1. A three-digit number XYZ is formed of three different non-zero digits X, Y, and Z. A new number is formed by rearranging the same three digits. What is the greatest possible difference between the two numbers? (For example, 345 could be rearranged into 435, for a difference of 435 – 345 = 90.)
- (A) 792
- (B) 293
- (C) 564
- (D) 984
- (E) 384
2. How many three-digit positive integers, with digits x, y, and z in the hundred’s, ten’s and unit’s place respectively exist such that x < y, z < y and x ≠ y, z < y and x ≠ 0?
- (A) 245
- (B) 285
- (C) 240
- (D) 320
- (E) 948
Math1089 
Problem #2) Solution :
B D=x cm;;
C o s
Or (A B ^2+ B D ^2-A D ^2)/(2 * A B* B D)=(A C ^2+C D ^2-A D ^2)/(2* A C * C D);
Or ( 1 7^2 +x ^2 -1 5 ^2/(2* 17 * x)=(1 7^2 + 4 ^2-1 5 ^2)/(2 * 17* 4);
Or (64+ x^2)/x= (80)/4=20;
Or 64+ x^2=20 *x; or x=4 or 16 ;
Admissible value of x=16B D= x cm=16 cm
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Thank you sir for another solution
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Please explain this answer.
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Solution of problem1.
First we will measure 3 conis each side.
Case1 – if balance is eqall we will measure two out of 3 coins putting one on each side, thus we will get the heavier one.
Case 2 – if one side is heavier them we will take 2 coins of the heavier side and measure by putting one each side, thus we can find the heavier one.
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Solution of problem1.
First we will measure 3 conis each side.
Case1 – if balance is eqall we will measure two out of 3 coins putting one on each side, thus we will get the heavier one.
Case 2 – if one side is heavier them we will take 2 coins of the heavier side and measure by putting one each side, thus we can find the heavier one.
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Thank you so much Sir
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Question 1 solution :
(c)
As 10 houses have less than 6rooms, they are to be excluded.
Given,
4 houses have more than 8 rooms .
Therefore, number of houses having either less than 6 or greater than 8 rooms = 10 + 4 = 14.
The remaining houses, that is 11 houses fulfill the above mentioned criteria.
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