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: $\frac{1043}{1100}$
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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 $\frac{1141}{1100}$.
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Question : The value of $0. \overline{57}-0.4 \overline{32}+0.3 \overline{5}$ is:
Option 1: $0.4 \overline{98}$
Option 2: $0.4 \overline{94}$
Option 3: $0. \overline{498}$
Option 4: $0. \overline{495}$
Question : The value of $11. \overline{4}+22.5 \overline{67}-33.5 \overline{9}$ is:
Option 1: $40. \overline{12}$
Option 2: $4. \overline{12}$
Option 3: $0.4\overline{12}$
Option 4: $0.04\overline{12}$
Question : The value of $0.4 \overline{6}+0.7 \overline{23}-0.3 \overline{9} \times 0. \overline{7}$ is:
Option 1: $0.\overline{97}$
Option 2: $0.\overline{57}$
Option 3: $0.\overline{77}$
Option 4: $0.\overline{87}$
Question : If $a: b: c=\frac{1}{4}: \frac{1}{3}: \frac{1}{2}$, then $\frac{a}{b}: \frac{b}{c}: \frac{c}{a}=$?
Option 1: 12 : 9 : 8
Option 2: 9 : 8 : 24
Option 3: 8 : 9 : 24
Option 4: 9 : 12 : 8
Question : If $\triangle A B C \sim \triangle F D E$ such that $A B=9 \mathrm{~cm}, A C=11 \mathrm{~cm}, D F=16 \mathrm{~cm}$ and $D E=12 \mathrm{~cm}$, then the length of $BC$ is:
Option 1: $5 \frac{3}{4} \mathrm{~cm}$
Option 2: $4\frac{3}{5} \mathrm{~cm}$
Option 3: $3\frac{5}{7} \mathrm{~cm}$
Option 4: $6\frac{3}{4} \mathrm{~cm}$
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