
Five years ago, Anjali’s age was twice Sarita’s age. Five years hence, their ages will be \[4:3\]. Anjali’s present age is:
A. \[15{\text{ years}}\]
B. \[10{\text{ years}}\]
C. \[12{\text{ years}}\]
D. \[16{\text{ years}}\]
Answer
457.2k+ views
Hint: Here we will solve this question by assuming the ages of a person by applying a rule that if a person’s present age is \[x\] then after \[n\] the number of years, that person’s age will be \[x + n\]. Also, before \[n\] the number of years the age will be \[x - n\].
Complete step-by-step answer:
Step 1: Assume that Sarita’s age is \[a\] and Anjali’s age is \[b\]. Since, five years ago Anjali’s age was twice of Sarita’s so, their age will be equals to as below:
\[b - 5 = 2\left( {a - 5} \right)\]
By solving the brackets in the RHS side of the above equation we get:
\[ \Rightarrow b - 5 = 2a - 10\]
By bringing \[5\] into the RHS side of the above equation, we get:
\[ \Rightarrow b = 2a - 5\]………………….. (1)
Step 2: After five years the age of Anjali and Sarita will be \[b + 5\] and \[a + 5\] respectively. But the ratio of ages after five years is \[4:3\] so, we get:
\[ \Rightarrow \dfrac{{b + 5}}{{a + 5}} = \dfrac{4}{3}\]
After bringing \[3\] into the LHS side of the equation and \[4\]in the RHS side we get:
\[ \Rightarrow 3\left( {b + 5} \right) = 4\left( {a + 5} \right)\]
By opening the brackets on both of the sides of the above equation we get:
\[ \Rightarrow 3b + 15 = 4a + 20\]
By substituting the value of \[b\] from the equation (1) we get:
\[ \Rightarrow 3\left( {2a - 5} \right) + 15 = 4a + 20\]
By solving the brackets in the LHS side of the above equation we get:
\[ \Rightarrow 6a - 15 + 15 = 4a + 20\]
By doing simple addition and subtraction on both the sides of the above equation:
\[ \Rightarrow 6a = 4a + 20\]
Bringing \[4a\] into the LHS side of the above equation we get:
\[ \Rightarrow 2a = 20\]
Taking \[2\] into the RHS side of the above equation and dividing it with \[20\], we get:
\[ \Rightarrow a = 10{\text{ years}}\] ………………………….. (2)
Step 3: By substituting the value of
\[a = 10\] from equation (2) to equation (1) we get:
\[ \Rightarrow b = 2\left( {10} \right) - 5\]
By solving the terms in the RHS side of the above expression we get:
\[ \Rightarrow b = 15{\text{ years}}\]
Option A is correct.
Note:
Students needs to remember some important formulas for solving these types of questions:
If you are assuming the present age of a person as \[x\] then his age after \[n\]years will be \[x + n\] years. If you are assuming the present age of a person as \[x\] then his age before \[n\] years will be \[x - n\] years. If you are assuming the present age of a person \[x\], then \[n\] times of present age will be \[nx\] years. If you are assuming the present age of a person \[x\], then \[\dfrac{1}{n}\] his present age will be \[\dfrac{x}{n}\] years.
Complete step-by-step answer:
Step 1: Assume that Sarita’s age is \[a\] and Anjali’s age is \[b\]. Since, five years ago Anjali’s age was twice of Sarita’s so, their age will be equals to as below:
\[b - 5 = 2\left( {a - 5} \right)\]
By solving the brackets in the RHS side of the above equation we get:
\[ \Rightarrow b - 5 = 2a - 10\]
By bringing \[5\] into the RHS side of the above equation, we get:
\[ \Rightarrow b = 2a - 5\]………………….. (1)
Step 2: After five years the age of Anjali and Sarita will be \[b + 5\] and \[a + 5\] respectively. But the ratio of ages after five years is \[4:3\] so, we get:
\[ \Rightarrow \dfrac{{b + 5}}{{a + 5}} = \dfrac{4}{3}\]
After bringing \[3\] into the LHS side of the equation and \[4\]in the RHS side we get:
\[ \Rightarrow 3\left( {b + 5} \right) = 4\left( {a + 5} \right)\]
By opening the brackets on both of the sides of the above equation we get:
\[ \Rightarrow 3b + 15 = 4a + 20\]
By substituting the value of \[b\] from the equation (1) we get:
\[ \Rightarrow 3\left( {2a - 5} \right) + 15 = 4a + 20\]
By solving the brackets in the LHS side of the above equation we get:
\[ \Rightarrow 6a - 15 + 15 = 4a + 20\]
By doing simple addition and subtraction on both the sides of the above equation:
\[ \Rightarrow 6a = 4a + 20\]
Bringing \[4a\] into the LHS side of the above equation we get:
\[ \Rightarrow 2a = 20\]
Taking \[2\] into the RHS side of the above equation and dividing it with \[20\], we get:
\[ \Rightarrow a = 10{\text{ years}}\] ………………………….. (2)
Step 3: By substituting the value of
\[a = 10\] from equation (2) to equation (1) we get:
\[ \Rightarrow b = 2\left( {10} \right) - 5\]
By solving the terms in the RHS side of the above expression we get:
\[ \Rightarrow b = 15{\text{ years}}\]
Option A is correct.
Note:
Students needs to remember some important formulas for solving these types of questions:
If you are assuming the present age of a person as \[x\] then his age after \[n\]years will be \[x + n\] years. If you are assuming the present age of a person as \[x\] then his age before \[n\] years will be \[x - n\] years. If you are assuming the present age of a person \[x\], then \[n\] times of present age will be \[nx\] years. If you are assuming the present age of a person \[x\], then \[\dfrac{1}{n}\] his present age will be \[\dfrac{x}{n}\] years.
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