8th Grade > Mathematics
ALGEBRAIC EXPRESSIONS AND IDENTITIES MCQs
:
B
Area of a rectangle = length × width
Area = 5xy×9y
= 45xy2
Hence the given statement is false.
:
C and D
Like terms are terms with the same variables raised to the same power. Coefficients of like terms need not be the same.
5q2p=5pq2.
Hence 5q2p and 12pq2 are both correct answers.
:
A
(t+s2)(t2−s)=[t(t2−s)+s2(t2−s)]
=t3−ts+s2t2−s3
=t3+s2t2−st−s3
:
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
:
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
:
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
:
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]
:
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)
:
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
:
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