(i) (-2mn) × (2m) 2 × (-3mn) = (-2mn) × 2 2 m 2 × (-3mn) = (- 2mn) × 4m 2 × (- 3mn)
= (-) (+)(-) (2 × 4 × 3) (m × m 2 × m) (n × n)
= +24 m 4 4n 2
(ii) (3x 2 y) × (-3xy 3 ) × (x 2 y 2 ) = (+) × (-) × (+) × (3 × 3 × 1)(x 2 × x × x 2 ) × (y × y 3 × y 2 )
= -9x 5 y 6
6xy × (-2x 2 ) = -12x 3 y
(ii) ______ × (-15m 2 n 3 p) = 45m 3 n 3 p 2
-3mp × (-15m 2 n 3 p) = 45m 3 n 3 p 2
(iii) 2y(5x 2 y – ____ + 3____ ) = 10x 2 y 2 – 2xy + 6y 3
2y(5x 2 y – x + 3y 2 ) = 10x 2 y 2 – 2xy + 6y 3
Speed of the car = (x + 30) km/hr.
Time = (y + 2) hours
Distance = Speed × time
= (x + 30) (y + 2) = x(y + 2) + 30(y + 2)
= (x) (y) + (x) (2) + (30) (y) + (30) (2)
= xy + 2x + 30y + 60
Distance covered = (xy + 2x + 30y + 60) km
Objective Type Questions
- A. mn 2 , 27
- B. m 2 n, 27
- C. m 2 n 2 , – 27
- D. mn 2 , – 27
(A) mn 2 , 27
- A. 6x 4 y 2
- B. 8x 2 y 2
- C. 6x 2 y
- D. -6x 2 y
(C) 6x 2 y
- A. a – b
- B. a + b
- C. a 2 – b
- D. (a + b) 2
(B) a + b
True(ii) 7ab 3 ÷ 14 ab = 2b 2
False
7y 2 – y 2 + 3y 2 = 10y 2 = (7 – 1 + 3)y 2
= (6 + 3)y 2
= 9y 2
(ii) 6xy + 3xy = 9x 2 y 2
6xy + 3xy = (6 + 3) xy
= 9 xy
(iii) m(4m – 3) = 4m 2 – 3
m(4m – 3) = m(4m) + m(-3)
= 4m 2 – 3m(iv) (4n) 2 – 2n + 3 = 4n 2 – 2n + 3
(4n) 2 – 2n + 3 = 16n 2 – 2n + 3
(v) (x – 2)(x + 3) = x 2 – 6
(x – 2)(x + 3) = x(x + 3) – 2 (x + 3)
= x(x) + (x) × 3 + (-2) (x) + (-2) (3)
= x 2 + 3x – 2x – 6
= x 2 + x – 6
(vi) -3p 2 + 4p – 7 = -(3p 2 + 4p – 7)
-3p 2 + 4p – 7 = -(3p 2 – 4p + 7)
(ii) A is true but B is false
Hint:
(i) Both A and B are true
Hint:
(2m – 5) – (5 – 2m) = 2m – 5 – 5 + 2m = 4m – 10
(2m – 5) + (2m – 5) = 4m – 10
Given side of the cube = (x + 1) cm
Volume of the cube = (side) 3 cubic units = (x + 1) 3 cm 3
We have (a + b) 3 = (a3 3 + 3a 2 b + 3ab 2 + b 3 ) cm 3
(x + 1) 3 = (x 3 + 3x 2 (1) + 3x(1) 2 + 1 3 )cm 3
Volume = (x 3 + 3x 2 + 3x + 1) cm 3
Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)
∴ Volume of the cuboid = (l × b × h) units 3
= (x + 2) (x – 1) (x – 3) units 3
We have (x + a)(x + b) (x+c) = x 3 + (a + b + c)x 2 + (ab + bc + ca)x + abc
∴ (x+2) (x- 1) (x-3) = x 3 + (2 – 1 – 3)x 2 + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2) (-1) (-3)
= x 3 – 2x 2 + (-2 + 3 – 6)x + 6
Volume = x 3 – 2x 3 – 5x + 6 units 3Objective Type Questions
- A. 8
- B. 3
- C. 2
- D. 1
(C) 2
Hint:
x 2 – y 2 = 16
(x + y) (x – y) = 16
8 (x – y) = 16
(x – y) = \(\frac { 16 }{ 8 }\) = 2
- A. a 2 – ab + b 2
- B. a 2 + ab + b 2
- C. a 2 + 2ab + b 2
- D. a2 2 – 2ab + b 2
(B) a 2 + ab + b 2
Hint:= a 2 + ab + b 2
- A. 15
- B. 18
- C. 62
- D. 72
(D) 72
Hint:
(a – b) = 3
(a – b) 2 = 3 2
a 2 + b 2 – 2ab = 9
a 2 + b 2 – 2(5) = 9
a 2 + b 2 = 9 + 10
a 2 + b 2 = 19
a 3 – b 3 = (a – b)(a 2 + ab + b 2 ) = 3(19 + 5)
= 3(24) = 72
- A. 3a(a + b)
- B. 3ab(a – b)
- C. -3ab(a + b)
- D. 3ab(a + b)
(D) 3ab(a + b)
Hint:
(a + b) 3 = a 3 + b 3 + 3a 2 b + 3ab 2
(a + b) 3 – 3a 2 b – 3ab 3 = a 3 + b 3
(a + b) 3 – 3ab(a + b) = a 3 + b 3
18xy – 12yz = (2 × 3 × 3 × y × x) – (2 × 2 × 3 × y × z)
Taking out the common factors 2, 3, y, we get
= 2 × 3 × y(3x – 2z) = 6y(3x – 2z)(ii) 9x 5 y 3 + 6x 3 y 2 – 18x 2 y
9x 5 + 6x 3 y 2 – 18x 2 y = (3 × 3 × x 2 × x 3 × y × y) + (2 × 3 × x 2 × x × y × y) – (2 × 3 × 3 × x 2 × y)
Taking out the common factors 3, x 2 , y, we get
= 3 × x 2 × y (3x 3 y 2 + 2xy – 6)
= 3x 2 y (3x 3 y 2 + 2xy – 6)
(iii) x(b – 2c) + y(b – 2c)
Taking out the binomial factor (b – 2c) from each term, we have
= (b – 2c)(x + y)
(iv)(ax + ay) + (bx + by)
