# Free Algebraic Expressions and Identities Subjective Test 01 Practice Test - 8th Grade

Subtract (ap+bq-cr) from (a+b+c)(p+q+r) .  [1 MARK]

#### SOLUTION

Solution : Concept: 1 Mark

(a + b + c)(p + q + r) - (ap + bq - cr)

= (ap + aq + ar + bp + bq + br + cp + cq + cr) - (ap + bq - cr)

Now,

= ap + aq + ar + bp + bq + br + cp + cq + cr - ap - bq + cr
= aq + ar + bp + br + cp + cq + 2cr

Add the following:

a(a+b),b(b+c),c(a+c) and a2+b2+c2  [2 MARKS]

#### SOLUTION

Solution :

Concept: 1 Mark
Application: 1 Mark

a(a+b)+b(b+c)+c(a+c)+a2+b2+c2

a2+ab+b2+bc+c2+ca+a2+b2+c2

2a2+2b2+2c2+ab+bc+ca

Using the identities of algebraic expressions , find

(i) 1002×1003

(ii) 97×103.   [2 MARKS]

#### SOLUTION

Solution :

Application: 1 Mark each

(i)1002×1003

=(1000+2)(1000+3)10002+(2+3)1000+(3)(2)

=1000000+5000+6

=1005006

(ii)97×103

=(1003)(100+3)

=[(100)2(3)2]

=

=9991

Find the volume of rectangular boxes with the following length, breadth and height respectively

(i) (a2,a3,a4)

(ii) (x2y2,xz,z4)
[2 MARKS]

#### SOLUTION

Solution :

Concept: 1 Mark
Application: 1 Mark

(i) Volume = a2×a3×a4=a9

(ii) Volume = x2y2×xz×z4=x3y2z5

Sandeep wants to buy 5 chocolates of Rs 10 each. He has to pay Rs 20 for parking also. If the price of chocolate is increased by Rs x, he decides to buy y less chocolates and he has to pay Rs z less for the parking. Write the arithmatic expression of the money spent by Sandeep.  [2 MARKS]

#### SOLUTION

Solution :

Concept: 1 Mark
Application: 1 Mark

Price of chocolate = (10 + x)

No. of chocolates = (5 – y)

Parking fees = (20 – z)

Total money spent =    (10 + x) (5 – y) + (20 – z)

Simplify the following:

(i) (1.5x4y)(1.5x+4y+3)

(ii) (x+y)(2x+y)+(x+2y)(xy)

(iii) (a2+5)(b3+3)+5
[3 MARKS]

#### SOLUTION

Solution :

Application: 1 Mark each

(i) (1.5x4y)(1.5x+4y+3)

2.25x2+6xy+4.5x6xy16y212y

2.25x216y2+4.5x12y

(ii) (x+y)(2x+y)+(x+2y)(xy)

(2x2+xy+2xy+y2)+(x2+xy2y2)

3x2+4xyy2

(iii) (a2+5)(b3+3)+5

a2b3+3a2+5b3+15+5

a2b3+3a2+5b3+20

Using the identities of algebraic expressions, Find :

(i) 1022

(ii) 532

(iii) 10012  [3 MARKS]

#### SOLUTION

Solution :

Application: 1 Mark each

Using the identity :
(a+b)2=a2+2ab+b2

(i) (102)2

(100+2)2

(100)2+(2)2+2(100)(2)

10000+4+400

10404

(ii) (53)2

(50+3)2

(50)2+(3)2+2(50)(3)

2500+9+300

2809

(iii) (1001)2

(1000+1)2

(1000)2+(1)2+2(1000)(1)

1000000+1+2000

1002001

Find the distance travelled with the following pairs of monomials as their speed and time respectively:

(i) (25a2b2,50b)

(ii) (12abc,23cde)

(iii)(5c2d2,4a2b2)  [3 MARKS]

#### SOLUTION

Solution :

Concept: 1 Mark
Application: 2 Marks

Distance Travelled

=Speed×Time

(i) Distance Travelled =25a2b2×50b=1250a2b3

(ii) Distance Travelled =12abc×23cde=276abc2de

(iii) Distance Travelled =5c2d2×4a2b2=20a2b2c2d2

Write the algebraic expression for:

(i) The volume of cube if its side(a) is reduced by 5.

(ii) Simple interest if time is reduced by 2 years and rate is increased by 5%.

