Download linear equations and simultaneous equations

LINEAR EQUATIONS
AND SIMULTANEOUS
EQUATIONS
Q.1. Solve the following linear equations:
(i) ( x − 4) ( x + 4) = ( x + 4) ( x − 7) + 33
3
4
(iii) ( x − 1) = ( x + 1) −
1
2
(ii) ( x − 2) ( x + 3) = x 2 − 4
(iv)
2
5
( x - 3) = 1 - (3 x - 4)
3
6
( x - 1) ( x + 1)
1
= 5- x
(vi) ( x − 3) − x − 3 = 2 ( x − 1)
3
2
3
Ans. (i) ( x − 4) ( x + 4) = ( x + 4) ( x − 7) + 33
(v)
⇒ x 2 − 4 x + 4 x − 16 = x 2 − 7 x + 4 x − 28 + 33
⇒ x 2 − 16 = x 2 − 3 x + 5 ⇒ x 2 − x 2 + 3 x = 5 + 16 ⇒ 3 x = 21
21
Hence, x =
=7
3
(ii) ( x − 2) ( x + 3) = x 2 − 4 ⇒ x 2 + 3 x − 2 x − 6 = x 2 − 4
⇒ x 2 + x − 6 = x 2 − 4 ⇒ x 2 + x − x 2 = −4 + 6
Hence, x = 2
3
1
(iii) ( x − 1) = ( x + 1) −
4
2
Multiplying both sides by 4, we get
3
1
4 ( x − 1) = ( x + 1) × 4 − × 4
4
2
⇒ 4 x − 4 = 3x + 3 − 2
⇒ 4 x − 3x = 3 − 2 + 4
Hence, x = 5
Math Class IX
1
Question Bank
2
5
( x − 3) = 1 − (3x − 4)
3
6
Multiplying both sides by 6, we get
2
5
6 × ( x − 3) = 1 × 6 − 6 × (3x − 4) ⇒ 4( x − 3) = 6 − 5 (3x − 4)
3
6
⇒ 4 x − 12 = 6 − 15 x + 20 ⇒ 4 x + 15 x = 6 + 20 + 12
38
=2
⇒ 19 x = 38 Hence, x =
19
x −1 x +1
(v)
−
=5− x
2
3
Multiplying both sides by 6, we get
( x − 1)
( x + 1)
− 6×
= 6 × (5 − x)
⇒ 6×
2
3
⇒ 3 ( x − 1) − 2 ( x + 1) = 6 (5 − x) ⇒ 3 x − 3 − 2 x − 2 = 30 − 6 x
⇒ 3 x − 2 x + 6 x = 30 + 3 + 2
35
⇒ 7 x = 35
⇒ x=
⇒ x =5
7
Hence, x = 5
1
( x − 3)
(vi) ( x − 3) − x − 3 = 2 ( x − 1) ⇒
− x − 3 = 2 ( x − 1)
3
3
Multiplying both sides by 3, we get
( x − 3)
3×
− 3 × x − 3 × 3 = 3 × 2 ( x − 1) ⇒ ( x − 3) − 3x − 9 = 6 x − 6
3
6
⇒ x − 3 − 3 x − 9 = 6 x − 6 ⇒ − 8 x = +6 ⇒ x =
−8
3
3
⇒ x=
⇒ x=−
−4
4
−3
Hence, x =
.
4
(iv)
Math Class IX
2
Question Bank
Q.2. Solve the following equations:
(i)
4
5 2
1
1
x+ − x−  =1
5
8 3
4
9
(ii)
( x − 2) 5 x
( x − 5)
+
=6−
3
2
6
( y − 2) 1
(2 y − 1)
+ = y−
4
3
3
(iv)
5y − 4 y − 3 y + 6
−
=
8
5
4
(iii)
4
5 2 
1 1
 x +  −  x −  =1
5
8 3 
4 9
4 (8 x + 5) 2 (4 x − 1) 10
−
=
⇒
5
8
3 4
9
8 x + 5 4 x − 1 10
−
=
⇒
10
6
9
Multiplying both sides by 90, we get
8x + 5
4 x − 1 10
90 ×
− 90 ×
= × 90 ⇒ 9 (8 x + 5) − 15 (4 x − 1) = 100
10
6
9
⇒
72 x + 45 − 60 x + 15 = 100 ⇒ 72 x − 60 x = 100 − 45 − 15 ⇒ 2 x = 40
40 10
1
1
= =3
Hence, x = 3 .
∴x=
12 3
3
3
( x − 2) 5 x
x−5
(ii)
+
=6−
3
2
6
Multiplying both sides by 6, we get
x−2
x−5
5x
+ 6× = 6×6 − 6×
6×
3
2
6
⇒
2 ( x − 2) + 15 x = 36 − ( x − 5) ⇒ 2 x − 4 + 15 x = 36 − x + 5
⇒
2 x + 15 x + x = 36 + 5 + 4 ⇒ 18 x = 45
45 5
1
∴x=
= =2 .
18 2
2
1
Hence, x = 2 .
2
( y − 2) 1
(2 y − 1)
(iii)
+ = y−
4
3
3
Multiplying both sides by 12, we get
y−2 1
2 y −1
12 ×
+ × 12 = 12 y − 12 ×
⇒
⇒ 3 ( y − 2) + 4 = 12 y − 4 (2 y − 1)
4
3
3
⇒
3 y − 6 + 4 = 12 y − 8 y + 4 ⇒ 3 y − 12 y + 8 y = 4 + 6 − 4
− y=6
⇒
Ans. (i)
Math Class IX
3
Question Bank
Hence,
y = −6
5y − 4 y − 3 y + 6
(iv)
−
=
8
5
4
Multiplying both sides by 40, we get
(5 y − 4)
( y − 3)
( y + 6)
40 ×
− 40 ×
= 40 ×
⇒ 5 × (5 y − 4) − 8 × ( y − 3) = 10 × ( y + 6)
8
5
4
⇒ 25 y − 20 − 8 y + 24 = 10 y + 60 ⇒ 25 y − 8 y − 10 y = 60 + 20 − 24
56
⇒ 7 y = 56
⇒ y=
⇒ y =8
7
Hence,
y = 8.
Q.3. If x = m + 1 and 1.5 +
3m + 2
= 0.5 (4 x − 3); find m.
