1. Identifying HCF and LCM in Word Problems
Application questions usually do not directly say “find the HCF” or “find the LCM”. The first task is to understand whether the problem requires the greatest common division or the earliest/smallest common multiple.
Think HCF when the problem asks for:
• Greatest possible equal measurement
• Maximum length of identical pieces
• Largest square tile
• Maximum number of identical groups
• Greatest divisor satisfying given conditions
Think LCM when the problem asks for:
• Least number divisible by several numbers
• Earliest time several events occur together
• Bells, lights or machines repeating simultaneously
• Minimum common quantity
• Smallest number satisfying several divisibility conditions
Exam Tip:
Do not decide only from the words “greatest” and “least”. Identify what quantity is being divided, grouped or repeated.
2. Greatest Measurement, Cutting and Tiling Problems
When different lengths, dimensions or quantities have to be divided into the largest possible equal units without any remainder, the required value is generally their HCF.
Core Rule:
Greatest possible equal measure = HCF of the given measurements.
Example — Greatest Measuring Length:
Find the greatest length that can exactly measure 4.2 m, 6.3 m and 8.4 m.
Convert into centimetres:
420 cm, 630 cm, 840 cm.
HCF=210 cm.
Therefore the greatest measuring length is:
210 cm = 2.1 m.
Example — Cutting into Largest Equal Pieces:
Three ropes of lengths 84 m, 126 m and 210 m are to be cut into equal pieces of maximum possible length.
Required length=HCF(84,126,210)
=42 m.
Largest Square Tile:
For a rectangular floor with integral dimensions, the side of the largest square tile that fits exactly is the HCF of the length and breadth, after converting both to the same unit.
Example — Square Tiles:
A rectangular floor is 8.4 m × 6.3 m.
Convert:
840 cm and 630 cm.
Largest square tile side:
HCF(840,630)=210 cm.
Number of tiles:
(840÷210)×(630÷210)
=4×3
=12 tiles.
Common Mistake:
For the largest square tile, do not take the LCM of the dimensions. The side must divide both dimensions exactly, so HCF is required.
3. Maximum Equal Grouping and Distribution
When different quantities must be distributed into the maximum number of identical groups with nothing left over, the number of groups is generally the HCF of the quantities.
Core Rule:
Maximum number of identical groups = HCF of the given quantities.
Example:
48 red balls, 72 blue balls and 120 green balls are to be packed into the maximum possible number of identical packets.
Number of packets:
HCF(48,72,120)=24.
Each packet contains:
48÷24=2 red balls
72÷24=3 blue balls
120÷24=5 green balls.
Therefore:
24 identical packets.
Exam Tip:
If the question asks for the maximum number of identical groups, first find the HCF. Then divide each quantity by the HCF to find the contents of each group.
4. Repeated Events, Bells, Lights and Machines
When several events repeat after fixed intervals and we need to know when they will occur together again, take the LCM of their intervals.
Core Rule:
Time after which all events occur together again = LCM of their intervals.
Example:
Three bells ring every 12 sec, 18 sec and 30 sec.
LCM(12,18,30)=180 sec.
Therefore they ring together again after:
180 sec = 3 minutes.
Applies To:
Bells, traffic lights, alarms, machines, rotating wheels, flashing lights, periodic signals and other repeating events.
Exam Trap:
If events have already started together, “when will they next occur together?” asks only for the LCM interval. Do not add the starting instant.
5. Counting How Many Times Events Occur Together
A different question asks how many times repeating events occur together during a given period.
If they start together at time 0:
Common interval = LCM of the individual intervals.
For a time period T that includes both the starting instant and the ending instant:
Number of common occurrences = floor(T ÷ LCM) + 1
Example:
Three bells ring at intervals of 8 sec, 10 sec and 12 sec and start together.
LCM=120 sec.
In 30 minutes:
30 min=1800 sec.
1800÷120=15.
Including the initial ringing at time 0:
15+1=16 times.
Common Mistake:
When the starting instant is included, forgetting the first simultaneous occurrence causes an answer smaller by 1.
Note:
Read the wording carefully. If the question counts only occurrences after the start, do not automatically add 1.
6. Least Number Divisible by Several Numbers
The smallest positive number exactly divisible by several given positive integers is their LCM.
