Prime Mathematics 8

Prime Mathematics Book - 8

Exercise -1.1(f)
1. (a) 5,1 (b) 10,3 (c) 6,2 (d) 15,7 (e) 13,6 (f) 16,7
2. (a) (x+13)(x−6) (b) (x−13)(x+2) (c) (y+5)y−3) (d) (a−16)(a+3) (e) (b+9)(2−b) (f) (m+15)(m−7)

(g) (p−12)(p+5) (h) (x+8)(x−7) (i) (x−9)(x+4)
3. (a) x(x−8)(x+1) (b) 4x(x−3)(x+1) (c) (p−11)(p+3) (d) (xy−13)(xy+9) (e) (a+b−7)(a+b+3)

(f) (p−2q+10)(p−2q−7) (g) m(m−9n)(m+4n)

Exercise - 1.1(g)
1. (a) (2x−3)(x−3) (b) (x+1)(3x+2) (c) (m+2)(3m+4) (d) (p−4)(7p−2) (e) (a−1)(3a−4) (f) (2y−1)(3y−7)

(g) (p−8)(2p−7) (h) (3p−2)(5p−1) (i) (2x+3)(3x+4)
2. (a) (a−b)(3a−10b) (b) (2m−3n)(2m−n) (c) (21x+4y)(x+y) (d) 3m(2m−n)(3m−7n)

(e) (2a−2b−7)(a−b−7) (f) (6a+3b+5)(4a+2b+3) (g) (m−2n−4)(7m−14n−2)
(h) (2x−y−2)(10x−5y−4)

Exercise - 1.1(h)
1. (a) (2x+7)(x−1) (b) 3(m+3)(m−1) (c) (x−3)(5x+1) (d) (2p−1)(5p+1) (e) 2(a+b)(3a−5b)

(f) (m−2n)(3m+5n) (g) (x−6)(3x+5) (h) (y+4z)(4y−3y)
2. (a) (3a+16)(a−1) (b) (3b−1)(4b+1) (c) (2x+5y)(5x−3y) (d) (x−7)(x+3) (e) y(29−y)

(f) (p+2q−2)(3p+6q+5) (g) (x−2y+4)(2x−4y−3) (h) (2x+3y+1)(10x+15y−14)

Exercise - 1.1(i) (Geometrical Concept of (a±b)3)

1. (a) x3+9x2+27x+27 (b) m3+6m2+12m+8 (c) a3−15a2+75a−125

(d) 8x3−36x2+54x−27 (e) 64−144a+108a2−27a3 (f) 8x3+36x2y+54xy2+27y3

(g) 27m3−108m2n+144mn2−64n3 (h) 1+9y+27y2+27y3 (i) 27x6+27x4y2+9x2y4+y6
(l) q3− 3 q+ 3 − 1
(j) m3−3m+ 3 − 1 (k) x3+3x+ 3 + 1
m m3 x x3 2 4q 8q3

(m) x3+8y3+a3+6x2y+12xy2+3a2x+3ax2+12ax+6a2y

(n) 27x3−8y3−z3−54x3y+36xy2−27x2z+36xyz−12y2z+9xz2−6yz2

2. (a) 27x3+54x2y+36xy2+8y3 (b) x6−3x4y+3x2y2−y3 (c) 64a3−48a2b+12ab2−b3

(d) 8a3 + 60a2 + 150a + 125 (e) 27a6 − 9a2 + 1 − 1 (f) m3 − 3m + 3n − n3
a2 27a6 n3 n m m3

3. (a) (3a+2b)3 (b) (2m−3n)3 (c) (4x+5y)3 4. (a) 144 (b) 80 5. (a) 136 (b) 335 6. 396

7. (a) (x−3)(x2+3x+9) (b) (2x+3y)(4x2−6xy+9y2) (c) (4p−6)(16p2+24p+36)

(d) (3m+5n)(9m2−15mn+25n2)

8. (a) 322 (b) 756 (c) 1692 (d) x3+3x 9. (a) 216 (b) 64 (c) 27 (d) 216

10. (a) −3yz(y+z) (b) 2y(3x2+y2) (c) 2z(3x2+z2) (d) x3+6xy2−y3 (e) 9x3−39x2y+57xy2−28y3

(f) x3+y3 (g) 7 (h) 4

296 12 Algebraic Expressions

Prime Mathematics Book - 8

Exercise - 1.2 (a) H.C.F.

1. (a) ab (b) ab (c) 3xy2 (d) abc (e) 2ax (f) 4m2np (g) a−2 (h) x+2 (i) p(p−q) (j) a−b (k) x+y
(l) x+5 (m) x+1 (n) x+3 (o) 1−n2

2. (a) x−6 (b) x+3 (c) x−y (d) a(a+b) (e) a−1 (f) 1 (g) x−3 (h) x−2 (i) m−2 (j) y+1 (k) b+2 (l) x+3

Exercise - 1.2 (b) (L.C.M.)
1. (a) 6x2y (b) 12ab (c) 3x3y3 (d) 6mn2 (e) 6p2q2 (f) 2x(x−2) (g) 10xy2 (h) 40x3y2 (i) 28a3b2
2. (a) 3(x2−1) (b) 2(2b+3)(2b−3) (c) x(x+y) (d) 2(a2−1)(a−1) (e) 5(x2−16) (f) xy(x−y)

(g) (x2−4)(x−3)(x−1) (h) (y+2)(3y+1)(2y−1) (i) (m−2n)2(m+2n) (j) (x−6)(x+8)(x−3)
3. (a) 24a2b2 (b) 30x2y2 (c) x(x−2)(x+2) (d) (a+b)2(a−b)(a2−ab+b2) (e) xy(x2−y2) (f) 2x(x2−1)

(g) (m−1)2(m+1)(m+2) (h) 2x(4−x2) (i) (a+2)2(a−2)(a+1) (j) (x−5)2(x+5)(x+4)
(k) 4(x2−4)(x−5)(x+1) (l) x2(x−7)(x+6)(x+2)(x−2) (m) (3m+5)(m−4)(2m+1)(2m−1)
(n) a(a2+a+1)(a2-a+1)

Exercise - 1.3(a) (Rational Algebraic Expressions)

1. (a) 3 (b) P (c) 5 (d) ±2 (e) −9 2. (a) 3 (b) 4 (c) y (d) 1 (e) ±4

3. (a) 4 (b) 2x (c) 2xb2 (d) 1 (e) a+b (f) −m (g) x+3 (h) x−5 (i) m
5x2 5y 7a 2 a−b n x+1 2 n−4

4. (a) a+b (b) 5(x−y) (c) 2 (d) m+3 (e) 5(y+3) (f) q2+q+1 (g) x−3 (h) − m+4 (i) y+4
a−b 7 m+3n m−3 4 −(q+1) x+2 4 y+2

(j) x−1 (k) p+4 (l) y−1
x+2 p−3 y+5

Exercise - 1.3(b)

1. (a) 10 (b) x (c) a (d) 9 (e) x (f) x+2 (g) 2 (h) −2 (i) 3 (j) 1 (k) 2y (l) 0
x x+2 m+n x−y
4x
2. (a) 2x (b) x2−3 (c) 5 (d) 1 (e) 3 (f) x+1 (g) a(a+2) (h) m (i) 5(3−m)
x−5 x+y x−3 m+3

(j) 3(p+q) (k) x−2y (l) 3x+4y

Exercise - 1.3(c)

