Huawei Placement Papers 2016-2017 - Interview Questions

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Huawei Placement Papers
Placement papers of Huawei 2016-2017. Learn and practice the placement papers and interview questions answers of Huawei and find out how much you score before you appear for your next interview and written test.

Huawei Placement Papers - Interview Questions:


1. B2CD, _____, BCD4, B5CD, BC6D

   
    A.B2C2D        B.BC3D        C.B2C3D        D.BCD7

    Answer: B

    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.

2. DEF, DEF2, DE2F2, _____, D2E2F3
   
    A.DEF3        B.D3EF3        C.D2E3F        D.D2E2F2

    Answer: D

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

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

    Answer: C

    Explanation:
    1 woman's 1 day's work=1/70 1 child's 1 day's work =1/140 (5 women + 10 children)'s day's work =5/70 +10/140=1/14 +1/14=1/7 5 women and 10 children will complete the work in 7 days.

4. 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. 375    B.Rs. 400    C.Rs. 600        D.Rs. 800

    Answer: B

    Explanation:
    C's 1 day's work = 1/3 - [1/6 + 1/8]= 1/3 - 7/24= 1/24 A's wages : B's wages : C's wages = 1/6: 1/8: 1/24= 4 : 3 : 1. C's share (for 3 days) = Rs.[3 x 1/24 x 3200] = Rs. 400.

5. 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.380        B.395        C.400        D.425

    Answer: C

    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.

6. The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?
   
    A.25% increase        B.50% increase        C.50% decrease        D.75% decrease

    Answer: B

    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%

7. 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 m2        B.2420 m2        C.2480 m2        D.2520 m2

    Answer: D

    Explanation:
    We 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.

8. 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.72        B.80        C.120        D.100

    Answer: D

9. 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 member    B.Error: Floating point formats not linked    C.No error    D.None of above

    Answer: B

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

10. 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 member    B.Error: Floating point formats not linked    C.No error    D.None of above

    Answer: B

    Explanation:
    At 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


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11. The fourth proportional to 5, 8, 15 is:
   
    A.18        B.24        C.19        D.20

    Answer: B

    Explanation:
    Let the fourth proportional to 5, 8, 15 be x. Then, 5 : 8 : 15 : x 5x = (8 x 15) x=(8*15)/5=24

12. 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 : 5        B.3 : 7        C.5 : 3        D.7 : 3

    Answer: C

    Explanation:
    Let 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.

13. 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 : 10    B.10 : 11 : 20    C.23:33:60    D.32 :43:53

    Answer: C

    Explanation:
    Let A = 2k, B = 3k and C = 5k. A's new salary = 115/ 100 of 2k =[115/ 100 x 2k]= 23k/ 10 B's new salary = 110/ 100 of 3k = [110/ 100x 3k] = 33k/ 10 C's new salary = 120/ 100 of 5k = [120/ 100 x 5k] = 6k New ratio = 23k : 33k :         6k = 23 : 33 : 60

14. 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.3/4        B.4/7        C.1/8        D.3/7

    Answer: B

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

15. 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.1/13        B.3/13        C.1/4        D.9/5

    Answer: B

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

16. 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 /20        B.29/ 34        C.47/ 100    D.13 /102

    Answer: D

    Explanation:
    Let 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

17. 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/ 22        B.3/ 22        C.2/ 91        D.2/ 81

    Answer: C

    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

18. 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 /13        B.2 /13        C.1 /26        D.1 /52

    Answer: C

    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

19. 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.27        B.33        C.49        D.55

    Answer: B

    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.

20. 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.50        B.100        C.150        D.200

    Answer: C

    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.


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21. What will be the output of the program ?

    #include int main()
    {
        char p[] = "%d\n";
        p[1] = 'c';
        printf(p, 65);
        return 0;
    }

   
    A.A        B.a        C.c        D.65

    Answer: A

    Explanation:
    Step 1: char p[] = "%d\n"; The variable p is declared as an array of characters and initialized with string "%d" Step 2: p[1] = 'c'; Here, we overwrite the second element of array p by 'c'. So array p becomes "%c". Step 3: printf(p, 65);         becomes printf("%c", 65); Therefore it prints the ASCII value of 65. The output is 'A'.

