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Section - I (Quantitative Ability)

This section contains 25 questions

Q. 1 The integers 1, 2,...., 40 are written on a blackboard. The following operation is then repeated 39 times. In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b-1 is written. What will be the number left on the board at the end?

1. 820
2. 821
3. 781
4. 819
5. 780

Ans. Here, in each step we are adding two number and reducing the sum by 1. So after 39 operations, we will have the sum of all the numbers from 1 to 40 reduced by 39. Hence the final number will be

Q. 2 What are the last two digits of ?

1. 21
2. 61
3. 01
4. 419
5. 81

Ans. The last two digits of a number is nothing but the remainder obtained when the number is divided by 100.This number leaves a remainder 1 when divided by 4 as well as 25. Hence the remainder obtained when this number is divided by 100 is also 1. Hence the last two digits of this number are 01.

Q. 3 If the root of the equation are three consecutive integers, then what is the smallest possible value of b?

1. 
2. -1
3. 0
4. 1
5. 

Ans. We know that the coefficient of x is sum of all the products which are obtained by taking every two of the three roots. Assuming the three consecutive roots to be (n-1), n and n + 1
we have, b=n(n - 1) + n(n + 1) + (n - 1) (n + 1)

Since, 0 minimum value of b is obtained for
Hence the minimu

Q. 4 A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?

1. 
2. 
3. 
4. 
5. 

Ans. From the data, we get a table

Quantity of Rice in the shop	Quantity of Rice bought	Quantity left
x

Directions for Questions 5 and 6

Let , where a, b and c are certain constants and . It is known that f(5)=-3f(2) and that 3 is a root of f(x)=0.

Q. 5 What is the other root of f(x)=0 ?

1. -7
2. -4
3. 2
4. 6
5. cannot be deterined

Ans. Given
3 is a root of f(x)
9a +3b + c = 0 .........(i)
also, f(5) = -3f(2).
25a + 5b + c = -3 (4a + 2b + c)
= -12a - 6b - 3c
37a + 11b + 4c = 0 .........(ii)
from (i) and (ii) a-b=0 a=b
Thus we get
Dividing f(x) by x - 3, we get c = -12a

f(x)=0 -4 is second root.

Q. 6. What is the value of a + b+ c ?

1. 9
2. 14
3. 13
4. 37
5. cannot be determined

Ans.. a + b + c = a + a - 12a = -10a
Since a is not explicitly given, we cannot get the value of a + b + c.

Q. 7. The number of common terms in the two sequences 17, 21, 25, ..., 417, and 16, 21, 26, ...., 466 is

1. 78
2. 19
3. 20
4. 77
5. 22

Ans.. %{font-family:verdana;font-size:13px}Let
and
So, term of are in the form
And term of are in the form
In order to have same terms, we should get 4n = 5m.
This happens only 20 times.
Thus, we get 21, 41, 61, ..., 401 i.e. 20 common terms.

Q. 8. How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0,1,2,3 and 4, if repetition of digits is allowed?

1. 499
2. 500
3. 375
4. 376
5. 501

Ans.. For number other than 4000:
1st digit = 3 possibilities
2nd digit = 5 possibilities
So, Total possible numbers = 15k + 1
The only option satisfying this is 376.

Directions for Questions 9 and 10

The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. there is also a prohibited region (D) in the town.

Q. 9.. Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is

1. 60
2. 75
3. 45
4. 90
5. 72

Ans.

Neelam has to take path XY
A to X = possibilities
Y to B =possibilities
In all 6 * 15 = 90 possibilities

Q. 10.. Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is

1. 1170
2. 630
3. 792
4. 1200
5. 936

Ans. From A to B = 90 paths
From B to C via N = 6 (an not via M)
From B to C via M = 7 paths
In all 90 * (6+7) = 1170 paths

Q. 11.. Let f(x) be a function satisfying f(x) f(y)=f(xy) for all real x,y. If f(2) = 4, then what is the value of ?

1. 0
2. 
3. 
4. 1
5. cannot be determined

Ans. We have f(x)f(y)=f(xy)
f(1) f(1)=f(1 * 1) =f(1)

f(1)= 0 or f(1)= 1
If f(1)= 0 then f(x)= 0 for any x x = x * 1
f(1)= 1
Now, f(2)= 4
So,

Q. 12. Suppose, the seed of any positive integer n is defined as follows:
seed(n)=n , if n < 10
=seed (s(n )), otherwise,
where s(n) indicates the sum of digits of n. for example,
seed(7)=7, seed(248)=seed(2 + 4 + 8) =seed(14)=seed(1 + 4)=seed(5)= 5 etc. How many positive integers n, such that n < 500, will have seed(n) = 9?

1. 39
2. 72
3. 81
4. 108
5. 55

Ans.. From the definition of "seed", it is clear that we have to count number of integers between 1 and 500, which are divisible by 9. The smallest is 9 and the largest is 495. 9 * 1 =9 and 9 * 55 = 495. Hence there are 55 such numbers.

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