Differentiation (Part-38) | S N De

 

$\,33.\,\,$ To verify the cash balance, the auditor of Jaya Bank Ltd. employs his assistant to count cash in hand of Rs. $\,4500\,$. At first he counts @ Rs. $\,150\,$ per minute for $\,10\,$ minutes only, at the end of which time he begins to count @ Rs. $\,2\,$ less every minute than he did in the previous minute. Ascertain how much time he will take to count Rs. $\,4500\,$?

Sol. By question, $\,\,10 \times 150 +[148+146+\cdots +n]=4500 \\ \Rightarrow 1500 +\frac n2[2 \times 148+(n-1)\cdot (-2)]\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=4500 \\ \Rightarrow n[148-(n-1)]=4500-1500 \\ \Rightarrow n(148-n+1)=3000 \\ \Rightarrow n(149-n)=3000 \\ \Rightarrow  n^2-149n+3000=0 \\ \Rightarrow n^2-125n-24n+3000=0\\ \Rightarrow  n(n-125)-24(n-125)=0 \\ \Rightarrow (n-125)(n-24)=0 \\ \Rightarrow  n=125,\,\,24.$

Now, for $\,n=125,\,\,$ let us compute the number of notes he'll count at $\,125\,$-th minute.

$\therefore \,t_{125}=148+(125-1)\cdot (-2)=-100,\,\,$ which is impossible.

Hence, the time he will take to count Rs. $\,4500\,$ is : $\,=10+24=34\,$ mins.

$\,34.\,$  A polygon has $\,25\,$ sides, the lengths of which starting from the smallest side are in A.P. If the perimeter of the polygon is $\,1100\,$ cm and the length of the largest side is $\,10\,$ times that of the smallest, find the length of the smallest side and the common difference of the A.P.

Sol. Let the sides of the polygon be $\,\,a,a+d,\cdots, a+(25-1)d=a+24d.$

So, $\,\,\frac{25}{2}[2a+24d]=1100 \\ \Rightarrow 25(a+12d)=1100  \\ \Rightarrow a+12d=\frac{1100}{25}  \\ \Rightarrow  a+12d=44 \rightarrow(1)$

Again, $\,\,10a=a+24d  \\ \Rightarrow 10a-a=24d  \\ \Rightarrow  9a=24d \\ \Rightarrow 3a=8d \rightarrow(2)$

From $\,(1)\,\,$ we get, $\,a+12d=44  \\ \Rightarrow 3(a+12d)= 3 \times 44  \\ \Rightarrow 8d+36d=3 \times 44 \,\,\,[\text{By (2)}] \\ \Rightarrow  44d=3 \times 44  \\ \Rightarrow d=3$

Now, from $\,(1),\,$ we get, $\,\,a+12 \times 3=44 \Rightarrow a=44-36=8.$

$\,35.\,\,$  A sets out from a certain place and travels $\,1\,$ mile the first day, $\,2\,$ miles the second day, $\,3\,$ miles the third day and so on. B sets out $\,5\,$ days after A and travels the same road @ $\,12\,$ miles a day. How far will A travel before he is overtaken by B? If they continue to travel, in what time will B be overtaken by A ?

Sol. A sets out from a certain place and suppose that after $\,n\,$ days A meets B. 

$\,\therefore 1+2+3+\cdots +n=(n-5)\times 12 \\ \Rightarrow \frac{n(n+1)}{2}=12(n-5) \\ \Rightarrow n^2+n=24(n-5) \\ \Rightarrow n^2-23n+120=0\\ \Rightarrow n^2-8n-15n+120=0\\ \Rightarrow n(n-8)-15(n-8)=0 \\ \Rightarrow (n-8)(n-15)=0 \\ \Rightarrow n= 8,15\rightarrow(1).$

So, first overtake happened after $\,8\,$ days whereas 2nd overtake happened after $\,15\,$days. 

Now, $\,1+2+3+\cdots+ 8=\frac{8\times (8+1)}{2}=36.$

So, A travels $\,36\,\,$ miles before he is overtaken by B.

If they continue to travel, after $\,15\,$ days B be overtaken by A .

$\,36.\,$  A man arranges to pay off a debt of Rs. $\,12000\,$ in $\,30\,$ annual instalments which form an A.P. When $\,20\,$ of the instalments are paid, he dies leaving a half of the debt unpaid. Find the value of the first instalment.

