Practice · Turn leading digits into an inequality · Sensim Math

Variant 1 easy

$x$ and $y$ are natural numbers with $10 < y < 20$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $1$ and the second digit after the decimal point is $1$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $10 < y < 20$ , has decimal $0.11\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $10 < y < 20$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.11\ldots$ (first digit $1$ , second digit $1$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.11$ means $0.11 \le \frac{x}{y} < 0.12$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $11$ and $19$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.11

is at least

0.11

and less than

0.12

. Multiply through by

y

.

\frac{11}{100}\,y \le x < \frac{12}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

11 \le y \le 19

, the smallest the lower bound can be is

\frac{11}{100} \times 11

and the largest the upper bound can be is

\frac{12}{100} \times 19

.

\frac{121}{100} \le x < \frac{228}{100}

The only whole number from

1.21

up to (not including)

2.28

is

x = 2

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 2

back:

0.11 \le \frac{2}{y} < 0.12

means

\frac{2}{0.12} < y \le \frac{2}{0.11}

.

\frac{50}{3} < y \le \frac{200}{11}

That is

16\frac{2}{3} < y \le 18\frac{2}{11}

, so with

10 < y < 20

the candidates are

17

and

18

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{2}{18}

reduces, so it is out. That leaves only

y = 17

.

x + y = 2 + 17 = 19

\frac{2}{17} = 0.1176\ldots

fits, and

17

is the biggest allowed denominator.

Answer: 19

4 · Reviewdoes it hold up?

Check: $\frac{2}{17} = 0.1176\ldots$ starts $0.11$ , and $\gcd(2, 17) = 1$ .

Another way: Systematic list: for each $y$ from $11$ to $19$ , the interval $[0.11y,\ 0.12y)$ contains a whole number only for $y = 17, 18$ (each gives $x = 2$ ); dropping $\frac{2}{18}$ leaves the same single fraction.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 2 easy

$x$ and $y$ are natural numbers with $15 < y < 25$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $0$ and the second digit after the decimal point is $8$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $15 < y < 25$ , has decimal $0.08\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $15 < y < 25$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.08\ldots$ (first digit $0$ , second digit $8$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.08$ means $0.08 \le \frac{x}{y} < 0.09$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $16$ and $24$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.08

is at least

0.08

and less than

0.09

. Multiply through by

y

.

\frac{8}{100}\,y \le x < \frac{9}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

16 \le y \le 24

, the smallest the lower bound can be is

\frac{8}{100} \times 16

and the largest the upper bound can be is

\frac{9}{100} \times 24

.

\frac{128}{100} \le x < \frac{216}{100}

The only whole number from

1.28

up to (not including)

2.16

is

x = 2

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 2

back:

0.08 \le \frac{2}{y} < 0.09

means

\frac{2}{0.09} < y \le \frac{2}{0.08}

.

\frac{200}{9} < y \le 25

That is

22\frac{2}{9} < y \le 25

, so with

15 < y < 25

the candidates are

23

and

24

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{2}{24}

reduces, so it is out. That leaves only

y = 23

.

x + y = 2 + 23 = 25

\frac{2}{23} = 0.0869\ldots

fits, and

23

is the biggest allowed denominator.

Answer: 25

4 · Reviewdoes it hold up?

Check: $\frac{2}{23} = 0.0869\ldots$ starts $0.08$ , and $\gcd(2, 23) = 1$ .

Another way: Systematic list: for each $y$ from $16$ to $24$ , the interval $[0.08y,\ 0.09y)$ contains a whole number only for $y = 23, 24$ (each gives $x = 2$ ); dropping $\frac{2}{24}$ leaves the same single fraction.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 3 easy

$x$ and $y$ are natural numbers with $25 < y < 35$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $0$ and the second digit after the decimal point is $9$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $25 < y < 35$ , has decimal $0.09\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $25 < y < 35$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.09\ldots$ (first digit $0$ , second digit $9$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.09$ means $0.09 \le \frac{x}{y} < 0.1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $26$ and $34$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.09

is at least

0.09

and less than

0.1

. Multiply through by

y

.