1. (a) 6y , 5ax (b) bcx , acy , abz (c) 2bc , 3acx , 5aby (d) 4bx , 9ay
3xy 3xy abc abc abc a2b2c2 a2b2c2 a2b2c2 6ab 6ab

(e) 5(a−b) , 2(a+b) (f) 3x(x+y) , 2y(x−y) , 5z (g) 3(a2+ab+b2) , 2 , 5(a−b)
a2−b2 a2−b2 x2−y2 x2−y2 x2−y2 a3−b3 a3−b3 a3−b3

(h) x ,2a1(−14−a22a), 3b(1+2a) (i) (x+1)a((xx++23))(x+3), (x+1)b((xx++21))(x+3), 2c(x+2)
1−4a2 1−4a2 (x+1)(x+2)(x+3)

(j) (x−1)((xx−−12))2(x−3), (x−1)((xx−−22))2(x−3), (x−3)2
(x−1)(x−2)(x−3)

22 Algebraic Expressions 297

Prime Mathematics Book - 8

Exercise - 1.3(d)

1. (a) 8+3x (b) 41x (c) 3m2 (d) 13x (e) 3n−2m (f) 35x−22 (g) 3a+2b (h) 31x
12 21 4 12 mn 55 24 7

(i) m−2n (j) 24−x2 (k) 2(2a2−b2) (l) 2b−3
n 4x 3ab ab

2. (a) 3x2−4y2 (b) 6+20x (c) 5(q4−p4) (d) (x+2y)(x−2y) (e) 4(m−1) (f) 8 (g) 3x+7y
12 5x2 pq y m(2−m) a2−4 x2−y2

(h) 7y−a (i) 3x+2y (j) a2−2ab−b2 (k) p−5 (l) x2−x+6
y2−a2 x2−4y2 a2−b2 5(p−1) 3(x2−9)

3. (a) 2x+1 (b) 2(x+3) (c) 16 (d) 14x+4 (e) 3x−4 (f) − 5
x2−1 x−6 (x−5)(x−3) 6 6(x+y) (x+2)(x−3)

4. (a) 4(x2−x+6) (b) b−4 (c) x−1 (d) x−3 (e) m−4
(x2−9)(x−1) b(b−2) x(x+4) x+1 (m+2)(m−1)

(f) 5y−3 (g) − 2
(y+1) (y−3) x2−4

(h) x+5 (i) a (j) x−y (k) 5 (l) 8x3
x+1 a2−1 x+y x+y x4−1

Exercise - 1.3(e)

1. (a) 3xy (b) 3a2 (c) 21x4z (d) 3x (e) ay
10ab b2 2y 2yz 9b3

2. (a) x (b) b2c(a−b) (c) y(x−y) (d) a−2 (e) 1 (f) 3a2
4y a(a2−ab+b2) x a−3 (a+1)2

3. (a) (a+b)2 (b) (x−2)(2x−1) (c) a−3 (d) a(a−2b) (e) x+y (f) a−2 (g) b(x+2) (h) a−3
(a−b)2 a−2 x(a+3x) x−y 2y c(x−2) a

Exercise - 1.3(f)

1. (a) ab2 (b) 5a (c) 3xy (d) x (e) x2y 2. (a) b(x+y) (b) a−1 (c) 1 (d) x(3y+4z)
2 8x 2ab 4y m a b x+4

(e) y2−x2 3. (a) a−2 (b) x−4 (c) x−3 (d) 2a−1 (e) x−5 (f) x−1
xy a−3 x(x−1) 3 a−2 x+4 x+1

( (4. (a) 4 (b)x2+1 4ab 2
x+1 (c) 1 (d) a2−b2

298 12 Algebraic Expressions

At the end of this unit the student will be able to Estimated periods -8
● know about the index (power) of base. Objectives:
● know about the laws of indices.
● know to simplify by using the law of indices. Index

5x4

Coefficient Base

xm × xn = xm + n

xm ÷ xn = xm - n

x0 = 1

Teaching Materials:

Chart of law of indices, tiles, local materials, etc.

Activities:

It is better to
● discuss about the coefficient, base and index of algebraic expressions.
● discuss about the laws of indices.
● use of laws of indices to simplify the algebraic expressions.

Prime Mathematics Book - 8 Indices Estimated period : 8

Unit - 1

Review Index

Let us see the following example and discuss: 4x7

x×x×x×x×x=?

Here, x is multiplied itself 5 times. So, we can Coefficient Base
write x × x × x × x × x = x5

In x5, 5 is called index of x and x is called the base.

Let us take an algebraic expression 4x7.

In 4x7, 4 is called coefficient of x7, x is called the base and 7 is called the index. Index
is also known as power or exponent.

If y is multiplied n times, then we can express it y×y×y×y×y×...... n times = yn

1.1 Laws of Indices

Law I: xm × xn = xm + n
e.g., x0 - x3-3 = (x×x)×(x×x×x×x) = x×x×x×x×x×x = x6 = x2 + 4
Hence, we add the indices when the algebraic terms with same base are

multiplied together. If x ≠ 0 and m and n are integers then xm × xn = x m + n.

Law II: xm ÷ xn = xm - n
x5
e.g., x5 ÷ x2 = x2 = x×x×x×x×x = x × x × x = x3 = x5 - 2
x×x

Hence, we subtract the index of divisor from the index of dividend when an
algebraic term is divided by another with same base.

If x ≠ 0, m > n and m & n both are integers then xm ÷ xn = xm - n.

Law III: Zero index x3 x×x×x
x3 x×x×x
e.g., x0 = x3-3 = x3 ÷ x3 = = =1

∴ x0 = 1, where x ≠ 0.

If the index of any nonzero base is zero, then the value is 1.

Law IV : Negative index

e.g., x2 ÷ x5 = x2 - 5 = x-3 ----------- (1)

x2 x×x
Again, x2 ÷ x5 = x5 = x×x×x×x×x (By law of multiplication)

11
= x×x×x = x3 ----------- (2)
From (1) and (2), we get x-3 = 1
1 x3

If x ≠ 0 and x-m, then x-m = xm .

300 Indices

Prime Mathematics Book - 8

Law V : (xm)n = xm × n = xmn = (xn)m
e.g., (x3)3 = x3 × x3 × x3 = x3 + 3 + 3 = x9 = x3 × 3
Similarly, (a2)5 = a2 × 5 = a10
If x ≠ 0 and m and n both are integers, then (xm)n = xm × n = xmn

Law VI : (xp yp)r = xpr yqr

Here, (x2y3)2 = x2y3 × x2y3 = x2 × x2 × y3 × y3 = x2 + 2y3 + 3 = x4y6 = x2 × 2y3 × 2

∴ (2a2b3)3 = 23(a2)3 (b3)3 = 8a6b9

If m, n ≠ 0, then (mn)p=mpnp

Again, x2 3 x2 x2 x2 x2 ×x2× x2
y3 y3 y3 y3 y3 ×y3 ×y3
= × × = = x2 +2+2 x6 x2 × 3
y3+3+3 = x9 = x3 × 3

m p mp
If m, n ≠ 0, then n =.
np

1

Law VII: Fractional index and radical sign. We take x n as the nth root of x.

1

n x = x n

x = x1 = (xn) 1 = n xn
n

15

Also, x5 = ( x5 ) 2 = x 2

∴ p xn = (xn) 1 n
p
= xp

Example 1: Simplify by using the law of indices.
(a) x3 × x2 × x5 (b) 52 × 55 × 5-3

Solution: (a) x3 × x2 × x5 (b) 52 × 55 × 5-3

= x3 + 2 + 5 = 52 + 5- 3

= x10 = 54 = 625

Example 2: Simplify by using the law of indices.

