\begin{array}
\text {Find each of the following products: ( 1-8 ) } \\
\text {In this excercise to find the product, we will use commutativity and associativity of multiplication along } \\
\text {with laws of exponents as per their applicability in the expression given in questions. } \\
\end{array}\begin{array}{l}
\text {Q1. } \quad 5 x^{2} \times 4 x^{3} \\
\end{array}\begin{array}{l}
\text {Sol. } \\
5 x^{2} \times 4 x^{3} \\
=5 \times 4 \times x^{2+3} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
=20 x^{5} \\
\end{array}\begin{array}{l}
\text {Q2. } \quad -3 a^{2} \times 4 b^{4} \\
\end{array}\begin{array}{l}
\text {Sol. } \\
-3 \times a^{2} \times 4 \times b^{4} \\
= -3 \times 4 \times a^{2} \times b^{4}
=-12 a^{2} b^{4} \\
\end{array}\begin{array}{l}
\text {Q3. } \quad (-5 x y) \times (-3 x^{2}yz) \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(-5 x y) \times (-3 x^{2}yz)\\
=(-5) \times (-3) \times x \times x^{2} \times y \times y \times z \\
=15 \times x^{1+2} \times y^{1+1} \times z \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
=15 x^{3} y^{2} z \\
\end{array}\begin{array}{l}
\text {Q4. } \quad \frac {1}{2} x y \times \frac {2}{3} x^{2} y z^{2} \\
\end{array}\begin{array}{l}
\text {Sol. } \\
\frac {1}{2} x y \times \frac {2}{3} x^{2} y z^{2} \\=\frac {1}{2} \times \frac {2}{3} \times x \times x^{2} \times y \times y \times z^{2} \\=\frac {1}{3}\times x^{1+2} \times y^{1+1} \times z^{2} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\=\frac {1}{3} x^{3} y^{2} z^{2} \\
\end{array}\begin{array}{l}
\text {Q5. } \quad (-\frac {7}{5} x y^{2} z) \times (\frac {13}{3} x^{2} y z^{2}) \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(-\frac {7}{5} x y^{2} z) \times (\frac {13}{3} x^{2} y z^{2}) \\=-\frac {7}{5} \times \frac {13}{3} \times x \times x^{2} \times y^{2} \times y \times z \times z^{2} \\=-\frac {91}{15} \times x^{1+2} \times y^{2+1} \times z^{1+2} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\=-\frac {91}{15} x^{3} y^{3} z^{3} \\
\end{array}\begin{array}{l}
\text {Q6 . } \quad (-\frac{24}{25} x^{3} z) \times(-\frac{15}{16} x z^{2} y) \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(-\frac{24}{25} x^{3} z) \times(-\frac{15}{16} x z^{2} y) \\=(-\frac{24}{25}) \times (-\frac{15}{16}) \times x^{3} \times x \times z \times z^{2} \times y \\=\frac{9}{10} \times(x^{3+1}) \times(z^{1+2}) \times y \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
=\frac{9}{10} x^{4} y z^{3} \\
\end{array}\begin{array}{l}
\text {Q7. } \quad (-\frac{1}{27} a^{2} b^{2}) \times (\frac{9}{2} a^{3} b^{2} c^{2}) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(-\frac{1}{27} a^{2} b^{2}) \times (\frac{9}{2} a^{3} b^{2} c^{2}) \\=(-\frac{1}{27}) \times (\frac{9}{2}) \times(a^{2} \times a^{3}) \times(b^{2} \times b^{2}) \times c^{2} \\=-\frac{1}{6} \times(a^{2+3}) \times(b^{2+2}) \times c^{2} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\=-\frac{1}{6} a^{5} b^{4} c^{2} \\\end{array}\begin{array}{l}
