edited by
528 views
0 votes
0 votes

$A$ can do a piece of work in $14$ days which $B$ can do in $21$ days. They begin together but $3$ days before the completion of the work, $A$ leaves off. The total number of days to complete the work is :

  1. $\frac{33}{5}$
  2. $8\frac{1}{2}$
  3. $\frac{51}{5}$
  4. $13\frac{1}{2}$
edited by

1 Answer

Best answer
1 votes
1 votes

Answer is C

x days (A and B work together) + 3 days (B alone worked) = 1

$x*(\frac{1}{14}+\frac{1}{21})+ 3*\frac{1}{21}=1$

On solving, we will get $5x = 36 \implies x =\frac{36}{5}$

Therefore, total number of days to complete the work = $\frac{36}{5}$ + 3 = $\frac{51}{5}$

selected by
Answer:

Related questions

0 votes
0 votes
1 answer
3
Lakshman Bhaiya asked Apr 3, 2020
610 views
The $L.C.M$ of $\left (x^{3}-x^{2}-2x \right)$ and $\left (x^{3}+x^{2} \right)$ is : $\left (x^{3}-x^{2}-2x \right)$$\left (x^{2}+x \right)$$\left (x^{4}-x^{3}-2x^{2} \ri...
0 votes
0 votes
1 answer
4
1 votes
1 votes
1 answer
5
Lakshman Bhaiya asked Apr 3, 2020
633 views
If $t^{2}-4t+1=0$, then the value of $\left[t^{3}+1/t^{3} \right]$ is :$44$$48$$52$$64$