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I think E

Stmt 1 : y+5 and x+5 ...but what could be x or what could be y ? Insufff

Stmt 2 : no. of apples > no of pears ok ...what about price ??

Insuff

y x y+5 x+5 No of Apples No of Pears Cost

1 3 6 8 8 7 8*6 < 7*8

2 3 7 8 8 7 56 = 56

E it is
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going with C

St1 Cost of apples x(y+5), Cost of pears y(x+5)
St2 x>y

Combined, from St1
Cost of apples = xy +5x --- 1
Cost of pears = xy +5y --- 2

Since xy is common the differentiator is 5x and 5y since , x>y it is obvious 5x>5y (x, y cannot be negative)
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C

St1: Cost of apples = x(y+5) = xy + 5x
Cost of pears = y(x+5) = xy + 5y :INSUFF

St2: x> y : Clearly INSUFF

Combined:
because x>y and both are +ve.

xy + 5x > xy + 5y is always TRUE: SUFF
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Answer: C
a = price of apples
p = price of pear

Need to find if
ax > py

S1: a = (y+5) p = (x+5)

Solving x > y
Not sufficient.

S2: a > p Not sufficient.

S1 & S2:
x > y, a > p => ax > py
Sufficient.
Answer: C



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