MATHS IN EVERYDAY LIFE Read the situation. Explain your choice.
4 CHAPTERS FREE40 original questions · worked solutions
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AMIT SIR · ORIGINAL DIGITAL PRACTICE
Class 10 Maths Competency-Based Practice
Apply Maths to real situations. Explore four free chapters of modelling, evidence-based reasoning and case-based practice, then check your explanation against the worked solution.
4free chapters40original questionsStep by stepworked solutions
Original competency-based practice by Amit Sir. Questions focus on modelling, interpreting evidence and justifying conclusions. Marks are suggested for practice, not an official CBSE marking scheme. This preview includes 10 selected questions in each of the four free chapters.
Choose HCF or LCM from the situation, test a claim and justify your decision.
Keep in mind: A model must match the condition: equal maximum grouping uses HCF; next common recurrence uses LCM.
10 questions · All topics · Free access, no purchase required
Grouping and designCase-based · 3 marks · practice
Q1. A lab has 96 red and 120 blue LEDs. Every kit must have the same red count and the same blue count, with no leftovers. A volunteer proposes 30 kits. (a) Explain why 30 is impossible. (b) Find the maximum possible number of kits and each kit’s contents.
Attempt on paper first. Include your reasoning and check the units or restrictions.
View answer & worked solution
30 is impossible; maximum 24 kits, each with 4 red and 5 blue LEDs.
96/30 and 120/30 must both be integers; 96/30 is not.
The greatest permitted kit count is HCF(96,120) = 24.
Divide each colour count by 24: 4 red and 5 blue.
Check: 24×4 = 96 and 24×5 = 120.
Common mistake: The number of kits must divide every colour quantity, not their sum.
SchedulingMCQ · 1 mark · practice
Q2. Two cleaning robots return to their docks every 16 and 24 minutes. Both dock at 10:00 a.m. Which is their next common docking time?
A. 10:08 a.m.
B. 10:40 a.m.
C. 10:48 a.m.
D. 11:04 a.m.
Attempt on paper first. Include your reasoning and check the units or restrictions.
View answer & worked solution
C. 10:48 a.m.
The next common return interval is LCM(16,24).
16 = 2⁴ and 24 = 2³×3, so LCM = 48 minutes.
Add 48 minutes to 10:00 a.m.
Common mistake: Adding the periods or taking their HCF does not give the next simultaneous return.
Testing a claimAssertion–Reason · 2 marks · practice
Q3. Assertion (A): No positive integer power of 14 ends in 0. Reason (R): A number ending in 0 has both 2 and 5 in its prime factorisation. Decide whether A and R are true and whether R explains A.
Attempt on paper first. Include your reasoning and check the units or restrictions.
View answer & worked solution
A and R are true, and R correctly explains A.
14ⁿ = 2ⁿ×7ⁿ has no factor 5.
A number ending in 0 is divisible by 10 = 2×5, so R is true.
The missing factor 5 makes A true, and R explains why.
Common mistake: Check the logical connection; two true statements need not always explain each other.
Checking dataReasoning · 3 marks · practice
Q4. A database records two positive integers with HCF 16, LCM 240 and one integer equal to 80. Is the record possible? Determine the other integer and verify the record.
Attempt on paper first. Include your reasoning and check the units or restrictions.
View answer & worked solution
Yes; the other integer is 48.
Use ab = HCF×LCM: 80b = 16×240, so b = 48.
80 = 2⁴×5 and 48 = 2⁴×3.
Their HCF is 16 and their LCM is 2⁴×3×5 = 240.
Both conditions are satisfied, so the record is possible.
Common mistake: A product match alone does not verify the HCF and LCM.
Tiling decisionsCase-based · 3 marks · practice
Q5. A designer needs the largest square tile that covers a 315 cm by 225 cm board exactly, without cutting. A supplier offers 30 cm tiles. Decide whether that option works, then determine the best tile size and number needed.
Attempt on paper first. Include your reasoning and check the units or restrictions.
