How To Solve Numbers, Decimal And Fractions Questions Quickly

How To Solve Quickly Number, Decimal and Fraction

It represents a part of a whole or more generally, any number of equal parts. If a unit is divided into any number of equal parts, then one or more of these parts is termed as a fraction of the unit.

The numerator in a fraction represents a number of equal parts and denominator (≠0) represents how many of those parts make up a unit or a whole.

The value of fraction (x/y) = 1, if numerator = denominator
The value of fraction is zero,if numerator=0 and denominator ≠0 
The value of fraction is infinity,if the denominator=0
The value of fraction remains unchanged,even if the numerator and denominator are multiplied or divided by same number

Pure recurring decimal: In a decimal fraction, if all the numbers after decimal point repeat, then it is called as pure recurring decimal. 
Mixed recurring decimal: In a decimal fraction, if some numbers are repetitive and some are not, then it is called as mixed recurring decimal. 

Basic Concept To Solve Number, Decimals and Fractions Questions Quickly

(Must Remember) 

1) (a – b)^{2} = (a^{2} + b^{2} – 2ab)
2) (a + b)^{2} = (a^{2} + b^{2} + 2ab)
3) (a + b) (a – b) = (a^{2} – b^{2} )
4) (a^{3} + b^{3}) = (a + b) (a^{2} – ab + b^{2})
5) (a^{3} – b^{3}) = (a – b) (a^{2} – ab + b^{2})
6) (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + bc + ca)
7) (a^{3} + b^{3} + c^{3} – 3abc) = (a + b + c) (a^{2}+ b^{2} + c^{2} – ab – bc – ac)

Now let us solve some of the questions to find some methods How To Solve Numbers And Fractions Question.

How To Solve Number, Decimals and Fractions Questions Quickly 

Question 1  : Convert 0.737373… into vulgar fraction?

Options

(a)  \frac{73}{99} \

(b)  \frac{77}{99} \

(c)  \frac{73}{90} \

(d)  \frac{73}{900} \

Correct Answer: Option A

Explanation:

In a decimal fraction, if there are n numbers of repeated numbers after a decimal point, then just write one repeated number in the numerator and in denominator take n number of nines equal to repeated numbers you observe after the decimal point.
0.737373… is written as  \frac{}{}0.\overline{73} .

Question 2 : Find the number of zeros in 2145 x 5234 

Options

(a) 145
(b) 234
(c) 10
(d) None of these.

Correct Options: A

Explanations

we have to look for  the pairs of 2x 5, so the maximum pairs we can form are 145 because we have the  maximum power  of 2 is 145 , so the number of zeros  in this case are 145.

Questions 3 : Find the number of zeros at end of 5 x 10 x 15 x 20 x 25 x 30 x 35…………………………………… x 240 x 245 x 250?

Options

(a) 30
(b) 47
(c) 50
(d) 48

Correct Options: B

Explanations

We can take 5 as common out of this expression and write it as
5(1 x 2 x 3 x 4……………………………………  x 49 x 50)
5(50!)
To find number of zeros we first find
Maximum power of 2 in 50!
[  \frac{50}{2} \ ] + [  \frac{50}{2^2} \ ] + [  \frac{50}{2^3} \ ] + [  \frac{50}{2^4} \ ] + [  \frac{50}{2^5} \ ] + ………. 
= 25 + 12 + 6 + 3 +1= 47
Maximum power of 5 in 50!

 \frac{50}{5} \ + [  \frac{50}{5^2} \ ] + [  \frac{50}{5^3} \ ]
= 10 + 2= 12

Thus maximum power of 5 present in given expression is 62 and maximum power of 2 present in given expression is 47. Hence number of zeros will be 47.

Question 4 : Find the value of \frac{5.45 + 6.23 * 9.7}{3.42 + 5.23 * 8.6}

A. 65.188 / 48.389
B. 65.881 / 48.398
C. 54.65 / 56.32
D. 55.654 / 65.456

Answer: A
Sol:
\frac{5.45 + 6.23 * 9.7}{3.42 + 5.23 * 8.6}

\frac{5.45 + 60.431}{3.42 + 44.978}

\frac{5.45 + 60.431}{3.42 + 44.978}

65.881 / 48.398

Question 5 : 30% of 40% of 980 + 6.35 × (72/8) = ?

A. 53.5
B. 57.39
C. 57.75
D. 45.25

Correct Option: B
30% of 40% of 980 + 6.35 × (72/8) = ?

\frac{30}{100}\times \frac{40}{100}* 980 +6.35\times9 = x

0.3 * 0.4 * 980 + 6.35 * 9= x 

117.6 + 57.15 = x

x = 174.75