IB ACIO Grade - II Solved Question Paper - 15th September 2013

Police and Security : IB ACIO Grade - II 2013

Solved Paper of Assistant Central Intelligence Officer (ACIO) Grade - II Examination 2013 held on 15th September 2013. Multiple choice Questions with Answers of Intelligence Bureau India Examination of year 2013. The paper-1 consist of 100 objective

1. If increasing 20 by P percent gives the same result as decreasing 60 by P percent, what is P percentage of 70?
[A] 50
[B] 140
[C] 14
[D] 35

Explanation:

$Increase\ 20\ by\ P\ percentage:\ 20+P\% of\ 20\ =\ 20 + \frac{20P}{100}= 20+ \frac{P}{5}$
$Decrease\ 60\ by\ P\ percentage:\ 60 - P\% of\ 60\ = 60 - \frac{60P}{100}=60 - \frac{3P}{5}$
$According\ to\ Question:\ 20 +\frac{P}{5} = 60 - \frac{3P}{5}$
$By\ solving\ we\ get\ P\ =\ 50$
$Now\ P\ percentage\ of\ 70\ means\ 50\%\ OF\ 70\ = \frac{50}{100}70\ =\ 35$
2. If republic day of India in 1980 falls on Saturday, X was born on March 3 1980 and Y is older to X by four days, then Y's birthday fell on?
[A] Thursday
[B] Friday
[C] Wednesday
[D] None of these

Explanation:

3. Find the missing number in the following series:
4,8,....,100,180,294
[A] 32
[B] 36
[C] 48
[D] 40

Explanation:

4. There are six tasks and six persons. Task 1 can not be assigned to either person 1 or 2; task 2 must be assigned to either person 3 or 4. Every person has to be assigned a task. In how many ways can this assignment be done?
[A] 144
[B] 180
[C] 192
[D] 360

Explanation:

Task 1 can not be assigned to either person 1 and 2 it means total number of ways in which task 1 can be assigned = 4 (i.e 6 - 2)
Task 2 can be assigned to either person 3 or 4, hence total number of ways in which task 2 can be assigned = 2 ways
Task 3 can be assigned in = 4 ways i.e 6 -4 as two tasks are already assigned.
Similarly Task 4, 5 and 6 can be assigned in 3,2,1 ways respectively.
Hence total number of ways in which this assignment can be done = 4 x 2 x 4 x 3 x 2 x 1 = 192 ways.
5. A number when divided by a divisor leaves a remainder of 24.When twice the original number is divided by the same divisor, the remainder is 11. What is the value of divisor?
[A] 13
[B] 59
[C] 35
[D] 37

Explanation:

Let X be the Number, D be the divisor and Q be the quotient, R is the remainder

Hence X = DQ+R

According to question:

X = DQ + 24 ...... (i)

Now twice the original number

2X = 2(DQ+24) [From eqn (i)]

It is divided by the same divisor D

[2(DQ+24)]/D = 2DQ/D+48/D

2DQ/D leaves 0 as remainder as D cancles Hence 48/D = 11 (According to question) So if we subtract 11 from 48 it becomes 37 (i.e 48-11=37)

Hence the correct Answer is 37
6. In how many ways can 15 people be seated around two round tables with seating capacities of 7 and 8 people?
[A] 15! / (8!)
[B] 7! * 8!
[C] (15C8) * 6! * 7!
[D] 15C8 * 8!

Answer: C. (15C8) * 6! * 7!

Explanation:

There are (n-1)! ways for the Circular arrangements of n people

Pick 8 people out of 15 in 15C8 ways

Now 8 people can be seated in (8-1)! and remaining 7 people can be seated in (7-1)! ways.

Hence required number of ways = 15C8 * 7! * 6!
7. What is the sum of all positive integers lying between 200 and 400 that are multiple of 7?
[A] 8729
[B] 8700
[C] 8428
[D] 8278

Explanation:

Integers between 200 and 400 which are multiple of 7 are :Â
203,210,......399
Â It is an AP series with a=203,d=7 and an=399Â
an= a+(n-1)d 399= 203+(n-1)7 399-203=(n-1)7 196=(n-1)7 196/7=n-1 28=n-1Â
n=29Â
Â So Total number of terms = 29Â
Â NowÂ
Â $sn=\frac{n}{2}[2a+(n-1)d] s29=\frac{29}{2}[2(203)+(29-1)7] =\frac{29}{2}[406+196] =\frac{29}{2} x 602 = 8729$

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8. The difference between the ages of two brothers is a prime number. Sum of their ages is also a prime number. If the elder brother is 28 years old, how many different values can the age of the younger brother take?
[A] 2
[B] 3
[C] 4
[D] 1

9. Find the least five digit number which on divided by 12,18,21,28 leaves the same remainder.
[A] 11019
[B] 10081
[C] 10059
[D] 10289