HARDEST LOGICAL REASONING QUIZ QUESTIONS

HARDEST LOGICAL REASONING QUIZ, This type of questions is best for all exams like IAS etc.
Answers will be posted after some attempts, First do it yourself.

1)What is the largest positive integer 'n' for which there is a unique integer 'k' such that 8/15 < n/(n + k) < 7/13 ?

2)A positive integer N satisfies the following two conditions:
1. It must be a 10 digit number with all ten digits distinct
2. It must be divisible by 11111.
How many such numbers are there?

3)N is a 80-digit +ve integer ( in the decimal scale ). All digits except the 44th digit (from the left are 2. If N is divisible by 13, find the 44th digit.

4)The entrance ticket at the Inox theatre in Chennai is worth Rs 250. The sale of tickets increased by 50% when the price of the ticket was lowered. However, the collections recorded a decrease of 17.5%.The price of the ticket is reduced by how much rupees?

5)A number is mistakenly divided by 5 instead of being multiplied by 5.Find the percentage change in the result due to this mistake.

6)(x^x)^x – x^(x^x) = 0
How many values of x are possible?

7)A fort has provision of food for 150 en for 45 days. After 10 days, 25 mens left the fort. The no. of days for which the remaining food will last is?

8)Ter is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men ll be catered to with the remaining meal?

9)40 cows eat 40 bags of husk in 40 days . In how many days 1 cow ll eat 1 bag of husk?

2 comments:

Manpreet said...

Please post the answers.

Vikas Ahlawat(Hi) said...

Answres
1.8/15 < n/(k + n) < 7/13

=> 13/7 < 1 + k/n < 15/8

=> 6/7 < k/n < 7/8

=> Making denominator of upper bound and lower bond same =>

48/56 < k/n < 49/56

=> 96/112 < k/n < 98/112

=> smallest integral permissible value of k = 97.

Now, obviously ...

n =112 because beside this, 97/(112 +/- 1) will not satisfy the condition.


2answer.
Since all 10 digit numbers with distinct digits are divisible by 9 so the number is divisible by 9. And since 11111 and 9 are relatively prime, then number should be divisible by 9*11111 = 99999.

Now, consider following number

abcdefghij and what number 99999 must be multiplied by. It must be
larger than abcde, because abcde*100000 = abcde00000 < abcdefghij. It
can't be abcde+2 (or larger), because:

99999*(abcde+2) = (100000-1)*(abcde+2) = abcde00000 + 199998 - abcde

Looking at the 10000's digit of that, 0 + 9 - e would not require a borrow, regardless of the value of e. Therefore the 100000's digit would be e + 1, which doesn't equal the 100000's digit of abcdefghij.

That implies that:

abcdefghij = 99999*(abcde+1) = (100000-1)*(abcde+1)

=> abcde00000 + fghij = abcde00000 + (99999 - abcde)

=> fghij = 99999 - abcde

which implies

f = 9 - a
g = 9 - b
h = 9 - c
i = 9 - d
j = 9 - e

Now a can not be 0, f can not be 9. => 9 possible values for a, => 8 possible values for b, Now, {a,b,g,f} are known, => 6 possible values of c and then h is known, => 4 possible values for d and then 'i' is known => 2 possible values for e and then j is known..

So in total 9*8*6*4*2 = 3456 such numbers are possible.


3. 6



4 Rs 112.50



5. let the number be 5x
now diff is given by (5*5x)-(5x/5) =25x-x = 24x
%change in the result = (diff/result)*100= (24x/25x)*100 =96


6. 3

7. 210,(Assume that every men eat equal quantity of food )
Now, 150 men can eat 45 days means 1 man can eat 6750 days.
after 10 days 1500 days of food will be finished.
remaining 5250 days of food is shared by 25 people

thus 5250/25=210


8. Let (120) men = (200) children= x . that means 150 children=90 men

now, x=(250) children+(men) y

but 120 men=90 men+men(y)

thus y=30 men


9. One cow eat 1 bag in 40 days. that means 40 cows need 40 bags for 40 days

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