1)The required number is 236 and the sum is 11.
It is given that the first two digits of the required number are prime numbers
i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third
digit is the multiplication of the first two digits. Thus, first two digits
must be either 2 or 3 i.e. 22, 23, 32 or 33 which means that there are four
possible numbers - 224, 236, 326 and 339.
Now, it is also given that - the difference between it's reverse and itself is
396. Only 236 satisfies this condition. Hence, the sum of the three digits is
11.
2)
A=2, B=1, C=9, D=4, E=3, F=8, G=6, H=5, I=7
Let's start with GHI = 9 * IE. Note that I appears on both the side. Also,
after multiplying IE by 9 the answer should have I at the unit's place. The
possible values of IE are 19, 28, 37, 46, 55, 64, 73, 82 and 91; out of which
only 64, 73 and 82 satisfies the condition. (as all alphabet should represent
different digits)
Now, consider DEF = 6 * IE. Out of three short-listed values, only 73
satisfies the equation. Also, ABC = 3 * IE is satisfied by 73.
Hence, A=2, B=1, C=9, D=4, E=3, F=8, G=6, H=5, I=7
219 438 657
--- = 73 --- = 73 --- = 73
3 6
3)
A, B & D are males; C is female. B is C's only son. A & D are C's brothers.
A(male) --- C(female) --- D(male)
|
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B(male)
Work out which relation can hold and discard the contradictory options.
From (2) and (4), D can not be a only daughter and have a sister (B or C).
Hence, D is A's brother i.e. D is a Male.
From (4), let's say that B is D's sister i.e. B is Female.
From (3), A is C's only son i.e. A is Male.
But D is A's brother which means that A is not C's only son. Hence, our
assumption was wrong.
Thus, C is D's sister i.e. C is Female. And B must be C's only son.
Now it is clear that D & B are Males and C is Female. A must be a Male as only
one of them is of opposite sex from each of the other three. And he is C & D's
brother.How are they related to each other?
4)
32 minutes 43.6 seconds
In twelve hours, the minute hand and the hour hand are together for 11 times.
It means that after every 12/11 hours, both the hands are together.
Similarly in twelve hours, the minute hand and the hour hand are exactly
opposite to each other for 11 times. It means that after every 12/11 hours,
both the hands are opposite.
Now, let's take an example. We know that at 12 both the hands are together and
at 6 both the hands are exactly opposite to each other.
After 6, both the hands are in opposition at [6+(12/11)] hours, [6+2*(12/11)]
hours, [6+3*(12/11)] hours and so on. The sixth such time is [6+6*(12/11)]
hours which is the first time after 12. Thus after 12, both the hands are
opposite to each other at 12:32:43.6
Hence, Dr. DoLittle takes 32 minutes and 43.6 seconds to reach home from the
clinic.
5)
Mr. Haani crossed 7 SlowRun Expresses during his journey.
Let's say that Mr. Haani travelled by SlowRun Express on Wednesday 10:00PM
from Mumbai. The first train he would have crossed is the one scheduled to
arrive at Mumbai at 11:30 PM the same day i.e. the one that left Bangalore at
10:00 PM on last Sunday.
Also, he would have crossed the last train just before reaching Bangalore on
Saturday.
6)
The cabins from left to right (1-6) are of Mr. Solanki, Mr. Sinha, Mr. Shaan,
Mr. Sharma, Miss Shudha and Miss Shalaka.
From (2), cabin number 5 is assigned to Miss Shudha.
As Miss Shudha is allergic to smoke and Mr. Sinha, Mr. Shaan & Mr. Solanki all
smoke, they must be in cabin numbers 1, 2 and 3 not necessarily in the same
order. Also, Miss Shalaka and Mr. Sharma must be in cabin 4 and 6.
From (3), Mr. Shaan must be in cabin 3 and Mr. Sharma must be in cabin 4.
Thus, Miss Shalaka is in cabin 6.
As Mr. Solanki needs silence during work and Mr. Shaan is in cabin 3 who often
talks to Mr. Sharma during work, Mr. Solanki must be in cabin 1. Hence, Mr.
Sinha is in cabin 2.
Thus, the cabins numbers are
1# Mr. Solanki,
2# Mr. Sinha,
3# Mr. Shaan,
4# Mr. Sharma,
5# Miss Shudha,
6# Miss Shalaka
7)
It is given that
2 wixsomes = 3 changs
8 wixsomes = 12 changs ----- (I)
Also, given that
4 changs = 1 plut
12 changs = 3 plutes
8 wixsomes = 3 plutes ----- From (I)
Therefore,
6 plutes = 16 wixsomes
8)
The required number is 236 and the sum is 11.
It is given that the first two digits of the required number are prime numbers
i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third
digit is the multiplication of the first two digits. Thus, first two digits
must be either 2 or 3 i.e. 22, 23, 32 or 33 which means that there are four
possible numbers - 224, 236, 326 and 339.
Now, it is also given that - the difference between it's reverse and itself is
396. Only 236 satisfies this condition. Hence, the sum of the three digits is
11.
9)
A simple one.
As it is given that only one of the four statement is correct, the correct
number can not appear in more than one statement. If it appears in more than
one statement, then more than one statement will be correct.
Hence, there are 4 marbles under each mug.
