Bank Math

Topics: Surds & Indices , Logarithim & Equation

Shohagh Sir
Facebook: M M Shohagh
Facebook Group: Bank Math Fighters
fwZ© ev Z‡_¨i Rb¨ †hvMv‡hvM Kiæb : 01708978397
Surds & Indices
Type-1(a)
1. If 1 =1, then 100 is:
A. 50 B. 25 C. 10 D. 1
2. 58=?
A. 0 B. 8 C. 40 D. 200
3. Which of the following are equal in value? [BHBFC (OF)-2017]
I. 14 II. 40 III. 41 IV. 0 4

A. II and III B. I and II C. III and IV D. I and IV

3 4
4. 17 +17 = ?
A. 177 B. 1712 C. 173(18) D. 2(174)+17
5. (81)4  (9)5 = ?
A. 9 B. 81 C. 729 D. 6581
6. (64) (8) =?
4 5

A. (8)8 B. (8)2 C. (8)12 D. None

7. (1000)12(10)30 = ?
A. (1000)2 B. 10 C. 100 D. (100)2
8. (3) (3) = ? (*)
8 4

A. (27)3 B. (27)5 C. (729)2 D. (729)3

9. (10)24(10)-21=? (*)
A. 3 B. 10 C. 100 D. 1000

Type-1(b)
 
2
6 3
1. What is the value 2 ?

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 A. 16 B. 12 C. 8 D. 4
1
1 2
3.   is equal to
2
1
A. B. 2 2 C. - 2 D. 2
2
2
3
4. The value of 27 lies between
A. 0 and 1 B. 1 and 2 C. 2 and 3 D. 3 and 4
4

 32  5
5. The value of   is
 243 
4 9 16 81
A. B. C. D.
9 4 81 16
2
 1 3
6. The value of    is
 216 
1 1
A. 36 B. -36 C. D. -
36 36
7. (0.04)1.5 =? [Janata Bank (EO) -2017 ]
A. 25 B. 250 C. 125 D. 625
5
  
5

5 3  3 
8. Simplified form of  x 5   is
 
  
 
1
A. B. x C. x-5 D. x5
x
2
6  3 67
3
9. Find the value of .
3
66
A. 6 B. 8 C. 11 D. 10
10. (256)0.16  (16)0.18
A. 4 B. 8 C. 12 D. 16
2 4
 
 1   1  3 3
11.     = ?
 216   27 
3 2 4 1
A. B. C. D.
4 3 9 8
1 4
12. 64 2   32 5 =?
1 3 1 3
A. B. C. D.
8 8 16 16
1
14. The value of 5  (125)0.25 is:
4

A. 5 B. 5 C. 5 5 D. 25

Type-02
1. 3x+3x+3x = ? [BKB-(SO)-2017]
A. 9x B. 27x2 C. 3 x+1
D. 3x 3

2. 53+53+53+53 + 5 3 = ? [Agrani Bank-(SO)-2017]

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 A. 2515 B. 252 C. 52 D. 515
3. 230 +230 +230 +230 = ?
A. 8120 B. 830 C. 232 D. 230
4. ? [BB-Officer-2022]
A. B. C. D.
5. 3 +320 +320 = ?
20

A. 920 B. 960 C. 360 D. 321

Type-03
1. If ax =b, by = c and, cz = a then the value of xyz is?
[PKB- Cash-2018] + [Janata- (Cash)-2020]
1
A. 0 B. 1 C. D. abc
abc
2. If px =q, qy = r and rz = p then the value of xyz is [Sonali Bank -SO--2019]
A. 1 B. 1 C. 0 D. pqr
a b c
3. If x = y , y = z and z = x then the value of abc is
[BHBFC -SO)-17, Rupali Bank-(Cash)-18 & BB-AD-18]
A. 1 B. 0 C. 0.5 D. Infinity
4. If ax =by =cz and b2 = ac, then y equals
xz xz xz 2 xz
A. B. C. D.
x z 2( x  z ) 2( z  x ) ( x  z)
n 1 n 3
p q
5. If      , Then, the value of n is- [PKB-(EO)-2019]
q p
1 7
A. 2 B. C. D. 1
2 2
x 1 x3
a b
6. If     , then the value of x is
b a
1 7
A. B. 1 C. 2 D.
2 2
rx  s px  q
 p q
7. If      , find value of x.
q  p
qs ps qr
A. B. C. D. None
pr qr ps
8. If (7a) (7b)= 7c/7d, what is d in terms of a, b and c? [PKB-Officer-2021]
A. 3/ab B. c-a-b C. a+b-c D. c-ab