\text {Q8. } \quad (-7 x y) \times (\frac{1}{4} x^{2}yz) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(-7 x y) \times(\frac{1}{4} x^{2} y z) \\=(-7 \times \frac{1}{4}) \times(x \times x^{2}) \times(y \times y) \times z \\=-\frac{7}{4} \times(x^{1+2}) \times(y^{1+1}) \times z \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\=-\frac{7}{4} x^{3} y^{2} z\\\end{array}\begin{array}{l}
\text {Q9. } \quad (7ab) \times (-5 ab^{2}c) \times (6 abc^{2}) \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(7ab) \times (-5 ab^{2}c) \times (6 abc^{2}) \\=7 \times (-5) \times 6 \times (a \times a \times a ) \times ( b \times b^{2} \times b ) \times ( c \times c^{2} ) \\= -210 \times a^{1+1+1} \times b^{1+2+1} \times c^{1+2} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
= -210 a^{3} b^{4} c^{3} \\\end{array}\begin{array}{l}
\text {Q10. } \quad (-5 a) \times(-10 a^{2}) \times(-2 a^{3}) \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(-5 a) \times(-10 a^{2}) \times(-2 a^{3}) \\=(-5) \times (-10) \times (-2) \times ( a \times a^{2} \times a^{3}) \\
=-100 \times a^{1+2+3} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
=-100 a^{6} \\
\end{array}\begin{array}{l}
\text {Q11. } \quad (-4 x^{2}) \times (-6 x y^{2}) \times (-3 y z^{2}) \\ \end{array}\begin{array}{l}
\text {Sol. } \\
(-4 x^{2}) \times (-6 x y^{2}) \times (-3 y z^{2}) \\=(-4) \times (-6) \times (-3) \times (x^{2} \times x ) \times (y^{2} \times y ) \times z^{2} \\= -72 \times x^{2+1} \times y^{2+1} \times z^{2} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\=-72 x^{3} y^{3} z^{2} \\
\end{array}\begin{array}{l}
\text {Q12. } \quad (-\frac{2}{7} a^{4}) \times (-\frac{3}{4} a^{2} b) \times (-\frac{14}{5} b^{2}) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(-\frac{2}{7} a^{4}) \times (-\frac{3}{4} a^{2} b) \times (-\frac{14}{5} b^{2}) \\=(-\frac{2}{7}) \times (-\frac{3}{4}) \times (-\frac{14}{5}) \times(a^{4} \times a^{2}) \times(b \times b^{2}) \\=-(\frac{2}{7} \times \frac{3}{4} \times \frac{14}{5}) \times(a^{4+2}) \times(b^{1+2}) \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\=-\frac{3}{5} a^{6} b^{3} \\
\end{array}\begin{array}{l}
\text {Q13. } \quad (\frac{7}{9} a b^{2}) \times (\frac{15}{7} a c^{2} b) \times (-\frac{3}{5} a^{2} c) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(\frac{7}{9} a b^{2}) \times (\frac{15}{7} a c^{2} b) \times (-\frac{3}{5} a^{2} c) \\=\frac{7}{9} \times \frac{15}{7} \times (-\frac{3}{5}) \times(a \times a \times a^{2}) \times(b^{2} \times b) \times(c^{2} \times c) \\=\frac{7}{9} \times \frac{15}{7} \times(-\frac{3}{5}) \times (a^{1+1+2}) \times(b^{2+1}) \times(c^{2+1}) \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\=- a^{4} b^{3} c^{3} \\
