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[Smart Math] Arithmetic Problem 23

Here’s and example of a SMART MATH problem for 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

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

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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