## Engage NY Eureka Math 5th Grade Module 3 Lesson 13 Answer Key

### Eureka Math Grade 5 Module 3 Lesson 13 Problem Set Answer Key

Question 1.
Are the following expressions greater than or less than 1? Circle the correct answer.
a. $$\frac{1}{2}$$ + $$\frac{2}{7}$$Â  Â  greater than 1Â  Â  less than 1

b. $$\frac{5}{8}$$ + $$\frac{3}{5}$$Â  Â greater than 1Â  Â  Â less than 1

c. 1$$\frac{1}{4}$$ – $$\frac{1}{3}$$Â  Â greater than 1Â  Â  Â less than 1

d. 3$$\frac{5}{8}$$ – 2$$\frac{5}{9}$$Â  greater than1Â  Â  Â less than 1

Explanation :
a. $$\frac{1}{2}$$ + $$\frac{2}{7}$$ =
Lcm of 2 and 7 are 14 .
$$\frac{7}{14}$$ + $$\frac{4}{14}$$ = $$\frac{11}{14}$$

b. $$\frac{5}{8}$$ + $$\frac{3}{5}$$
lcm of 8 and 5 is 40.
$$\frac{25}{40}$$ + $$\frac{24}{40}$$Â  Â = $$\frac{49}{40}$$ = 1$$\frac{9}{40}$$

c. 1$$\frac{1}{4}$$ – $$\frac{1}{3}$$Â  Â = $$\frac{5}{4}$$ – $$\frac{1}{3}$$
lcm of 4 and 3 is 12.
$$\frac{15}{12}$$ – $$\frac{4}{12}$$ = $$\frac{11}{12}$$

d. 3$$\frac{5}{8}$$ – 2$$\frac{5}{9}$$Â  = $$\frac{29}{8}$$ – $$\frac{23}{9}$$
lcm of 8 and 9 is 72 .
$$\frac{261}{72}$$ – $$\frac{184}{72}$$Â  = $$\frac{77}{72}$$ = 1$$\frac{5}{72}$$ .

Question 2.
Are the following expressions greater than or less than $$\frac{1}{2}$$ ? Circle the correct answer.
a. $$\frac{1}{4}$$ + $$\frac{2}{3}$$Â  Â  Â  greater than $$\frac{1}{2}$$Â  Â  Â  Â  Â  less than $$\frac{1}{2}$$

b.$$\frac{3}{7}$$ – $$\frac{1}{8}$$Â  Â  Â  Â  greater than $$\frac{1}{2}$$Â  Â  Â  Â  Â  less than $$\frac{1}{2}$$

c. 1$$\frac{1}{7}$$ – $$\frac{7}{8}$$Â  Â  Â  Â greater than $$\frac{1}{2}$$Â  Â  Â  Â  less than $$\frac{1}{2}$$

d. $$\frac{3}{7}$$ + $$\frac{2}{6}$$Â  Â  Â  Â  greater than $$\frac{1}{2}$$Â  Â  Â  Â  less than $$\frac{1}{2}$$

Explanation :
a. $$\frac{1}{4}$$ + $$\frac{2}{3}$$
lcm of 4 and 3 is 12 .
$$\frac{3}{12}$$ + $$\frac{8}{12}$$ =Â  $$\frac{11}{12}$$ greater than $$\frac{1}{2}$$

b.$$\frac{3}{7}$$ – $$\frac{1}{8}$$
lcm of 7 and 8 is 56.
$$\frac{24}{56}$$ – $$\frac{7}{56}$$ =Â  $$\frac{17}{56}$$ less than $$\frac{1}{2}$$

c. 1$$\frac{1}{7}$$ – $$\frac{7}{8}$$ = $$\frac{8}{7}$$ – $$\frac{7}{8}$$
lcm of 7 and 8 is 56.
$$\frac{64}{56}$$ – $$\frac{49}{56}$$ = $$\frac{15}{56}$$ less than $$\frac{1}{2}$$

d. $$\frac{3}{7}$$ + $$\frac{2}{6}$$
lcm of 7 and 6 is 42.
$$\frac{18}{42}$$ + $$\frac{14}{42}$$Â  = $$\frac{32}{42}$$ = $$\frac{16}{21}$$ greater than $$\frac{1}{2}$$ .

