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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 2 · Set 1 · Q36
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,…
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 ₹ $\displaystyle 1,18,000$ by paying every month, starting with the first instalment of ₹ $\displaystyle 1{,}000$ and he increases the instalment by ₹ $\displaystyle 100$ every month. Based on the information given above, answer the following questions :
(i)
Find the amount paid by him in the $\displaystyle 30^{\text {th }}$ instalment.
(ii)
If the total number of instalments is $\displaystyle 40$, what is the amount paid in the last instalment ?
(iii)
(a)
What amount does he still have to pay after the $\displaystyle 30^{\text {th }}$ instalment ?
OR
(b)
Find the ratio of the tenth instalment to the last instalment.
Marking-scheme solution
\(\displaystyle \mathrm{a}=1000, \mathrm{~d}=100\)
(i)
\[\begin{aligned}
a_{30} & =1000+29(100) \\
& =3900
\end{aligned}
\]
Amount paid in \(\displaystyle 30^{\text {th }}\) instalment \(\displaystyle =₹ 3900\)
(ii)
\[\begin{aligned}
a_{40} & =1000+39(100) \\
& =4900
\end{aligned}
\]
Amount paid in last instalment \(\displaystyle =₹ 4900\)
(iii)
\[\text { (a) } \begin{aligned}
S_{30} & =\frac{30}{2} \times[2(1000)+29 \times 100] \\
& =73500
\end{aligned}
\]
Amount still he has to pay \(\displaystyle =118000-73500=₹ 44500\)
(iii)(b)
\[\text { (b) } \frac{a_{10}}{a_{40}}=\frac{1000+900}{1000+3900}
\]
\[=\frac{1900}{4900}
\]
\(\displaystyle =\frac{19}{49}\)
The ratio is $\displaystyle 19$:$\displaystyle 49$
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CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.