Core Rule:
Least positive number exactly divisible by a, b, c, ... = LCM(a,b,c,...)
Example:
Find the least number divisible by 18, 24 and 30.
LCM(18,24,30)=360.
Exam Tip:
If an extra condition such as “perfect square”, “perfect cube” or “divisible by another number” is added, first obtain the basic LCM and then satisfy the additional condition.
7. Least Number to Add or Subtract for Exact Divisibility
If a given number must be made divisible by several divisors, first find their LCM.
Let:
L = LCM of the required divisors
N = given number
r = remainder when N is divided by L
Least number to subtract:
r
Least number to add:
L−r, if r≠0.
If r=0, nothing needs to be added.
Example:
What least number should be added to 758 so that the result is divisible by 8, 12 and 18?
LCM(8,12,18)=72.
758=72×10+38.
Required addition:
72−38=34.
Related Form:
If the least number N becomes exactly divisible by several divisors after adding x, then:
N+x = a common multiple.
For the basic least positive solution, usually start from their LCM:
N = LCM−x,
provided this satisfies the wording and positivity requirements.
8. Remainder-Based HCF Applications
Many competitive-exam HCF questions ask for the greatest divisor that leaves specified remainders. The essential idea is to remove the remainders first.
Same Known Remainder r:
Greatest divisor of numbers a, b, c leaving remainder r in each case:
HCF(a−r, b−r, c−r)
Example:
Find the greatest number that divides 1780 and 2452 leaving remainder 4 in each case.
1780−4=1776
2452−4=2448
HCF(1776,2448)=48.
Same Unknown Remainder:
If the same remainder is left but its value is not given, take the HCF of the pairwise differences.
Example:
Find the greatest number that divides 391, 527 and 731 leaving the same remainder.
Differences:
527−391=136
731−527=204
731−391=340
HCF(136,204,340)=68.
Different Remainders:
If a divisor leaves remainders r₁, r₂, r₃ from numbers a, b, c respectively, calculate:
HCF(a−r₁, b−r₂, c−r₃)
Necessary Check:
The required divisor must be greater than every remainder given in the question.
Common Mistake:
Do not take the HCF of the original numbers when non-zero specified remainders are given. Subtract the remainders first.
9. Remainder-Based LCM Applications
LCM is used when a number leaves the same remainder on division by several divisors.
General Form:
If N leaves remainder r when divided by a, b and c, then:
N−r is divisible by all of them.
Therefore:
N = k × LCM(a,b,c) + r
Important Mathematical Note:
If no lower-bound or extra condition is given, the remainder r itself can technically satisfy the congruence when r is smaller than all divisors. Competitive-exam questions normally include wording or an additional condition that determines the intended larger value.
Example:
A number leaves remainder 3 when divided by 8, 16, 18, 20 and 25.
LCM(8,16,18,20,25)=3600.
Therefore all such numbers have the form:
3600k+3.
Exam Strategy:
If an additional condition says the required number must also be divisible by another number, test values of k or solve the resulting divisibility condition.
10. Unit Conversion in HCF-LCM Applications
Measurements must be expressed in the same unit before taking HCF or LCM.
Examples of Required Conversion:
metre → centimetre
hour → minute
minute → second
kilogram → gram
Example:
4 m 90 cm = 490 cm
3 m 85 cm = 385 cm
Only after conversion should HCF be calculated.
Exam Trap:
Never calculate the HCF of 4.90 m and 385 cm as if the numerical values 4.90 and 385 were in the same unit.
11. Fast Decision Strategy
| Question Pattern | Use |
|---|
| Greatest equal measure or piece | HCF |
| Largest square tile | HCF |
| Maximum identical groups | HCF |
| Greatest divisor leaving remainders | HCF after remainder adjustment |
| Least exactly divisible number | LCM |
| Events occurring together again | LCM of intervals |
| Least addition/subtraction for common divisibility | LCM + remainder method |
| Same remainder on division by several numbers | Number of form k×LCM+r |
Exam Tip:
Translate the story into a divisibility statement before beginning arithmetic. This prevents most HCF-LCM application errors.
12. Common Exam Traps
Trap 1:
Using LCM when the largest equal measure or piece is required.