\end{array}\begin{array}{l}
\text {Q14. } \quad (\frac{4}{3} u^{2} v w) \times (-5 u v w^{2}) \times (\frac{1}{3} v^{2} w u) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(\frac{4}{3} u^{2} v w) \times (-5 u v w^{2}) \times (\frac{1}{3} v^{2} w u) \\=(\frac{4}{3}) \times (-5) \times (\frac{1}{3}) \times(u^{2} \times u \times u) \times(v \times v \times v^{2}) \times(w \times w^{2} \times w) \\=-\frac{20}{9} \times(u^{2+1+1}) \times(v^{1+1+2}) \times(w^{1+2+1}) \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\=-\frac{20}{9} u^{4} v^{4} w^{4}\\
\end{array}\begin{array}
\text {Q15. } \quad (0.5 x) \times (\frac{1}{3} x y^{2} z^{4}) \times (24 x^{2} y z) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(0.5 x) \times (\frac{1}{3} x y^{2} z^{4}) \times (24 x^{2} y z) \\=0.5 \times \frac{1}{3} \times 24 \times (x \times x \times x^{2}) \times (y^{2} \times y) \times (z^{4} \times z) \\=[\frac{1}{2} \times \frac{1}{3} \times 24] \times(x^{1+1+2}) \times(y^{2+1}) \times(z^{4+1}) \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\=4 x^{4} y^{3} z^{5} \\
\end{array}\begin{array}{l}
\text {Q16. } \quad (\frac{4}{3} p q^{2}) \times (-\frac{1}{4} p^{2} r) \times (16 p^{2} q^{2} r^{2}) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(\frac{4}{3} p q^{2}) \times (-\frac{1}{4} p^{2} r) \times (16 p^{2} q^{2} r^{2}) \\=\frac{4}{3} \times (-\frac{1}{4}) \times 16 \times (p \times p^{2} \times p^{2}) \times (q^{2} \times q^{2}) \times (r \times r^{2}) \\=-\frac{16}{3} \times (p^{1+2+2}) \times (q^{2+2}) \times(r^{1+2}) \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\=-\frac{16}{3} p^{5} q^{4} r^{3} \\
\end{array}\begin{array}{l}
\text {Q17. } \quad (2.3 xy) \times (0.1 x) \times (0.16) \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(2.3 xy) \times (0.1 x) \times (0.16) \\=2.3 \times 0.1 \times 0.16 \times x \times x \times y \\
=0.0368 \times (x^{1+1}) \times y \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
=0.0368 x^{2} y \\
\end{array}\begin{array}{l}
\text {Express each of the following products as a monomials and verify the result in each case for x=1 : (Q18 – Q21) } \\
\end{array}\begin{array}{l}
\text {Q18. } \quad (3 x) \times (4 x) \times (-5 x) \\
\end{array}\begin{array}
\text {Sol. } \\
(3 x) \times (4 x) \times (-5 x) = 3 \times 4 \times-5 \times x \times x \times x \\
=-60 \times x^{1+1+1} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
=-60 x^{3} \\\therefore \quad (3 x) \times (4 x) \times (-5 x) =-60 x^{3} \\
\text {Substitue x =1 in both LHS and RHS expressions, we got} \\\text { LHS } = (3 x) \times (4 x) \times (-5 x) = (3 times 1 ) \times (4 \times 1) \times (-5 \times 1) = -60 \\\text {And, RHS } = -60 x^{3} =-60 \times 1^{3} = -60 \\\text {As LHS = RHS for x =1, hence result is verifed.}\end{array}\begin{array}
