DXC Quants Surds & Indices Quiz 3

Explanation:

=(0.98)^{3} + (0.02)^{3} + 3 \times0.98 \times 0.2 – 1
= 0.941192 + 0.000008+0.0588– 1
= 1 – 1 = 0

Explanation:

Expression = 0.65\times0.65 + 0.35\times 0.35 + 2\times0.35\times0.65
= (0.65 + 0.35)^{2}
= (1)^{2}= 1

Explanation:

\frac{2.4 \times 10^{3}}{8 \times 10^{-2}}
=\frac{24 \times10^{3}}{8 \times 10 \times 10^{-2}}
=\frac{24 \times 10^{3} = 3 \times 10^{4}

Explanation:

=[3– 4(3 – 4)^{–1}]^{ –1}
=[3– 4(-1)^{–1}]^{ –1}
=[\frac{(3-4)}{{–1}}]^{ –1}
= [3 + 4]^{-1}
= (7)^{-1} = 1/7

Explanation:

a = 55, b = 17, c = – 72
a + b + c = 55 + 17 – 72 = 0
\therefore a^{3} + b^{3} + c^{3}– 3abc = 0

Explanation:

= (0.2)^{3} \times 200 +2000 of (0.2)^{2}
= (0.2)^{3} \times 200 + (2000 \times 0.2 \times 0.2)
= \frac{0. 2 \times 0. 2 \times 0. 2\times 200}{2000\times 2\times2 \times 10 }
= 2/100
= 1/50

Explanation:

= \frac{(75.8)^{2}-( 35. 8)^{2}}{40}
= \frac{(75.8+ 35. 8)(75.8- 35.8)}{40}
= 111.6 * \frac{40}{40}
= 111.6

Explanation:

1^{ 3} + 2^{ 3} + ..... + 10^{ 3} = 3025
\therefore 2^{ 3} + 4^{ 3} + ...... + 20^{ 3}
= 2^{ 3} (1 + 2^{ 3} + ...... + 10^{ 3} )
= 8 * 3025 = 24200

Explanation:

(d^{ s + t} ÷ d^{ s} ) ÷ d^{ t}
= d ^{s + t – s} ÷ d ^{t}
= d ^{t} ÷ d ^{t} = 1

Explanation:

(2^{51} + 2^{52} + 2^{53} + 2^{54} + 2^{55})
= 2^{51} (1 + 2 + 22 + 23 + 24 )
= 2^{51} (1 + 2 + 4 + 8 + 16)
= 2^{51}\times 31
= 2^{49} \times 4 \times31
= 2^{49}\times 124