Write 4 + 7/100 + 9/10000 as a decimal.

Align the place values: 4 + 0.07 + 0.0009 = 4.0709.

What is the difference between 0.(7), where 7 repeats, and 0.7?

The difference is 7/9 − 7/10 = 7/90.

If a = 0.85 and b = 0.15, evaluate (a² + b²)² − (a² − b²)².

Use (u + v)² − (u − v)² = 4uv with u = a² and v = b²: 4 × (0.85 × 0.15)² = 0.065025.

Which statement is true?

0.375 < 5/13 < 0.39. 5/13 is about 0.384615, so it lies between 0.375 and 0.39.

By what number must 7.35 be divided to obtain 0.0735?

The decimal point moves 2 places left, so the divisor is 100.

A parcel weighs 3.65 kg including packaging of 0.275 kg. What is the total mass of the contents of 8 identical parcels, in kg?

Contents per parcel = 3.65 − 0.275 = 3.375 kg; total = 3.375 × 8 = 27 kg.

Data sufficiency: How many litres of petrol did Aman buy? I. His bill was ₹971.25 after a reduction of ₹4.50 per litre from the displayed rate of ₹97 per litre. II. After he used 0.4 of the petrol bought, 6.3 litres remained.

I gives 971.25/(97 − 4.50) = 10.5 litres. II gives 6.3/(1 − 0.4) = 10.5 litres, so each statement alone is sufficient.

Evaluate 0.32 ÷ 0.008 × 0.15.

Perform division and multiplication from left to right: 0.32 ÷ 0.008 = 40, then 40 × 0.15 = 6.

A number becomes 0.6 times its original value. What is its percentage decrease?

Decrease = (1 − 0.6) × 100 = 40%.

If 0.001875 = p/q in lowest terms, find q − p.

0.001875 = 1875/1000000 = 3/1600, so q − p = 1600 − 3 = 1597.

If x = 0.12(3), where only 3 repeats, find the smallest positive integer k such that kx is an integer.

x = (123 − 12)/900 = 37/300. Since 37 and 300 are coprime, the smallest k is 300.

Find x: (7.84 − x) ÷ 0.16 = 28.5.

7.84 − x = 28.5 × 0.16 = 4.56, so x = 7.84 − 4.56 = 3.28.

Which number is greater than 0.4 and less than 0.41?

81/200 = 0.405, and 0.4 < 0.405 < 0.41.

Evaluate 4.8 × 1.25 − 2.4 × 0.75.

4.8 × 1.25 = 6 and 2.4 × 0.75 = 1.8, so the difference is 4.2.

Rounded to the nearest tenth, what is (39.8 × 0.602) ÷ 1.21?

The quotient is about 19.8013, so rounding to the nearest tenth gives 19.8.

Two couriers charge ₹42.75 and ₹51.15 for the first kg. Their charges rise by ₹3.60 and ₹2.40 respectively for every additional kg. At what total mass, in kg, are their charges equal?

The initial gap is 51.15 − 42.75 = 8.4 and closes by 3.60 − 2.40 = 1.2 per extra kg. Extra mass = 8.4/1.2 = 7 kg, so total mass = 1 + 7 = 8 kg.

How many digits are in the shortest repeating block of the decimal expansion of 1/11?

1/11 = 0.(09), so the shortest repeating block is 09 and contains 2 digits.

A fraction has value 0.48. Its numerator is increased by 25% and its denominator by 20%. What is the new value?

New value = 0.48 × 1.25/1.20 = 0.48 × 25/24 = 0.5.

If a/b = 0.6 and b is nonzero, find (5a − 2b) ÷ (3a + b).

Divide numerator and denominator by b: (5 × 0.6 − 2)/(3 × 0.6 + 1) = 1/2.8 = 5/14.

Which fraction has the greatest shortfall from 1?

