[s2If current_user_can(access_s2member_level1)]
The proper fraction of \(3\dfrac{1}{8}\%\) is …
(A) \(\dfrac{1}{16}\)
(B) \(\dfrac{1}{32}\)
(C) \(\dfrac{7}{12}\)
(D) \(\dfrac{3}{25}\)
The proper fraction of \(3\dfrac{1}{8}\%\) is …
(A) \(\dfrac{1}{16}\)
(B) \(\dfrac{1}{32}\)
(C) \(\dfrac{7}{12}\)
(D) \(\dfrac{3}{25}\)
Solution: B
\begin{equation*} \begin{split} 3\frac{1}{8}\%& = \left(\frac{24}{8} + \frac{1}{8}\right)\%\\\\ 3\frac{1}{8}\%& = \frac{25}{8}\div 100\\\\ 3\frac{1}{8}\%& = \frac{25}{800}\\\\ 3\frac{1}{8}\%& = \frac{25 \div \color {red} 25}{800 \div \color {red} 25}\\\\ 3\frac{1}{8}\%& = \frac{1}{32}\\\\ \end{split} \end{equation*}
How much is \(\dfrac{3}{8}\) of 2000?
(A) 500
(B) 650
(C) 700
(D) 750
How much is \(\dfrac{3}{8}\) of 2000?
(A) 500
(B) 650
(C) 700
(D) 750
Solution: D
\begin{equation*} \begin{split} \frac{3}{8}\times 2000& = \frac{6000}{8}\\\\ \frac{3}{8}\times 2000& = 6000 \div 8\\\\ \frac{3}{8}\times 2000& = 750 \end{split} \end{equation*}
Arrange \(0.5, \:30\%, \: 200‰, \: \dfrac{2}{7}\) in ascending order
(A) \(0.5, \: 30\%, \: 200‰, \:\dfrac{2}{7}\)
(B) \(\dfrac{2}{7}, \:0.5, \:30\%, \:200‰\)
(C) \(30\%, \:0.5, \: 200‰, \:\dfrac{2}{7}\)
(D) \(200‰, \: \dfrac{2}{7}, \: 30\%, \:0.5\)
Arrange \(0.5, \:30\%, \: 200‰, \: \dfrac{2}{7}\) in ascending order
(A) \(0.5, \: 30\%, \: 200‰, \:\dfrac{2}{7}\)
(B) \(\dfrac{2}{7}, \:0.5, \:30\%, \:200‰\)
(C) \(30\%, \:0.5, \: 200‰, \:\dfrac{2}{7}\)
(D) \(200‰, \: \dfrac{2}{7}, \: 30\%, \:0.5\)
Solution: D
\begin{equation*} \begin{split} &\text{Convert into decimal form}\\\\ &30\% = 30 \div 100 = 0.3\\\\ &200‰ = 200 \div 1000 = 0.2\\\\ &\frac{2}{7}= 2 \div 7 = 0.286\\\\\\ &\text{In ascending order :}\\\\ & 0.2\:\:\: 0.286\:\:\:0.3\:\:\:0.5\\\\ & \text{or}\\\\ &200‰,\:\dfrac{2}{7},\:30\%,\: 0.5 \end{split} \end{equation*}
Arrange \(\dfrac{2}{5}, \dfrac{3}{4}, \dfrac{5}{6}, \dfrac{1}{3}\) in descending order
(A) \(\dfrac{5}{6}, \dfrac{1}{3}, \dfrac{2}{5}, \dfrac{3}{4}\)
(B) \(\dfrac{5}{6}, \dfrac{3}{4}, \dfrac{2}{5}, \dfrac{1}{3}\)
(C) \(\dfrac{2}{5}, \dfrac{5}{6}, \dfrac{3}{4}, \dfrac{1}{3}\)
(D) \(\dfrac{1}{3}, \dfrac{3}{4}, \dfrac{5}{6}, \dfrac{2}{5}\)
Arrange \(\dfrac{2}{5}, \dfrac{3}{4}, \dfrac{5}{6}, \dfrac{1}{3}\) in descending order
(A) \(\dfrac{5}{6}, \dfrac{1}{3}, \dfrac{2}{5}, \dfrac{3}{4}\)
(B) \(\dfrac{5}{6}, \dfrac{3}{4}, \dfrac{2}{5}, \dfrac{1}{3}\)
(C) \(\dfrac{2}{5}, \dfrac{5}{6}, \dfrac{3}{4}, \dfrac{1}{3}\)
(D) \(\dfrac{1}{3}, \dfrac{3}{4}, \dfrac{5}{6}, \dfrac{2}{5}\)
Solution: B
Find the Least Common Multiple of the denominators
The LCM of 5, 4, 6, and 3 is 60