Type-04
1. 4n = 64 n‡j n Gi gvb wb‡Pi †KvbwU n‡e? [Sonali Bank SO-2014]
A. 0 B. 1 C. 3 D. 4
2. If 2n =64, what will be the value of n?
[Janata Bank (EO)-2017 ,Janata Bank (AEO-Teller)-2019]
A. 2 B. 4 C. 6 D. 12
3. If 3n = 729, the value of n is: ( 3n = 729 n‡j, n Gi gvb KZ?)
[Agrani Bank-SO-2017]
A. 6 B. 8 C. 12 D. 10

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4. If 4x+1 =32, then x = ? [BKB-(Cash)-2017]
3 2
A. 2 B. 3 C. D.
2 3
5. For what value of x is 82x4 =16x? [Agrani Bank-(SO)-2017]
A. 2 B. 3 C. 4 D. 6
6. If 162x+4 =43x+3 then x = ? [BKB (Officer)-2017]
A. 5 B. 1 C. 13/5 D.1
7. If 82x+3 =23x+6 then x =?
A. 1 B. 1 C. 1/2 D. 2
8. For what value of x is 82x4 =16x?
[PKB-Cash-2018,Combined-4 Banks (Off)-2019]
A. 2 B. 3 C. 4 D. 6
9. If 32a+b = 16a+2b, then a = ?
A. b B. b+2 C. 2b D. 3b
n (n3)
10. if 5 = 3125, then the value of 5 is [Agrani Bank-SO-17]
A. 25 B. 125 C. 625 D. 1625
11. If 2n = 128, then (2n1) (5n2) = ?
A. 2(105) B. 5(105) C. 2(106) D. 2(56)
12. Given (‡`qv Av‡Q) 2 =32, What is the value of 3x+2?
x+3

A.4 B. 27 C. 81 D. 125
1
13. If 22x1 = x3 , Then the value of x is. [Rupali & Janata Bank -(Off)-2019]
8
A. 3 B. 2 C. 0 D. 2
7
1
14. If x 2  then the value of X is? [Sonali Bank Officer -2018]
128
A. 8 B. 4 C. 4 D. 2
15. If then [Combined 6 Banks -AP-2021]
a. b. c. d.

16. If = 0.008, then find [PKB-Cash-2021]

a. 0.75 b. -0.75 c. 0.25 d. 0.0

Type-05
1. If 3 x  2 3 , what is the value of x? [BKB (SO)-2017]
A. 3 B. 1.33 C. 2 D. 3 2
2. 50 20 is 10 times larger than 110 , where x is:
7 7 x 8
[Agrani Bank-(SO)-2017]
A. 13 B. 6 C. 21 D. 29
3. If x = 4, then  2 2 x +2?
A. 14 B. 8 C. 2 D. 0
3 2
4. If x = 2 then the value of x +27x +243x +631 is [Rupali & Janata Bank -(Off.)-2019]
A. 1211 B. 1231 C. 1321 D. 1233

5. If ‘m’ and ‘n’ are whole numbers such that mn = 121, the value of (m1)n+1 is?
[Agrani Bank-SO -17, Combined 5 Banks -Cash-19]
A. 21 B. 10 C. 100 D. 1000

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Type-06
1. If 2x-1+2x+1 =1280, then find the value of x.
A. 9 B. 8 C. 7 D. 10
n+4 n+2
2. If 2 -2 =3, then n is equal to
A. 0 B. 2 C. -1 D. -2
n-1 n+1
3. If 2 +2 =320, then n is equal to
A. 6 B. 8 C. 5 D. 7
4. If 3x-3x-1=18, then the value of xx is
A. 3 B. 8 C. 27 D. 216
a a+1 a+2
5. If 4 + 4 = 4 -176, what is the value of a?
A. 2 B. 4 C. 6 D. 8

