#### Wipro NLTH Registration 2020 | Wipro NLTH Syllabus | Wipro NLTH Coding Questions | Wipro NLTH Aptitude Questions

**Wipro NLTH Registration 2020:** Elite National Level Talent Hunt 2021 (NLTH) is a fresher’s hiring initiative to attract the best of 2021 engineering talent across the country. The objective of this initiative is to enable an equal opportunity for employment to the most deserving talent across all engineering streams in India. An enthusiastic engineering candidate should not miss this opportunity to be part of an exciting journey with Wipro!

**Company: Wipro India**

**Website: Wipro**

**Wikipedia: Wipro Wiki**

**Name Of The Event: Elite National Level Talent Hunt 2021 (NLTH)**

**Position: Project Engineer**

**Degree Needed: B.E/B.Tech/M.E/M.Tech**

**Passout Year: 2021**

**Salary: Best In Industry**

**Job Location: Across India**

**Experience Level: 0 – 2 Years**

**Jobs By Location And Passout Year:**

Pune | Hyderabad | Delhi |

Mumbai | Coimbatore | Noida |

Bangalore | Chennai | Off Campus |

Nagpur | Gurgaon | Walk In |

2015 Batch | 2018 Batch | 2021 Batch |

2016 Batch | 2019 Batch | 2022 Batch |

2017 Batch | 2020 Batch | 2023 Batch |

**Wipro NLTH 2020 Syllabus:**

**There will be four round of Wipro NLTH 2020: **

**Aptitude Test**– Logical Ability, Quantitative Ability , English (verbal) Ability.

Duration: 48 mins**Written Communication Test**– Essay writing. Duration: 20 mins**Online Programming Test**– Two programs for coding.

Duration: 60 mins

Candidate can chose any one of programming language for coding test: Java, C, C++ or Python

**Wipro NLTH Question Paper**

**1. Wipro NLTH Aptitude Questions**

**1.**

**A student scores 55% marks in 8 papers of 100 marks each. He scores 15% of his total marks in English. How much does he score in English?**

a) 55**b) 66**

c) 77

d) 44

e) None of these **Answer: Option b**

Given that a student scores 55% marks in English in 8 papers of 100 marks each.

Hence, his total marks = (55 x 800)/100 = 440

15% of his 440 marks = 440*(15/100) = 66 = Marks scored in English.

**2.**

**A software engineer creates a LAN game where an 8 digit code made up of 1,2,3,4,5,6, 7,8 has to be decided on as a universal code. There is a condition that each number has to be used and no number can be repeated. What is the probability that the first 4 digits of the code are even number?****a) 1/70**

b) 1/840

c) 1/8

d) 1/40320**Answer: Option a**

Total no of possible ways are 8!

The total number of possible ways to find even numbers at the last 4 digits are 4!

The total number of possible ways to find odd numbers at the last 4 digits are 4!

So, the probability that the first 4 digits of the code are even number =4!*4!/8! = 1/70

**3.**

**A car is 250 meters behind the bus. The car and bus are moving at a speed of 60 km/hr and 35 km/hr respectively. In what time will the car be ahead of the bus by 250 meters.**

a)37 secs

b)48 secs**c)72 secs**

d)68 secs **Answer: Option c**

Relative speed = 60-35 = 25 Km/hr.

For the car to be ahead of the bus by 250 meters, it needs to cover = 250+250 =500m = 0.5kms.

Time taken to cover 0.5 km = 0.5/25 =1/50 hrs = 72 secs

**4.**

**Square of 2 more than a 2 digit number is multiplied and % by 2 and 5 respectively. If twice of the result is equal to 500 then find the number.**

a) 12 days**b) 24 days**

c) 36 days

d) 48 days**Answer:Option b**

A’s 1 day work = 1/30.

A’s 3 day work = 3/30

(A+B+C) 1 day work = (1/30+1/45+1/90) = 1/15

4 days work = 3/30+1/15 = 1/6.

Total work will be done in= 4*6 = 24 days

**5.**

**A starts a business with Rs.3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2:3. What is Bs contribution to the Capital?**

a) 5000

b) 7500**c) 9000**

d)9500**Answer:Option c**

Let x be the capital invested by B.

Profit will be divided in the ratio of 2:3 = (12*3500) : 7x = 6000 : x.

2/3 = 6000/x

x = 9000

**6.**

**What will be the value of x^(1/2).x^(1/4).x^(1/8)…………to infinity.**

a) x^2

b) x**c) x^(3/2)**

d) x^3 **Answer:Option c**

The term is x ^ (1/2 + 1/4 + 1/8 + …). Sum of infinite series with first term 1/2 and common ratio of 1/2 is 1. So the answer is ‘x’.

**7.**

**For the word SURITI, if you arrange the letters in dictionary order then what is its rank?**

a) 234

b) 235**c) 236**

d) 237**Answer:Option c**

I = 5! =120

R= 5!/2 =60

SI = 4! =24

SR = 4!/2 = 12

ST = 4!/2 = 12

SUI = 3! = 6

= 234 Words

So, words with SUR begins from 235. SURIIT is the 235th word and SURITI is the 236rd word.

