(Algebra)The products of x and \(\frac {7}{15}\), as well as of x and \(\frac{41}{12}\) are positive integers.

[Max@Math Revolution]

(Algebra) The products of x and \(\frac{7}{15}\), as well as of x and \(\frac{41}{12}\) are positive integers. What is the smallest possible value of x?

A. \(\frac{7}{60}\)

B. \(\frac{60}{7}\)

C. \(\frac{7}{30}\)

D. \(\frac{30}{7}\)

E. \(\frac{15}{7}\)

[Scott@TargetTestPrep]

(Algebra) The products of x and \(\frac{7}{15}\), as well as of x and \(\frac{41}{12}\) are positive integers. What is the smallest possible value of x?

A. \(\frac{7}{60}\)

B. \(\frac{60}{7}\)

C. \(\frac{7}{30}\)

D. \(\frac{30}{7}\)

E. \(\frac{15}{7}\)

Solution:


None of the answer choices produce an integer when multiplied by 41/12. If 42/12 = 7/2 was meant instead, then 30/7 would have been the answer since the LCM of the denominators 15 and 2 is 30 and since both numerators cancel the denominator of 30/7 when multiplied.

Answer: None

[Max@Math Revolution]

Solution:

We have to find the smallest possible value of x such that x * \(\frac{7}{15}\) and x * 4\(\frac{1}{12}\) are positive integers.


=> x *\(\frac{7}{15}\)= p where p is a positive integer and x *\(\frac{49}{12}\) = q where q is a positive integer.

x = \(\frac{15p}{7}\) ---------------(1) and x = \(\frac{12q}{49}\)---------------(2)

Dividing equation (1) by (2), we get:

=> \(\frac{x}{x} = \frac{\frac{15p}{7}}{\frac{12q}{49}}\)

=> \(1=\frac{\frac{15p}{7}}{\frac{12q}{49}}\)

=> 1 = \(\frac{15p}{7}\) *\(\frac{49}{12q}\)

=> 1 = \(\frac{35p}{4q}\)

=> \(\frac{q}{p}\) = \(\frac{35p}{4}\)

Hence, p = 4 and q = 35.

x = \(\frac{15 * 4}{7}\) = \(\frac{60}{7}\)

Therefore, B is the correct answer.

Answer B

[deloitte247]

$$x\cdot\frac{7}{15}=\frac{7}{15}x\ and\ x\cdot\frac{41}{12}=\frac{41}{12}x$$
$$What\ is\ the\ smallest\ possible\ value\ of\ x?$$
$$going\ through\ all\ available\ options\ one\ after\ the\ other$$
$$if\ x=\frac{7}{60}\ \ for\ \ \frac{7}{15}x$$
$$\frac{7}{15}\cdot\frac{7}{60}=\frac{14}{900}$$
$$this\ is\ not\ an\ integer\ so\ x\ cannot\ be\ \frac{7}{60}$$
$$if\ x=\frac{60}{7}\ \ for\ \ \frac{7}{15}x$$
$$\frac{7}{15}\cdot\frac{60}{7}=4$$
$$if\ x=\frac{60}{7\ }\ \ for\ \ \frac{41}{12}x$$
$$\frac{41}{12}\cdot\frac{60}{7}=\frac{205}{7}$$
$$this\ is\ not\ an\ integer\ so\ x\ cannot\ be\ \frac{60}{7}$$
$$if\ x=\frac{7}{30}\ \ for\ \ \frac{7}{15}x$$
$$\frac{7}{15}\cdot\frac{7}{30}=\frac{14}{450}$$
$$this\ is\ not\ an\ integer\ so\ x\ cannot\ be\ \frac{7}{30}$$
$$if\ x=\frac{30}{7}\ \ for\ \ \frac{7}{15}x$$
$$\frac{7}{15}\cdot\frac{30}{7}=2$$
$$if\ x=\frac{30}{7\ }\ \ for\ \ \frac{41}{12}x$$
$$\frac{41}{12}\cdot\frac{30}{7}=\frac{205}{14}$$
$$this\ is\ not\ an\ integer\ so\ x\ cannot\ be\ \frac{30}{7}$$
$$if\ x=\frac{15}{7}\ \ for\ \ \frac{41}{12}x$$
$$\frac{41}{12}\cdot\frac{15}{7}=\frac{615}{84}$$
$$this\ is\ not\ an\ integer\ so\ x\ cannot\ be\ \frac{15}{7}$$

Note: There is no correct answer in the available options. Hence, we cannot ascertain a definite answer