## HUAWEI Written Test: Aptitude, Technical; Questions and Answers with Explanations PDFs

Please Like and/ Share and/ G+1. Hi, welcome to nrrbeassistance.blogspot.com. Dear viewer, especially for you, here is the HUAWEI Written Test: Aptitude, Reasoning, English, Technical; Questions and Answers with Explanations.

1. B2CD, _____, BCD4, B5CD, BC6D        A.B2C2D        B.BC3D        C.B2C3D        D.BCD7

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

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

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

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

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

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

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: D9. 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        A.Error: invalid structure member    B.Error: Floating point formats not linked    C.No error    D.None of above

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?<p> </p>    #include int main()    {        struct emp        {            char name[20];            float sal;        };        struct emp e[10];        int i;        for(i=0; i        A.Error: invalid structure member    B.Error: Floating point formats not linked    C.No error    D.None of above

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

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

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

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

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 : 6014. 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

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

Explanation:     Clearly, there are 52 cards, out of which there are 12 face cards. P (getting a face card)=12/52=3/1316. 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

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

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/ 9118. 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

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

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

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.

21. What will be the output of the program ?<p> </p>    #include int main()    {        char p[] = "%d\n";        p[1] = 'c';        printf(p, 65);        return 0;    }        A.A        B.a        C.c        D.65

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 ?<p> </p>    #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

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<p> </p>    #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<p> </p>    #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

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?<p> </p>    #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

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?<p> </p>    #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

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?<p> </p>    #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

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?<p> </p>    #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

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 ?<p> </p>    #include    #include    int main()    {    printf("%d\n", strlen("123456"));    return 0;    }        A.6        B.12        C.7        D.2

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 ?<p> </p>    #include    int main()    {        int arr[5], i=0;        while(i        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,

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.

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