Taking at ‘a’ from the first term and ‘b’ from the second term we have
(ax + ay )+ (bx + by) = a(x + y) + b(x + y)
Now taking out the binomial factor (x + y) from each term
= (x + y) (a + b)(v) 2x 2 (4x – 1) – 4x + 1
Taking out -1 from last two terms
2x 2 (4x – 1) – 4x + 1 = 2x 2 (4x – 1) – 1 (4x – 1)
Taking out the binomial factor 4x – 1, we get
= (4x – 1) (2x 2 – 1)
(vi) 3y(x – 2) 2 – 2(2 – x)
3y(x – 2) 2 – 2(2 – x) = 3y(x – 2)(x – 2) – 2( -1)(x – 2)
[∵ Taking out – 1 from 2 – x]
= 3y(x – 2)(x – 2) + 2(x – 2)
Taking out the binomial factor x – 2 from each term, we get
= (x – 2) [3y(x – 2) + 2]
(vii) 6xy – 4y 2 + 12xy – 2yzx
= 6xy + 12xy – 4y 2 – 2yzx [∵ Addition is commutative]
= (6 × x × y) + (2 × 6 × x × y) + (-1) (2) (2) y + y) + ((-1) (2) (y) (z) (x))
Taking out 6 x x x y from first two terms and (-1) × 2 × y from last two terms we get
= 6 × x × y(1 + 2) + (-1) (2) y [2y + zx]
= 6 × y(3) – 2y(2y + zx)
= (2 × 3 × 3 × x × y) – 2xy(2y + zx)
Taking out 2y from two terms
= 2y(9x – (2y + zx))
= 2y (9x – 2y – xz)(viii) a 3 – 3a 2 + a – 3
a 2 – 3a 2 + a – 3 = a 2 (a – 3) + 1(a – 3) [:Groupingthetermssuitably]
= (a – 3) (a 2 + 1)
(ix) 3y 3 – 48y
3y 2 – 48y = 3 × y × y 2 – 3 × l6 × y
Taking out 3 × y
= 3y(y 2 – 16) = 3y(y 2 – 4 2 )
Comparing y 2 – 4 2 with a 2 – b 2
a = y, b = 4
a 2 – b 2 = (a + b) (a – b)
y 2 – 4 2 = (y + 4) (y – 4)
∴ 3y(y 2 – 16) = 3y(y + 4)(y – 4)
(x) ab 2 – bc 2 – ab + c 2
ab 2 – bc 2 – ab + c 2
Grouping suitably
ab 2 – bc 2 – ab + c 2 = b (ab – c 2 ) – 1 (ab – c 2 )
Taking out the binomial factor ab – c 2 = (ab – c 2 ) (b – 1)
x 2 + 14x + 49 = x 2 + 14x + 72
Comparing with a 2 + 2ab + b 2 = (a + b) 2 we have a = x and b = 7
⇒ x 2 + 2(x)(7) + 7 2 = (x + 7) 2
∴ x 2 + 14x + 49 = (x + 7) 2
(ii) y 2 – 10y + 25
y 2 – 10y + 25 = y 2 – 10y + 5 2
Comparing with a 2 – 2ab + b 2 = (a – b) 2 we get a = y; b = 5
⇒ y 2 – 2(y) (5) + 5 2 = (y – 5) 2
∴ y 2 – 10y + 25 = (y – 5) 2
(iii) c 2 – 4c – 12
This is of the form ax 2 + bx + c
Where a = 1, b = -4 c = -12, x = c
Now the product ac = 1 × – 12 = -12 and the sum b = -4
Product = – 72
Sum = 1
1 × (-12) = -12
1 + (-12) = -11
2 × (-6) = – 12
2 + (-6) = – 4
∴ The middle term – 4c can be written as 2c – 6c
∴ c 2 – 4c – 12 = c 2 + 2c – 6c – 12
= c(c + 2) -6 (c + 2)Taking out (c + 2)
⇒ (c + 2)(c – 6)
∴ c 2 – 4c – 12 = (c + 2)(c – 6)(iv) m 2 + m – 72
m 2 + m – 72
This is of the form ax + bx + c
where a = 1, b = 1, c = -72
Product = – 72
Sum = 1
1 × -72 = – 72
1 + (-72) = -71
2 × – 36 = – 72
2 + (-36) = – 34
3 × (-24) = – 72
3 + (-24) = – 21
4 × (-18) = -72
4 + (-18) = – 14
6 × (-12) = -72
6 + (-12) = – 6
8 × (-9) = -72
8 + (-9) = – 1
9 × (-8) = – 72
9 + (-8) = 1
Product a × c = 1 × -72 = -72
Sum b = 1
The middle term m can be written as 9m – 8m
m 2 + m – 72 = m 2 + 9m – 8m – 72
= m(m + 9) – 8(m + 9)Taking out (m + 9)
= (m + 9)(m – 8)
∴ m 2 + m – 72 = (m + 9)(m – 8)
(v) 4x 2 – 8x + 3
4x 2 – 8x + 3
This is of the form ax 2 + bx + c with a = 4 b = -8 c = 3
Product ac = 4 × 3 = 12
sum b = -8
Product = 12
Sum = -8
(-1) × (-12) = 12
(-1) + (-12) = – 13
(-2) × (-6) = 12
(-2) + (-6) = – 8
The middle term can be written as – 8x = – 2x – 6x
4x 2 – 8x + 3 = 4x 2 – 2x – 6x + 3
= 2x (2x – 1) – 3 (2x – 1)
= (2x – 1)(2x – 3)
4x 2 – 8x + 3 = (2x – 1) (2x – 3)
For ₹ 10 ab the number of note books can buy = 1.Number of note book he can buy = \(\frac { 1 }{ 2 }\)a + 2b + 4
Given Number of rooms = x + 2
Number of rooms × Number of outlets = amount of wire.
