## Huawei Placement Papers

Huawei Verbal Ability Questions

Huawei previous years Technical questions

Huawei Placement Paper August 03, 2006

Huawei Written Test Paper 2015

Huawei Placement Questions 2014

Huawei Placement Paper 2012

Huawei Aptitude Questions

Huawei Written Exam January 01, 2005

Huawei Written Test 2013

Huawei Placement paper 2011

Huawei Arithmetic Ability

## Placement Papers for All Companies

3i infotech

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CGI Group

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Cisco system

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CMC Limited

CMS

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Emphasis

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GlobalEdge

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Google

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Headstrong

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KPIT

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Persistent

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ZTE

# Huawei Placement Paper

Posted on :15-02-2016

**Q1. B2CD, _____, BCD4, B5CD, BC6D**

A. B2C2DB. BC3DC. B2C3DD. BCD7

**B**

__ANS:__:Explanation

Because the letters are the same, concentrate on the number series, which is a simple 2, 3, 4, 5, 6 series, and follows each letter in order.

**Q2. DEF, DEF2, DE2F2, _____, D2E2F3**

A. DEF3B. D3EF3C. D2E3FD. D2E2F2

**D**

__ANS:__:Explanation

*In this series, the letters remain the same: DEF. The subscript numbers follow this series: 111, 112, 122, 222, 223, 233, 333, ...*

**Q3. 10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?**

A. 3B. 5C. 7D. Cannot be determined

**C**

__ANS:__:Explanation

1 womans 1 days work=1/70 1 childs 1 days work =1/140 (5 women + 10 children)s days work =5/70 +10/140=1/14 +1/14=1/7 5 women and 10 children will complete the work in 7 days.

**Q4. A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?**

A. Rs. 375B. Rs. 400C. Rs. 600D. Rs. 800

**B**

__ANS:__:Explanation

Cs 1 days work = 1/3 - [1/6 + 1/8]= 1/3 - 7/24= 1/24 As wages : Bs wages : Cs wages = 1/6: 1/8: 1/24= 4 : 3 : 1. Cs share (for 3 days) = Rs.[3 x 1/24 x 3200] = Rs. 400.

**Q5. The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:**

A. 380B. 395C. 400D. 425

**C**

__ANS:__:Explanation

Let the numbers be x and y. Then, xy = 9375 and x/y=15 Xy/(x/y)=9375/15 y2 = 625. y = 25. x = 15y = (15 x 25) = 375. Sum of the numbers = x + y = 375 + 25 = 400.

**Q6. The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?**

A.25% increaseB.50% increaseC.50% decreaseD.75% decrease

**B**

__ANS:__:Explanation

Let original length = x and original breadth = y. Original area = xy. New length =x/2 New breadth = 3y. New area=[(x/2)*3y]=(3/2)xy Increase % =[(1/2)xy*(1/xy)*100]%=50%

**Q7. The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:**

A.1520 m2B.2420 m2C.2480 m2D.2520 m2

**D**

__ANS:__:ExplanationWe have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103. Solving the two equations, we get: l = 63 and b = 40. Area = (l x b) = (63 x 40) m2 = 2520 m2.

**Q8. In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years age which is 48. What is the total number of students in the school?**

A.72B. 80C.120D.100

**D**

__ANS:__**Q9. Point out the error in the program?**

#include int main(){struct emp{char name[20];float sal;};struct emp e[10];int i;for(i=0; i<=9; i++)scanf("%s %f", e[i].name, &e[i].sal);return 0;}

A. Error: invalid structure memberB.Error: Floating point formats not linkedC.No errorD.None of above

**B**

__ANS:__:Explanation

At run time it will show an error then program will be terminated. Sample output: Turbo C (Windows) Sample 12.123 scanf : floating point formats not linked Abnormal program termination

**Q10. Point out the error in the program?**

#include int main(){struct emp{char name[20];float sal;};struct emp e[10];int i;for(i=0; i<=9; i++)scanf("%s %f", e[i].name, &e[i].sal);return 0;}

A.Error: invalid structure memberB.Error: Floating point formats not linkedC.No errorD.None of above

