Free Algebraic Expressions and Identities Subjective Test 01 Practice Test - 8th Grade
Question 1
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
Question 2
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
Question 3
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
=(100−3)(100+3)
=[(100)2−(3)2]
=[10000−9]
=9991
Question 4
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
Question 5
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)
Question 6
Simplify the following:
(i) (1.5x−4y)(1.5x+4y+3)
(ii) (x+y)(2x+y)+(x+2y)(x−y)
(iii) (a2+5)(b3+3)+5 [3 MARKS]
SOLUTION
Solution :Application: 1 Mark each
(i) (1.5x−4y)(1.5x+4y+3)
2.25x2+6xy+4.5x−6xy−16y2−12y
⇒2.25x2−16y2+4.5x−12y
(ii) (x+y)(2x+y)+(x+2y)(x−y)
(2x2+xy+2xy+y2)+(x2+xy−2y2)
3x2+4xy−y2
(iii) (a2+5)(b3+3)+5
a2b3+3a2+5b3+15+5
⇒a2b3+3a2+5b3+20
Question 7
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
Question 8
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
Question 9
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
=(a−5)3
Here a is the initial length of the side of the cube.
(ii) Simple interest
=[P×(R+5)×(T−2)]100P=Principal Amount
R=rate
T=Time
(iii) Distance = (s+5) (t+10)s=speed
t=time
Question 10
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+4y−24 and 10x2y2+12x2y−5xy2
(iii) 4x2+5x−16,5x3−2x2+24
(iv) 10x2y2+12x2y,8xy−5x and 5xy+8x
Question 11
Using the identites of algebriac expression , find the following :
(i) 403×397
(ii) 8.22
(iii) 9.7×9.8
(iv) 22.92−7.12 [4 MARKS]
SOLUTION
Solution :Application: 1 Mark each
(i) 403×397
=(400+3)(400−3)
=[(400)2−(3)2]
=160000−9
=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) 22.92−7.12
=(22.9+7.1)(22.9−7.1)
=(30)(15.8)
=474
Question 12
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)
Question 13
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 -: 4x2,5xyzBinomials -: 5xy2+6xyz,x+y
Trinomials -: a+b+c,5x2+6x+25
Question 14
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; 5x−3y; 4x2y2+4x2y+5y+2
(iii) 5x−4y; 4a2+5b2; 10y+2
Question 15
Add the following:
(i) 5a2b2+10ab−12a2+15b−20 and 34ab+24a2+22b+24
(ii) abc+6bcd+abd−9ab+4bc−9a−43 and 4acd+10ac−54b−32c
(iii) 5bc+9ad−3c−12 and 24−5c−5b+2ab [4 MARKS]
SOLUTION
Solution :Concept: 1 Mark
Application: 1 Mark each
(i) 5a2b2+10ab−12a2+15b−20+34ab+24a2+22b+24
=5a2b2+44ab+12a2+37b+4
(ii) abc+6bcd+abd−9ab+4bc−9a−43+4acd+10ac−54b−32c
=abc+6bcd+abd−9ab+4bc−9a−43+4acd+10ac−54b−32c
(iii) 5bc+9ad−3c−12+24−5c−5b+2ab
=5bc+9ad−8c+12−5b+2ab