Trap 2:
Using HCF when the earliest repeated common event is required.
Trap 3:
Forgetting to convert all measurements to the same unit.
Trap 4:
For a “times together” question, forgetting whether the starting occurrence must be counted.
Trap 5:
Subtracting the same remainder from only one number instead of every relevant number.
Trap 6:
In different-remainder questions, subtracting one common remainder instead of each specified remainder.
Trap 7:
Accepting a divisor that is smaller than or equal to one of the stated remainders.
Trap 8:
Assuming LCM+r is always automatically the least mathematical solution without checking the wording and additional conditions.
13. Quick Revision
Remember:
• Greatest equal measure → HCF.
• Maximum equal piece length → HCF.
• Largest square tile side → HCF of dimensions.
• Maximum identical groups → HCF of quantities.
• Events together again → LCM of intervals.
• Least exactly divisible number → LCM.
• If events start together, count the initial occurrence when the question includes it.
• For divisibility adjustment, first find the LCM.
• Least subtraction from N = remainder of N÷LCM.
• Least addition = LCM−remainder when remainder is non-zero.
• Greatest divisor leaving known remainder r → HCF of numbers after subtracting r.
• Same unknown remainder → HCF of pairwise differences.
• Different remainders → subtract the corresponding remainder from each number before taking HCF.
• Same-remainder LCM problems have the form k×LCM+r.
• Always convert measurements into a common unit before calculation.
14. Verified Previous-Year Questions
SSC GD PYQ
Greatest Measurement
24 November 2021 · Shift I
Q1. Find the greatest possible length that can exactly measure 4 m 90 cm, 3 m 85 cm and 12 m 95 cm.
A. 35 cm
B. 25 cm
C. 20 cm
D. 45 cm
Correct Answer: A. 35 cm
Convert into centimetres:
4 m 90 cm=490 cm
3 m 85 cm=385 cm
12 m 95 cm=1295 cm.
Required greatest length:
HCF(490,385,1295)=35 cm.
SSC GD PYQ
Repeated Events
31 January 2023 · Shift I
Q2. Three bells begin tolling together and then toll at intervals of 8 sec, 10 sec and 12 sec. How many times do they toll together in 30 minutes, including the initial toll?
A. 14
B. 17
C. 16
D. 18
Correct Answer: C. 16
LCM(8,10,12)=120 sec.
30 minutes=1800 sec.
Common intervals within 30 minutes:
1800÷120=15.
Including the initial toll:
15+1=16 times.
SSC GD PYQ
Same Remainder
26 November 2021 · Shift III
Q3. What is the greatest number that leaves the same remainder when it divides 265, 580 and 825?
A. 35
B. 55
C. 25
D. 45
Correct Answer: A. 35
The remainder is the same but unknown, so take differences:
580−265=315
825−580=245
825−265=560.
HCF(315,245,560)=35.
SSC GD PYQ
Least Number & LCM
3 March 2019 · Shift III
Q4. What is the least positive number which, when increased by 8, becomes exactly divisible by 4, 5, 6 and 7?
A. 322
B. 312
C. 412
D. 422
Correct Answer: C. 412
LCM(4,5,6,7)=420.
If the required number is increased by 8, it becomes 420.
Required number:
420−8=412.
15. Practice MCQs
Practice MCQ
Q1. What is the greatest length that can exactly measure 4.2 m, 6.3 m and 8.4 m?
A. 1.4 m
B. 1.8 m
C. 2.1 m
D. 4.2 m
Correct Answer: C. 2.1 m
Convert to 420,630 and840 cm. HCF=210 cm=2.1 m.
Practice MCQ
Q2. A rectangular floor is 8.4 m long and 6.3 m wide. What is the minimum number of largest possible equal square tiles required to cover it completely?
A. 6
B. 8
C. 12
D. 16
Correct Answer: C. 12
Largest tile side=HCF(840,630)=210 cm. Number=(840/210)×(630/210)=4×3=12.
Practice MCQ
Q3. 48 red, 72 blue and 120 green balls are packed into the maximum number of identical packets with none left. How many packets can be made?
A. 12
B. 18
C. 24
D. 36
Correct Answer: C. 24
Maximum packets=HCF(48,72,120)=24.