Sol. $\,\frac{30}{2}[2a+(30-1)d]=12000 \\ \Rightarrow 15(2a+29d)=12000 \\ \Rightarrow  2a+29d=\frac{12000}{15}=800\rightarrow(1) $

Again, $\,\,\frac{20}{2}[2a+(20-1)d]=12000/2 \\ \Rightarrow 10(2a+19d)=6000 \\ \Rightarrow  2a+19d=600 \rightarrow(2)$

Now, subtracting $\,(2)\,$ from $\,(1)\,$ we get,  

$10d=800-600 \Rightarrow d=200/10=20.$

Now, from $\,(2)\,$ we get, $\,2a+ 19 \times 20=600 \\ \Rightarrow 2a=600-380 \\ \Rightarrow  2a=220 \\ \Rightarrow  a=\frac{220}{2}=110.$

$\,37.\,$ A man divides Rs. $\,24000\,$ among his four sons in such a way that the amounts received by the four sons are in A.P. The ratio of the product of the amounts received by the $\,1\,$st and the $\,3\,$rd sons to the product of the amounts received by $\,2\,$nd and $\,4\,$th sons is $\,7:15.\,$ Calculate the amounts received by each son.

Sol. Let the  amounts received by four sons of the man are $\,a-3d,\,a-d,\,a+d,\,a+3d\,$ respectively.

$\,\therefore (a-3d)+(a-d)+(a+d)+(a+3d)=24000 \\ \Rightarrow 4a=24000 \\ \Rightarrow  a=\frac{24000}{4}\\ \Rightarrow  a=6000.$

Again, $\,\,\frac{(a-3d)(a+d)}{(a-d)(a+3d)}=\frac{7}{15} \\ \Rightarrow  15(a^2+ad-3ad-3d^2)\\=7(a^2+3ad-ad-3d^2)\\ \Rightarrow  15(a^2-2ad-3d^2)=7(a^2+2ad-3d^2)\\ \Rightarrow  15a^2-30ad-45d^2=7a^2+14ad-21d^2 \\ \Rightarrow  8a^2-44ad-24d^2=0 \\ \Rightarrow 4(2a^2-11ad-6d^2)=0\\ \Rightarrow 2a^2-11ad-6d^2=0 \\ \Rightarrow 2a^2-12ad+ad-6d^2=0 \\ \Rightarrow  2a(a-6d)+d(a-6d)=0 \\ \Rightarrow  (a-6d)(2a+d)=0 \\ \Rightarrow  d=\frac a6 \,\quad[\because 2a+d \neq 0] \rightarrow(1)$

So, by $\,(1),\,$ we get, $\,\,d=a/6=6000/6=1000.$

Hence, the  amounts received by four sons of the man are 

$\,a-3d=6000-3 \times 1000=3000, \\ a-d=6000-1000=5000, \\ a+d=6000+1000=7000, \\ a+3d=6000+ 3\times 1000=9000.$

To download full PDF solution of Arithmetic Progression (SEQUENCE AND SERIES) of S.N.De Math solution, click here.

$\,38.\,$  Two posts are offered to a man. In the first one, the starting salary is Rs. $\,1200\,$ and increases annually by Rs.$\, 80;\,$ in the second, the salary commences at Rs. $\,850\,$ and increases annually by Rs.$\,120.\,$ The man decides to accept the post that will give him better income in the first 16 years. Which post will he accept? Justify your answer.

In the $\,1\,$st case, $\,a=1200 \times 12=14400,\,\,d=12\times 80=960.$

So, total income in the first $\,16\,$ years , 

$S_1=\frac{16}{2}[2 \times 14400+(16-1)\times 960]\\~~~~=345600.$

In the $\,2\,$nd case, $\,a=850 \times 12=10200,\,\,d=12 \times 120=1440.$

So,  total income in the first $\,16\,$ years , 

$S_2=\frac{16}{2}[2 \times 10200+(16-1)\times 1440]\\~~~~=336000.$

Hence, $\,\,S_1>S_2,\,\,$ and so he will accept first post.

$\,39.\,$ The salary of a man starts at Rs. $\,800\,$ and increases annually by Rs. $\,20\,$ until a maximum of Rs. $\,1120\,$ is reached. Find the total amount of money earned by the man when he serves $\,(i) 15\,$ years $\,(ii)\,\, 22\,$ years.

Sol. Here, $\,a=800 \times 12=9600,\,d=20 \times 12=240.$ Suppose that after $\,n\,$ years monthly salary of rs. $\,1120\,$ is reached.

Now, according to the problem, 

$\, 1120 \times 12=9600+(n-1)\times 240 \\ \Rightarrow 1120=800+(n-1)\times 20 \\ \Rightarrow 1120-800=(n-1) \times 20 \\ \Rightarrow n-1=\frac{320}{20} \\ \Rightarrow n=16+1=17.$

$\,(i)\,\,S_{15}=\frac{15}{2}[ 2\times 9600 +(15-1) \times 240]\\~~~~~~~~~~~~=169200. \\ (ii)\,S_{22}=\frac{17}{2}[2 \times 9600+(17-1)\times 240]\\~~~~~~~~~~~~+5\times 12 \times 1120\\~~~~~~~~~~~~=263040.$