Question

Read the given data carefully and answer the following questions.
On a particular day, 792 people went to gym by cycle, bike and car in three different shifts. Ratio of number of people who went by cycle, bike and car is 34:55:43 respectively. Morning Shift: Total number of people who went by cycle is 30 less than number of people who went by bike which is 30 less than number of people who went by car. Afternoon Shift : Ratio of number of people who went by cycle and bike is 8:11 respectively and number of people who went by cycle is 33.33% more than number of people who went by car. Evening Shift: Ratio of number of people who went by cycle and car is 6:7 respectively and total number of people who went by bike is 12 more than twice the number of people who went by cycle.

find the ratio of number of people who went by bike in evening and in morning respectively?​

Answer 1

Answer:

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Answer 2

Given the data about people going to gym in three shifts, find ratio of number of people who went by bike in evening and in morning

Explanation:  

1. given the total number of people on a particular day, $T=792$      
2. let the total number of people out these going by cycle, bike and car be 'y' ,'b' and 'c' respectively and their ratio $34:55:43$. Then we have,    [tex]y=34n,\ \ \ b=55n,\ \ \ c=43n\\ ->34n+55n+43n=792 \\ ->n=6\\ ->y=204,\ \ \ b=330,\ \ \ c=258[/tex]
3. MORNING: let the people going by cycle, bike, and car be denoted by $y_1, b_1\ and \ c_1$ respectively. Then we have,                                                              [tex]y_1=b_1-30,\ \ \ \ \ b_1=c_1-30\\ ->c_1-y_1=60[/tex]     ------(a)
4. AFTERNOON:  let the people going by cycle, bike, and car be denoted by $y_2, b_2\ and \ c_2$ respectively. Then we have,                                  [tex]\frac{y_2}{b_2}=\frac{8}{11},\ \ \ \ \ y_2=\frac{2c_2}{3} \\ ->c_2-y_2=-\frac{4b_2}{11}[/tex]                       -------(b)
5. EVENING: let the people going by cycle, bike, and car be denoted by $y_3, b_3\ and \ c_3$ respectively. Then we have,                                                               $\frac{y_3}{c_3}=\frac{6}{7},\ \ \ \ \ b_3=2y_3+12 \\->c_3-y_3=\frac{b_3-12}{12}$   -------(c)
6. now we have ,                                                                                                   $y_1+y_2+y_3=204 -(i)\ \ \ \ b_1+b_2+b_3=330 -(ii)\ \ \ \ c_1+c_2+c_3=258 -(iii)$
7. on doing (iii)-(i) we get,  $(c_1-y_1)+(c_2-y_2)+(c_3-y_3)=54$        
8. from (a)(b) and (c) we get,                                                                                           [tex]60-\frac{4b_2}{11} +\frac{b_3-12}{12}= 54 \\ 792-48b_2+b_3-132=0\\ 660+b_3=48b_2[/tex]  
9. from (ii) we get,     $660+b_3=48(330-b_1-b_3)$                                                      $->\frac{b_3}{b_1}=\frac{7}{11}$   ------(ANSWER)