\text {Q19. } \quad (4 x^{2}) \times (-3 x) \times (\frac{4}{5} x^{3}) \\
\end{array}\begin{array}
\text {Sol. } \\(4 x^{2}) \times (-3 x) \times (\frac{4}{5} x^{3}) \\=(4 \times (-3) \times \frac{4}{5}) \times (x^{2} \times x \times x^{3}) \\=-\frac{48}{5} \times (x^{2+1+3}) \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
=-\frac{48}{5} x^{6} \\
\therefore \quad (4 x^{2}) \times (-3 x) \times (\frac{4}{5} x^{3}) = -\frac{48}{5} x^{6} \\
\text {Substitue x =1 in both LHS and RHS , we got} \\\text { LHS } =(4 x^{2}) \times(-3 x) \times(\frac{4}{5} x^{3})
=(4 \times 1^{2}) \times(-3 \times 1) \times(\frac{4}{5} \times 1^{3}) \\
=4 \times(-3) \times \frac{4}{5} =-\frac{48}{5} \\\text {And, RHS } = -\frac{48}{5} x^{6} \\
=-\frac{48}{5} \times 1^{6} = -\frac{48}{5}
\text {As LHS = RHS for x =1, hence result is verifed.}\end{array}\begin{array}{l}
\text {Q20. } \quad (5 x^{4}) \times (x^{2})^{3} \times (2 x)^{2} \\
\end{array}\begin{array}{l}
\text {Sol. } \\(5 x^{4}) \times (x^{2})^{3} \times (2 x)^{2} \\
=(5 x^{4}) \times(x^{6}) \times(2^{2} \times x^{2}) \\
=(5 \times 2^{2}) \times(x^{4} \times x^{6} \times x^{2}) \\
=20 \times (x^{4+6+2}) \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
=20 x^{12} \\\therefore \quad (5 x^{4}) \times (x^{2})^{3} \times (2 x)^{2} = =20 x^{12} \\
\text {Substitue x =1 in both LHS and RHS , we got} \\\text { LHS } =(5 x^{4}) \times(x^{2})^{3} \times(2 x)^{2} \\
=(5 \times 1) \times(1^{6}) \times(2)^{2} \\
=5 \times 1 \times 4 =20 \\\text {And, RHS }=20 x^{12} \\
=20 \times 1^{12} =20 \times 1 =20 \\
\text {As LHS = RHS for x =1, hence result is verifed.}\end{array}\begin{array}{l}
\text {Q21. } \quad (x^{2})^{3} \times (2 x) \times (-4 x) \times 5 \\
\end{array}\begin{array}{l}
\text {Sol. } \\(x^{2})^{3} \times (2 x) \times (-4 x) \times 5 \\
=(x^{6}) \times(2 x) \times(-4 x) \times 5 \\
=(2 \times (-4) \times 5) \times (x^{6} \times x \times x) \\
=(2 \times(-4) \times 5) \times (x^{6+1+1}) \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
=-40 x^{8} \\ \\\therefore \quad (x^{2})^{3} \times(2 x) \times(-4 x) \times 5=-40 x^{8} \\
\text {Substitue x =1 in both LHS and RHS , we got} \\\text { LHS } =(x^{2})^{3} \times(2 x) \times(-4 x) \times 5 \\
=(1^{2})^{3} \times(2 \times 1) \times(-4 \times 1) \times 5 \\
=1 \times 2 \times (-4) \times 5 =-40 \\\text {And, RHS } =-40 x^{8} \\
=-40 \times 1^{3} = -40 \\
\text {As LHS = RHS for x =1, hence result is verifed.}
\end{array}\begin{array}{l}
\text {Q22. Write down the product of } -8 x^{2} y^{6} \text { and }-20 x y \text {Verify the product for x=2.5, y=1 }\\
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {We have to find the product of } -8 x^{2} y^{6} \text { and }-20 x y \\
-8 x^{2} y^{6} \times (-20 x y) \\
=(-8 ) \times (-20) \times (x^{2} \times x ) \times (y^{6} \times y) \\
