Maria needs to buy shoes for $50, jeans for $40, and a shirt for $20. She has a coupon that gives her a discount of x percent on
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45 dimes and 93 nickels were in bank
<em><u>Solution:</u></em>
Let "n" be the number of nickels
Let "d" be the number of dimes
We know that,
value of 1 nickel = $ 0.05
value of 1 dime = $ 0.10
<em><u>Given that There are three more than twice as many nickels as there are dimes</u></em>
Number of nickels = 3 + 2(number of dimes)
n = 3 + 2d ---- eqn 1
<em><u>Also given that coin bank that excepts only nickels and dimes contains $9.15</u></em>
number of nickels x value of 1 nickel + number of dimes x value of 1 dime = 9.15
0.05n + 0.10d = 9.15 ---- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
Substitute eqn 1 in eqn 2
0.05(3 + 2d) + 0.10d = 9.15
0.15 + 0.1d + 0.10d = 9.15
0.2d = 9
<h3>d = 45</h3>From eqn 1
n = 3 + 2(45)
n = 3 + 90 = 93
<h3>n = 93</h3>Thus 45 dimes and 93 nickels were in bank
Answer:
a) Number of projects in the first year = 90
b) Earnings in the twelfth year = $116500
Total money earned in 12 years = $969000
Step-by-step explanation:
Given that:
Number of projects done in fourth year = 129
Number of projects done in tenth year = 207
There is a fixed increase every year.
a) To find:
Number of projects done in the first year.
This problem is nothing but a case of arithmetic progression.
Let the first term i.e. number of projects done in first year =
Given that:
Formula for term of an Arithmetic Progression is given as:
Where will represent the number of projects increased every year.
and is the year number.
Subtracting (2) from (1):
By equation (1):
<em>Number of projects in the first year = 90</em>
<em></em>
<em>b) </em>
Number of projects in the twelfth year =
Each project pays $500
Earnings in the twelfth year = 233 500 = $116500
Sum of an AP is given as:
It gives us the total number of projects done in 12 years = 1938
Total money earned in 12 years = 500 1938 = $969000