In the following question, which one set of letters when sequentially placed at the gaps in the given letter series shall complete it? JK_MJ_LM_KL_

Let us assume the one adult fare be a and one child fare be c.
According to question,
Bus fare of one Adult is six times the fare of one child.
a = 6c ........................(1)
If an adult's bus fare is 114.
a = 114
put the value of a in equation (1). We will get
6c = 114
? c = 114/6
? c = 19
Now we have to calculate amount paid by 4 adults and 5 children for Bus fare
Bus fare of 4 adults and 5 child = 4a + 5c
Put the value of a and c in above equation. We will get
? Bus fare of 4 adults and 5 child = 4 x 114 + 5 x 19
? Bus fare of 4 adults and 5 child = 456 + 95
? Bus fare of 4 adults and 5 child =?/- 551

(x - 1) (2x - 5) = 0 ? x = 1, 5/2
So, its roots are real.

[a + (1/a)]² = 3
Taking square roots both sides, we get
a + (1/a) = ?3
On cubing both sides, we
[a + (1/a)]³ = (?3)³
? a³ + 1/a³ + 3.a.1/[a(a + 1/a)] = 3 ?3
? a³ + 1/a³ + 3?3 = 3?3
? a³ + 1/a³ = 0

Given quadratic equation is
7y² - 50y + k = 0
If one root is 7, then it will satisfy the equation i.e putting y = 7 in equation
7 x (7)² - 50 x 7 + k = 0
? 7 x 49 - 350 + k = 0
? 343 - 350 + k = 0
? k = 7

Given equation is
x² + 2(k - 4)x + 2k = 0
On comparing with ax² + bx + c = 0
Here, a = 1, b = 2(k -4), c = 2k
Since, the root are equal, we have D = 0.
b² - 4ac = 0
? 4(k - 4)² - 8k = 0
4(k² + 16 - 8k) - 8k = 0
? 4k² + 64 - 32k - 8k = 0
? 4k² - 40k + 64 = 0
? k² - 10k + 16 = 0
? k² - 8k - 2k + 16 = 0
? k(k - 8) -2 (k - 8) = 0
? (k - 8) (k - 2) = 0
Hence, the value of k 8 or 2.

Let ? and ? be the roots of the quadratic equation x² + px + q = 0
Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e products of roots
= q = ? . ? = 6 x 2 = 12 ...(i)
and B starts with a wrong value of q and gets the roots as 2 and - 9. But this time p is correct i.e., sum of roots
= p = ? + ? = - 9 + 2 = - 7 ..(ii)
(? - ?)² = (? + ?)² - 4? ?
= (-7)² - 4.12 = 49 - 48 = 1
[from Eqs. (i) and (ii)]
? ? - ? = 1 ..(iii)
Now, from Eqs. (ii) and (iii) , we get
? = - 3 and ? = - 4
which are correct roots .

Let the original price of be N .
? At discount of 20% article costs ? 596
? 596 = (80 x N)/100
? N = (596 x 100)/80
? N = 745
? original price = ? 745

Let CP of the goods = ? N
? Marked price of the goods = N x {(100 + 30)/100} = ? 13N/10

Now, SP of the goods = (13N/10) x (100 - 15/100)
= (13N/10) x (85/100) = ? 221N/200

? Profit = {(221 x N)/200} - N = ? 21N/200

Hence, profit per cent = [{21N/200} /N] x 100 % = 2100/200 = 10.5 %

? MP of one dozen of pairs of socks = ? 80
? SP of one dozen of pairs of socks
=80 x (100 - 10)/100 = 80 x 90/100= ? 72

Hence, required number of pairs of socks
purchased for ? 24 = (12 x 24)/72 = 4

? S.P of the notebook = ?457
? Commission on one notebook = ? (4 x 457)/100
and commission on 10 notebooks = 10 x (4 x 457)/100 = ? 18280
Now,S.P of the pencil box = ? 80
? Commission on 1 pencil box = ? (80 x 20)/100
Commission on 6 pencil boxes = (80 x 20 x 6)/100 = ? 96

Hence ,total commission of 1 day = (182.80 + 96 ) = ? 278.80
Thus, total commission of 2 weeks = 278.80 x 14 = ? 3903.20