1. Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Sample Gateway Problems: . . Working with Fractions and the Order of Operations Without Using a Calculator

2. NOTE: Gateway problems 1 & 2 on adding and subtracting fractions can both be done using the same set of steps.
Adding fractions and subtracting fractions both require finding a least common denominator (LCD), which is most easily done by factoring the denominator (bottom number) of each fraction into a product of prime numbers (a number that can be divided only by itself and 1.)

3. Sample Problem #1: Adding fractions
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Step 1: Factor the two denominators into prime factors, then
write each fraction with its denominator in factored form:
10 = 2∙5 and 35 = 5∙7, so = ∙5 5∙7
Step 2: Find the least common denominator (LCD):
LCD = 2∙5∙7

4. Sample Problem #1 (continued)
Step 3: Multiply the numerator (top)and denominator of each fraction by the factor(s) needed to turn each denominator into the LCD.
LCD = 2∙5∙ ∙ ∙
2∙5∙ ∙7∙2
Step 4: Multiply each numerator out, leaving the denominators in factored form, then add the two numerators and put them over the common denominator.
= = (note that 5∙7∙2 = 2∙5∙7 by
2∙5∙7 5∙7∙ ∙5∙ ∙5∙7 the commutative property)
Step 5: Now factor the numerator, then cancel any common factors that appear in both numerator and denominator. Once you multiply out any remaining factors, the result is your simplified answer.
= = 5∙5 = 5∙5 = =
2∙5∙ ∙5∙7 2∙5∙ ∙ /

5. Full Solution to Sample Problem #1:
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Full Solution to Sample Problem #1: 5 14
Here is the work we expect to see on your worksheet:
10 = 2∙5 and 35 = 5∙7, so = , and LCD = 2∙5∙7
∙ ∙7
= 3∙ ∙2 = = = 5∙5 = 5∙5 = = 5 2∙5 5∙7 2∙5∙7 5∙7∙2 2∙5∙7 2∙5∙ ∙5∙ ∙5∙7 2∙5∙7 2∙ /

6. Sample Problem #2: Subtracting fractions
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Step 1: Factor the two denominators into prime factors, then
write each fraction with its denominator in factored form:
14 = 2∙7 and 35 = 5∙7, so
2∙ ∙7
Step 2: Find the least common denominator (LCD):
LCD = 2∙7∙5

7. Sample Problem #2 (continued)
Step 3: Multiply the numerator and denominator of each fraction by the factor(s) needed to turn each denominator into the LCD: form:
LCD = 2∙7∙ ∙ ∙2
2∙7∙ ∙7∙2
Step 4: Multiply out the numerators, leaving the denominators in factored form, then add the two numerators and put them over the common denominator.
= =
2∙5∙7 5∙7∙ ∙5∙ ∙5∙7
Step 5: Now factor the numerator, then cancel any common factors that appear in both numerator and denominator. Once you multiply out any remaining factors, the result is your simplified answer.
21 = 3∙7 = 3∙7 = =
2∙5∙ ∙5∙7 2∙5∙ ∙ /

8. Full Solution to Sample Problem #2:
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Full Solution to Sample Problem #2: 10
Here is the work we expect to see on your worksheet:
14 = 2∙7 and 35 = 5∙7, so = , and LCD = 2∙5∙7
∙7 5∙7
= 5∙ ∙2 = = = 3∙7 = 3∙7 = = ∙7 5∙7 2∙7∙ ∙7∙ ∙5∙7 2∙5∙7 2∙5∙7 2∙5∙ ∙5∙ ∙ /

9. NOTE: Gateway problems 3 & 5 on multiplying and dividing fractions can both be done using the similar steps.
Neither multiplying fractions nor dividing fractions requires finding an LCD. These kinds of problems can be most easily done by factoring both the numerator (top number) and denominator of both fractions into a product of prime numbers, and then canceling any common factors (numbers that appear on both the top and the bottom.)

