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8th Grade > Mathematics

ALGEBRAIC EXPRESSIONS AND IDENTITIES MCQs

Total Questions : 88 | Page 4 of 9 pages
Question 31. Give three examples of the expressions: 
(i) containing one variable 
(ii) containing two variables  
(iii) containing two terms.  [4 MARKS]
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Concept: 1Mark
Application: 1Mark each

(i)4x;5x+2;4x2+4x+2
(ii)x+y;5x3y;4x2y2+4x2y+5y+2
(iii)5x4y;4a2+5b2;10y+2
Question 32. Subtract (ap+bq-cr) from (a+b+c)(p+q+r) .  [1 MARK]
 Discuss Question

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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 33. Add the following:
a(a+b),b(b+c),c(a+c) and a2+b2+c2  [2 MARKS]
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Concept: 1Mark
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 34. Find the volume of rectangular boxes with the following length, breadth and height respectively
(i) (a2,a3,a4)
(ii) (x2y2,xz,z4)  
 [2 MARKS]
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Concept: 1Mark
Application: 1 Mark
(i) Volume = a2×a3×a4=a9
(ii) Volume = x2y2×xz×z4=x3y2z5
Question 35. 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]
 Discuss Question

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Concept: 1Mark
Application: 2Marks
Distance Travelled
=Speed×Time
(i) Distance Travelled =25a2b2×50b=1250a2b3
(ii) Distance Travelled =12abc×23cde=276abc2de
(iii) Distance Travelled =5c2d2×4a2b2=20a2b2c2d2
Question 36. Using the identities of algebraic expressions, Find :
(i) 1022
(ii) 532
(iii) 10012  [3 MARKS]
 Discuss Question

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Application: 1Mark 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 37. 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]
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Concept: 1Mark
Application: 3Marks
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 38. 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]
 Discuss Question

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Concept: 1Mark
Application: 1Mark 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
Question 39. Find the value of 362352=
___
 Discuss Question

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Usingthe Identity(x+a)(xa)=x2a2
We Put x = 36 and a = 35
(36 + 35) (36- 35) = (71) (1) = 71
Question 40. Simplify (xy+yz)22x2y2z and find it's value when x=1,y=1 and z=2
  1.    3
  2.    4
  3.    -3
  4.    0
 Discuss Question
Answer: Option C. -> -3
:
C
(xy+yz)2 is in the form of(a+b)2 where a=xy and b=yz.
Using (a+b)2=a2+b2+2ab,
(xy+yz)2=x2y2+y2z2+2xzy2.
Therefore,
(xy+yz)22x2y2z=x2y2+y2z2+2xzy22x2y2z
Substitutingthe values of x,y and z in the above expression, we get
x2y2+y2z2+2xzy22x2y2z
(1)2(1)2+(1)2(2)2+2(1)(2)(1)22(1)2(1)2(2)
1 +4 - 4 - 4 = -3

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