Change each fraction (using equivalent fractions) to make their denominators the same as the least common denominator
\begin{equation*} \begin{split} &\frac{24}{60}\:\:\:\:\:\frac{45}{60}\:\:\:\:\:\frac{50}{60}\:\:\:\:\:\frac{20}{60}\\\\\\ &\text{So, the descending order is:}\\\\ &\frac{50}{60}\:\:\:\:\:\frac{45}{60}\:\:\:\:\:\frac{24}{60}\:\:\:\:\:\frac{20}{60}\\\\ &\text{or}\\\\ &\frac{5}{6}\:\:\:\:\:\frac{3}{4}\:\:\:\:\:\frac{2}{5}\:\:\:\:\:\frac{1}{3}\\\\ \end{split} \end{equation*}
Two fractions that can be inserted between \(\dfrac{1}{6} \text{ and }\dfrac{1}{5}\) are …
(A) \(\dfrac{17}{90} \text { and } \dfrac{16}{45}\)
(B) \(\dfrac{16}{90}\text { and }\dfrac{8}{45}\)
(C) \(\dfrac{16}{90}\text { and }\dfrac{17}{90}\)
(D) \(\dfrac{17}{30}\text { and }\dfrac{8}{30}\)
Two fractions that can be inserted between \(\dfrac{1}{6} \text{ and }\dfrac{1}{5}\) are …
(A) \(\dfrac{17}{90} \text { and } \dfrac{16}{45}\)
(B) \(\dfrac{16}{90}\text { and }\dfrac{8}{45}\)
(C) \(\dfrac{16}{90}\text { and }\dfrac{17}{90}\)
(D) \(\dfrac{17}{30}\text { and }\dfrac{8}{30}\)
Solution: C
Find the Least Common Multiple of the denominators
The LCM of 5 and 6 is 30
\begin{equation*} \begin{split} &\frac{1 \times \color {red} 5}{6 \times \color {red} 5} = \frac{5}{30}\\\\ &\frac{1 \times\color {red} 6}{5 \times \color {red} 6} = \frac{6}{30}\\\\ \end{split} \end{equation*}
\begin{array}{ccccc} &\dfrac{5}{30} & \:\:\dotso\:\:&\:\:\dotso\:\:&\dfrac{6}{30}\\\\ &\dfrac{5\times \color {red} 3}{30\times \color {red} 3} & \:\:\dotso\:\:&\:\:\dotso\:\:&\dfrac{6\times \color {red} 3}{30 \times \color {red} 3}\\\\ &\dfrac{15}{90}& \:\:\dotso\:\:&\:\:\dotso\:\:&\dfrac{18}{90}\\\\ &\dfrac{15}{90}&\color {red} \dfrac{16}{90}&\color {red}\dfrac{17}{90}&\color{black}\dfrac{18}{90}\\\\ \end{array}
\begin{equation*} \begin{split} &\text{Two fractions that can be inserted are:}\\\\ &\frac{16}{90}\:\:\text{and}\:\:\frac{17}{90} \end{split} \end{equation*}
Three fractions that can be inserted between \(\dfrac{4}{11}\text{ and }\dfrac{2}{5}\) are …
(A) \(\dfrac{20}{55}, \dfrac{21}{55}, \dfrac{22}{55}\)
(B) \(\dfrac{21}{55}, \dfrac{22}{55}, \dfrac{21}{55}\)
(C) \(\dfrac{41}{110}, \dfrac{42}{110}, \dfrac{43}{110}\)
(D) \(\dfrac{42}{110}, \dfrac{43}{110}, \dfrac{44}{110}\)
Three fractions that can be inserted between \(\dfrac{4}{11}\text{ and }\dfrac{2}{5}\) are …
(A) \(\dfrac{20}{55}, \dfrac{21}{55}, \dfrac{22}{55}\)
(B) \(\dfrac{21}{55}, \dfrac{22}{55}, \dfrac{21}{55}\)
(C) \(\dfrac{41}{110}, \dfrac{42}{110}, \dfrac{43}{110}\)
(D) \(\dfrac{42}{110}, \dfrac{43}{110}, \dfrac{44}{110}\)
Solution: C