10. Sample Problem #3: Multiplying fractions
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Step 1: Factor both the numerators and denominators into
prime factors, then write each fraction in factored form:
First fraction: 39= 3∙13 and 50 = 2∙5∙5
Second fraction: 15= 3∙5 and 26 = 2∙13
So you can write 39 • as 3∙13 • 3∙
∙5∙ ∙13
NOTE: You do NOT need an LCD when multiplying fractions.

11. Sample Problem #3 (continued)
Step 2: Now just cancel any common factors that appear in both numerator and denominator. Once you multiply out any remaining factors, the result is your simplified answer.
3∙13 • 3∙5 = 3∙3 =
2∙5∙ ∙ ∙5∙
/ /
/ /
NOTE: It is much easier to factor first and then cancel, rather than multiplying out the numerators and denominators and then trying to simplify the answer (especially if you aren’t using a calculator!) If you multiplied first, you’d have gotten
585 , which would be nasty to simplify by hand…
1300

12. Full Solution to Sample Problem #3:
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Full Solution to Sample Problem #3: 20
Here is the work we expect to see on your worksheet:
39 • = 3∙13 • 3∙5 = 3∙13 • 3∙5 = 3∙3 =
∙5∙ ∙ ∙5∙ ∙ ∙5∙
/ /

13. Sample Problem #5: Dividing fractions
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Step 1: Multiply the first fraction by the ÷ 21 = 45 • 26
reciprocal of the second fraction
(i.e. flip the second fraction upside down and change ÷ to • .)
Step 2: Factor both the numerators and denominators into
prime factors, then write each fraction in factored form:
First fraction: = 3∙3∙5 and 13 = 13 (prime)
Second fraction: = 2∙13 and 21 = 3∙7
So you can write 45 • as 3∙3∙5 • 2∙ ∙7

14. Sample Problem #5 (continued)
NOTE: You do NOT need an LCD when dividing fractions.
Step 3: Now just cancel any common factors that appear in both numerator and denominator. Once you multiply out any remaining factors, the result is your simplified answer.
3∙3∙5 • 2∙13 = 3∙5∙2 = 30 ∙
/ /
/ /
NOTE: Once again, it is much easier to factor first and then cancel, rather than multiplying out the numerators and denominators and then trying to simplify the answer (especially if you aren’t using a calculator!) If you multiplied first, you’d have gotten , which would be pretty hard to simplify by hand.
273

15. Full Solution to Sample Problem #5:
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Full Solution to Sample Problem #5: 7 20
Here is the work we expect to see on your worksheet:
45 ÷ 21 = 45 • 26 = 3∙3∙5 • 2∙13 = 3∙3∙5 • 2∙13
∙ ∙7 .
= 3∙5∙2 = 30 1∙
/ /
/ /

16. NOTE: Gateway problems 4 & 6 using mixed numbers both start with the same step.
A mixed number consists of an integer part and a fraction part. We want to covert the mixed number into an improper fraction, This is done by multiplying the integer part by the denominator of the fraction part, then adding that product to the numerator of the fraction and putting that sum over the original denominator.

17. Sample Problem #4: Multiplying mixed numbers
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Step 1: Convert the mixed number into
an improper fraction: (Note that ) .
So becomes , which we can then
solve the same way we did problem #3.

18. Sample Problem #4 (continued)
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Step 2: Factor both the numerators and denominators into
prime factors, then write each fraction in factored form:
First fraction: and 3 are both prime
Second fraction: 6 = 2∙3 and 7 is prime
So you can write 17 ∙ 6 as 17 ∙ 2∙3 .
Step 3: Now just cancel any common factors that appear in both
numerator and denominator. Once you multiply out any remaining
factors, the result is your simplified answer. /
17 ∙ 2∙3 = 17∙2 =

19. Full Solution to Sample Problem #4:
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Full Solution to Sample Problem #4:
Here is the work we expect to see on your worksheet: /

