4 Marks Questions

Question 14 Marks

Complete the following example using Nityasam.
$\begin{array}{c}(i)(2 a-7 b)^3=(2 a)^3+(-7 b)^3+3(2 a)(-7 b)(\_\_\_\_\_\_) \\ =\_\_\_\_\_\_+\left(-343 b^3\right)+(-42 a b)(2 a-7 b) \\ =8 a^3-\_\_\_\_\_\_\_-84 a^2 b+\_\_\_\_\_\_\end{array}$
$\begin{array}{c}(ii) 95 \times 105=(100+\_\_\_\_\_\_)\left(100-\_\_\_\_\_\_\right) \\ =(100)^2-(\_\_\_\_\_\_)^2 \\ =\_\_\_\_\_\_-\_\_\_\_\_\_\\=9975\end{array}$

Answer

$\begin{aligned}(i)(2 a-7 b)^3 & =(2 a)^3+(-7 b)^3+3(2 a)(-7 b)(2 a-7 b) \\ & =8 a^3-343 b^3-42 a b(2 a-7 b) \\ & =8 a^3-343 b^3-84 a^2 b+294 a b^2\end{aligned}$
$\begin{array}{c}(ii)95 \times 105=(100+5)(100-5) \\ =(100)^2-(5)^2 \\ =10000-25 \\ =9975\end{array}$

View full question & answer→

Question 24 Marks

Complete the following example using Nityasam.
$\begin{array}{c}(i)(4 x+3 y)^3=(4 x)^3+(3 y)^3+3(4 x)(3 y)\left(\_\_\_\_\_\_\right) \\ =\_\_\_\_\_\_+27 y^3+36 x y(4 x+3 y) \\ =64 x^3+\_\_\_\_\_\_+144 x^2 y+\_\_\_\_\_\_\end{array}$
$\begin{array}{c}(ii) 106 \times 94=(100+\_\_\_\_\_\_)\left(100-\_\_\_\_\_\_\right) \\ =(100)^2-(\_\_\_\_\_\_)^2 \\ =\_\_\_\_\_\_-\_\_\_\_\_\_\\=9964\end{array}$

Answer

$\begin{array}{c}(i)(4 x+3 y)^3=(4 x)^3+(3 y)^3+3(4 x)(3 y)(4 x+3 y) \\ =64 x^3+27 y^3+36 x y(4 x+3 y) \\ =64 x^3+27 y^3+144 x^2 y+108 x y^2\end{array}$
$\begin{array} {c}(ii)106 \times 94=(100+6)(100-6) \\ =(100)^2-(6)^2 \\ =10000-36 \\ =9964\end{array}$

View full question & answer→

Question 34 Marks

Complete the following example using Nityasam.
$\begin{array}{c}(i)(5 x+2 y)^3=(5 x)^3+(2 y)^3+3(5 x)(2 y)(\_\_\_\_\_\_) \\ =\_\_\_\_\_\_+8 y^3+30 x y(5 x+2 y) \\ =125 x^3+\_\_\_\_\_\_+150 x^2 y+\_\_\_\_\_\_\end{array}$
$\begin{array}{c}(ii) 108 \times 92=(100+\_\_\_\_\_\_)\left(100-\_\_\_\_\_\_\right) \\ =(100)^2-(\_\_\_\_\_\_)^2 \\ =\_\_\_\_\_\_-\_\_\_\_\_\_\\=9936\end{array}$

Answer

$\begin{array}{c}(i)(5 x+2 y)^3=(5 x)^3+(2 y)^3+3(5 x)(2 y)(5 x+2 y) \\ =125 x^3+8 y^3+30 x y(5 x+2 y) \\ =125 x^3+8 y^3+150 x^2 y+60 x y^2\end{array}$
$\begin{array} {c}(ii)108 \times 92=(100+8)(100-8) \\ =(100)^2-(8)^2 \\ =10000-64 \\ =9936\end{array}$

View full question & answer→

Question 44 Marks

Complete the following example using Nityasam.
$\begin{array}{c}(i)(2 x+5 y)^3=(2 x)^3+(5 y)^3+3(2 x)(5 y)\left(\_\_\_\_\_\_\right) \\ =\_\_\_\_\_\_+125 y^3+30 x y(2 x+5 y) \\ =8 x^3+\_\_\_\_\_\_+60 x^2 y+\_\_\_\_\_\_\end{array}$
$\begin{array}{l}(ii) 112 \times 88=\left(100+\_\_\_\_\_\_\right)\left(100-\_\_\_\_\_\_\right) \\ =(100)^2-\left(\_\_\_\_\_\_\right)^2 \\ =\_\_\_\_\_\_-\_\_\_\_\_\_\ \\ =9856\end{array}$

Answer

$\begin{array}{c}(i)(2 x+5 y)^3=(2 x)^3+(5 y)^3+3(2 x)(5 y)(2 x+5 y) \\ =8 x^3+125 y^3+30 x y(2 x+5 y) \\ =8 x^3+125 y^3+60 x^2 y+150 x y^2\end{array}$
$\begin{array}{c}(ii)112 \times 88=(100+12)(100-12) \\ =(100)^2-(12)^2 \\ =10000-144 \\ =9856\end{array}$

View full question & answer→

Question 54 Marks

Complete the following example using Nityasam.
$\begin{aligned}(i)(3 a+4 b)^3 & =(3 a)^3+\left(\_\_\_\_\_\_\right) +3(3 a)(4 b)\\ & =27 a^3+\left(\_\_\_\_\_\_\right)+36 a b(3 a+4 b) \\ & =27 a^3+64 b^3+\left(\_\_\_\_\_\_\right)+144 a b^2 \end{aligned}$
$\begin{array}{c}(ii) 104 \times 96=\left(100+\_\_\_\_\_\_\right)\left(100-\_\_\_\_\_\_\right) \\ =(100)^2-\left(\_\_\_\_\_\_\right)^2 \\ =\left(\_\_\_\_\_\_\right)-16 \\ =9984\end{array}$

Answer

$\begin{array}{c}(i)(3 a+4 b)^3=(3 a)^3+(4 b)^3+3(3 a)(4 b)(3 a+4 b) \\ =27 a^3+64 b^3+36 a b(3 a+4 b) \\ =27 a^3+64 b^3+108 a^2 b+144 a b^2\end{array}$
$\begin{array}{c}(ii)100 \times 96=(100+0)(100-4) \\ =(100)^2-(4)^2 \\ =10000-16 \\ =9984\end{array}$

View full question & answer→

Maths 4 Marks Questions

Test yourself on this topic