**8.**

**There are two sections in a question paper each contains five questions. A student has to answer 6 questions. The maximum number of questions that can be answered from any section is 4. How many ways he can attempt the paper?**

a) 50

b) 100

c) 120**d) 200**

**Answer:Option d**

Possible ways in which he can attempt 6 questions are

5C4*5C2 = 50

5C3*5C3 = 100

5C2*5C4 = 50

50+100+50 = 200

**9.**

**If 5 men take an hour to dig a ditch, then how long should 12 men take a dig to the ditch of the same type?****a) 25 min**

b) 30 min

c) 28 min

d) 20 min **Answer:Option a**

M1H1 = M2H2

5*60 = 12*H2

H2 = 25 min

**10.**

**Sameer plants 7225 plants so that there are as many rows as there are trees in a row. How many trees are there in a row?**

(a) 75

(b) 95**(c) 85**

(d) 65

2. Wipro NLTH Verbal Questions

2. Wipro NLTH Verbal Questions

**1.**

**In the question, a part of the sentence is italicized. Alternatives to the italicized part are given which may improve the sentence. Choose the correct alternative. In case no improvement is needed.**

She *gave *most of her time to music.

a) Spent

b) lent**c) devoted**

d) No improvement

**2.**

**Select the word or phrase that is closest in meaning to the given word – NONCHALANCE**

a) Neutrality**b) Indifference**

c) All-knowing

d) Ignorance

e) timeliness

**3.**

**Find the synonym of the word – PERSISTENT**

a) Armageddon

b) Flagrant**c) Tenacious**

d) Inconsistent

**4.**

**Choose the correct alternative****Flow : River :: Stagnant : ?**

a) rain

b) stream**c) pool**

d) canal

**5.**

**GIGANTIC (opposite)**

a) Huge

b) Invisible

c) Zero**d) Tiny**

**6.**

**Archaic (opposite)****a) Fresh**

b) Modern

c) Ancient

d) Present

**7.**

**Castigate (opposite)**

a) Trape

b) Discard**c) Complement**

d) Berate

**8.**

**Credulous (Meaning)**a) Credible

b) Discipline**c) Gullible**

d) Incredible

**9.**

**Unyoke (Meaning)**a) Merge

b) Amalgamate**c) Split**

d) Federate

**10.**

**Altercation (Meaning)**

a) flexibility**b) animosity**

c) incompatibility

d) Concord

**3. Wipro NLTH Coding Questions****1. Program to find the GCD or HCF of two numbers**

Program:

include

include

int main()

{

int a,b,gcd;

printf(“\nEnter two numbers : “);

scanf(“%d %d”,&a,&b);

int i;

for(i = 1; i <= a && i <= b; i++)

{

if((a % i == 0) && (b % i == 0))

{

gcd = i;

}

}

printf(“\nGCD of %d and %d is %d “,a,b,gcd);

printf(“\n”);

return 0;

}**Output:**

Input- Enter two numbers:20 28 Output- GCD of 4

**2.**

**Program to check whether the number is Armstrong or not**Program:

include

include

int main()

{

int number, temp, remainder, result = 0, n = 0 ;

printf(“Enter an integer: “);

scanf(“%d”, &number);

temp = number;

while (temp != 0)

{

temp /= 10;

++n;

}

temp = number;

while (temp != 0)

{

remainder = temp%10;

result += pow(remainder, n);

temp /= 10;

}

if(result == number)

printf(“%d is an Armstrong number\n”, number);

else

printf(“%d is not an Armstrong number\n”, number);

return 0;

}

**3.**

**Program to convert a number from decimal to binary using***for*loop

Program:

include

long int decimal_to_binary(int n)

{

long int binary = 0;

int remainder, i, flag = 1;

for(i = 1; n != 0; i = i * 10)

{

remainder = n % 2;

n /= 2;

binary += remainder * i;

}

return binary;

}

int main()

{

int n;

printf(“Enter a decimal number: “);

scanf(“%d”, &n);

printf(“Equivalent binary number: %d\n”, decimal_to_binary(n));

return 0;

}

Output:

Input- Enter a decimal number:255

Output- Equivalent binary number: 11111111

**4. Program to Reverse a Number in C**

Program:

include

int main()

{

int n, rev = 0, rem;

printf(“\nEnter a number : “);

scanf(“%d”, &n);

printf(“\nReversed Number : “);

while(n != 0)

{

rem = n%10;

rev = rev*10 + rem;

n /= 10;

}

printf(“%d\n”, rev);

return 0;

}

Output:

Enter a number : 12345

Reversed Number : 54321

**5.**

**Program to remove vowels from a string**Program:

include

int check_vowel(char);

int main()

{

char s[100], t[100];

int c, d = 0;

gets(s);

for(c = 0; s[c] != ‘\0’; c++)

{

if(check_vowel(s[c]) == 0)

{

t[d] = s[c];

d++;

}

}

t[d] = ‘\0’;

strcpy(s, t);

printf(“%s\n”, s);

return 0;

}

int check_vowel(char ch)

{

if (ch == ‘a’ || ch == ‘A’ || ch == ‘e’ || ch == ‘E’ || ch == ‘i’ || ch == ‘I’ || ch ==’o’ || ch==’O’ || ch == ‘u’ || ch == ‘U’)

return 1;

else

return 0;

}**Output:**

Input- qwertyuiop Output- qwrtyp

**6. Program to convert a number from binary to octal**

Program:

include

include

int binary_to_octal(long int binary)

{

int octal = 0, decimal = 0, i = 0;

while(binary != 0)

{

decimal += (binary%10) * pow(2,i);

++i;

binary/=10;

}

i = 1;

while (decimal != 0)

{

octal += (decimal % 8) * i;

decimal /= 8;

i *= 10;

}

return octal;

}

int main()

{

long int binary;

printf(“\nEnter a binary number: “);

scanf(“%lld”, &binary);

printf(“\nOctal Equivalent : %d\n”, binary_to_octal(binary));

return 0;

}

Output:

Input- Enter a binary number:1111 Output- Octal Equivalent :17

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