1)6^4 2)60 3)20 4)21 5)64 km/hr 6)26.66 % 7)52 8)3 9)80 10)31 11)30 12)17 13)1 14)68 15)2
SOLUTIONS: 1)6*6*6*6
2)Consider the center of the two concentric circles as 0. Now Triagle OBC would be an equilateral triangle.(2*30) Now Consider the triangle OAD, With Triangle OAB and Triangle OCD you would get Angle OAD=120
Therefore Angle AQD=120/2= 60
3)Number of Minutes in 4 hours = 240 All alarms are set with the integral muntiples of minutes.
We are basicallly looking for the number of factors of 240.
240=2^4 * 3 * 5
Number of Factors = 5 * 2 * 2 =20
Therefore Maximum num of Clocks in the room= 20
4)Any number such that its last two digits are 0 would be divisible by 25(25*4). Therefore we have to take care of the last two digits only to get the remainder.
The Muntiplication of 21,we would always get the Last to digits as 21 41 61 81 01 and then it would start repeating.
This has cyclicity of 5
Consider 21^21 This has cyclicity 21mod5=1
Therefore 21.
So everytime we raise a power of 21 we would get a remainder 21 21*21=(rem)21 If we repeat this 21 times the final remainder would be 21.
5)Total travelling Time of Bus B = 6:30 am to 9:30 pm = 15 hours Total distance it had reduced ,between Bus A and B = 5 *15( 5 kms every hour) =75 km At 9:30 pm the distance between A and B.=21.
So at Morning from 5:00 am to 6:30 am (1 and 1/2 hours). The distance Bus A had travelled = 75+21 = 96 kms Distance Travelled by A in 1 hour = 64 km/hr
6) Let P,A and G be the respective volumes of Poison,Alcohol and Glycerine in V1,V2,V3 Total Poison in V2= 2P/9[ 2/3(P/3)] + P/81[1/3(1/3(1/3(P/3)))] =19/81
Total Alcohol in V2= 2A/3 + A/27[ 1/3(1/3(A/3))] 19/27
Therefore Ratio of Poison and Alcohol in V2
= { (19/81)P/(19/27)A)
But P:A = 4:5 Therefore (27*4) / (81*5) = 26.66 %
7)When 2 Students, A and B were absent No. of Bands left = 102 If there are x + 2 students in the class,
These 102 bands include x bands which student A would tie to the Class x bands which the class would tie to A.
x bands which Student B would tie to the Class x bands which the class would tie to B.
1 band which A ties to B. 1 band which B ties to A. =4x +2 bands =102
Therefore x= 25, Total students in the class =27.
No one bands left when only 1 student A is absent(B turns up) = 26 bands which B ties to rest of the class 26 bands which rest of the class ties to A.
=52
8)1. 37,1,1, 2. 33,3,3, 3. 21,15,3
9)To divide by 999 means to divide into groups such that each group has 999,Whatever is left is the remainder. Let us Suppose that We Divide this Number 123456789123456789341 by 1000.
Now this would leave a remainder 341.Now number of Small parts of 1000 Formed is 123456789123456789 Now if we want the Groups to be of 999,Each of 123456789123456789 groups of 1000 will give us 1 member each. Along with this we have 341.
Now we again divide What we have received 123456789123456789 members into group of 1000. 123456789123456 groups of 1000 Formed at the same time 789 members are left.
If we again want this to be divided into groups of 999 instead of 1000 We will get 123456789123456 .Along with this we will get the 789 Who could not be divided.
If we follow this process in recursion( Beleive me if we make a program using recursive function to find a solution to this ,it is a cakewalk). The final group of 1000 members that we would get will be 123. So we again take 1 member out from these 123 groups of 1000 members each.
Finally we have been able to sort out the whole lot that we have into groups of 999.. Now let us also sort out into GROUPS of 999, the members that remained in each of the Rounds.
341 + 789 + 456 + 123 + 789 + 456 + 123 _____
3077 ____
Now if we divide these 3077 into groups of 999, We could form 3 groups of 999, But still 80 members will be left.
Which is the remainder.
10)Is very easy to understand if you draw the diagram, Break Everything into Right Angled Triangle and Equilateral Triangles
E * * 60 * * * * * * * * * A * 2 * * 2 Z 5 * 60 * D ******************************* * ******* * X *60 *90 * * Y * * * * * * * * 4 * * 5 * * **************30 C B
Consider the equilateral triangle formed by the ED X
Now,the lengh of one of the arms of this equilateral triangle EDX is DZ + ZY + YX = 2 + 5 + 2 = 9
Therefore ED = EX = 9. But EX = EA + AX SO EA= 9  2 = 7
Now we have EA = 7 ED = 9 AB = 4 BC = 5 CD = 6
Therefore perimeter of ABCDE = 31
11) Let the number be abcd.
Now a+b+c+d=18 b+c = a+d b+d = 5 ( a + c )
a+c + 5(a+c) =6(a+c) = 18 Therefore a+c =3 b+d =15 Possible vaules of b And d
b d __ ___ 9 6 8 7 7 8 6 9
Therefore sum of all the possible values is 30.
12) D is one of the vertices of the triangle and it is fixed. Now we want to have two lines from D on BA and AC such that they should be of minimum length. The minimum length can be achieved if the two sides of the triangle FDE coincide with XD AND DY. So the three sides of the Triangle DFE would be of lengths 2.5,6 and The diagonal XY of the rectangle XDYA 13 is eleminated as the perimeter has to be more than 6 + 6+ 2.5 =14.5 Therefore Option 2) 17 cms. So the
13) 1 (17805,17803)
14)To find the Heaviest box the weight of the other boxes should be as much close to each other such that the difference between the Heaviest and the one less than the Heaviest is the maximum.
Consider the Sum of weights of the 2 lightest Boxes = 122 ,So the lightest box cannot be 61 kg,It has to be less than that. If the lightest box is 60,the second box would be 62,Similarly considering the Sum of weights the next boxes would be 64 and 65.. Now comparing all the Sum of boxes , The heaviest box would be of weight =68 Kg. Therfore 60+62 =122 60+64 =124 62+64 =126 60+68=128 60+65 = 125 62+65=127 65+64 =129 62+68 =130 64+68=132 65+68 =133 All Comply.
So Heaviest box =68 kg.
15)Solving the two equations we will get x^2  5x + 4 = 0
No of positive values of X. , x=1 and x=2
