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# Accenture Aptitude Test Papers

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## Total Qs: 145+

NA
SHSTTON
45
Solv. Corr.
69
Solv. In. Corr.
114
Attempted
1 M:43 S
Avg. Time

11 / 145

Choose the correct option.

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

A7

B6

C5

D9

| | Important Formulas | Topic: | Asked In Accenture |

Explanation:

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

Tags: Accenture

NA
SHSTTON
33
Solv. Corr.
93
Solv. In. Corr.
126
Attempted
3 M:39 S
Avg. Time

12 / 145

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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.

A4 Km.

B4.5 Km.

C3 Km.

D3.75 Km.

| | Important Formulas | Topic: | Asked In Accenture |

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.

Tags: Accenture

NA
SHSTTON
57
Solv. Corr.
80
Solv. In. Corr.
137
Attempted
0 M:38 S
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13 / 145

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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:

A2 Kg.

B4 Kg.

C8 Kg.

D3 Kg.

| | Important Formulas | Topic: | Asked In Accenture |

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.

Tags: Accenture

NA
SHSTTON
42
Solv. Corr.
99
Solv. In. Corr.
141
Attempted
1 M:42 S
Avg. Time

14 / 145

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

A182

B184

C216*9!

DNone of these

| | Important Formulas | Topic: | Asked In Accenture |

Explanation:

Here is no explanation for this answer

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NA
SHSTTON
88
Solv. Corr.
31
Solv. In. Corr.
119
Attempted
0 M:24 S
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15 / 145

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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?

A90

B100

C110

D105

| | Important Formulas | Topic: | Asked In |

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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NA
SHSTTON
24
Solv. Corr.
82
Solv. In. Corr.
106
Attempted
1 M:27 S
Avg. Time

16 / 145

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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?

A4 times the draining rate

B3 times the draining rate

C2.5 times the draining rate

D2 times the draining rate

| | Important Formulas | Topic: | Asked In Accenture |

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.
(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.

the fill rate is 4 times the drain rate.

Tags: Accenture

NA
SHSTTON
55
Solv. Corr.
113
Solv. In. Corr.
168
Attempted
1 M:9 S
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The ratio of two numbers is 3:4 and their HCF is 4.Their LCM is:

A12

B16

C24

D48

| | Important Formulas | Topic: | Asked In |

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

Tags: Accenture

NA
SHSTTON
38
Solv. Corr.
72
Solv. In. Corr.
110
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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?

A1/4

B1/3

C1/2

D2/3

| | Important Formulas | Topic: | Asked In Accenture |

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

Tags: Accenture

NA
SHSTTON
59
Solv. Corr.
32
Solv. In. Corr.
91
Attempted
1 M:0 S
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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?

A300

B400

C420

D480

| | Important Formulas | Topic: | Asked In Accenture |

Explanation:

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

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NA
SHSTTON
31
Solv. Corr.
67
Solv. In. Corr.
98
Attempted
2 M:1 S
Avg. Time

20 / 145

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Three runners A, B and C run a race, with runner A finishing 12 meters ahead of runner B and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race?

A36 mtrs.

B48 mtrs.

C60 mtrs.

D72 mtrs.

| | Important Formulas | Topic: | Asked In Accenture |

Explanation:

Lets assume distance of race = x mtrs.
Then when A finishes x m , B has run (x- 12)mtrs and C has run (x-18) mtrs.
=> at this point B is 6 m ahead of C. Now to finish race b needs to run another 12 m,
=> he runs another 12 m. when B finishes race he is 8 m ahead of C.
so last 12 m B has run, C has run 10 m.as speeds are constant,
=> x-12/ x-18 = 12/10 => x = 48 mtrs.