(x + 2) × Number of outlets = 4x 2 + 11x + 6
Number of outlets = \(\frac{4 x^{2}+11 x+6}{x+2}\) … (1)
Now factorising 4x 2 + 11x + 6 which is of the form ax 2 + bx + c with a = 4 b = 11 c = 6.
The product a × c = 4 × 6 = 24
sum b = 11
Product = 24
Sum = 11
1 × 24 = 24
1 + 24 = 25
2 × 12 = 24
2 + 12 = 14
3 × 8 = 24
3 + 18 = 11The middle term 11x can be written as 8x + 3x
∴ 4x 2 + 11 x + 6 = 4x 2 + 8x + 3x + 6
= 4x(x + 2) + 3 (x + 2)
4x 2 + 11x + 6 = (x + 2)(4x + 3)
Now from (1) the number of outlets∴ Number of outlets = 4x + 3
Given length of the room = x + 4 .
Area of the room = x 2 + 6x + 8
Length × breadth = x 2 + 6x + 8
breadth = \(\frac{x^{2}+6 x+8}{x+4}\) ….. (1)
Factorizing x 2 + 6x + 8, it is in the form of ax 2 + bx + c
Where a =1 b = 6 c = 8.
The product a × c = 1 × 8 = 8
sum = b = 6
Product = 8
Sum = 6
1 × 8 = 8
1 + 8 = 9
2 × 4 = 8
2 + 4 = 6The middle term 6x can be written as 2x + 4x
∴ x 2 + 6x + 8 = x 2 + 2x + 4x + 8
= x(x + 2) + 4(x + 2)
x 2 + 6x + 8 = (x + 2)(x + 4)
Now from (1)∴ Width of the room = x + 2
We have (x + a)(x + b) = x 2 + (a + b)x + ab
56 = 7 × 8
∴ y 2 + (7 + 8)x + 56 = (y + 7) (y + 8)
7
Hint:
Given, x + 5 = 12
x = 12 – 5 = 7 (by transposition method)
Value of x is 7
11
Hint:
Given, y – 9 = (-5) + 7
y – 9 = 7 – 5 (re-arranging)
y – 9 = 2
∴ y = 2 + 9 = 11 (by transposition method)
7
Hint:
Given, 8m = 56
Divided by 8 on both sides
\(\frac{8 \times m}{8}=\frac{56}{8}\)
∴ m = 7
15
Hint:
Given, \(\frac{2 p}{3}\) = 10
Multiplying by 3 on both sidesDividing by 2 on both sides∴ p = 15
one
True
False
[Linear equation in one variable has only one variable with power one – correct statement]
Expanding the bracket,
-3 × 4 + (-3) × 9 = 21
∴ -12x + (-27) = 21
– 12x – 27 = 21
Transposing – 27 to other side, it becomes +27
– 12 x = 21 + 27 = 48
∴ – 12x = 48 ⇒ 12x = – 48
Dividing by 12 on both sides(ii) 20 – 2 (5 – p) = 8Expanding the bracket,
20 – 2 × 5 – 2 × (-p) = 8
20 – 10 + 2p = 8
(- 2 × – p = 2p)
10 + 2p = 8 transposing lo to other side
2p = 8 – 10 = – 2
∴ 2p = – 2
∴ p = – 1
(iii) (7x – 5) – 4(2 + 5x) = 10(2 – x)Expanding the brackets,
7x – 5 – 4 × 2 – 4 × 5x = 10 × 2 + 10 × (-x)
7x – 5 – 8 – 20x = 20 – 10x
7x – 13 – 20x = 20 – 10x
Transposing 10x & – 13, we get
7x – 13 – 20x + 10x = 20
7x – 20x + 10x = 20 + 13, Simplifying,
– 3x = 33
3x = – 33
x = \(\frac{-33}{3}\) = – 11
x = – 11
\(-\frac{b}{a}\)
Hint:
ax + b = 0
ax = – b
∴ x = \(-\frac{b}{a}\)
Positive
Hint:
Since a & b are positive integers, b
The solution to the equation ax = b is x = \(\frac{b}{a}\) is also positive.
30
Hint:
Let the number be x.
As per question, when one sixth of number is subtracted from itself it gives 25
40°
Hint:
Given angles are in the ratio 2:3:4
Let the angles be 2x, 3x & 4x
Since sum of the angles of a triangle is 180°,
We get 2x + 3x + 4x = 180
∴ 9x = 180
∴ x = \(\frac{180}{9}\) = 20°
∴ The angles are 2x = 2 × 20 = 40°
3x = 3 × 20 = 60°
4x = 4 × 20 = 80°
∴ Difference between greatest & smallest angle is
80° – 40° = 40°
b = 9
Hint:
Given equation is a + b = 23, a = 14
14 + b = 23
∴ b = 23 – 14 = 9
b = 9
True
Hint:
Let the number be ‘y’
∴ Sum of number & two times that number is 48
Can be written as y + 2y = 48 – True
True
Hint:
5 (3x + 2) = 3 (5x – 7) is a linear equation in one variable – ‘x’ – True
False
Hint:
One third of number is 10 less than original number.