22. What will be the output of the program ?

    #include
    #include
    int main()
    {
        char str1[20] = "Hello", str2[20] = " World";
        printf("%s\n", strcpy(str2, strcat(str1, str2)));
        return 0;
    }

   
    A.Hello        B.World        C.Hello World        D.WorldHello

    Answer: C

    Explanation:
    Step 1: char str1[20] = "Hello", str2[20] = " World"; The variable str1and str2 is declared as an array of characters and initialized with value "Hello" and " World" respectively. Step 2: printf("%s\n", strcpy(str2, strcat(str1, str2))); => strcat            (str1, str2)) it append the string str2 to str1. The result will be stored in str1. Therefore str1 contains "Hello World". => strcpy(str2, "Hello World") it copies the "Hello World" to the variablestr2. Hence it prints "Hello World".

23. Point out the error in the program

    #include int main()
    {
        int i;
        #if A
        printf("Enter any number:");
        scanf("%d", &i);
        #elif B
        printf("The number is odd");
        return 0;
    }

   
    A.Error: unexpected end of file because there is no matching #endif        B.The number is odd    C.Garbage values        D.None of above
   
    Answer: A

    Explanation:
    The conditional macro #if must have an #endif. In this program there is no#endif statement written.

24. Point out the error in the program

    #include
    #define SI(p, n, r) float si; si=p*n*r/100;
    int main()
    {
        float p=2500, r=3.5;
        int n=3;
        SI(p, n, r);
        SI(1500, 2, 2.5);
        return 0;
    }

   
    A.26250.00 7500.00    B.Nothing will print        C.Error: Multiple declaration of si    D.Garbage values

    Answer: C

    Explanation:
    The macro #define SI(p, n, r) float si; si=p*n*r/100; contains the error. To remove this error, we have to modify this macro to #define SI(p,n,r) p*n*r/100

25. What will be the output of the program?

    #include int main()
    {
        float d=2.25;
        printf("%e,", d);
        printf("%f,", d);
        printf("%g,", d);
        printf("%lf", d);
        return 0;
    }

   
    A.2.2, 2.50, 2.50, 2.5    B.2.2e, 2.25f, 2.00, 2.25        C.2.250000e+000, 2.250000, 2.25, 2.250000        D.error

    Answer: C

    Explanation:
    printf("%e,", d); Here '%e' specifies the "Scientific Notation" format. So, it prints the 2.25 as 2.250000e+000. printf("%f,", d); Here '%f' specifies the "Decimal Floating Point" format. So, it prints the 2.25 as 2.250000. printf("%g,", d); Here         '%g' "Use the shorter of %e or %f". So, it prints the 2.25 as 2.25. printf("%lf,", d); Here '%lf' specifies the "Long Double" format. So, it prints the 2.25 as 2.250000.

26. What will be the output of the program?

    #include
    #include
    int main()
    {
        float n=1.54;
        printf("%f, %f\n", ceil(n), floor(n));
        return 0;
    }

   
    A.2.000000, 1.000000    B.1.500000, 1.500000    C.1.550000, 2.000000    D.1.000000, 2.000000

    Answer: A

    Explanation:
    ceil(x) round up the given value. It finds the smallest integer not < x. floor(x) round down the given value. It finds the smallest integer not > x. printf("%f, %f\n", ceil(n), floor(n)); In this line ceil(1.54) round up the 1.54 to 2 and floor(1.54)         round down the 1.54 to 1. In the printf("%f, %f\n", ceil(n), floor(n)); statement, the format specifier "%f %f" tells output to be float value. Hence it prints 2.000000 and 1.000000.

27. What will be the output of the program?

    #include
    int main()
    {
        float a=0.7;
        if(a < 0.7f)
        printf("C\n");
        else
        printf("C++\n");
        return 0;
    }
   

    A.C        B.C++        C.Compiler error        D.Non of above

    Answer: B

    Explanation:
    if(a < 0.7f) here a is a float variable and 0.7f is a float constant. The float variable a is not less than 0.7f float constant. But both are equal. Hence the ifcondition is failed and it goes to else it prints 'C++'

28. What will be the output of the program?

    #include
    int main()
    {
        float f=43.20;
        printf("%e, ", f);
        printf("%f, ", f);
        printf("%g", f);
        return 0;
    }

   
    A.4.320000e+01, 43.200001, 43.2        B.4.3, 43.22, 43.21    C.4.3e, 43.20f, 43.00        D.Error

    Answer: A

    Explanation:
    printf("%e, ", f); Here '%e' specifies the "Scientific Notation" format. So, it prints the 43.20 as 4.320000e+01. printf("%f, ", f); Here '%f' specifies the "Decimal Floating Point" format. So, it prints the 43.20 as 43.200001. printf("%g, ", f);     Here '%g' "Use the shorter of %e or %f". So, it prints the 43.20 as 43.2.