Type-07
1. If a and b are positive real numbers, then (a3b)5 =? [BKB-(SO)-2017]
A. 0 B. 1 C. -32 D. -1
2. (2x )  x is equal to
1 2  5
[Sonali Bank Officer -2018]
A. 2x2 B. 4x C. 4x3 D. 4x7
3. (2x1)2  x5 is equal to
A. 2x2 B. 4x C. 4x2 D. 4x3
1
4. 4
x  x 4 Gi mwVK gvb †KvbwU? [Sonali Bank -SO-2014]
1
A. x B. x C. D. 1
x4

 
1
17 18 4
5. If x and y are non-negative, simplify 81.x y [BKB-(SO)-2017]
17 9 17 7 15 9 17 9
A. 81x 4 y 2 B. 9x 4 y 2 C. 3x 4 y 2 D. 3x 4 y 2

64x 
2

6.  27a 3
3 3 =?
9ax 9 9 3 2 2
A. B. C. D. x a
16 16ax 16 x 2 a 2 4
7. If a, b, c are real numbers, then the value of a 1b . b 1c . c 1a . is
1
A. abc B. abc C. D. 1
abc

Type-08
2 n  2 n1
1. Which one is true for the mathematical expression ?
2 n1  2 n
3 2
A. B. C. 1 D. 230
2 3
2n  4  2  2n -3
2. +2 is equal to
2  2(n 3)
9   1
A. 2 n 1 B.   2 n  C.   2 n1  D. 1
8   8

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 2n  4  2(2n )
3. when simplified is
2(2n  3 )
1 7
A. 2n 1  B. -2n+1 C. 1-2n D.
8 8
2n 1  2n
4. The value of n 4 is
2  2 n 1
1 2 1 5
A.  B. C. D.
36 3 13 13

Type-09
(a 2  b 2  ab) (b 2  c 2  bc) (c2  a 2  ca)
 xa   xb   xc 
1.  b    c    a 
x  x  x 
A. 1 B. 2 C. 3 D. 4
(a b ) (b c) (c a)
x a
x b
 x  c
2.  b  . c  . a  =?
x  x   
x
A. 0 B. xabc C. xa+b+c D. 1
 1   1   1 
     
 xa   ab   xb   bc   xc   ca 
3.  b  . c  . a  =?
x  x  x 
1
1
A. 1 B. x abc C. x D. None
(ab  bc  ca)
b c a c  a b a b c
 xb   xc   xa 
4.  c  . a  . b  =?
x  x  x 
A. xabc B. 1 C. xab+bc+ca D. xa+b+c
1 1 1
 bc ac  a-b  ac ba  b-c  abbc  c-a
5. The value of  x  . x  . x  is
     
A. 1 B. a C. b D. c

Type-10
1. If 3(x-y) = 27 and 3(x+y) = 243, then x is equal to
A. 0 B. 2 C. 4 D. 6
2. If 32x-y =3x+y = 27 , the value of y is
1 1 3 3
A. B. C. D.
2 4 2 4
x+y x-y
3. If 4 = 1 and 4 =4, then the values of x and y respectively are
1 1 1 1 1 1 1 1
A.  and B.  and - C. and - D. and
2 2 2 2 2 2 2 2

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 Logarithim
Type-01
ab
log a  1 ,  b log a  b
a a
log a

1. KZ? [Sonali & Janata Bank- Officer IT-2020]

a. 1 b. 2 c. 3 d. 4
2. The value of log 2
32 is:
5 1
A. B. 5 C. 10 D.
2 10
3. log 3
81  ? [BKB-Cash-2017]
A. 9 B. 7 C. 6 D. 8
4. The value of log10 (0.0001) is
1 1
A. B.  C.  4 D. 4
4 4

Type-02
If log a  y , than x  a y
x

1. If logx144=4, then x = ? [Sonali & Janata Bank- Officer IT-2020 & PKB-Officer-2021]
A. 2 3 B. 4 C. 3 D. 3 2
2. If logx625 = 4, then x = ? [Combined 7 banks & 1 FI-SO-2021]
A. 15 B. 5 C. 4 D. 25
1
3. log x 9  2 then x = ? [BDBL-(SO)-2017]
1 1
A. B. C. 3 D. 3
3 3
1
4. log x 4  2 , then x = ? [Rupali Bank -Cash-2018]
1 1
A.  B. C. 2 D. 17
2 2
5. If log3 x = 2, then x is equal to:
A. 9 B. 6 C. 8 D. 1/9
6. If logx 4= 0.4, then the value of x is:
A.1 B. 4 C. 16 D. 32