\frac{9}{100}\,y \le x < \frac{10}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

26 \le y \le 34

, the smallest the lower bound can be is

\frac{9}{100} \times 26

and the largest the upper bound can be is

\frac{10}{100} \times 34

.

\frac{234}{100} \le x < \frac{340}{100}

The only whole number from

2.34

up to (not including)

3.4

is

x = 3

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 3

back:

0.09 \le \frac{3}{y} < 0.1

means

\frac{3}{0.1} < y \le \frac{3}{0.09}

.

30 < y \le \frac{100}{3}

That is

30 < y \le 33\frac{1}{3}

, so with

25 < y < 35

the candidates are

31

,

32

,

33

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{3}{33}

reduces, so it is out. That leaves

y = 31

and

y = 32

; the larger sum uses

y = 32

.

x + y = 3 + 32 = 35

\frac{3}{32} = 0.0937\ldots

fits, and

32

is the biggest allowed denominator.

Answer: 35

4 · Reviewdoes it hold up?

Check: $\frac{3}{32} = 0.0937\ldots$ starts $0.09$ , and $\gcd(3, 32) = 1$ . The other survivor $\frac{3}{31} = 0.0967\ldots$ gives only $34$ .

Another way: Systematic list: for each $y$ from $26$ to $34$ , the interval $[0.09y,\ 0.1y)$ contains a whole number only for $y = 31, 32, 33$ (each gives $x = 3$ ); dropping $\frac{3}{33}$ leaves the same two fractions.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 4 easy

$x$ and $y$ are natural numbers with $25 < y < 35$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $1$ and the second digit after the decimal point is $3$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $25 < y < 35$ , has decimal $0.13\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $25 < y < 35$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.13\ldots$ (first digit $1$ , second digit $3$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.13$ means $0.13 \le \frac{x}{y} < 0.14$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $26$ and $34$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.13

is at least

0.13

and less than

0.14

. Multiply through by

y

.

\frac{13}{100}\,y \le x < \frac{14}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

26 \le y \le 34

, the smallest the lower bound can be is

\frac{13}{100} \times 26

and the largest the upper bound can be is

\frac{14}{100} \times 34

.

\frac{338}{100} \le x < \frac{476}{100}

The only whole number from

3.38

up to (not including)

4.76

is

x = 4

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 4

back:

0.13 \le \frac{4}{y} < 0.14

means

\frac{4}{0.14} < y \le \frac{4}{0.13}

.

\frac{200}{7} < y \le \frac{400}{13}

That is

28\frac{4}{7} < y \le 30\frac{10}{13}

, so with

25 < y < 35

the candidates are

29

and

30

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{4}{30}

reduces, so it is out. That leaves only

y = 29

.

x + y = 4 + 29 = 33

\frac{4}{29} = 0.1379\ldots

fits, and

29

is the biggest allowed denominator.

Answer: 33

4 · Reviewdoes it hold up?

Check: $\frac{4}{29} = 0.1379\ldots$ starts $0.13$ , and $\gcd(4, 29) = 1$ .

Another way: Systematic list: for each $y$ from $26$ to $34$ , the interval $[0.13y,\ 0.14y)$ contains a whole number only for $y = 29, 30$ (each gives $x = 4$ ); dropping $\frac{4}{30}$ leaves the same single fraction.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 5 medium

$x$ and $y$ are natural numbers with $30 < y < 40$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $0$ and the second digit after the decimal point is $7$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $30 < y < 40$ , has decimal $0.07\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $30 < y < 40$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.07\ldots$ (first digit $0$ , second digit $7$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.07$ means $0.07 \le \frac{x}{y} < 0.08$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $31$ and $39$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.07

is at least

0.07

and less than

0.08

. Multiply through by

y

.

\frac{7}{100}\,y \le x < \frac{8}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

31 \le y \le 39

, the smallest the lower bound can be is

\frac{7}{100} \times 31

and the largest the upper bound can be is

\frac{8}{100} \times 39

.