Find the Least Common Multiple of the denominators
The LCM of 11 and 5 is 55
\begin{equation*} \begin{split} &\frac{4 \times \color {red} 5}{11 \times \color {red} 5} = \frac{20}{55}\\\\ &\frac{2 \times \color {red} 11}{5 \times \color {red} 11} = \frac{22}{55}\\\\ \end{split} \end{equation*}
\begin{array}{cccccc} \dfrac{20}{55}&\:\:\dotso\:\:&\:\:\dotso\:\:&\:\:\dotso\:\:&\dfrac{22}{55}\\\\ \dfrac{20\times \color {red} 2}{55\times\color {red} 2}&\:\:\dotso\:\:&\:\:\dotso\:\:&\:\:\dotso\:\:&\dfrac{22\times\color {red} 2 }{55 \times \color {red} 2 }\\\\ \dfrac{40}{110}&\:\:\dotso\:\:&\:\:\dotso\:\:&\:\:\dotso\:\:&\dfrac{44}{110}\\\\ \dfrac{40}{110}&\color {red} \dfrac{41}{110}&\color {red}\dfrac{42}{110}&\color {red}\dfrac{43}{110}&\color{black}\dfrac{44}{110}\\\\ \end{array}
\begin{equation*} \begin{split} &\text{Three fractions that can be inserted are:}\\\\ &\frac{41}{110}\:\:\frac{42}{110}\:\:\text{and}\:\:\frac{43}{110} \end{split} \end{equation*}
\(\left[\dfrac{1}{2} – \dfrac{2}{5}\left(\dfrac{7}{8} – \dfrac{1}{4}\right)-\dfrac{4}{5}\div \dfrac{1}{5}\right] = \dotso\)
(A) \(-3\dfrac{1}{4}\)
(B) \(-3\dfrac{3}{4}\)
(C) \(3\dfrac{1}{4}\)
(D) \(3\dfrac{3}{4}\)
\(\left[\dfrac{1}{2} – \dfrac{2}{5}\left(\dfrac{7}{8} – \dfrac{1}{4}\right)-\dfrac{4}{5}\div \dfrac{1}{5}\right] = \dotso\)
(A) \(-3\dfrac{1}{4}\)
(B) \(-3\dfrac{3}{4}\)
(C) \(3\dfrac{1}{4}\)
(D) \(3\dfrac{3}{4}\)
Solution: B
\begin{equation*} \begin{split} &\left[\frac{1}{2} – \frac{2}{5}\left(\frac{7}{8} – \frac{2}{8}\right) – \frac{4}{5} \div \frac{1}{5}\right]\\\\ &\left(\frac{1}{2} – \frac{2}{5}\times \frac{5}{8}- \frac{4}{\cancel {5}} \times \frac{\cancel {5}}{1}\right)\\\\ &\left(\frac{1}{2} – \frac{\cancelto {1}{2}}{\cancel {5}}\times \frac{\cancel {5}}{\cancelto{4}{8}} – 4\right)\\\\ &\left(\frac{1}{2} – \frac{1}{4} – 4\right)\\\\ &\left(\frac{2}{4} – \frac{1}{4} – 4\right)\\\\ &\left(\frac{1}{4} – \frac{16}{4}\right)\\\\ &-\frac{15}{4}\\\\ &-3\frac{3}{4} \end{split} \end{equation*}
\(4\dfrac{2}{5} \times \left(-\dfrac{5}{11}\right)\div \dfrac{1}{4} = \dotso\)
(A) −8
(B) −2
(C) 2
(D) 8
\(4\dfrac{2}{5} \times \left(-\dfrac{5}{11}\right)\div \dfrac{1}{4} = \dotso\)
(A) −8
(B) −2
(C) 2
(D) 8
Solution: A
\begin{equation*} \begin{split} & \left(\frac{20}{5}+\frac{2}{5}\right)\times \left(-\frac{5}{11}\right)\times \frac{4}{1}\\\\ & \frac{22}{5} \times \left(-\frac{5}{11}\right)\times 4\\\\ & \frac{\cancelto {2}{22}}{\cancel{5}} \times (-\frac{\cancel {5}}{\cancelto {1}{11}})\times 4\\\\ & -8 \end{split} \end{equation*}
\(\left(\dfrac{1}{2}\right)^5 \div \left(\dfrac{1}{2}\right)^3 \times \left(\dfrac{1}{2}\right)^2 = \dotso\)
