GGK technologies placement papers
Hi this is the over all information about the placement papers of all Top MNC Companies. Here we are going to update the latest placement papers conducted in the All MNC companies. The ggk technologies placement papers are provided here for the candidates to download. GGK conducts the placements for the fresher as well as for the experience candidates to recruit the eligible and talented candidates to work in the organization. GGK conducts selection process in 3 rounds. Those rounds are written test, technical round and HR interview. All the candidates who are attending the GGK placements selection process felling difficult in the written test only. Especially in aptitude section. So the candidates who are preparing for the GGK placements are searching for the ggk technologies placement papers. For all those candidates we are providing the ggk technologies previous written test papers. The candidates also download these ggk technologies placement papers pdf by clicking the link provided below. The GGK Technologies Placement Papers contains all types of questions so that you can get an idea which type the questions will be. so we hope that this GGK Technologies Placement Papers will help you to get good job placement in good companies
GGK Technologies Placement Papers
GGK Technologies Placement Papers contains questions in different sections like Aptitude, reasoning, basic programing languages questions and English grammar questions. By seeing this GGK Technologies Placement Papers applicants can get an idea of what type of questions are asked in the written test so applicants by seeing this GGK Technologies Placement Papers can practice the papers as years wise. We are providing all latest ggk technologies previous written test papers for the candidates reference so all candidates who are preparing for the GGK placements are refer this GGK Technologies Placement Papers it will helpful while writing the online test. For candidates reference we are providing some sample question below in ggk technologies previous placement papers. Candidates can refer these ggk technologies placement papers pdf download for your future reference. Applicants please download the GGK Technologies Placement Papers as given below. Down we provide GGK Technologies Placement Papers links so that you can directly download the GGK Technologies Placement Papers
About ggk technologies
GGK Technologies is a consulting and IT services company with world a decade of proven experience providing commitments. GGK the commitments made to its customers. The GGK has made commitments to its employees. The GGK has made commitments to stakeholder’s ecosystem.
GGK Tech custom built Business Intelligence for your company. Using technologies that suits your needs, we provide you with commercial information action to improve the quality of decision making.
GGK includes more mature DevOps improves IT performance. Our customers have improved their IT performance by deploying frequency, the deadlines for time changes to recover say.
GGK helps customers define the overall test strategy. We help to build from start to finish under test automation (data driven and keyword driven), to accelerate the time to market with our test accelerator.
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Some ggk technologies Placement Papers Questions
- How many 6 digit even numbers can be formed from digits 1, 2, 3, 4, 5, 6, and 7 so that the digit should not repeat and the second last digit is even?
If the we have to form even numbers, units digit must be 2, 4, 6. i.e., 3 ways. Also 5th digit should be even. So it can be filled in 2 ways. Now remaining 5 digits can be filled in 5! Ways. So total 5! × 3 × 2 = 720 ways.
- The five tyres of a car (four road tyres and one spare) were used equally in a journey of 40,000 kms. The number of kms of use of each tyre was
Total kilometers travelled by 4 tyre = 40000 × 4 = 1,60,000. This has to be share by 5 tyres. So each tyre capacity = 1,60,000 / 5 = 32, 000. You have a doubt, after we travel 32,000 km, we are left with 4 worn tyes and one new tyre. But If the tyres are rotated properly after each 8000 km, all the tyres are equally used.
- In a group of five families, every family is expected to have a certain number of children, such that the number of children forms an arithmetic progression with a common difference of one, starting with two children in the first family. Despite the objection of their parents, every child in a family has as many pets to look after as the number of offsprings in the family. What is the total number of pets in the entire group of five families.
As the number of children are in arithmetic progression starting with 2, the five families have 2, 3, 4, 5, 6 kids respectively. As each children has kept the pets equal to the number of kids in the family, each family has n2 pets. So total = 22+32+42 +52 +62 = 90
- According to the stock policy of a company, each employee in the technical division is given 15 shares of the company and each employee in the recruitment division is given 10 shares. Employees belonging to both committees get 25 shares each. There are 20 employees in the company, and each one belongs to at least one division. The cost of each share is $10. If the technical division has 15 employees and the recruitment division has 10 employees, then what is the total cost of the shares given by the company?
We have to use addition formula n(A∪B) = n(A)+n(B)−n(A∩B)
20 = 15 + 10 – x
x = 5
So total shares given to only technical = 10 × 15 = 150
Shares given to only Recruitment = 5 × 10 = 50
Share given to Technical as well as recruitment people = 5 × 25 = 125
Total shares = 150 + 50 + 125 = 325.
Total value = 325 × 10 = 3250
- The average marks of 3 students A, B and C is 60. When another student D joins the group, the new average becomes 56 marks. If another student E, who has 3 marks more than D, joins the group, the average of the 4 students B, C, D and E becomes 55 marks. How many marks did A get in the exam?
Given that A + B + C = 60 × 3 = 180
A + B + C + D = 56 × 4 = 224
Therefore, D = 44
E = 44 + 3 = 47
Given, B + C + D + E = 55 × 4 = 220
B + C + 44 + 47 = 220
⇒ B + C = 220 – 91 = 129
So A + 129 = 180 ⇒ A = 51
- What is the number of ways of expressing 3600 as a product of three ordered positive integers (abc, bca etc. are counted as distinct). For example, the number 12 can be expressed as a product of three ordered positive integers in 18 different ways.
3600 = 24 × 32 × 52
Let abc = 24 × 32 × 52
We have to distribute four 2’s to three numbers a, b, c in 4+3−1C3−1=6C2 = 15 ways.
Now two 3’s has to be distributed to three numbers in 2+3−1C3−1=4C2 = 6 ways
Now two 5’s has to be distributed to three numbers in 2+3−1C3−1=4C2 = 6 ways
Total ways = 15 × 6 × 6 = 540
- There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible?
- None of these
If left and right digits are fixed with 5 and 6, then the remaining 5 places can be filled by remaining 8 digits in 8P5 = 6720 ways.
- A certain sum of money is sufficient to pay either George’s wages for 15 days or Mark’s wages for 10 days. For how long will it suffice if both George and Mark work together?
Let the money to be paid = 30 rupees. Then George daily wage = 30/15 = 2, and Mark daily wage = 30/10 = 3.
If both are working, then 5 rupees to be paid. So given sum is sufficient for 30 / 5 = 6 days.
- The remainder when m + n is divided by 12 is 8, and the remainder when m – n is divided by 12 is 6. If m > n, then what is the remainder when mn divided by 6?
m + n = 12a + 8 ⇒ (m+n)2=144a2+192a+64 – – – (1)
m – n = 12b + 6 ⇒ (m−n)2=144b2+144b+36 – – – (2)
(1) – (2) ⇒ 4mn = 144a2+192a−144b2−144b+28
mn = 36a2+48a−36b2−36b+7
Now mn is divided by 6, all the terms except 7 gives 0. So 7 divided by 6, remainder = 1
- There is a set of 36 distinct points on a plane with the following characteristics:
* There is a subset A consisting of fourteen collinear points.
* Any subset of three or more collinear points from the 36 are a subset of A.
How many distinct triangles with positive area can be formed with each of its vertices being one of the 36 points? (Two triangles are said to be distinct if at least one of the vertices is different)
The given data indicates that 14 points are collinear and remaining 22 points are non collinear.
A triangle can be formed by taking 1 points from 14 and 2 points from 22 (or) 2 points from 14 and 1 points from 22 (or) 3 points from 22
⇒ 14C1×22C2+14C2×22C1+22C3= 6776
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