Q.A home loan of ₹6,00,000 is sanctioned at 12% per annum, to be repaid over 36 monthly instalments. Find the EMI, and the total interest paid over the loan's tenure.

Apply the EMI formula with the given home-loan numbers, then use $n\times\text{EMI}-P$ for the total interest.

$(1.01)^{36}=1.430769$.

$\text{EMI}=\dfrac{6{,}00{,}000\times0.01\times1.430769}{0.430769}\approx19{,}928.59$

Total interest $\approx36\times19{,}928.59-6{,}00{,}000\approx1{,}17{,}429.24$

Here $P=₹6{,}00{,}000$, annual rate $12\%$ so $r=0.01$, tenure $n=36$ months.

Step 1 — Compute $(1+r)^n$. $(1.01)^{36} = 1.430769$.

Step 2 — Apply the EMI formula.

Step 3 — Total repaid and total interest.

Comparing this loan's EMI to the smaller-principal Worked Examples above and concluding "the EMI should be roughly proportional to the principal" is only approximately true, since the tenure here is also different (36 months, not 12 or 24) — the EMI depends on all three of $P$, $r$ and $n$ together, not on $P$ alone.