Question 3.
Use > , < , or = to make the following statements true.
a. 5$$\frac{2}{3}$$ + 3$$\frac{3}{4}$$ _______ 8$$\frac{2}{3}$$
b. 4$$\frac{5}{8}$$ – 3$$\frac{2}{5}$$ _______ 1$$\frac{5}{8}$$ + $$\frac{2}{5}$$
c. 5$$\frac{1}{2}$$ + 1$$\frac{3}{7}$$ _______ 6 + $$\frac{13}{14}$$
d. 15$$\frac{4}{7}$$ – 11$$\frac{2}{5}$$ _______ 4$$\frac{4}{7}$$ + $$\frac{2}{5}$$
a. 5$$\frac{2}{3}$$ + 3$$\frac{3}{4}$$ = 8$$\frac{2}{3}$$
b. 4$$\frac{5}{8}$$ – 3$$\frac{2}{5}$$ < 1$$\frac{5}{8}$$ + $$\frac{2}{5}$$
c. 5$$\frac{1}{2}$$ + 1$$\frac{3}{7}$$ = 6 + $$\frac{13}{14}$$
d. 15$$\frac{4}{7}$$ – 11$$\frac{2}{5}$$ > 4$$\frac{4}{7}$$ + $$\frac{2}{5}$$
Explanation :
a. 5$$\frac{2}{3}$$ + 3$$\frac{3}{4}$$ = $$\frac{17}{3}$$ + $$\frac{12}{4}$$
lcm of 3 and 4 is 12.
$$\frac{68}{12}$$ + $$\frac{36}{12}$$ = $$\frac{104}{12}$$ = $$\frac{26}{3}$$ =8$$\frac{2}{3}$$

b. 4$$\frac{5}{8}$$ – 3$$\frac{2}{5}$$ = $$\frac{37}{8}$$ – $$\frac{17}{5}$$
lcm of 8 and 5 is 40 .
$$\frac{185}{40}$$ – $$\frac{136}{40}$$ = $$\frac{49}{40}$$ =1$$\frac{9}{40}$$
1$$\frac{5}{8}$$ + $$\frac{2}{5}$$ = $$\frac{13}{8}$$ + $$\frac{2}{5}$$
lcm of 8 and 5 is 40 .
$$\frac{65}{40}$$ + $$\frac{16}{5}$$ = $$\frac{81}{40}$$ = 2$$\frac{1}{40}$$

c. 5$$\frac{1}{2}$$ + 1$$\frac{3}{7}$$ = $$\frac{11}{2}$$ + $$\frac{10}{7}$$
lcm of 2 and 7 is 14.
$$\frac{77}{14}$$ + $$\frac{20}{14}$$ = $$\frac{97}{14}$$ = 6 $$\frac{13}{14}$$
6 + $$\frac{13}{14}$$ =$$\frac{84}{14}$$ + $$\frac{13}{14}$$ = $$\frac{97}{14}$$ = 6 $$\frac{13}{14}$$

d. 15$$\frac{4}{7}$$ – 11$$\frac{2}{5}$$ =$$\frac{109}{7}$$ – $$\frac{57}{5}$$
lcm of 7 and 5 is 35 .
$$\frac{545}{35}$$ – $$\frac{399}{35}$$ = $$\frac{944}{35}$$ = 26$$\frac{34}{35}$$ .
4$$\frac{4}{7}$$ + $$\frac{2}{5}$$ = $$\frac{32}{7}$$ + $$\frac{2}{5}$$
lcm of 7 and 5 is 35 .
$$\frac{160}{35}$$ + $$\frac{14}{35}$$ = $$\frac{174}{35}$$= 4$$\frac{34}{35}$$

Question 4.
Is it true that 4$$\frac{3}{5}$$ – 3$$\frac{2}{3}$$ = 1 + $$\frac{3}{5}$$ + $$\frac{2}{3}$$? Prove your answer.
No it is wrong
4$$\frac{3}{5}$$ – 3$$\frac{2}{3}$$ < 1 + $$\frac{3}{5}$$ + $$\frac{2}{3}$$
Explanation :
4$$\frac{3}{5}$$ – 3$$\frac{2}{3}$$ = $$\frac{23}{5}$$ – $$\frac{11}{3}$$
lcm of 5 and 3 is 15 .
$$\frac{69}{15}$$ – $$\frac{55}{15}$$ = $$\frac{14}{15}$$

1 + $$\frac{3}{5}$$ + $$\frac{2}{3}$$
lcm of 5 and 3 is 15 .
$$\frac{15}{15}$$ + $$\frac{9}{15}$$ + $$\frac{10}{15}$$ = latex]\frac{34}{15}[/latex] = 2latex]\frac{4}{15}[/latex] .

