# Huawei Placement Paper

Posted on :15-02-2016

Q1. B2CD, _____, BCD4, B5CD, BC6D

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

ANS: 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.

Q2. DEF, DEF2, DE2F2, _____, D2E2F3

A. DEF3
B. D3EF3
C. D2E3F
D. D2E2F2

ANS: D

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

ANS: C

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

ANS: B

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. 380
B. 395
C. 400
D. 425

ANS: 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.

Q6. 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

ANS: 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%

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

ANS: 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.

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.72
B. 80
C.120
D.100

ANS: D

Q9. Point out the error in the program?

#include int main()
{
struct emp
{
char name;
float sal;
};
struct emp e;
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

ANS: B

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;
float sal;
};
struct emp e;
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

ANS: 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

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

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

ANS: 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

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

ANS: 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.

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

ANS: C

Explanation:

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

ANS: B

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

ANS: B

Explanation:

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

ANS: 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

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

ANS: 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

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

ANS: 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

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.27
B.33
C.49
D.55

ANS: 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.

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.50
B.100
C.150
D.200

ANS: 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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