5
Ans. x = m + 1
(Given)
3m + 2
15 3m + 2 1
= 0.5 (4 x − 3) ⇒ +
= (4 x − 3)
Then, 1.5 +
5
10
5
2
3 3m + 2 1
3 3m + 2 1
+
= [4 (m + 1) − 3] ⇒ +
= (4m + 4 − 3)
⇒
2
5
2
2
5
2
3 3m + 2 4m + 1
15 + 6 m + 4 20m + 5
+
=
=
⇒
⇒
2
5
2
10
10
⇒ 15 + 4 − 5 = 20 m − 6m ⇒ 14 = 14m
14
⇒ m = =1
14
Hence, m = 1.
Q.4. Find the value of a , if x = 0.5 is a solution of equation ax 2 + ( a − 1) x + 3 = a .
1
1
Ans. x = 0.5 =
(Given) x = is a solution of the given equation
2
2
2
ax + (a − 1) x + 3 = a
Putting the value of x in the given equation, we have
2
a a −1
1
1
a   + (a − 1) × + 3 = a ⇒
+
+3= a
2
4
2
2
Multiplying both sides by 4, we get
a 4 × (a − 1)
+ 3 × 4 = 4a
⇒ 4× +
4
2
⇒ a + 2 a − 2 + 12 = 4 a ⇒ a + 2a − 4 a = 2 − 12
⇒ − a = −10
Hence,
a = 10
Math Class IX
4
Question Bank
Q.5. If x (5 − a ) = 10 − x 2 and x = 2, find the value of ‘a’.
Ans.
...(i)
x (5 − a ) = 10 − x 2
...(ii)
x=2
Putting the value of x from eqn. (ii) in (i), we get
2 (5 − a ) = 10 − (2) 2
10 − 2a = 10 − 4 ⇒ − 2a = 10 − 4 − 10
4
⇒ − 2 a = −4 ⇒ 2 a = 4 ⇒ a =
2
Hence, a = 2.
Q.6. If (2ax + 1) (3 x + 1) = 6a ( x + 1) and x = 1 , find the value of ‘a’.
Ans. (2ax + 1) (3 x + 1) = 6a ( x + 1)
...(i)
...(ii)
x =1
Putting the value of x from eqn. (ii) in (i), we get
⇒ ( 2a × 1 + 1) (3 × 1 + 1) = 6a (1 + 1)
⇒ ( 2a + 1) (3 + 1) = 6a (2) ⇒ (2 a + 1) (4) = 12a
⇒ 8a + 4 = 12 a ⇒ 8a − 12a = −4 ⇒ − 4a = −4 ⇒ 4a = 4
4
⇒ a = =1
4
Hence, a = 1.
⇒
⇒
Q.7. Find the value of x , if: m =
1 − 3x
1 − 2x
; n=
and 3 ( n − 2m ) + 1 = 0.
5
3
1 − 3x
...(i)
5
1 − 2x
n=
...(ii)
3
...(iii)
3 ( n − 2m ) + 1 = 0
Putting the values of m and n in (iii), we get
1 − 2 x
 1 − 3x 
 (1 − 2 x ) (2 − 6 x) 
⇒ 3
− 2×
−
+1 = 0
 + 1 = 0 ⇒ 3 
5 
 5 
 3
 3
 5 (1 − 2 x) − 3 (2 − 6 x ) 
 5 − 10 x − 6 + 18 x 
⇒ 3
+1 = 0 ⇒ 3 

 + 1 = 0
15
15



3 [5 − 6 − 10 x + 18 x ] + 15
= 0 ⇒ 3 [−1 + 8 x ] + 15 = 0 × 15
⇒
15
⇒ − 3 + 24 x + 15 = 0 ⇒ − 3 + 15 + 24 x = 0
⇒ 12 + 24 x = 0 ⇒ 24 x = −12
Ans. m =
Math Class IX
5
Question Bank
− 12
1
⇒ x=−
24
2
Find a number such that one-fifth of it is less than one-fourth of it by 3.
Let the number be x.
According to the given problem, we get
5x − 4 x
x x
x x
x
=3⇒
= −3 ⇒ − =3 ⇒
=3
5 4
4 5
20
20
∴ x = 3 × 20 = 60
Hence, number is 60.
Find three consecutive natural numbers whose sum is 120.
Let first natural number be x. Second natural number be x + 1
and third natural number be x + 2 .
According to the given problem, we get
x + x + 1 + x + 2 = 120 ⇒ 3 x + 3 = 120
⇒ 3 x = 120 − 3 = 117
117
= 39
⇒ x=
3
Hence, first natural number = 39 and other numbers = 40, 41
Thus, numbers are 39, 40 and 41.
In a class there are x seats. If each student occupies one seat, 9 students
remain standing and if two students occupy one seat, 7 seats are left
empty. Find number of seats in the class.
Number of seats = x
(Given)
Case I: When I student occupy 1 seat 9 students are left standing
Thus number of students = x + 9
Case II: When 2 students occupy 1 seat, 7 seats are left
Thus number of seats occupied = x − 7
Number of students = 2 ( x − 7)
According to the given problem,
x + 9 = 2 ( x − 7) ⇒ x + 9 = 2 x − 14 ⇒ x − 2 x = −14 − 9
⇒ − x = −23 ⇒ x = 23
Hence number of seats are 23.
⇒
Q.8.
Ans.
Q.9.
Ans.
Q.10.
Ans.
x=
Q.11. Divide 32 into two parts such that if the larger is divided by the smaller,
the quotient is 2 and the remainder is 5.
Ans. Let larger part be x.
Math Class IX
6
Question Bank
Q.12.
Ans.
Q.13.
Ans.
Dividend = quotient × divisor + remainder
∴
x = 2 × (32 − x ) + 5 ⇒ x = 64 − 2 x + 5
⇒ x + 2 x = 69 ⇒ 3 x = 69
69
= 23
⇒ x=
3
Thus larger part = 23 and smaller part = 32 − 23 = 9.
The sum of two consecutive odd natural numbers is 140. Find the bigger
number.
Let the two consecutive odd numbers be 2 x + 1 and 2 x + 3.
According to the given problem,
2 x + 1 + 2 x + 3 = 140
⇒ 4 x + 4 = 140 ⇒ 4 x = 140 − 4
136
= 34
⇒ 4 x = 136 ⇒ x =
4
Since odd numbers are 2 x + 1 and 2 x + 3 i.e. 2 × 34 + 1 and 2 × 34 + 3 i.e. 69
and 71.