(a) y7 ÷ y2 (b) 27x5 ÷ 9x3 (c) ax-1 ÷ a2x-4

Solution: (a) y7 ÷ y2 (b) 27x5 ÷ 9x3 (c) ax-1 ÷ a2x-4

27x5 = a(x-1) - (2x-4)
= y7-2 = 9x3 = ax-1-2x+4

= y5 = 3x5-3

= 3x2 = a-x+3 = a3-x

Indices 301

Prime Mathematics Book - 8 (b) (2x2y3)3
(b) (2x2y3)3
Example 3: Simplify: (a) (a2b)5
Solution: (a) (a2b)5 = (2)3(x2)3(y3)3
= (a2)5(b)5
= 8x6y9
= a10b5

a2b 3 (b) px-y × py-z × pz-x
Example 4: Simplify: (a) a3b2

Solution: (a) 3

a2b
(b) px-y × py-z × pz-x

a3b2

= px - y + y - z + z - x [ xm × xn = xm + n ]∴
= p0 = 1.
(a2b)3 x n xn
= (a3b2)3 y = yn
∴
(a2)3(b)3
= (a3)3(b2)3 ∴[ (ab)m = am bm ]
∴
a6b3 ∴[ (xm)n = xmn ]
= a9b6 ∴

∴ xm
= a6-9 b3-6 [ xn = xm-n ]

= a-3b-3

11 1
= a3 . b3 [ x-n = ]
xn

1
= a3b3

Example 5: Simplify: 5 a4 ÷­ 5 a-1

Solution: Here,

= 5 a4 ­÷ 5 a-1 11 n p = (p) 1 ]
n
= (a4)5 ÷ (a-1)5 [

= a 4 ÷ a- 1 = (a) 4 + 1 = 4+1 = (a) 5 = a.
5 5 5 5 5
(a) 5

Exercise 1.1

1. Simplify by using the law of indices :

(a) a2 × a7 (b) a6 × a-9 (c) x3 × x4 × x-2

(d) 83 × 8-2 × 85 (e) (a3b) × (ab) × (a2b2) (f) 3y2 × (-2y3)

302 13 Indices

2. Simplify by using the law of indices : Prime Mathematics Book - 8

(a) y7 ÷ y3 (b) 15 x 5 ÷ 5 x 2 (c) - 30m6 ÷ 5m-2

(d) -125x7 ÷ (-25x-3) (e) qm - 2 ÷ q2m - 5

3. Simplify by using the law of indices:

(a) (x3)4 (b) (5a2)4 (c) (x3y2)3 (d) (-7q2)4
(h) (4m3)3 × (3m4)4
(e) (m2n3p-2)3 2x2 3 (g) (xy2)3 × xy
(f)

3y

4. Simplify: (b) (3x2y)4 (c) (-2x3y3)2 (x3y2)-2
(a) (3a)2 × (2a)3
(e) 3qn-1 pm × (2q2n+1 p-2)
(d) 2m2a × (m2n2)-2

5. Evaluate: (b) 3 x3y6z-9 (c) 3 xy2 × 3 x2y (d) ab-1 × ab-3
(a) 3 m6
8 - 2 23×42
(e) 3 27-1 3 (i) 27
44×55
(f) 27 (g) 4 (81)-3
(j) 252×162
(k) 26+n × 42n-3
24n+1 × 23n-1

6. Simplify: (b) 3x+2 - 3x+1
(a) 5m+3 - 5m 3x.6

5m.31 4x+2 - 48.4x-1
7x+1 - 7x (d) 4n.8
(c)
(b) (am-n)m+n×(an-p)n+p÷(am-p)m+p
7x.3

7x+2 - 28 × 7x-1
(e) 7x × 45

7. Simplify by using the law of indices:
(a) (xa-b)c×(xb-c)a×(xc-a)b

xa a2+ab+b2 xb b2+bc+c2 xc c2+ca+a2 xm+n+2 × xm+n+2
(c) xb × × (d) xm+n
xc xa

Indices 303

Prime Mathematics Book - 8

(e) xa-b+1 × xb-c+1 × xc-a+1 (f) (xa+b)2.(xb+c)2.(xc+a)2
x2 (xa.xb. xc)6

(g) xp+q×xq+r ×xr+p h) �xxba�4 × �xxbc�4× �xxac�4
x2p+2q+2r 11 x y 11 y z 11 z x
j)
(i) �aayx�1�xy. �aayz�1�yz. �aaxz�1�zx �x x1 1 �� .. �x y 1 1 � � .. �x x1 1 � �
x y x z x x

(k) � × � × �


8. (a) If x = 2, y = 3 and w = 3, what will be the value of xz+w yz-w ?

xz-w yz+w

(b) If a = 2, b = 1 and c = 3, what will be the value of 4a0b4c-2 ?
axy × bx+y × cx-y

(c) If a = 4, b = 5, c = 2, x = 2 and y = 3, what will be the value of ax+y × bx-y ×cxy ?

304 13 Indices

Prime Mathematics Book - 8

Unit Revision Test (Indices)

Indices 1. Simplify by using the law of indices:

(a) x2 × x7 (b) (m3b) × (mb) × (m2b2) (c) z7 ÷ z3
3

(d) -125x7 ÷ (-25x - 3) (e) (m2n3p - 2)3 (f) 2a2
3b

(g) 3an - 1bm × (2a2n + 1 b - 2)

2. Evaluate : (b) 3 mn2 × 3 m2n c) 4 (16)-3
(a) 3 a3b6c-9


44×55

(d) 252×162

3. (a) If m = 2, n = 1 and p = 3, what will be the value of 4m0 n4 p-2 ?
axy × bx+y × cx-y

(b) If a = 2, b = 3, c = 1, x = 1 and y = 2, what will be the value of ax+y × bx-y ×cxy ?

4. Simplify: na a2+ab+b2 nb b2+bc+c2 nc c2+ca+a2
(a) (ya - b)c × (yb - c)a × (yc - a)b (b) nb × ×
nc na

Answers

Exercise - 1.1 Indices

1. (a) a9 (b) 1 (c) x5 (d) 86 (e) a6b4 (f) −6y5 2. (a) y4 (b) 3x3 (c) −6m8 (d) 5x10 (e) q3−m
3. (a) x12 (b) a3 x9y6 (d) 2401q8
625a8 (c) (e) m6n9 (f) 8x6 (g) x4y7 (h) 5184m16n9
p6 27y3

4. (a) 72a5 (b) 81x8y4 (c) 4y2 (d) 2m2a−4 (e) 6q3npm−2
n4
5. (a) m2 (b) xy2 (c) xy (d) a (e) 1 (f) 9 (g) 1 (h) 128 (i) 5 (j) 1
z3 b2 3 4 27 27 22n

6. (a) 1 (b) 1 (c) 1 (d) 1 (e) 1

7. (a) 1 (b) 1 (c) 1 (d) xm+n+4 (e) x (f) 1 (g) 1 (h) 1 (i) 1 (j) 1 (k) 1 8. (a) 256 (b) 4 (c) 15625
32
6561 9

Indices / Answers 305

At the end of this unit the student will be able to Estimated periods - 12
Objectives:
● solve linear equation in one variable.
● translate the word problem into linear equation in one variable and solve it.
● represent inequality in a number line.
● translate the word problem into inequality form and solve.
● solve the linear equations of two variables by graphical method.
● know about quadratic equation.
● solve the quadratic equation by factor method.

y 3x - 4y = 5
(1, 5) (7, 4)

5

4

3 (2, 3)

2

1 (3, 1)

x’ 123456 789 x
-2 -1 0
-1 (4, -1)
2x + y = 7
-2

(-1, -2)

Teaching Materials: y'

Chart of algebraic equation and inequality, tiles, local materials, graph paper, etc.