\frac{217}{100} \le x < \frac{312}{100}

The only whole number from

2.17

up to (not including)

3.12

is

x = 3

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 3

back:

0.07 \le \frac{3}{y} < 0.08

means

\frac{3}{0.08} < y \le \frac{3}{0.07}

.

\frac{75}{2} < y \le \frac{300}{7}

That is

37\frac{1}{2} < y \le 42\frac{6}{7}

, so with

30 < y < 40

the candidates are

38

and

39

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{3}{39}

reduces, so it is out. That leaves only

y = 38

.

x + y = 3 + 38 = 41

\frac{3}{38} = 0.0789\ldots

fits, and

38

is the biggest allowed denominator.

Answer: 41

4 · Reviewdoes it hold up?

Check: $\frac{3}{38} = 0.0789\ldots$ starts $0.07$ , and $\gcd(3, 38) = 1$ .

Another way: Systematic list: for each $y$ from $31$ to $39$ , the interval $[0.07y,\ 0.08y)$ contains a whole number only for $y = 38, 39$ (each gives $x = 3$ ); dropping $\frac{3}{39}$ leaves the same single fraction.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 6 medium

$x$ and $y$ are natural numbers with $30 < y < 40$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $0$ and the second digit after the decimal point is $6$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $30 < y < 40$ , has decimal $0.06\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $30 < y < 40$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.06\ldots$ (first digit $0$ , second digit $6$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.06$ means $0.06 \le \frac{x}{y} < 0.07$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $31$ and $39$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.06

is at least

0.06

and less than

0.07

. Multiply through by

y

.

\frac{6}{100}\,y \le x < \frac{7}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

31 \le y \le 39

, the smallest the lower bound can be is

\frac{6}{100} \times 31

and the largest the upper bound can be is

\frac{7}{100} \times 39

.

\frac{186}{100} \le x < \frac{273}{100}

The only whole number from

1.86

up to (not including)

2.73

is

x = 2

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 2

back:

0.06 \le \frac{2}{y} < 0.07

means

\frac{2}{0.07} < y \le \frac{2}{0.06}

.

\frac{200}{7} < y \le \frac{100}{3}

That is

28\frac{4}{7} < y \le 33\frac{1}{3}

, so with

30 < y < 40

the candidates are

31

,

32

,

33

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{2}{32}

reduces, so it is out. That leaves

y = 31

and

y = 33

; the larger sum uses

y = 33

.

x + y = 2 + 33 = 35

\frac{2}{33} = 0.0606\ldots

fits, and

33

is the biggest allowed denominator.

Answer: 35

4 · Reviewdoes it hold up?

Check: $\frac{2}{33} = 0.0606\ldots$ starts $0.06$ , and $\gcd(2, 33) = 1$ . The other survivor $\frac{2}{31} = 0.0645\ldots$ gives only $33$ .

Another way: Systematic list: for each $y$ from $31$ to $39$ , the interval $[0.06y,\ 0.07y)$ contains a whole number only for $y = 31, 32, 33$ (each gives $x = 2$ ); dropping $\frac{2}{32}$ leaves the same two fractions.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 7 medium

$x$ and $y$ are natural numbers with $35 < y < 45$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $1$ and the second digit after the decimal point is $4$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $35 < y < 45$ , has decimal $0.14\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $35 < y < 45$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.14\ldots$ (first digit $1$ , second digit $4$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.14$ means $0.14 \le \frac{x}{y} < 0.15$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $36$ and $44$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.14

is at least

0.14

and less than

0.15

. Multiply through by

y

.

\frac{14}{100}\,y \le x < \frac{15}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

36 \le y \le 44

, the smallest the lower bound can be is

\frac{14}{100} \times 36

and the largest the upper bound can be is

\frac{15}{100} \times 44

.

\frac{504}{100} \le x < \frac{660}{100}

The only whole number from

5.04

up to (not including)

6.6

is

x = 6

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 6

back:

0.14 \le \frac{6}{y} < 0.15

means

\frac{6}{0.15} < y \le \frac{6}{0.14}

.