10)
aswer is 7
11)
The cicumference of the earth is
= 2 * PI * r
= 2 * PI * 6400 km
= 2 * PI * 6400 * 1000 m
= 2 * PI * 6400 * 1000 * 100 cm
= 1280000000 * PI cm
where r = radius of the earth, PI = 3.141592654
Hence, the length of the thread is = 1280000000 * PI cm
Now length of the thread is increasd by 12 cm. So the new length is =
(1280000000 * PI) + 12 cm
This thread will make one concentric circle with the earth which is slightly
away from the earth. The circumfernce of that circle is nothing but
(1280000000 * PI) + 12 cm
Assume that radius of the outer circle is R cm
Therefore,
2 * PI * R = (1280000000 * PI) + 12 cm
Solving above equation, R = 640000001.908 cm
Radius of the earth is r = 640000000 cm
Hence, the thread will be separatedfrom the earth by
= R - r cm
= 640000001.908 - 640000000
= 1.908 cm
12)
At the beginning of the 11th year, there would be 1,024,000 rabbits.
At the beginning, there were 1000 rabbits. Also, there were 4000 rabbits at
the beginning of third year which is equal to 2828Z. Thus, Z = 4000/2828 i.e.
1.414 (the square root of 2)
Note that 2828Z can be represented as 2000*Z*Z (Z=1.414), which can be further
simplified as 1000*Z*Z*Z*Z
Also, it is given that at the end of 6 months, there were 1000Z rabbits.
It is clear that the population growth is 1.414 times every six months i.e. 2
times every year. After N years, the population would be 1000*(Z^(2N)) i.e.
1000*(2^N)
Thus, at the beginning of the 11th year (i.e. after 10 years), there would be
1000*(2^10) i.e. 1,024,000 rabbits.
13)
The man drunk 190oz of water.
It is given that the man has 20oz bottle of sprite. Also, he will drink 1oz on
the first day and refill the bottle with water, will drink 2oz on the second
day and refill the bottle, will drink 3oz on the third day and refill the
bottle, and so on till 20th day. Thus at the end of 20 days, he must have
drunk (1 + 2 + 3 + 4 + ..... +18 + 19 + 20) = 210oz of liquid.
Out of that 210oz, 20oz is the sprite which he had initially. Hence, he must
have drunk 190oz of water.ed
14)
There are 5 such expressions.
99 + (9/9) = 100
(99/.99) = 100
(9/.9) * (9/.9) = 100
((9*9) + 9)/.9 = 100
(99-9)/.9 = 100
15)
answer is 7/9
Assume that number of boys graduated from City High School = B
Therefore, number of girls graduated from City High School = 2*B
It is given that 3/4 of the girls and 5/6 of the boys went to college
immediately after graduation.
Hence, total students went to college
= (3/4)(2*B) + (5/6)(B)
= B * (3/2 + 5/6)
= (7/3)B
Fraction of the graduates that year went to college immediately after
graduation
= [(7/3)B] / [3*B]
= 7/9
Therefore, the answer is 7/9
16)
The mule was carrying 5 sacks and the donkey was carrying 7 sacks.
Let's assume that the mule was carrying M sacks and the donkey was carrying D
sacks.
As the donkey told the mule, "If you gave me one of your sacks I'd have double
what you have."
D + 1 = 2 * (M-1)
D + 1 = 2M - 2
D = 2M - 3
The donkey also said, "If I give you one of my sacks we'd have an even
amount."
D - 1 = M + 1
D = M + 2
Comparing both the equations,
2M - 3 = M + 2
M = 5
Substituting M=5 in any of above equation, we get D=7
Hence, the mule was carrying 5 sacks and the donkey was carrying 7 sacks.
17) Person A came in first.
Let's assume that the distance between start and the point is D miles.
Total time taken by Person A to finish
= (D/20) + (D/20)
= D/10
= 0.1D
Total time taken by Person B to finish
= (D/10) + (D/30)
= 2D/15
= 0.1333D
Thus, Person A is the Winner.
18)
Assume that initial there were 3*X bullets.
So they got X bullets each after division.
All of them shot 4 bullets. So now they have (X - 4) bullets each.
But it is given that, after they shot 4 bullets each, total number of bullets
remaining is equal to the bullets each had after division i.e. X
Therefore, the equation is
3 * (X - 4) = X
3 * X - 12 = X
2 * X = 12
X = 6
Therefore the total bullets before division is = 3 * X = 18
19)The sum of the digits of D is 1.
Let E = sum of digits of D.
It follows from the hint that A = E (mod 9)
consider, (" ^ " means power)
A = 1999 ^ 1999
< 2000 ^ 2000
= 2 ^ 2000 * 1000 ^ 2000
= 1024 ^ 200 * 10 ^ 6000
< 10 ^ 800 * 10 ^ 6000
= 10 ^ 6800
i.e. A < 106800
i.e. B < 6800 * 9 = 61200
i.e. C < 5 * 9 = 45
i.e. D < 2 * 9 = 18
i.e. E <= 9
i.e. E is a single digit number.
Also,
1999 = 1 (mod 9)
so 19991999 = 1 (mod 9)
Therefore we conclude that E=1.
20)
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with uniform speed.
Also, they both moved for the identical amount of time. Hence, the ratio of
the distance they covered - while person moving forward and backword - are
equal.
Let's assume that when the last person reached the first person, the platoon
moved X meters forward.
Thus, while moving forward the last person moved (50+X) meters whereas the
platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-X)] X meters
whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
