1) Using two 2's and two 3's and using a maximum of three mathematical signs, symbols, can you have a result in between 14 and 15 ? concatanation ( clubbing of digits ) allowed.
Choices: a) ( 23 + 3! ) / 2 = 14.5 b) ( 33 * 2! ) - 5 = 12.5 c) ( 23 + 3! ) / 2 = 14.5 d) ( 32 - 2! ) / 3 = 15.4
Solution: The correct answer is : ( 23 + 3! ) / 2 = 14.5
2) A tap can fill the tank in 15 minutes and another can empty it in 8 minutes. If the tank is already half full and both the taps are opened together, the tank will be.
Choices: a) emptied in 8 min b) emptied in 12 min c) filled in 12 min d) filled in 8 min
Solution: The correct answer is : emptied in 8 min.
Explanation:
Rate of waste pipe being more the tank will be emptied when both taps are opened.
Net emptying workdone in 1min =(1/8 -1/16)= 1/16
So full tank will be emptied in 16 min
Half tank will be emptied in 8 minutes.
3) 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?
Choices: a) 79ml b) 81ml c) 72ml d) 91ml
Solution: The correct answer is : 81 ml.
Explanation:
Milk = (729 * (7/9))=567ml
Water = (729-567)= 162ml
Let water to be added be x ml 567/(162+x) = 7/3 1701 = 1134 + 7x x = 81ml
4) A man can row 5 kmph in still water. If the river is running at 1kmph, it takes him 75 minutes to row to a place and back. How far is the place?
Choices: a) 2.5 km b) 3 Km c) 4 Km d) 5 Km
Solution: The correct answer is :3 Km.
Explanation:
Speed downstream = (5+1)km/hr = 6 km/hr Speed upstream = (5-1)km/hr = 4 km/hr Let the required distance be x km x/6 + x/4 = 75/60 2x+3x = 15 x = 3km .
5) 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.
Choices: a) 2,6,8,20,65,150 b) 2,3,9,27,81,162 c) 2,4,10,25,97,185
Solution: The correct answer is : 2,3,9,27,81,162.
Explanation:
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
6) Joe has two sons and he says the fourth power of the age number one of his two sons when added to 4 which is the age of his other son is a prime number.
Sue, Joe's wife makes the following statements :
1. her youngest son is born on a tuesday
2. her eldest son fell ill on his 7th birthday
3. sum of the present ages of her two sons is divisible by 2
Now if only one of the three statements made by Sue is true, how old is Joe and Sue's other son ?
Choices: a) Sum of the present ages of her two sons is divisible by 2 b) Her eldest son fell ill on his 7th birthday c) Her youngest son is born on a tuesday d) None of the above
Solution: The correct answer is : Her youngest son is born on a tuesday.
Explanation:
If x is even then y=x^4+4 is even and cannot be prime. Then 3. is false and x is odd.
If x ends in 1, 3, 7 or 9 then x^4 ends in 1 and y=x^4+4 ends in 5. If x>1 then y>5 and cannot be prime. Then either x=1 or x ends in 5.
A quick test shows that fourth powers of numbers in human age range ending in 5 are not primes:
nn^4+4factors
562917*37
1550629197*257
25390629577*677
35150062913*115433
45410062913*315433
5591506292917*3137
651785062917*1050037
753164062953*596993
855220062913*4015433
958145062913*6265433
Then x=1, y=5, her eldest son fell ill on his 7th birthday is false, her youngest son is born on a tuesday is the true statement.
6) what is the 28383rd term in the series 1234567891011121314............
Choices: a) 3 b) 4 c) 7 d) 9
Solution: The correct answer is : 3.
Explanation:
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 devided 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.
7) 'Peerlike fleshy fruit, if you don't care
it's not important, it's the truth bare.
can you frame an idiom with the idea above ?'
Choices: a) A BLESSING IN DISGUISE b) NOT CARE A FIG c) CHOW DOWN
Solution: The correct answer is :
NOT CARE A FIG
8) Find the largest two and three digit integer value of 'X' with last digit ( at the unit place ) as 0 ( zero ) and a,b,c and d being non negative single digit integers ( including '0' ).
The additional condition now is : all them ( a,b,c,d ) are diffrent ( non repeating ).
Choices:
a) Two digits:62
Three digits:542
The additioanl case:
Two digits:59
Three digits:559
b) Two digits:90
Three digits:712
The additioanl case:
Two digits:60
Three digits:660
c) Two digits:85
Three digits:867
The additioanl case:
Two digits:74
Three digits:774
Solution:
The correct answer is :
Two digits:90
Three digits:712.
Explanation:
2^a can contribute 1
3^b can contribute 1, 3, 9, 27, or 81
5^c can contribute 1, 5, or 25
7^d can contribute 1, 7, or 49
81 can be used in 90 = 1 + 81 + 1 + 7, the best we can do
Three digits:
2^a can contribute 1
3^b can contribute 1, 3, 9, 27, 81, 243, or 729
5^c can contribute 1, 5, 25, 125, or 625
7^d can contribute 1, 7, 49, or 343
729 can be used in 780 = 1 + 729 + 1 + 49
(a higher power of 5 would require 7^d = 5 mod 10)
625 can be used in 970 = 1 + 1 + 625 + 343
(constructing 990 would require 3^b + 7^d = 364)
(constructing 980 would require 3^b + 7^d = 354)
Without 729 or 625, the highest possible is 712 = 1 + 243 + 125 + 343.
The additional condition is :
The additional condition now is :
*all them ( a,b,c,d ) are diffrent ( non repeating ).*
Two digits:
2^a can contribute 1
5^c + 7^d can contribute 5 + 49 or 7 + 25
3^b can contribute 27 or 81
60 can be constructed as 1 + 27 + 7 + 25
Three digits:
2^a can contribute 1
5^c can contribute 5, 25, 125, or 625
3^b + 7^d must contribute 4 (mod 10)
3^b can contribute 3, 9, 27, 81, 243, or 729
7^d can contribute 7, 49, or 343
550 can be constructed as 1 + 81 + 125 + 343
660 can be constructed as 1 + 27 + 625 + 7