40 < y \le \frac{300}{7}

That is

40 < y \le 42\frac{6}{7}

, so with

35 < y < 45

the candidates are

41

and

42

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{6}{42}

reduces, so it is out. That leaves only

y = 41

.

x + y = 6 + 41 = 47

\frac{6}{41} = 0.1463\ldots

fits, and

41

is the biggest allowed denominator.

Answer: 47

4 · Reviewdoes it hold up?

Check: $\frac{6}{41} = 0.1463\ldots$ starts $0.14$ , and $\gcd(6, 41) = 1$ .

Another way: Systematic list: for each $y$ from $36$ to $44$ , the interval $[0.14y,\ 0.15y)$ contains a whole number only for $y = 41, 42$ (each gives $x = 6$ ); dropping $\frac{6}{42}$ leaves the same single fraction.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 8 medium

$x$ and $y$ are natural numbers with $35 < y < 45$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $0$ and the second digit after the decimal point is $9$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $35 < y < 45$ , has decimal $0.09\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $35 < y < 45$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.09\ldots$ (first digit $0$ , second digit $9$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.09$ means $0.09 \le \frac{x}{y} < 0.1$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $36$ and $44$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.09

is at least

0.09

and less than

0.1

. Multiply through by

y

.

\frac{9}{100}\,y \le x < \frac{10}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

36 \le y \le 44

, the smallest the lower bound can be is

\frac{9}{100} \times 36

and the largest the upper bound can be is

\frac{10}{100} \times 44

.

\frac{324}{100} \le x < \frac{440}{100}

The only whole number from

3.24

up to (not including)

4.4

is

x = 4

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 4

back:

0.09 \le \frac{4}{y} < 0.1

means

\frac{4}{0.1} < y \le \frac{4}{0.09}

.

40 < y \le \frac{400}{9}

That is

40 < y \le 44\frac{4}{9}

, so with

35 < y < 45

the candidates are

41

,

42

,

43

,

44

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{4}{42}

and

\frac{4}{44}

reduce, so they are out. That leaves

y = 41

and

y = 43

; the larger sum uses

y = 43

.

x + y = 4 + 43 = 47

\frac{4}{43} = 0.0930\ldots

fits, and

43

is the biggest allowed denominator.

Answer: 47

4 · Reviewdoes it hold up?

Check: $\frac{4}{43} = 0.0930\ldots$ starts $0.09$ , and $\gcd(4, 43) = 1$ . The other survivor $\frac{4}{41} = 0.0975\ldots$ gives only $45$ .

Another way: Systematic list: for each $y$ from $36$ to $44$ , the interval $[0.09y,\ 0.1y)$ contains a whole number only for $y = 41, 42, 43, 44$ (each gives $x = 4$ ); dropping $\frac{4}{42}$ and $\frac{4}{44}$ leaves the same two fractions.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 9 hard

$x$ and $y$ are natural numbers with $40 < y < 50$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $1$ and the second digit after the decimal point is $1$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $40 < y < 50$ , has decimal $0.11\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $40 < y < 50$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.11\ldots$ (first digit $1$ , second digit $1$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.11$ means $0.11 \le \frac{x}{y} < 0.12$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $41$ and $49$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.11

is at least

0.11

and less than

0.12

. Multiply through by

y

.

\frac{11}{100}\,y \le x < \frac{12}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

41 \le y \le 49

, the smallest the lower bound can be is

\frac{11}{100} \times 41

and the largest the upper bound can be is

\frac{12}{100} \times 49

.

\frac{451}{100} \le x < \frac{588}{100}

The only whole number from

4.51

up to (not including)

5.88

is

x = 5

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 5

back:

0.11 \le \frac{5}{y} < 0.12

means

\frac{5}{0.12} < y \le \frac{5}{0.11}

.

\frac{125}{3} < y \le \frac{500}{11}

That is

41\frac{2}{3} < y \le 45\frac{5}{11}

, so with

40 < y < 50

the candidates are

42

,

43

,

44

,

45

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{5}{45}

reduces, so it is out. That leaves

y = 42

,

y = 43

,

y = 44

; the larger sum uses

y = 44

.

x + y = 5 + 44 = 49

\frac{5}{44} = 0.1136\ldots

fits, and

44

is the biggest allowed denominator.

