Arithmetic Smart Math

[Smart Math] Arithmetic Problem 17

Here’s and example of a SMART MATH problem for ARITHMETIC.



\frac{\frac{1}{4}\times \left( \frac{1}{6}-\frac{1}{48} \right)}{\frac{1}{4}-\left( \frac{1}{6}-\frac{1}{48} \right)}\times \frac{\left( \frac{1}{4}-\frac{1}{6}-\frac{1}{48} \right)}{\left( \frac{1}{4}\times \frac{1}{6}-\frac{1}{48} \right)}=?

  1. \frac{1}{2}
  2. \frac{3}{2}
  3. \frac{21}{20}
  4. \frac{20}{21}
  5. \frac{2}{3}

The Usual Method

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First simplify each of the terms as follows:

\frac{1}{4}\times \left( \frac{1}{6}-\frac{1}{48} \right)=\frac{1}{4}\times \left( \frac{7}{48} \right)=\frac{7}{48\times 4}

\frac{1}{4}-\left( \frac{1}{6}-\frac{1}{48} \right)=\frac{1}{4}-\frac{7}{48}=\frac{5}{48}


\frac{1}{4}\times \frac{1}{6}-\frac{1}{48}=\frac{1}{24}-\frac{1}{48}=\frac{1}{48}

Hence the expression becomes:

\frac{\frac{7}{48\times 4}}{\frac{5}{48}}\times \frac{\frac{3}{48}}{\frac{1}{48}}=\frac{\frac{7}{4}\times 3}{5}=\frac{21}{20}

(Ans: 3)

Estimated Time to arrive at the answer = 75 seconds.

Using Technique

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If you would observe each component keenly, you would observe that except in the first component i.e. \frac{1}{4}\times \left( \frac{1}{6}-\frac{1}{48} \right), you get 4 x 48 as denominator. In all other components, you get only 48 as the denominator. It is important to note that in order to know the denominator, you do not have to ‘solve’ each component rather you can simply get it from the LCM of 4, 6 and 48.

Thus you will have a picture in your mind that would look as shown below:

\frac{\frac{something}{48\times 4}}{\frac{something}{48}}\times \frac{\frac{something}{48}}{\frac{something}{48}}

Canceling all 48s, we get \frac{something}{4}. This means that in the final answer, you should have 4 or a multiple of it in the denominator. The only option satisfying this condition is option ‘3’ i.e. \frac{21}{20}

(Ans: 3)

Estimated Time to arrive at the answer = 15 seconds.
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