# Question Bank - Equations - Set 2

• Write equation in this way as 2b^2 – 10ab + (17a^2 - 6a +2) = 0
Find discriminant D = b^2 – 4ac ( ax^2 + b x +c = 0 )
D = (-10a)^2 – 4*2 (17^2 -6a+2)= -4(3a-2)^2 , this is negative unless a= 2/3 , now solving for b , it is 5/3
(a,b)= (2/3,5/3)

• Q12) Find the sum of digits in the product 199999997 * 200000003 ?

• Use identity ( a-b) (a+b)= a^2 –b^2 , take a= 2 * 10^8 , b = 3 , so the product is
39999…(15 times)1
Sum = 3 + 15 * 9 + 1 = 139

• Q13) A three digit natural number xyz is written side by side to form a six digit natural number n = xyzxyz . then it must be divisible by ?

• N = abc * (1001)
It must be divisible by 7, 11,13 because 7 * 11 * 13= 1001
Also it must be divisible by factor of abc

• Q14) If (x-7)^{ (x^2 - 29x + 154)/(x^2 - 12x + 32)} = 1 , how many real values of x satisfy the given equation ?

• (x-70^{(x-7)(x-22)/(x-8)(x-4)} , now let we take f(x) = (x-7) and G(x) = (x-7)(x-22)/ (x-8)(x-4) , now we take three cases .
i) If f(x) = 1 and g(x) can be anything
so x = 8 , for this value g(x) can not be defined
ii) if f(x)=-1 , g(x) is an even , then x = 6 , g(x) is defined
iii) if f(x) does not zero and g(x) is zero ,
(x-7)(x-22) = 0 => x=7,22 , but for x=7 the eqn is not defined , so eqn is defined for two reals x i.e 6, 22

• Q15) Find the number of integral solutions for (a, b, c) for if a^2 = bc + 4 and b^2 = ca + 4

• OA - 12 triplets are possible.

• Q16) Find a natural number n for which 3^9 + 3^12 + 3^15 + 3^n is a perfect cube

• First of all we will try to find out cube
Now, 3^9+ 3^12 +3^15 +3^n = 3^9( 1 + 3^3 + 3^6 +3^n-9 )
= (3^3)^3 [ 1+ 3.3^2 + 3.(3^2)^2 +(3^2)^3 + 3^(n-9) – 3(3^2)^2]
= (3^3)^3[(1+3^2)^3]
Since, 3^(n-9) – 3^5 = 0
3^(n-9) = 3^5
n - 9 = 5
so, n = 14
hence given number is a pefect cube for n = 14

• Q17) If a, b, c, d, e are positive real numbers such that a+b+c+d+e = 8 and a^2 + b^2 + c^2 + d^2 + e^2 = 16 find range of e.

• Concept - We know that these of type of questions can be solved by inequality.
If a, b, c, d are positive real numbers then [( a+ b+ c + d)/4]^2 ≤ (a^2 +b^2 + c^2 + d^2)/4 ... eqn(1) { by Techebycheff’s Inequality}
Now, a + b + c + d + e = 8
a + b + c + d = 8 - e
a^2 + b^2 + c^2 + d^2 = 16 – e^2
putting above value in equation (1)
(8-e)^2/ 16 ≤ ( 16 – e^2)/ 4
On solving this ,
e(5e – 16) ≤ 0
Using number line rule .
0 ≤ e ≤ 16/5

• Q18) Find the period of f(x) = Sin^4 x + cos^4 x

• Since Sin x and Cos x both has a period Pi, but f(x) is even function and Sin x and Cos x are complementary.
So f(x) has period = 1/2 {LCM of (Pi, Pi) } = Pi/2

• Q19) How many values of x in -19 < x < 98 satisfy Cos^2 x + 2Sin^2 x = 1

• Cos^2 x + 2Sin^2 x = 1
We know , Cos^2 x + Sin^2 x = 1
On subtracting we get ,
Sin^2 x = 0
Sin x = 0 then x = nPi {general solution}
So, -19 < n Pi < 98
Dividing by Pi we get, -6.04 < n < 31.19
So, total number solutions equal to n = 38

• Q20) The cost of 3 Pizzas, 5Samosas, 1 order of milk ata a restaurant is 23.5 rupees . In the same restaurant the cost of 5 pizzas, 9 samosas, and 1 order of milk is 39.50 rupees. What is the cost of 2 pizzas, 2 samosas, 2 milk at restaurant.

• Let, Pizza cost = P, Samosa cost = S, Milk cost = M
Now, Making equation
3P + 5S + M = 23.5 …… (A)
5P + 9S + M = 39.5 …… (B )
Multiplying equation (A) by 2 & subtracting equation (B )
P + S + M = 7.5
So required answer = 2 * 7.5 = 15

Second approach - Solve equation (A) & (B ) by elimination method and get values of P & S .
It is bit lengthy .
So this method will give you exact value of P , S & M .
Try yourself .

• Q21) Find all Positive integers x &y satisfying 1/root(y) + 1/root(x) = 1/root(50)

63

60

1

61

63

61

3

61