= 160 \times x^{2+1} \times y^{6+1} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\
=160 x^{3} y^{7} \\\therefore \quad -8 x^{2} y^{6} \times (-20 x y) = 160 x^{3} y^{7} \\
\text {Substitue } x=2.5=\frac{25}{10}=\frac{5}{2} \text {, y=1 in both LHS and RHS , we got} \\\text { LHS } = -8 x^{2} y^{6} \times (-20 x y) \\
=-8 \times (2.5)^{2} \times 1^{6} \times (-20 \times 1 \times 2.5) \\=-8 \times (\frac{5}{2})^{2} \times 1^{6} \times (-20 \times 1 \times \frac{5}{2}) \\
= -8 \times \frac{25}{4} \times (-20) \times \frac{5}{2} \\
= 2500 \\\text { RHS } = 160 x^{3} y^{7} \\
=160 \times (\frac{5}{2})^{3} \times (1)^{7} \\
= 160 \times \frac{125}{8} =2500 \\
\text {As LHS = RHS for x =1, hence result is verifed.}
\end{array}\begin{array}{l}
\text {Q23. Evaluate } (3.2 x^{6} y^{3}) \times (2.1 x^{2} y^{2}) \text {when x=1 and y=0.5 } \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(3.2 x^{6} y^{3}) \times (2.1 x^{2} y^{2}) \\
=3.2 \times 2.1 \times x^{6} \times x^{2} \times y^{3} \times y^{2} \\
=6.72 \times x^{6+2} \times y^{3+2} \\
=6.72 x^{8} y^{5} \\
\text {Now substituting x=1 and y=0.5 in the expression we will get } \\6.72 x^{8} y^{5} \\
=6.72 \times 1^{8} \times 0.5^{5} \\
=6.72 \times 1^{8} \times (\frac{1}{2} )^{5} \\
=\frac{672}{100} \times \frac{1}{32} \\
=\frac{21}{100}=0.21 \\
\end{array}\begin{array}{l}
\text {Q24. Find the value of } (5 x^{6}) \times (-1.5 x^{2} y^{3}) \times(-12 x y^{2}) \text { when x=1, y=0.5 } \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(5 x^{6}) \times (-1.5 x^{2} y^{3}) \times(-12 x y^{2}) \\=5 \times (-1.5) \times (-12) \times (x^{6} \times x^{2} \times x) \times (y^{3} \times y^{2}) \\
=90 \times x^{6+2+1} \times y^{3+2} \\
=90 x^{9} y^{5} \\ \\\text {Now substituting x=1 and y=0.5 in the expression we will get } \\90 x^{9} y^{5}
=90 \times (1)^{9} \times (0.5)^{5} \\
=90 \times 1 \times (\frac{1}{2} )^{5} \\
=\frac {45}{16} \\
\end{array}\begin{array}{l}
\text {Q25. Evaluate } (2.3 a^{5} b^{2}) \times (1.2 a^{2} b^{2}) \text {when a=1 and b=0.5 }\\
\end{array}\begin{array}
\text {Sol. } \\
(2.3 a^{5} b^{2}) \times (1.2 a^{2} b^{2}) \\=2.3 \times 1.2 \times (a^{5} \times a^{2}) \times (b^{2} \times b^{2}) \\=2.76 \times a^{5+2} \times b^{2+2} \\
=2.76 a^{7} b^{4} \\ \\\text {Now substituting a=1 and b=0.5 in the expression we will get } \\2.76 a^{7} b^{4} \\
=2.76 \times (1)^{7} \times (0.5)^{4} \\=\frac{276}{100} \times 1 \times (\frac{1}{2})^{4} \\=\frac{276}{100} \times 1 \times \frac{1}{16}\\
=\frac{69}{400} \\
\end{array}\begin{array}{l}
\text {Q26. Evaluate } (-8 x^{2} y^{6}) \times (-20 x y) \text {for x=2.5 and y=1 } \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(-8 x^{2} y^{6}) \times (-20 x y)
= -8 \times (-20) \times (x^{2} \times x) \times (y^{6} \times y) \\
=160 \times x^{2+1} \times y^{6+1} \\