Activities:

It is better to
● discuss the way of solving the equation of one variable.
● discuss the way of representing an inequality of one variable in a number line.
● discuss about to draw the graph of linear equation of two variables.
● discuss about the solution of linear equations of two variables by graphical method.
● discuss about the solution of quadratic equation of one variables.

Unit - 1 Prime Mathematics Book - 8

Equation, Inequalities & Graph Estimated period : 12

Review:

In the previous classes we have already discussed about the equation. Let us revise it in
brief by using some examples. What is the number in which 5 is added, the sum is 14? In
this sentence, the number is unknown. So, we suppose the number be x. Then the given
sentence can be written mathematically as: x + 5 = 14.

This mathematical sentence is true by putting the value of x = 9 otherwise it is false
by putting other values of x. Thus, an open mathematical sentence which contains an
equal sign (=) is called an equation.

The above equation x + 5 = 14 is true only the value of x is 9. So, 9 is called the solution
or a root of the equation x + 5 = 14.

1.1 Linear equation of one variable:

We have already discussed about the linear equation of one variable in the previous
classes too. Try to answer of the following questions.

What is linear equation ?
What is a variable ?
What is the degree of x in x + 5 = 14 ?
In x + 5 = 14, x is a variable. The power of x is 1. So, the degree of x is 1.
If the power of the variable in an equation is 1, then the equation is called a linear

equation. It is also called the first degree equation. The linear equation which contain
only one variable is called linear equation of one variable. x+5=14 is an example of it.
You have the idea of the process for solving the linear equation of one variable. Here,
again we discuss the solution of the linear equation of one variable in the following
examples.

Example 1: Solve and check the answer:

(a) 4x - 5 = 23 (b) 5x - 3 = 2x + 18

Solution: (a) Here, 4x - 5 = 23 to cheek the answer:

or, 4x = 23 + 5 [transpose -5 from left to right side]

or, 4 x - 5 = 23
or, 4x = 28
or, 4 × 7 - 5 = 23
or, 28 - 5 = 23

28
or, x= 4 [dividing both sides by 4]

\ x = 7 or, 23 = 23 [ which is true]

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(b) Here, 5x - 3 = 2x + 18

or, 5x - 2x = 18 + 3[transpose 2x from right to left side and -3 from left to right side]

or, 3x = 21
to cheek the answer:

21 or, 5x - 3 = 2x +18
or, x = 3 [dividing both sides by 3] or, 5 × 7 -3 = 2 × 7 + 18
or, 35 - 3 = 14 + 18
or, x = 7
or, 32 = 32 [which is true]

x +2 3
Example 2: Solve: (a) x-2 = 4 (b) x + 33% of x = 266

Solution:

x+2 3
(a) Here, x-2 = 4

or, 4(x + 2) = 3(x - 2) [by cross multiplication]

or, 4x + 8 = 3x - 6 or, 4x - 3x = - 6 - 8 or, x = - 14

(b) Here, x + 33% of x = 266 or, 133x = 266
100
or, x + 33 × x = 266
100 or, 133x = 266×100

33x or, x = 266×100
or, x + 100 = 266 133

or, 100x + 33x = 266 or, x = 200
100


Example 3: Solve: 3y - 1 2y - 3 y - 16
4 − 5=
10
Solution: Here, 3y-1 − 2y-3 y-16
4 5= 10
or, 10(7y + 7) = 20(y + 16)
5(3y-1) - 4(2y-3) y-16
or, 20 = 10 or, 70y + 70 = 20y + 320

15y-5 - 8y+12 y-16 or, 70y - 20y = 320 - 70

or, 20 = 10 or, 50y = 250

7y+7 y-16 or, y = 250
50
or, 20 = 10
∴y=5

308 14 Equation, Inequalities & Graph

Prime Mathematics Book - 8

Exercise 1.1

1. Solve and check the answer:

(a) x + 31 = - 5 (b) 9x - 19 = 8 (c) 2y - 2 = y + 4
(d) 8y + 9 = 10 (f) 3y - 13 = 8
(e) x + 1 = 6
7 (c) 4x - 10 = 3x - 7
(f) 15(y - 2) - 4(y + 2) = 6
2. Solve:

(a) 5x - 6 = 3x + 10 (b) y + 4 = 13 - 2y

(d) 17 + 20y = 5 + 8y (e) 8x - 3(x - 2) = 36

(g) (x + 3) (x + 2) = x2 +3x +20

3. Solve and check your answer.

(a) 2 = 1 (b) 5-4x = 2 (c) 4x-3 = 5
2x-3 3x-4 3-4x 7 2x+1 2

(d) 3x+8 = x+4 (e) 2x+3 =5 y-12 = 6
7 2 7 2 (f) y-10 5

(g) 4+3x = 1 (h) x-4 =3 (i) 3 - x = x+4
5+4x 2 x-2 7 9

4. Solve:

(a) x + x = x + 1 x+1 x+2 =5
2 (b) 3 + 5

(c) 2m + m =3 (d) 2y 6- 3y
m-2 m+2 x-1 3y+2 = y-2

(e) 2x+3 − 2x-2 = 12 (f) 2x - 3x = -1
2 3 x+1 x-1

5. Solve: b) 15% of y = 12 (c) x + 11% of x = 333 (d) y + 15% of y = 92
(a) 5% of x = 15

1.2 Word Problems leading to linear Equations with one variable:

To solve the given word problem, we first write the given problem in a mathematical
open sentence as an equation. For the equation, we consider the unknown quantity as a
variable like x, y, z, ..... . Then find the value of the variable by solving the equation as the
same process as in the previous exercise. Let us discuss the solution of the word problem of
one variable in the following examples.

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Example 4: If three-fifth of a number increased by two is equal to 17, find the number.

Solution: let the number be x. or, 3x = 17 - 2
5
Then, its three-fifth = 3 of x = 3x .
5 5 or, 3x = 15
5
3x is increased by two.
5 or, 3x = 15 × 5

So, it becomes 3x + 2 or, x = 15 × 5
5 3

Now, ∴ x = 25

3x + 2 = 17
5

Hence, the required number is 25.

Example 5: The ratio of two numbers is 4:5. If the sum of two numbers is 981, find the

numbers.

Solution: Let a number be x.
According to question, the sum of two numbers is 981. So, the other number

is 981 - x
Now,

x : (981 - x) = 4:5

or, x = 4
981-x 5

or, 5x = 4(981 - x)

or, 5x = 3924 - 4x or, 5x+4x = 3924 or, 9x = 3924 or, x = 3924 ∴ x = 436
Hence, one number = x = 436 9

Other number = 981 - x = 981 - 436 = 545

Example 6: Divide 48 into two parts, so that one part may be three-fifth of the other.

Solution: let one part of 48 be x. Then other part of 48 is 48-x.

According to question

or, x = 3 of (48-x)
5

or, x = 3(48 - x)
5
144
or, 5x = 144 - 3x or, 5x + 3x = 144 or, 8x = 144 or, x = 8 ∴ x = 18

Hence, two parts of 48 are 18 and 48-18 = 30.

