{"id":9326,"date":"2022-08-26T07:23:34","date_gmt":"2022-08-26T07:23:34","guid":{"rendered":"https:\/\/www.prepbytes.com\/blog\/?p=9326"},"modified":"2022-09-13T13:21:21","modified_gmt":"2022-09-13T13:21:21","slug":"accenture-aptitude-preparation-for-freshers-on-campus","status":"publish","type":"post","link":"https:\/\/prepbytes.com\/blog\/accenture-aptitude-preparation-for-freshers-on-campus\/","title":{"rendered":"Accenture Aptitude Preparation for Freshers On-Campus"},"content":{"rendered":"

\"\"<\/p>\n

In this article, we have provided Accenture Aptitude Questions<\/a> and Answers For Freshers (2022 and 2023 Batches).
\nApplicants can gain more knowledge by practicing all the Accenture Questions and Answers from this section. According to our analysis most of the students are facing more difficulty in the Aptitude section.
\nFor those types of candidates, this article helps a lot.
\nPlease note that these are Accenture model questions but not the exact one. By practicing the sample Accenture Aptitude Questions and Answers with Solutions, you were able to clear all topics involved in the aptitude section.<\/p>\n

You can now get one step closer to your dream job and streamline your placement preparation journey with PrepBytes Placement Programs.<\/b><\/a>
\n<\/strong><\/p>\n

Accenture Aptitude questions and answers for freshers<\/h3>\n

Accenture’s most repeated aptitude questions with solutions are given here.<\/p>\n

    \n
  1. \n

    4 women and 5 men working together can do 3 times the work done by 2 women and one man together. Calculate the work of a man to that of a woman.<\/strong>
    \nA. 1:1
    \nB. 3:2
    \nC. 1:2
    \nD. 2:1
    \nAnswer – A. 1:1<\/strong>
    \nExplanation:
    \nGiven
    \n4w + 5m = 3 * (2w + m)
    \ni.e. 2w = 2m
    \nso the ratio of work done by man to woman is 1:1.<\/p>\n<\/li>\n

  2. \n

    Manoj can do a work in 20 days, while Chandu can do the same work in 25 days. They started the work jointly. A few days later Suresh also joined them and thus all of them completed the whole work in 10 days. All of them were paid a total of Rs.1000. What is the share of Suresh?<\/strong>
    \nA. 100
    \nB. 300
    \nC. 200
    \nD. 400
    \nAnswer \u2013 A. 100<\/strong>
    \nExplanation:
    \nEfficiency of Manoj = 5%
    \nThe efficiency of Chandu = 4%
    \nThey will complete only 90% of the work = [(5+4) 10] =90
    \nRemaining work was done by Suresh = 10%.
    \nShare of Suresh = 10\/100 <\/em> 1000 = 100.<\/p>\n<\/li>\n

  3. \n

    While calculating the weight of a group of men, the weight of 63 kg of one of the members was mistakenly written as 83 kg. Due to this the average of the weights increased by half kg. What is the number of men in the group?<\/strong>
    \nA. 25
    \nB. 20
    \nC. 40
    \nD. 60
    \nAnswer – C. 40<\/strong>
    \nExplanation:
    \nIncrease in marks lead to an increase in average by 1\/2
    \nSo (83-63) = x\/2
    \nx = 40
    \nTherefore, the number of men in the group are 40.<\/p>\n<\/li>\n

  4. \n

    In a group of 8 boys, 2 men aged at 21 and 23 were replaced, two new boys. Due to this the average cost of the group increased by 2 years. What is the average age of the 2 new boys?<\/strong>
    \nA. 17
    \nB. 30
    \nC. 28
    \nD. 23
    \nAnswer – B. 30<\/strong>
    \nExplanation:
    \nAccording to the given data
    \nAverage of 8 boys increased by 2, this means the total age of boys increased by 8 * 2 = 16 yrs
    \nSo sum of ages of two new boys = 21+23+16 = 60
    \nAverage of these = 60\/2 = 30.<\/p>\n<\/li>\n

  5. \n

    A Boat takes a total 16 hours for traveling downstream from point A to point B and coming back to point C which is somewhere between A and B. If the speed of the Boat in Still water is 9 Km\/hr and the rate of stream is 6 Km\/hr, then what is the distance between A and C?<\/strong>
    \nA. 60 Km
    \nB. 90 Km
    \nC. 30 Km
    \nD. Cannot be determined
    \nAnswer – D. Cannot be determined<\/strong><\/p>\n<\/li>\n

  6. \n

    A Boat going upstream takes 8 hours 24 minutes to cover a certain distance, while it takes 5 hours to cover 5\/7 of the same distance running downstream. Then what is the ratio of the speed of the boat to the speed of water current?<\/strong>
    \nA. 11:5
    \nB. 11:6
    \nC. 11:1
    \nD. 6:5
    \nAnswer – C. 11:1<\/strong>
    \nExplanation:
    \n(S-R) 42\/5 = (S+R) <\/em> 7
    \nS:R = 11:1.<\/p>\n<\/li>\n