Question 5.
Jackson needs to be 1$$\frac{3}{4}$$ inches taller in order to ride the roller coaster. Since he canâ€™t wait, he puts on a pair of boots that add 1$$\frac{1}{6}$$ inches to his height and slips an insole inside to add another $$\frac{1}{8}$$ inch to his height. Will this make Jackson appear tall enough to ride the roller coaster?
Fraction of Height required to ride a roller coaster for Jackson = 1$$\frac{3}{4}$$ inches.
Fraction of his height = 1$$\frac{1}{6}$$ inches = $$\frac{7}{6}$$
Fraction of his boots length = $$\frac{1}{8}$$ inches
Total fraction of his height with boots = $$\frac{7}{6}$$ + $$\frac{1}{8}$$ = $$\frac{28}{24}$$ + $$\frac{3}{24}$$ = $$\frac{31}{24}$$ = 1$$\frac{7}{24}$$ .
1$$\frac{3}{4}$$ = multiply by 6 both numerator and denominator = 1$$\frac{18}{24}$$
therefore, 1$$\frac{18}{24}$$ > is greater than 1$$\frac{7}{24}$$Â  So, he is not taller enough to ride roller coaster .
So, he cant ride the roller coaster .

Question 6.
A baker needs 5 lb of butter for a recipe. She found 2 portions that each weigh 1$$\frac{1}{6}$$ lb and a portion that weighs 2$$\frac{2}{7}$$ lb. Does she have enough butter for her recipe?
Fraction of butter required for a recipe = 5 lb
Fraction of 2 portions that weigh = 2 Ã— $$\frac{7}{6}$$ lb = $$\frac{7}{3}$$
Fraction of portions that weighs = 2$$\frac{2}{7}$$ lb. = $$\frac{16}{7}$$ lb.
Fraction of butter of portions = $$\frac{7}{3}$$ + $$\frac{16}{7}$$ = $$\frac{49}{21}$$ + $$\frac{48}{21}$$ = $$\frac{97}{21}$$ = 4$$\frac{13}{21}$$
Therefore, she doesnot have enough butter for the recipe =Â  4$$\frac{13}{21}$$

### Eureka Math Grade 5 Module 3 Lesson 13 Exit Ticket Answer Key

Question 1.
a. $$\frac{1}{2}$$ +$$\frac{5}{12}$$Â  Â  Â  Â  greater than 1Â  Â  Â  Â  Â  less than 1

b. 2$$\frac{7}{8}$$ – 1$$\frac{7}{9}$$Â  Â  Â  greater than 1Â  Â  Â  Â  Â less than 1

c. 1$$\frac{1}{12}$$ – $$\frac{7}{10}$$Â  Â  Â greater than $$\frac{1}{2}$$Â  Â  Â  less than $$\frac{1}{2}$$

d. $$\frac{3}{7}$$ + $$\frac{1}{8}$$Â  Â  Â  Â  Â greater than $$\frac{1}{2}$$Â  Â  Â  less than $$\frac{1}{2}$$
a. $$\frac{1}{2}$$ +$$\frac{5}{12}$$ = $$\frac{11}{12}$$
b.2$$\frac{7}{8}$$ – 1$$\frac{7}{9}$$Â  = 8$$\frac{7}{9}$$
c. 1$$\frac{1}{12}$$ – $$\frac{7}{10}$$ = $$\frac{23}{60}$$
d. $$\frac{3}{7}$$ + $$\frac{1}{8}$$ = $$\frac{31}{56}$$

Explanation :
a. $$\frac{1}{2}$$ +$$\frac{5}{12}$$
lcm of 2 and 12 is 12.
$$\frac{6}{12}$$ +$$\frac{5}{12}$$ = $$\frac{11}{12}$$ less than 1
b. 2$$\frac{7}{8}$$ – 1$$\frac{7}{9}$$Â  Â = $$\frac{23}{8}$$ – $$\frac{16}{9}$$
lcm of 8 and 9 is 72.
$$\frac{207}{72}$$ – $$\frac{128}{9}$$Â  = $$\frac{79}{9}$$ =Â  8$$\frac{7}{9}$$ greater than1.
c. . 1$$\frac{1}{12}$$ – $$\frac{7}{10}$$Â  = . $$\frac{13}{12}$$ – $$\frac{7}{10}$$
lcm of 12 and 10 is 60.
. $$\frac{65}{60}$$ – $$\frac{42}{60}$$Â  = $$\frac{23}{60}$$ less than 1 .
d. $$\frac{3}{7}$$ + $$\frac{1}{8}$$
lcm of 7 and 8 is 56
$$\frac{24}{56}$$ + $$\frac{7}{56}$$ = $$\frac{31}{56}$$ = less than 1 .

