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Numerical Ability
Arithmetic
Three generous friends, each with some money, redistribute the money as follows: Sandra gives enough money to David and Mary to double the amount of money each has. David then gives enough to Sandra and Mary to double their amounts. Finally, Mary gives enough to Sandra and David to double their amounts. If Mary had 11 rupees at the beginning and 17 rupees at the end, what is the total amount that all three friends have?
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- answer will be 71. it is because mary at start had 11 rupees which got doubled 22 again doubled to 44 and now she had rupees 17 so she gave 44-17=27 rupees to friends to double the rupees so friend already had 27 rupees so total rupees 27+27+17=71
- 9 years agoHelpfull: Yes(37) No(9)
- 71 is answer.
let sandra, david and mary each has s, d and 11(given) respectively.
After first distribution
David has d+d=2d, marry has 11+11=22 and sandra has s-d-11.
After second distribution,
sandra has 2*(s-d-11) , mary has 2*22=44 and david has 2d-(s-d-11)-22=3d-s-11.
After third distribution,
sandra has 2*2(s-d-11), david has 2*(3d-s-11) and mary has 44-2(s-d-11)-(3d-s-11)=77-s-d
It is given that finally Mary has 17 rs. So,
77-s-d=17
=>s+d=60
=> s+d+11(Mary's money)=60+11=71. - 9 years agoHelpfull: Yes(35) No(0)
- In the starting Mary has 11 r . First sandra gives money to mary her money becomes double 22 r. then david gives her money then again her money double 44r. At last
she gives mones and left with 17 rs.
44-17=27rs (this money is of sandra and david before last step)
and mary has 44 rs before last step.
so total money they have 44+27=71rs - 9 years agoHelpfull: Yes(7) No(3)
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