1) Eight friends Harsha, Fakis, Balaji, Eswar, Dhinesh, Chandra, Geetha, and Ahmed are
sitting in a circle facing the center.
Balaji is sitting between Geetha and Dhinesh. Harsha is third to the left of Balaji and
second to the right of Ahmed. Chandra is sitting between Ahmed and Geetha and Balaji
and Eshwar are not sitting opposite to each other. Who is third to the left of Dhinesh?
Answer: Fakis
Explanation: Arranging the friends as per the question statement we can arrive at the following
diagram
Ahmed
Fakis Chandra
Harsha Geetha
Eswar Balaji
Dhinesh
Hence correct answer is Fakis
2) A fast typist can type some matter in 2 hours and a slow typist can type the same in 3
hours. If both type jointly, in how much time will they finish?
Answer: 1 hr 12 min
Explanation : The fast typist's work done in 1 hr = 1/2
The slow typist's work done in 1 hr = 1/3
If they work jointly, work done in 1 hr = 1/2+1/3 = 5/6
So, the work will be completed in 6/5 hours. i.e., 1+1/5 hours = 1hr 12 min
3) Today is 4.11.09. Keeping that figure 41109 in mind, i have arrived at the following
sequence: 2, 1, 9, 5, _ Which of the following four numbers can fill the dash?
Options
a) 7
b) 65
c) 4563
d) 262145
Answer : 41109
1^4 + 1 = 2
1^1 + 0 = 1
0^1 + 9 = 9
9^0 + 4 = 5
So next is 4^9 + 1 = 262145
4) Let S be a Set of some positive integral numbers; with an average of 47; and containing
the number 83. The numbers may or may not be distinct .However ; when the number 83 is
removed ; the Avg drops to 46 .What is the largest number that can be possibly contained
in that Set ?
Solution:
Let S be the sum of that set of n positive integers. S/n = 47
(S - 83)/(n - 1) = 46
Solving the above 2 equations, we get S = 1739; n = 37.
This set of 37 positive integers contains 83. To get to the expected answer, we have to
Suppose that 35 of the remaining 36 integers has a value of 1 each (least +ve integer).
Thus, the largest possible integer in the set = 1739 - 83 - 35*1 = 1621
5) Sum of squares of two numbers 'x' and 'y' is less than or equal to 100. How many sets
of integer solutions of 'x', 'y' is possible?
Answer: Total 317 solutions
Explanation:
x = 0, |y| <= 10 -> 21 solutions
|x| = 1, |y| <= 9 -> 38 solutions
|x| = 2, |y| <= 9 -> 38 solutions
|x| = 3, |y| <= 9 -> 38 solutions
|x| = 4, |y| <= 9 -> 38 solutions
|x| = 5, |y| <= 8 -> 34 solutions
|x| = 6, |y| <= 8 -> 34 solutions
|x| = 7, |y| <= 7 -> 30 solutions
|x| = 8, |y| <= 6 -> 26 solutions
|x| = 9, |y| <= 4 -> 18 solutions
|x| = 10, y = 0 -> 2 solutions
6) There is a unique number of which the square and the cube together use all ciphers
from 0 up to 9 exactly once. Which number is this?
Answer: The number is 69.
Explanation:
69^2=4761 and 69^3=328509..
7) You are standing next to a well, and you have two jugs. One jug has a content of 3
litres and the other one has a content of 5 litres. How can you get just 4 litres of water
using only these two jugs?
Solution:
Fill 3 litre jug pour to 5 litre jug
Fill again 3 litre jug and add to 5 litre jug then 1 litre will be there in 3 litre jug
Pour all water outside from 5 litre jug
Fill 1 litre water from 3 litre jug to 5 litre jug
Fill 3 litre jug and add to 5 litre jug making it 4 litres of water.
8) Blocks are chosen randomly on a chessboard. What is the probability that they are in
the same diagonal?
Answer:
There are total of 64 blocks on a chessboard. So 3 blocks can be chosen out
of 64 in 64C3 ways.
So the sample space is = 41664
There are 2 diagonal on chessboard each one having 8 blocks. Consider one of them.
3 blocks out of 8 blocks in diagonal can be chosen in 8C3 ways.
But there are 2 such diagonals, hence favourables = 2 * 8C3 = 2 * 56 = 112
The required probability is
= 112 / 41664
= 1 / 372
= 0.002688
9) What is the area of the triangle ABC with A(e,p) B(2e,3p) and C(3e,5p)?
where p = PI (3.141592654)
Answer:
A tricky ONE.
