8th Grade > Mathematics
ALGEBRAIC EXPRESSIONS AND IDENTITIES MCQs
Total Questions : 88
| Page 4 of 9 pages
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Concept: 1Mark
Application: 1Mark each
(i)4x;5x+2;4x2+4x+2
(ii)x+y;5x−3y;4x2y2+4x2y+5y+2
(iii)5x−4y;4a2+5b2;10y+2
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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
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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
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Concept: 1Mark
Application: 1 Mark
(i) Volume = a2×a3×a4=a9
(ii) Volume = x2y2×xz×z4=x3y2z5
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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
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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
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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)
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Concept: 1Mark
Application: 1Mark 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
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Usingthe Identity(x+a)(x−a)=x2−a2
We Put x = 36 and a = 35
(36 + 35) (36- 35) = (71) (1) = 71
Answer: Option C. -> -3
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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)2−2x2y2z=x2y2+y2z2+2xzy2−2x2y2z
Substitutingthe values of x,y and z in the above expression, we get
x2y2+y2z2+2xzy2−2x2y2z
(−1)2(1)2+(1)2(2)2+2(−1)(2)(1)2−2(1)2(1)2(2)
1 +4 - 4 - 4 = -3
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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)2−2x2y2z=x2y2+y2z2+2xzy2−2x2y2z
Substitutingthe values of x,y and z in the above expression, we get
x2y2+y2z2+2xzy2−2x2y2z
(−1)2(1)2+(1)2(2)2+2(−1)(2)(1)2−2(1)2(1)2(2)
1 +4 - 4 - 4 = -3