Question
Ajay goes to his office from his house at a speed of 18
km/hr and returns to his home from his office at a speed of 36 km/hr and he takes 8 hours in all. If the distance of his friend's house from his office is 15% more than the distance of his house from his office, find the distance of his house to his friend's house. (Assuming the office lies between Ajay's house and his friend's house)Solution
Let the distance of his house to his office = D km T = D/S 8 = D/18 + D/36 8 = (2D + D)/36 8┬а┬а= 3D/36 D = 96 The distance from his house to his office = 96 km The distance from his office to his friend's house = 96┬а km ├Ч 115% = 110.4 km The distance from his house to his friend's house = 96 + 110.4 = 206.4 km
" рдЕрдХрд╛рд░рдг рд╢рдмреНрдж рдХрд╛ рд╡рд┐рд▓реЛрдорд╛рд░реНрдереА рд╢рдмреНрдж рд╣реИ :
Addressee рдХреЗ рд▓рд┐рдП рд╕рд╣реА рдкрд╛рд░рд┐рднрд╛рд╖рд┐рдХ рд╢рдмреНрдж рд╣реИ ?
рдХреНрд╖рдгрд┐рдХ ' рдХрд╛ рд╡рд┐рд▓реЛрдо рд╢рдмреНрдж рдХреНрдпрд╛ рд╣реЛрдЧрд╛ ?
тАШрдХрдерд╛' рдХрд╛ рдмрд╣реБрд╡рдЪрди рдХреНрдпрд╛ рд╣реЛрдЧрд╛?
рдХрд┐рд╕ рдХреНрд░рдорд╛рдВрдХ рдореЗрдВ тАШрдкрд░рд┐рдорд╛рдг тАУ рдкрд░рд┐рдгрд╛рдотАЩ рд╢рдмреНрджрд░ рдпреБрдЧреНрдорд╛ рдХрд╛ рд╕рд╣реА рдЕрд░реНя┐╜...
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'рдкреАрддрд╛рдВрдмрд░' рдореЗрдВ рдХреМрди рд╕рд╛ рд╕рдорд╛рд╕ рд╣реИ?
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тАШрдЖрдХрд╛рдВрдХреНрд╖рд╛тАЩ рдХрд╛ рдкрд░реНрдпрд╛рдпрд╡рд╛рдЪреА рдХреМрди-рд╕рд╛ рд╣реИ ?┬а