Algebra – Typical Problems
Consecutive Numbers
• Two consecutive integers add to 𝟏𝟓𝟏 . What are the two numbers?
1st Number : 𝒙
2nd Number : 𝒙 + 𝟏
Equation : (𝒙) + (𝒙 + 𝟏) = 𝟏𝟓𝟏 .
Solution : 𝟐𝒙 + 𝟏 = 𝟏𝟓𝟏


𝟐𝒙 = 𝟏𝟓𝟎
𝒙 = 𝟕𝟓 , 𝒙 + 𝟏 = 𝟕𝟔 .
Check : 𝟕𝟓 + 𝟕𝟔 = 𝟏𝟓𝟏 
• Three consecutive integers add to 𝟐𝟎𝟔𝟏 . What are the numbers?
(𝒙) + (𝒙 + 𝟏) + (𝒙 + 𝟐) = 𝟐𝟎𝟔𝟏

𝟑𝒙 + 𝟑 = 𝟐𝟎𝟔𝟏
𝟑𝒙 = 𝟐𝟎𝟓𝟖

𝒙 = 𝟔𝟖𝟔 , 𝒙 + 𝟏 = 𝟔𝟖𝟕 , 𝒙 + 𝟐 = 𝟔𝟖𝟖
Check : 𝟔𝟖𝟔 + 𝟔𝟖𝟕 + 𝟔𝟖𝟖 = 𝟐𝟎𝟔𝟏 
• Two consecutive odd integers add to 𝟏𝟔𝟐 . What are the numbers?
(𝒙) + (𝒙 + 𝟐) = 𝟏𝟔𝟒

𝟐𝒙 = 𝟏𝟔𝟐

𝒙 = 𝟖𝟏 , 𝒙 + 𝟐 = 𝟖𝟑
Check : 𝟖𝟏 + 𝟖𝟑 = 𝟏𝟔𝟒 
• Three consecutive even integers add to 𝟏𝟔𝟖 . What are the numbers?
(𝒙) + (𝒙 + 𝟐) + (𝒙 + 𝟒) = 𝟏𝟔𝟖
𝟑𝒙 = 𝟏𝟔𝟐


𝟑𝒙 + 𝟔 = 𝟏𝟔𝟖

𝒙 = 𝟓𝟒 , 𝒙 + 𝟐 = 𝟓𝟔 , 𝒙 + 𝟒 = 𝟓𝟖
Check : 𝟓𝟒 + 𝟓𝟔 + 𝟓𝟖 = 𝟏𝟔𝟖 
• Find the three consecutive odd integers such that :
The sum of the first, twice the second, and three times the third is 𝟕𝟎 .
𝒙 + 𝟐(𝒙 + 𝟐) + 𝟑(𝒙 + 𝟒) = 𝟕𝟎

