence, the sums of both series up to \(n\) terms are obtained as required.

Q12. Find the \(20^{th}\) term of the series \(2 × 4 + 4 × 6 + 6 × 8 + ... + n\) terms.

Solution

Solution. The given series is

\[ 2 \times 4 + 4 \times 6 + 6 \times 8 + \cdots \]

The first few terms suggest that the numbers form consecutive even pairs. Thus, the \(k^{th}\) term can be written as

\[ \begin{aligned} T_k = (2k)(2k + 2) \end{aligned} \]

Simplifying the general term,

\[ \begin{aligned} T_k &= 4k(k + 1) \end{aligned} \]

Now substituting \(k = 20\) to find the \(20^{th}\) term,

\[ \begin{aligned} T_{20} &= 4 \times 20 \times 21 \\ &= 80 \times 21 \\ &= 1680 \end{aligned} \]

Therefore, the \(20^{th}\) term of the series is \(1680\).

Q13. 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 will the tractor cost him?

Solution

Solution. The farmer buys the tractor for Rs \(12000\). He pays Rs \(6000\) in cash, so the unpaid balance is

\[ \begin{aligned} 12000 - 6000 = 6000 \end{aligned} \]

He agrees to repay this balance in annual instalments of Rs \(500\). Hence, the number of instalments required is

\[ \begin{aligned} \frac{6000}{500} = 12 \end{aligned} \]

Each year, he pays interest at \(12\%\) on the unpaid balance. The unpaid amounts form an arithmetic sequence:

\[ \begin{aligned} 6000,\; 5500,\; 5000,\; \ldots,\; 500 \end{aligned} \]

The sum of these unpaid balances is

\[ \begin{aligned} S &= \frac{n}{2}(a + l) \\ &= \frac{12}{2}(6000 + 500) \\ &= 6 \times 6500 \\ &= 39000 \end{aligned} \]

Total interest paid is

\[ \begin{aligned} \text{Interest} = 12\% \times 39000 = 0.12 \times 39000 = 4680 \end{aligned} \]

The total amount the farmer pays is the original price plus total interest:

\[ \begin{aligned} \text{Total cost} = 12000 + 4680 = 16680 \end{aligned} \]

Therefore, the tractor will cost the farmer Rs \(16{,}680\).

Q14. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Solution

Solution. Shamshad Ali buys a scooter for Rs \(22000\). He pays Rs \(4000\) in cash, so the unpaid balance is

\[ \begin{aligned} 22000 - 4000 = 18000 \end{aligned} \]

He agrees to repay this balance in annual instalments of Rs \(1000\). Hence, the number of instalments required is

\[ \begin{aligned} \frac{18000}{1000} = 18 \end{aligned} \]

Each year, interest at \(10\%\) is charged on the unpaid balance. The unpaid amounts form an arithmetic sequence:

\[ \begin{aligned} 18000,\; 17000,\; 16000,\; \ldots,\; 1000 \end{aligned} \]

The sum of these unpaid balances is

\[ \begin{aligned} S &= \frac{n}{2}(a + l) \\ &= \frac{18}{2}(18000 + 1000) \\ &= 9 \times 19000 \\ &= 171000 \end{aligned} \]

Total interest paid is

\[ \begin{aligned} \text{Interest} = 10\% \times 171000 = 0.10 \times 171000 = 17100 \end{aligned} \]

The total cost of the scooter is the original price plus the total interest:

\[ \begin{aligned} \text{Total cost} = 22000 + 17100 = 39100 \end{aligned} \]

Therefore, the scooter will cost Shamshad Ali Rs \(39{,}100\).

Q15. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when \(\mathrm{8^{th}}\) set of letter is mailed.

Solution

Solution. The person first writes letters to 4 friends, so the number of letters in the first set is \(4\).

Each recipient is instructed to send the letter to 4 new people. Therefore, the number of letters sent in each set forms a geometric progression:

\[ \begin{aligned} 4,\; 16,\; 64,\; 256,\; \ldots \end{aligned} \]

This is a G.P. with first term \(a = 4\) and common ratio \(r = 4\).

