CAT Questions

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Given that If Theresa's daughter is my daughter's mother

It has 2 ways of Answers such that it depends on the sex of me. That is Male or Female.

1) If I am a Male :

My daughter's mother is my wife.

My wife is Teresa's daughter

=> I am Teresa's Son-in-law.

2) If I am a Female :

My daughter's mother is myself.

Myself Teresa's daughter.

=> I am Teresa's Daughter.

But in the given options only Son-in-law is available and daughter is not given.

So, Son-in-law is correct.

There is two sequence.
S(odd) P(Even) I(odd) D(even) E(odd) R(even) M(odd) A(even) N(odd)
For odd sequence, we want to write next alphabet of that
and for even, we want to write previous alphabet of that.

TOJCFQNZO is the Answer.

If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are k, l, m, n & o respectively, then
k + l + m + n = 58 x 4 = 232 ---- (i) &
l + m + n + o = 60 x 4 = 240 ---- (ii)
Subtracting (i) from (ii), we get
o-k = 8 ---(iii)
Given, k/o = 8/9 ----(iv)
So from (iii) & (iv), we get
a = 64, e = 72
Therefore, number of visitors on 5th day is 72.

Let the present ages of Arun and Harish be 4x and 3x years respectively.

Then, 4x + 6 = 26 => x = 5

Harish's age = 3x = 15 years.

Let the principle be Rs. P
As the amount double itself the interest is Rs. P too
So P = P x r x 15/100
=> r = 100/15 = 20/3 % = 6.66 %.

The smallest unit of data storage in a computer is called Bit (Binary Digit). Half a byte (four bits) is called a Nibble.

1 bit (binary digit) = the value of 0 or 1

8 bits = 1 byte

1024 bytes = 1 kilobyte

1024 kilobytes = 1 megabyte

1024 megabytes = 1 gigabyte

1024 gigabytes = 1 terabyte

1024 terabytes = 1 petabyte

Hence, Computer storage units smallest to largest i.e, in ascending order is

 1 bit  <  1 byte  <  1 KB  <  1 MB  <  1 GB  <  1TB  <  1 PB.

Let there are L men

job requires LK man hours.

job completed in first 4 hrs = L x 4 = 4L

job completed in next 4 hrs = 4 x L/2 = 2L

job completed in next 4 hrs = 4 x L/4 = L

job completed in last 4 hrs = 4 x L/8 = L/2

4L + 2L + L + L/2 = KL

K = 7+1/2 = 7.5 hours.

Let work done by 1 man in i day be m

and Let work done by 1 boy in 1 day be b

From the given data,

4(5m + 3b) = 23

20m + 12b = 23....(1)

2(3m + 2b) = 7

6m + 4b = 7 ....(2)

By solving (1) & (2), we get

m = 1, b = 1/4

Let the number of required boys = n

6(7 1 + n x 1/4) = 45

=> n = 2.