WILL GIVE BRAINLIEST
2 answers:
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The answers are
D --> Boron family
H --> Halogens
A --> Alkali metals
I --> Noble gases
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<u>Answer:</u>
<em>Mathew invested</em><em> $600 and $2400</em><em> in each account.</em>
<u>Solution:</u>
From question, the total amount invested by Mathew is $3000. Let p = $3000.
Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’
So, the amount invested in second account = 3000 – P
Step 1:
Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest earned in first account for one year,
Where
p = amount invested in first account
n = number of years
r = rate of interest
hence, by using above equation we get as,
----- eqn 1
Step 2:
Mathew has paid 8% interest in second account. Let us calculate the simple interest earned in second account,
Step 3:
Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)
Step 4:
Total simple interest = simple interest on first account + simple interest on second account.
Hence we get,
By substituting eqn 1 , 2, 3 in eqn 4
12000=3P + 24000 - 8P
5P = 12000
P = 2400
Thus, the value of the variable ‘P’ is 2400
Hence, the amount invested in first account = p = 2400
The amount invested in second account = 3000 – p = 3000 – 2400 = 600
Hence, Mathew invested $600 and $2400 in each account.