=160 x^{3} y^{7} \\\text {Now substituting x=2.5 and y=1 in the expression we will get } \\=160 \times (2.5)^{3} \times (1)^{7} \\
=160 \times \frac {25}{10}^{3} \times 1 \\
=160 \times \frac {5}{2}^{3} = 160 \times \frac {125}{8} \\
=2500 \\
\end{array}\begin{array}{l}
\text {Express each of the following products as a monomials and verify the result for x=1, y=2 : (27 -31)}\\
\end{array}\begin{array}{l}
\text {Q27. } \quad (-x y^{3}) \times (y x^{3}) \times (x y) \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(-x y^{3}) \times (y x^{3}) \times (x y) \\
=-(x \times x^{3} \times x ) \times (y^{3} \times y \times y \\
=-x^{1+3+1} \times y^{3+1+1} \\
=-x^{5} y^{5} \\ \\\therefore \quad (-x y^{3}) \times (y x^{3}) \times (x y) =-x^{5} y^{5} \\
\text {Substitue x=1, y=2 in both LHS and RHS , we got} \\\text { LHS } = (-x y^{3}) \times (y x^{3}) \times (x y)
=- 1 \times 2^{3} \times 2 \times 1^{3} \times 1 \times 2 \\
=- 8 \times 2 \times 2 =-32 \\\text {And, RHS } = -x^{5} y^{5}= – 1^{5} \times 2^{5} = -32 \\\text {As LHS = RHS for x =1 and y =2 , hence result is verifed.}\end{array}\begin{array}{l}
\text {Q28. } \quad (\frac{1}{8} x^{2} y^{4}) \times (\frac{1}{4} x^{4} y^{2}) \times (x y) \times 5 \\
\end{array}\begin{array}{l}
\text {Sol. } \\
(\frac{1}{8} x^{2} y^{4}) \times (\frac{1}{4} x^{4} y^{2}) \times (x y) \times 5 \\=(\frac{1}{8} \times \frac{1}{4} \times 5) \times(x^{2} \times x^{4} \times x) \times(y^{4} \times y^{2} \times y) \\=(\frac{1}{8} \times \frac{1}{4} \times 5) \times(x^{2+4+1}) \times(y^{4+2+1}) \\=\frac{5}{32} x^{7} y^{7} \\\therefore \quad (\frac{1}{8} x^{2} y^{4}) \times (\frac{1}{4} x^{4} y^{2}) \times (x y) \times 5 =\frac{5}{32} x^{7} y^{7} \\
\text {Substitue x=1, y=2 in both LHS and RHS , we got} \\\text { LHS } =(\frac{1}{8} x^{2} y^{4}) \times(\frac{1}{4} x^{4} y^{2}) \times(x y) \times 5 \\=(\frac{1}{8} \times(1)^{2} \times(2)^{4}) \times(\frac{1}{4} \times(1)^{4} \times(2)^{2}) \times(1 \times 2) \times 5 \\=(\frac{1}{8} \times 1 \times 16) \times(\frac{1}{4} \times 1 \times 4) \times(1 \times 2) \times 5 \\
=2 \times 1 \times 2 \times 5 =20 \\\text {And, RHS } = \frac{5}{32} x^{7} y^{7} \\
=\frac{5}{32}(1)^{7}(2)^{7} \\
=\frac{5}{32} \times 1 \times 128 =20 \\\text {As LHS = RHS for x =1 and y =2 , hence result is verifed.}
\end{array}\begin{array}{l}
\text {Q29. } \quad (\frac{2}{5} a^{2} b) \times (-15 b^{2} a c) \times (-\frac{1}{2} c^{2}) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(\frac{2}{5} a^{2} b) \times (-15 b^{2} a c) \times (-\frac{1}{2} c^{2}) \\=[\frac{2}{5} \times(-15) \times(-\frac{1}{2})] \times(a^{2} \times a) \times(b \times b^{2}) \times(c \times c^{2}) \\=(\frac{2}{5} \times (-15) \times (-\frac{1}{2}) \times(a^{2+1}) \times(b^{1+2}) \times(c^{1+2}) \\