Given 3 points are collinear. Hence, it is a straight line.
Hence area of triangle is 0.
10) Silu and Meenu were walking on the road.Silu said, "I weigh 51 Kgs. How much do you
weigh?" Meenu replied that she wouldn't reveal her weight directly as she is overweight.
But she said, "I weigh 29 Kgs plus half of my weight”. How much does Meenu weigh?
Answer
Meenu weighs 58 Kgs.
It is given that Meenu weighs 29 Kgs plus half of her own weight. It means that
29 Kgs is the other half. So she weighs 58 Kgs.
Solving mathematically, let's assume that her weight is X Kgs.
X = 29 + X/2
2*X = 58 + X
X = 58 Kgs
11)Consider the sum: ABC + DEF + GHI = JJJ .If different letters represent different digits,
and there are no leading zeros, what does J represent?
Answer
The value of J must be 9.Since there are no leading zeros, J must be 7, 8, or 9. (JJJ = ABC
+ DEF + GHI= 14? + 25? + 36? = 7??)Now, the remainder left after dividing any number by 9
is the same as the remainder left after dividing the sum of the digits of that number by 9.
Also, note that 0 + 1 + ... + 9 has a remainder of 0 after dividing by 9 and JJJ has a
remainder of 0, 3, or 6. The number 9 is the only number from 7, 8 and 9 that leaves a
remainder of 0, 3, or 6 if you remove it from the sum 0 + 1 + ... + 9. Hence, it follows that J
must be 9.
12) Using two 2's and two 3's and using a maxim of three mathematical signs, symbols,
can you have a result in between 14 and 15? Concatenation (clubbing of digits) allowed.
Solution:
( 23 + 3! ) / 2 = 14.5
18) What is the 28383rd term in the series 1234567891011121314............
a) 3
b) 4
c) 7
d) 9
Solution:
there are 9 no. of single digit
there are 180 no. of double digit
there are 2700 no. of three digit
now total 2889 no. till 999
remaining no. are 25494 that is divided by 4 and the q is 6373 with reminder of 2 so 28381
is 6373+999=737(2)
n next no is 7(3)73
so ans is 3
13) a*b*c*d*e + b*c*d*e*f + a*c*d*e*f + a*b*d*e*f + a*b*c*e*f + a*b*c*d*f = a*b*c*d*e*f and
a,b,c,d,e and f are all positive nonrepeating integers then solve a,b,c,d,e, and f.
Solution :
Start with 1/2 + 1/2, then progressively split the last part x
into 2x/3 + x/3. This gives the following progression:
2,2
2,3,6
2,3,9,18
2,3,9,27,54
2,3,9,27,81,162
14) 729 ml of a mixture contains milk and water in ratio 7:2. How much of the water is to be
added to get a new mixture containing half milk and half water?
(i) 79 ml
(ii) 81 ml
(iii) 72 ml
(iv) 91 ml
Solution:
Milk = (729 * (7/9))=567ml
Water = (729-567)= 162ml
Let water to be added be x ml l 567/(162+x) = 7/3 1701 = 1134 + 7x x = 81ml
15) If one-seventh of a number exceeds its eleventh part by 100 then the number is…
(i) 770
(ii) 1100
(iii) 1825
(iv) 1925 Solution:
Let the number be x. Then X/7 - x/11 =100 11x-7x = 7700 x=1925.
16) If 1.5x=0.04y then the value of (y-x)/(y+x) is
(i) 730/77
(ii) 73/77
(iii) 7.3/77
(iv) None
Solution:
x/y = 0.04/1.5 = 2/75
So (y-x)/(y+x) = (1 - x/y)/(1 + x/y) = (1 - 2/75)/ (1 + 2/75) = 73/77.
17) The smallest number which when diminished by 3 is divisible by 21, 28, 36 and 45 is...
(i) 869
(ii) 859
(iii) 4320
(iv) 1263
Solution:
The required number = l.c.m. of (21, 28, 36 ,45)+3=1263
18) If x and y are the two digits f the number 653xy such that this number is divisible by 80,
then x+y is equal to:
(i) 2
(ii) 3
(iii) 4
(iv) 6
Solution:
Since 653xy is divisible by 2 as well as by 5, so y = 0
Now 653x0 is divisible by 8 so 3x0 is also divisible by 8.
By hit and trial x=6 and x+y = 6
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