Hence, bigger odd number is 71.
10 years ago, a man was six times as old as his daughter. After 10 years, he
will be twice as old as his daughter. Find their present ages.
Case I: 10 years ago,
Let age of daughter be x years then age of father be 6 x years
∴ Present age of daughter = ( x + 10) years
Present age of father = (6 x + 10) years
Case II: 10 years later,
Age of daughter = ( x + 10 + 10) = x + 20 years
and age of father = (6 x + 10 + 10) = (6 x + 20) years
According to the given problem.
6 x + 20 = 2 ( x + 20)
⇒ 6 x + 20 = 2 x + 40 ⇒ 6 x − 2 x = 40 − 20
⇒ 4 x = 20
20
Thus, x =
=5
4
∴ Present age of daughter = ( x + 10) years = 5 + 10 = 15 years
and present age of father = 6 x + 10 = 6 × 5 + 10 = 30 + 10 = 40 respectively.
Hence, present ages of father and daughter are 40 years and 15 years.
Math Class IX
7
Question Bank
Q.14. A number consists of two digits. The digit at ten’s place is twice the digit at
units place. The number formed by reversing the digits is 27 less than the
original number. Find the original number.
Ans. Let digit at unit’s place be x
then digit at ten’s place be 2 x
Thus, number = x + (10 × 2 x ) = x + 20 x = 21x
The number formed by reversing the digit the unit’s digit = 2 x and ten’s digit = x
Thus, number = 2 x + 10 × x = 2 x + 10 x = 12 x
According to the given problem
− 27
=3
12 x = 21x − 27 ⇒ 12 x − 21x = −27 ⇒ − 9 x = −27 ⇒ x =
−9
Hence, original number = 21x = 21 × 3 = 63
Q.15. The denominator of a fraction is 4 more than its numerator. If 1 is
subtracted from both the numerator and the denominator, the fraction
1
becomes . Find the original fraction.
2
Ans. Let numerator of the fraction be x then its denominator be x + 4
x
∴ Fraction =
x+4
According to the given problem,
x −1
1
x −1 1
= ⇒
=
x + 4 −1 2
x +3 2
⇒ 2x − 2 = x + 3
By cross multiplication, we get
2x − x = 3 + 2 ⇒ x = 5
x
5
5
Hence, fraction =
=
=
x+4 5+4 9
Q.16. The height of a triangle is 3 cm more than its base. If the area of the
triangle is 104 cm 2 , find the lengths of its base and height.
Ans. Let base b of a triangle (b) be x cm then height (h) be ( x + 3) cm
and area of triangle (A) = 104 cm2
According to the given problem
1
1
b × h = A ⇒ x × ( x + 3) = 104
2
2
⇒ x ( x + 3) = 208 ⇒ x 2 + 3 x − 208 = 0
⇒ x 2 + 16 x − 13x − 208 = 0
Math Class IX
{∵ 3 = 16 − 13 and − 208 = 16 × (−13)}
8
Question Bank
⇒ x ( x + 16) − 13 ( x + 16) = 0 ⇒ ( x + 16) ( x − 13) = 0
If x + 16 = 0, then x = −16 which is not possible as it is negative
If x − 13 = 0 then x = 13
Hence, Base = 13 cm and height = 13 + 3 = 16 cm
Q.17. The perimeter of a rectangular park is 80 m. If the length of the park be
decreased by 2 m and breadth increased by 2 m, the area will be increased
by 36 m 2 . Find the original length and breadth of the park.
Ans. Perimeter of a rectangular park = 80 m (Given)
⇒ 2 (length + breadth) = 80 m
Let length of park be x m then breadth be (40 − x ) m
and area of rectangular park = lb = x (40 − x) m 2
According to the given problem,
( x − 2) [40 − x + 2] = x (40 − x) + 36
⇒
( x − 2) (42 − x) = 40 x − x 2 + 36 ⇒ 42 x − x 2 − 84 + 2 x = 40 x − x 2 + 36
42 x − x 2 + 2 x − 40 x + x 2 = 36 + 84 ⇒ 44 x − 40 x = 120
120
= 30
⇒ 4 x = 120 ∴ x =
4
Hence, length of park is 30 m and breadth of park is 10 m.
Q.18. The total cost of a desk and a chair is Rs 477. If the desk costs 12% more
than the chair, find the cost of each.
Ans. Total cost of desk and chair is Rs 477
Let cost of the chair be x
x × (100 + 12) 112
then cost of desk =
=
x
100
100
According to the given problem,
112
x+
x = 477 ⇒ 100 x + 112 x = 47700
100
47700
= 225
⇒ 212 x = 47700 ⇒ x =
212
Hence, cost of chair is Rs 225.
and cost of desk is Rs 252.
⇒
Math Class IX
9
Question Bank
Q.19. Amit walks from his house to school at a speed of 3 kmph and returns
back at a speed of 4 kmph. If he takes 42 minutes for the whole journey,
find the distance between his house and the school.
Ans. Let the distance between school and house be x km.
42
Total time taken = 42 minutes =
hours
60
According to given problem,

x x 42
Distance
since, Time =
+ =

Speed 
3 4 60

4 x + 3x 42
7 x 42
42 12 6
=
=
⇒
⇒
⇒ x= × =
12
60
12 60
60 7 5
6
Hence, distance between his house and school is km = 1.2 km.
5
Q.20. A boy has x coins of 50 paise each, 2x coins of 25 paise each, 4 x coins of 10
paise each and 8x coins of 5 paise each. Find the number of 5 paise coins if
the value of all the coins is Rs 9.
Ans. Number of 50 paise coins be x
Number of 25 paise coins be 2 x
Number of 10 paise coins be 4 x
Number of 5 paise coins be 8 x
But, total value of coins = Rs 9
Now according to the given problem
x × 50 + 2 x × 25 + 4 x × 10 + 8 x × 5 = 9 × 100
⇒ 50 x + 50 x + 40 x + 40 x = 900 ⇒ 180 x = 900
900
=5
⇒ x=
180
Hence, number of 5 paise coins = 8 x = 8 × 5 = 40 .
Q.21. Solve the following pairs of linear (simultaneous) equations by the method
of elimination by substitution.
