surds and indices Model Questions & Answers, Practice Test for ibps po prelims 2023
ibps po prelims 2023 SYLLABUS WISE SUBJECTS MCQs
Number System
Simplification
Surds & Indices
Ratio & Proportion
Profit & Loss
Time & Work
Simple Interest & Compound Interest
Mensuration: Area & Volumes
$(0.125)^3 ÷ (0.25 )^2 × (0.5)^2 = (0.5)^{?-3}$
Answer: (d)
$(0.125)^3 ÷ (0.25)^2 × (0.5)^2 = (0.5)^{? – 3}$
⇒$(0.5)^{3 × 3} ÷ (0.5)^{2 × 2} × (0.5)^2 = (0.5)^{? – 3}$
⇒$(0.5)^9 ÷ (0.5)^4 × (0.5)^2 = (0.5)^{? – 3}$
⇒$(0.5)^{9 – 4 + 2} = (0.5)^{? – 3}$⇒9 – 4 + 2 = ? – 3
∴? = 10
Simplify : $[5(8^{1/3}+27^{1/3})^3]^{1/4}$
Answer: (c)
$[5(8^{1/3}+27^{1/3})^3]^{1/4}=[5((2^3)^{1/3}+(3^3)^{1/3})^3]^{1/4}$
$[5(2+3)^3]^{1/4}=[5(5)^3]^{1/4}$
$[5^4]^{1/4}=5$
$17^{3.5} × 17^{7.3} ÷ 17^{4.2} = 17^{?}$
Answer: (c)
${17^{3.5}×17^{7.3}}/{17^{4.2}} = 17^{?}⇒17^{?} = 17^{3.5 + 7.3 - 4.2} = 17^{6.6}$
∴ ? = 6.6
If $27^k$ = , $9/3^k$,then value of $1/k^2$ is
Answer: (a)
If $27^k=9/{3^k}$
$⇒3^{3k}=9/{3^k}⇒3^{4k}=9 [a^m× an=a^{m+n}]$
$⇒9^{2k}=9⇒k=1/2 [a^m=a^n then m=n]$
$ ⇒1/k^2=4$
When simplified the product $(1+1/2) (1+1/3) (1+1/4)….. (1+1/n)$ becomes
Answer: (c)
$(1+1/2) (1+1/3) (1+1/4)…… (1+1/n)$
$3/2×4/3×5/4×.......×{n+1}/n$
$={n+1}/2$
ibps po prelims 2023 IMPORTANT QUESTION AND ANSWERS
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