If logx   =  , then x is equal to:

9 1
7.
 16  2
3 3 81 256
A.  B. C. D.
4 4 256 81

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Type-03
01. Log 36/Log 6 -------? [Agrani Bank-(Cash)-2017,BB-(Officer)-2018]
A. 5 B. 8 C. 3 D. 2
log 8
02. is equal to:
log8
1 1 1 1
A. B. C. D.
8 4 2 8
03. 3log102 + log105 =? [Combined 7 banks-SO-2021]
5 8 2
A. log10   B. log10   C. log10   D. log10 (40 )
8 5 5

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 Written part
Previous year question
1. What is the value of
2  4.2  ? x4 x 1
[Combined 5 Banks- Cash -2022]
2 x2  2
Solution:
 =

2 x 4  4.2 x1 2 x  2 4  4  2 x  21 2 x (16  8) 8
   4 (Ans)
2 x2  2 2 x  22  2 2x  2 2

2. 2 n1  2 n1  320 n‡j n Gi gvb KZ? [Bangladesh Banks- Officer- 2020]
5
n
2
Solution: 2 n1  2 n1  320   2 n  2  320  2 n    320  2 n  128  2 n  2 7  n  7
2 2

3. If and are whole numbers such that = 121, then the value of is
[Combined 9 Banks- Officer 2022]
Solution: Given that, m  121  m  11  m  11, n  2
n n 2

 (m  1) n1  (11  1) 21  10 3  1000 Ans

4 p
 16 pq  2 2 p
4. 3 q
 4  pq [BB AD-written-2022]
8 q
4 2
pq

4 p
 16 pq  2 2 p
Solution: 3 q
 4  pq
8 q
4 2 pq

2 2 p
 2 4 pq  2 2 p
= 3 q
 2 2 pq
2 3 q
2 2 pq
2
2 p  4 pq  2 p  2 pq
2
=
2 3 q  2 pq 3 q
2 2 p  2 pq2 p
=
2 2 pq
2 2 pq
= 2 pq = 1 (Ans)
2

5. If 10000 = 10p+r and 100 = 10p-r then find the value of p and r? [BKB -SO-2011 (Written)]
Solution:
10000 = 10p+r ⇒104 =10p+r ⇒4 = p+r p+r = 4 .......... (i)
Again, 100 = 10p-r  102=10p-r p-r = 2 ......... (ii)
Adding (i) & (ii) we get 2p = 6 p = 3
From (i) we have, p+r = 4 ⇒3+r=4 r=1
Required value p = 3, r=1 Ans: 3 & 1

6. If a = xyp-1, b = xyq-1, c = xyr-1 and p+q+r = 3 Then prove that aq-r. br-p. cp-q =1
[ABL –Cash- Written-18 & RBL-Cash- Written-2018]
Solution:
a = xyp-1
aq-r = {xy(p-1)}(q-r) =xq-r y (p-1) (q-r)
And b = xyq-1

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 br-p = xr-p .y(q-1) (r-p)
Again, c  xy r 1
cp-q = x(p-q) . y(r-1) (p-q)

L.H.S = aq-r.br-p. c(p-q)

= xq-r .y(p-1)(q-r) .x r-p . y(q-1)(r-p) . x(p-q) . y(r-1) (p-q)
= {xq-r. xr-p. x p-q} {y(p-1) (q-r)  y(q-1)(r-p)  y(r-1) (p-q)}
= (xq-r+r-p+p-q)  (ypq-pr-q+r+qr-pq-r+p + pr-qr-p+q)
=x.y
= 11
= 1 = R.H.S.
L.H.S. = R.H.S (proved)