View answer & worked solution
30 cm tiles do not work; use 45 cm tiles, 35 tiles in total.
315/30 and 225/30 are not integers, so the offer does not fit exactly.
The largest square side is HCF(315,225) = 45 cm.
The board fits 7 tiles along one side and 5 along the other.
Total = 7×5 = 35 tiles.
Common mistake: A tile with smaller area can still fail to divide the board’s lengths.
Prime powersMCQ · 1 mark · practice
Q6. A machine accepts only a positive integer that is divisible by both 8 and 18 and is a perfect square. What is the smallest valid input?
A. 72
B. 144
C. 288
D. 576
Attempt on paper first. Include your reasoning and check the units or restrictions.
View answer & worked solution
B. 144
LCM(8,18) = 2³×3² = 72.
To make all prime exponents even, multiply by 2.
144 = 12² is the smallest valid input.
Common mistake: The smallest common multiple need not be a perfect square.
Irrationality in measurementReasoning · 3 marks · practice
Q7. A square sign has area 5 m². A contractor claims its exact side length is 2.236 m. Explain the difference between an exact value and an approximation, and prove the exact side is irrational.
Attempt on paper first. Include your reasoning and check the units or restrictions.
View answer & worked solution
Exact side √5 m; 2.236 m is an approximation.
The positive side length s satisfies s² = 5, so s = √5.
Suppose √5 = p/q in lowest terms. Then p² = 5q², so 5 divides p; write p = 5k.
Also 2.236² = 4.999696, not 5; therefore 2.236 is not exact.
Common mistake: A calculator display with finitely many digits is not the exact irrational value.
Grouping constraintsCase-based · 3 marks · practice
Q8. A library has 84 story books and 126 science books. It forms identical mixed boxes with no books left. Each box must contain at least 4 books. Find the largest possible number of boxes and the contents of each.
Attempt on paper first. Include your reasoning and check the units or restrictions.
View answer & worked solution
42 boxes, each containing 2 story books and 3 science books.
The possible box counts are common divisors of 84 and 126.
The maximum is HCF(84,126) = 42.
Each box has 84/42 = 2 story and 126/42 = 3 science books.
That is 5 books per box, satisfying the minimum of 4.
Common mistake: After using HCF, still check the extra condition about the contents.
Divisibility logicMCQ · 1 mark · practice
Q9. Two positive integers have HCF 15. Which proposed LCM is impossible?
A. 45
B. 90
C. 100
D. 150
Attempt on paper first. Include your reasoning and check the units or restrictions.
View answer & worked solution
C. 100
The HCF must divide the LCM.
45, 90 and 150 are multiples of 15; 100 is not.
Hence 100 cannot be the LCM of such a pair.
Common mistake: A larger number is not automatically a possible LCM.
Synchronized cyclesCase-based · 4 marks · practice
Q10. A garden sprinkler runs every 20 minutes and a misting system every 28 minutes. Both start at 8:00 a.m. (a) Find their next common start. (b) How many common starts occur from 8:00 a.m. through 5:20 p.m., including both endpoints?
Attempt on paper first. Include your reasoning and check the units or restrictions.
View answer & worked solution
Next common start: 10:20 a.m.; 5 common starts in the given interval.
LCM(20,28) = 2²×5×7 = 140 minutes.
8:00 a.m. plus 140 minutes is 10:20 a.m.
The interval to 5:20 p.m. is 560 minutes = 4×140.
Common starts are at elapsed minutes 0,140,280,420,560: five starts.
Common mistake: The number of events includes the starting event; it is one more than the number of complete intervals.
What comes after the free preview?
The full book is planned to include the remaining chapters listed below. These are outside this free preview; join Notify me above for publication updates.
Arithmetic ProgressionsTrianglesCoordinate GeometryIntroduction to TrigonometryApplications of TrigonometryCirclesAreas Related to CirclesSurface Areas and VolumesStatisticsProbability
