Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 November 13th, 2018, 12:24 AM #1 Senior Member   Joined: Aug 2014 From: India Posts: 486 Thanks: 1 How to solve time and work problem? Ajay and Sunil together can complete a piece of work in 10 days, Sunil and Sanjay in 15 days and Sanjay and Ajay in 20 days. They worked together for 6 days, and then Ajay leaves. Sunil and Sanjay worked together for 4 more days, and Sunil leaves. How long will Sanjay take to complete the work? A) Ajay and Sunil complete the work in 10 days; Aj + Su = 1/10 Sunil and Sanjay complete the work in 15 days; Su + Sa = 1/15 Sanjay and Ajay complete the work in 20 days; San + Aj = 1/25 Ajay, Sanjay and Sunil one day work if they worked together; 2Su + 2Aj + 2San = 1/10 + 1/15 + 1/25 Su + Aj + San = 1/2[1/10 + 1/15 + 1/25] Su + Aj + San = 1/2[13/60] Su + Aj + San = 13/120 Sunil, Sanjay and Ajay worked together for 6 days = 13/120*6 = 13/20 Su + Aj + San = 13/20 Sunil and Sanjay worked for 4 days after Ajay left = 1/15*4 = 4/15 Remaining work left for Sanjay alone = 1 – [13/20 + 4/15] → 1 - (39+16)/60 → 1 - 55/60 = 5/60 = 1/12 I got answer 12 days but right answer is 10. Where I am doing wrong? November 13th, 2018, 12:44 AM #2 Senior Member   Joined: Apr 2014 From: UK Posts: 967 Thanks: 344 Your 3rd equation looks wrong: "Sanjay and Ajay complete the work in 20 days; San + Aj = 1/25" Thanks from romsek November 13th, 2018, 01:00 AM #3 Senior Member   Joined: Sep 2015 From: USA Posts: 2,629 Thanks: 1469 You made a typo on the 3rd line. You typed 25 instead of 20 and this error trickled down through the calculations. I suspect if you correct this and redo your working you'll get 10 days. November 13th, 2018, 06:37 AM   #4
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Quote:
 Originally Posted by Ganesh Ujwal Ajay and Sunil together can complete a piece of work in 10 days, Sunil and Sanjay in 15 days and Sanjay and Ajay in 20 days.
...and use less confusing variables; example:

a = Ajay, b = Sunil, c = Sanjay
a and b : 10
b and c : 15
a and c : 20 November 13th, 2018, 12:18 PM   #5
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Quote:
 Originally Posted by Ganesh Ujwal Ajay and Sunil together can complete a piece of work in 10 days, Sunil and Sanjay in 15 days and Sanjay and Ajay in 20 days. They worked together for 6 days, and then Ajay leaves. Sunil and Sanjay worked together for 4 more days, and Sunil leaves. How long will Sanjay take to complete the work? A) Ajay and Sunil complete the work in 10 days; Aj + Su = 1/10 Sunil and Sanjay complete the work in 15 days; Su + Sa = 1/15 Sanjay and Ajay complete the work in 20 days; San + Aj = 1/25 Ajay, Sanjay and Sunil one day work if they worked together; 2Su + 2Aj + 2San = 1/10 + 1/15 + 1/25 Su + Aj + San = 1/2[1/10 + 1/15 + 1/25] Su + Aj + San = 1/2[13/60] Su + Aj + San = 13/120 Sunil, Sanjay and Ajay worked together for 6 days = 13/120*6 = 13/20 Su + Aj + San = 13/20 Sunil and Sanjay worked for 4 days after Ajay left = 1/15*4 = 4/15 Remaining work left for Sanjay alone = 1 – [13/20 + 4/15] → 1 - (39+16)/60 → 1 - 55/60 = 5/60 = 1/12 I got answer 12 days but right answer is 10. Where I am doing wrong?
You will do MUCH better if you "name things" that you do not know. By naming things, I mean to assign a unique letter to each unknown number.

$\text {Fraction of work done by Ajay in one day } = x.$

$\text {Fraction of work done by Sunil in one day } = y.$

$\text {Fraction of work done by Sanjay in one day } = z.$

You have three unknowns. Therefore you need three independent and consistent equations involving those unknowns. So translate what you are told into equations.

$10x + 10y = 1 \implies x = \dfrac{1 - 10y}{10}.$

$15y + 15z = 1 \implies z = \dfrac{1 - 15y}{15}.$

$20x + 20z = 1 \implies$

$20 * \dfrac{1 - 10y}{10} + 20 * \dfrac{1 - 15y}{15} = 1 \implies$

$30 * \left ( 20 * \dfrac{1 - 10y}{10} + 20 * \dfrac{1 - 15y}{15} \right ) = 30 * 1 \implies 60(1 - 10y) + 40(1 - 15y) = 30 \implies$

$60 - 600y + 40 - 600y = 30 \implies 1200y = 70 \implies y = \dfrac{7}{120} \implies$

$x = \dfrac{1 - 10 * \dfrac{7}{120}}{10} = \dfrac{\dfrac{12}{12} - \dfrac{7}{12}}{10} = \dfrac{5}{12} * \dfrac{1}{10} = \dfrac{5}{120}.$

$y = \dfrac{7}{120} \implies z = \dfrac{1 - 15 * \dfrac{7}{120}}{15} = \dfrac{\dfrac{8}{8} - \dfrac{7}{8}}{15} = \dfrac{1}{8} * \dfrac{1}{15} = \dfrac{1}{120}.$

Before proceeding to the next step, check your work so far.

$10 * \dfrac{5}{120} + 10 * \dfrac{7}{120} = \dfrac{50}{120} + \dfrac{70}{120} = \dfrac{120}{120} = 1.$ OK

$15 * \dfrac{7}{120} + 15 * \dfrac{1}{120} = \dfrac{105}{120} + \dfrac{15}{120} = \dfrac{120}{120} = 1.$ OK

$20 * \dfrac{5}{120} + 20 * \dfrac{1}{120} = \dfrac{100}{120} + \dfrac{20}{120} = \dfrac{120}{120} = 1.$ OK

It checks.

Now can you finish it? Tags problem, solve, time, work ,

### ajay and sunil can do a piece of work in 10 days.sunil and sanjay in 15 days ,sanjay and ajay in 20 days..they together work at it for 6 days and then ajay leaves and sunil and sanjay go on together for 4 days more ..if sunil then leaves how long will san

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