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Person |
No. of days they worked |
Percentage of work done to complete the project |

A | 8 | 20% |

B | 3 | 10% |

C | 6 | 25% |

D | 15 | 30% |

E | 6 | 15% |

Important for :

1

B

A does 20% work in 8 days.So, 100% work in 100 × | 8 | = 40 days |

20 |

B does 10% work in 3 days

So, 100% work in 100 × | 3 | = 30 days |

10 |

1 day work of A and B together is:-

So, | 1 | + | 1 | = | 7 |

40 | 30 | 120 |

So in 10 days they completed 7/12 part of the work

Now, C completed 25% = | 1 | of work |

4 |

So now remaining work

= 1 – | ( | 7 | + | 1 | ) | = | 1 |

12 | 4 | 6 |

F complete | 1 | work in 16 days, |

6 |

So complete work in 96 days.

Hence, option (B) is correct.

2

D

A’s part of work = 20% = | 1 |

5 |

So, G did 1/5 of work and whole work in 30 days,

⇒ | 1 | work in | 1 | × 30 = 6 days |

5 | 5 |

Now, B also worked for 6 days.

B can complete 10% of work in 3 days.

So, B can complete the whole work in taken time

= 100 × | 3 | |

10 |

So, in 6 days,

B completed | 6 | = | 1 | of work |

30 | 5 |

Now, remaining work

= 1 – | ( | 1 | + | 1 | ) | = | 3 |

5 | 5 | 5 |

Now, E can complete 15% of work in 6 days.

So, E can complete the whole work in taken time

= 100 × | 6 | |

15 |

M can complete the work in 1/4

= | 1 | × 40 |

4 |

So, M completed 3/5 work in taken time

= | 3 | × 10 = 6 days |

5 |

Hence, total number of days

= 6 + 6 + 6

= 18

Therefore, option D is correct.

3

B

people equally divided the work so each did 1/5 work now

A does 1/5^{th} work in 8 days

B does 1/10th (10%) work in 3 days

⇒ 1/5^{th} of work in 6 days

⇒ 1/5

C does 1/4

⇒ 1/5

D does 3/10^{th} (30%) work in 15 days

⇒ 1/5^{th} work in 10 days

⇒ 1/5

E does 3/20^{th} (15%) work in 6 days

⇒ 1/5^{th} work in 8 days

⇒ 1/5

Hence, total work completed in

= 8 + 4.8 + 6 + 10 + 8

= 36.8 days

Therefore, option B is correct.

4

C

B can complete 10% of work in 3 days.So, B can complete the whole work in

100 × | 3 | = 30 days |

10 |

As P is 20% more efficient than B

⇒ P can complete the work in 25 days

C can complete 25% of work in 6 days.

So, C can complete the whole work in

100 × | 6 | = 24 days |

25 |

As Q is 60% more efficient than B

⇒ Q can complete the work in 15 days

Now, P & Q worked for 5 days,

⇒ | 5 | + | 5 | = | 8 |

25 | 15 | 15 |

Remaining work = 1 – | 8 | = | 7 | |

15 | 15 |

D can complete 30% of work in 15 days.

So, D can complete the whole work in

100 × | 15 | = 50 days |

30 |

So, D does 7/15

50 × | 7 | = 23 | 1 | days |

15 | 3 |

Hence, option C is correct.

5

C

Let us assume he buys n goods.

Total CP = 20n

Total SP = 2 + 4 + 6 + 8 ….n terms

Total SP should be at least 40% more than total CP

2 + 4 + 6 + 8 ….n terms ≥ 1.4 × 20 n

2 (1 + 2 + 3 + ….n terms) ≥ 28n

Sum of n – terms = | {n (n + 1)} |

2 |

n (n + 1) ≥ 28n

n^{2} + n ≥ 28n

n^{2} – 27n ≥ 0

n ≥ 27

He should sell a minimum of 27 goods.

Hence, option (C) is correct.

Hence, option (C) is correct.