=3 a^{3} b^{3} c^{3} \\\text {In this case, result cannot be verified for x=1 and y=2 as x and y does not exists in the expression.}
\end{array}\begin{array}{l}
\text {Q30. } \quad (-\frac{4}{7} a^{2} b) \times (-\frac{2}{3} b^{2} c) \times (-\frac{7}{6} c^{2} a) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(-\frac{4}{7} a^{2} b) \times (-\frac{2}{3} b^{2} c) \times (-\frac{7}{6} c^{2} a) \\=(-\frac{4}{7}) \times (-\frac{2}{3}) \times (-\frac{7}{6}) \times(a^{2} \times a) \times(b \times b^{2}) \times(c \times c^{2}) \\=(-\frac{4}{7}) \times(-\frac{2}{3}) \times(-\frac{7}{6}) \times(a^{2+1}) \times(b^{1+2}) \times(c^{1+2}) \\=-\frac{4}{9} a^{3} b^{3} c^{3} \\ \\
\text {In this case, result cannot be verified for x=1 and y=2 as x and y does not exists in the expression.} \\
\end{array}\begin{array}{l}
\text {Q31. } \quad (\frac{4}{9} a b c^{3}) \times (-\frac{27}{5} a^{3} b^{2}) \times (-8 b^{3} c) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(\frac{4}{9} a b c^{3}) \times (-\frac{27}{5} a^{3} b^{2}) \times (-8 b^{3} c) \\=[(\frac{4}{9}) \times(-\frac{27}{5}) \times(-8)] \times(a \times a^{3}) \times(b \times b^{2} \times b^{3}) \times(c^{3} \times c) \\=[(\frac{4}{9}) \times(-\frac{27}{5}) \times(-8)] \times(a^{1+3}) \times(b^{1+2+3}) \times(c^{3+1}) \\=\frac{96}{5} a^{4} b^{6} c^{4} \\ \\\text {In this case, result cannot be verified for x=1 and y=2 as x and y does not exists in the expression.} \\
\end{array}\begin{array}{l}
\text {Evaluate each of the following when x=2, y=-1 } \\
\end{array}\begin{array}
\text {Q32. } \quad (2 x y) \times (\frac{x^{2} y}{4}) \times (x^{2}) \times (y^{2}) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(2 x y) \times (\frac{x^{2} y}{4}) \times (x^{2}) \times (y^{2}) \\=(2 \times \frac{1}{4}) \times(x \times x^{2} \times x^{2}) \times(y \times y \times y^{2}) \\=(2 \times \frac{1}{4}) \times(x^{1+2+2}) \times(y^{1+1+2}) \\
=\frac{1}{2} x^{5} y^{4} \\
\text { On substituting x=2 and y=-1 in the expression, we will get } \\\frac{1}{2} x^{5} y^{4} \\
=\frac{1}{2} \times (2)^{5} \times (-1)^{4} \\
=\frac{1}{2} \times 32 \times 1 = 16 \\
\end{array}\begin{array}{l}
\text {Q33. } \quad (\frac{3}{5} x^{2} y) \times (-\frac{15}{4} x y^{2}) \times (\frac{7}{9} x^{2} y^{2}) \\
\end{array}\begin{array}{l}
\text {Sol. } \\(\frac{3}{5} x^{2} y) \times (-\frac{15}{4} x y^{2}) \times (\frac{7}{9} x^{2} y^{2}) \\=(\frac{3}{5} \times(-\frac{15}{4}) \times \frac{7}{9}) \times(x^{2} \times x \times x^{2}) \times(y \times y^{2} \times y^{2})$=(\frac{3}{5} \times(-\frac{15}{4}) \times \frac{7}{9}) \times(x^{2+1+2}) \times(y^{1+2+2}) \\=-\frac{7}{4} x^{5} y^{5} \\\text { On substituting x=2 and y=-1 in the expression, we will get } \\-\frac{7}{4} x^{5} y^{5} \\
=-\frac{7}{4}(2)^{5}(-1)^{5} \\
=(-\frac{7}{4}) \times 32 \times(-1) =56 \\
\end{array}
InfiniteGrades
Learners today, leaders tomorrow