CAT old paper set-1

CAT 2008 paper
Question:-
Option (A)
2 ≤ x ≤ 6
Option (B)
5 ≤ x ≤ 8
Option (C)
9 ≤ x ≤ 12
Option (D)
11 ≤ x ≤ 14
Option (E)
13 ≤ x ≤ 18
Correct Option:
B
Question Solution:
The initial quantity of rice is x kg.
The first customer buys half the total rice in the store, and another half kg.
∴ Rice purchased by the first customer =
∴ Remaining rice = x -
Now, the second customer buys half of this, and another half kg.
∴ Rice purchased by the second customer =
∴ Remaining rice =
Now, the third customer buys half the remaining rice, and another half kg.
∴ Rice purchased by the third customer =
Since after this purchase, there is no rice left in the store, we conclude that:

∴ x = 7
Hence, option 2.
Question:-
Option (A)
−7
Option (B)
−4
Option (C)
2
Option (D)
6
Option (E)
cannot be determined
Correct Option:
B
Question Solution:
∵ 3 is a root of f(x) = 0,
∴ 9a + 3b + c = 0 ......(i)
Also, f(5) = −3f(2)
∴ 25a + 5b + c = −3(4a + 2b + c)
∴ 37a + 11b + 4c = 0 .......(ii)
On solving equations (i) and (ii), we get, a – b = 0
∴ a = b
We know that sum of the roots of a quadratic equation (ax2 + bx + c = 0) is –b/a
∴ 3 + other root = -1
∴ Other root = −4
Hence, option 2.
Question:-
Option (A)
9
Option (B)
14
Option (C)
13
Option (D)
37
Option (E)
cannot be determined
Correct Option:
E
Question Solution:
The roots at f(x) = 0 are 3 and −4
∴ The equation can be written as (x – 3)(x + 4) = 0
Or, x2 − x + 12 = 0
The co-efficient of x2 is 1 here, but all equations which are multiple of this equation will also have same roots.
For example, 10(x2 − x + 12) = 0 will also have same roots
∴ (a + b + c) cannot be determined uniquely.
Hence, option 5.
Question:-
Option (A)
78
Option (B)
19
Option (C)
20
Option (D)
77
Option (E)
22
Correct Option:
C
Question Solution:
The first sequence can be written as 17, 17 + 4, 17 + 8, ....... , 417 and second sequence can be written as 16, 16 + 5, 16 + 10, ....... , 466
The common difference for the first sequence is 4 and that for the second sequence is 5 and both the sequences have 21 as the first common term.
∴ Common terms are 21, 21 + L, 21 + 2L, .....
[Here, L = LCM of 4 and 5 = 20]
∴ Common terms are 21, 21 + 20, 21 + 40, .....
The common terms have a common difference of 20 and first term as 21.
∵ 417 − 21 = 396 and 396/20 = 19.8,
∴ 19 terms are common, other than 21.
∴ The total number of terms which are common to both the sequences = 19 + 1 = 20
Hence, option 3.
Question:-
Option (A)
60
Option (B)
75
Option (C)
45
Option (D)
90
Option (E)
72
Correct Option:
D
Question Solution:

From point A to E, there are 6 ways to reach with the minimum distance travelled.
Here E to F is the shortest distance because the third side of a triangle is always less than the sum of the other two sides.
From point F to B, there are 15 ways to reach with the minimum distance travelled.
∴ There are a total of 15 x 6 = 90 paths possible
Hence, option 4.
Question:-
Option (A)
1170
Option (B)
630
Option (C)
792
Option (D)
1200
Option (E)
936
Correct Option:
A
Question Solution:
From point A to B, there are 90 paths possible with the minimum distance travelled.
Then from B to C, there are 13 paths possible with the minimum distance travelled.
∴ Overall there are 90 × 13 = 1170 paths possible
Hence, option 1.