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59 Analytical Questions And Answers

2⟩ A Father, son and grandson are walking in the park. A man approaches them and asks for their age. The Father replies, "My son is as many weeks as my grandson is in days, and my grandson is as many months old as I am in years. We are all 100 years together.60,35,5?

The detail solution to this question is like this.

Let's Say father's age in year is Y yr. Son's age in weeks is x week, which if converted to year is 7x/365yr. Now as per the question,

The Father replies, "My son is as many weeks as my grandson is in days

so grandson's age in days is X days.,which if converted to year is x/365 yr.

So 1st equation: y+7x/365+x/365 = 100....................(1)

Father says My grandson is as many months old as I am in years.

means, grandson is y months old,b'coz father is y yrs old.So equating grandson's age in days(x days) from above statement with his age in months (y month),both converted in years, gives us

x/365 = y/12.......................(2)

solving these two equations gives y=60 in years,that is father' age.grandson is as many months old as father is in years.So grandson's age is 60 months,i.e 5yrs,which converted to days is 1825 days.As son's age is as many weeks as grandson's age is in days.So son's age is 1825 weeks,that is converted in yrs is 1825*7/365=35yrs.

father 60yrs

son 35yrs

Grandson 5yrs

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5⟩ What is the next Sequence?1. 7,6,8,5,3,7,?2. 13 9,18,21,25,20,?3. 3,3,4,8,10,36,?4. 15,30,28,25,20,34,28,?

the digit count of the sequence isof the form..... (ie)1+5=6 3+0=3 2+8=1 etc.

6 3 1 7 2 7 1 .....so the remaining two nos. should have the count 3 and 6 respectively.....

moreover the flow of the sequence is "the second no. is greater than first and all other are less upto the no. with count 2(i.e) middle no.

and the no. next to middle no. (i.e)34 is greater than 20...

so all others should be less than 34....

on keeping all these in mind... v ll have 21 : 2+1=3

15 : 1+5=6

so the sequence is 15 30 28 25 20 34 28 21 15

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9⟩ A camel must travel 15 miles in order to reach the nearest city. She have 45 bananas with her but can carry only 15 at a time. Also she eat 1 banana /mile. Then the maximum no of bananas that can be transported to the city?a) 4 b) 6 c) 8 d) 12

6 Bananas can be transported to the city.

The camel carries 15 bananas for a kilometer and comes back (15 bananas becomes 13). It carries the second set of 15 bananas for a kilometer and comes back (15 bananas becomes 13). It carries the third set of 15 bananas for a kilometer (15 bananas becomes 14). Now we have 40 bananas which has to be carried for 14 kilometers.

At the end of the second kilometer the total number of bananas becomes 35.

At the end of the third kilometer the total number of bananas becomes 30.

At the end of the fourth kilometer the total number of bananas becomes 25.

At the end of the fifth kilometer the total number of bananas becomes 20.

At the end of the Sixth kilometer the total number of bananas becomes 15.

The camel carries the remaining 15 bananas for the last 9 more kilometers; and by the time it reaches its destination, it has consumed 9 bananas -- The remaining is 6 Bananas

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15⟩ Suppose If there are 30 cans out of them one is poisoned if a person tastes very little he will die within 14 hours so if there are mice to test and 24 hours to test, how many mice are required to find the poisoned can?

Pick 5 mice

Assign binary codes to 30 cans. you need 5 bits to represent all 30. Pick 5 mice to represent each bit. Have a mice taste all the cans for which the corresponding bit is one. After 14 hours, some will die and some wont. Set 1 for the former and 0 for the latter. The corresponding bit representation is the poisoned can.

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