Bunuel wrote:
10 students totally donated 28 clothes to an institution. If the range of the numbers of the clothes denoted by each student is 2, how many students denoted exactly 3 clothes?
I. 0
II. 5
III. 9
A. I
B. I and II only
C. II and III only
D. I an III only
E. I, II and III
Can it be all 10? NO, then we would require total to be 30.
Then, can it be 9? Yes, it could be with the tenth donating only 1. 9*3+1*1=28.
There is NO other way the range could be min 1 and max 3....
The range 2 can further give us the min as 2 and max as 4.
So, 2x+3y+4z = 30......This tells us that y or number of students donating 3 has to be even.
so y CANNOT be 5.
Can we have 2x+3*0+4z=30, where z=10-x. => 2x+40-4x=28 or x=6 and z=4.
Hence, 0 is possible.
D
ALGEBRAIC
range is 1 to 2......1a+2b+3c=28 and a+b+c=10
10-b-c+2b+3c=28........2c+b=18
So, b is even.
b=0.....2c=18 or c=9.......a+2b+3c = a+0+3*9=28....a=1......Possible
b=2.....2c=16 or c=8.......a+2b+3c = a+2*2+3*8=28....a=0......Not possible as range will become 1.
Thus, any other value will make x negative, and hence, not possible
range is 2 to 4......2x+3y+4z=28 and x+y+z=10
20-2y-2z+3y+4z=28........2z+y=8
So, y is even.
y=0.....2z=8 or z=4.......2*6+0+4*4 =28.........Possible
y=2.....2z=6 or z=3.......2*5+3*2+4*3=28.........Possible
Similarly for y=4, z=2 and x=4 and for y=6, z=1 and x=3.
Possible values for number of students who donated 3 shirts are 0, 2, 4, 6 and 9.