On simplification, \(\frac{(625)^{6.25}\times (\sqrt5)^{10.4}}{(\sqrt5)^{54}\times (5)^{1.2}\times (25)^{0.5}}\) reduces to:

This question was previously asked in
DSSSB PGT Chemistry (Female) Official Paper (Held On: 06 Jul, 2018 Shift 1)
View all DSSSB PGT Papers >
  1. √5
  2. 5
  3. 5√5
  4. 25

Answer (Detailed Solution Below)

Option 2 : 5
Free
DSSSB PGT Hindi Full Test 1
1.9 K Users
300 Questions 300 Marks 180 Mins

Detailed Solution

Download Solution PDF

Given:

Expression: \( \frac{(625)^{6.25} × (\sqrt5)^{10.4}}{(\sqrt5)^{54} × 5^{1.2} × (25)^{0.5}} \)

Formula used:

Power of a power: \((a^m)^n = a^{m×n}\)

Roots as fractional exponents: \(\sqrt5=5^{1/2},\;25=5^2\)

Calculations:

\(625=5^4\)

\((5^4)^{6.25}=5^{4×6.25}=5^{25}\)

\((\sqrt5)^{10.4}=(5^{1/2})^{10.4}=5^{10.4/2}=5^{5.2}\)

⇒ Numerator = \(5^{25}×5^{5.2}=5^{30.2}\)

\((\sqrt5)^{54}=5^{54/2}=5^{27}\)

\((25)^{0.5}=(5^2)^{0.5}=5^{2×0.5}=5^1\)

⇒ Denominator exponent = 27 + 1.2 + 1 = 29.2 ⇒ Denominator = \(5^{29.2}\)

⇒ Fraction = \(5^{30.2−29.2}=5^1\) = 5

∴ The expression simplifies to 5.

Latest DSSSB PGT Updates

Last updated on Jul 15, 2025

-> The DSSSB PGT Application Form 2025 has been released. Apply online till 7 August.

-> The DSSSB PGT Notification 2025 has been released for 131 vacancies.

-> Candidates can apply for these vacancies between 8th Juy 2025 o 7th August 2025.

-> The DSSSB PGT Exam for posts under Advt. No. 05/2024 and 07/2023 will be scheduled between 7th to 25th July 2025.

-> The DSSSB PGT Recruitment is also ongoing for 432 vacancies of  Advt. No. 10/2024.

-> The selection process consists of a written examination and document verification..

-> Selected Candidates must refer to the DSSSB PGT Previous Year Papers and DSSSB PGT Mock Test to understand the trend of the questions.

Get Free Access Now
Hot Links: teen patti all app teen patti gold new version teen patti 51 bonus teen patti wealth teen patti - 3patti cards game downloadable content