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 Equation (For Written only)
1. If , find the value of x ? [Combined 8 bank-officer-written-2022]
Solution:
x3 – 8x2 + 17x – 10 0
 x3 – x2 – 7x2 + 7x + 10x – 10 = 0
 x2(x – 1) – 7x(x – 1) + 10(x – 1) = 0
 (x – 1) (x2 – 7x + 10) = 0
 (x – 1) (x2 – 5x – 2x + 10) = 0
 (x – 1) (x – 5) (x – 2) = 0
x – 1 = 0 Or, x – 5 = 0 or x – 2 = 0
 x = 1 or 5 or 2 Ans: 1 or 2 or 5

2. Find the value of ‘x’ if (2x2-1) = (3x3-2x) [BB-AD-Written-2004 ]

Solution:
Given that, (2x2 -1) = (3x2-2x)
⇒2x1 = 3x2 - 2x2
⇒2x-1 = x2 ⇒ x2-2x+1 =0 ⇒ (x-1)2 = 0 ⇒x-1 = 0  x=1 Ans: 1

10
3. Find the value of ‘a’ if (a-3) = [BB -AD- Written-2004]
a
Solution:
10
(a-3) =
a
⇒ a2-3a = 10
⇒ a2-3a -10 = 0
⇒ a2-5a + 2a -10 = 0
⇒ a(a-5) + 2(a-5) = 0 ⇒ (a-5) (a+2) = 0
So, a = 5 or, a = -2
Ans: 5 Or, -2

4. x Gi gvb †ei Kiæb: [BRC- Written-1986]

1 1

x 2
2
1 x
Solution:
1 1 1 1 1 1 1 x 1
 ⇒  ⇒  ⇒ 
x 2 2  2 x  x 2 2  3 x 2 2  3x 2
2
1 x 1 x 1 x
⇒ 2-2x = 2-3x ⇒ -2x+3x= 2-2 x = 0 Ans: 0

2x  2x
5. =2. Find the value of x. [BB–Officer- Written-2015]
2x  2x
Solution:
2x  2x
=2
2x  2x
2  x  2  x  2  x  2  x 2 1
⇒ 
2  x  2  x  2  x  2  x 2 1

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 2 2x 3
⇒ 
2 2x 1
2
 2 x 
⇒    (3) 2

 2x 
2x 16 8 8
⇒  9 ⇒ 2+x = 18-9x ⇒10x =16 ⇒x = x = Ans:
2x 10 5 5

6. If , calculate the value of x. [Combined 5 Banks – Cash-2022]

x 2  7 x  12
 3  x 2  7 x  12  3x 2  12  2 x 2  7 x  0  x(2 x  7)  0
x 4
2

7
x =0 x =
2
3x 5x  2 4x  5
7. mij Ki:   [Karmahgangsthan Bank –SO- Written--2013]
4 6 8
Solution:
3x 5x  2 4x  5
 
4 6 8
9x  10x  4 4x  5
⇒ 
12 8
19x  4 4x  5
⇒ 
12 8
19x  4 4x  5
⇒ 
3 2
⇒ 38x-8 = 12x+15
⇒ 26x = 23
23 23
x = Ans: x =
26 26
3 6 18
8. Solve the equation:   . [Janata Bank - Written-FA-2015 ]
x  1 2x  1 3x  1
Solution:
3 6 18
 
x  1 2x  1 3x  1
3 6 9 9
⇒   
x  1 2x  1 3x  1 3x  1
3 9 9 6
⇒   
x  1 3x  1 3x  1 2x  1
3(3x  1)  9( x  1) 9( 2x  1)  6(3x  1)
⇒ 
( x  1)(3x  1) (3x  1) ( 2x  1)
9x  3  9x  9 18x  9  18x  6
⇒ 
( x  1) (3x  1) (3x  1) ( 2x  1)
6 3
⇒ 
( x  1) (3x  1) (3x  1)(2 x  1)
2 1
⇒ 
x  1 2x  1

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 3 3
⇒ x +1 = 4x 2 ⇒5x=-3 x =  Ans: 
5 5