(A) \(\dfrac{1}{16}\)
(B) \(\dfrac{1}{8}\)
(C) \(\dfrac{1}{4}\)
(D) \(\dfrac{1}{2}\)
\(\left(\dfrac{1}{2}\right)^5 \div \left(\dfrac{1}{2}\right)^3 \times \left(\dfrac{1}{2}\right)^2 = \dotso\)
(A) \(\dfrac{1}{16}\)
(B) \(\dfrac{1}{8}\)
(C) \(\dfrac{1}{4}\)
(D) \(\dfrac{1}{2}\)
Solution: A
\begin{equation*} \begin{split} & \frac{1^5}{2^5}\div \frac{1^5}{2^3}\times\frac{1^5}{2^2}\:\:\:\:\:\color {red}\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}\\\\ & \frac{1}{32}\div \frac{1}{8}\times \frac{1}{4}\\\\ & \frac{1}{\cancelto {4}{32}}\times \frac{\cancelto {1}{8}}{1}\times \frac{1}{4}\\\\ & \frac{1}{4}\times \frac{1}{4}\\\\ & \frac{1}{16} \end{split} \end{equation*}
\(2\dfrac{1}{2} \times 0.25 + 1\dfrac{1}{8} \div \dfrac{3}{4} = \dotso\)
(A) \(\dfrac{4}{5}\)
(B) \(1\dfrac{5}{16}\)
(C) \(1\dfrac{7}{8}\)
(D) \(2\dfrac{1}{8}\)
\(2\dfrac{1}{2} \times 0.25 + 1\dfrac{1}{8} \div \dfrac{3}{4} = \dotso\)
(A) \(\dfrac{4}{5}\)
(B) \(1\dfrac{5}{16}\)
(C) \(1\dfrac{7}{8}\)
(D) \(2\dfrac{1}{8}\)
Solution: D
\begin{equation*} \begin{split} & \left(\frac{4}{2} + \frac{1}{2}\right)\times \frac{25}{100} + \left(\frac{8}{8} + \frac{1}{8}\right) \div \frac{3}{4}\\\\ & \frac{5}{2}\times \frac{1}{4} + \frac{9}{8}\div \frac{3}{4}\\\\ & \frac{5}{8} + \frac{\cancelto {3}{9}}{\cancelto{2}{8}}\times \frac{\cancelto{1}{4}}{\cancelto {1}{3}}\\\\ & \frac{5}{8} + \frac{3}{2}\\\\ & \frac{5}{8} + \frac{12}{8}\\\\ & \frac{17}{8}\\\\ & \frac{16}{8} + \frac{1}{8}\\\\ & 2\frac{1}{8} \end{split} \end{equation*}
\(1 + \dfrac{2}{1 + \dfrac{1}{1 + \dfrac{2}{1+\dfrac{1}{2}}}}=\dotso\)
(A) \(2\dfrac{1}{5}\)
(B) \(2\dfrac{2}{5}\)
(C) \(2\dfrac{3}{5}\)
(D) \(2\dfrac{4}{5}\)
\(1 + \dfrac{2}{1 + \dfrac{1}{1 + \dfrac{2}{1+\dfrac{1}{2}}}}=\dotso\)
(A) \(2\dfrac{1}{5}\)
(B) \(2\dfrac{2}{5}\)
(C) \(2\dfrac{3}{5}\)
(D) \(2\dfrac{4}{5}\)
Solution: B
\begin{equation*} \begin{split} & 1 + \frac{2}{1+\frac{1}{1+\frac{2}{1+\frac{1}{2}}}}\\\\ & 1 + \frac{2}{1+\frac{1}{1+\frac{2}{\frac{3}{2}}}}\\\\ & 1 + \frac{2}{1+\frac{1}{1 + 2\times{\frac{2}{3}}}}\\\\ & 1 + \frac{2}{1+\frac{1}{1 + \frac{4}{3}}}\\\\ & 1 + \frac{2}{1+\frac{1}{\frac{7}{3}}}\\\\ & 1 + \frac{2}{1 + \frac{3}{7}}\\\\ & 1 + \frac{2}{\frac{10}{7}}\\\\ & 1 + 2\times\frac{7}{10}\\\\ & 1 + \frac{14}{10}\\\\ & 1 + \frac{10}{10} + \frac{4}{10}\\\\ & 1 + 1 + \frac{4}{10}\\\\ & 2 + \frac{2}{5}\\\\ & 2 \frac{2}{5} \end{split} \end{equation*}
Anita’s recipe for cookies calls for \(2\dfrac{3}{4}\) of a cup of sugar. If Anita needs to make half the amount of cookies, how many cups of sugar she need?
