DSQ: Understanding decimals - GMAT Quantitative

Assume both statements to be true. A multiple of both 2 and 3 is also a multiple of 6, so both denominators are multiples of 6. Examine these two examples:

In both examples, the conditions of the problem, including both statements, are met, but in only one is the sum equivalent to a terminating decimal or integer.

Assume Statement 1 alone. Examine these two examples:

Both fit the main condition and that of Statement 1, but in only one case is the sum equal to a terminating decimal or integer.

Assume Statement 2 alone. A necessary and sufficient condition for a lowest-terms fraction to be equivalent to a terminating decimal is that its only prime factors are 2 and 5. By Statement 2, both denominators have 2 as their only prime factor, so both can be represented by terminating decimals; their product must be terminating as well.

Assume both statements. Examine these two examples; both fit the main condition and those given in both statements.

In only one case is the product equivalent to an integer or a terminating decimal.

Assume Statement 1 alone. Rewrite the fractions as follows:

, where - that is, their lowest-terms representations.

By Statement 1, , so . Therefore,

Since is in lowest terms, is not a multiple of 3, so the 3 must remain in the denominator, even if the and the can be reduced. Since a fraction which, in lowest terms, has a denominator with any prime factor other than 2 or 5 cannot be represented by an integer or a terminating decimal, Statement 1 proves that has a repeating decimal as its equivalent.

A similar argument can be used to demonstrate that Statement 2 proves the same of .

Assume Statement 1 alone. Examine these examples:

In both cases, the main conditon and that of Statement 1 are satisfied, but in only one case can the result be represented by a terminating decimal or integer.

A similar argument shows Statement 2 alone to be insufficient:

Assume both statements. Then rewrite the fractions as follows:

, where - that is, their lowest-terms representations.

By Statement 1, ; by Statement 2, . Therefore,

,

and their product is

,

a terminating decimal.

Assume Statement 1 alone. A necessary and sufficient condition for a lowest-terms fraction to be equivalent to a terminating decimal is that its only prime factors are 2 and 5. Statement 1 says this about both fractions; the sum of terminating decimals is itself an integer or a terminating decimal.

Statement 2 alone provides insuffiicient information. Examine these examples:

Both fit the conditions of the main premise and Statement 2, but in only one case is the sum equal to a terminating decimal.

We have one equation and two unknowns. To find x, we need to find y.

I) Tells us that y is .79 therefore we can plug this value into our equation and solve for x.

Plugging in the y value from Statement I and using algebraic operations to solve for x.

Multiply first term.

Subtract 4.74 from both sides.

Divide by -13.

II) Is irrelevant.

Therefore Statement I) alone is sufficient to answer the question but Statement II) alone is not.

Assume Statement 1 alone. Examine these examples:

In both cases, the main conditon and that of Statement 1 are satisfied, but in only one case can the result be represented by a terminating decimal or integer.

A similar argument shows Statement 2 alone to be insufficient:

Assume both statements. Then rewrite the fractions as follows:

, where - that is, their lowest-terms representations.

By Statement 1, ; by Statement 2, . Therefore,

,

and their product is

,

a terminating decimal.

A fraction in lowest terms can be expressed as a terminating decimal if and only if its denominator has no prime factors other than 2 or 5; the numerator is irrelevant. Statement 1 is unhelpful since it only gives information about the numerator. If Statement 2 is assumed - that is, if we know the denominator has 7 as a prime factor - we know that the decimal representation of is repeating.

Assume both statements to be true. A multiple of both 2 and 3 is also a multiple of 6, so both denominators are multiples of 6. Examine these two examples:

In both examples, the conditions of the problem, including both statements, are met, but in only one is the sum equivalent to a terminating decimal or integer.

Assume Statement 1 alone. Examine these two examples:

Both fit the main condition and that of Statement 1, but in only one case is the sum equal to a terminating decimal or integer.

Assume Statement 2 alone. A necessary and sufficient condition for a lowest-terms fraction to be equivalent to a terminating decimal is that its only prime factors are 2 and 5. By Statement 2, both denominators have 2 as their only prime factor, so both can be represented by terminating decimals; their product must be terminating as well.

Assume both statements. Examine these two examples; both fit the main condition and those given in both statements.

In only one case is the product equivalent to an integer or a terminating decimal.

Assume Statement 1 alone. Rewrite the fractions as follows:

, where - that is, their lowest-terms representations.

By Statement 1, , so . Therefore,

Since is in lowest terms, is not a multiple of 3, so the 3 must remain in the denominator, even if the and the can be reduced. Since a fraction which, in lowest terms, has a denominator with any prime factor other than 2 or 5 cannot be represented by an integer or a terminating decimal, Statement 1 proves that has a repeating decimal as its equivalent.

A similar argument can be used to demonstrate that Statement 2 proves the same of .

Assume Statement 1 alone. A necessary and sufficient condition for a lowest-terms fraction to be equivalent to a terminating decimal is that its only prime factors are 2 and 5. Statement 1 says this about both fractions; the sum of terminating decimals is itself an integer or a terminating decimal.

Statement 2 alone provides insuffiicient information. Examine these examples:

Both fit the conditions of the main premise and Statement 2, but in only one case is the sum equal to a terminating decimal.

We have one equation and two unknowns. To find x, we need to find y.

I) Tells us that y is .79 therefore we can plug this value into our equation and solve for x.

Plugging in the y value from Statement I and using algebraic operations to solve for x.

Multiply first term.

Subtract 4.74 from both sides.

Divide by -13.

II) Is irrelevant.

Therefore Statement I) alone is sufficient to answer the question but Statement II) alone is not.

A fraction in lowest terms can be expressed as a terminating decimal if and only if its denominator has no prime factors other than 2 or 5; the numerator is irrelevant. Statement 1 is unhelpful since it only gives information about the numerator. If Statement 2 is assumed - that is, if we know the denominator has 7 as a prime factor - we know that the decimal representation of is repeating.

Assume both statements to be true. A multiple of both 2 and 3 is also a multiple of 6, so both denominators are multiples of 6. Examine these two examples:

In both examples, the conditions of the problem, including both statements, are met, but in only one is the sum equivalent to a terminating decimal or integer.

Assume Statement 1 alone. Examine these two examples:

Both fit the main condition and that of Statement 1, but in only one case is the sum equal to a terminating decimal or integer.

Assume Statement 2 alone. A necessary and sufficient condition for a lowest-terms fraction to be equivalent to a terminating decimal is that its only prime factors are 2 and 5. By Statement 2, both denominators have 2 as their only prime factor, so both can be represented by terminating decimals; their product must be terminating as well.

Assume both statements. Examine these two examples; both fit the main condition and those given in both statements.

In only one case is the product equivalent to an integer or a terminating decimal.