310 14 Equation, Inequalities & Graph

Prime Mathematics Book - 8

Example 7: Ten years ago the age of the father was three times the age of the son. If the
difference of their ages was 20, find the present of the son.

Solution: Let the present of the son be x years.

Ten years ago, the age of the son was (x - 10) years.

Ten years ago, the age of the father = 3 times the age of the son

= 3(x - 10) = 3x - 30
According to question,

(3x - 30) - (x - 10) = 20

or, 3x - 30 - x + 10 = 20

or, 3x - x - 20 = 20

or, 2x - 20 = 20

or, 2x = 20 + 20 or, 2x = 40 or, x = 40 ∴ x = 20
2

Hence, the present age of the son is 20 years.

Exercise 1.2

1. The sum and difference of two numbers are 65 and 9 respectively. Find the numbers.

2. If one fourth of a number increased by three is equal to 5, find the number.

3. The sum of two numbers is 30. If one number is more than other by 6, find the
numbers.

4. The difference of 3 of a number and 1 of the same number is 7. Find the number.
4 6

5. The ratio of two numbers is 5:7. If the sum of two numbers is 996, find the numbers.

6. The length of a rectangle is 8cm more than its breadth. If the perimeter of the
rectangle is 56cm, find its area.

7. The number of girl students of class '8' in a school is more than the number of boys
by 7. If the total students in the class is 39, find the numbers of boys and girls.

8. Divide 75 into two parts, So that three times one part may be double of the other.

9. Five years ago the age of the father was two times the age of his son. If the sum of
their ages was 45, find their present ages.

10. A father is four times as old as his daughter. In 16 years he will only be twice
as old, find their ages.

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1.3 Liner Inequality with single variables:

Revision:

As open mathematical sentences line x ≤ 3 or 2x + 1≥ 5 is called an inequality. You
have already studied about the inequality in the previous classes. The process to solve
inequalities of one variable have already discussed in the classes 6 and 7. The following
basic rules are used to solve the inequalities.

(i) The same number or expression can be added to or subtracted from both sides of
an inequality. For example; if x < a, then x + b < a + b or x − b < a − b.

(ii) Both sides of an inequality can be multiplied or divided by the same positive

number. For example; if x > a, then x × b > a × b or x > a .
b b

(iii) If both sides of an inequality is multiplied or divided by the same negative number,
then the sign of the inequality is reversed.

For example; if x < a, then x × − b > a × − b or x > a .
-b -b

(iv) If b < x + 2, then x + 2 > b.

(v) If x − b > a, then x > a + b.

Study the solution of the inequalities in the following examples.

Example 8: Solve the inequalities and show the solution in the number line.

(a) 5x−2 > 8 (b) 7x−2(x−3) ≤ 16

Solution : (a) Here,

5x − 2 > 8 or, 5x > 10 ∴x>2
or, 5x − 2 + 2 > 8 + 2 or, 5x > 10 5 5

The solution in the set = { 3, 4, 5, ....} The solution in the number line is



-6 -5 -4 -3 -2 -1 01 2
-7 or, 5x + 6 − 6 ≤ 16 - 6 or, 5x ≤ 10 3

(b) Here,
7x − 2(x − 3) ≤ 16

or, 7x − 2x + 6 ≤ 16 or, 5x ≤ 10
∴ x ≤ 2 5 5

The solution in the set = {2, 1, 0, -1, ...} The solution in the number line is



-7 -6 -5 -4 -3 -2 -1 0 1 2 3

312 14 Equation, Inequalities & Graph

Prime Mathematics Book - 8

Example 9: Solve the given inequality and show the solution in the set form and in the
number line. −7 < 2x + 5 ≤ 1

Solution: Here, −7 < 2x + 5 ≤ 1

or, −7−5 < 2x + 5 − 5 ≤ 1 − 5

or, −12 < 2x ≤ − 4

or, −12 < 2x ≤ −4 or, −6 < x ≤ − 2
2 2 2

From the above inequality, we get

−6 < x means x > −6

∴ The solution in the set form is {−5, −4, −3, −2, −1, 0, 1, 2, .....}

2nd part is x ≤ −2

∴ The solution set is {−2, −3, −4, .....} The solution in the number line is



-7 -6 -5 -4 -3 -2 -1 0 1 2 3

Example 10: Mohan wants to buy some copies at Rs.20 each and 5 pens at Rs.30 each.
How many copies can he buy if he has Rs.270 with him?

Solution: Let the number of copies can he buy be x.

Then, the cost of x copies and 5pens = Rs.20x + Rs.30 × 5

= Rs.(20x +150)

Since the total cost should not exceed Rs.270, because mohan has Rs.270 only
So, 20x + 150 ≤ 270

or, 20x +150 −150 ≤ 270 −150

or, 20x ≤ 120 or, x ≤ 120 ∴ x≤6
20

Hence, Mohan can buy maximum 6copies.

Example 11: If the value of x is 2 or more than 2 in y = 7x − 9, what will the value of y ?

Solution: Here,

y = 7x − 9 and x ≥ 2

Then, y ≥ 7×2 − 9 or, y ≥ 14 − 9 ∴y≥5

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Prime Mathematics Book - 8

Exercise 1.3

1. Solve the following inequalities and show in number line.

(a) 3x + 2 > 8 (b) 3y − 2 < 2y (c) 5 + 4x ≤ 29
(d) 5x + 2(3x − 10) ≤ x (e) 4x + 3 ≥ 23 (f) 3y + 2(2y − 9) ≤ y

2. Solve the following inequalities and show in number line.

(a) 3x − 1 > 2x (b) x −3 ≥ 3− x (c) x−2 < 3x−1
4 2 3 2 2 3 4

(d) x+4 −x ≥ x − 4 (e) 0.3 x ≥ 0.8 − 0.1x
6 2

3. Write the solution in set form and show in number line.

(a) −5 ≤ x < 2 (b) 4 ≤ x < 9 (c) −11 ≤ 3x − 2 ≤ −5

(d) −2 ≥ x − 1 < 5 (e) x+4 − x >x−4 − 142 >x −4
6 3 10

4. What will be the value of y in y = 4x + 5 if x ≥ 2 ?

5. What is the value of y in y=7x−3 if x ≤ 2 ?

6. If x < 3,what is the value of y in x + 9y − 18 = 0 ?

7. For the equation 7x + 5y + 9 = 0, (b) What is the value of x if y > 8 ?
(a) What is the value of y if x ≤ 3 ?
(c) What is the value of y if x ≥ −7 ?

8. Mina wants to buy a pen for Rs.50 and some copies at Rs.12 each. If she has Rs.150,
find the maximum number of copies she can buy.

9. If 13 is subtracted from four times a number, the difference becomes greater than or
equal to 3. Write down an inequality of this statement and solve it. Also show your
solution in the number line.

10. A man ordered a packet of biscuits which cost Rs.42 and some cups of tea worth
Rs.15 each. If he had Rs.150 only, how many cups of tea at most did he order?

11. If 8 is added to three times a number, the result is less than 29. What is that number?

12. The length and breadth of the given rectangle A (2x+7) units D
are (2x + 7) units and (x + 1) units respectively.
If the perimeter of the rectangle is more than (x+1) units
70 units, write down an inequality of this
statement and solve it.

BC

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Prime Mathematics Book - 8

1.4 Graphic solution of two variabled linear Equations

We have already discussed about the graph of linear equation with one variable and two
variables in class7 (VII). The graph of the linear equation is always a straight line.