(i) x + y = 7, 5 x + 12 y = 7
(ii) 8 x + 5 y = 9, 3 x + 2 y = 4
(iii) 2 x - 3 y = 7, 5 x + y = 9
(iv) 5 x + 4 y - 4 = 0, x - 20 = 12 y
Ans. (i) x + y = 7
...(i)
...(ii)
5 x + 12 y = 7
From eqn. (i),
...(iii)
y =7−x
Putting the value of y in eqn. (ii), we get
5 x + 12 (7 − x) = 7
Math Class IX
10
Question Bank
⇒ 5 x + 84 − 12 x = 7 ⇒ 5 x − 12 x = 7 − 84
⇒ − 7 x = −77 ⇒ x = 11
∴ Putting the value of x in eqn. (iii), we get
y = 7 − 11 = −4
Hence, x = 11, y = −4
(ii) 8 x + 5 y = 9
...(i)
...(ii)
3x + 2 y = 4
From eqn (i),
9 − 8x
5 y = 9 − 8x ⇒ y =
5
Putting the value of y in eqn. (ii), we get
 9 − 8x 
3x + 2 
=4
 5 
⇒ 15 x + 18 − 16 x = 20 ⇒ 15 x − 16 x = 20 − 18
⇒ − x = 2 ⇒ x = −2
∴ Putting the value of x in eqn. (iii), we get
9 − 8 ( −2 )
9 + 16
=5
y=
⇒ y=
5
5
Hence,
x = −2, y = 5.
(iii) We have 2 x − 3 y = 7
...(i)
5x + y = 9
...(ii)
From eqn. (ii),
...(iii)
y = 9 − 5x
Putting the value of y in eqn. (i), we get
2 x − 3 (9 − 5 x) = 7 ⇒ 2 x − 27 + 15 x = 7 ⇒ 2 x + 15 x = 7 + 27
34
=2
⇒ 17 x = 34 ⇒ x =
17
∴ Putting the value of x in eqn. (iii), we get
y = 9 − 5 × 2 = −1
Hence, x = 2, y = −1
(iv) 5 x + 4 y − 4 = 0
...(i)
...(ii)
x − 20 = 12 y
x = 12 y + 20
...(iii)
From eqn. (ii),
Putting the value of x in eqn. (i), we get
5 (12 y + 20) + 4 y − 4 = 0 ⇒ 60 y + 100 + 4 y − 4 = 0 ⇒ 64 y + 96 = 0
Math Class IX
11
Question Bank
−96
3
=−
64
2
Putting the value of y in eqn. (iii), we get
 −3
x = 12 × 
 + 20 = −18 + 20 = 2
 2 
−3
Hence, x = 2, y =
.
2
Q.22. Solve the following pairs of equations :
⇒ y=
(i) 7 x + 6 y = 71, 5 x - 8 y = -23
x 7
+ =4
3 4
3 2
1
2
(iv) x - y + 1 = 0, y - x = 4
5 3
3
5
...(i)
(ii) 3 x - y = 23,
x-3 y-7
=
, 11 x = 13 y
4
2
Ans. (i) 7 x + 6 y = 71
...(ii)
5 x − 8 y = −23
Multiplying eqn. (i) by 4 and eqn. (ii) by 3, we get
...(iii)
28 x + 24 y = 284
...(iv)
15 x − 24 y = −69
Adding eqn. (iii) and eqn. (iv), we get
215
=5
43 x = 215 ⇒ x =
43
Putting x = 5 in eqn. (i), we get
7 × 5 + 6 y = 71 ⇒ 35 + 6 y = 71 ⇒ 6 y = 71 − 35
36
=6
⇒ 6 y = 36 ⇒ y =
6
Hence, x = 5 and y = 6.
(ii)
3x − y = 23
...(i)
x y
+ =4
3 4
x
y
⇒ 12 × + 12 × = 12 × 4
3
4
Multiplying both sides by 12, we get
4 x + 3 y = 48
Multiplying eqn. (i) be 3, we get
...(iii)
9 x − 3 y = 69
...(iv)
4 x + 3 y = 48
Adding eqn. (iii) and eqn. (iv), we get
(iii)
Math Class IX
12
Question Bank
9 x − 3 y = 69
4 x + 3 y = 48
13 x = 117
117
x=
=9
⇒
13
Putting the value of x in eqn. (i), we get
3 × 9 − y = 23 ⇒ 27 − y = 23 ⇒ − y = 23 − 27
⇒ − y = −4 ⇒ y = 4
Hence,
x = 9 and y = 4.
x−3 y −7
=
(iii)
4
2
2 ( x − 3) = 4 ( y − 7)
By cross-multiplication, we get
2 x − 6 = 4 y − 28
⇒ 2 x − 4 y = −28 + 6 ⇒ 2 x − 4 y = −22 ...(i)
...(ii)
11x = 13 y ⇒ 11x − 13 y = 0
Multiplying eqn. (i) by 11 and eqn. (ii) by 2, we get
...(iii)
22 x − 44 y = −242
...(iv)
22 x − 26 y = 0
Subtracting eqn. (iv) from (iii), we get
22 x − 44 y = −242
22 x − 26 y = 0
− 18 y = −242
121
Putting y =
in eqn. (i), we get
9
121
2x − 4 ×
= −22
9
484
484
= −22 ⇒ 2 x = −22 +
⇒ 2x −
9
9
− 198 + 484
286
⇒ 2x =
⇒ 2x =
9
9
289
143
8
⇒ x=
⇒ x=
⇒ x = 15
2×9
9
9
8
4
Hence,
x = 15 and y = 13
9
9
Math Class IX
13
Question Bank
3
2
x − y +1 = 0
5
3
Multiplying both sides by 15, we get
3
2
15 × x − 15 × y + 15 × 1 = 0 × 15
5
3
⇒ 9 x − 10 y + 15 = 0
...(i)
⇒ 9 x − 10 y = −15
1
2
y+ x=4
3
5
Multiplying both sides by 15, we get
1
2
15 × y + 15 × x = 15 × 4
3
5
⇒ 5 y + 6 x = 60
...(ii)
⇒ 6 x + 5 y = 60
Multiplying eqn. (i) by 1 and eqn. (ii) by 2, we get
...(iii)
12 x + 10 y = 120
...(iv)
9 x − 10 y = −15
Adding eqn. (iii) and (iv), we get
12 x + 10 y = 120
(iv)
9 x − 10 y = −15
21x = 105
105
=5
21
Putting x = 5 in eqn. (i), we get
9 × 5 − 10 y = −15
⇒ 45 − 10 y = −15 ⇒ − 10 y = −15 − 45
⇒ − 10 y = −60 ⇒ 10 y = 60
60
⇒ y=
⇒ y =6
10
Hence,
x = 5 and y = 6 .
