The IBM Written Exam consists of two sections: Quantitative Ability and Reasoning Skills. Find the complete IBM Recruitment Process, here. Below you will find IBM aptitude questions based on previous IBM Aptitude questions asked. Read on to discover the topics of the questions commonly asked in IBM Aptitude Questions for freshers.

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## IBM Aptitude Syllabus

Topic | Difficulty Level | No of Questions |
---|---|---|

Ratio & Proportions | Easy | 1-2 |

Number System | Easy | 1-2 |

Profit & Loss | Medium | 2-3 |

Percentages | Medium | 2-3 |

Probablity | Medium | 2-3 |

Permutation & combination | Hard | 2-3 |

Speed Time & Distance | Hard | 2-3 |

A total of 18 questions to be solved in 38 minutes.

To know more about IBM Aptitude Syllabus, click here.

## IBM Aptitude Questions

Try IBM’s free aptitude questions. The questions will be based on the logic building, general aptitude and mind puzzles. IBM aptitude questions should be attempted laying emphasis on getting shortlisted by IBM

**1. If Rs 20/- is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive? **

**A) 12****B) 15****C) 20****D) 16**

Ans: Rs 12

Exp:

Ratio of A & B is 42:28 = 3:2

Then A receive = 20 x 3/5 = 12

**2. A man reaches his office two hours late travelling at 50 km/hr. if he increases his speed to 60km/hr, he is late by 1 hour. find the distance he has to travel to reach his office and speed required to reach the office in time **

**A) 310****B) 300****C) 200****D) 205**

Ans.let distance be d km & time reqd be t hour then

d/50 = (t + 2)

d/60 = (t + 1)

=> 50(t + 2) = 60(t + 1)

=> 5t + 10 = 6t + 6

=> t = 4

put t = 4 in eqn d/50 = (t + 2) => d = 50(4 + 2) = 300

d = 300 km & t = 4 hour

**3. In a simultaneous throw of two dice, what is the probability of getting a total of 10 or 11? **

**A. 1/4 ****B. 1/6 ****C. 7/12 ****D. 5/36 **

Ans. two dice are thrown the outcomes are 62 = 36 ways

getting of a total 10 or 11 are (4,6), (5,5), (5,6) (6,4), (6,5)

i.e., 5 ways

so the probability of getting a total of 10 or 11 is = 5/36

option 4 is the answer

**4. The printed price on a book is 400, a bookseller offers a 10% discount on it. If he still earns a profit of 12&, the CP of the book is **

**a) 280 ****b) 352 ****c) 360 ****d) 300 **

Ans. S.P = 360 Rs.

If Profit = 12% then 1.12 x C.P = S.P

1.12 x C.P = 360

C.P = 321.43 Rs.

And If Profit = 12 Rs. then C.P = 348 Rs.

As there is ambiguity in the question.

**5. Printer A prints 8192 characters per min and printer B prints 13862 characters per min four characters are equal to one word. Printer A starts at 7:15 am while Printer B starts at 7:29 am then at what time both will have same no of words printed.**

**A) 7:43 AM****B) 7:47 AM****C) 7:48 AM****D) 7:49 AM**

Ans. four character are equal to one word”

8192/4 x (t+14) = 13862/4 x t

8192 x (t+14) = 13862 x t yields the same result

t = 20.2272 minutes = about 20 min 14 sec

7:49:14 am

**6. 10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work? **

**A. 90 ****B. 145 ****C. 150 ****D. 225 **

Ans. 1 man’s 1-day work =1/100

let 1 woman’s 1-day work = 1/x

so

(1/10) + (15/x) = (1/6)

x = 225

**7. If a salesman’s average is a new order every other week, he will break the office record of the year. However, after 28 weeks, he is six orders behind schedule. In what proportion of the remaining weeks does he have to obtain a new order to break the record?**

**A. 1/2****B. 1/4 ****C. 3/2 ****D. 3/4 **

Ans. Number of weeks in a year = 52

Total orders in a year = 52/2 = 26 (Because every other weeks means, alternate)

so after 28 weeks no. of orders should be 28/2 = 14

Originally completed orders in 28 weeks = 14 – 6 = 8 orders (Because 6 orders behind)

next 24 (52- 28) weeks, orders needed to collect = 26 – 8 = 18

So, required proportion for remaining weeks = 18/24 = 3/4

**8. If 19 and 1140 are the respective HCF and LCM of two numbers, which are greater than 19 then what will be the possible number of such pair?**

**A. 90 ****B. 60 ****C. 30 ****D. 10**

Ans. Product of HCF and LCM = product of the numbers

Then, product of the numbers = 19 x 1140

Let 19a and 19b be the numbers.

19a x 19b = 19 x 1140

ab = 19 x 1140 / 19 x 19 = 60

If ab = 60 then (a,b) = (1,60), (2,30), (3,20), (4,15), (5,12) and (6,10).

Since a and b are co-primes then (a,b) = (1,60), (4,15) and (5,12)

Hence the number of such pairs = 3a

**9. Foreign language broadcast records last 90Mins on each of two sides if it takes 5hrs to translate one hour of broadcast how long will it take to translate 16 full records?**

**a)256 ****b)26 ****c)48 ****d)144 ****e)188 **

Ans. Foreign language broadcast records last 90 Mins on each of two sides.

so total 3 hr recording in 1 full record.

so 48 hr recording in 16 full records.

it takes 5 hrs to translate one hour of the broadcast

so it will take 48 x 5 = 240 hrs to translate 16 full records.

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