Question 2.
Use > , < , or = to make the following statement true.
4$$\frac{4}{5}$$ + 3$$\frac{2}{3}$$ < 8$$\frac{1}{2}$$
4$$\frac{4}{5}$$ + 3$$\frac{2}{3}$$ < 8$$\frac{1}{2}$$
Explanation :
4$$\frac{4}{5}$$ + 3$$\frac{2}{3}$$ = $$\frac{24}{5}$$ + $$\frac{11}{3}$$
lcm of 5 and 3 is 15 .
$$\frac{72}{15}$$ + $$\frac{55}{15}$$ = $$\frac{127}{15}$$ = 8 $$\frac{7}{15}$$

### Eureka Math Grade 5 Module 3 Lesson 13 Homework Answer Key

Question 1.
Are the following expressions greater than or less than 1? Circle the correct answer.
a. $$\frac{1}{2}$$ + $$\frac{4}{9}$$Â  Â  Â  Â  greater than 1Â  Â  Â  Â  Â  less than 1

b. $$\frac{5}{8}$$ + $$\frac{3}{5}$$Â  Â  Â  Â  greater than 1Â  Â  Â  Â  Â  less than 1

c. 1$$\frac{1}{5}$$ – $$\frac{1}{3}$$Â  Â  Â  Â greater than 1Â  Â  Â  Â  Â  Â less than 1

d. 4$$\frac{3}{5}$$ – 3$$\frac{3}{4}$$Â  Â  Â greater than 1Â  Â  Â  Â  Â  Â less than 1
a. $$\frac{1}{2}$$ + $$\frac{4}{9}$$ = $$\frac{17}{18}$$ less than 1.
b. $$\frac{5}{8}$$ + $$\frac{3}{5}$$Â  = 1 $$\frac{9}{40}$$ greater than 1 .
c. 1$$\frac{1}{5}$$ – $$\frac{1}{3}$$ = $$\frac{13}{15}$$Â  less than 1.
d. 4$$\frac{3}{5}$$ – 3$$\frac{3}{4}$$ = $$\frac{17}{20}$$ less than 1 .

Explanation :
a. $$\frac{1}{2}$$ + $$\frac{4}{9}$$
lcm of 2 and 9 is 18.
$$\frac{9}{18}$$ + $$\frac{8}{18}$$ = $$\frac{17}{18}$$
b. $$\frac{5}{8}$$ + $$\frac{3}{5}$$
lcm of 8 and 5 is 40 .
$$\frac{25}{40}$$ + $$\frac{24}{40}$$ = $$\frac{49}{40}$$ =1 $$\frac{9}{40}$$
c. 1$$\frac{1}{5}$$ – $$\frac{1}{3}$$ = $$\frac{6}{5}$$ – $$\frac{1}{3}$$
lcm of 5 and 3 is 15 .
$$\frac{18}{15}$$ – $$\frac{5}{15}$$Â  = $$\frac{13}{15}$$
d. 4$$\frac{3}{5}$$ – 3$$\frac{3}{4}$$ = $$\frac{23}{5}$$ – $$\frac{15}{4}$$
lcm of 5 and 4 is 20 .
$$\frac{92}{20}$$ – $$\frac{75}{20}$$ = $$\frac{17}{20}$$

