Bank Math Master

Question: If 4x² - px + 16 is a square number, then p =?

Solution:
4x² - px + 16
= (2x)² - 2. 2x. 4 + 4² - px + 16x
= (2x - 4)² - px + 16x 

Now,
- px + 16x = 0 [ the expression is a square number]
⇒ px = 16
∴ p = 16 

Question: In a certain language PENSION is coded as NEISNOP, how is FOLLAGE coded in that code?

Solution: 
PENSION ⇔ NEISNOP
P ⇒ E ⇒ N ⇒ S ⇒ I ⇒ O ⇒ N 
1 ⇒ 2 ⇒ 3 ⇒ 4 ⇒ 5 ⇒ 6 ⇒ 7
N ⇒ E ⇒ I ⇒ S ⇒ N⇒ O ⇒ P
Here,
Letters of 1st and 7th position inter-change their position.
Letters of 3rd and 5th position also inter-change their position.
And the letters of even number position remain same.

F ⇒ O ⇒ L ⇒ L ⇒ A ⇒ G ⇒ E 
1 ⇒ 2 ⇒ 3 ⇒ 4 ⇒ 5 ⇒ 6 ⇒ 7
E ⇒ O ⇒ A ⇒ L ⇒ L ⇒ G ⇒ F

∴ FOLLAGE ⇔ EOALLGF

Question: If A is to the south of B and C is to east of B, in what direction is A with respect to C?

Solution:

Question: If x² - 3x + 1 = 0, what is the value of x² - 1/x²?

Solution:
Given that,
x² - 3x + 1 = 0
⇒ x² + 1 = 3x
⇒ x + 1/x = 3 [Divided by x]

Now, 
x - 1/x = √{(x + 1/x)² - 4.x.(1/x)}
= √{3² - 4}
= √(9 - 4)
= √5

So,
x² - 1/x² = (x + 1/x)(x - 1/x)
= 3√5

Question: If X and Y are prime numbers then which of the following cannot be the sum of X and Y?

Solution:
The prime numbers which are less than 23 ⇒ 2, 3, 5, 7, 11, 13, 17, 19

Now,
5 = 2 + 3 
10 = 3 + 7 
26 = 7 + 19 

But we can not write 23 as sum of two different prime numbers.

Question: Having scored 117 runs in the 19th innings Tamim increase his average score by 5. What will be his average score after the 19th innings?

Solution: 
Let, 
Tamim's average score after the 19th innings is x
∴ Tamim's total score after 19th innings = 19x

Tamim's Total score after 18th innings = 19x - 117
∴ Tamim's average score after the 18th innings was (19x - 117)/18

ATQ,
(19x - 117)/18 = x - 5
⇒ 19x - 117 = 18x - 90
⇒ 19x - 18x = 117 - 90
∴ x = 27

His average score is 27

Question: If Z = 52 and BAT = 46, then ACT will be equal to-

Solution:

Here,
The Position number of Z is 26 which is re-write as 26 × 2 = 52.
Now,
BAT = (2 × 2) + (1 × 2) + (20 × 2) 
= 4 + 2 + 40
= 46

Similarly,
ACT = (1 × 2) + (3 × 2) + (20 × 2)
= 2 + 6 + 40
= 48 

Question: If x + 1/x = √7, then x - 1/x =?

Solution: 
Given that,
x + 1/x = √7

We know that,
x - 1/x = √{(x + 1/x)² - 4. x.(1/x)}
= √{(√7)² - 4}
= √(7 - 4)
= √3

Question: If the word NATION is written as OCWMTT, how COUNTRY would be written in that code?

Solution:

NATION = OCWMTT

Now,

COUNTRY = DQXRYXF

Question: If a = 5 + √2, then find the value of a²?

Solution: 
a = 5 + √2
a² = (5 + √2)²
= 5² + 2 × 5 × √2 + (√2)²
= 25 + 10√2 + 2
= 27 + 10√2

Question: If m = 7 -  4√3, then √m + 1/√m =?

Solution:
Given that,
m = 7 -  4√3
⇒ m = 4 - 4√3 + 3
⇒ m = (2)² - 2 . 2 . √3 + (√3)²
⇒ m = (2 - √3)²
∴  √m = 2 - √3

∴ 1/√m = 1/(2 - √3)
= {1(2 + √3)}/{(2 - √3)(2 + √3)}
= (2 + √3)/{2² - (√3)²}
= (2 + √3)/(4 - 3)
= 2 + √3

∴ √m + 1/√m = 2 - √3 + 2 + √3
= 4

Question: Six persons A, B, C, D, E, and F are standing in a circle. B is between F and C; A is between E and D; F is to the left of D. Who is between A and C?

Solution: 

Question: If a + 1/a = 2, Then a³ + 1/a³ =?

Solution:
Given that,
a + 1/a = 2

∴ a³ + 1/a³ = (a + 1/a)³ - 3. a. (1/a)(a + 1/a)
= 2³ - 3 × 2
= 8 - 6
= 2 

Question: Which number will complete the series: 1, 3, 7, 15, 31, 63, __?

Solution:
3 - 1 = 2
7 - 3 = 4 = 2 × 2
15 - 7 = 8 = 4 × 2
31 - 15 = 16 = 8 × 2
63 - 31 = 32 = 16 × 2

∴ The next number of 63 will be 63 + 32 × 2
= 63 + 64
= 127

Question: In a code PREMIER is written as XOILSIO, ANTAGONAISE is written as MQNMZBQMSWI, then how can REPORT be written in the same code?

Solution:
PREMIER is written as XOILSIO
P → X
R → O
E → I
M → L
I → S
E → I
R → O

ANTAGONAISE is written as MQNMZBQMSWI
A → M
N → Q
T → N
A → M
G → Z
O → B
N → Q
A → M
I → S
S → W
E → I

REPORT
R → O
E → I
P → X
O → B
R → O
T → N
∴ REPORT can be written as OIXBON.

Question: Four girls are sitting on a bench to be photographed. Seema is to the left of Rani. Mary is to the right of Rani. Rita is between Rani and Mary. Who would be second from the left in the photograph?

Solution:
Seema is to the left of Rani.
Seema ⇔ Rani

Mary is to the right of Rani.
Rani ⇔ Mary

Rita is between Rani and Mary.
Rani ⇔ Rita ⇔ Mary

∴ Seema ⇔ Rani ⇔ Rita ⇔ Mary.

Rani would be second from the left in the photograph.