In this grade (class 8), we are discussing the solution of two variables linear equations
by using graphical method. Let us discuss about that in the following example.

Example 11: Mohan and Sohan want to distribute 7 pens among them. If Mohan gets 3

pens more than Sohan, how many pens each will get ?

Solution: Here,
Let the number of pens Mohan will get be x and the number of pens Sohan will

get be y.
Then,

∴ x + y = 7...... (i) the table from above equation is x 2 5 6 0
y 5 2 1 7

The above table shows that the total number of pens Mohan and Sohan will
get is 7.

Again, if Mohan will take 3 pens more than Sohan, the difference of the
number of pens which Mohan and Sohan get is 3.

∴ x − y = 3............. (ii)
The solution of above question can be shown in the following table.

x 5 637
y 2 304

Equation (i) and equation (ii) both are linear equation with variables x and y.
So, when we draw the graph of these two equations in the same graphs, we
will get two straight lines which intersected at a point (5, 2). So, the point
(5, 2) is the solution of both equations (i) and (ii) because the point (5, 2)
satisfies both equations.

[Any two linear equations in x and y which intersect each other at a point in
the graph are called the simultaneous equation.]

Example 11: Solve graphically: 2x + y = 7; 3x − 4y = 5.
Solution: Here,

2x + y = 7..................(i)
3x − 4y = 5.................(ii)
From equation (i),
2x + y = 7
or, y = 7 − 2x

Equation, Inequalities & Graph 315

Prime Mathematics Book - 8

x 0 124 3
y 7 5 3 −1 1

The points from the above table are (0, 7), (1, 5), (2, 3), (4, −1) and (3, 1).

Plotting these points in the graph, we get a straight line l1.
Again, from equation (ii),

3x − 4y = 5

or, x = 5 + 4y
3

x 3 −1 7 −5
y 1 −2 4 −5

The points from the above table are (3, 1), (−1, −2), (7, 4) and (−5, −5).
Plotting these points in the graph, we get a straight line l2.

y 3x - 4y = 5
(1, 5) (7, 4)

5

4

3 (2, 3)

2

1 (3, 1)

x’ 123456 789 x
-2 -1 0
-1 (4, -1)
2x + y = 7
-2

(-1, -2)

y'



In the graph, two straight lines l1 and l2of 2x + y = 7 and 3x − 4y = 5 are intersected
at (3,1).

Therefore, the required solution of the given two equations 2x + y = 7 and 3x
− 4y = 5 is x = 3 and y = 1.

Note: To plot the graph at least three points are necessary.

316 14 Equation, Inequalities & Graph

Prime Mathematics Book - 8

Example 12: The difference of two numbers is 5. Two times the greater number is 13

more than smaller number. Find the numbers by using graph.

Solution: Let the greater number be x and smaller number be y.

Then, according to question, y

x − y = 5.......(i) 9 (7, 4)
and 2x = y + 13.....(ii)
From equation(i), 8
x − y = 5
or, x = y+5 7 (0, 7)

6

(2, 5)

5

4

x 5 74 8 3 (6, 3)
y 0 2 −1 3 2 (5, 2)
1
x’ (6, 1)
-2 -1 0
The points from the above -1 (3, 0) (7, 0) x
table are (5, 0), (7, 2), (4, -2
123456 789

x−y=3 x+y=7

−1) and (8, 3). y'
From equation(ii),

2x = y + 13

y+13 x 7 86 5
or, x = 2 y 1 3 −1 −3

The points from the above table are (7, 1), (8, 3), (6, −1) and (5, −3).



The points obtained y
from two equations are
plotted in the graph. 5 (8, 3)
We get two straight
lines which intersect 4 (7, 2)
each other at (8,3). (7, 1)
∴ x=8 and y=3 are the 3
solution of the two (5, 0)
equations. 2
123456 789 x
1
(4, -1) (6, -1)
x’ x-y=5
-2 -1 0 2x = y + 13
-1

-2

Hence, greater number y' (6, -1)
is 8 and smaller

number is 3.

Equation, Inequalities & Graph 317

Prime Mathematics Book - 8

Exercise 1.4

1. Solve graphically:

(a) x + y = 4 (b) 3x + y = 11 (c) 2x + y = 5 (d) x + 5y = 9
x − y = 2 x−y=1 3y + 4x = 6 2x − 3y = 5
(e) 3x + y = 8 (f) x + y = 13
2x + y = 7 2x − y = 8 (g) 4x + 2y = 2 (h) x − y+ 2 = 0
x − 3y = 11 3x − 2y = 0

2. Write the equations of the following problems and solve graphically:

(a) The sum of two numbers is 17 and their difference is 7.
(b) The difference of two number is 3. Two times of the greater number is equal

to 3 times of the smaller number.
(c) The sum of two numbers is 12. The greater number is three times the smaller

number.
(d) The cost of 4 oranges and 3 apples is Rs.41.The cost of 2 oranges and 6 apples

is Rs.52. Find the cost of 1 orange and 1 apple.
(e) The cost of 3 pens and 4 copies is Rs.200 and the cost of 5 pens and 2 copies is

Rs.240. Find the cost of each pen and copy.
(f) Anima is 4 years elder than Adrash 2 years ago, the age of Anima was twice the

age of Adrash, find their present ages.
(g) The age of a father is 3 years more than three times the age of the daughter. If

the difference of the ages of them is 37 years, find their present ages.
Request to teacher: In order to give extra knowledge, you can teach about

elimination and substitution method.

1.5 Quadratic Equation

Let us take some equations as follows: 4x2 − 10x = 0............ (iii)
4x − 9 = 11 ............(i)

x2 − 16 = 0 ............(ii) 2x2 − 5x + 3 = 0............ (iv)

In the above all four equations, there is single variable x. In equation (i), the power
of variable x is 1. So, equation (i) is called first degree or linear equation. About such
equation, we already discuss in the previous classes. But in equations (ii), (iii) and (iv),
the highest power of variable x is 2. So, these three equations are called second degree
equation. The second degree equation is called a quadratic equation. The equations (ii),
(iii) and (iv) are the quadratic in variable x. The general form of the quadratic equation
is ax2 + bx + c = 0, where a ≠ 0. There are two types quadratic equation.

(a) Pure quadratic equation (b) Adfected quadratic equation
(a) Pure quadratic equation:

If the quadratic equation contains the terms with variable 2 degree only, it is called
a Pure quadratic equation. For examples; x2 − 25 = 0, 3x2 − 12 = 0 etc.

318 14 Equation, Inequalities & Graph

Prime Mathematics Book - 8

(b) Adfected quadratic equation:

If the quadratic equation contains the terms with variable 2 degree and one degree
also, it is called adfected quadratic equation.

For examples, 3x2 − 4x = 0, 2x2 − 7x = 0 etc.

We can solve a quadratic equation by different methods. But in this class (grade 8)
we discuss only one method that is factorization method. In this method, we use the
following property of numbers:

If a × b = 0, then either a = 0 or b = 0 or both a and b = 0.

We use the following points to solve the given quadratic equations by factor method.

(i) Remove brackets or fractions, if any.
(ii) Simplify and reduce the equation to the form ax2 + bx + c = 0.
(iii) Factorize the left side into two factors.
(iv) Use the property as given above and write each factor equal to 0.

(v) Solve both linear equations and get two values of the given variable. The values
of the variable is also called the roots of the equation.

