A man invested 10000 rupees, split into two
schemes, at annual rates of interest 8% and 9%.
After one year he got 875 rupees as interest from
both. How much did he invest in each?
can u solve in algebric way?
Answers
Answer:
hlloo frnds!!!
Step-by-step explanation:
Let x be the amount invested at rate of 8% .Hence 10000‐ x will be the amount invested at 9%
=> we know SI = ptr/100 = x *1*8/100
=>and other SI = (10000‐x) *1*9/100
=> he got 875Rs as interest on both investments i.e,
=> ((8x/100) + (10000‐x )*9/100) = 875
=> on solving this we get x = 2500 I.e, 2500 invested at the rate of 8% and 10000‐ x = 7500 is the amount invested at the rate of 9%
hope it helps you!!!
please mark me as brainlist
Answer:
hlloo frnds!!!
Step-by-step explanation:
Let x be the amount invested at rate of 8% .Hence 10000‐ x will be the amount invested at 9%
=> we know SI = ptr/100 = x *1*8/100
=>and other SI = (10000‐x) *1*9/100
=> he got 875Rs as interest on both investments i.e,
=> ((8x/100) + (10000‐x )*9/100) = 875
=> on solving this we get x = 2500 I.e, 2500 invested at the rate of 8% and 10000‐ x = 7500 is the amount invested at the rate of 9%
hope it helps you!!!
please mark me as brainlist