The product (1−1/2)(1−1/3)(1−1/4)(1−1/5)...

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The product
$$\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right).........\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)$$
equals...

$$A:\frac{1}{10}$$

$$B:\frac{1}{20}$$

$$C:\frac{3}{20}$$

$$D:\frac{3}{10}$$

$$E:\frac{7}{20}$$

The OA is B.

Can any expert help me with this PS question please? Thanks.
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by GMATGuruNY » Sun Oct 22, 2017 3:57 am
LUANDATO wrote:The product
$$\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right).........\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)$$
equals...

$$A:\frac{1}{10}$$

$$B:\frac{1}{20}$$

$$C:\frac{3}{20}$$

$$D:\frac{3}{10}$$

$$E:\frac{7}{20}$$
In the solution below, all of the values in red CANCEL OUT.

$$\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right).........\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)$$

= (1/2)(2/3)(3/4)....(17/18)(18/19)(19/20).

= 1/20.

The correct answer is B.
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by Scott@TargetTestPrep » Wed Nov 20, 2019 6:02 pm
BTGmoderatorLU wrote:The product
$$\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right).........\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)$$
equals...

$$A:\frac{1}{10}$$

$$B:\frac{1}{20}$$

$$C:\frac{3}{20}$$

$$D:\frac{3}{10}$$

$$E:\frac{7}{20}$$

The OA is B.

Can any expert help me with this PS question please? Thanks.

Solving the product for the 1st four parentheses, we have:

(1/2) x (2/3) x (3/4) x (4/5) x ... x (18/19) x (19/20)

After cross canceling, we see that only the numerator of the first fraction and the denominator of the last fraction remain. Thus, the product of the entire expression is 1/20.

Answer: B

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