@IIM-A2013
Assuming weight of bucket = b, capacity of bucket = w (b and w are integers)
So, b + w/2 = 4b/3 => w = 2b/3 => b is divisible by 3
Taking b = 3x and w = 2x (x is an integer)
Take y = 50% of Weight of bucket when it is filled to capacity
= 1/2 * (b + w)
= (1/2) * (3x + 2x)
= (1/2) * 5x
= 5x/2
=> x = 2y/5. For x to be an integer, y has to be divisible by 5 => (b) 5kg

When the sun ray's inclination from 30 degree to 60 degree,the length of the shadow of a tower decreases by 60m..Find the height of the tower..


Statistics for road use in a certain county show that in the past year, there were 32
accidents per 100,000 miles driven on rural roads and 18 accidents per 100,000 miles
driven on city roads. Combined statistics for both rural and city roads show that there
were 24 accidents per 100,000 miles driven. Let x be the total number of accidents on
rural roads and y be the total number of accidents on city roads. The value of x/y is
(a) 3/2 (b) 4/3 (c) 5/4 (d) 5/3 (e) 7/5

I don't have the OA for both..

x/r = 32
y/c = 18
32r+18c = 24r+24c
8r = 6c
r/c = 3/4
x/y * 18/32 = 3/4 => x/y = 3*32 / 4*18 = 4/3. Is this correct?

I just groaned you for this post , was my first groan ( and could have been yours too), but took it back just because we are peer at this thread. Relax Jatin ain't that big thing.


The value of 1.1!+2.2!+3.3!+--------+n.n! is
(a) (n+1)!
(b) (n+1)!+1
(c) (n+1)!-1
(d) none of these

n*n! = (n+1-1) * n! = (n+1)! - n!
So, sigma (n*n!)
= sigma ((n+1)!) - sigma(n!)
= (n+1)! - 1!

OA :

option A
An equivalence relation is reflexive, symmetric and
transitive.

Suja

I get that. I did not read the question correctly, so I pointed out the 'correct' statement instead of the incorrect statement.
How is 'R union S' an equivalence relation?

slammed

If R = {(1, 1), (2, 2), (1, 2), (2, 1), (3, 3)} and
S = {(1, 1), (2, 2), (2, 3), (3, 2), (3, 3)} are two relations in the
set X = {1, 2, 3}, the incorrect statement is:
(A) R and S are both equivalence relations
(B) R S is an equivalence relations
(C) R^1 S^1 is an equivalence relations
(D) R S is an equivalence relations

This would be R intersection S as an equivalence relation (If I've got the equivalence relation definition correct). Right?

Three circles A, B and C have a common centre O.
A is the inner circle, B middle circle and C is outer circle. The
radius of the outer circle C, OP cuts the inner circle at X and
middle circle at Y such that OX = XY = YP. The ratio of the
area of the region between the inner and middle circles to
the area of the region between the middle and outer circle
is:
(A)
1/
3 (B)
2/
5
(C)
3/
5 (D)
1/
5

Suja

The Big Bang Theory

OX = XY = YP
=> a = b-a = c-b
=> b = 2a & c = a + b = 3a

Now,
Answer = pi* (4-1)*a^2 / pi* (9-4) * a^2 = 3/5

Itna sannata!! :lookround::lookround:
In an examination, the average marks obtained by
students who passed was x%, while the average of those
who failed was y%. The average marks of all students taking
the exam was z%. Find in terms of x, y and z, the percentage
of students taking the exam who failed.

p*x + f*y = (p+f)*z
=> p(x-z) = f(z-y)
=> p/f = (z-y)/(x-z)
=> (p+f)/f = (z-y+x-z)/(x-z)
= (x-y)/(x-z)
=> f/(p+f) = (x-z)/(x-y)

Happy Birthday Shivani. Have a great day!

N should be expressed only as the product of its prime factors to calculate the number of factors.
N = 45^2 = 3^4*5^2
number of factors = 5*3 = 15 not a prime number.
So, N cannot be 45^2. like wise for many other numbers that you have considered.
m+1 = 3 => m = 2 => N = 2^2, 3^2, 5^2, 7^2, 11^2, 13^2, 17^2, 19^2, 23^2, 29^2, 31^2, 37^2, 41^2, 43^2 => 14 numbers
m+1 = 5 => m = 4 => N = 2^4,3^4,5^4 => 3 numbers
m+1 = 7 => m = 6 => N = 2^6 , 3^6 => 2 numbers
m+1 = 11 => m = 10 => N = 2^10 => 1 numbers

My bad. I got carried away. Here is the modified solution:

If a number expressed as a product of its prime factors is 2^m * 3^n....., then the number of factors is (m+1)(n+1)...
=> if the number of factors is prime, then (m+1)(n+1)... is prime
=> There is only one prime factor of the number with a power m such that:
(m+1) is prime
m+1 = 2 => m = 1 => N is prime. Do not consider as we have to count only composite numbers.
m+1 = 3 => m = 2 => N = 2^2 to 45^2 (in the given range)
Considering only the primes from 2 to 45: 2,3,5,7,11,13,17,19,23,29,31,37,41,43 => 14 numbers.

m+1 = 5 => m = 4 => N = 2^4 to 6^4 (in the given range)
Considering only the primes from 2 to 6: 2,3,5 => 3 numbers

m+1 = 7 => m = 6 => N = 2^6 to 3^6 (in the given range) => 2 numbers
m+1 = 11 => m = 10 => N = 2^10 (in the given range) => 1 numbers

Total = 14 + 3 + 2 + 1 = 20 numbers. I hope this is correct.

How many composite integers from 1 to 2100 have a prime number of divisors (positive integers)?


my take :-any composite number can be represented as product if its prime factors:-
so all the numbers b/w 1 to 2100 except prime numbers.
for even:-4,6,8,10.....2100
total even number:-1049
for odd:- 9,15,21,25,27.......
all odd mutiples of 9,.18.......2097 (total233)
if i will find like this for 15 ,21 and 25..it may be possible that they all contain common terms how to delete them

please check it and tell my mistakes too..

Okay. This is my take. If a number expressed as a product of its prime factors is 2^m * 3^n....., then the number of factors is (m+1)(n+1)...
=> if the number of factors is prime, then (m+1)(n+1)... is prime
=> There is only one prime factor of the number with a power m such that:
(m+1) is prime
m+1 = 2 => m = 1 => N is prime. Do not consider as we have to count only composite numbers.
m+1 = 3 => m = 2 => N = 2^2 to 45^2 (in the given range) => 44 numbers
m+1 = 5 => m = 4 => N = 2^4 to 6^4 (in the given range) => 5 numbers
m+1 = 7 => m = 6 => N = 2^6 to 3^6 (in the given range) => 2 numbers
m+1 = 11 => m = 10 => N = 2^10 (in the given range) => 1 numbers

Total = 44 + 5 + 2 + 1 = 52 numbers. Is this correct?