(A) \(1\dfrac{1}{8}\) cups of sugar
(B) \(1\dfrac{3}{8}\) cups of sugar
(C) \(1\dfrac{1}{2}\) cups of sugar
(D) \(1\dfrac{5}{8}\) cups of sugar
Anita’s recipe for cookies calls for \(2\dfrac{3}{4}\) of a cup of sugar. If Anita needs to make half the amount of cookies, how many cups of sugar she need?
(A) \(1\dfrac{1}{8}\) cups of sugar
(B) \(1\dfrac{3}{8}\) cups of sugar
(C) \(1\dfrac{1}{2}\) cups of sugar
(D) \(1\dfrac{5}{8}\) cups of sugar
Solution: B
\begin{equation*} \begin{split} & \frac{1}{2}\times 2\frac{3}{4}\\\\ & \frac{1}{2}\times \left(\frac{8}{4} + \frac{3}{4}\right)\\\\ & \frac{1}{2}\times \frac{11}{4}\\\\ & \frac{11}{8}\\\\ & \frac{8}{8} + \frac{3}{8}\\\\ & 1 + \frac{3}{8}\\\\ & 1\frac{3}{8} \end{split} \end{equation*}
Mr.Paul uses \(\dfrac{2}{5}\) of his land for growing apples, \(\dfrac{1}{4}\) for pineapple and \(\dfrac{1}{3}\) for guavas and the remaining 10 ha for mangoes. What the total area of his land?
(A) 600 ha
(B) 700 ha
(C) 800 ha
(D) 900 ha
Mr.Paul uses \(\dfrac{2}{5}\) of his land for growing apples, \(\dfrac{1}{4}\) for pineapple and \(\dfrac{1}{3}\) for guavas and the remaining 10 ha for mangoes. What the total area of his land?
(A) 600 ha
(B) 700 ha
(C) 800 ha
(D) 900 ha
Solution: A
\begin{equation*} \begin{split} \text{The land used for growing mangoes is}& = 1 – \frac{2}{5} – \frac{1}{4} – \frac{1}{3}\\\\ \text{The land used for growing mangoes is}& = \frac{60}{60} – \frac{24}{60} – \frac{15}{60} – \frac{20}{60}\\\\ \text{The land used for growing mangoes is}&= \frac{1}{60}\:\text{parts} = 10\:\text{ ha}\\\\\\ \text{1 parts (total of land)}& = \frac{60}{1}\times 10 \:\text{ ha}\\\\ \text{1 parts (total of land)}& = 600\:\text{ ha} \end{split} \end{equation*}
A cylinder was \(\dfrac{3}{4}\) full of water when it contained 276 liters. How many liters would it contains when \(\dfrac{1}{4}\) full?
(A) 92 liters
(B) 184 liters
(C) 368 liters
(D) 446 liters
A cylinder was \(\dfrac{3}{4}\) full of water when it contained 276 liters. How many liters would it contains when \(\dfrac{1}{4}\) full?
(A) 92 liters
(B) 184 liters
(C) 368 liters
(D) 446 liters
Solution: A
\begin{equation*} \begin{split} &\frac{3}{4}\:\text{part of water} = 276 \:\text{liters}\\\\\\ &\text{1 parts of water (full water)} = \frac{4}{\cancelto{1}{3}}\times \cancelto{92}{276} \:\text{liters}\\\\ &\text{1 parts of water (full water)} = 368\:\text{liters}\\\\\\ &\frac{1}{4}\:\text{part of water} = \frac{1}{4}\times 368\:\text{liters}\\\\ &\frac{1}{4}\:\text{part of water} = 368 \div 4 \:\text{liters}\\\\ &\frac{1}{4}\:\text{part of water} = 92 \:\text{liters} \end{split} \end{equation*}
\(\dfrac{5}{8}\) of students in a class 7B are the girls. If the number of boys in 7B is 15, then the total number of students in a class 7B is …
(A) 10
(B) 20
(C) 30
(D) 40
\(\dfrac{5}{8}\) of students in a class 7B are the girls. If the number of boys in 7B is 15, then the total number of students in a class 7B is …
(A) 10
(B) 20
(C) 30
(D) 40
Solution: D
\begin{equation*} \begin{split} \text{The boys in a class}& = 1 – \frac{5}{8}\:\text {parts}\\\\ \text{The boys in a class}&= \frac{8}{8} – \frac{5}{8}\:\text {parts}\\\\ \text{The boys in a class}& = \frac{3}{8} \:\text {parts} = 15 \:\text{students}\\\\\\ \text{1 parts (total students)}& = \frac{8}{\cancelto {1}{3}}\times \cancelto {5}{15} \:\text{students}\\\\ \text{1 parts (total students)}& = 40\:\text{students} \end{split} \end{equation*}
Shane got salary Rp4.500.000,00 per month, \(\dfrac{1}{6}\) of the salary used to pay the rent, 40% part his food and the rest for other purposes. How many rupiahs that are used for the other purposes?
