**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

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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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