Let number be ‘x’. Therefore let us frame the equation
\(\frac{x}{3}\) = x – 10
∴ x = 3x – 30
3x – x = 30
2x = 30
x = 15 is the solution
Let the numbers be x & y
Given that one number is 7 times the other & that the difference is 18.
Let x = 7y
also, x – y = 18 (given)
Substituting for x in the above
We get 7y – y = 18
∴ 6y = 18
y = \(\frac{18}{6}\) = 3
∴ x = 7y = 7 × 3 = 21
The number are 3 & 21
Given sum of three consecutive odd numbers is 75
Odd numbers are 1,3,5,7,9, 11, 13
∴ The difference between 2 consecutive odd numbers is always 2. or in other words, if one odd number is x, the next odd number would be x + 2 and the next number would be x + 2 + 2x + 4
i.e x + 4
Since sum of 3 consecutive odd nos is 75
∴ x + x + 2 + x + 4 = 75
∴ 3x + 6 = 75 ⇒ 3x = 75 – 6
∴ 3x = 69
x = \(\frac{69}{3}\) = 23
∴ The odd numbers are 23, 23 + 2, 23 + 4
i.e 23, 25, 27
∴ Largest number is 27.
Let length & breadth of rectangle be ‘l’ and ‘b’ respectively
Given that length is \(\frac{1}{3}\) of breadth,
∴ l = \(\frac{1}{3}\) × b ⇒ l = \(\frac{b}{3}\) ⇒ b = 3l ……. (1)Also given that perimeter is 64 m
Perimeter = 2 × (l + b)
2 × 1 + 2 × b = 64
Substituting for value of b from (1), we get
2l + 2(3l) = 64
∴ 2l + 6l = 64
8l = 64
∴ l = \(\frac{64}{8}\) = 8m
b = 3l = 3 × 8 = 24m
Ienglh l = 8 m in & breadth b = 24 m
Let the number of ₹ 5 notes be ‘x’
And number of ₹ 10 notes be ‘y’
Total numbers of notes is x + y = 90 (given)
The total value of the notes is 500 rupees.
Value of one ₹ 5 rupee note is 5
Value of x ₹ 5 rupee notes is 5 × x = 5x
∴ Value of y ₹ 10 rupee flotes is 10 × y = lOy
∴ The total value is 5x + 10y which is 500
∴ we have 2 equations:
x + y = 90 ….(1)
5x + 10y = 500 ….(2)
Multiplying both sides of(1) by 5, we get
5 × x + 5 × y = 90 × 5
5x + 5y = 450
Subtracting (3) from (2), we get∴ y = \(\frac{50}{5}\) = 10
Substitute y = 10 in equation (1)
x + y = 90 ⇒ x + 10 = 90 ⇒ x = 90 – 10 ⇒ x = 80
There are ₹5 denominations are 80 numbers and ₹10 denominations are 10 numbers
Let present ages of Thenmozhi & Murali be ‘t’ & ‘m’
Given that at present
Then mozhi’s age is 5 years more than Murali
∴ t = m + 5 …… (1)
5 years ago, Thenmozhi’s age would be t – 5
& Murali’s age would be m – 5
Ratio of their ages is given as 3 : 22(t – 5) = 3(m – 5)
2 × t – 2 × 5 = 3 × m – 3 × 5 ⇒ 2t – 10 = 3m – 15
Substituting for t from (1)
2(m + 5) – 10 = 3m – 152m = 3m – 15
3m – 2m = 15
m = 15
t = m + 5 = 15 + 5 = 20
∴ Present ages of Thenmozhi & Murali are 20 & 15
Let the units/digit of a number be ‘u’ & tens digit of the number be ‘t’
Given that sum of it’s digits is 9
∴ t + u = 9 ……. (1)
If 27 is subtracted from original number, the digits are interchanged
The number is written as 10t + u
[Understand: Suppose a 2 digit number is 21
it can be written as 2 × 10 + 1
∴ 32 = 3 × 10 + 2
45 = 4 × 10 + 5
tu = t × 10 + u = 1ot + u]
Given that when 27 is subtracted, digits interchange
10t + u – 27 = 10u + t (number with interchanged digits)
∴ By transposition & bringing like variables together
10t – t + u – 10u = 27
∴ 9t – 9u = 27
Dividing by ‘9’ throughout , we get
\(\frac{9 t}{9}-\frac{9 u}{9}=\frac{27}{9}\) ⇒ t – u = 3 ……. (2)
Solving (1) & (2):t = 6 substitute in (1)
t + u = 9 ⇒ 6 + u = 9 ⇒ u = 9 – 6 = 3
Hence the number is 63.
Let the numerator & denominator be ‘n’ & ‘d’
Given that denominator exceeds numerator by 8
∴ d = n + 8 ……. (1)
If numerator increased by 17 & denominator decreased by 1,
it becomes (n + 17) & (d – 1), fraction is \(\frac{3}{2}\).2(n + 17) = 3(d – 1)
2n + 2 × 17 = 3d – 3∴ 34 + 3 = 3d – 2n
∴ 3d – 2n = 37 …….. (2)
Substituting eqn. (1) in (2), we get,
3 × (n + 8) – 2n = 37
3n + 3 × 8 – 2n = 37∴ n = 37 – 24 = 13
d = n + 8 = 13 + 8 = 21
The fraction is \(\frac{n}{d}=\frac{13}{21}\)
Let the distance to be covered by train be ‘d’Case 1:
If speed = 60km/h
The time taken is 15 minutes more than usual (t + \(\frac{15}{60}\))
Let usual time taken be ‘t’ hrs.
Caution: Since speed is given in km/hr, we should take care to maintain all units such as time should be in hour and distance should be in kin.