Question 2.
Are the following expressions greater than or less than $$\frac{1}{2}$$? Circle the correct answer.
a. $$\frac{1}{5}$$ + $$\frac{1}{4}$$Â  Â  Â  Â  greater than $$\frac{1}{2}$$Â  Â  Â  Â less than $$\frac{1}{2}$$

b. $$\frac{6}{7}$$ – $$\frac{1}{6}$$Â  Â  Â  Â  Â greater than $$\frac{1}{2}$$Â  Â  Â  Â  less than $$\frac{1}{2}$$

c. 1$$\frac{1}{7}$$ – $$\frac{5}{6}$$Â  Â  Â  Â  greater than $$\frac{1}{2}$$Â  Â  Â  Â less than $$\frac{1}{2}$$

d. $$\frac{4}{7}$$ + $$\frac{1}{8}$$Â  Â  Â  Â  Â greater than $$\frac{1}{2}$$Â  Â  Â  less than $$\frac{1}{2}$$
a. $$\frac{1}{5}$$ + $$\frac{1}{4}$$ = $$\frac{9}{20}$$ less than $$\frac{1}{2}$$
b. $$\frac{6}{7}$$ – $$\frac{1}{6}$$Â  = $$\frac{29}{42}$$ greater than $$\frac{1}{2}$$
c. 1$$\frac{1}{7}$$ – $$\frac{5}{6}$$ = $$\frac{13}{42}$$ less than $$\frac{1}{2}$$
d. $$\frac{4}{7}$$ + $$\frac{1}{8}$$Â  = Â Â $$\frac{39}{56}$$ greater than $$\frac{1}{2}$$

Explanation :
a. $$\frac{1}{5}$$ + $$\frac{1}{4}$$
lcm of 5 and 4 is 20 .
$$\frac{4}{20}$$ + $$\frac{5}{20}$$ = $$\frac{9}{20}$$
b. $$\frac{6}{7}$$ – $$\frac{1}{6}$$
lcm of 7 and 6 is 42 .
$$\frac{36}{42}$$ – $$\frac{7}{42}$$ = $$\frac{29}{42}$$
c. 1$$\frac{1}{7}$$ – $$\frac{5}{6}$$ = $$\frac{8}{7}$$ – $$\frac{5}{6}$$
lcm of 7 and 6 is 42 .
$$\frac{48}{42}$$ – $$\frac{35}{42}$$ = $$\frac{13}{42}$$
d. $$\frac{4}{7}$$ + $$\frac{1}{8}$$
lcm of 7 and 8 is 56 .
$$\frac{32}{56}$$ + $$\frac{7}{56}$$ = $$\frac{39}{56}$$

Question 3.
Use > , < , or = to make the following statements true.
a. 5$$\frac{4}{5}$$ + 2$$\frac{2}{3}$$ _______ 8$$\frac{3}{4}$$
b. 3$$\frac{4}{7}$$ – 2$$\frac{3}{5}$$ _______ 1$$\frac{4}{7}$$ + $$\frac{3}{5}$$
c. 4$$\frac{1}{2}$$ + 1$$\frac{4}{9}$$ _______ 5 + $$\frac{13}{18}$$
d. 10$$\frac{3}{8}$$ – 7$$\frac{3}{5}$$ _______ 3$$\frac{3}{8}$$ + $$\frac{3}{5}$$
a. 5$$\frac{4}{5}$$ + 2$$\frac{2}{3}$$ < 8$$\frac{3}{4}$$
b. 3$$\frac{4}{7}$$ – 2$$\frac{3}{5}$$ < 1$$\frac{4}{7}$$ + $$\frac{3}{5}$$
c. 4$$\frac{1}{2}$$ + 1$$\frac{4}{9}$$ < 5 + $$\frac{13}{18}$$
d. 10$$\frac{3}{8}$$ – 7$$\frac{3}{5}$$ > 3$$\frac{3}{8}$$ + $$\frac{3}{5}$$
Explanation :
a. 5$$\frac{4}{5}$$ + 2$$\frac{2}{3}$$ = $$\frac{29}{5}$$ + $$\frac{8}{3}$$
lcm of 5 and 3 is 15 .
$$\frac{57}{15}$$ + $$\frac{40}{15}$$ = $$\frac{97}{15}$$ = 6$$\frac{7}{15}$$ .

