Home / iGCSE Mathematics (0580) :E2.4 Use and interpret positive, negative and zero indices.iGCSE Style Questions Paper 2

iGCSE Mathematics (0580) :E2.4 Use and interpret positive, negative and zero indices.iGCSE Style Questions Paper 2

Question

Simplify.

$( \mathbf{a} )$ $\frac {32g^{16}}{16g^{8}}$

$( \mathbf{b} )$ $( 625k^8) ^{\frac {1}{4}}$

▶️Answer/Explanation

[(a)] $2g^8$ (final answer)
[(b)] $5k^2$ (final answer)

(a)

$
\frac{32g^{16}}{16g^8}
$

Dividing the coefficients
$
\frac{32}{16} = 2
$

dividing powers
$
\frac{g^{16}}{g^8} = g^{16-8} = g^8
$

$
2g^8
$

(b)
$(625k^8)^{\frac{1}{4}}$

$
= 625^{\frac{1}{4}} \times (k^8)^{\frac{1}{4}}
$
$
625^{\frac{1}{4}} = 5
$
$
(k^8)^{\frac{1}{4}} = k^{8 \times \frac{1}{4}} = k^2
$
$
5k^2
$

Question

Factorise completely.

$4x^2y-5xy^2$

▶️Answer/Explanation

$xy(4x – 5y)$ final answer

Both terms have a common factor of \( xy \). So

$
= xy(4x – 5y)
$
The fully factorised expression is

$
\mathbf{xy(4x – 5y)}
$

Question

Calculate \(\sqrt{17.8}-1.3^{2.5}\)

Answer/Explanation

2.29 or 2.292

 

 

Question

(a) \(3^x = \sqrt[4]{3^5}\)
Find the value of x.
(b) Simplify \((32y^{15})^{\frac{2}{5}}\)

Answer/Explanation

Ans:

(a) \(\frac{5}{4}\) oe
(b) \(4y^6\)

Question

Simplify
(a) $\left(\frac{p^4}{16}\right)^{0.75}$,
(b) $3^2 q^{-3} \div 2^3 q^{-2}$.

▶️Answer/Explanation

(a)$\left(\frac{p^4}{16}\right)^{0.75} = \frac{p^{4 \times 0.75}}{16^{0.75}}$
Simplifying the exponents:
$\frac{p^3}{16^{0.75}}$
To simplify further, we can evaluate the exponent $16^{0.75}$:
$16^{0.75} = \sqrt[4]{16^3} = \sqrt[4]{4096} = 8$
Substituting this back into the expression:
$\frac{p^3}{8}$
Therefore, $\left(\frac{p^4}{16}\right)^{0.75}$ simplifies to $\frac{p^3}{8}$.
(b) To simplify $3^2 q^{-3} \div 2^3 q^{-2}$, we can simplify each term separately and then perform the division:
$3^2 = 9$
$q^{-3} = \frac{1}{q^3}$
$2^3 = 8$
$q^{-2} = \frac{1}{q^2}$
Substituting these simplified terms back into the expression:
$9 \cdot \frac{1}{q^3} \div 8 \cdot \frac{1}{q^2}$
Applying the division rule for exponents (subtracting exponents when dividing):
$\frac{9}{8} \cdot \frac{1}{q^3} \cdot q^2$
Simplifying:
$\frac{9}{8q}$
Therefore, $3^2 q^{-3} \div 2^3 q^{-2}$ simplifies to $\frac{9}{8q}$.

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