Given that in case 1, it takes 15 min. more
15m = \(\frac{15}{60}\) hr = \(\frac{1}{4}\)hr.
∴ Substituting in formula,Since usually it takes ‘t’ hr, but when running at 60 k, it kes 15 min (\(\frac{1}{4}\)hr) extra.
Multiplying by 60 on both sides
d = 60 × t + 60 × \(\frac{1}{4}\) = 6ot + 15 …… (1)
Case 2:
Speed is given as 85 km/h
Time taken is only 4 min (\(\frac{4}{60}\)hr) more than usual time
∴ time taken = (t + \(\frac{1}{15}\)) hr. \(\left(\frac{4}{60}=\frac{1}{15}\right)\)
Using the formula,Multiplying by 85 on both sides
\(\frac{d}{85}\) × 85 = 85 × t + 85 × \(\frac{1}{15}\)
∴ d = 85t + \(\frac{17}{3}\) ….. (2)
From (1) & (2), we will solve for ‘r’
Equating & eliminating ‘d’ we get∴ By transposing, we getSubstituting this value of ‘t’ in eqn. (1), we get
d = 60t + 15
= 60 × \(\frac{28}{75}\) + 15 = \(\frac{1680}{75}\) + 15 = 22.4 + 15
= 37.4 km
- A. 15
- B. 20
- C. 25
- D. 40
(b) 20
Hint:
Let number be ‘x’
∴ half of number is \(\frac{x}{2}\)
Sum of number and it’s half is given by
x + \(\frac{x}{2}\) = 30 [Multiplying by 2 on both sides]
2x + x = 30 × 2
3x = 60
x = \(\frac{60}{3}\) = 20
- A. 62°
- B. 72°
- C. 78°
- D. 68°
(a) 62°
As per property of A. exterior angle is equals to sum of interior opposite angles
Let the other interior angle to be found be ‘x’
∴ x + 58 = 120°
∴ x = 120 – 58 = 62°
- A. 50000
- B. 30000
- C. 10000
- D. 5000
(c) 10000
Hint:
Let sum of money be P’
Time period (n) is given as 1 yr.
Rate of simple interest (r) is given as 5% p.a
∴ As per formula for simple interest.
S.I = \(\frac{\mathrm{P} \times r \times n}{100}=\frac{\mathrm{P} \times 5 \times 1}{100}\) = 500
∴ P × 5 × n = 500 × 100
∴ p = \(\frac{500 \times 100}{5}\) = 100 × 100 = 10,000
- A. 6
- B. 2
- C. 4
- D. 8
(C) 4
Hint:
Product of LCM & HCF of 2 numbers is always product of the numbers. [this is property]
Product of LCM & HCF is given as 24
∴ Product of the 2 nos. is 24
Given one number is 6. Let other number be ‘x’
∴ 6 × x = 24
x = \(\frac{24}{6}\) = 4
- A. x
- B. x + 1
- C. x + 2
- D. x – 1
(D) x – 1
Hint:
The 3 consecutive numbers are: x – 1, x, x + 1
Origin (0,0)
negative
Y-axis
Zero
X
True
Hint:
(-10, 20)
x = -10, y = 20
∴ (-10, 20) lies in second quadrant – True
True
Hint:
(-9, 0) on x – axis. Y- coordinate is always zero.
∴ (-9, 0) lies on x axis – True
False
Hint:
Coordinate of origin is (0, 0), not (1, 1). Hence – False
If X & y coordinate are positive – I quad
If x is positive,y is negative – IV quad
If x is negative, y is positive – II quad
If both are negative, then – III quad
Origin (0,0)
Hint:
[When we substitute x = 0 in equation, y also becomes zero. (0,0) is a solution]
4, -4
Hint:
x = 4 is a line parallel to the y – axis and
y = -4 is a line parallel to the x – axis. The point of intersection is a point that lies on both lines & which should satisfy both the equations. Therefore, that point is (4, -4)
On the x-axis 1 cm = _________ units
y-axis 1 cm = _________ units3 units, 25 units
Hint:
With reference to given graph,
On the x – axis. 1 cm = 3 units
y axis, 1 cm = 25 units
True
Hint:
The points (1, 1), (2, 2), (3, 3) all satisfy the equation y = x which is straight line. Hence, it is true
False
Hint:
y = -9x substituting for x as zero, we get y = -9 × 0 = 0
∴ for x = 0, y = 0. Which means line passes through (0, 0), hence statement is false.
Given a line intersects the axis at (2, 0) & (0, 2)
Let line intercept form be expressed as
ax + by = 1 Where a & b are the x & y intercept respectively.
Since the intercept points are (2. 0) & (0, 2)
a = 2, b = 2
∴ 2x + 2y = 1
When the point (2. 2) is considered & substituted in the equation
2x + 2y = 1, we get
2 × 2 + 2 × 2 = 4 ≠ 1
∴ the point (2. 2) does not satisfy the equation. Therefore the line does not pass through (2, 2)
Alternatively graphical methodas we can see the line doesn’t pass through (2, 2)
Line passes through (4, – 2)
y – axis intercept point – (0, 2) using 2 point formula.Any point in II quadrant will have x as negative & y as positive.
So let us take x value as – 2
∴ -2 + y = 2
∴ y = 2 + 2 = 4
∴ Point in II Quadrant is (-2, 4)
Plotting the points on a graph (approximately)
Steps:
Plot P, Q, R approximately on a graph.
As it is a rectangle, RS should be parallel to PQ & QR should be paraHel to PS
S should lie on the straight line from R parallel to x-axis & straight line from P parallel to y-axis
Therefore, we get S to be (5, -4)
[Note: We don’t need graph sheet for approximate plotting. This is just for graphical understanding]
In a trapezium. there are 2 opposite sides that are parallel. The other opposite sides are non-parallel.
