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Placement Questions & Answers :: Accenture

Company: Accenture

Total Qs: 103+

Website: https://www.accenture.com

Total View: 36.7K


Not Attempted

11. At 10:00 am 2 trains started travelling towards each other from station 287 miles apart they passed each other at 1:30 pm the same dayy .if average speed of the faster train exceeded by 6 miles /hr what is speed of faster train in miles/hr:

View Answer | Submit Your Solution | Important Formulas | Topic: Problems on Trains | Asked In |

Answer: Option C

Explanation:

Lets assume the speed of slower train = x miles/hrs.
So, speed of faster train is = (x+6) miles/hrs.
Given, passed each other at 1:30 PM i.e after 3 1/2 hrs.
Both train travelling towards each other so total relative speed = x+(x+6) = (2x+6)
So, 287/(2x+6) = 7/2 => 574 = 14x + 42
=> 14x = 542 => x = ~38

So spee of faster train = (x+6) miles/hrs. = 44 miles/hrs.

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Not Attempted

12. Find the approximate value of the following equation.
6.23% of 258.43 - ? + 3.11% of 127 = 13.87

View Answer | Submit Your Solution | Important Formulas | Topic: Simplification | Asked In |

Answer: Option B

Explanation:

[(6.23/100)*258.43]-'X'+[(3.11/100)*127]=13.87
16.100189-X+3.9497=13.87
X=6.179889

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Not Attempted

13. A cycled from P to Q at 10 kmph and returned at the rate of 9 kmph. B cycled both ways at 12 kmph. In the whole journey B took 10 minutes less than A. Find the distance between P and Q.

View Answer | Submit Your Solution | Important Formulas | Topic: Time and Distance | Asked In |

Answer: Option D

Explanation:

Lets assume distance between P and Q = d km.
Time taken by A in both side = d/10 + d/9 = 19d/90 hrs.
Time taken by B in both side = 2d/12 = d/6 hrs.
B tool 10 min. or 1/6 hrs. less than A.
So, 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
=> d = 15/4 km = 3.75 km.

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Not Attempted

14. An alloy of zinc and copper contains the metals in the ratio 5 : 3. The quantity of zinc to be added to 16 kg of the alloy so that the ratio of the metal may be 3 : 1 is:

View Answer | Submit Your Solution | Important Formulas | Topic: Alligation or Mixture | Asked In |

Answer: Option C

Explanation:

Lets assume quantity of zinc to added = x kg.
zinc quantity in alloy = 5/8*16=10 Kg.
and copper quantity = 3/8*16=6 kg.
Alloy new ratio of zinc and copper = 3:1
Zinc quantity in alloy => (10+x)/6=3/1 => 10 + x = 18 => x = 8 kg.

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Not Attempted

15. How many words can be formed using the letters of the word "CORRESPONDANCE" if the consonants are always written together?

View Answer | Submit Your Solution | Important Formulas | Topic: Permutations & Combinations | Asked In |

Answer: Option D

Explanation:

Here is no explanation for this answer

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Not Attempted

16. A gets Rs.33 when a sum of money was distributed among A, B and C in the ratio 3:2:5, What will be the sum of money?

View Answer | Submit Your Solution | Important Formulas | Topic: Partnership | Asked In |

Answer: Option C

Explanation:

Lets assume sum of money = Rs. x ;
A gets Rs. 33 when sum distributed in the ratio of 3:2:5
so, 33=x*3/10
=> x=110

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Not Attempted

17. It takes 30 minutes to empty a half-full tank by draining it at a constant rate. It is decided to simultaneously pump water into the half-full tank while draining it. What is the rate at which water has to be pumped in so that it gets fully filled in 10 minutes?

View Answer | Submit Your Solution | Important Formulas | Topic: Pipes & Cistern | Asked In |

Answer: Option A

Explanation:

The equation to use is RT = Q
Q is equal to 1/2 the tank.
T is equal to 30 minutes
R is derived from the equation as follows:
RT = Q
R*30 = 1/2
R = 1/60
the tank is being drained at the rate of 1/60 of the tank per minute.
at that rate, the remaining 1/2 of the tank will be drained in 30 minutes.

the rate of filling and the rate of draining are opposing forces so you need to subtract one from the other to get the rate of filling.

you want to fill 1/2 the tank while at the same time you are draining the tank.
the combined formula would be as follows:
(RF - RD) * 10 = 1/2
you know RD is 1/60 because we just solved for that.
your formula becomes:
(RF - 1/60) * 10 = 1/2
simplify this to get:
10*RF - 10/60 = 1/2
add 10/60 to both sides of this equation to get:
10*RF = 1/2 + 10/60 which becomes:
10*RF = 30/60 + 10/60 which becomes:
10*RF = 40/60.
divide both sides of this equation by 10 to get:
RF = 4/60.

4/60 is 4 times 1/60.

fill rate is 4 times the drain rate.

you start at 1/2 a tank which is the same as 3/6 of a tank.
in 10 minutes you will have drained 10*1/60 = 10/60 = 1/6 more of the tank.
in the same 10 minutes you will have filled 10*4/60 = 40/60 = 4/6 of the tank.
sum of fill and drain is equal to 3/6 - 1/6 + 4/6 which is equal to 6/6 which is equal to 1.

at the end of the 10 minutes, the tank is full.

you could have solved this another way as well.
you still had to find the drain rate which is equal to 1/60 of the tank per minute.

in 10 minutes, you will have drained 1/6 more of the tank.
the tank is now 1/2 - 1/6 = 3/6 - 1/6 = 2/6 full.

in order to fill the tank in the same 10 minutes, you have to fill 4/6 of the tank.

the same formula is used.
RT=Q
10R = 4/6
R = 4/60.

your drain rate was 1/60
your fill rate is 4/60.
the fill rate is 4 times the drain rate.

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Not Attempted

18. The ratio of two numbers is 3:4 and their HCF is 4.Their LCM is:

View Answer | Submit Your Solution | Important Formulas | Topic: H.C.F and L.C.M | Asked In |

Answer: Option D

Explanation:

Given Two No, ratio = 3:4 and their HCF = 4
So, No. = 3*4 =12 and 4*4=16
LCM of 12,16 = 48

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Not Attempted

19. A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?

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Answer: Option B

Explanation:

Consider a solution = 1 ltr
X is the certain quantity which has to be replaced
Now,
40%of (1-x)+(25%of x)=35/100
=> 40/100 * (1-x) + 25x/100 = 35/100
=> 40/100 - 40x/100 + 25x/100 = 35/100
=> 15x/100 = 5/100
=> x = 1/3

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Not Attempted

20. A store owner is packing small radios into larger boxes that measure 25 in. by 42 in. by 60 in. If the measurement of each radio is 7 in. by 6 in. by 5 in., then how many radios can be placed in the box?

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Answer: Option A

Explanation:

No of radios that can be placed in the box = (25*42*60)/(7*6*5)=300

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Here is the list of questions asked in Accenture Aptitude Test Question with Answers Accenture Mock Test page 2. Practice Accenture Written Test Papers with Solutions and take Q4Interview Accenture Online Test Questions to crack Accenture written round test. Overall the level of the Accenture Online Assessment Test is moderate. Only those candidates who clear the written exam will qualify for the next round, so practic all the questions here and take all the free tests before going for final selection process of Accenture