Question

$\mathfrak{\huge{\underline{Question:}}}$

Read carefully,Think logically,Analyse correctly and solve the following twisted problem :D


Question 1 : Me,a pickle lover said,'During last 5 days,I have eaten 100 pickles.' Then I explained,'Everyday,I ate six pickles more than the previous day.' If after 5 days,my total would be 100,how many pickle did I had on the first day?
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Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

I will be solving it in 2 ways :-

1) Arithmetic Progression

GIVEN :-

The given information forms a question related to AP.

$\tt Given \begin{cases} \sf{S_n = 100\:pickles} \\ \sf{n=5\: days} \\ \sf{d=6\:pickles} \end{cases}$

TO FIND :-

a = ?

SOLUTION :-

We know that,

$S_n = \dfrac {n}{2} ( 2a + (n-1)d )$

Putting in the given values,

$\leadsto 100 = \dfrac {5}{2} ( 2a + (5-1)6 )$

$\leadsto 100 = \dfrac {5}{2} ( 2a + (4)6 )$

$\leadsto 100 \times \dfrac{2}{5} = ( 2a + (4)6 )$

$\leadsto \dfrac{\cancel {200}\:\:40}{\cancel {5}} = ( 2a + (4)6 )$

$\leadsto 40 = ( 2a + 24 )$

$\leadsto 40 = ( 2a + 24 )$

$\leadsto 40 - 24 = ( 2a )$

$\leadsto 16 = ( 2a )$

$\leadsto a = 8\:pickles$

Hence, first day you " The pickle lover " ate 8 pickles.

2)Linear Equation

Let the pickles you had on the first day = x.

So, the pickles you had on the second day = x + 6.

On the third day = x + 12.

On the fourth day = x + 18.

And On the fifth day = x + 24.

It is given that the total pickles you had all the five days = 100.

So, x + x + 6 + x + 12 + x + 18 + x + 24 = 100.

5x + 60 = 100

5 ( x + 12 ) = 100

x + 12 = 100/5

x + 12 = 20

x = 20 - 12

x = 8 pickles is your answer.

________________

Answer 2

AnswEr :

$\underline{\bf{\dag} \:\mathfrak{Arithmetic \:Progression \:Approach :}}$

$\textsf{Let the Pickle consumed by you on First Day} \\ \textsf{be \bf{x.} \textsf{You are eating 6 more pickles more than}} \\ \textsf{the previous day.}$

$\begin{aligned}\textsf{First Day, Second Day, Third Day...so on}\\\textsf{x, (x + 6), (x + 6 + 6)...so on}\\\textsf{x, (x + 6), (x + 12)...so on}\\\star\:\boxed{\textsf{This is in Arithmetic Progression}}\end{aligned}$

$\bf{ Given}\begin{cases}\textsf{First Term (a) = x}\\\textsf{Common Difference (d) = 6}\\ \textsf{Sum of Terms \sf(S_n) = 100}\\\textsf{Number of Terms (n) = 5}\end{cases}$

$\rule{100}{2}$

For any Arithmetic Progression ( AP ), the sum of n terms is Given by :

$\bf{\dag}\quad\large\boxed{\sf S_n = \dfrac{n}{2}\bigg(a + l\bigg)}$

where :

• n = no. of terms
• a = First Term
• l = Last Term

• Since, the last term is also the nth term :

◗ l = a + (n – 1)d

• Substituting for l in the formula for sum :

$\longrightarrow \tt S_n = \dfrac{n}{2}\bigg(a + [a + (n -1)d]\bigg)\\\\\\\longrightarrow \tt S_n = \dfrac{n}{2}\bigg(2a+(n-1)d\bigg)\\\\\\\longrightarrow \tt 100 = \dfrac{5}{2}\bigg((2 \times x)+(5-1)\times 6\bigg)\\\\\\\longrightarrow \tt \cancel\dfrac{100 \times 2}{5} =[( 2x + (4 \times 6)] \\ \\\\\longrightarrow \tt40 = 2x + 24 \\ \\ \\\longrightarrow \tt40 - 24 = 2x \\ \\\\\longrightarrow \tt16 = 2x \\ \\ \\\longrightarrow \tt \cancel\dfrac{16}{2} = x \\ \\ \\\longrightarrow \large \red{\boxed{\tt x = 8}}$

⠀

∴ She Had 8 Pickles on the first day.

$\rule{300}{2}$

$\underline{\bf{\dag} \:\mathfrak{Linear \: Equation \:Approach :}}$

$\textsf{As the Pickles are increasing each day linearly} \\ \textsf{till last day. So we can say third day is Average} \\ \textsf{of Total Pickles upon Total Days.}$

$:\implies\sf{Average\:Day = \dfrac{Total \:Pickles}{Total \:days}} \\\\\\:\implies\sf{Third \:Day = \dfrac{Total \:Pickles}{Total \:days}} \\\\\\:\implies\sf{Third \:Day = \dfrac{100}{5}} \\ \\ \\:\implies\sf{Third\:Day = 20}$

$\textsf{As you are eating 6 more pickles each day,} \\ \textsf{then Pickles Consumed on First Day will be :} \\\\\longrightarrow \sf First \:Day = Third \:Day - 6 - 6 \\ \\\longrightarrow \sf First \:Day = 20 - 6 - 6 \\ \\\longrightarrow \sf First \:Day = 20 - 12 \\ \\\longrightarrow \boxed{\sf First \:Day = 8}$

⠀

∴ She Ate 8 Pickles on the very first day.