HUAWEI Written Test: Aptitude, Technical; Questions and Answers with Explanations PDFs
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: 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
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?<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
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
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 : 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
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=3/1316. 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
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/ 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
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.
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
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 ?<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
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<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
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?<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
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?<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
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?<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
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?<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
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 ?<p> </p> #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 ?<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,
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.
Link: Download PDF | Learn More
Updates: Don’t miss, what you want? So kindly, 1. Like our Facebook, 2. Subscribe by E-mail, 3. Follow our Google+, 4. Twitter 5. YouTube, 6. Download our Toolbar, 7. Install our App and 8. Others.
Help: If you have any suggestions, requests and problems, please comment below or ask us here.
Help: If you have any suggestions, requests and problems, please comment below or ask us here.
Alerts: Dear viewer, “Hard work never fails”. Along with hard work include smart work for your goal. Your hard work and our smart work, guarantee success. To get in touch with you, we have created a WhatsApp Group named “Question Answer Explanation”. Kindly send a message to +91-7893356131 as “add to Question Answer Explanation" to add you to this group.
Encourage us: Please don’t forget to G+1 and/ Like and/ Share and/ Recommend… thank you and all the best / best of luck for your bright future.
Our motto: Serve poor and needy whole heartedly – Nareddula Rajeev Reddy (NRR) and his team.
Comments
Post a Comment
* Dear viewer, ask us what you want..?