BANK & INSURANCE (NEW PATTERN QUADRATIC EQUATION)

Total Questions: 15

11. Direction (11-12): Study the following information carefully and answer the questions given beside.

(i) a² - (8²)a/2 + 25 × 8 = a
(ii) b² - 2Kb - 5K = 0
(iii) c² - 121 = (K + 1)c + 29

One root of the equation (i) is 2K (where K is an integer)

What is the sum of the smaller root of equation (ii) and the larger root of equation (iii)?

Correct Answer: (a) 13
Solution:

(i) a² − (8²)a/2 + 25 × 8 = a
a² − 32a − a + 200 = 0
a = 25, 8
As K is an integer 2K = 8
K = 4

(ii) b² − 2Kb − 5K = 0
b² − 8b − 20 = 0
b = 10, −2

(iii) c² − 121 = (K + 1)c + 29
c² − 121 = 5c + 29
c² − 5c − 150 = 0
c = 15, −10

11) Reqd. difference = 15 + (−2) = 13

12. What is the sum of all the larger roots of all three equations?

Correct Answer: (a) 50
Solution:

(i) a² − (8²)a/2 + 25 × 8 = a
a² − 32a − a + 200 = 0
a = 25, 8
As K is an integer 2K = 8
K = 4

(ii) b² − 2Kb − 5K = 0
b² − 8b − 20 = 0
b = 10, −2

(iii) c² − 121 = (K + 1)c + 29
c² − 121 = 5c + 29
c² − 5c − 150 = 0
c = 15, −10

12) Reqd. sum = 25 + 10 + 15 = 50

13. One of the roots of equation a² - ma + 50 = 14 is 4 and others roots is n.

Find sum of roots of equation 5a² - (m - n) × 6a + 100 = 33?

Correct Answer: (a) 4/5
Solution: Sum of roots of equation a² − ma + 50 = 14 or a² − ma + 36 = 0 is m and products of roots are 36.
One root is 4 so, another root is 36/4 = 9 = n
So, m = 9 + 4 = 13
So, sum of roots 5a² − (13 − 9) × 6a + 63 = 0 is
[−(−4)/5] = 4/5

14. Direction (14-15):

Equation I: y³ + Ky = R(y² + 1.25)
Equation II: (n - a)ᵃ⁺¹ = K + R + a

a - 1, a and 2a + 1 are the roots of equation I, where 'a' is a natural number.

If one root of the equation 2y² - 29y + z = 0 is n/a, find the other root.

Correct Answer: (b) 12  
Solution:

y³ + Ky = R (y² + 1.25)

y³ − Ry² + Ky − 1.25R = 0
a − 1 + a + 2a + 1 = R
4a = R

a = 0.25R
(a − 1) × a × (2a + 1) = 1.25R
(a − 1)(2a + 1) × 0.25R = 1.25R
2a² + a − 2a − 1 = 5
2a² − a − 6 = 0
2a² − 4a + 3a − 6 = 0
a = 2, −1.5 (As a is a natural number, it cannot be −1.5)
a = 2
R = 4a = 8

y³ − 8y² + Ky − 10 = 0
Putting y = 2
2³ − 8 × 2² + 2K − 10 = 0
K = 17

(n − a)^(a + 1) = K + R + a
(n − 2)³ = 17 + 8 + 2
(n − 2)³ = 27
n = 5

14): one root = n/a = 5/2

other root = 29/2 − 5/2 = 12

15. If there are (K + 3) students in a class and 25 % of them are girls, find the difference between the number of boys and girls.

Correct Answer: (b) R + a  
Solution:

y³ + Ky = R (y² + 1.25)

y³ − Ry² + Ky − 1.25R = 0
a − 1 + a + 2a + 1 = R
4a = R

a = 0.25R
(a − 1) × a × (2a + 1) = 1.25R
(a − 1)(2a + 1) × 0.25R = 1.25R
2a² + a − 2a − 1 = 5
2a² − a − 6 = 0
2a² − 4a + 3a − 6 = 0
a = 2, −1.5 (As a is a natural number, it cannot be −1.5)
a = 2
R = 4a = 8

y³ − 8y² + Ky − 10 = 0
Putting y = 2
2³ − 8 × 2² + 2K − 10 = 0
K = 17

(n − a)^(a + 1) = K + R + a
(n − 2)³ = 17 + 8 + 2
(n − 2)³ = 27
n = 5

15) K + 3 = 17 + 3 = 20

  Boys : Girls = 75 : 25 = 3 : 1

  Difference between boys and girls

  = 20 × (3 − 1) / (3 + 1) = 10

  R + a = 8 + 2 = 10