Infosys Last year’s placement papers

Hello friend I want Last year’s placement papers of Infosys, so please provide me the t Last year’s placement papers of Infosys?

[Saksham]

Hello brother as you ask for Last year’s placement papers of Infosys, so on your demands I am providing you the paper here

1.Nithin was counting down from 32. Sumit was counting upwards the numbers starting from 1 and he was calling out only the odd numbers. What common number will they call out at the same time if they were calling at the same speed?

A. 19 B. 21 C.22 D.They will not call out the same numberE. None of these.
Ans: D
Explanation:
Nithin: 32 31 30 29 28 27 26 25 24 23 22 21 20…..Sumit: 1 3 5 7 9 11 13 15 17 19 21 23 25…Clearly it is seen that they never call out same nmber at the same time.


2.Radha moves towards South-east a distance of 7 km, then she moves towards West and travels a distance of 14m. From here, she moves towards North-west a distance of 7 m and finally she moves a distance of 4 m towards East and stod at that point. How far is the starting point from where she stood?

A. 3 m B. 4m C.8 m D10 m E.11 mAns

Explanation: The movements of Radha are shown as below:


Clearly Radha’s distance from starting point O = OD = (OC –CD) =(AB-CD) = (14-4) m = 10 m



3. In a certain office, 1/3 of the workers are women, ½ of the women are married and 1/3 of the married women have children. If 3/4 th of the men are married and2/3 rd of the married men have children, what part of workers are without children?A. 5/18 B. 4/9 C. 11/18 D17/18 E.17/36Ans: C
Explanation: Let total no. of workers be x
Number of women = x/3 and number of men = 2x/3Women married=1/2 * x/3 = x/6
Women having children = 1/3 *x/6 =x/18
Married Men =3/4 *2x/3 = x/2
Men having Children =2/3 * x/2 = x/3
Workers with children = x/3 + x/18 = 7x/18
Hence, workers without children = x- 7x/18 = 11x/18



4. A, P, R, X, S and Z are sitting in a row. S and z are in the centre. A and P are at the ends. R is sitting to the left of A. Who is to the right of P?A. P B. A C.X D.S E.ZAns: C
Explanation : The seating arrangement is as follows :P X S Z R AHence , right of P is X.


5. Introducing a boy, a girl said, "He is the son of the daughter of the father of my uncle." How is the boy related to the girl?

A. Brother B. Nephew C. Uncle D. son in law E. grand fatherAns: A
Explanation:
The father of the boy's uncle → the grandfather of the boy and daughter of the grandfather → sister of father.


Directions (Questions 11-12): Each Question Given Below has a problem and two statements numbered I and II giving certain Information. You have to decide if the information given in the statements is sufficient for answering the problem. Indicate your answer as(i) if data in statement I alone are sufficient to answer the question;(ii) if data in statement II alone are sufficient to answer the question;
(iii) if data either in I or II alone are sufficient to answer the question;
(iv) if the data even in both the statements together are not sufficient to answer the question;
(v) if the data in both the statements are needed.


6. Is Anil taller than Sachin?I. Dinesh is of the same height as Arun and Sachin.
II. Sachin is not shorter than Dinesh.
A. i B. iii C. ii D. v E. ivAns: A
Explanation: From statement I, we can conclude that Dinesh, Arun and Sachin are of same height. So, Arun is not taller than Sachin. So, only statement I is sufficient to answer the question.


7. In a certain code language, ‘13’ means ‘stop smoking’ and ‘59’ means ‘injurious habit’. What is the meaning of ‘9’ and ‘5’ respectively in that code?I. ‘157’ means ‘stop bad habit’
II. ‘839’ means ‘smoking is injurious’.
A. ii B. iii C. v D. iv E. i Ans: BExplanation: ‘59’ means ‘injurious habit’ and ‘157’ means ‘stop bad habit’. Hence common letter ‘5’ denotes ‘habit’. Hence ‘9’ is obviously ‘injurious’. So I alone can be sufficient. Also, ‘59’ is ‘injurious habit’ and ‘839’ is ‘smoking is injurious’ from which it can be implied that ‘9’ is ‘injurious’. Hence II alone can also be sufficient. Hence either I or II alone can be sufficient.

Directions (13- 15): In the following problem, there is one question and three statements I, II and III below the question. You have to decide whether the data given in the statements is sufficient to answer the question. Read all the statements carefully and find out the probable pair which can be sufficient to answer the question.