(A) Rp1.502.500,00
(B) Rp1.750.000,00
(C) Rp1.850.000,00
(D) Rp1.950.000,00
Shane got salary Rp4.500.000,00 per month, \(\dfrac{1}{6}\) of the salary used to pay the rent, 40% part his food and the rest for other purposes. How many rupiahs that are used for the other purposes?
(A) Rp1.502.500,00
(B) Rp1.750.000,00
(C) Rp1.850.000,00
(D) Rp1.950.000,00
Solution: D
\begin{equation*} \begin{split} \text{Rent}& = \frac{1}{6}\:\text{parts}\\\\ \text{Food}& = 40\%\\\\ \text{Food}& = \frac{40}{100}\\\\ \text{Food}& = \frac{40 \div \color {red} 20 }{100 \div \color {red} 20}\\\\ \text{Food}& = \frac{2}{5}\:\text{parts}\\\\\\ \text{For other purposes}& = 1 – \frac{1}{6} – \frac{2}{5}\:\text{parts}\\\\ \text{For other purposes}& = \frac{30}{30} – \frac{5}{30} – \frac{12}{30}\:\text{parts}\\\\ \text{For other purposes}& = \frac{13}{30}\:\text{parts}\\\\\\ \text{For other purposes}& = \frac{13}{30}\times \text{Rp4.500.000,00}\\\\ \text{For other purposes}& = \text{Rp1.950.000,00} \end{split} \end{equation*}
Thrice a number is 84 more than \(\dfrac{2}{3}\)of the same number. What is the number?
(A) 32
(B) 34
(C) 35
(D) 36
Thrice a number is 84 more than \(\dfrac{2}{3}\)of the same number. What is the number?
(A) 32
(B) 34
(C) 35
(D) 36
Solution: D
\(\text{Let the unknown number be x}\)
\begin{equation*} \begin{split} 3x & = 84 + \frac{2}{3}x\\\\ 3x – \color{red} \frac{2}{3}x\color{black} & = 84 + \frac{2}{3}x – \color{red}\frac{2}{3}x\:\:\:\:\:\color{blue}\text{subtract both sides by} \:\frac{2}{3}x\\\\ \frac{9}{3}x – \frac{2}{3}x & = 84\\\\ \frac{7}{3}x & = 84\:\:\:\:\color{blue}\text{ divide both sides by}\:\frac{7}{3}\\\\ \frac{\cancel {\frac{7}{3}}}{\cancel {\frac{7}{3}}}x & = \frac{84}{\frac{7}{3}}\\\\ x &= \cancelto {12}{84} \times \frac{3}{\cancelto{1}{7}}\\\\ x &= 36 \end{split} \end{equation*}
A container is \(\dfrac{1}{8}\) filled with orange juice. When 200 ml of orange juice is added into the container, it becomes \(\dfrac{3}{4}\) full. The capacity of the container is …
(A) 320 ml
(B) 354 ml
(C) 450 ml
(D) 500 ml
A container is \(\dfrac{1}{8}\) filled with orange juice. When 200 ml of orange juice is added into the container, it becomes \(\dfrac{3}{4}\) full. The capacity of the container is …
(A) 320 ml
(B) 354 ml
(C) 450 ml
(D) 500 ml
Solution: A
\begin{equation*} \begin{split} \text{The fraction of orange juice added into the container}& = \frac{3}{4} – \frac{1}{8}\\\\ \text{The fraction of orange juice added into the container}& = \frac{6}{8} – \frac{1}{8}\\\\ \text{The fraction of orange juice added into the container}& = \frac{5}{8}\:\text{volume of the container} = 200 \:\text{ml}\\\\\\ \text{So, the capacity of the container}& = \frac{200}{\frac{5}{8}}\\\\ \text{So, the capacity of the container}& =\cancelto {40}{200}\times \frac{8}{\cancelto {1}{5}}\\\\ \text{So, the capacity of the container}& = 320\:\text{ml} \end{split} \end{equation*}
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