The compound interest received on investing Rs. 10,000 for 2 years at 'y'% p.a., compounded annually is Rs. 3,689. The simple interest earned on investing Rs. 'K' for 2 years at (y + 3)% p.a. is Rs. 181 more than the compound interest received on investing Rs. (K + 400) for 2 years at (y − 2)% p.a., compounded annually. Find the value of 'K'. 

Compound interest received on investing Rs. 10,000
= Rs. [10000 × {1 + (y/100)}² − 10000]
So, [10000 × {1 + (y/100)} ² − 10000] = 3689
Or, 10000 × {(100 + y)/100}² = 10000 + 3689 = 13689
Or, {(100 + y)/100}² = 13689 ÷ 10000 = 1.3689
So, {(100 + y)/100} = √1.3689 = 1.17
So, 100 + y = 117
So, y = 17
Simple interest received on investing Rs. 'K'   = K × (17 + 3) × 2 ÷ 100 = Rs. 0.4K
Compound interest received on investing Rs. (K + 400)
= (K + 400) × {1 + (15/100)}² − (K + 400)
= (K + 400) × 1.3225 − (K + 400)
= Rs. (0.3225K + 129)
According to the question,
0.4K = (0.3225K + 129) + 181
Or, 0.0775K = 310
So, K = 310 ÷ 0.0775 = 4000