Example 13: Solve: (a) 25x2 − 16 = 0 (b) 3x2 − 15x = 0

Solution: (a) 25x2 − 16 = 0

or, (5x)2 − (4)2 = 0

or, (5x + 4)(5x − 4) = 0 [∴ a2 − b2 = (a + b)(a − b)]

Either 5x + 4 = 0

OR, 5x − 4 = 0 [If a and b are two factors in a × b = 0, then a = 0 or b = 0]

or, 5x = − 4 or, 5x = 4

or, x = −54 or, x = 4
5
-4 4
∴ The required values of x are 5 or 5

(b) 3x2 −15x = 0

or, 3x(x − 5) = 0

Either 3x = 0 Or, x − 5 = 0

or, 3x = 0 or, x = 5
0 or, x = 5

or, x= 3
or, x = 0

∴ The required values of x is 0 or 5.

Equation, Inequalities & Graph 319

Prime Mathematics Book - 8

x+2 1 1
Example 14: Solve : (a) x2 − 12x + 35 = 0 (b) 6 − x+2 = 6
Solution: Here,

(a) x2 − 12x + 35 = 0

or, x2 − (7 + 5)x + 35 = 0 [1 × 35 = 35 so, 7 × 5 = 35; 7 + 5 = 12]

or, x2 − 7x − 5x + 35 = 0

or, x(x − 7) − 5(x − 7) = 0

or, (x − 7)(x − 5) = 0

Either x − 7 = 0 OR, x − 5 = 0

or, x = 7 or, x = 5

∴ The required values of x is 7 or 5.

(b) x + 2 −x 1 2 = 1
6 + 6

or, (x + 2)2 − 6 = 1
6 (x + 2) 6

or, x2 + 4x + 4− 6 = 1
6(x + 2) 6

or, x2 + 4x − 2 = x + 2

or, x2 + 4x − x − 2 − 2 = 0

or, x2 + 3x − 4 = 0

or, x2 + (4 − 1)x − 4 = 0

or, x2 + 4x − 1x − 4 = 0

or, x(x + 4) − 1(x + 4) = 0

or, (x − 1)(x + 4) = 0

Either x − 1 = 0 OR, x + 4 = 0

or, x = 1 or, x = −4

∴ The required values of x is 1 or −4.

Exercise 1.5

1. Solve: (b) y2 − 49 = 0 (c) x2 − 4 = 21 (d) 25x2 = 1
(a) x2 − 9 = 0 (g) 3x + 9x2 = 0 (h) 121x2 = 100
(e) 2x2 − 72 = 0 (f) 2x2 − x = 0
(k) 3y2 = 4y (l) 7x2 − 5x = 0
(i) 4y2 − 9y = 0 (j) x2 − 9 = 0
4

320 14 Equation, Inequalities & Graph

2. Solve: (b) x2 − x − 2 = 0 Prime Mathematics Book - 8
(a) x2 − 2x + 1 = 0 (e) y2 − 34 + 15y = 0
(d) 3x2 − 2x − 1 = 0 (h) 7a2 + 13a − 2 = 0 (c) x2 − 10x − 24 = 0
(g) 2x2 + 11x + 12 = 0 (k) 10x2 + 19x + 6 = 0 (f) y2 − 5y − 6 = 0
(j) 2y2 + 3y − 9 = 0 (i) a2 − 8a + 15 = 0
(l) 3z2 − 3z = 18

3. Solve:

(a) (x + 1)2 − 9 = 0 (b) (y + 6)2 = 36 (c) (x − 3)2 = (x − 3)(2x + 1)

(d) 5y(3y − 2) = 2(3y − 2) (e) (2a − 1)2 −10a = 5a + (a − 3)2

4. Solve:

(a) x+1 = 7x − 3 (b) y+1 = 3y−7 (c) 6x − 5 +5= x
2 3x y−1 2y−5 x+5 3

(d) z+2 − 4−z 1 (e) x + 3 + 4(x − 6) = 3
z−1 2z =2 3 x − 3 x+6

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Prime Mathematics Book - 8

Area Revision Test (Equation, Inequalities & Graph)

1. Solve: (b) x+2 = 5 (c) 4x - 3 (x - 1) = 15
(a) 5x + 7 = 27 5

2. Solve : (b) 2a
(a) 2x a+1 3a = -1
x-2 + x =3 - a-1
x+2

3. (a) The sum and difference of two numbers are 70 and 20 respectively. Find the
numbers.

(b) Three years ago the age of the father was 3 times the age of his son. If the sum
of their ages was 44, find their present ages.

4. Solve the following inequality and show in the number line:

(a) 4x + 2(3x-7) ≤ 3x (b) y+4 − y >y − 4 − 142 >y− 4
6 3 10

5. Solve graphically:

(a) 2x + y =7 (b) 5x - 2y = 7

x - 2y = -4 3x + y = 2

6. Solve: (b) 7x2 + 13x - 2 = 0
(a) 16x2 - 9 = 0

7. Solve:

(a) y+2 − 4−y =2 1 a+2 − 1 = 1
y−1 2y 3 (b) 6 a+2 6

322 14 Equation, Inequalities & Graph

Prime Mathematics Book - 8

Answers

1 Exercise - 1.1
8
1. (a) −36 (b) 3 (c) 6 (d) (e) 41 (f) 7 2. (a) 8 (b) 3 (c) 3 (d) −1 (e) 6 (f) 4 (g) 7

3. (a) 5 (b) 29 (c) − 11 (d) −12 (e) 16 (f) 0 (g) − 3 (h) 8 (i) 1
4 20 2 2 16

4. (a) 2 ( b) 8 (c) − 6 (d) − 2 (e) − 9 ( f) − 1
5

5. (a) 360 (b) 80 (c) 300 (d) 80

Exercise - 1.2

1. 37,28 2. 8 3. 12,18 4. 12 5. 415,581 6. 180 cm2 7. boys−16, girls−23 8. 30,45

9. 35 years, 20 years 10. 32 years, 8 years

1. (a) x > 2 x>2 Exercise - 1.3 y<2 3
(b) y<2 -1 0 1 2
-3 -2 -1 0 1 2 3

(c) x≤6
-3 -2
x≤6

-3 -2 -1 0 1 2 3 4 5 6 7
(d) x≤2 (e) x≤5

x≤2 x≤5

--22 --11 00 11 22 33 4 5 6
y≤3
(f) y≤3

-2 -1 0 1 2 3 4 x>6

2. (a) x>6

-2 -1 0 1 2 3 4 5 6 7

(b) x≥6 x>−1 (c) x>−1

x≥6

-2-2 -1-1 0 0 1 1x≤32.5 2 3 4 5 67 x≥2

(d) x≤3.5 (e) x≥2

-2 -1 0 1 2 3 4 -2 -1 0 1 2 3 4

Answers 323

Prime Mathematics Book - 8
3. (a) {−5, −4, ......., 1}

-6 -5 -4 -3 -2 -1 0 1 2 3

(b) {4, 5, 6, 7, 8} -1 0 1 2 3 4 5 6 7 8 9
(c) {−3, −2, −1} -4 -3 -2 -1 0 1 2

(d) {..., −1, 0, 1, ....., 5}

-2 -1 0 1 2 3 4 5 6

4. y≥13 5. y≤11 6. y< 5 7. (a) y≤6 (b) x>−7 (c) y≥8 8. 8 copies
3

9. 4x−13≥3, -2 -1 0 1 2 3 4 5 10. 7 cups 11. less than 7 12. x<9

1 Exercise - 1.4
2
1. (a) x = 3, y = 1 (b) x = 3, y = 2 (c) x = 4 , y = −4 (d) x = 4, y = 1 (e) x = 1, y = 5 (f) x = 7, y = 6
(g) x = 2, y = −3
(h) x = 4, y = 6