The shortfalls from 1 are 1/18, 1/24, 1/32 and 1/40 for denominators 18, 24, 32 and 40. The greatest is 1/18, so the fraction is 17/18.

What is 5 − 0.0007?

Write 5 as 5.0000, then subtract: 5.0000 − 0.0007 = 4.9993.

A student multiplies a positive number by 0.04 instead of dividing it by 0.04. The incorrect result is what fraction of the correct result?

Incorrect/correct = (0.04x)/(x/0.04) = 0.04² = 0.0016 = 1/625.

Data sufficiency: What is the value of a positive decimal x? I. x has exactly two digits after the decimal point, both nonzero and 0.4 < x < 0.5. II. In lowest terms, x = p/20 for a positive integer p coprime to 20, and 0.4 < x < 0.5.

Statement II alone is sufficient; I alone is not.. Statement I permits values such as 0.41 and 0.42. Statement II gives 8 < p < 10, so p = 9 and x = 9/20 = 0.45; II alone is sufficient.

Evaluate (2.4² + 0.7² − 2 × 2.4 × 0.7) ÷ 0.17.

The numerator is (2.4 − 0.7)² = 1.7² = 2.89; then 2.89 ÷ 0.17 = 17.

Evaluate 1/(0.2 × 0.3) + 1/(0.3 × 0.4) + 1/(0.4 × 0.5).

Use 1/[a(a + 0.1)] = 10[1/a − 1/(a + 0.1)]. The sum telescopes to 10(1/0.2 − 1/0.5) = 10(5 − 2) = 30.

Topic

Decimal Fractions MCQ – Part 4 - Topic Guide and MCQs

Q: How long is the Decimal Fractions MCQ – Part 4 quiz?

Decimal Fractions MCQ – Part 4 runs 25 untimed questions - about 25 min at a steady pace. The set clears at 40% and you can retake it as often as you like.

Q: What kind of questions does Decimal Fractions MCQ – Part 4 ask?

The set answers that: each question carries four options and a full explanation, so you learn the wording an examiner uses, not just the fact behind it.

Q: Where else can I practise Quantitative Aptitude?

Open the quiz page above, or start from the Quantitative Aptitude subject page if you want to drill neighbouring lanes in the same sitting.

Q: Do my scores save?

Scores, bookmarks and streaks need the optional Google sign-in. Opening the set and answering questions does not.

Part of Quantitative Aptitude > 03 Decimal Fractions - 25 questions, about 25 min, with an explanation after every answer.

Question count

about 25 min

Estimated time

Mixed

Difficulty

03 Nov 2026

Last updated

Quick answer

Decimal Fractions MCQ – Part 4 is a topic inside Quantitative Aptitude (03 Decimal Fractions). It runs 25 untimed MCQs - roughly 25 min - and the review explains every answer once you submit.

Topic introduction

What Decimal Fractions MCQ – Part 4 covers

The shape of the set before you start it.

Decimal Fractions MCQ – Part 4 sits inside Quantitative Aptitude, under the 03 Decimal Fractions lane, and the portal keeps it as one self-contained set: 25 questions, four options each, no clock and a written explanation behind every answer.

Attempt it once to find the gaps, read why each distractor is wrong, then retake the same set a few days later. The questions you miss twice are the ones worth writing down.

Curated by Gurpreet Singh · Punjab Police SI, Rank 2 (713/800) · Founder, House of Aspirants · Read his story →

Question bank

Decimal Fractions MCQ – Part 4: all 25 questions

Read the question, pick an option, then open the answer.