⇒ x=
Math Class IX
14
Question Bank
Q.23. Solve the following pairs of equations
(i) x + y = 2 xy, x – y = 6 xy
(ii)
a b
ab 2 a 2 b
– = 0,
+
= ( a 2 + b2 )
x y
x
y
Ans. (i) x + y = 2 xy
x − y = 6 xy
Dividing both sides by xy, we get
x
y 2 xy
x
y 6 xy
+
=
⇒
− =
xy xy xy
xy xy xy
1 1
...(i)
+ =2
y x
1 1
...(ii)
− =6
y x
Adding eqn. (i) and eqn. (ii), we get
1 1
+ =2
y x
1 1
− =6
y x
2
= 8 ⇒ 8y = 2
y
2 1
y= =
⇒
8 4
Putting the value of y in eqn. (i), we get
1
1
1 1
+ = 2 ⇒ 4 + = 2 ⇒ = 2 − 4 = −2
1 x
x
x
4
1
−1
1
Hence, x =
and y = .
⇒ − 2x = 1 ⇒ x = −
2
2
4
a b
(ii)
...(i)
− =0
x y
ab 2 a 2b
+
= (a 2 + b 2 )
x
y
...(ii)
Multiplying (i) by b 2 and (ii) by 1, we get
Math Class IX
15
Question Bank
ab 2 b 3
...(iii)
−
=0
x
y
Subtracting eqn. (ii) from (iii), we get
ab 2 b3
−
=0
x
y
ab 2 a 2b
+
= (a 2 + b2 )
( −) x ( −) y
(−)
−
b 3 a 2b
−
= −( a 2 + b 2 )
y
y
b 2
b
b
a 2 + b2
2
2
2
= −1 ⇒ = 1 ⇒ y = b
⇒ − (b + a ) = −(a + b ) ⇒ − = − 2
2
y
y
y
a +b
Putting the value of y in eqn. (i), we get
a b
a
a
− = 0 ⇒ −1 = 0 ⇒ = 1
x b
x
x
⇒ x=a
Hence,
x = a and y = b.
Q.24. If 7 x = 10 y + 4 and 12 x + 18 y = 1 , find the values of (4 x + 6 y ) and (8 y + x ).
Ans. 7 x = 10 y + 4
7 x − 10 y = 4
...(i)
⇒
12 x + 18 y = 1
...(ii)
Multiplying (i) by 9 and eqn. (ii) by 5,
...(iii)
63x − 90 y = 36
...(iv)
60 x + 90 y = 5
Adding eqn. (iii) and eqn. (iv), we get
63x − 90 y = 36
60 x + 90 y = 5
123 x = 41
41 1
=
⇒ x=
123 3
Putting the value of x in eqn. (i), we get
1
7 × − 10 y = 4
3
Math Class IX
16
Question Bank
7
7 5
− 10 y = 4 ⇒ − 10 y = 4 − =
3
3 3
5
1
=−
⇒ y=−
3 × 10
6
1
1
 1 4
(i) 4 x + 6 y = 4 × + 6 ×  −  = − 1 =
3
3
 6 3
and
 1 1 −4 1 −3
(ii) 8 y + x = 8 ×  −  + =
= −1.
+ =
3 3 3
 6 3
⇒
Q.25. The sides of an equilateral triangle are ( x + 3 y ) cm, (3 x + 2 y − 2) cm and
1


 4 x + y + 1  cm. Find the length of each side.
2


Ans. ∵ triangle is an equilateral, so its sides are equal.
1
∴ x + 3 y = 3x + 2 y − 2 = 4 x + y + 1
2
Taking x + 3 y = 3 x + 2 y − 2
⇒ 3x + 2 y − x − 3 y = 2
...(i)
⇒ 2x − y = 2
Again taking
1
1
3x + 2 y − 2 = 4 x + y + 1 ⇒ 3x + 2 y − 4 x − y = 1 + 2
2
2
3
⇒ −x+ y =3
2
...(ii)
⇒ − 2x + 3 y = 6
Adding eqn. (i) and eqn. (ii), we get
8
2y = 8 ⇒ y = = 4
2
1
1
 1 4
(i) 4 x + 6 y = 4 × + 6 ×  −  = − 1 =
3
3
 6 3
and
 1 1 −4 1 −3
(ii) 8 y + x = 8 ×  −  + =
+ =
= −1.
3 3 3
 6 3
Putting the value of y in eqn. (i), we get
2x − 4 = 2
⇒ 2x = 2 + 4 ⇒ 2x = 6
Thus x = 3
Math Class IX
17
Question Bank
Thus sides of an equilateral triangle are
1


(3 + 3 × 4), (3 × 3 + 2 × 4 − 2),  4 × 3 + × 4 + 1 = 15 cm, 15 cm, 15 cm.
2


Q.26. Of the two numbers, 4 times the smaller one is less than 3 times the larger
one by 6. Also, the sum of the numbers is larger than 6 times their
difference by 5. Find the numbers.
Ans. Let the larger number be x and the smaller number be y
According to the given problem
3x − 4 y = 6
...(i)
x + y − 6 ( x − y) = 5 ⇒ x + y − 6 x + 6 y = 5
...(ii)
⇒
− 5x + 7 y = 5
Multiply eqn. (i) by 7 and eqn. (ii) by 4, we get
...(iii)
21x − 28 y = 42
...(iv)
− 20 x + 28 y = 20
Adding eqn. (iii) and (iv), we get
21x − 28 y = 42
− 20 x + 28 y = 20
x = 62
Putting the value of x in eqn. (i), we get
3 × 62 − 4 y = 6
− 180
= 45
−4
Hence, larger number = 62 and smaller number = 45.
Q.27. If from twice the greater of the two numbers, 45 is subtracted, the result is
the other number. If from twice the smaller number, 21 is subtracted, the
result is the greater number. Find the numbers.
Ans.
