CAT Old Paper - Number System

CAT Old Paper - Number System
Question:-
Option (A)
2
Option (B)
4
Option (C)
0
Option (D)
1
Option (E)
3
Correct Option:
D
Question Solution:
Let the four digit number be xxyy.
=> 1000x + 100x + 10 y + y
=> 1100x + 11y
=> 11 (100x + y)
If this has to be a perfect square 100 x + y must be of the form 11p where p is a perfect square.
∴100x + y = 11p
=> y = 11p – 100 x
As x & y are single digit positive integers we will have to look for the solution which satisfies this equation. If p = 16
y = 11 x 16 – 100 x
y = 156 – 100 x
If we take x = 15 then only we get y as 6. But in that case x is two digit no. which is contradiction.
p = 25
y = 11 x 25 – 100 x = 275 – 100 x
here at the max x can be 2 but in that case y is 75 which is not possible. Similarly checking for other perfect square
we get solution only at p = 64
y = 11 x 64 – 100 x = 704 – 100 x
If we put x = 7 we get y = 4. Hence the only no. which satisfies this is 7744.
Question:-
Option (A)
22 yr
Option (B)
21 yr
Option (C)
25 yr
Option (D)
24 yr
Option (E)
23 yr
Correct Option:
D
Question Solution:
Ten years ago the total age of 8 members = 231
Today the age of 8 people = 231 + 80 = 311
Total age after 3 years when a man of 60 year died = 311 + 24 – 60 = 275
Again after 3 year, the total age when one more member died = 275 + 24 – 60 = 239
So the current total age = 239 – 48 = 191
∴ The average age = 191/8 = 24 yr (approx)
Question:-
Option (A)
April 11
Option (B)
May 20
Option (C)
April 10
Option (D)
June 30
Option (E)
May 21
Correct Option:
B
Question Solution:
After 100 days the price of Darzeeling tea is constant equal to 100 + 0.10 n = 100 + 10 = 110 price, of Ooty tea on
100th day is 89 + 0.15 (100) = 104.
Now the price of Darzeeling tea is constant so the price will be equal when 89 + 0.15 n = 110
=> n = 21/0.15 = 140
On 140th day the prices will be equal i.e. 20th May.
Question:-
Option (A)
k(n – 1)
Option (B)
n(k – 2)
Option (C)
k(n – 2)
Option (D)
(n – 1)(k – 1)
Option (E)
n(k – 1)
Correct Option:
E
Question Solution:
If the total number of players in each of the n teams be ‘k’, then total players in the tournament will be nk. But as one
player is common in following pairs i.e., T1 and T2, T2 and T3 ..... Tn and T1, therefore we can say that there are n players who are common. Therefore total number of players participating
nk – n = n(k–1).