5x  2 2x  1
9. Simplify:  [Agrani Bank (SO-Auditor)- Written-2018]
x2  x  20 x2  4x  5
Solution:
5x  2 2x  1

x  x  20 x  4x  5
2 2

5x  2 2x  1
⇒ 2  2
( x  5x  4x  20) x  5x  x  5
5x  2 2x  1
⇒ 
x( x  5)  4( x  5) x( x  5)  1( x  5)
5x  2 2x  1
⇒ 
( x  5)( x  4) ( x  5)( x  1)
(5x  2) ( x  1)  ( 2x  1) ( x  4)
⇒
( x  5)( x  4) ( x  1)
5 x 2  5 x  2 x  2  2 x 2  8x  x  4
⇒
( x  5)( x  4) ( x  1)
7 x 2  14x  2 7x2  14x  2
⇒ Ans:
( x  5)( x  4) ( x  1) (x  5)(x  4) (x  1)
xa xb xc
10. x Gi gvb wbY©q Kiæb:   =3 [BKB -Cash- Written-- 2012]
bc ca ab
Solution:
x a x b x c
⇒   3
bc ca ab
x a xb x c
⇒ 1  1 1  0
bc ca ab
x a bc x bca x ca b
⇒   0
bc ca ab
x  (a  b  c) x  (a  b  c) x  (a  b  c)
⇒   0
bc ca ab
 1 1 1 
⇒ x  (a  b  c)    0
bc ca a b
 1 1 1 
⇒ x  (a  b  c)  0 [ where,      0]
bc c  a a b
 x = a+b+c Ans: a+b+c
2
 x  3  x  3
11. Solve the equation: 2    7  +6 = 0 [BKB-Officer- Written-2017]
 x 3  x 3
Solution:
 x 3
Let,  = a
 x 3
2
 x  3  x  3
 2   7 + 6 = 0
 x 3  x 3
⇒ 2a2 -7a+6=0
⇒ 2a2 -4a-3a+6=0

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 ⇒ 2a(a-2) -3(a-2)=0
⇒ (a-2) (2a-3) =0
Now, either
x3
a-2 = 0 ⇒ a = 2 ⇒ =2 ⇒2x-6 = x+3 x = 9
x 3
3
or, 2a-3 = 0 ⇒ 2a=3 a =
2
x 3 3
Now,  ⇒ 3x-9 = 2x+6  x = 15 Ans: 9 or 15
x 3 2

x y x y
12. If   1 and   1 Find the value of x and y [[Agrani Bank-Cash- Written-2018 ]
2 3 3 2
Solution:
x y 3x  2 y
Here,  = 1 ⇒ =1 3x+2y = 6 ......... (i)
2 3 6
x y 2 x  3y
Again,  =1 ⇒ = 1 2x+3y = 6 ................. (ii)
3 2 6
By (i)  3and (ii)  2 then subtracting (ii) from (i) we get,
9x+6y = 18
4x+6y = 12
6
5x= 6 x=
5
Putting the Value of x in (i)
6 18 12 6
⇒ 3  2 y  6 ⇒ 2y = 6 - ⇒ 2y =  y
5 5 5 5
6 6 6 6
So, (x, y) =  ,  Ans:  , 
5 5 5 5

13. Solve: x2- yx= 7, y2 +xy =30 [Sonali Bank (SO) 2018 (Written)]
Solution:
7
Given, x2-yx =7, x(x-y) = 7 ⇒x-y =
x
7
y = x- .................... (i)
x
Again, y2 + xy = 30 ..............(ii)
Now,
x2-xy = 7
y2  xy  30
[by adding]
x 2  y2  37
2
 7
⇒x +  x   =37
2
 x
2
7 7
⇒ x + x – 2.x.    =37
2 2
x x
49
⇒2x2 -14+ 2  37
x
2x  49
4
⇒ = 51
x2

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 ⇒2x4 +49 = 51x2
⇒2x4-51x2 +49 = 0
⇒2x4 -49x2- 2x2 +49 = 0
⇒x2(2x2-49) – 1 (2x2-49) = 0
⇒(2x2 -49) (x2 -1) = 0
 2x2-49 = 0  x2-1 = 0
49 7
Or, x2 = x=  Or, x2 = 1 x =± 1
2 2
7 7
When x = , When x = - ,
2 2
7 7 7 7
Y=  Y==  
2 7 2 7
2 2
7 72 5 7 72 5
=  2  =  2 
2 2 2 2 2 2

7
Again, When x =1, When x = - ,
2
7 7
Y = 1 Y = = 1
1 1
 6 6

 7 5   7 5 
Ans: (x, y) =  ,  Or,   ,  or (1, 6) or (-1, 6)
 2 2  2 2

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