power indices and surds Model Questions & Answers, Practice Test for ssc chsl tier 1 2024

Answer: (d)

We know that, $(42)^2 = 1764 and (43)^2$ = 1849

But 1780 lies between 1764 and 1849.

Now, smallest number = 1849 – 1780 = 69

Answer: (c)

x = 2 + $2^{2/3} + 2^{1/3} +⇒(x – 2) = (2^{2/3} + 2^{1/3})$ ...(i)

on cutting both sides, we get.

$(x - 2)^3 = (2^{1/3} + 2^{2/3})^3$

$x^3 - 8 - 6x^2 + 12x = 2 + 2^2 + 3.2^{1/2}.2^{2/3}(2^{1/3} + 2^{2/3})$.

$x^3 - 8 - 6x^2$ + 12x = 6 + 6(x – 2)

$x^3 –2 – 6x^2$ + 6x = 0

∴ $x^3 – 6x^2$ + 6x = 2

Answer: (d)

a = $xy^{P–1}⇒a^{q – r} = (xy^{P – 1}) = x^{(q – r)}.y^{(P – 1)(q – r)}$

b = $yz^{q–1}⇒b^{r–p} = (yz^(q–1))^(r–p) = y^{(r–P)}.z^{(q–1)(r–p)}$.

c = $zx^{r–1}⇒c^{p–q} = (zx^(r–1))^{P–q} = z^{(p–q)}.x^{(r–1)(p–q)}$.

Now, $a^{q–r}, b^{r–p}.c^{p–q} = (x^{(q–r)}.y^{(pq–pr–q+r)}).(y^{(r–p)}.z^{(qr– pq–r+p)}).(.z^{(p–q)}.x^{(rp –qr –p + q)})$

= $x^(q– r + rp – p + q– qr).y^(pq–pr–q + r + r – p).z^(qr –pq –r + P + P –q)$

Hence, correct option is (d)

Answer: (c)

Number in ascending orders are

$2222^2 < 222^{22} < 22^{222} < 2^{2222}$

Answer: (d)

2 + $√{2 + √{2 + √{2 + .....}}}$ = x

⇒$√{2 + √{2 + √{2 + ...}}}$ = x - 2

Squaring both side

2 + $√{2 + √{2 + √{2 + .....}}} = (x - 2)^2$

⇒x = $x^2$ – 4x + 4

⇒$x^2$ – 5x + 4 = 0

⇒$x^2$ – 4x – x + 4 = 0

⇒x (x – 4) – 1 (x + 4) = 0

⇒(x – 1) (x – 4) = 0

∴ x = 1, 4

As x always take greater than 2

∴ x = 4