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[Smart Math] Arithmetic Problem 23

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

Arithmetic

Problem

Find the smallest from: \frac{3}{7}, \frac{7}{20}, \frac{13}{25}, \frac{11}{20} and \frac{15}{23}.

  1. \frac{3}{7}
  2. \frac{7}{20}
  3. \frac{13}{25}
  4. \frac{11}{20}
  5. \frac{15}{23}

The Usual Method

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Equate denominators by finding LCM.

LCM = 7 x 25 x 4 x 23 = 16100

Now correspondingly multiply numerators with the same multiple that is used to get the denominator = 16100.

\therefore \frac{3}{7}=\frac{3}{7}\times \frac{25\times 4\times 23}{25\times 4\times 23}=\frac{6900}{16100}

\therefore \frac{7}{20}=\frac{7}{20}\times \frac{5\times 7\times 23}{5\times 7\times 23}=\frac{5635}{16100}

\therefore \frac{13}{25}=\frac{13}{25}\times \frac{4\times 7\times 23}{4\times 7\times 23}=\frac{8372}{16100}

\therefore \frac{11}{20}=\frac{11}{20}\times \frac{5\times 7\times 23}{5\times 7\times 23}=\frac{8855}{16100}

\therefore \frac{15}{23}=\frac{15}{23}\times \frac{4\times 7\times 25}{4\times 7\times 25}=\frac{10500}{16100}

As can be seen the smallest is \frac{5635}{16100} = \frac{7}{20}

(Ans: 2)

Estimated Time to arrive at the answer = 150 seconds.

Using Technique

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Between \frac{7}{20} and \frac{11}{20}, \frac{7}{20} is the smaller one. This eliminates option ‘4’.

Also, \frac{7}{20} = \frac{7\times 5}{20\times 5}=\frac{35}{100}=0.35

Similarly, \frac{3}{7} = \frac{3\times 14}{7\times 14}=\frac{42}{98}\approx 0.43

Since, 0.35 < 0.43, eliminate option ‘1’.

Similarly, \frac{13}{25} = \frac{13\times 4}{25\times 4}=\frac{52}{100}=0.52

Since, 0.35 < 0.52, eliminate option ‘3’.

Similarly, \frac{15}{23} = \frac{15\times 4}{23\times 4}=\frac{60}{92}\approx 0.65

Since, 0.35 < 0.65, eliminate option ‘5’.

Hence answer will be the left out option ‘2’.

(Note: In this case we are trying to bring the denominators close to 100 to get the decimal value of the fraction, so that it can directly be compared)

(Ans: 2)

Estimated Time to arrive at the answer = 30 seconds.
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