# What Is 70 As A Fraction

What Is 70 As A Fraction – Surds and Fractions Question Solution for SSC CGL Set 70 explains how to solve all 10 Surds and Fractions questions quickly. But do the test first.

SSC CGL 10 Question Answers and Fractions Solution: Time to solve is 12 minutes. Problem 1.

## What Is 70 As A Fraction

The product of two fractions is \$displaystylefrac\$ and their quotient is \$displaystylefrac\$ . The larger of the two fractions,

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This solution is fast and completely mind-blowing. In this case the fractions must be treated as single numbers, which is a bit of a mental hurdle.

The difference between the denominator and the numerator is exactly 1. In this case, the fraction with the larger numerator is larger.

If the difference between the denominator and numerator of two fractions is exactly the same, the fraction with the larger numerator is the larger fraction.

Therefore, if the denominator, \$b gt d\$, or the numerators \$a gt c\$, the first fraction is greater than both, and vice versa.

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With a numerator difference equal to the denominator, whichever denominator or numerator is greater, the corresponding fraction is greater.

The problem involves comparisons of fractions between lower and upper range limits with selected values. Without a proper strategy, this process takes time.

If the numerator of the two fractions to be compared is a factor of the second numerator, multiply the first numerator (and also the first denominator) by an appropriate factor to make the second numerator equal. Now with even numbers, the

For example, if we compare the fraction \$displaystylefrac\$ to \$displaystylefrac\$ and find the first numerator 7 as a factor of the second numerator 35, we multiply both the first numerator and the first denominator \$display style. frac=5\$ To convert a fraction,

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Since the two numerators are now equal, since the second denominator 43 is smaller than the first 45, the second fraction is larger than the first (dividing the same number by a smaller number gives a larger result).

This is a rarely used comparison of fractions by matching the numerator, but it is just as effective as a frequently used comparison of fractions by matching the denominator.

If the two fractions to be compared are \$displaystylefrac\$ and \$displaystylefrac\$ , multiply the numerator and denominator of the first fraction by \$displaystylefrac=6\$ ,

Now 24 with both denominators, since the first number 18 is smaller than the second 19, the second fraction is larger.

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When you cannot use the faster methods of equalizing the numerator or denominator, you must follow the traditional direct method.

And multiply the numerators by these special factors. Whichever result is larger leads to a larger fraction.

In this case, the unique factor of 18 that is not in 24 is 3, and the unique factor of 24 that is not in 18 is 4.

This is a method derived from the method of addition and subtraction of fractions, where we find the LCM of both denominators.

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Returning to our problem, the first option in this case is not only the largest of the four options (and therefore should be compared to the upper range limit first), it is also easily compared using numerator equalization.

Then we start comparing the next smallest option, \$displaystylefrac\$ or \$displaystylefrac\$ to \$displaystylefrac\$ . This time by direct comparison we find that the fourth choice is larger than the upper bound of the \$displaystylefrac\$ interval.

If we now drop the upper bound of the range, we take the lower bound \$displaystylefrac\$ and compare it to the lowest of the four options  \$displaystylefrac\$ or \$displaystylefrac\$ . Since this second option is smaller than the lower bound of the array, this leaves the third option \$displaystylefrac\$ as the only remaining option, and thus the answer is discarded.

– Strategic selection of pairs of comparison fractions to reduce computations – Rank comparison strategy – Example of analytical procedure.

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Although all processing was mental, without strategic selection of different pairs for comparison, time would have been greater.

When the upper and lower values ​​of a range are given and we need to find which of the given four option values ​​are in the range, we try to find the options outside the range using

We must place one of the four given values ​​in a range of two values, first mentally arranging the four optional values ​​in a sorted order (ascending or descending). In this case, the option values ​​are very conveniently sorted and we identify the pattern.

Step Two: First Comparison: Compare the highest range limit value with the second largest option value (not the largest option value)

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If the result of the comparison returns the option value smaller, then in this particular case, the particular option value will be the answer,

An MCQ problem should not have more than one answer, i.e. more than one option value is less than the upper range but greater than the lower range.

In fact, the second largest option value \$displaystylefrac\$ is greater than the upper limit of the range \$displaystylefrac\$ and is therefore out of range. Only two possible options remain.

The second lowest option value should be compared to the lower limit of the array, but the convenience of comparing \$displaystylefrac\$ or \$displaystylefrac\$ to \$displaystylefrac\$ is much greater than comparing \$. \$displaystylefrac\$ with displaystylefrac \$.

### As A Fraction Is (a) 1 /40 (b) 1/ 60 (c)

Now we can compare the highest limit value with the next lower option value or use the lower limit value in comparison. We take the second step as a convenience to compare \$displaystylefrac\$ or \$displaystylefrac\$ with \$displaystylefrac\$ .

It is a systematic process and gives you a quick result with a maximum of two comparisons.

If the difference between a positive proper fraction and the reciprocal of the same fraction is \$displaystylefrac\$ , then the fraction is:

The main challenge of this problem is that from the problem statement, “the fraction is subtracted from the reciprocal of the fraction” and not the other way around.

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The sum of the three fractions is \$2displaystylefrac\$. Dividing the largest fraction in \$displaystylefrac\$ by the smallest results in \$displaystylefrac\$ greater than the middle fraction. Then the smaller fraction is:

Among the option values, the second option has 7 in the denominator. By MCM addition, no fraction has 7 as its denominator. A divisor value does not contain a 5 in the denominator, so a small fraction cannot contain a 5 in the numerator. Most likely the third option,

If we subtract this from the sum of three we get, \$displaystylefrac\$ is the sum of the first and third fractions. Subtracting the first value again, we get the third fraction \$displaystylefrac\$ .

Many times, by analyzing the relationships between given values, factors in a set of choice values ​​can be eliminated as numerator or denominator, resulting in a quick solution.

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Whenever we find an expression that originates under a square root, we know that we must square the expression that originates under the square root. This problem has three such expressions.

Sums in brackets are the third denominator (after subtracting square roots). We rationalize to form a simplified expression,

Since \$sqrt\$ is a surd term, it must be equal to the surd term of LHS (irrationals and rationals cannot be added, even if we represent a sum, we cannot arrive at any result by doing addition).

Ignoring the first two terms, we decide to find a suitable pattern for the last two terms and combine them first,

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We temporarily substitute, for convenience, \$sqrt=p\$ and \$sqrt=q\$. It is a component expression substitution and converts the given expression into simpler form,