  7. \n

    A Boat takes 128 min less to travel 48 Km downstream than to travel the same distance upstream. If the speed of the stream is 3 Km\/hr. Then Speed of Boat in still water is?<\/strong>
    \nA. 12 Km\/hr
    \nB. 15 Km\/hr
    \nC. 6 Km\/hr
    \nD. 9 Km\/hr
    \nAnswer – A. 12 Km\/hr<\/strong>
    \nExplanation:
    \n32\/15 = 48(1\/s-3 \u2013 1\/s+3)
    \ns= 12
    \nTherefore, Speed of Boat in still water is 12 Km\/hr.<\/p>\n<\/li>\n

  8. \n

    An alloy contains Brass, Iron, and Zinc in the ratio 2:3:1 and another contains Iron, zinc, and lead in the ratio 5:4:3. If equal weights of both alloys are melted together to form a third alloy, then what will be the weight of lead per kg in the new alloy?<\/strong>
    \nA. 1\/4
    \nB. 41\/7
    \nC. 1\/8
    \nD. 51\/9
    \nAnswer – C. 1\/8<\/strong>
    \nExplanation:
    \nShortcut:
    \nIn the first alloy,
    \n2:3:1 =6 * 2
    \n5:4:3 =12
    \nMultiply 2 to make it equal,
    \n4:6:2
    \n5:4:3
    \nAdding all,
    \n4:11:6:3=24
    \n3\/24=1\/8.<\/p>\n<\/li>\n

  9. \n

    A milkman mixes 6 liters of free tap water with 20litres of pure milk. If the cost of pure milk is Rs.28 per liter the % Profit of the milkman when he sells all the mixture at the cost price is?<\/strong>
    \nA. 30%
    \nB. 16(1\/3)%
    \nC. 25%
    \nD. 16.5%
    \nAnswer – A. 30%<\/strong>
    \nExplanation:
    \nProfit=28 6=728
    \nCp= 28 <\/em> 20=560
    \nProfit = 168 * 100\/560=30%<\/p>\n<\/li>\n

  10. \n

    144 liters of the mixture contains milk and water in the ratio 5: 7. How much milk needs to be added to this mixture so that the new ratio is 23: 21 respectively?<\/strong>
    \nA. 40 liters
    \nB. 28 liters
    \nC. 32 liters
    \nD. 36 liters
    \nAnswer – C. 32 liters<\/strong>
    \nExplanation:
    \n144 == 5:7
    \n60: 84
    \nNow == 21 = 84
    \n23 = 92
    \n92-60 = 32.<\/p>\n<\/li>\n

  11. \n

    A shopkeeper bought 30kg of rice at Rs.75 per kg and 20 kg of rice at the rate of Rs.70. per kg.If he mixed the two brands of rice and sold the mixture at Rs.80 per kg. Find his gain.<\/strong>
    \nA. Rs.350
    \nB. Rs.550
    \nC. Rs.420
    \nD.Rs.210
    \nAnswer – A. Rs.350<\/strong>
    \nExplanation:
    \nCP = 30 75 + 20 <\/em> 70 = 2250 + 140 = 3650
    \nSP =80 * (30+20) = 4000.
    \nHence, Gain = 4000-3650 = 350.<\/p>\n<\/li>\n

  12. \n

    Cost price of 80 notebooks is equal to the selling price of 65 notebooks. The gain or loss % is?<\/strong>
    \nA. 32%
    \nB. 42%
    \nC. 27%
    \nD. 23%
    \nAnswer – D. 23%<\/strong>
    \nExplanation:
    \n% = [80 \u2013 65\/65] 100
    \n= 15 <\/em> 100\/65 = 1500\/65
    \n= 23.07 = 23% profit
    \nTherefore, the gain percentage is 23%.<\/p>\n<\/li>\n

  13. \n

    Eight years ago, Pranathi\u2019s age was equal to the sum of the present ages of her one son and one daughter. Five years hence, the respective ratio between the ages of her daughter and her son will be 7:6. If Pranathi\u2019s husband is 7 years elder to her and his present age is three times the present age of their son, what is the present age of the daughter?<\/strong>
    \nA. 19 years
    \nB. 27 years
    \nC. 15 years
    \nD. 23 years
    \nAnswer – D. 23 years<\/strong>
    \nExplanation:
    \nP – 8 = S + D \u2014(1)
    \n6D + 30 = 7S + 35 \u2014(2)
    \nH = 7 + P
    \nH = 3S
    \n3S = 7 + P \u2014(3)
    \nSolving equation (1),(2) and (3) D = 23
    \nTherefore, the present age of the daughter is 23 years.<\/p>\n<\/li>\n

  14. \n

    Shas married 8 year ago. Today her age is 9\/7 times that of marriage. At present his son\u2019s age is 1\/6th of her age. What was her son\u2019s age 3 year ago?<\/strong>
    \nA. 4 yr
    \nB. 2 yr
    \nC. 3 yr
    \nD. 5 yr
    \nAnswer – B. 2 yr<\/strong>
    \nExplanation:
    \nLet us assume that Sravan\u2019s age 8 year ago = x
    \nPresent age = x + 8
    \nx + 8 = 9\/7 x
    \n7(x + 8)= 9x
    \nx = 28; 28 + 8 = 36
    \nSon\u2019s age = 1\/6 * 36 = 6
    \nSon\u2019s age 4 year ago = 6-4 =2<\/p>\n<\/li>\n