20. Sample Problem #6: Dividing mixed numbers
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Step 1: Convert the mixed numbers into improper fractions:

21. Sample Problem #6 (continued)
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Step 2: Factor both the numerators and denominators into
prime factors, then write each fraction in factored form:
First fraction: 50 = 2∙5∙5 and 7 is prime
Second fraction: is prime and 25 = 5∙5
So you can write 50 • 2 as 2∙5∙5 • ∙5
Step 3: Now just cancel any common factors that appear in
Both numerator and denominator. Once you multiply out any
remaining factors, the result is your simplified answer.
2∙5∙5 • 2 = 2∙2 = 4 ∙
/ /

22. Full Solution to Sample Problem #6:
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Full Solution to Sample Problem #6:
Here is the work we expect to see on your worksheet:
/ /

23. NOTE: Gateway problems 7 & 8 both require using the order of operations.
First, calculate expressions within grouping symbols
(parentheses, brackets, braces,absolute values, fraction bars).
If there are nested sets of grouping symbols, start with the innermost ones first and work your way out.
Exponential expressions – left to right
Multiplication and division – left to right
Addition and subtraction – left to right

24. “Please excuse my dear Aunt Sally”
Order of operations memory device:
“Please excuse my dear Aunt Sally”
1. Please (Parentheses)
2. Excuse (Exponents)
3. My Dear (Multiply and Divide)
4. Aunt Sally (Add and Subtract)
… or just remember PEMDAS

25. Sample Problem # 7: Order of Operations
Strategy: Calculate out the entire top expression and then the entire bottom
expression, using the order of operations on each part. Then simplify the resulting fraction, if necessary.
TOP EXPRESSION: 24 – 4(7 + 2)
Step 1: Parentheses: – 4(7 + 2) = 24 – 4(9)
Step 2: Exponents: – 4(9) = 2•2•2•2 – 4(9) = 16 – 4(9)
(because 2•2•2•2 = 4•2•2 = 8•2 = 16)
Step 3: Multiply/Divide: 16 – 4(9) = 16 – 4•9 = 16 – 36
Step 4: Add/Subtract: – 36 = -20

26. Now calculate the bottom expression: 2(6+2) + 4
Step 1: Parentheses: 2(6+2) + 4 = 2(8) + 4
Step 2: Exponents: There aren’t any in this part.
Step 3: Multiply/Divide: 2(8) + 4 = 2•8 + 4 =
Step 4: Add/Subtract: = 20
Now put the top over the bottom and simplify the resulting fraction:
TOP = 24 – 4(7 + 2) = -20 = = -1
BOTTOM (6+2)

27. Full Solution to Sample Problem #7:
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Full Solution to Sample Problem #7:
Here is the work we expect to see on your worksheet:
24 – 4(7 + 2) = 24 – 4(9) = 16 – 4(9) = 16 – 36 = -20 = -1 = -1
2(6+2) (8)

28. Sample Problem # 8: Order of Operations
Strategy: Deal with the expressions inside the grouping symbols (parentheses, brackets) first, starting with the innermost set (-3 + 6). STEP 1: (inside the parentheses) 3[17 + 5(-3 + 6) - 10] = 3[17 + 5(3) - 10] STEP 2: (inside the brackets; multiply first, then add and subtract) 3[17 + 5(3) -10] = 3[17 + 5•3 -10] = 3[ ] = 3[ ] = 3[ ] = 3[22] STEP 3: Do the final multiplication: 3[22] = 3•22 = 66

29. Full Solution to Sample Problem #8:
Math TLC (Math 010 and Math 110) How to Solve Gateway Problems
Full Solution to Sample Problem #8:
Here is the work we expect to see on your worksheet:
3[17 + 5(-3 + 6) - 10] = 3[17 + 5(3) - 10] =
3[ ] = 3[ ] = 3[22] = 66