Now, let us approximately plot the points for our understanding
[no need of graph sheet]Plot the points (0, 6), (6, 0), (-3, 0) & (0, -3)
Join (0, 6) & (6, 0)
Join (-3,0) & (0, – 3)
We find that the lines formed by joining the points are parallel lines.
So, for forming a trapezium, we should join (0, 6), (-3, 0) & (0, -3), (6, 0)
Equation of line joining 2 points by 2 point formula is given byCross multiplying, we getTransposing the variables, we get
11 x + 5 y = 35 – 33 = 2
11 x + 5y = 2 – Line 1
Similarly, we should find out equation of second line∴ 9y – 54 = 13x – 52
∴ 9y – 13x = 2 – Line 2
For finding point of intersection, we need to solve the 2 line equation to find a point that will satisfy both the line equations.
∴ Solving for x & y from line 1 & line 2 as below
11x + 5y = 2 ⇒ multiply both sides by 13,
11 × 13x + 5 × 13y = 26 …….. (3)
Line 2: 9y – 13x = 2 ⇒ multiply both sides by 11
9 × 11y – 13 × 11x = 22 ……… (4)∴ 164 y = 48
∴ y = \(\frac{48}{164}=\frac{12}{41}\)
Substituting this value ofy in line I we get
11 x + 5 y = 2
11 x + 5 × \(\frac{12}{41}\) = 2
11 x = 2 – \(\frac{60}{41}=\frac{82-60}{41}=\frac{22}{41}\)
∴ x = \(\frac{2}{41}\)
[∴ Point of intersection is \(\left(\frac{2}{41}, \frac{12}{41}\right)\)]
To find point of intersection of the lines with the axis, we should substitute values & check
Line 1: 11 x + 5 y = 2
Point of intersection of line with x – axis, i.e y coordinate is ‘0’
∴ put y = 0 in above equation
∴ 11 x – 5 × 0 = 2
∴ 11x + 0 = 2
∴ x = \(\frac{2}{11}\)
∴ [Point is \(\left(\frac{2}{11}, 0\right)\)]
Similarly, Point of intersection of line with y – axis is when x-coordinate becomes ‘0’
∴ put x = 0 in above equation
∴ 11 × 0 + 5y = 2
∴ 0 + 5y = 2
y = \(\frac{2}{5}\)
∴ [Point is \(\left(0, \frac{2}{5}\right)\)]
Similarly for line 2,
9y – 13x = 2
For finding x intercept, i.e point where line meets x axis, we know that y coordinate becomes ‘0’
∴ Substituting y = 0 in above eqn. we get
9 × 0 – 13x = 2
∴ 0 – 13x = 2
∴ x = \(\frac{-2}{13}\)
∴ [Point: \(\left(\frac{-2}{13}, 0\right)\)]
Similarly for y – intercept, x – coordinate becomes ‘0’,
∴ Substituting for x = 0 in above equation, we get
9 y – 13 × 0 = 2
9y – 0 = 2
9y = 2
y = \(\frac{2}{9}\)
[Point \(\left(0, \frac{2}{9}\right)\)]
To draw graph, we need to find out some points.
(i) y = – 3x
for y = -3x, let us first substituting values & check
put x = 0
y = 3 × 0 = 0
∴ (0,0) is a point
put x = 1
y = -3 × 1 = – 3
∴ (1, – 3) is a point
If join these 2 points, we will get the line
(ii) y = x – 4
for y = x – 4
put x = 0
y = 0 – 4 = – 4
∴ (0, – 4) is a point
x = 4
y = 4 – 4 = 0
∴ (4, 0) is a point
(iii) y = 2x + 5
for y = 2x + 5
put x = – 1
y = 2(-1) + 5 = – 2 + 5 = 3
∴ (-1, 3) is a point
put x = – 2
y = 2(-2) + 5 = – 4 + 5 = 1
∴ (-2, 1) is a point
Now let us plot the points & join them on graph
Here what we know
a + b + c = 58 (sum of three numbers is 58)
Let the first number be b ‘x’
b = a + 3 (the second number is three times of of the first \(\frac{2}{5}\) number)
b = 3 × \(\frac{2}{5}\)x \(\frac{6}{5}\)x
Third number = x – 6
Sum of the numbers is given as 58.
∴ x + \(\frac{6}{5}\)x + (x – 6) = 58
Multiplying by 5 throughout, we get
5 × x + 6x + 5 × (x – 6) = 58 × 5
5x + 6x + 5x – 30 = 290
∴ 16x = 290 + 30
∴ 16x = 320
∴ x = \(\frac{320}{16}\)
x = 20
1 st number = 203 rd number = 24 – 16 = 14
Let angle ∠A be a°
Given that ∠B = \(\frac{2}{3}\) × ∠A = \(\frac{2}{3}\)a
& given ∠C = ∠A + 20 = a + 20
Since A, B & C are angles of a triangle, they add up to 180° (∆ property)
∴∠A + ∠B + ∠C = 180°
⇒a + \(\frac{2}{3}\)a + a + 20 = 180°
\(\frac{3 a+2 a+3 a}{3}\) + 20 = 180°
\(\frac{8 a}{3}\) = 180 – 20 = 160
∴ a = \(\frac{160 \times 3}{8}\) = 60°∠C = 80°
Given that 5y – 2 & 4y + 9 are the equal sides of an isosceles triangle.
∴ The 2 sides are equal∴5y – 4y = 9 + 2 (by transposing)
∴ y = 11
∴ 1 st side = 5y – 2 = 5 × 11 – 2 = 55 – 2 = 53
2 nd side = 53 .
