Question : If $\mathrm{A}=0.3 \overline{12}, \mathrm{~B}=0.4 \overline{15}$ and $\mathrm{C}=0.30 \overline{9}$, then what is the value of $A + B + C$?
Option 1: $\frac{1141}{1100}$
Option 2: $\frac{1097}{1100}$
Option 3: $\frac{1211}{1100}$
Option 4: $\frac{1043}{1100}$
Correct Answer: $\frac{1141}{1100}$
Solution :
Given: $\mathrm{A}=0.3 \overline{12}, \mathrm{~B}=0.4 \overline{15}$ and $\mathrm{C}=0.30 \overline{9}$.
Let $xyz$ be a three-digit number after the decimal.
Then, if the first two digits of the calculation have a bar i.e. $0.x\overline{yz}=\frac{xyz–x}{990}$.
Also, if there is a bar on just one digit i.e. $0.xy\overline{z}=\frac{xyz–xy}{900}$.
$\mathrm{A}=0.3 \overline{12}$
⇒ $A=\frac{312–3}{990}=\frac{309}{990}$
$\mathrm{~B}=0.4 \overline{15}$
⇒ $B=\frac{415–4}{990}=\frac{411}{990}$
$\mathrm{C}=0.30 \overline{9}$
⇒ $C=\frac{309–30}{900}=\frac{279}{900}$
The value of $A + B + C=\frac{309}{990}+\frac{411}{990}+\frac{279}{900}$.
$=\frac{720}{990}+\frac{279}{900}=\frac{7200+3069}{9900}=\frac{10269}{9900}=\frac{1141}{1100}$
Hence, the correct answer is $\frac{1141}{1100}$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
Upcoming Exams
Trending Articles around SSC CGL
