Latest Infosys Aptitude Question SOLUTION: a sixth grade teacher asked her students to draw rectangles with positive integer length and a perimeter of 42.the difference between the largest and smallest possible ares of the

• Placement Papers
• /
• Infosys
• /
• Logical Reasoning
• /
• Direction Sense

Infosys Question Papers: Infosys Interview Experience(10) INFOSYS Company Profile

(#M40013511) INFOSYS Logical Reasoning question Keep an EYE

a sixth grade teacher asked her students to draw rectangles with positive integer length and a perimeter of 42.the difference between the largest and smallest possible ares of the rectangles that the students could have come up with is?

Asked In Infosys praveen (11 years ago)

Solved Mayur Read Solution (5)

Is this Puzzle helpful?   (8)   (2) Submit Your Solution Direction Sense

Read Solution (5) :

Mayur  (11 years ago)

sum of length and breadth will be 21 units . 42/2=21
area will be max when lxb = 11x10=110 sq units
area will be min when lxb = 20x1=20 sq units.

.the difference between the largest and smallest possible ares of the rectangles that the students could have come up with= 110-20=90 sq units

Is this solution Helpfull? Yes (47) | No (3)  

Anuj Katiyal  (11 years ago)

agree with mayur

Is this solution Helpfull? Yes (4) | No (3)  

T G  (10 years ago)

2(l+b)=42
l+b=21
l=20b=1
area=20(lowest)
if l=10
b=11
area=110(highest)

Is this solution Helpfull? Yes (3) | No   

Abhradeep Ganguli  (10 years ago)

smallest area will be for sides 1 and 20, area=20 sq. units. and largest area will be for sides 10,11. area=110 sq. units.

Is this solution Helpfull? Yes (2) | No (1)  

DEEPA RAM CHOUDHARY  (9 years ago)

2(a+b)=42 then
a+b =21
for lowest a=1and b=20 or a=20 and b=1

then area is 20
for highest max vale of both a anb b then
a=10 or 11 and b= 11 or 10 then area is 110

Is this solution Helpfull? Yes (1) | No   

Please Login to Read Solution.