Let larger number be x
and smaller number be y
2 x − 45 = y
...(i)
⇒ 2 x − y = 45
2 y − 21 = x
...(ii)
⇒ x − 2 y = −21
Adding eqn. (i) and eqn. (ii), we get
2 x − y = 45
⇒ 186 − 4 y = 6 ⇒ − 4 y = 6 − 186 = −180 ⇒ y =
x − 2 y = −21
Math Class IX
18
Question Bank
3 x − 3 y = 24
3 x − 3 y = 24
⇒ x− y =8
Subtracting (ii) from (i), we get
2 x − y = 45
...(iii)
−( x − 2 y = −21)
x + y = 66
Again adding eqn. (iii) and (iv), we get
x + y = 66
x− y = 8
2 x = 74
74
∴
x=
= 37
2
Subtracting eqn. (iii) from (iv), we get
x + y = 66
x− y = 8
58
= 29
2
Hence, larger number = 37 and smaller number = 29.
2 y = 58
∴
y=
Q.28. If the numerator of a fraction is decreased by 1, its value becomes
if its denominator is increased by 7 its value becomes
Ans.
Let the fraction be
1
. Find the fraction.
4
x
, then
y
x −1 1
=
y
3
x
1
nd
=
y+7 4
∴ From eqn. (i), we get
x −1 1
= ⇒ 3 ( x − 1) = y
y
3
⇒ 3x − 3 = y
and from eqn. (ii),
Math Class IX
1
; and
3
...(i)
...(ii)
...(iii)
19
Question Bank
x
1
= ⇒ 4× x = y + 7
y+7 4
...(iv)
⇒ 4x = y + 7
Putting the value of y in eqn. (iv), we get
⇒ 4 x = 3x − 3 + 7 ⇒ 4 x − 3 x = 4 ⇒ x = 4
Putting the value of x in eqn. (iii), we get
y = 3x − 3
⇒ y = 3 × 4 − 3 ⇒ y = 12 − 3 ⇒ y = 9
x 4
Hence, = .
y 9
Q.29. If the numerator of a fraction is increased by 2 and denominator is
2
decreased by 1, it becomes . If the numerator is increased by 1 and
3
1
denominator is increased by 2, it becomes . Find the fraction.
3
x
Ans. Let the fraction be , then
y
x+2 2
=
...(i)
and
y −1 3
x +1 1
=
...(ii)
y+2 3
∴ From eqn. (i), we get
x+2 2
= ⇒ 3 ( x + 2) = 2 ( y − 1)
y −1 3
⇒ 3 x + 6 = 2 y − 2 ⇒ 3 x − 2 y = −2 − 6
...(iii)
⇒ 3 x − 2 y = −8
x +1 1
From eqn. (ii),
=
y+2 3
⇒ 3 ( x + 1) = 1 ( y + 2) ⇒ 3x + 3 = y + 2 ⇒ 3x − y = −3 + 2
⇒ 3x − y = −1
or y = 3 x + 1
...(iv)
Putting the value of y in eqn. (iv) in eqn. (iii), we get
3 x − 2 (3 x + 1) = −8 ⇒ 3 x − 6 x − 2 = −8
Math Class IX
20
Question Bank
⇒ − 3 x = −8 + 2 ⇒ − 3 x = −6 ⇒ x =
∴ y = 3× 2 +1 ⇒ y = 6 + 1 = 7
x 2
Thus, =
y 7
Hence, the required fraction is
−6
=2
−3
x 2
= .
y 7
Q.30. Five years ago, A’s age was four times the age of B. Five years hence, A’s
age will be twice the age of B. Find their present ages.
Ans. Let, present age of A be x years and present age of B be y years
5 years ago, age of A = ( x − 5) years
5 years ago, age of B = ( y − 5) years
∴ x − 5 = 4 ( y − 5) ⇒ x − 5 = 4 y − 20
⇒ x − 4 y = −20 + 5
...(i)
⇒ x − 4 y = −15
5 years hence, age of A = ( x + 5) years
5 years hence, age of B = ( y + 5) years
(Given)
∴ x + 5 = 2 ( y + 5)
⇒ x + 5 = 2 y + 10
...(ii)
⇒ x − 2 y = 10 − 5
x − 2y = 5
Subtracting eqn. (ii) from (i), we get
x − 4 y = −15
⇒
x − 2y = 5
− 2 y = −20
2 y = 20
20
= 10
2
Putting the value of y in eqn. (i), we get
x − 4 × 10 = −15 ⇒ x = −15 + 40 ⇒ x = 25
Hence, present age of A is 25 years and B is 10 years.
⇒ y=
Math Class IX
21
Question Bank
Q.31. There are two examination halls A and B. If 12 pupils are sent from A to
B, the number of pupils in each halls becomes the same. If 11 pupils are
sent from B to A, then the number of pupils in A is double their number in
B. Find the number of pupils in each room.
Ans. Let number of pupils in examination hall A be x. and number of pupils in
examination hall B be y.
According to the given problem,
x − 12 = y + 12
...(i)
⇒ x − y = 12 + 12 = 24
⇒ x + 11 = 2 ( y − 11) ⇒ x + 11 = 2 y − 22
...(ii)
⇒ x − 2 y = −22 − 11 = −33
Subtracting eqn. (ii) from eqn. (i), we get
x − y = 24
x − 2 y = −33
y = 57
Putting y = 57 in eqn. (i), we get
∴ x − 57 = 24 ⇒ x = 24 + 57 = 81
Hence, number of pupils in examination hall A = 81.
Q.32. The area of a rectangle gets reduced by 8 m2 , if its length is reduced by 5
m and breadth increased by 3 m. If we increase the length by 3 m and
breadth by 2 m, the area is increased by 74 m 2 . Find the length and
breadth of the rectangle.
Ans. Let length of rectangle be x m and breadth of rectangle be y m
and area of rectangle = xy m 2
Case I:
( x − 5) ( y + 3) = xy − 8
⇒ xy + 3x − 5 y − 15 = xy − 8
⇒ 3 x − 5 y = −8 + 15 = 7
...(i)
⇒ 3x − 5 y = 7
Case II:
( x + 3) ( y + 2) = xy + 74
⇒ xy + 2 x + 3 y + 6 = xy + 74
...(ii)
⇒ 2 x + 3 y = 74 − 6 = 68
Multiplying eqn. (i) by 3 and eqn. (ii) by 5, we get
...(iii)
9 x − 15 y = 21
Math Class IX
22
Question Bank
10 x + 15 y = 340
...(iv)
Adding eqn. (iii) and eqn. (iv), we get
9 x − 15 y = 21
10 x + 15 y = 340
19 x = 361
361
x=
= 19
⇒
19
Putting the value of x in eqn. (i), we get
3 × 19 − 5 y = 7 ⇒ 57 − 5 y = 7 ⇒ − 5 y = 7 − 57 = −50 ⇒ y =
− 50
= 10
−5
Hence, length of rectangle = 19 m
and breadth of rectangle = 10 m
Q.33. A man sold a chair and a table for Rs 2178, thereby making a profit of
12% on the chair and 16% on the table. By selling them for Rs 2154, he
gains 16% on the chair and 12% on the table. Find the cost price of each.
