Algebra Exercise : Q.1 to Q.6 , Q.5 and Q.7 , Q.8 to Q.14 , Q.15 to Q.18 , Q.19 to Q. 21
Answer 11.
Salary increased by 8%
After increasing the salary = $ 237.60
That means $237 . 60 is 108% of original salary
So, 108 % = 237 .60
100 % = ?
= (100 x 237. 60) /108
= 220
Hence before 8% increase the salary was $ 220.
Answer 13
Interest for first year = $ 256
Suppose Pat invests $ ‘m’ amount at 10% simple interest & $ ‘n’ amount at 8% interest
m +n = 3000 ………………………Equation 1
m = 3000 – n
Now investment for first year will be
( 10 /100) x m + (8/100) x n = 256 ……………….Equation 2
Put the value of ‘m’ from Equation 1
(10/100) (3000-n) + (8/100) x n = 256
Multiply by 100 both side
10(3000-n) + 8n = 256 x 100
30000 – 10n + 8n = 25600
-2n = – 4400
2n = 4400
n = 4400/2
n = 2200
m + n = 3000
m = 3000 – 2200
m = 800
Hence Pat will invest $ 800 with 10% interest and $ 2200 at 8% interest
Answer 14