𝒙 + 𝟐𝒙 + 𝟒 + 𝟑𝒙 + 𝟏𝟐 = 𝟕𝟎

𝟔𝒙 + 𝟏𝟔 = 𝟕𝟎

𝟔𝒙 = 𝟓𝟒

𝒙 = 𝟗 , 𝒙 + 𝟐 = 𝟏𝟏 , 𝒙 + 𝟒 = 𝟏𝟑 .
Check : (𝟗) + 𝟐(𝟏𝟏) + 𝟑(𝟏𝟑) = 𝟗 + 𝟐𝟐 + 𝟑𝟗 = 𝟕𝟎 
Number Problems
• Two times some number is the same as four less than three times the same number.
? ( )
𝟐𝒙 = 𝟑𝒙 − 𝟒 
𝒙 = 𝟒 . Check : 𝟐(𝟒)
𝟑 𝟒 −𝟒
 𝟖=𝟖 
=
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Algebra – Typical Problems
The Distance Formula
𝒅=𝒓∙𝒕,
where
𝒅 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 , 𝒓 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑠𝑝𝑒𝑒𝑑 , 𝒕 = 𝑡𝑖𝑚𝑒 .
• A boat travels 𝟏𝟐 𝑘𝑚/ℎ𝑟 in still water. How long will it take to cruise 𝟐𝟓 𝑘𝑚
upstream, if the river’s current is 𝟑 𝑘𝑚/ℎ𝑟 ?
Since the current is opposite the direction of travel, the speed of the boat is
reduced. 𝒓 = 𝟗 𝑘𝑚/ℎ𝑟 . 𝒕 =
𝒅
𝒓
𝟐𝟓 𝑘𝑚
𝟕
= 𝟗 𝑘𝑚/ℎ𝑟 = 𝟐 𝟗 ℎ𝑟 .
• Two trains, 𝟏𝟎𝟎 𝑚𝑖 apart, are traveling towards each other on the same track.
One of the trains is traveling at 𝟓𝟎 𝑚𝑝ℎ and the other at 𝟔𝟎 𝑚𝑝ℎ . How long
before they hit?
We can think of this as one train traveling at the combined speed of 𝟏𝟏𝟎 𝑚𝑝ℎ
for a distance of 𝟏𝟎𝟎 𝑚𝑖 . 𝒕 =
In minutes,
𝟔
𝟏𝟏
𝑚𝑖𝑛 ×
𝟏𝟎
𝟏𝟏
ℎ𝑟 ×
𝟔𝟎 𝑠𝑒𝑐
𝟏 𝑚𝑖𝑛
=
𝟔𝟎 𝑚𝑖𝑛
𝟏 ℎ𝑟
𝟑𝟔𝟎
𝟏𝟏
=
𝟔𝟎𝟎
𝟏𝟏
𝒅
𝒓
𝟏𝟎𝟎 𝑚𝑖
𝟏𝟎
= 𝟏𝟏𝟎 𝑚𝑖/ℎ𝑟 = 𝟏𝟏 ℎ𝑟 .
𝟔
min = 𝟓𝟒 𝟏𝟏 𝑚𝑖𝑛 .
𝟖
𝑠𝑒𝑐 = 𝟑𝟐 𝟏𝟏 𝑠𝑒𝑐 .
It will take almost 𝟓𝟒 𝑚𝑖𝑛 𝟑𝟑 𝑠𝑒𝑐 .
• Fred and Barney are traveling the same distance in different cars. Fred is driving
𝟏𝟎 𝑚𝑝ℎ faster than Barney. It takes Fred 𝟓 ℎ𝑟𝑠 and Barney 𝟔 ℎ𝑟𝑠 .
a. What are their respective speeds?
b. How far did they travel?
We need to use two instances of the distance formula, one for each driver :
𝒅𝑭 = 𝒓𝑭 ∙ 𝒕𝑭 and 𝒅𝑩 = 𝒓𝑩 ∙ 𝒕𝑩 . Their distances are the same, so
𝒓 𝑭 ∙ 𝒕𝑭 = 𝒓 𝑩 ∙ 𝒕𝑩 .
Barney’s speed : 𝒙 𝑚𝑝ℎ
Fred’s speed :
𝒙 + 𝟏𝟎 𝑚𝑝ℎ
𝟔𝒙 = 𝟓(𝒙 + 𝟏𝟎)  𝟔𝒙 = 𝟓𝒙 + 𝟓𝟎 
𝒙 = 𝟓𝟎 𝑚𝑝ℎ .
Barney’s speed : 𝒙 = 𝟓𝟎 𝑚𝑝ℎ .
Fred’s speed :
𝒙 + 𝟏𝟎 = 𝟓𝟎 + 𝟏𝟎 = 𝟔𝟎 𝑚𝑝ℎ .
Distance : 𝟓𝟎 𝑚𝑝ℎ × 𝟔 ℎ𝑟 = 𝟑𝟎𝟎 𝑚𝑖 . Check : 𝟔𝟎 𝑚𝑝ℎ × 𝟓 ℎ𝑟 = 𝟑𝟎𝟎 𝑚𝑖 .
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Algebra – Typical Problems
Simple Interest
𝑰 = 𝑷𝒓𝒕 , where 𝑰 = 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑒𝑎𝑟𝑛𝑒𝑑 𝑜𝑟 𝑝𝑎𝑖𝑑 , 𝒓 = 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒 , and
𝒕 = 𝑡𝑖𝑚𝑒 𝑜𝑓 𝑙𝑜𝑎𝑛 𝑜𝑟 𝑖𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 .
𝟏
• Mike put $𝟓𝟎𝟎 in a savings account with a simple interest rate of 𝟏 𝟐 % .
a. How much interest did he interest did he earn in one year?
𝑰 = 𝑷𝒓𝒕 = (𝟓𝟓𝟎)(𝟎. 𝟎𝟏𝟓)(𝟏) = $𝟖. 𝟐𝟓 .
b. How much did he have in his account at the end of one year?
$𝟓𝟎𝟎 + $𝟖. 𝟐𝟓 = $𝟓𝟎𝟖. 𝟐𝟓 .
Accumulated Amount – The Accumulated Amount is the Principle plus the Interest :
𝑨 = 𝑷 + 𝑰 = 𝑷 + 𝑷𝒓𝒕 = 𝑷(𝟏 + 𝒓𝒕) .
• Melinda had $𝟏𝟑, 𝟓𝟏𝟓 in an account after one year. The interest rate was 𝟐% .
a. How much did she originally deposit?
𝑨
𝑷 = 𝟏+𝒓𝒕 =
𝟏𝟑,𝟓𝟏𝟓
𝟏.𝟎𝟐
= $𝟏𝟑, 𝟐𝟓𝟎 .
• Jill borrowed $𝟏𝟎𝟎𝟎 with simple interest. She didn’t pay anything for two years,
at which time she owed $𝟏𝟓𝟎𝟎 .
a. What was the interest rate ?
Since 𝑨 = 𝑷 + 𝑷𝒓𝒕 , 𝑷𝒓𝒕 = 𝑨 − 𝑷 , and 𝒓 =
𝑨−𝑷
𝑷∙𝒕
𝟓𝟎𝟎
= 𝟐𝟎𝟎𝟎 = 𝟎. 𝟐𝟓 = 𝟐𝟓% .
Averages
𝑺𝒖𝒎 𝒐𝒇 𝑽𝒂𝒍𝒖𝒆𝒔
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 = 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑽𝒂𝒍𝒖𝒆𝒔
.
• Tom got scores on his 1st four tests of 𝟔𝟒 , 𝟕𝟐 , 𝟓𝟖 , and 𝟔𝟖 , each out of
𝟏𝟎𝟎 points . There is one more test, and he needs an average of 𝟕𝟎 to pass
the class.
a. How many points does he need on the 5 th test?
He needs at least a 𝟕𝟎 average :
𝟔𝟒 + 𝟕𝟐 + 𝟒𝟖 + 𝟔𝟖 + 𝒙
𝟓
≥ 𝟕𝟎

𝟐𝟓𝟐 + 𝒙 ≥ 𝟑𝟓𝟎 
𝒙 ≥ 𝟗𝟖 .
b. Can he pass?
Yes, but it is not likely that he will get a 𝟗𝟖 , considering his other scores.
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