Ans. Let CP of chair be Rs x and CP of table be Rs y.
Case I:
Gain on chair = 12%
and gain on table = 16%
x (100 + 12) 112
=
∴ SP of chair =
x
100
100
y (100 + 16) 116
=
y
and SP of table =
100
100
Case II:
Gain on chair = 16% and gain on table = 12%
x (100 + 16) 116
∴ SP of chair =
=
x
100
100
y (100 + 12) 112
and SP of table =
=
y
100
100
According to the given problem,
112
116
x+
y = 2178
100
100
...(i)
⇒ 112 x + 116 y = 217800
116
112
and
x+
y = 2154
100
100
...(ii)
⇒ 116 x + 112 y = 215400
Adding eqn. (i) and eqn. (ii), we get
Math Class IX
23
Question Bank
112 x + 116 y = 217800
116 x + 112 y = 215400
228 x + 228 y = 433200
⇒ x + y = 1900
Subtracting eqn. (ii) from eqn. (i), we get
112 x + 116 y = 217800
...(iii)
116 x + 112 y = 215400
− 4 x + 4 y = 2400
⇒ x − y = −600
Adding eqn. (iii) and eqn. (iv), we get
x + y = 1900
...(iv)
x − y = −600
1300
= 650
2
Subtracting eqn. (iii) from (iv), we get
x − y = −600
2 x = 1300 ⇒
x=
x + y = 1900
− 2500
= 1250
−2
Hence, CP of chair is Rs 650 and CP of table is Rs 1250.
Q.34. A and B together can do a piece of work in 6 days. If A’s one day’s work
1
be 1 times the one day’s work of B, find in how many days, each alone
2
can finish the work.
1
Ans. (A + B)’s one days work =
6
Let A can do the work in = x days and B can do the work in = y days
1
1
∴ A’s 1 day’s work = and B’s 1 day’s work =
x
y
According to the given problem
1 3 1 3
...(i)
= × =
x 2 y 2y
1 1 1
...(ii)
+ =
x y 6
− 2 y = −2500 ⇒ y =
Math Class IX
24
Question Bank
1
in eqn. (ii), we get
x
3 1 1
5 1
30
= 15
+ = ⇒
= ⇒ 2 y = 30 ⇒ y =
2
2y y 6
2y 6
Putting the value of y in eqn. (i), we get
1
3
1
=
=
⇒ x = 10
x 2 × 15 10
Hence, A can do the work in = 10 days and B can do the work in = 15 days.
Draw the graph of: (i) 3 x + 2 y = 0 (ii) x − 5 y + 4 = 0
(i) 3 x + 2 y = 0
−2 y
x=
⇒ 3 x = −2 y ⇒
3
If x = 0 then y = 0
2
If y = 3, then x = − × 3 = −2
3
If y = 6, then
2
x = − × 6 = −4
3
The table of 3 x + 2 y = 0 is
0
−2
−4
0
3
6
Putting the value of
Q.35.
Ans.
x
y
Plotting the above points and
joining them we get the line AB.
Which is the required graph.
Math Class IX
25
Question Bank
(ii) x − 5 y + 4 = 0 ⇒ x + 4 = 5 y
x+4
or 5 y = x + 4 ∴ y =
5
1+ 4 5
= =1
If x = 1, then y =
5
5
−4+4 0
If x = −4, then y =
= =0
5
5
6 + 4 10
If x = 6, then y =
= =2
5
5
The table of x − 5 y + 4 = 0 is
x
1
6
−4
y
1
0
2
Plotting the above points and
joining them we get the line AB,
which is the required graph.
Q.36. Draw the graph of the equation
4 x + 3 y + 6 = 0. From the graph, find:
(i)
y1 , the value of y , when x = 6.
(ii) y2 , the value of y , when x = −6.
Ans. 4 x + 3 y + 6 = 0
− (3 y + 6)
⇒ 4 x = −3 y − 6 ⇒ x =
4
If y = 2, then
− (3 × 2 + 6) − 12
x=
=
= −3
4
4
If y = −2, then
− [3 × (−2) + 6] − [−6 + 6]
x=
=
=0
4
4
If y = −6, then
 3 × (−6) + 6  −(−18 + 6)
x = −
 =
4
4

−(−12) 12
=
= =3
4
4
Math Class IX
26
Question Bank
The table of 4 x + 3 y + 6 = 0
x
0
3
−3
y
2
−6
−2
Plotting the above points, we get the required graph.
From the graph,
(i) If x = 6 , then value of y i.e. y1 will be = −10
(ii) If x = −6, then value of y i.e. y2 will be = 6.
Q.37. Draw the graph of the equation 4 x + 3 y + 6 = 0 . From the graph find:
(i)
y1 , the value of y , when x = 12
(ii) y2 , the value of y , when x = −6
Ans. 4 x + 3 y + 6 = 0
⇒ 3 y = −4 x − 6
− 4x − 6
⇒ y=
3
The table of 4 x + 3 y = −6 is
x
0
3
−3
y
2
−
6
−2
Plotting the above points as
shown in the table, we get the
required graph.
From graph
(i) the value of y when x = 12 is
− 18.
(ii) the value of y when x = −6 is 6.
Math Class IX
27
Question Bank
Q.38. Solve, graphically, the following pairs of equations:
(i) x – 5 = 0, y + 4 = 0
(ii) 2 x + y = 23, 4 x – y = 19
(iii) 2 x – 3 y = 6, 2 x + y + 10 = 0
Ans. (i) x − 5 = 0 ⇒ x = 5
y + 4 = 0 ⇒ y = −4
The graph is shown in the
adjoining figure then, x = 5
and y = −4.