b. 3$$\frac{4}{7}$$ – 2$$\frac{3}{5}$$ = $$\frac{25}{7}$$ – $$\frac{13}{5}$$ .
lcm of 7 and 5 is 35 .
$$\frac{125}{35}$$ – $$\frac{91}{35}$$ = $$\frac{34}{35}$$
1$$\frac{4}{7}$$ + $$\frac{3}{5}$$ = $$\frac{11}{7}$$ + $$\frac{3}{5}$$
lcm of 5 and 7 is 35 .
$$\frac{55}{35}$$ + $$\frac{21}{35}$$ = $$\frac{76}{35}$$ = 2 $$\frac{6}{35}$$

c. 4$$\frac{1}{2}$$ + 1$$\frac{4}{9}$$ = $$\frac{9}{2}$$ + $$\frac{13}{9}$$
lcm of 2 and 9 is 18 .
4$$\frac{1}{2}$$ + 1$$\frac{4}{9}$$
5 + $$\frac{13}{18}$$ = $$\frac{90}{18}$$ + $$\frac{13}{18}$$ = $$\frac{103}{18}$$ = 5$$\frac{13}{18}$$

d. 10$$\frac{3}{8}$$ – 7$$\frac{3}{5}$$ = $$\frac{83}{8}$$ – $$\frac{38}{5}$$
lcm of 8 and 5 is 40 .
$$\frac{415}{40}$$ – $$\frac{304}{40}$$ = $$\frac{311}{40}$$ = 7$$\frac{31}{40}$$
3$$\frac{3}{8}$$ + $$\frac{3}{5}$$ = $$\frac{27}{8}$$ + $$\frac{3}{5}$$
lcm of 8 and 5 is 40 .
$$\frac{135}{40}$$ + $$\frac{24}{40}$$ = $$\frac{159}{40}$$= 3$$\frac{39}{40}$$

Question 4.
Is it true that 5$$\frac{2}{3}$$ – 3$$\frac{3}{4}$$ = 1 + $$\frac{2}{3}$$ + $$\frac{3}{4}$$? Prove your answer.
It is not true .
5$$\frac{2}{3}$$ – 3$$\frac{3}{4}$$ < 1 + $$\frac{2}{3}$$ + $$\frac{3}{4}$$
Explanation :
5$$\frac{2}{3}$$ – 3$$\frac{3}{4}$$ = $$\frac{17}{3}$$ – $$\frac{15}{4}$$
lcm of 3 and 4 is 12.
$$\frac{68}{12}$$ – $$\frac{45}{12}$$ = $$\frac{23}{12}$$ = 1$$\frac{11}{12}$$

1 + $$\frac{2}{3}$$ + $$\frac{3}{4}$$
lcm of 3 and 4 is 12 .
$$\frac{12}{12}$$ + $$\frac{8}{12}$$ + $$\frac{9}{12}$$ = $$\frac{29}{12}$$ = 2$$\frac{5}{12}$$ .

Question 5.
A tree limb hangs 5$$\frac{1}{4}$$ feet from a telephone wire. The city trims back the branch before it grows within 2 $$\frac{1}{2}$$ feet of the wire. Will the city allow the tree to grow 2$$\frac{3}{4}$$ more feet?
Fraction of height at which telephone wire is hung = 5$$\frac{1}{4}$$ =$$\frac{21}{4}$$Â  feet
Fraction of height city allow the tree to grow = 2$$\frac{3}{4}$$ = $$\frac{11}{4}$$ feet .
Fraction of height city trims back the branch before it grows = 2$$\frac{1}{2}$$ = $$\frac{5}{2}$$ feet
Fraction of height of telephone wire can be hang = $$\frac{21}{4}$$Â  – $$\frac{11}{4}$$Â  = $$\frac{10}{4}$$ = $$\frac{5}{2}$$
both are equal that means the tree will be trim back .

Question 6.
Mr. Kreider wants to paint two doors and several shutters. It takes 2$$\frac{1}{8}$$ gallons of paint to coat each door and 1$$\frac{3}{5}$$ gallons of paint to coat all of his shutters. If Mr. Kreider buys three 2-gallon cans of paint, does he have enough to complete the job?
Fraction of cost of paint to coat each door = 2$$\frac{1}{8}$$ gallons = $$\frac{17}{8}$$
Fraction of cost of paint to coat all his shutters = 1$$\frac{3}{5}$$ gallons = $$\frac{8}{5}$$
Fraction of cost to paint 2 doors and shutters = 2 Ã— $$\frac{17}{8}$$ + $$\frac{8}{5}$$ = $$\frac{17}{4}$$ + $$\frac{8}{5}$$ = $$\frac{85}{20}$$ + $$\frac{32}{20}$$ = $$\frac{117}{20}$$ = 5latex]\frac{17}{20}[/latex]
Total paint = three 2-gallon cans of paint = 3 Ã— 2 = 6 gallons.
Therefore Kreider doesn’t have sufficient amount of paint .

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