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Assistant Central Intelligent Officer GradeII (Exec) in Intelligence Beuro was conducted on 23.09.2012. Detailed solutions for the math and logic related problems are provided here:
Assistant Central Intelligent Officer GradeII (Exec) in Intelligence Beuro was conducted on 23.09.2012. Detailed solutions for the math and logic related problems are provided here:
1.
Find the last two digits of :
15*37*63*51*97*17
(i)35 (ii)45 (iii)55 (iv)85
Solution:
Taking two numbers at a
time and finding last two digits of their product by crisscross method.
Last two digits of
15*37=> 7*5=35, take 5 as units digit. 7*1=7 add this to 3 to get 10,
consider 0 here and add it to unit digit of 3*5=15 ie., 5 to get 5 as tens
digit ===> 55 last two digits
Last two digits of 63*51
===> 13
Last two digits of 97*17
===> 49
Now we have come down to
three numbers: 55,13,49
Now take 55 and 13 at a
time to get the last two digits of their product. The result is : 15
Now we have come down to
two numbers: 15,49 and the final result is :35
2.
A man decides to travel 80
kilometers in 8 hours partly by foot and partly on a bicycle. If his speed on
foot is 8 km/hr and on bicycle 16 km/hr, what distance would he travel on foot?
(i)20 (ii)30 (iii)48 (iv)60
Solution:
Let the distance
travelled by foot be f and that by bicycle be b => f+b = 80 –(1)
Time taken for foot ride
= f/8, as the speed is given as 8 km/hr
Time taken for bicycle
ride = b/16, as the speed is given as 16 km/hr
Total time taken = f/8 +
b/16 = 8 –(2)
Solving (1) and (2), we
get f = 48
3.
Due to 25% increase in the price
of rice of rice per kilogram, a person is able to purchase 20 kilograms less
for Rs400. What is the increased price of rice per kilogram?
(i)Rs 5 (ii) Rs 6 (iii)
Rs 10 (iv) Rs 4
Solution:
Arithmetic
Method:
“25% increase in price”
=> Let x be the increased price per kilogram. As the price increase is 25%,
for an original price of 4x there is an increase of x per kilogram.
“a person is able to purchase
20 kilograms less for Rs400” => for a Rs 400 worth rice there is a price
hike of 20x.
Now compare the two
statements:
For 400 he is loosing 20x
due to price hike
For 4x there is an
increase of x => for 5x he is loosing x
By proportion, 5x/400 =
x/20x => x = 4
Algebraic
Method:
Let x be original price
per kilogram. With an increase of 25%, the new price per kilogram is 1.25x.
For Rs400, he could have
purchased 400/x for the original price but he can purchase 400/1.25x now. The
difference is 20 kilogram. => 400/x – 400/1.25x = 20
=> 1/x(11/1.25) =
20/400=1/20 =>1/x(14/5)=1/20 => x = 4
4.
There is a 7digit telephone
number with all different digits. If the digit at extreme right and extreme
left are 5 and 6 respectively, find how many such telephone numbers are
possible?
(i)120 (ii) 100000 (iii)
6720 (iv) 30240
Solution:
There are 5 more digits
are to be selected and arranged from 8 digits (0,1,2,3,4,7,8,9).
As this involves
selection and arrangement, it is a nPr problem but not nCr.
8P5 = 8!/3! = 8*7*6*5*4 =
6720
5.
A speaks the truth 3 out of 4
times, and B 5 out of 6 times. What is the probability that they will
contradict each other in stating the same fact?
(i)2/3 (ii) 1/3 (iii)
5/6 (iv) 1/21
Solution:
Probability that A speaks
truth = A = 3/4
Probability that A speaks
false = A’ = 13/4 = 1/4
Probability that B speaks
truth = B = 5/6
Probability that B speaks
false = B’ = 15/6 = 1/6
Probability that they
will contradict each other in stating the same fact = AB’+A’B
=3/4 * 1/6 + 1/4*5/6 =
8/24 = 1/3
6.
A circle is inscribed inside an
equilateral triangle touching all the three sides. If the radius of the circle
is 2cm, then find the area of the triangle?
(i)15√3 (ii) 18√3 (iii)
12√2 (iv) 12√3
Solution:
It is an incircle. For an
equilateral triangle incentre and centroid coincide. Centroid divides the
median in the ration 2:1. AG:GD = 2:1. Given inradius is 2, which is nothing
but GD.
GD = 2 => AG = 4 =>
AD = AG+GD = 6.
AD is height of the
triangle. AD = h = 6
For an equilateral
triangle height h = √3/2 a , where a is side
=> a = 2h/√3 = 12/√3
Area of triangle = ½*a*h
= ½*12/√3*6 = 12√3
7.
Three bells chime at intervals