3 rd side = 2y + 5 = 2 × 11 + 5 = 22 + 5 = 27
Perimeter is the sum of all 3 sides
∴ P = 53 + 53 + 27 = 133 units
Let the 3 consecutive integers be ‘x’, ‘x + 1’ & ‘x + 2’
Given that when multiplied by 2, 3 & 4 respectively & added up, we get 74Simplifying the equation, we get
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 63
9x = 63 ⇒ x = \(\frac{63}{9}\) = 7
First number = 7
Second numbers = x + 1 ⇒ 7 + 1 = 8
Third numbers = x + 2 ⇒ 7 + 2 = 9
∴ The numbers are 7, 8 & 9
Let the number of students in each bus be ‘x’
∴ number of students in 6 buses = 6 × x = 6x
A part from 6 buses, 7 students went in van
A total number of students is 331
∴ 6x + 7 = 331
∴ 6x = 331 – 7 = 324
∴ x = \(\frac{324}{6}\) = 54
∴ There are 54 students in each bus.
Let vendor have ‘p’ number of pencils & ‘b’ number of ball pens
Given that total number of items is 22
∴ p + b = 22
Pencils are sold for ₹ 15 each & ball pens for ₹ 20 each
total sale amount = 15 × p + 20 × b
= 15p + 20b which is given to be 380.
∴ 15p + 20b = 380
Dividing by 5 throughout,
\(\frac{15 p}{5}+\frac{20 b}{5}\) = \(\frac{380}{5}\) ⇒ – 3p + 4b = 76
Multiplying equation (1) by 3 we get
3 × p + 3 × b = 22 × 3
⇒ 3p + 3b = 66
Equation (2) – (3) gives∴ b = 10
∴ p = 12
He sold 12 pencils
(i) y = x
(ii) y = 2x,
(iii) y = 3x
(iv) y = 5x(i) y = x
When x = 1, y = 1
x = 2, y = 2
x = 3, y = 2
(ii) y = 2x
When x = 1, y = 2
x = 2, y = 4
x = 3, y = 6
(iii) y = 3x
When x = 1, y = 3
x = 2, y = 6
x = 3, y = 9
(i) y = 5x
When x = 1, y = 5
x = 2, y = 10
x = 3, y = 15
When we plot the above points & join the points to form line, we notice that the lines become progressively steeper. In other words, the slope keeps increasing.
4 terms
(ii) m 2 n 2 c 2
1 term(iii) a 2 b 2 c – ab 2 c 2 + a 2 bc 2 + 3abc
4 terms
(iv) 8x 2 – 4xy + 7xy 2
3 terms
Numerical co efficient in 2x 2 is 2
Numerical co efficient in -5xy is -5
Numerical co efficient in 6y 2 is 6
Numerical co efficient in 7x is 7
Numerical co efficient in -10y is -10
Numerical co-efficient in 9 is 9
(ii) \(\frac{x}{3}+\frac{2 y}{5}\) – xy + 7
Numerical co efficient in \(\frac{x}{3}\) is \(\frac{1}{3}\)
Numerical co efficient in \(\frac{2 y}{5}\) is \(\frac{2}{5}\)
Numerical co efficient in – xy is – 1
Numerical co efficient in 7 is 7
2x + 6y + 9x – 2y
= 2x + 9x + 6y – 2y
= (2 + 9) x + (6 – 2)y
= 11 x + 4 y
6 mn – (-2mn) = 6mn + (+ 2mn)
= (6 + 2)mn
= 8mn
Quantity of oil in the second tin = 3x 2 + 6x – 5 litres.
Quantity of oil added = x + 7 litres
∴ Total quantity of oil in the second tin
= (3x 2 + 6x – 5) + (x + 7)litres
= 3x 2 + (6x + x) + (-5 + 7) = 3x 4 + (6 + 1)x + 2
= 3x 2 + 7x + 2litres
Quantity of oil sold = x 2 + 6 litres
∴ Quantity of oil left in the second tin
= (3x 2 + 7x + 2) – (x 2 + 6) = (3x 2 – x 2 ) + 7x + (2 – 6)
= (3 – 1)x 2 + 7x + (-4) = 2x 2 + 7x – 4
Quantity of oil left = 2x 2 + 7x – 4 litres
Think (Text Book Page No. 77)
No, This statement is not true. Because Polynomials contain only whole numbers as the powers of their variables. But an algebraic expression may contains fractions and negative powers on their variables.
Eg. 2y 2 + 5y -1 – 3 is a an algebraic expression. But not a polynomial.Try These (Text Book Page No. 78)
Find the product of
(i) 3ab 2 , – 2a 2 b 3
(3ab 2 ) × (- 2a 2 b 3 ) = (+) × (-) × (3 × 2) × (a × a 2 ) × (b 2 × b 3 )
= – 6a 3 b 5
(ii) 4xy, 5y 2 x, (-x 2 )
(4xy) × (5y 2 x) × (-x 2 ) = (+) × (+) × (-) × (4 × 5 × 1) × (x × x × x 2 ) × (y × y 2 )
= -20x 4 y 3
(iii) 2m, – 5n, – 3p
(2m) × (-5n) × (-3p) = (+) × (-) × (-) × (2 × 5 × 3) × m × n × p
= + 30 mnp
= 30 mnpThink (Text Book Page No. 79)
why 3 + (4x – 7y) ≠ 12 x – 21 y ?
Addition and multiplication are different 3 + (4x – 7y) = 3 + 4x – 7y
We can add only like terms.
Try These (Text Book Page No. 79)
Side of a square = x – 2
∴ Area = Side × Side
= (x – 2)(x – 2) = x(x – 2) – 2 (x – 2)
= x(x) + (x) (-2) + (-2)(x) + (-2) (-2)
= x 2 – 2x – 2x + 4 .