Write 4 + 7/100 + 9/10000 as a decimal.
1. 4.079
2. 4.0709
3. 4.0079
4. 4.709
Show answer

Answer: (B) 4.0709

Explanation: Align the place values: 4 + 0.07 + 0.0009 = 4.0709.
What is the difference between 0.(7), where 7 repeats, and 0.7?
1. 1/90
2. 7/9
3. 7/100
4. 7/90
Show answer

Answer: (D) 7/90

Explanation: The difference is 7/9 − 7/10 = 7/90.
If a = 0.85 and b = 0.15, evaluate (a² + b²)² − (a² − b²)².
1. 0.2601
2. 0.0325125
3. 0.065025
4. 0.65025
Show answer

Answer: (C) 0.065025

Explanation: Use (u + v)² − (u − v)² = 4uv with u = a² and v = b²: 4 × (0.85 × 0.15)² = 0.065025.
Which statement is true?
1. 0.375 < 5/13 < 0.39
2. 0.39 < 5/13 < 0.375
3. 0.375 < 0.39 < 5/13
4. 5/13 < 0.375 < 0.39
Show answer

Answer: (A) 0.375 < 5/13 < 0.39

Explanation: 5/13 is about 0.384615, so it lies between 0.375 and 0.39.
By what number must 7.35 be divided to obtain 0.0735?
1. 0.01
2. 10
3. 1000
4. 100
Show answer

Answer: (D) 100

Explanation: The decimal point moves 2 places left, so the divisor is 100.
A parcel weighs 3.65 kg including packaging of 0.275 kg. What is the total mass of the contents of 8 identical parcels, in kg?
1. 29.2
2. 27
3. 26.8
4. 31.4
Show answer

Answer: (B) 27

Explanation: Contents per parcel = 3.65 − 0.275 = 3.375 kg; total = 3.375 × 8 = 27 kg.
Data sufficiency: How many litres of petrol did Aman buy? I. His bill was ₹971.25 after a reduction of ₹4.50 per litre from the displayed rate of ₹97 per litre. II. After he used 0.4 of the petrol bought, 6.3 litres remained.
1. Each statement alone is sufficient.
2. Statement II alone is sufficient; I alone is not.
3. Both statements together are sufficient, but neither alone is sufficient.
4. Statement I alone is sufficient; II alone is not.
Show answer

Answer: (A) Each statement alone is sufficient.

Explanation: I gives 971.25/(97 − 4.50) = 10.5 litres. II gives 6.3/(1 − 0.4) = 10.5 litres, so each statement alone is sufficient.
Evaluate 0.32 ÷ 0.008 × 0.15.
1. 60
2. 800/3
3. 6
4. 0.6
Show answer

Answer: (C) 6

Explanation: Perform division and multiplication from left to right: 0.32 ÷ 0.008 = 40, then 40 × 0.15 = 6.
A number becomes 0.6 times its original value. What is its percentage decrease?
1. 60
2. 6
3. 40
4. 4
Show answer

Answer: (C) 40

Explanation: Decrease = (1 − 0.6) × 100 = 40%.
If 0.001875 = p/q in lowest terms, find q − p.
1. 1597
2. 797
3. 1599
4. 1603
Show answer

Answer: (A) 1597

Explanation: 0.001875 = 1875/1000000 = 3/1600, so q − p = 1600 − 3 = 1597.
If x = 0.12(3), where only 3 repeats, find the smallest positive integer k such that kx is an integer.
1. 30
2. 300
3. 900
4. 1000
Show answer

Answer: (B) 300

Explanation: x = (123 − 12)/900 = 37/300. Since 37 and 300 are coprime, the smallest k is 300.
Find x: (7.84 − x) ÷ 0.16 = 28.5.
1. 4.56
2. 12.4
3. 3.38
4. 3.28
Show answer

Answer: (D) 3.28

Explanation: 7.84 − x = 28.5 × 0.16 = 4.56, so x = 7.84 − 4.56 = 3.28.
Which number is greater than 0.4 and less than 0.41?
1. 81/200
2. 41/100
3. 9/20
4. 2/5
Show answer

Answer: (A) 81/200

Explanation: 81/200 = 0.405, and 0.4 < 0.405 < 0.41.
Evaluate 4.8 × 1.25 − 2.4 × 0.75.
1. 0.42
2. 7.8
3. 4.2
4. 3.6
Show answer