of 48,60 and 90 minutes respectively. If all the three bells chime together at
10 AM, at what time will all the three chime again that day?
(i)1 PM (ii) 2 PM (iii) 8 PM (iv)
10 PM
Solution:
This is an LCM problem.
The three bells chime together in a regularly interval of LCM of 48,60,90 ie.,
720 minutes=12 Hrs. So after 10 AM, the next term will be at 10PM.
8.
After striking the floor, a ball
rebounds to 4/5^{th} of height from which it has fallen. Find the total
distance that it travels before coming to rest if it has been gently dropped
from a height of 120 meters?
(i)540(ii) 960 (iii) 1080 (iv)
1120
Solution:
After the drop, it
travels down for h meters and rebounds back to 4h/5 and travels down 4h/5 and
rebounds back 4/5(4h/5) etc...
The total distance it
travels = h + 4h/5+ 4h/5+ 4/5(4h/5)+ 4/5(4h/5) + 4/5*4/5(4h/5)+...
S= h + 2*4h/5 +
2(4/5)(4h/5) + ....
= h + 2{4h/5 + 4/5(4h/5)
+ ....}
= h + 2(Sum of infinite
GP with initial term 4h/5 and common ratio 4/5)
Formula: Sum of Infinite
GP with initial term ‘a’ and common ratio ‘r’ is a/(1r)
S = h + 2{(4h/5)/(14/5)}
= h + 8h = 9h = 9*120 =
1080
Note: two things required
to solve this problem are,
first deciding how much
distances it travels before and after each rebound
second formula for
infinite GP.
9.
A toy weighing 24 grams of an
alloy of two metals is worth Rs174/, but if the weights of the two metals be
interchanged, the toy would be worth Rs 162/. If the price of one metal be Rs
8 per gram, find the price of other metal used to make the toy?
(i)Rs 10/gram(ii) Rs 6/gram (iii) Rs 4/gram (iv) Rs 5/gram
Solution:
Logic:
If the weight of one
metal be A and pergramprice be x and the weight of other metal be B and
pergramprice by y, then the total cost is Ax+By. If the weights are
interchanged, the total cost becomes Ay+Bx. So here the point is if we add up
both these expressions, it becomes Ax+By+Ay+Bx and be factored as (A+B)(x+y).
As the total weight is given, it can be substituted in A+B and the problem can
be easily solved.
Now let us do the
problem:
The pergram price of one
metal is given as 8. So let x=8.
Total weight of alloy =
A+B = 24
8A+By = 174;
yA+8B = 162;
add up the two
equations=> 8A+By+yA+8B = 336 => 8(A+B)+y(A+B) = 336
=> (8+y)(A+B) = 336
=> 8+y = 336/24 = 14 => y = 6
10.
Indira is three times older than
Yogesh while Zaheer is half the age of Wahida. If Yogesh is older than Zaheer,
then which of the following statements can be inferred?
(i)Yogesh is older than
Wahida(ii) Indira is older than Wahida
(iii) Indira may be
younger than Wahida (iv) None of the
above
Solution:
I = 3Y => I > Y;
Z = W/2 => W > Z;
Y > Z
One thing surely we can
infer here is: I > Z. (as I > Y and Y > Z)
I > Z and W > Z:
from these two statements, we can’t infer about relation b/w I and W.
W > Z and Y > Z:
from these two statements, we can’t infer about relation b/w Y and
None of the given
statements can be inferred. Answer is (iv)
11.
Four sisters – Suvarna,Tara,Uma
and Vibha are playing a game such that the looser doubles the money of each of
the of the other players from her share.They played four games and each sister
lost one game in alphabetical order. At the end of fourth game each sister had
Rs.32. How much money did Suvarna start with?
(i)Rs. 60(ii) Rs. 34(iii)
Rs. 66 (iv) Rs. 28
Solution:
I like this problem. It
is based on logic.
At the end of four games,
each had Rs.32 => total amount with all the four is 4*32 = 128
“The point is even if
some money changes hands with the outcome of each game, the total money (Rs128)
with all the four does not vary at any point”.
As the games were lost by
the four sisters in alphabetical order, Suvarna lost the first game and after
that in the remaining three games her money got doubled three times ie., “The
point here is she lost only in the first game and from this point
onwards, her money got multiplied 8 times after completion of the
remaining 3 games”.
Let the money that
Suvarna initially had with be ‘X’. So the total amount with all the other three
sisters is 128X before starting the first game. After first game, in
which Suvarna lost, Suvarna must be left with X(128X) as she needed to
give (128X) to make double the amounts with the other three.
Suvarna's money

