A does 7/10 of a piece of work in 14 days; he then calls in B, and they finish the work in 2 days. How long would B take to do the work alone ?
Answers
Answer:
Radius, r = 4, and center (h, k) = (-2, 3).
We know that the equation of a circle with centre (h, k) and radius r is given as
→ (x – h)² + (y – k)² = r² \: \: \: ….(1)→(x–h)²+(y–k)²=r²….(1)
Now, substitute the radius and center values in (1), we get
Therefore, the equation of the circle is
→ (x + 2)²+ (y – 3)² = (4)²→(x+2)²+(y–3)²=(4)²
→ x²+ 4x + 4 + y² – 6y + 9 = 16→x²+4x+4+y²–6y+9=16
Now, simplify the above equation, we get:
→ x² + y²+ 4x – 6y – 3 = 0→x²+y²+4x–6y–3=0
Thus, the equation of a circle with center (-2, 3) and radius 4 is :
→ x² + y²+ 4x – 6y – 3 = 0→x²+y²+4x–6y–3=0
Answer:
A completes 7/10 of a work in 14 days, means A's daily performance is...
7/10 divided by 14 = 1/20 part of the work or A can complete the work in 20 days.
After 14 days the unfinished work left is...
1 - 7/10 = 3/10 part and A and B jointly completed this part in 2 days.
A had contributed 1/20 x 2 = 2/20 or 1/10 part and B's contribution is... The rest i.e..
3/10 - 1/10 =2/10 in 2 days or B's daily contribution is 1/10 part of the work. Therefore B Alone could have completed the work in...
10 days. Answer.
Note : For A the reamaining 3/10 part of the work would have taken 3/10 by 1/20 = 6 days more. Whereas by the help of B this part was completed in 2 days and rate of performance of B is obviously DOUBLE of A. Hence B could do the work within HALF of Timeframe for A, i.e 20/2=10 answer.