8. Five persons --- A, B, C, D and E are sitting in a row. Who is sitting in the middle?I. B is in between E and C.II. B is to the right of E.III. D is in between A and E.I and II together B. II and III together C. I and III togetherD. I, II and III together E. Data insufficient.Ans: D
Explanation:
From I, the order is E, B, C or C, B, E.
From II, the order is E, B.
From III, the order is A, D, E.
Combining all the three, we get the order as: A, D, E, B, C.
Clearly, E is sitting in the middle.
Hence all the three statements are required.


9. Four Subjects --- Physics, Chemistry, Mathematics and Biology were taught in four consecutive periods of one hour each starting from 8.00 a.m. At what time was the Chemistry period scheduled?I. Mathematics period ended at 10.00 am which was preceded by Biology.
II. Physics was scheduled in the last period.
III. Mathematics period was immediately followed by Chemistry.
Only I B. Only I or II C. Only II D. II and III together.E. I and II together or I and III together
Ans: E
Explanation: From I and II we conclude that Mathematics period began at9.00 a.m., Biology period began at 8.00 a.m. and Physics period began at 11.00 a.m. So, the Chemistry period began at 10.00 a.m.
From I and III, it is clearly seen that Mathematics period ended at 10.00 a.m. followed by Chemistry to start at 10.00 a.m.


10. How many sons does Sharma have? I. Saurav and Aditya are brothers of Sonali.II. Ayesha is sister of Sharmila and Aditya.
III. Ayesha and Sonali are daughters of Sharma.
A. I and II only. B. II and III together. C. I, II and III togetherD. I, II, III together are not sufficient E. I and III togetherAns: C
Explanation: From I, Saurav, Aditya and Sonali are siblings. From II, Ayesha, Sharmila and Aditya are siblings. It implies that Saurav, Aditya, Ayesha, Sharmila and Sonali are siblings. This is supported by III.

Directions (Questions 16-20): The following table shows the number of new employees added to different categories of employees in a company and also the no of employees from these categories who left the company ever since the foundation of the company in 1995.


11. During the period of 1995 and 2000, the total no of operators who left the company is what percent of the total number of Operators who joined the company?A. 19% B. 21% C. 27% D. 29% E. 32% Ans: DExplanation: Total no. of operators who left the company during 1996 to 2000
= (104 + 120 + 100 + 112 + 144) = 580.
Total No. of Operators who joined the company during 1996 to 2000
= (880 + 256 + 240 + 208 + 192 + 248) = 2024.
Hence, required Percentage = (580/2024 * 100) = 28.66% = 29%


12. For which of the following categories the percentage increase in the number of employees working in the company from 1996 to 2000 was maximum?A. Managers B. Technicians C. Operators D. AccountantsE. Peons.
Ans: A
Explanation:
No. of managers in 1995 = 760
No. of managers by 2000 = (760 + 280 + 179 + 145 + 160 + 193) – (120 + 92 + 88 + 72 + 96) = 1252.
Hence, percentage of increase = (1262-760)/760 * 100 = 64.74%Similarly we can calculate for the rest of employees.


13. What is the difference between total number of Technicians added to the company and total number of Accountants added to the company during the year 1996 to 2000 at the maximum?A. 128 B. 112 C. 96 D. 88 E. 72Ans: D
Explanation: (272 + 240 + 236 + 256 + 288) – (200 + 224 + 248 + 272 + 260) = 88


14. What was the total no. of peons working in the company in the year 1999?A. 1312 B. 1192 C.1088 D.968 E.908Ans: B
Explanation: (820 + 184 + 152 + 196 + 224) – (96 + 88 + 80 + 120) = 1192


15. What is the pooled average of all employees in the year 1997?A. 1325 B. 1285 C. 1265 D. 1235 E. 1195Ans: E
Explanation:
Managers: (760 + 280 + 179) – (120 + 92) = 1007
Technicians: (1200 + 272 + 240) – (120 + 128) = 1464
Operators: (880 +256 +240) – (104 + 120) = 1152
Accountants: (1160 +200 +224) – (100 + 104) = 1380
Peons: (820 + 184 + 152) – (96 + 88) = 972
Hence pooled average of 5 categories = (1007 + 1464 + 1152 + 1380 + 972)/5 = 1195