HI there..
ANS-I C, let four balls be A,B,C,&D. S (A+B+C+D)/4 = 20;
now the median weight of the balls is 20 , so B+C?)/2=20 i.e. B+C=40 and A+D= 40, and as per the second statement no ball weighs 20 pounds hence...from above two eqns we can safely say that by combining two statements...we can get the answer ....
So i think the answer should be C....
ANS-II. my take on the question is B,we can solve the question by using only the seconsd statement... As the average is 32 so we can solve for x. and x =32...so the mode of the would be = 32...
what are the answers.,...
regrds
Jakhar
Quote:
Originally Posted by paki
Guys need your help in the following DS problems. Pls. include explanations!
I)
The average (arithmetic mean) weight of four balls is 20 pounds.
How many of the balls weigh more than 20 pounds?
1) The median weigh of the balls is 20 pounds
2) None of the balls weighs 20 pounds
II)
The high temperatures recorded for each day of a period of 5 consecutive days in February were 34, 30, x, 32, and x degrees Fahrenheit. What was the mode of these temperatures?
1) The median of the five temperatures is equal to the mode
2) The average(arithmetic mean) of the five temperatures is 32
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