Aptitude Discussion

Q. |
A man arranges to pay off a debt of Rs 3600 by 40 annual installments which are in A.P. When 30 of the installments are paid he dies leaving one-third of the debt unpaid. The value of the $8^{th}$ installment is: |

✖ A. |
Rs 35 |

✖ B. |
Rs 50 |

✔ C. |
Rs 65 |

✖ D. |
Rs 70 |

**Solution:**

Option(**C**) is correct

Let the first installment be '$ a $' and the common difference between any two consecutive installments be '$ d $'

Using the formula for the sum of an A.P.

\[S=\dfrac{n}{2}[2a+(n−1)d]\]

We have,

\begin{align*}

3600&= \dfrac{40}{2}[2a+(40−1)d]\\

\Rightarrow 180&=2a+39d \tag{1} \label{eq1}\\

\text{And }2400&= \dfrac{30}{2}[2a+(30−1)d] \\

\Rightarrow 160&=2a+29d \tag{2} \label{eq2}

\end{align*}

On solving both the equations we get:

$ d=2 $ and $ a=51 $

**Note:** To check out the calculations, check comment by **Anita.**

Value of $8^{th}$ installment;

$=51+(8−1)2$

$= \text{Rs. } 65$

**Priyanka**

*()
*

It came by solving the two equations arrived at the solution,

$180=2a+39d $ -------- (1)

$160=2a+29d $ -------- (2)

Now, $(1)*29-(2)*39$

$180*29=2a*29+39d*29 $

-$(160*39=2a*39+29d*39)$

$=>180*29- 160*39=2a*29-2a*39$

$=> -1020=-20a$

$=> a=51$

It's not that difficult if you give it a proper try.

Although it would be easier to find $d$ first and then find the value of $a$.

To find $d$, just do $(2)-(1)$,

$=>(180-160)=2a+39d -(2a+29d)$

$=> 20=10d$

$=>d=2$

Now, put the value of $d$ in $(1)$,

$180=2a+39*2$

$a=\dfrac{180-78}{2}$

$=\dfrac{102}{2}$

$=51$

**Priyanka**

*()
*

2400 = 30/2 [2a + ( 30 - 1 )d ]

how 2400 came??

please explain

Since the man dies after paying 30 instalments and leaving one-third of the debt unpaid.

Total debt to be paid = 3600

Debt remaining= $\frac{1}{3}$ of 3600

$=1200$

So, debt paid in the 30 installments,

$=3600-1200$

$=2400$

Which takes to the below equation,

$2400= \dfrac{30}{2}[2a+(30−1)d]$

Hope this helps.

How $a = 51$ ,came???