Total Money with other three


At the start

X

128X

After 1st game

X(128X) = 2X128

2(28X)

After 2nd game

2(2X128)

Not required to find

After 3rd game

4(2X128)

Not required to find

After 4th game

8(2X128)

Not required to find

Finally the amount left
with Suvarna is 8(2X128). But it is given that at the end of four games, each
had Rs.32. It means Suvarna ended up with Rs 32.
=> 8(2X128) = 32
=> X = 66
12.
It was Saturday on 17^{th}
December, 1982. What will be the day on 22^{nd} December, 1984?
(i)Monday(ii) Tuesday(iii)
Wednesday (iv) Sunday
Solution:
Here
we have to check for any leap years. 1984 is a leap year.
From
17^{th} December, 1982 to 16^{th} December,1983 => 365 days
From
17^{th} December, 1983 to 16^{th} December,1984 => 366 days
as Feb’1984 has 29 days instead of 28
From
17^{th} December, 1984 to 22^{nd} December,1984 => 6 days
22^{nd}
Dec’1984 is (365+366+6)h day counting from 17^{th} Dec’1982 => 737^{th}
day
As on week consists of 7 days, we want remainder
when 737 is divided by 7 => it is 2
=> that means 22^{nd}
Dec’1984 is 2^{nd} day ahead of
some weeks of gap counting from 17^{th} Dec’1982
=>
if 17^{th} Dec’1982 is Saturday, 21^{st} Dec’1984 is Sunday and
22^{nd} Dec’1984 is Monday.
13.
One term in the following number
series is not correct. 15,16,22,29,45,70
Find out the wrong term?
(i)16(ii) 22(iii) 45 (iv) 70
Solution:
Find
out the consecutive differences:
=>
1,6,7,16,25
It seems like series of
squares of consecutive numbers excepting 6 and 7. The difference 6 is from
2216 and the difference 7 is from 2922. Had it been 20 instead of 22 in the
given series of numbers, the differences will be in the following way:
1,4,9,16,25.
This is perfect. So the wrong number is 22.
14.
A man is facing west. He turns
45^{0} in the clockwise direction and another 180^{0} in the
same direction and then 270^{0} in the anticlockwise direction. Which
direction is he facing now?
(i)SouthWest(ii) South(iii)
NorthWest (iv) West
Solution:
Let
us represent clockwise as negative and anticlockwise as positive.
The
total angle turns out to be = 45180+270 = +45^{0}
Originally he is facing
West and the final turnout angle is 45^{0} positive ie., 45^{0}
anticlockwise => 45^{0} towards South => the final position is :
he is looking towards SouthWest.
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Good Evening.The solutions given above are quite useful.May I know how to do multiplication of three numbers,as in the same pattern of first question?
ReplyDeleteexample: 123*678*469*120*375*194
Just consider the last two digits of each number while doing it, no matter how many digits each number contains.
Deletereally thanks alot... i was not able to do it alone....
ReplyDeletewow!!!
ReplyDeletedis is vry useful
ReplyDeletethanks for the solutions
ReplyDeletethanks a ton sir....!!!!!
ReplyDelete