A fisherman's day is rated as good if he catches 9 fishes ,fair if 7 fishes and bad if 5 fishes .He catches 53 fishes in a week n had all good,faiur n bad days in the week.So how many good ,fair n bad days did the fisher man had in the weekAns:4 good,1 fair n 2 bad daysplz explain the procedure

balachandra

• Nov 29th, 2006

 

Simple Silly question

Go to river catch fish

4*9=36

7*1=7

2*5=10

36+7+10=53...

Sumeet

• Dec 29th, 2006

 

Hi Mr. BalachandraI think you worked on the solution provided above. Plz imagine the solotion is not provided then?Imagine that u dont have any option to choose. Then how to come to the conclusion?

Sumeet Chakraborty

• Dec 29th, 2006

 

I have another possibility:3 Good: 3*9=273Fair: 3*7=211 Bad : 1*5=5The total becomes: 53

pradeep reddy

 Profile Answers by pradeep reddy 

• Jan 4th, 2007

 

its depends on options of that question

suganya_m11

 Profile Answers by suganya_m11 

• Feb 12th, 2007

 

I think u r missing some statement here.the procedure will be take no of good days as x,fair days is y ,bad days is z.then x+y+z=7and9x+7y+5z=53one more equation must be there to get a exact and unique answer.So i think a statement must be missing here.

sankar reddy

• Feb 20th, 2007

 

simple

             take what is given 53
                
               good days means --- 9 fishes so 53/9=4(remainder=17)
          if u assume 5 then there is no chance for bad days.
              
               fair days means ----- 7 fishes so remaining 17 --- 17/7=1(remainder=10)
         if u assume 2 then there is no chance for bad days.
  
             bad days means -------5 fishes so remaining 10---10/5=2days.

   Ans: 4 good ,1 fair, 2bad. ==== total 7 days.

sanjeev

• Feb 27th, 2007

 

since 7 can be formed by 3 numbers in only one way =4+2+1.
so this is the only way and by hit & trial method it can be determined easily.
4*9+1*7+2*5.

rajkumar

• Mar 4th, 2007

 

----no need of third eq
----hit and trial not giving proper ans



x+y+z=7--------- eq1
9*x+7*y+5*z=53 -------eq2
multiply eq 1 by 9,
9*x+9*y+9*z=35 -------------eq3
from eq2 and eq3
2*y+4*z=10-----eq4
since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z = 2 or 1 from eq 4
for first  y=1,z=2 then from eq1 x= 4
so 9*4+1*7+2*5=53.... satisfied
now for second y=3 z=1 then from eq1 x=3
so 9*3+3*7+1*5=53 ......satisfied


so finally there are two solution of this question

(x,y,z)=(4,1,2) and (3,3,1)...

if u ppl like my ans reply on raj_er04@yahoo.co.in

rajkumar from sastra here ............... thakns

RAHUL UBHALE

• Aug 26th, 2014

 

3*9=27
3*7=21
1*5=5

21+27+5=53

so
good days=3
fair days=3
bad days=1