Equations and Expressions

Hello Samyak , 

Let the four digit number be aabb 

1100a + 11b = x²

Clearly , x is a multiple of 11 

so , 1100a + 11b = (11y)²

100a+ b = 11y²

a + b mod 11 = 0 

b = 11-a

100a +11 -a = 11y² 

99a +11 = 11y² 

9a +1 = y² 
Only possibility , a = 7, 

So , b = 4. 

aabb = 7744

Alternate Approach : 

We know 12² ends with __44, i.e last 2 digits repeating

So the square must be (50k - 12)² or (50k + 12)²

12² = 144 
38² = 1444
62² = 3844
88² = 7744 

Hence , 7744. 

p² + q² = ( p+ q)² - 2pq 

We know , sum of the roots = -  ( Coefficient of x) / ( Coefficient of x²) 

So , p + q = ( a -2) 

and the product of the roots = ( constant term )/ ( coefficient of x² )

=> pq = - ( a +1)

p² + q² = (a -2)² + 2( a + 1) 

= a² + 4 - 4a + 2a + 2 

= a² -2a + 6 

= ( a -1)² + 5 . 

minimum value of ( a -1)² is zero. Hence the minimum possible value of p² + q² is 5. 

Option (D) 

 answer 101?

open all the brackets with + sign and solve it will give you x<=36

open all the brackets with -ve sign and solve it will give you x>=(-64)

total integer values =36+64 +1              1 is for 0

=101 (answr)

9 is answer?

Let the total rings be n
1st  Nazgul recieved => 100 + (n - 100)/10 rings.  

2nd Nazgul recieved=> 200 + 1/10(n - 300 - (n- 100)/10) rings . 

since these both are equal, equate it.

n = 8100

so 1st Nazgul recieved = 100 + 8000/10 = 900

Since all person received same amount Hence , 8100/900 = 9 Nazguls

Hello Richa, 

Let the present age of B be x years.  

So ,when A was 3/5 of x. That time B was 5/7 of 28 = 20. 

Now , 

28 - 3/5 x = x -20 

{ Difference should be constant } 

x = 30 . 

27th

Solution to the 27th Question : 

-2 , -1 , 1 and 2 are the roots of the equation so 

f(x)=k•(x+2)(x+1)(x-1)(x-2).

f(p)=k(p+2)(p+1)(p)(p-1)(p-2)/p

=>  f(p) × p is a product of 5 consecutive numbers.

So,must be divisible by 5.

Now p, p+1, p+2 are 3 consecutive numbers,so should be divisible by 3,will say the same for p, p-1, p-2. So, divisible by 3² i.e 9. 

p-1, p+1 are consecutive even numbers,must be divisible by 8(Because 2 × 2²).

Hence , divisible by 5, 9 and 8

Largest Integer =5 × 8 × 9 = 360.

Option (4) 

28th

x² + 2y² + 4z² + 3 ( xy + yz + zx) = 16 ... ( 1) 

y² + 2z² + 4x² + 3 ( xy + yz + zx) = 16 ... (2) 

z² + 2x² + 4y² + 3 ( xy + yz + zx) = 16 ... (3) 

On adding (1) , (2) and (3) 

7 ( x² + y² + z²) + 9 ( xy + yz + zx) = 48

x² + y² + z² =3

xy+yz+zx=3

x=y=z=+1 or-1

so only 2 solutions are possible 

hence option 2

for a cubic polynomial ax³ + bx²+cx+d=0 

three roots can be described as 

p +q+r=-b/a    

pq+qr+pr= c/a

pqr=-d/a

so in given equation 

p+q+r=0.................(1)

pq+qr+pr=a.................(2)

squaring (1) 

p² +q²+ r²  +2(pq+qr+rp)= 0

p² +q²+ r²  =-2(pq+pr+qr)

from (2)

p² +q²+ r²  =-2a

a=-(p² +q²+ r²  )/2

since roots are integers and squaring them will always give a positive value 

hence a will always be negative integer

so 

answer is option (3)

answer is 4th option?

There is no formula for solving cubic polynomial you have to factorise them if possible or you have to look for the sum and product of roots here it can be factorised easily 

Got it. thanks alot 🙂