
Two kinds of tea were prepared. In the first, 30 grams of sugar was mixed with 190 grams of tea. In the second, 40 grams of sugar was mixed with 270 grams of tea. If the two kinds of tea are mixed together, determine the percentage of tea in the mixture.
Answer
580.2k+ views
Hint: Assume the mass of tea as \[{{m}_{1}}\] and \[{{m}_{2}}\] and that of sugar as \[{{m}_{3}}\] and \[{{m}_{4}}\]. Use the relation: - percentage of tea in the mixture = \[\dfrac{{{m}_{1}}+{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\times 100\%\], substitute the given values of \[{{m}_{1}},{{m}_{2}},{{m}_{3}}\] and \[{{m}_{4}}\] from the given question to get the required answer.
Complete step by step answer:
Let us assume that in the two kinds of tea, first and second kind contains \[{{m}_{1}}\] and \[{{m}_{2}}\] mass of tea respectively and \[{{m}_{3}}\] and \[{{m}_{4}}\] mass of sugar respectively. Therefore, according to question: -
(i) For first kind: -
Mass of tea = \[{{m}_{1}}\] = 190 grams
Mass of sugar = \[{{m}_{3}}\] = 30 grams
(ii) For second kind: -
Mass of tea = \[{{m}_{2}}\] = 270 grams
Mass of sugar = \[{{m}_{4}}\] = 40 grams
Now, we know that the percentage of an object in a given substance is the ratio of mass of that object to the total mass of substance, multiplied by 100.
\[\Rightarrow \] % of object = (mass of object/ total mass of substance) \[\times \] 100%
\[\Rightarrow \] % of tea in the mixture
= (mass of tea in the mixture / total mass of mixture) \[\times \] 100%
=\[\dfrac{{{m}_{1}}+{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\times 100\%\]
Substituting the values of \[{{m}_{1}},{{m}_{2}},{{m}_{3}}\] and \[{{m}_{4}}\], we get,
% of tea in the mixture \[=\dfrac{190+270}{190+270+30+40}\times 100\%\]
\[\begin{align}
& =\dfrac{460}{460+70}\times 100\% \\
& =\dfrac{460}{530}\times 100\% \\
& =86.79\% \\
\end{align}\]
Note: One may note that, you can apply another way to get the answer. We can find the percentage of sugar in the mixture by using the relation: - \[\dfrac{{{m}_{3}}+{{m}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\times 100\] and then subtract this from 100 to get the percentage of tea in the mixture. This approach can be useful when we have to find the percentage of both components, i.e. tea and sugar.
Complete step by step answer:
Let us assume that in the two kinds of tea, first and second kind contains \[{{m}_{1}}\] and \[{{m}_{2}}\] mass of tea respectively and \[{{m}_{3}}\] and \[{{m}_{4}}\] mass of sugar respectively. Therefore, according to question: -
(i) For first kind: -
Mass of tea = \[{{m}_{1}}\] = 190 grams
Mass of sugar = \[{{m}_{3}}\] = 30 grams
(ii) For second kind: -
Mass of tea = \[{{m}_{2}}\] = 270 grams
Mass of sugar = \[{{m}_{4}}\] = 40 grams
Now, we know that the percentage of an object in a given substance is the ratio of mass of that object to the total mass of substance, multiplied by 100.
\[\Rightarrow \] % of object = (mass of object/ total mass of substance) \[\times \] 100%
\[\Rightarrow \] % of tea in the mixture
= (mass of tea in the mixture / total mass of mixture) \[\times \] 100%
=\[\dfrac{{{m}_{1}}+{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\times 100\%\]
Substituting the values of \[{{m}_{1}},{{m}_{2}},{{m}_{3}}\] and \[{{m}_{4}}\], we get,
% of tea in the mixture \[=\dfrac{190+270}{190+270+30+40}\times 100\%\]
\[\begin{align}
& =\dfrac{460}{460+70}\times 100\% \\
& =\dfrac{460}{530}\times 100\% \\
& =86.79\% \\
\end{align}\]
Note: One may note that, you can apply another way to get the answer. We can find the percentage of sugar in the mixture by using the relation: - \[\dfrac{{{m}_{3}}+{{m}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\times 100\] and then subtract this from 100 to get the percentage of tea in the mixture. This approach can be useful when we have to find the percentage of both components, i.e. tea and sugar.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