Answer: (C) 4.2

Explanation: 4.8 × 1.25 = 6 and 2.4 × 0.75 = 1.8, so the difference is 4.2.
Rounded to the nearest tenth, what is (39.8 × 0.602) ÷ 1.21?
1. 20.0
2. 1.98
3. 198.0
4. 19.8
Show answer

Answer: (D) 19.8

Explanation: The quotient is about 19.8013, so rounding to the nearest tenth gives 19.8.
Two couriers charge ₹42.75 and ₹51.15 for the first kg. Their charges rise by ₹3.60 and ₹2.40 respectively for every additional kg. At what total mass, in kg, are their charges equal?
1. 6
2. 8
3. 9
4. 7
Show answer

Answer: (B) 8

Explanation: The initial gap is 51.15 − 42.75 = 8.4 and closes by 3.60 − 2.40 = 1.2 per extra kg. Extra mass = 8.4/1.2 = 7 kg, so total mass = 1 + 7 = 8 kg.
How many digits are in the shortest repeating block of the decimal expansion of 1/11?
1. 4
2. 2
3. 1
4. 3
Show answer

Answer: (B) 2

Explanation: 1/11 = 0.(09), so the shortest repeating block is 09 and contains 2 digits.
A fraction has value 0.48. Its numerator is increased by 25% and its denominator by 20%. What is the new value?
1. 0.576
2. 0.6
3. 0.46
4. 0.5
Show answer

Answer: (D) 0.5

Explanation: New value = 0.48 × 1.25/1.20 = 0.48 × 25/24 = 0.5.
If a/b = 0.6 and b is nonzero, find (5a − 2b) ÷ (3a + b).
1. 7/5
2. 5/4
3. 5/14
4. 1/4
Show answer

Answer: (C) 5/14

Explanation: Divide numerator and denominator by b: (5 × 0.6 − 2)/(3 × 0.6 + 1) = 1/2.8 = 5/14.
Which fraction has the greatest shortfall from 1?
1. 17/18
2. 39/40
3. 23/24
4. 31/32
Show answer

Answer: (A) 17/18

Explanation: The shortfalls from 1 are 1/18, 1/24, 1/32 and 1/40 for denominators 18, 24, 32 and 40. The greatest is 1/18, so the fraction is 17/18.
What is 5 − 0.0007?
1. 4.993
2. 4.9997
3. 4.99993
4. 4.9993
Show answer

Answer: (D) 4.9993

Explanation: Write 5 as 5.0000, then subtract: 5.0000 − 0.0007 = 4.9993.
A student multiplies a positive number by 0.04 instead of dividing it by 0.04. The incorrect result is what fraction of the correct result?
1. 1/250
2. 1/625
3. 1/2500
4. 1/25
Show answer

Answer: (B) 1/625

Explanation: Incorrect/correct = (0.04x)/(x/0.04) = 0.04² = 0.0016 = 1/625.
Data sufficiency: What is the value of a positive decimal x? I. x has exactly two digits after the decimal point, both nonzero and 0.4 < x < 0.5. II. In lowest terms, x = p/20 for a positive integer p coprime to 20, and 0.4 < x < 0.5.
1. Statement II alone is sufficient; I alone is not.
2. Each statement alone is sufficient.
3. Statement I alone is sufficient; II alone is not.
4. Both statements together are sufficient, but neither alone is sufficient.
Show answer

Answer: (A) Statement II alone is sufficient; I alone is not.