2. (a) x = 12, y = 5 (b) x = 9, y = 6 (c) x = 9, y = 3 (d) orange= Rs.5, apple = Rs.10

(e) pen = Rs.40, copy = Rs.20 (f) Anima = 8years, Adrash = 4years

(g) father = 54 years, daughter = 17 years

Exercise - 1.5

1. (a) ±3 (b) ±7 (c) ±5 (d) ± 1 (e) ±6 (f) 0, 1 (g) 1 (h) ± 10 (i) 0, 2 1 (j) ±6 1 (l) 0, 5
2. (a) 1 (b) −1, 2 (c) −2, 12 (d) − 5 (e) −17, 2 2 0, − 3 11 4 (k) 0, 1 3 7
1 ,1 (h) 1
3 (f) −1, 6 (g) − 1 1 , −4 −2, 7
2

(i) 3, 5 (j) −3, 1 1 (k) − 2 , −1 1 (l) −2, 3
2 5 2

3. (a) 2, −4 (b) 0, −12 (c) 3,−4 (d) 2 , 2 (e) −2, 4
3 5 3

4. (a) 2 , 3 (b) 3, 4 (c) −2, 30 (d) − 4 , 3 (e) 6, 12.
3 5

324 Answers

Prime Mathematics Book - 8

Model Question Set

Attempt all the questions.

Group-"A" [10 × 1 = 10] 3x°
1. a) Find the value of x x

b) If the value of π=3.14, find the circumference of the
circle having radius 4.5cm.

2. a) Find the distance between the points A(4, -7) and B(-1, 5)

b) Write the direction SE in three figure bearing.

3. a) If A = {2, 4, 6, 8, 10} and B = {4, 8, 10, 12, 14}, find A∩B.

b) Convert 19410 into quinary.
4. a) Find the median of the data: 35, 46, 68, 34, 49, 38, 56, 86 b) Factorize: 9x2 - y2

5. a) using the rules of indices, simplify: 12x7 ÷ 3x4 b) solve: 9x - 19 = 8.

Group-"B" [17 × 2 = 34] A
6. a) From the given figure, find the value of x and y. 44 0
b) Find the capacity of cistern of length 18m,
x yC
breadth 6m and height 15m. B
c) Find the cube of: x2 -y
7. a) From the given figure, find the value of x. 440
x°

560

A

4 cm (x + 2) cm
b) From the given figure, find the value of x and y. N
M 2.5 cm 3 cm
y
x

B 8 cm C

A 860 4.2 C D
cm (x + 5) cm 540

c) If ∆ ABC ~ ∆ DEF, find the value of x.

400 E 5.6 cm 860 F
B

Model Questions: 325

Prime Mathematics Book - 8

8. a) The circumference of a circle is 65cm, find its area.

b) Calculate the total surface area of the given prism. 6cm

c) What do you mean by Pythagorean triple? Check whether 8cm 18cm
the triplet 17,15,8 is a Pythagorean triplet or not.

9. a) If A={x : x is a factor of 6} and B= {x : x ∈ N, x ≤ 5} list the elements of A ∩ B.

b) Simplify: 2.5×105+6.7×106−2.3×104

c) What should be subtracted from the terms of the ratio 2:5 to make it 5:2?

10. a) If ∑fx = 2000 + 55a and N = 70 + a and X = 50, find the value of a.

b) Simplify: (xa−b)a+b×(xb−a)b+a c) Simplify: 5 a4 ÷ 5 a−1

21
11. a) Solve: 2n−3 = 3n−4

b) Solve the given inequality and show the solution set in a number liner. 3x − 1 > 2x
4 2 3

Group-"C" [14 × 4 = 56]

12. Out of 100 students 60 participated in sports and 50 participated in music. How many students
participated in both? Also find the number of students who participated only in music.

13. Factorise: 6x2 + 17x + 12

14. Find the H.C.F. of: a2 + 2a − 3, a3 − 3a + 2 and a2 − 1

2x−1 − x−2
15. Simplify: x2+4x x2+2x−8

16. Solve graphically: 3x + y = 11; x − y = 1

17. Monthly income of a family is shown in the table given below. Present the data in a pie-chart.

Source House rent Business Agriculture Pension Salary
18000
Income(Rs.) 8000 12000 7000 15000

18. Volume of a cubical box is 729cm3, find its total surface area.

19. Simplify: 80 ÷ 4[400 ÷ 4{7 + (19 + 8 − 24)}]

20. What will be the charge to be paid after allowing 3% discount and levying 13% VAT in the bill of
electricity of Rs. 1600?

326 Model Questions:

Prime Mathematics Book - 8

21. 20 men can complete a work in 24 days. In how many days can 15 men finish the work?
22. In what time will the amount on Rs. 2400 at the rate of 6% per year be Rs. 3000?
23. Reflect ∆ ABC on x-axis where co-ordinates of the ∆ ABC are A(6, −2), B(−2, −2) and C(2, 5) and

plot the ∆ ABC and image ∆ A1B1C1 on the same graph.
24. Verify experimentally that exterior angle of a triangle is equal to the sum of the interior non

adjacent angles.
25. With the given measures construct a rectangle ABCD where AB = 6.4cm and ∠ABC = 30°

Model Questions: 327

Lower Secondary Level Final Examination Prime Mathematics Book - 8

Time
Allocations
Total Marks
Total No.
Questions
Sub: Mathematics Specification Grid-2069 Class:-8

Level of Objectives Knowledge & Understanding Skills Problem 12 40 Minutes
Solving
3 5 Minutes
S.N. Areas Types of Questions Very Short Short Short Longs Longs 6 11 Minutes
5 9 Minutes
Topics No.of Marks No.of Marks No.of Marks No.of Marks No.of Marks 7 13 Minutes
7
ques. ques. ques. ques. ques.
38 Minutes
1. Line & Angles 11 12 14
7 13 Minutes
2. Triangle, Parallelogram and Polygon 12 28 29 51 Minutes
100 180 Minutes
1 GEOMETRY 3. Similarity and Congruency 24 10

4. Circle 11 12

5. Solid Shapes 12

2 CO-ORDINATE GEOMETRY 6. Co- ordinates 11 12 2

3 MENSURATION 7. Perimeter, Area and Volume 12 1 42

4 TRANSFORMATION 8. Transformation 14 2
9. Bearing and Scale Drawing
11

5 SET 10. Set 11 12 1 43 Grid

11. Whole Number

12. Integer 11 1214 3
13. Rational Number

6 ARITHMETIC 14. Real Number 12
15. Ratio, Proportion & Percentage

16. Profit & Loss 1 4 4
17. Unitary Method 1 4

18. Simple Interest 14

7 STATISTICS 19. Statistics 11 1214 3

20. Algebraic Expression 11 1 2 3 12

8 ALGEBRA 21. Indices 11 24 12

22. Equation, Inequality & Graphs 11 2414

Total 10 10 3 6 14 28 10 40 4 16 41

Question Group, No. of Question, Mark & Time allocation

Question Group No. of questions Mark for each question Total Marks Time Allocation Remarks
Group-A(Very-Short) 18 Minutes
Group-B(Short) 10 1 10 62 Minutes
Group-C(Long) 100 Minutes
17 2 34

14 4 56