= x 2 – 4x + 4 units square
Length of the rectangle = y + 4
breadth of the rectangle = y – 3
Area of the rectangle = length x breadth
= (y + 4)(y – 3) = y 2 + (4 + (-3))y + (4)(-3)
= y 2 + y – 12Try These (Text Book Page No. 91)
Expand :
(i) (x + 5) 3
Comparing (x + 5) 3 with (a + b) 3 , we have a = x and b = 4.
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
(x + 5) 3 = x 3 + 3x 2 (5) + 3(x)(5) 2 + 5 3
= x 3 + 15x 2 + 75x + 125
(ii) (y – 2) 3
Comparing (y – 2) 3 with (a – b) 3 we have a = y b = z
(a – b) 3 = a 3 – 3a 2 b + 3ab 2 – b 3
(y – 2) 2 = y 3 – 3y 2 (2) + 3y(2) 2 + 2 3
= y 3 – 6y 2 + 12y + 8
(iii) (x + 1)(x + 4)(x + 6)
Comparing (x + 1)(x + 4)(x + 6) with (x + a)(x + b)(x + c) we have
a = 1 b = 4 and c = 6
(x + a)(x + b)(x + c) = x 3 + (a + b + c)x 2 + (ab + bc + ca)x + abc
= x 3 + (1 + 4 + 6)x 2 + (1) (4) + (4) (6) + (6) (1)x + (1) (4) (6)
= x 3 + 11x 2 + (4 + 24 + 6)x + 24
= x 3 + 11x 3 + 34x + 24Try These (Text Book Page No. 94)
Find the factorsThink (Text Book Page No. 94)
x 2 – 4(x – 2) = (x 2 – 4)(x – 2) Is this correct? If not correct it.
(3a) 2 = 3 2 a 2 = 9a 2
x 2 – 4 (x – 2) = x 2 – 4x + 8Try These (Text Book Page No. 95)
Convert to linear equations:
Given that on subtracting 8 from product of 5 and a, we get 32
∴ 5 × x – 8 = 32
∴ 5x – 8 = 32
Sum of 3 consecutive integers is 78
Let integer be bx
∴ x + (x + 1) + (x + 2) = 78
∴ x + x + 1 + x + 2 = 78
∴ 3x + 3 = 78
Let cost of one book be ‘x’
∴ Given that 200 – 7 × x = 60
∴ 200 – 7x = 60
Let base angles each be equal to x & vertex bottom angle is 80°. Applying triangle property, sum of all angles is 180°
∴ x + x + 80 = 180°
∴ 2x + 80 = 180°
Let ∠B = b
Given ∠A = 10° + ∠B = 10 + b
Also given that ∠C = 3 × ∠A = 3 × (10 + b) = 30 + 3b
Sum of the angles = 180°
∠A + ∠B + ∠C = 180°
10 + b + b + 30 + 3b = 180°
∴ 5b + 40 = 180°Think (Text Book Page No. 101)
Can you get more than one solution for a linear equation?
Yes, we can get. Consider the below line or equation.
x + y = 5
here,when x = 1, y = 4
when x = 2, y = 3
x = 3, y = 2
x = 4, y = 1
Hence, we get multiple solutions for the saine linear equation.
Try These (Text Book Page No. 101)
Identify which among the following are linear equations.
(i) 2 + x = 10
2 + x = 10
⇒ x = \(\frac{10}{2}\) = 5
(ii) 3 + x = 5
3 + x ⇒ 5
x = 5 – 3 = 2
(iii) x – 6 = 10
x – 6 = 10
x = 10 + 6 = 16
(iv) 3x + 5 = 2
⇒ 3x + 5 = 2
3x = 2 – 5 = -3(v) \(\frac{2 x}{7}\) = 3
⇒ 2x = 3 × 7 = 21
x = \(\frac{2 1}{2}\)
(vi) – 2 = 4m – 6
⇒ -2x = 4m – 6
– 2 + 6 = 4m
4 = 4m
m = \(\frac{4}{4}\) = 1
(vii) 4(3x – 1) = 80
⇒ 4(3x – 1) = 80
12x – 4 = 80
12x = 80 + 4 = 84
x = \(\frac{84}{12}\) = 7
(viii) 3x – 8 = 7 – 2x
⇒ 3x – 8 = 7 – 2x
3x + 2x = 7 +8 = 15
5x = 15
x = \(\frac{15}{5}\) = 3
(ix) 7 – y = 3(5 – y)
⇒ 7 – y = 3(5 – y)
7 – y = 15 – 3y
3y – y = 15 – 7
2y = 8
y = \(\frac{8}{2}\) = 4
(x) 4(1 – 2y) – 2(3 – y) = 0
⇒ 4(1 – 2y) – 2(3 – y) = 0
4 – 8y – ó – 2y = 0
– 2 – 6y = 0
6y = -2
y = \(\frac{-2}{6}=\frac{-1}{3}\)Think (Text Book Page No. 102)
Not be any change in the solution
When an equation is multiplied or divided by 2 different numbers on either side, there will be a change in the equation & accordingly, solution will also change.
Think (Text Book Page No. 104)
Suppose we take the second piece to be x and the first piece to be (200 – x), how will the steps vary ? Will the answer be different?
Let 2 nd piece be ‘x’ & 1 st piece is 200 – x
Given that 1st piece is 40 cm smaller than hence the other piece
∴ 200 – x = 2 × x – 40∴ 200 + 40 = 2x + x
240 = 3x
∴ x = \(\frac{240}{3}\) = 80
∴ 1 st piece = 200 – x = 200 – 80 = 120 cm
2 nd piece = x = 80 cm
The answer will not changeThink (Text Book Page No. 109)
If instead of (4,3), we write (3,4) and tn to mark it, will it represent ‘M’ again?
Let 3, 4 be M. when we mark, we find that it is a different point and not ‘M’
Try These (Text Book Page No. 111)