Practice MCQ
Q4. Three alarms repeat every 12 sec, 18 sec and 30 sec. If they sound together now, after how long will they next sound together?
A. 90 sec
B. 120 sec
C. 150 sec
D. 180 sec
Correct Answer: D. 180 sec
LCM(12,18,30)=180 sec.
Practice MCQ
Q5. Three bells ring at intervals of 8 sec, 10 sec and 12 sec and begin together. How many times will they ring together in 30 minutes, including the start?
A. 15
B. 16
C. 17
D. 18
Correct Answer: B. 16
LCM=120 sec. In1800 sec there are15 complete common intervals. Including the start gives 16.
Practice MCQ
Q6. Find the least positive number exactly divisible by 18, 24 and 30.
A. 180
B. 240
C. 360
D. 720
Correct Answer: C. 360
Required number=LCM(18,24,30)=360.
Practice MCQ
Q7. What least number must be added to 758 so that the result is divisible by 8, 12 and 18?
A. 24
B. 30
C. 34
D. 38
Correct Answer: C. 34
LCM=72. 758 leaves remainder38 on division by72. Required addition=72−38=34.
Practice MCQ
Q8. Find the greatest number that divides 275, 455 and 635 leaving remainder 5 in each case.
A. 45
B. 60
C. 90
D. 135
Correct Answer: C. 90
Subtract5: 270,450,630. HCF=90.
Practice MCQ
Q9. Find the greatest number that divides 391, 527 and 731 leaving the same remainder in each case.
A. 34
B. 51
C. 68
D. 102
Correct Answer: C. 68
Differences are136,204 and340. Their HCF=68.
Practice MCQ
Q10. What is the greatest number that divides 465, 724 and 983 leaving remainders 3, 4 and 5 respectively?
A. 6
B. 8
C. 12
D. 18
Correct Answer: A. 6
Subtract corresponding remainders: 462,720,978. Their HCF=6, which is greater than every stated remainder.
1. Word Problems में HCF या LCM पहचानना
Application questions में अक्सर सीधे “HCF निकालिए” या “LCM निकालिए” नहीं लिखा होता। पहले यह समझना आवश्यक है कि question greatest common division चाहता है या earliest/smallest common multiple.
HCF के बारे में सोचें जब question पूछे:
• Greatest possible equal measurement
• Maximum length के identical pieces
• Largest square tile
• Maximum number of identical groups
• दी गई conditions satisfy करने वाला greatest divisor
LCM के बारे में सोचें जब question पूछे:
• कई numbers से divisible least number
• कई events का earliest common time
• Bells, lights या machines का फिर एक साथ होना
• Minimum common quantity
• कई divisibility conditions satisfy करने वाली smallest number
Exam Tip:
केवल “greatest” या “least” शब्द देखकर HCF-LCM decide न करें। देखें कि quantity divide, group या repeat किस प्रकार हो रही है।
2. Greatest Measurement, Cutting एवं Tiling Problems
जब अलग-अलग lengths या dimensions को बिना remainder के largest possible equal units में divide करना हो, तो सामान्यतः HCF उपयोग किया जाता है।
Core Rule:
Greatest possible equal measure = Given measurements का HCF।
उदाहरण — Greatest Measuring Length:
4.2 m, 6.3 m एवं8.4 m को exactly measure करने वाली greatest length ज्ञात करें।
Centimetres में:
420 cm, 630 cm, 840 cm।
HCF=210 cm।
इसलिए greatest length=
210 cm = 2.1 m।
उदाहरण — Largest Equal Pieces:
84 m, 126 m एवं210 m लंबी ropes को maximum possible equal lengths में काटना है।
Required length=HCF(84,126,210)
=42 m।
Largest Square Tile:
Rectangular floor को largest equal square tiles से exactly cover करने के लिए tile की side = floor की length और breadth का HCF।
उदाहरण — Square Tiles:
Floor=8.4 m × 6.3 m।
840 cm तथा630 cm।
Largest tile side=HCF(840,630)=210 cm।
Tiles की संख्या:
(840÷210)×(630÷210)
=4×3
=12 tiles।
