father is 5 times faster than son. father completes work in 40days before son.if both of them work together when will work get complete?
Answers
Suppose time taken by father to complete the work = x days
Because father is 5 times faster than son so son would be taking 5 times of the time than his father
Time taken by son to complete the work = 5x days
According to question Father completes his work 40 days before the time his son completes the work
So times taken by Son = time taken by father and 40 more days
====> 5x = x + 40
- =======> 5x - x = 40
========> 4x = 40
==========> x= 40/4 = 10 days
So times taken by father to complete the work = 10 days
and times taken by son to complete the work = 5x = 50 days
One days work of father = 1/ times taken to complete the work = 1/10
One day work of son = 1/ time taken to complete the work = 1/50
one day work when ( father and son ) both work = ( 1/10 ) + ( 1/50 ) = ( 6/50)
times taken to complete the work when both work together = 50/6 days
Answer = 50/6 days .
Answer:
Step-by-step explanation:
GIVEN THAT
FATHER IS 5 TIMES FASTER THAN HIS SON.
IF SON'S SPEED IS ASSUMED AS "S", THEN FATHER'S SPEED IS "5S".
AND ALSO GIVEN THAT FATHER COMPLETES THE WORK IN 40 DAYS BEFORE SON.
i.e., IF SON HAS COMPLETED THE WHOLE WORK IN 'X' DAYS, THEN THE FATHER HAS COMPLETED IN (X-40)DAYS.
SX=5S*(X-40);
SX=5SX-200S;
i.e., X=5X-200;
=>5X-X=200
=> 4X=200
=>X=50 [i.e.,(200/4)];
i.e., TIME TAKEN FOR THE SON TO COMPLETE HIS WORK IS 50 DAYS AND HIS FATHER IS (X-40)=(50-40)=10 DAYS.
L.C.M OF 10 AND 50 IS 50, WHICH IS THE TOTAL WORK DIVIDED INTO 50 PARTS.
i.e., SON CAN COMPLETE 1 PART OF WORK IN A DAY AND FATHER COMPLETES 5 PARTS OF WORK IN A DAY(AS CONDITION SATISFIED)
NOW, IF BOTH TOGETHER WORKS, THEY CAN COMPLETE 6 PARTS OF WORK IN A DAY i.e., TOTAL NO. OF DAYS REQUIRED = 50/6 = (25/3)==>8(1/3) DAYS.