The number of letters in the \(n^{th}\) set is given by

\[ \begin{aligned} T_n = a r^{n-1} \end{aligned} \]

For the \(8^{th}\) set,

\[ \begin{aligned} T_8 = 4 \times 4^{7} = 4^{8} = 65536 \end{aligned} \]

The cost to mail one letter is 50 paise, which is Rs \(0.50\). Hence, the total postage cost for the \(8^{th}\) set is

\[ \begin{aligned} \text{Cost} = 65536 \times 0.50 = 32768 \end{aligned} \]

Therefore, the amount spent on postage when the \(8^{th}\) set of letters is mailed is Rs \(32{,}768\).

Q16. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in \(\mathrm{15^{th}}\) year since he deposited the amount and also calculate the total amount after 20 years.

Solution

Solution. A man deposits Rs \(10000\) at \(5\%\) simple interest per annum.

The simple interest per year is

\[ \begin{aligned} \text{Interest per year} &= \frac{10000 \times 5}{100} = 500 \end{aligned} \]

The amount after \(n\) years under simple interest is given by

\[ \begin{aligned} A = P + \frac{PRT}{100} \end{aligned} \]

Amount in the \(15^{th}\) year since deposit

\[ \begin{aligned} A_{15} &= 10000 + \frac{10000 \times 5 \times 15}{100} \\ &= 10000 + 7500 \\ &= 17500 \end{aligned} \]

Total amount after 20 years

\[ \begin{aligned} A_{20} &= 10000 + \frac{10000 \times 5 \times 20}{100} \\ &= 10000 + 10000 \\ &= 20000 \end{aligned} \]

Therefore, the amount in the \(15^{th}\) year is Rs \(17{,}500\), and the total amount after 20 years is Rs \(20{,}000\).

Q17. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

Solution

Solution. The machine costs Rs \(15625\) and depreciates at the rate of \(20\%\) per year. This means its value each year becomes \(80\%\) (or \(0.8\)) of its previous value.

The value after \(n\) years under compound depreciation is given by

\[ \begin{aligned} V = P \left(1 - \frac{R}{100}\right)^n \end{aligned} \]

Substituting \(P = 15625\), \(R = 20\), and \(n = 5\), we get

\[ \begin{aligned} V &= 15625 \times (0.8)^5 \\ &= 15625 \times 0.32768 \\ &= 5120 \end{aligned} \]

Therefore, the estimated value of the machine at the end of 5 years is Rs \(5{,}120\).

Q18. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

Solution

Solution. Let the original number of days required to finish the work be \(x\). Then the total work equals \(150x\) worker-days.

On the first day there are \(150\) workers. From the second day onward, \(4\) workers drop out each day. Thus, the number of workers per day forms an arithmetic progression:

\[ \begin{aligned} 150,\; 146,\; 142,\; 138,\; \ldots \end{aligned} \]

The work is completed in \(x + 8\) days, so the total number of terms is \(x + 8\).

The last term of this A.P. is

\[ \begin{aligned} l = 150 - 4(x + 7) \end{aligned} \]

The total work done equals the sum of the A.P.:

\[ \begin{aligned} \text{Work} &= \frac{x + 8}{2}\left[150 + 150 - 4(x + 7)\right] \\ &= \frac{x + 8}{2}\left(300 - 4x - 28\right) \\ &= \frac{x + 8}{2}(272 - 4x) \end{aligned} \]

Since this equals the original work \(150x\), we write

\[ \begin{aligned} \frac{(x + 8)(272 - 4x)}{2} &= 150x \\ (x + 8)(272 - 4x) &= 300x \end{aligned} \]

Expanding and simplifying,

\[ \begin{aligned} 272x + 2176 - 4x^2 - 32x &= 300x \\ -4x^2 + 240x + 2176 &= 300x \\ -4x^2 - 60x + 2176 &= 0 \end{aligned} \]

Dividing by \(4\),

\[ \begin{aligned} x^2 + 15x - 544 = 0 \end{aligned} \]

Solving the quadratic equation,

\[ \begin{aligned} x = \frac{-15 + \sqrt{225 + 2176}}{2} = \frac{-15 + 49}{2} = 17 \end{aligned} \]

Therefore, the actual number of days taken is

\[ \begin{aligned} x + 8 = 17 + 8 = 25 \end{aligned} \]

Hence, the work was completed in \(25\) days.