# Placement Questions & Answers :: Accenture

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31. In an exam, Ajith, Sachu, Karna, Saheep and Ramesh scored an average of 39 marks. Saheep scored 7 marks more than Ramesh. Ramesh scored 9 fewer than Ajith. Sachu scored as many as Saheep and Ramesh scored.Sachu and Karna scored 110 marks them. If Ajith scores 32 marks then how many marks did Karna score?

**Answer:**** Option A **

**Explanation:**

Lets assume marks of Ajith, Sachu, Karna, Saheep and Ramesh respectively u, v, w, x, y

So, as per the question:

z+7=x ---(i)

u- 9=z ---(ii)

x+ z=v ---(iii)

v+w=110 ---(iv)

Given, u=32 ---- (v)

By solving eq. (i), (ii), (iii), (iv) and (v)

w=57

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32. The average number of visitors of a library in the first 4 days of a week was 58. The average for the 2nd, 3rd, 4th and 5th days was 60.If the number of visitors on the 1st and 5th days were in the ratio 7:8 then what is the number of visitors on the 5th day of the library?

**Answer:**** Option A **

**Explanation:**

If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are a, b, c, d & e respectively, then

a+b+c+d=58*4=232 ----(i)

b+c+d+e=60*4=240 ----(ii)

Subtracting eq. (i) from (ii)

e-a=8 ---(iii)

Given, a/e=7/8 ---(iv)

So from eq. (iii) & (iv)

a=56, e=64

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33. Find the smallest number which leaves 22,35, 48 and 61 as remainders when divided by 26, 39, 52 and 65 respectively.

**Answer:**** Option B **

**Explanation:**

26-22=4

39-35=4

52-48=4

65-61=4

LCM of (26,39,52,65)=780

So smallest number = (780-4) = 776

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34. A constructor estimates that 3 people can paint Mr. khans house in 4 days. If he uses 4 people instead of 3,how long will they take to complete the job?

**Answer:**** Option C **

**Explanation:**

Lets assume day required to complete the job = x day

Men Day

---------------------

3(u) 4(d)

4 x

x/4 = 3/4

=> x = 3 days.

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35. In a party attended by 11 persons, each clinch their glass with every other. How many glass clinches?

**Answer:**** Option A **

**Explanation:**

Total no. of person = 11

Total no. of glass clinches = n(n-1)/2

=11*10/2 = 55

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36. Two circles touch each other externally. The distance between their centres is 14 cm and the sum of their areas is 130 cm^2. Find their radii.

**Answer:**** Option C **

**Explanation:**

Lets assume the radius of 1st circle = x cm.

so, the radius of 2nd circle = (14-x)cm.

(pi*x*x)+(pi*(14-x)*(14-x))=130

By solving above eq.

x = 11 or 3

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37. A vessel is filled to its capacity with pure milk. Ten litres are withdrawn from it and replaced by water. This procedure is repeated again. The vessel now has 32 litres of milk. Find the capacity of the vessel (in litres).

**Answer:**** Option B **

**Explanation:**

Lets assume the capacity of the vessel = x litres.

x (1- 10/x)^2 = 32

x^2 +100-20x = 32x

x^2 +100 - 52x= 0

x^2 - 50x - 2x + 100 = 0

=> (x-2) (x-50) = 0

=> x=2 & x=50

As 10 litres is withdrawn so vessel capacity will be 50 litres.

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38. Two pipes can fill a tank in 10 hours and 12 hours respectively while a third, pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?

**Answer:**** Option D **

**Explanation:**

Given, All the pipes are operate simultaneously

=> Tank filled in 1 hrs = 1/10+1/12-1/20 = 16/120 = 2/15

So, time taken to fill the tank = 15/2 = 7.5 hrs.

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39. If the sales tax reduced from 3 1/2% to 3 1/3%, then what difference does it make to a person who purchases an article with market price of Rs. 8400 ?

**Answer:**** Option C **

**Explanation:**

Tax reduced from 3 1/2 or 7/2% to 3 1/3% or 10/3%.

So, the difference in tax = (7/2 - 10/3)% = 1/6%.

=> 1/6 % of Rs. 8400 = 8400*1/6% = 8400* (1/6 * 1/100) = Rs. 14.

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40. In a stream running at 2 kmph,a motar boat goes 6 km upstream and back again to the starting point in 33 minutes. Find the speed of the motarboat in still water.

**Answer:**** Option B **

**Explanation:**

Lets assume speed of boat in still water = x km/hr

So boat downstream speed = (x+2) km/hr.

boat upstream speed = (x-2) km/hr.

6/(x-2)+6/(x+2)=33/60 => 6/(x-2)+6/(x+2)=11/20

=> 120*[(x-2) + (x+2)] = 11*x^2 -44

=> 240 x = 11*x^2-44

=> 11* x^2 -240x -44 = 0 ---------(1)

By solving eq. (1)

x = 22 km/hr.

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Here is the list of questions asked in Accenture Aptitude Test Question with Answers Accenture Mock Test page 4. 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