Explanation: Statement I permits values such as 0.41 and 0.42. Statement II gives 8 < p < 10, so p = 9 and x = 9/20 = 0.45; II alone is sufficient.
Evaluate (2.4² + 0.7² − 2 × 2.4 × 0.7) ÷ 0.17.
1. 1.7
2. 10
3. 17
4. 170
Show answer

Answer: (C) 17

Explanation: The numerator is (2.4 − 0.7)² = 1.7² = 2.89; then 2.89 ÷ 0.17 = 17.
Evaluate 1/(0.2 × 0.3) + 1/(0.3 × 0.4) + 1/(0.4 × 0.5).
1. 300
2. 20
3. 3
4. 30
Show answer

Answer: (D) 30

Explanation: Use 1/[a(a + 0.1)] = 10[1/a − 1/(a + 0.1)]. The sum telescopes to 10(1/0.2 − 1/0.5) = 10(5 − 2) = 30.

Why it matters

Why Decimal Fractions MCQ – Part 4 is worth the hours

Where the marks actually come from.

Decimal Fractions MCQ – Part 4 is factual rather than analytical: you either recall it in seconds or you lose the mark to somebody who did. That makes it one of the cheapest sections to convert - a handful of short sittings usually settles the whole cluster.

Exam relevance

Exams where Decimal Fractions MCQ – Part 4 matters

Derived from the subject and lane this topic belongs to.

Punjab Police Constable

The highest-volume recruitment on this list. Build your base with Punjab GK and current affairs, then drill reasoning until speed is automatic.

Punjab Police ASI

Assistant Sub-Inspector papers separate candidates on English and Punjabi accuracy as much as on general awareness. Practise language daily.

Punjab Police Sub-Inspector (SI)

The most competitive paper here. Polity, history and quant reasoning decide the rank - so take full timed mocks before the exam.

Punjab Police Intelligence Assistant

Data handling, mental ability and computer awareness carry most of the weight. Use short topic sets and check every explanation.

Punjab Police Jail Warder

A tighter syllabus: general awareness, Punjabi and basic reasoning. Perfect for a daily quiz habit - short sessions, high consistency.

Patwari

Revenue work starts with Punjab geography, history and current affairs. Add quant for the sections that reward calculation speed.

Excise Inspector

General awareness and English carry this paper. Track Punjab policy news daily and revise it as MCQs, not as notes.

Naib Tehsildar

Land records, Punjab history and polity, plus data interpretation. A strong polity base is the fastest score to build.

Clerk

Grammar, comprehension, typing and basic computer knowledge. Accuracy beats speed here - review every wrong answer.

PSPCL jobs

Recruitment by the Punjab State Power Corporation Limited. Technical aptitude sits alongside general awareness and reasoning.

Punjab Police Head Constable

The communication and clerical tier of Punjab Police. Punjabi, English and general knowledge carry the paper, and computer awareness is worth real marks.

Staff Selection Commission

The central recruitment ladder most aspirants attempt alongside a state paper. Arithmetic, reasoning and English decide the score; general awareness decides the cutoff.

Railway Recruitment

A high-volume, high-speed paper. Arithmetic, reasoning and general awareness asked at a pace that punishes anyone still working out method during the attempt.

Banking Exams

Quantitative aptitude and reasoning at speed, with a separate English section and a general awareness paper that moves with the financial news cycle.

Expected questions

Expected questions from Decimal Fractions MCQ – Part 4

What an examiner usually reaches for in this cluster.

Expect one-line recall questions - definitions, stand-for expansions, identifiers and the ordering of things. Practise the wording as well as the concept: these papers test terminology, and the set above uses the same four-option shape they do.

Daily habit

Keep the Expected MCQs habit running

20 fresh mixed questions a day, plus the current-affairs sets examiners keep reusing.

Daily Quiz Expected MCQs

Previous year questions

Previous Year Questions

How this topic behaves in real papers.

Past papers reuse clusters like this almost verbatim across years and posts. Run the same set until every distractor is familiar - previous-year style questions stop being recall and start being free marks.

Where the previous-year sets live

The portal's previous-year practice runs through the General Knowledge lane and the daily mixed quiz.

Previous Year Questions Daily Quiz

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Take the Decimal Fractions MCQ – Part 4 quiz

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Questions

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