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Your elder brother wants to buy a car and plans to take a loan from a bank for his car. He repays his total loan of ₹ $1,18,000$ by paying every month, starting with the first instalment of ₹ $1,000$ and he increases the instalment by ₹ $100$ every month.
Based on the information given above, answer the following questions :
(i) Find the amount paid by him in the $30^{th}$ instalment.
(ii) If the total number of instalments is $40$, what is the amount paid in the last instalment?
(iii) (a) What amount does he still have to pay after the $30^{th}$ instalment?
OR
(iii) (b) Find the ratio of the tenth instalment to the last instalment.
Based on the information given above, answer the following questions :
(i) Find the amount paid by him in the $30^{th}$ instalment.
(ii) If the total number of instalments is $40$, what is the amount paid in the last instalment?
(iii) (a) What amount does he still have to pay after the $30^{th}$ instalment?
OR
(iii) (b) Find the ratio of the tenth instalment to the last instalment.
Show SolutionHide Solution↓
$a = 1000, d = 100$
(i) $a_{30} = 1000 + 29(100)$ (1 Mark)
$= 3900$
Amount paid in $30^{th}$ instalment = ₹ $3900$
(ii) $a_{40} = 1000 + 39(100)$ (1 Mark)
$= 4900$
Amount paid in last instalment = ₹ $4900$
(iii) (a) $S_{30} = \frac{30}{2} \times [2(1000) + 29 \times 100]$ (1 Mark)
$= 73500$
Amount still he has to pay = $118000 – 73500 = \text{Rs}44500$ (1 Mark)
OR
(iii) (b) $\frac{a_{10}}{a_{40}} = \frac{1000+900}{1000+3900}$ (1 Mark)
$= \frac{1900}{4900}$
$= \frac{19}{49}$ (1 Mark)
The ratio is $19:49$
(i) $a_{30} = 1000 + 29(100)$ (1 Mark)
$= 3900$
Amount paid in $30^{th}$ instalment = ₹ $3900$
(ii) $a_{40} = 1000 + 39(100)$ (1 Mark)
$= 4900$
Amount paid in last instalment = ₹ $4900$
(iii) (a) $S_{30} = \frac{30}{2} \times [2(1000) + 29 \times 100]$ (1 Mark)
$= 73500$
Amount still he has to pay = $118000 – 73500 = \text{Rs}44500$ (1 Mark)
OR
(iii) (b) $\frac{a_{10}}{a_{40}} = \frac{1000+900}{1000+3900}$ (1 Mark)
$= \frac{1900}{4900}$
$= \frac{19}{49}$ (1 Mark)
The ratio is $19:49$