**B**

__ANS:__:ExplanationAt run time it will show an error then program will be terminated. Sample output: Turbo C (Windwos) c:>myprogram Sample 12.123 scanf : floating point formats not linked Abnormal program termination

**Q11. The fourth proportional to 5, 8, 15 is:**

A.18B.24C.19D.20

**B**

__ANS:__:ExplanationLet the fourth proportional to 5, 8, 15 be x. Then, 5 : 8 : 15 : x 5x = (8 x 15) x=(8*15)/5=24

**Q12. If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number?**

A. 2 : 5B. 3 : 7C. 5 : 3D. 7 : 3

**C**

__ANS:__:ExplanationLet 40% of A=2/3 B Then,40 A/100=2B/3 => 2A/5=2B/3 => A/B-[2/3 - 5/2]=5/3p A : B = 5 : 3.

**Q13. The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?**

A. 3 : 3 : 10B. 10 : 11 : 20C. 23:33:60D. 32 :43:53

**C**

__ANS:__:ExplanationLet A = 2k, B = 3k and C = 5k. As new salary = 115/ 100 of 2k =[115/ 100 x 2k]= 23k/ 10 Bs new salary = 110/ 100 of 3k = [110/ 100x 3k] = 33k/ 10 Cs new salary = 120/ 100 of 5k = [120/ 100 x 5k] = 6k New ratio = 23k : 33k : 6k = 23 : 33 : 60

**Q14. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?**

A.3/4B.4/7C.1/8D.3/7

**B**

__ANS:__:Explanation

Let number of balls = (6 + 8) = 14. Number of white balls = 8. P (drawing a white ball)=8/14=4/7

**Q15. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card?**

A.1/13B.3/13C.1/4D.9/5

**B**

__ANS:__:ExplanationClearly, there are 52 cards, out of which there are 12 face cards. P (getting a face card)=12/52=3/13

**Q16. Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:**

A.3 /20B.29/ 34C.47/ 100D.13 /102

**D**

__ANS:__:ExplanationLet S be the sample space. Then, n(S) = 52C2 = 52 * 51/ (2*1)= 1326. Let E = event of getting 1 spade and 1 heart. n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = (13C1 x 13C1) = (13 x 13) = 169. P(E) =n (E)/n(S) = 169/ 1326= 13/102

**Q17. A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:**

A.1/ 22B.3/ 22C.2/ 91D.2/ 81

**C**

__ANS:__Explanation:Let S be the sample space. Then, n(S)= number of ways of drawing 3 balls out of 15 = 15C3 = (15 x 14 x 13)/ (3 x 2 x 1) = 455. Let E = event of getting all the 3 red balls. n(E) = 5C3 = 5C2 = (5 x 4)/ (2 x 1)= 10. P(E) = n(E) / n(S)= 10/ 455 = 2/ 91

**Q18. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:**

A.1 /13B.2 /13C.1 /26D.1 /52

**C**

__ANS:__Explanation:Here, n(S) = 52. Let E = event of getting a queen of club or a king of heart. Then, n(E) = 2. P(E) =n(E)/ n(S)= 2/ 52= 1/ 26

**Q19. Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:**

A.27B.33C.49D.55

**B**

__ANS:__Explanation:Let the numbers be 3x and 5x. Then ,(3x-9)/(5x-9)=12/13 23(3x - 9) = 12(5x - 9) 9x = 99 x = 11. The smaller number = (3 x 11) = 33.

**Q20. In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?**

A.50B.100C.150D.200

**C**

__ANS:__Explanation:Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively. Then, sum of their values=Rs[(25x/100)+(10*2x)/100+(5*3x)/100]=Rs 60x/100 60x/100=30 =>x=(30*100)/60=50 Hence, the number of 5 p coins = (3 x 50) = 150.

## Huawei Placement Papers

➤ Huawei Verbal Ability Questions

➤ Huawei previous years Technical questions

➤ Huawei Placement Paper August 03, 2006

➤ Huawei Written Test Paper 2015

➤ Huawei Placement Questions 2014

➤ Huawei Placement Paper 2012

➤ Huawei Aptitude Questions

➤ Huawei Written Exam January 01, 2005

➤ Huawei Written Test 2013

➤ Huawei Placement paper 2011

➤ Huawei Arithmetic Ability

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