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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.

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

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

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

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

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’.

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.

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

a) 25 min
b) 30 min
c) 28 min
d) 20 min

Answer:Option a

M1H1 = M2H2
5*60 = 12*H2
H2 = 25 min

(a) 75
(b) 95
(c) 85
(d) 65

She gave most of her time to music.
a) Spent
b) lent
c) devoted
d) No improvement

a) Neutrality
b) Indifference
c) All-knowing
d) Ignorance
e) timeliness

a) Armageddon
b) Flagrant
c) Tenacious
d) Inconsistent

a) rain
b) stream
c) pool
d) canal

a) Huge
b) Invisible
c) Zero
d) Tiny

a) Fresh
b) Modern
c) Ancient
d) Present

a) Trape
b) Discard
c) Complement
d) Berate

a) Credible
b) Discipline
c) Gullible
d) Incredible

a) Merge
b) Amalgamate
c) Split
d) Federate

a) flexibility
b) animosity
c) incompatibility
d) Concord

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

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;
}

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

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

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

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