  15. \n

    The respective ratio between the present age of Mani and Dheeraj is x : 42. Mani is 8 years younger than Murali. Murali\u2019s age after 8 years will be 33 years. The difference between Dheeraj\u2019s and Mani\u2019s age is the same as the present age of Murali. What is the value of x?<\/strong>
    \nA. 18
    \nB. 10
    \nC. 16
    \nD. 17
    \nAnswer – D. 17<\/strong>
    \nExplanation:
    \nMurali\u2019s age after 8 years = 33 years
    \nMurali\u2019s present age = 33 \u2013 8= 25 years
    \nMani\u2019s present age = 25 \u2013 8 = 17 years
    \nDheeraj\u2019s present age = 17 + 25 = 42 years
    \nRatio between Mani and Dheeraj = 17: 42
    \nX = 17.<\/p>\n<\/li>\n

  16. \n

    Revanth\u2019s present age is three times his son\u2019s present age and 4\/5th of his father\u2019s present age. The average present age of all of them is 62 years. What is the difference between the Revanth\u2019s son\u2019s present age and Revanth\u2019s father\u2019s present age?<\/strong>
    \nA. 64 years
    \nB. 69 years
    \nC. 66 years
    \nD. 62 years
    \nAnswer – C. 66 years<\/strong>
    \nExplanation:
    \nPresent age of Revanth is = 4\/5x
    \nPresent age of Revanth\u2019s father is = 4\/15x
    \nRatio = 15: 12 : 4
    \nDifference between the Revanth\u2019s son\u2019s present age and Revanth\u2019s father\u2019s present age = 62\/31 3(15 \u2013 4).
    \n= 2<\/em>3*11 = 66 years.<\/p>\n<\/li>\n

  17. \n

    36% of 945 \u2013 26% of 765 + 17.7 =?<\/strong>
    \nA. 167
    \nB. 187
    \nC. 159
    \nD. 143
    \nAnswer – C. 159<\/strong>
    \nExplanation: 340.2 \u2013 198.9 =141.3+17.7 = 159<\/p>\n<\/li>\n

  18. \n

    If tan \u03b8 + cot \u03b8 = 2, then the value of tan2\u03b8 + cot2\u03b8 is<\/strong>
    \nA. 2
    \nB. 1
    \nC. \u27132
    \nD. 0
    \nAnswer: A. 2<\/strong>
    \nExplanation: tan \u03b8 + cot \u03b8 = 2 , On squaring both sides,
    \n(tan \u03b8 + cot \u03b8)2 = 4,
    \n\u21d2 tan2\u03b8 + cot2\u03b8 + 2 tan \u03b8 cot \u03b8 = 4,
    \n\u21d2 tan2\u03b8 + cot2\u03b8 = 4 \u2013 2 = 2 , [tan \u03b8 . cot \u03b8 = 1].<\/p>\n<\/li>\n

  19. \n

    The average age of a man and his son is 28 years. The ratio of their ages is 3 :1 respectively. What is the man’s age?<\/strong>
    \nA. 44
    \nB. 42
    \nC. 38
    \nD. 30
    \nAnswer: B. 42 Years<\/strong>
    \nExplanation: Total sum of man’s age & his son’s age =28 \u00d7 2 = 56 Now, the Ratio of their ages is 3 : 1.Therefore, Man’s age = (3\/4) \u00d7 56 = 42.
    \nSo, the correct answer is option B.<\/p>\n<\/li>\n

  20. \n

    A, B, C and D are four consecutive odd numbers and their average is 42. What is the product of B and D?<\/strong>
    \nA. 1860
    \nB. 1890
    \nC. 1845
    \nD. 1677
    \nAnswer: C. 1845<\/strong>
    \nExplanation: As there are As diff. The average should lie between B and C so B is 41 & C is 43 so D must be 45 as we have to find the product of B and D so it would be 1845.<\/p>\n<\/li>\n<\/ol>\n

    This article tried to discuss the Accenture Aptitude Questions and Answers for Freshers. Hope this blog helps you understand and solve the problem. To Practice more problems you can check out <\/a> at Prepbytes<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

    In this article, we have provided Accenture Aptitude Questions and Answers For Freshers (2022 and 2023 Batches). Applicants can gain more knowledge by practicing all the Accenture Questions and Answers from this section. According to our analysis most of the students are facing more difficulty in the Aptitude section. For those types of candidates, this […]<\/p>\n","protected":false},"author":52,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[147],"tags":[],"class_list":["post-9326","post","type-post","status-publish","format-standard","hentry","category-interview-questions"],"yoast_head":"\nAccenture Aptitude Preparation for Freshers On-Campus<\/title>\n<meta name=\"description\" content=\"This article tried to discuss the Accenture Aptitude Questions and Answers for Freshers, hope this blog helps you understand and solve the problems in the interviews.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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