**1**. Let *P* be a prime number greater than 37. Then the largest number that will always divide (*P* – 1) × (*P* + 1) is

- (
**A**) 32 - (
**B**) 16 - (
**C**) 8 - (
**D**) 4 - (
**E**) 24

**2**. In the given figure, *AL* is perpendicular to *AB* and *AL* || *BK* || *CI* || *DG* || *EF*. Also *AE* || *KG* || *JF* and *AL* = *AB* = 2*BK* = *BC* = *CI* = 4*CD* = 3*IJ*. Then the ratio of the areas of the quadrilaterals *ABKL* and *IJKG* is

- (
**A**) 3 : 1 - (
**B**) 36 : 19 - (
**C**) 19 : 12 - (
**D**) 22 : 13 - (
**E**) None of these

**3**. How many functions are there from the set {1, 2, . . . , *k*} to the set {1, 2, . . . , *n*}?

- (
**A**) *kn* - (
**B**) *n*^{k} - (
**C**) *n*^{k }^{− 1} - (
**D**) *k*^{n }^{− 1} - (
**E**) *k*^{n}

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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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