Answer: 49

4 · Reviewdoes it hold up?

Check: $\frac{5}{44} = 0.1136\ldots$ starts $0.11$ , and $\gcd(5, 44) = 1$ . The other survivors $\frac{5}{42} = 0.1190\ldots$ and $\frac{5}{43} = 0.1162\ldots$ give only $47$ and $48$ .

Another way: Systematic list: for each $y$ from $41$ to $49$ , the interval $[0.11y,\ 0.12y)$ contains a whole number only for $y = 42, 43, 44, 45$ (each gives $x = 5$ ); dropping $\frac{5}{45}$ leaves the same three fractions.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 10 hard

$x$ and $y$ are natural numbers with $40 < y < 50$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $0$ and the second digit after the decimal point is $7$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $40 < y < 50$ , has decimal $0.07\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $40 < y < 50$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.07\ldots$ (first digit $0$ , second digit $7$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.07$ means $0.07 \le \frac{x}{y} < 0.08$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $41$ and $49$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.07

is at least

0.07

and less than

0.08

. Multiply through by

y

.

\frac{7}{100}\,y \le x < \frac{8}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

41 \le y \le 49

, the smallest the lower bound can be is

\frac{7}{100} \times 41

and the largest the upper bound can be is

\frac{8}{100} \times 49

.

\frac{287}{100} \le x < \frac{392}{100}

The only whole number from

2.87

up to (not including)

3.92

is

x = 3

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 3

back:

0.07 \le \frac{3}{y} < 0.08

means

\frac{3}{0.08} < y \le \frac{3}{0.07}

.

\frac{75}{2} < y \le \frac{300}{7}

That is

37\frac{1}{2} < y \le 42\frac{6}{7}

, so with

40 < y < 50

the candidates are

41

and

42

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{3}{42}

reduces, so it is out. That leaves only

y = 41

.

x + y = 3 + 41 = 44

\frac{3}{41} = 0.0731\ldots

fits, and

41

is the biggest allowed denominator.

Answer: 44

4 · Reviewdoes it hold up?

Check: $\frac{3}{41} = 0.0731\ldots$ starts $0.07$ , and $\gcd(3, 41) = 1$ .

Another way: Systematic list: for each $y$ from $41$ to $49$ , the interval $[0.07y,\ 0.08y)$ contains a whole number only for $y = 41, 42$ (each gives $x = 3$ ); dropping $\frac{3}{42}$ leaves the same single fraction.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 11 hard

$x$ and $y$ are natural numbers with $55 < y < 65$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $0$ and the second digit after the decimal point is $4$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $55 < y < 65$ , has decimal $0.04\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $55 < y < 65$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.04\ldots$ (first digit $0$ , second digit $4$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.04$ means $0.04 \le \frac{x}{y} < 0.05$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $56$ and $64$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.04

is at least

0.04

and less than

0.05

. Multiply through by

y

.

\frac{4}{100}\,y \le x < \frac{5}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

56 \le y \le 64

, the smallest the lower bound can be is

\frac{4}{100} \times 56

and the largest the upper bound can be is

\frac{5}{100} \times 64

.

\frac{224}{100} \le x < \frac{320}{100}

The only whole number from

2.24

up to (not including)

3.2

is

x = 3

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 3

back:

0.04 \le \frac{3}{y} < 0.05

means

\frac{3}{0.05} < y \le \frac{3}{0.04}

.

60 < y \le 75

That is

60 < y \le 75

, so with

55 < y < 65

the candidates are

61

,

62

,

63

,

64

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{3}{63}

reduces, so it is out. That leaves

y = 61

,

y = 62

,

y = 64

; the larger sum uses

y = 64

.

x + y = 3 + 64 = 67

\frac{3}{64} = 0.0468\ldots

fits, and

64

is the biggest allowed denominator.

