Exercise 11.1

1. Find the product of the following pairs:

(i) 6, 7k

Solution:

6 × 7k =

(ii) -3l, -2m

Solution:

-3l × -2m =

(iii) −5t2, −3t2

Solution:

−5t2 × −3t2 = 15t413t4

(iv) 6n, 3m

Solution:

6n × 3m =

(v) −5p2, -2p

Solution:

−5p2 × -2p = 10p315p3

2. Complete the table of the products.

Solution:

Instruction

3. Find the volumes of rectangular boxes with given length, breadth and height in the following table.

Solution:

4. Find the product of the following monomials

(i) xy, x2y, xy, x

Solution:

xy × x2y × xy × x = x5y2x4y2

(ii) a, b, ab, a3b, ab3

Solution:

a × b × ab × a3b × ab3 = a6b6a6b5

(iii) kl, lm, km, klm

Solution:

kl × lm × km × klm = k3l3m2k2l3m2

(iv) pq, pqr, r

Solution:

pq × pqr × r = p2q2r2p2q3r2

(v) -3a, 4ab, -6c, d

Solution:

-3a × 4ab × -6c × d = 72a2bcd72a2bd

5. If A = xy, B = yz and C = zx, then find ABC = ..........

Solution:

ABC = xy × yz × zx = x1+1 × y1+1 × z1+1 = x2y2z2x3y3z3

6. If P = 4x2, T = 5x and R = 5y, then PTR/100 = ..........

Solution:

PTR/100 = 4x2×5x×5y100 = (100x3y100x3)/100

= x3y10x3y

7. Write some monomials of your own and find their products.

Solution:

Let's consider the following monomials:

Monomial 1: 2ab2

Monomial 2: −3a2bc

Monomial 3: 4c2

Product = (2ab2) × (−3a2bc) × (4c2) = (2 × -3 × 4) × a1+2 × b2+1 × c1+2 = −24a3b3c3−24a3b2c3

Therefore, the product of these monomials is −24a3b3c3.