29. If the size of an integer is 4 bytes, What will be the output of the program ?

    #include
    #include
    int main()
    {
    printf("%d\n", strlen("123456"));
    return 0;
    }
   

    A.6        B.12        C.7        D.2

    Answer: A

    Explanation:
    The function strlen returns the number of characters int the given string. Therefore, strlen("123456") contains 6 characters. The output of the program is "6".

30. What will be the output of the program ?

    #include
    int main()
    {
        int arr[5], i=0;
        while(i<5)
        arr[i]=++i;
        for(i=0; i<5; i++)
        printf("%d, ", arr[i]);
        return 0;
    }

   
    A.1, 2, 3, 4, 5,    B.Garbage value, 1, 2, 3, 4,        C.0, 1, 2, 3, 4,        D.2, 3, 4, 5, 6,

    Answer: B

    Explanation:
    Since C is a compiler dependent language, it may give different outputs at different platforms. We have given the TurboC Compiler (Windows) output. Please try the above programs in Windows (Turbo-C Compiler) and Linux (GCC Compiler),     you will understand the difference better.



Huawei Placement Paper Syllabus:

Quantitative Aptitude Syllabus:

  • Probability
  • Permutations & Combinations
  • Algebra
  • Averages
  • Time Speed & Distance
  • Time & Work
  • Profit & Loss
  • Ratio & Proportion
  • Simple & Compound Interest
  • Percentage
  • Number Series
  • Mixtures & Alligations
  • Simplification
  • Number System
  • Heights and Distances
  • Geometry & Mensuration
  • Data Sufficiency
  • Logarithms
  • Progressions
  • LCM and HCL
  • Pipes and Cisterns
  • Partnership
  • Boats and Streams
  • Areas, Volumes

Reasoning Syllabus:

  • Number Series
  • Letter Series
  • Anologies
  • Puzzles
  • Syllogisms
  • Binary Logic
  • Clocks & Calendars
  • Cubes & Dice
  • Classification 
  • Blood Relations
  • Coding-Decoding
  • Data Sufficiency
  • Seating Arrangement
  • Venn Diagrams
  • Problem Solving
  • Coded Inequalities
  • Double Lineup
  • Logical Deductions
  • Routes & Networks
  • Grouping & Selections
  • Evaluating Course of Action
  • Statements and Conclusions
  • Mathematical and Computer Operations
  • Critical Reasoning
  • Inferences
  • Situation Reaction Test
  • Decision Making
  • Symbols and Notations
  • Direction Sense Test
  • Logical Sequence Of Words
  • Assertion and Reason
  • Verification of Truth of the Statement
  • Statements and Assumptions
  • Data Interpretation

Verbal Ability Syllabus:

  • Synonyms
  • Antonyms
  • Sentence Completion
  • Spelling Test
  • Passage Completion
  • Sentence Arrangement
  • Idioms and Phrases
  • Para Completion
  • Error Correction (Underlined Part)
  • Fill in the blanks
  • Synonyms
  • Prepositions
  • Active and Passive Voice
  • Spotting Errors
  • Substitution
  • Transformation
  • Sentence Improvement
  • Joining Sentences
  • Error Correction (Phrase in Bold)
  • Articles
  • Gerunds
  • Identify the Errors
  • Plural Forms
  • Odd Words
  • Prepositions
  • Suffix
  • Tense
  • Adjectives
  • Homophones
  • Identify the Sentences
  • Nouns
  • Prefix
  • Sentence Pattern
  • Tag Questions
Dear readers, the syllabus and Huawei placement papers provided here are just for information purpose only. Sometimes companies may change their syllabus and exam pattern. So Please check official company website for latest syllabus.



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