Answer: 67

4 · Reviewdoes it hold up?

Check: $\frac{3}{64} = 0.0468\ldots$ starts $0.04$ , and $\gcd(3, 64) = 1$ . The other survivors $\frac{3}{61} = 0.0491\ldots$ and $\frac{3}{62} = 0.0483\ldots$ give only $64$ and $65$ .

Another way: Systematic list: for each $y$ from $56$ to $64$ , the interval $[0.04y,\ 0.05y)$ contains a whole number only for $y = 61, 62, 63, 64$ (each gives $x = 3$ ); dropping $\frac{3}{63}$ leaves the same three fractions.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Variant 12 hard

$x$ and $y$ are natural numbers with $60 < y < 70$ . The fraction $\frac{x}{y}$ is in lowest terms and is less than $1$ . When it is written as a decimal, the first digit after the decimal point is $0$ and the second digit after the decimal point is $5$ . Find the greatest possible value of $x + y$ .

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction $\frac{x}{y}$ in lowest terms, less than $1$ , with $60 < y < 70$ , has decimal $0.05\ldots$ . Find the largest possible $x + y$ .

Givens

$x$ , $y$ are natural numbers and $60 < y < 70$ .
$\frac{x}{y} < 1$ and $\frac{x}{y}$ is in lowest terms.
The decimal starts $0.05\ldots$ (first digit $0$ , second digit $5$ ).

Unknowns

The greatest possible value of $x + y$ .

Constraints

Starting with $0.05$ means $0.05 \le \frac{x}{y} < 0.06$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Turn the digit condition into an inequality. Since $y$ is trapped between $61$ and $69$ , the inequality first forces the numerator, then gives a short list of denominators to test for lowest terms.

3 · Execute4 carry out the plan

1Turn the digits into an inequality

#13 Convert to Algebra 7.EE.B.4

A decimal that begins

0.05

is at least

0.05

and less than

0.06

. Multiply through by

y

.

\frac{5}{100}\,y \le x < \frac{6}{100}\,y

The digit facts become bounds on

x

.

2Pin down the numerator

#13 Convert to Algebra 8.NS.A.1

With

61 \le y \le 69

, the smallest the lower bound can be is

\frac{5}{100} \times 61

and the largest the upper bound can be is

\frac{6}{100} \times 69

.

\frac{305}{100} \le x < \frac{414}{100}

The only whole number from

3.05

up to (not including)

4.14

is

x = 4

.

3Find the allowed denominators

#13 Convert to Algebra 7.EE.B.4

Put

x = 4

back:

0.05 \le \frac{4}{y} < 0.06

means

\frac{4}{0.06} < y \le \frac{4}{0.05}

.

\frac{200}{3} < y \le 80

That is

66\frac{2}{3} < y \le 80

, so with

60 < y < 70

the candidates are

67

,

68

,

69

.

4Keep lowest terms and maximize

#3 Eliminate Possibilities 6.NS.B.4

\frac{4}{68}

reduces, so it is out. That leaves

y = 67

and

y = 69

; the larger sum uses

y = 69

.

x + y = 4 + 69 = 73

\frac{4}{69} = 0.0579\ldots

fits, and

69

is the biggest allowed denominator.

Answer: 73

4 · Reviewdoes it hold up?

Check: $\frac{4}{69} = 0.0579\ldots$ starts $0.05$ , and $\gcd(4, 69) = 1$ . The other survivor $\frac{4}{67} = 0.0597\ldots$ gives only $71$ .

Another way: Systematic list: for each $y$ from $61$ to $69$ , the interval $[0.05y,\ 0.06y)$ contains a whole number only for $y = 67, 68, 69$ (each gives $x = 4$ ); dropping $\frac{4}{68}$ leaves the same two fractions.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Relating a rational number to its decimal expansion and digit bounds.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the inequalities for $x$ and $y$.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the greatest common factor to keep only fractions in lowest terms.

💡Takeaway. Leading digits are really a pair of inequalities -- squeeze with them.

Turn leading digits into an inequality · 12 practice problems

Generated variants — 12

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