(iii) Distance travelled if speed is increased by 5 units and time by 10 units.  [3 MARKS]

#### SOLUTION

Solution :

Concept: 1 Mark each

(i) Volume

=(a5)3

Here a is the initial length of the side of the cube.

(ii) Simple interest

=[P×(R+5)×(T2)]100

P=Principal Amount

R=rate

T=Time

(iii) Distance = (s+5) (t+10)

s=speed

t=time

Construct:

(i) 2 binomials with 3 variables.

(ii)  3 polynomials with 3 terms and 2 variables.

(iii)  2 trinomials with x as the only variable.

(iv) 3 binomials with x and y as variable.    [4 MARKS]

#### SOLUTION

Solution :

Application: 1 Mark each

(i) xy+xyz,4x2y+5xyz

(ii) 4ab2+5ab+10a,5xy+4y24 and 10x2y2+12x2y5xy2

(iii) 4x2+5x16,5x32x2+24

(iv) 10x2y2+12x2y,8xy5x and 5xy+8x

Using the identites of algebriac expression , find the following :

(i) 403×397

(ii) 8.22

(iii) 9.7×9.8

(iv) 22.927.12   [4 MARKS]

#### SOLUTION

Solution :

Application: 1 Mark each

(i) 403×397

=(400+3)(4003)

=[(400)2(3)2]

=1600009

=159991

(ii) 8.22

=(8+0.2)2

=(8)2+(0.2)2+2(8)(0.2)

=64+0.04+3.2

=67.24

(iii) 9.7×9.8

=(9+0.7)(9+0.8)

=(9)2+(0.7+0.8)9+(0.7)(0.8)

=81+13.5+0.56

=95.06

(iv) 22.927.12

=(22.9+7.1)(22.97.1)

=(30)(15.8)

=474

A rectangular box having length, breadth and height as (a+b+c),(a2+b2) and (2a+2b+2c) respectively. Find the volume of the box.  [4 MARKS]

#### SOLUTION

Solution :

Concept: 1 Mark
Application: 3 Marks

Total volume =

(a+b+c)(a2+b2)(2a+2b+2c)

=(a3+ab2+a2b+b3+ca2+cb2)(2a+2b+2c)

=(2a4+2a3b+2a3c+2a2b2+2ab3+2ab2c+2a3b+2a2b2+2a2bc+2ab3+2b4+2b3c+2a3c+2a2bc+2a2c2+2ab2c+2b3c+2b2c2)

=(2a4+4a3b+4a3c+4a2b2+4ab3+4ab2c+4a2bc+2b4+4b3c+2a2c2+2b2c2)

Classify the following polynomials as monomials, binomials, trinomials.

(i) 4x2

(ii) 5xy2+6xyz

(iii) 5xyz

(iv) a+b+c

(v) 5x2+6x+25

(vi) x+y  [4 MARKS]

#### SOLUTION

Solution :

Concept: 1 Mark
Application: 0.5 Mark each

Monomials -: 4x2,5xyz

Binomials -: 5xy2+6xyz,x+y

Trinomials -: a+b+c,5x2+6x+25

Give three examples of the expressions:

(i) containing one variable

(ii) containing two variables

(iii) containing two terms.  [4 MARKS]

#### SOLUTION

Solution :

Concept: 1 Mark
Application: 1 Mark each

(i) 4x; 5x+2; 4x2+4x+2

(ii) x+y; 5x3y; 4x2y2+4x2y+5y+2

(iii) 5x4y; 4a2+5b2; 10y+2

Add the following:

(i) 5a2b2+10ab12a2+15b20 and 34ab+24a2+22b+24

(ii) abc+6bcd+abd9ab+4bc9a43 and 4acd+10ac54b32c

(iii) 5bc+9ad3c12 and 245c5b+2ab  [4 MARKS]

#### SOLUTION

Solution :

Concept: 1 Mark
Application: 1 Mark each

(i) 5a2b2+10ab12a2+15b20+34ab+24a2+22b+24

=5a2b2+44ab+12a2+37b+4

(ii) abc+6bcd+abd9ab+4bc9a43+4acd+10ac54b32c

=abc+6bcd+abd9ab+4bc9a43+4acd+10ac54b32c

(iii) 5bc+9ad3c12+245c5b+2ab

=5bc+9ad8c+125b+2ab