(ii)
2 x + y = 23
⇒ y = 23 − 2 x
If x = 10, then
y = 23 − 2 × 10 = 3
If x = 15, then
y = 23 − 2 × 15 = −7
If x = 5, then
y = 23 − 2 × 5 = 13
The table of y = 23 − 2 x is
x
5
10
15
y
13
3
−7
Also, 4 x − y = 19
⇒ y = 4 x − 19
If x = 3, then
y = 4 × 3 − 19 = −7
If x = 4, then y = 4 × 4 − 19 = −3
If x = 6, then y = 4 × 6 − 19 = 5
The table of y = 4 x − 19 is
x
3
4
6
y
5
−7
−3
Plotting the points given in the
tables, we get the required graph.
Hence, x = 7 and y = 9.
Math Class IX
28
Question Bank
(iii) 2 x − 3 y = 6
⇒ 2x = 3y + 6
3y + 6
⇒ x=
2
If y = 0, then
3× 0 + 6 6
= =3
x=
2
2
If y = −2, then
3 × (−2) + 6 − 6 + 6
x=
=
=0
2
2
If y = −4, then
3(−4) + 6 − 12 + 6 − 6
x=
=
=
= −3
2
2
2
Table for 2 x + y + 10 = 0
x
3
0
−3
y
0
−2
−4
Plotting the points as shown in the table we get the required graph.
Again, 2 x + y + 10 = 0
2 x = −( y + 10)
( y + 10)
x=−
2
− (0 + 10) − 10
=
If y = 0, then x =
= −5
2
2
− (−2 + 10) − 8
If y = −2, then x =
=
= −4
2
2
− (−4 + 10) − 6
If y = −4, then x =
=
= −3
2
2
Hence,
x
−5
−3
−4
y
0
−2
−4
We see that the graph of two lines intersect each other at the point ( −3, − 4).
Hence, solution is x = −3 and y = −4.
Math Class IX
29
Question Bank
Q.39. Solve graphically the following equations: 3 x + 5 y = 12; 3 x − 5 y + 18 = 0
(Plot only three points per line)
Ans. 3 x + 5 y = 12
⇒ 3 x = 12 − 5 y
12 − 5 y
⇒ x=
3
If y = 0, then
12 − 5 × 0
=4
x=
3
If y = 3, then
12 − 5 × 3 − 3
x=
=
= −1
3
3
If y = 6, then
12 − 5 × 6
18
x=
= − = −6
3
3
The table of 3 x + 5 y = 12 is
x
4
−6
−1
y
0
3
6
And, 3 x − 5 y + 18 = 0
⇒ 3x = 5 y − 18
5 y − 18
⇒ x=
3
5 × 0 − 18
If y = 0, then x =
= −6
3
5 × 6 − 18 12
If y = 6, then x =
= =4
3
3
5 × 3 − 18 − 3
=
If y = 3, then x =
= −1
3
3
The table of 3 x − 5 y = −18 is
x
4
−6
−1
y
0
6
3
Plotting the points given in the above tables, we get the graph of the two lines.
The graph of the two lines intersect at the point (−1, 3). Hence, x = −1 and y = 3.
Math Class IX
30
Question Bank
Q.40. Draw the graphs of 2 x − y − 1 = 0 and 2 x + y = 9 on the same axis. Write
down the co-ordinates of the point of intersection of the two lines.
Ans. 2 x − y − 1 = 0 ⇒ 2 x = y + 1
y +1
⇒ x=
2
If y = 1, then
1+1 2
= =1
x=
2
2
If y = 3, then
3 +1 4
= =2
x=
2
2
If y = −1, then
−1+1
x=
=0
2
Table of 2 x = y + 1
X
1
2
0
Y
1
3
−1
Again 2 x + y = 9
⇒ y = 9 − 2x
If x = 1, then y = 9 − 2 × 1 = 9 − 2 = 7
If x = 2, then y = 9 − 2 × 2 = 9 − 4 = 5
If x = 3, then y = 9 − 2 × 3 = 9 − 6 = 3
Table of y = 9 − 2 x
X
1
2
3
Y
7
5
3
Plot the points given in above tables. We see that the graph of the lines
5

intersect each other at the point  , 4. 
2

Math Class IX
31
Question Bank
Q.41. Use graph paper for this question.
(i) Draw the graphs of x + y + 3 = 0 and 3 x − 2 y + 4 = 0 on the same axes.
(ii) Write down the co-ordinates of the point of intersection of the lines.
Ans.
x + y + 3 = 0 ⇒ x = −( y + 3)
If y = 0, then x = −(0 + 3) = −3
If y = −1, then x = −(−1 + 3) = −(2) = −2
If y = −3, then x = −(−3 + 3) = 0
Table of x = −( y + 3)
0
x
−3
−2
y
0
−3
−1
Again, 3 x − 2 y + 4 = 0
⇒ 3x = 2 y − 4
2y − 4
x=
3
If y = −1, then
2 × (−1) − 4
x=
3
−2 − 4 −6
=
=
= −2
3
3
If y = 2, then
2× 2 − 4 4 − 4
x=
=
=0
3
3
If y = 5, then
2 × 5 − 4 10 − 4 6
x=
=
= =2
3
3
3
Table of 3 x − 2 y + 4 = 0
0
2
x
−2
y
2
5
−1
Plot the points in two tables. From the graph we see that the graph of the lines
intersect at (−2, − 1).
Math Class IX
32
Question Bank
Q.42. Find, graphically, the vertices of the triangle whose sides have the
equations 2 y − x = 8; 5 y − x = 14 and y − 2 x = 1 respectively. Take
1 cm = 1 unit on both the axes.
Ans. 2 y − x = 8
x = 2y −8
⇒
If y = 1, then x = 2 × 1 − 8 = 2 − 8 = −6
If y = 3, then x = 2 × 3 − 8 = 6 − 8 = −2
If y = 4, then x = 2 × 4 − 8 = 8 − 8 = 0
The table of x = 2 y − 8 is
X
0
−6
−2
Y
1
3
4
Again, 5 y − x = 14
⇒
x = 5 y − 14
The table of 5 y − x = 14 is
X
1
−9
−4
Y
1
2
3
y − 2x = 1
⇒ y = 1+ 2 x
If x = 2, then
y = 1+ 2× 2 = 5
If x = −2, then
y = 1 + 2 × ( −2 ) = −3
If x = 0, then y = 1 + 0 = 1
The table of y = 1 + 2 x is
X
2
0
−2
Y
5
1
−3
Plotting the points given in the
above tables, we get the required graph.
From the graph the vertices of the
triangle are A ( −4, 2), B (2, 5) and C (1, 3).
Math Class IX
33
Question Bank