Common Mistake:
Largest square tile के लिए dimensions का LCM न लें। Tile side को दोनों dimensions को divide करना है, इसलिए HCF चाहिए।
3. Maximum Equal Grouping एवं Distribution
जब अलग-अलग quantities को बिना remainder के maximum number of identical groups में बाँटना हो, तो groups की maximum संख्या सामान्यतः quantities का HCF होती है।
Core Rule:
Maximum number of identical groups = Given quantities का HCF।
उदाहरण:
48 red, 72 blue तथा120 green balls को maximum identical packets में pack करना है।
Packets की संख्या:
HCF(48,72,120)=24।
प्रत्येक packet में:
2 red
3 blue
5 green balls।
अतः:
24 identical packets।
Exam Tip:
Maximum identical groups पूछे जाने पर पहले HCF निकालें और फिर each quantity को HCF से divide करके per-group contents निकालें।
4. Repeated Events, Bells, Lights एवं Machines
जब अलग-अलग events fixed intervals पर repeat होते हैं और पूछा जाता है कि वे फिर एक साथ कब होंगे, तो intervals का LCM लिया जाता है।
Core Rule:
सभी events के फिर एक साथ होने का समय = उनके intervals का LCM।
उदाहरण:
तीन bells 12 sec, 18 sec एवं30 sec के intervals पर बजती हैं।
LCM(12,18,30)=180 sec।
वे फिर एक साथ बजेंगी:
180 sec = 3 minutes बाद।
यह Rule लागू होता है:
Bells, traffic lights, alarms, machines, rotating wheels, flashing lights तथा periodic signals पर।
Exam Trap:
यदि events अभी together start हुए हैं और पूछा गया है “next time कब together होंगे?”, तो answer केवल LCM interval होगा।
5. Events कितनी बार Together होंगे?
यदि events time 0 पर साथ शुरू हों:
Common interval = Individual intervals का LCM।
यदि starting तथा ending instants दोनों count किए जा रहे हों:
Number of common occurrences = floor(T ÷ LCM) + 1
उदाहरण:
तीन bells 8 sec, 10 sec एवं12 sec के intervals पर बजती हैं और together start करती हैं।
LCM=120 sec।
30 min=1800 sec।
1800÷120=15।
Initial ringing भी count करने पर:
15+1=16 times।
Common Mistake:
Starting occurrence शामिल होने पर उसे count न करना answer को1 कम कर देता है।
Note:
Question wording ध्यान से पढ़ें। यदि केवल start के बाद की occurrences पूछी गई हों, तो automatically +1 न करें।
6. कई Numbers से Divisible Least Number
Core Rule:
a, b, c, ... से exactly divisible least positive number = LCM(a,b,c,...)
उदाहरण:
18,24 एवं30 से divisible least positive number:
LCM(18,24,30)=360।
Exam Tip:
यदि “perfect square”, “perfect cube” या कोई अतिरिक्त divisibility condition दी हो, तो पहले basic LCM निकालें और फिर additional condition satisfy करें।
7. Exact Divisibility के लिए Least Addition या Subtraction
मान लें:
L = Required divisors का LCM
N = Given number
r = N को L से divide करने पर remainder
Least number to subtract:
r
Least number to add:
L−r, यदि r≠0।
यदि r=0 है तो कुछ add करने की आवश्यकता नहीं।
उदाहरण:
758 में least कितना add करें ताकि result 8,12 एवं18 से divisible हो?
LCM=72।
758=72×10+38।
Required addition:
72−38=34।
Related Form:
यदि किसी least number N में x add करने पर वह कई divisors से exactly divisible हो जाता है, तो:
N+x = common multiple।
Basic least positive solution के लिए सामान्यतः LCM से शुरू करते हैं:
N=LCM−x,
बशर्ते question की बाकी conditions satisfy हों।
8. Remainder-Based HCF Applications
Same Known Remainder r:
a, b, c को divide करने पर प्रत्येक case में remainder r छोड़ने वाली greatest divisor:
HCF(a−r, b−r, c−r)
उदाहरण:
1780 और2452 को divide करने पर प्रत्येक case में remainder4 छोड़ने वाली greatest number ज्ञात करें।
1776 तथा2448 का HCF=48।
Same Unknown Remainder:
यदि same remainder है लेकिन उसका value नहीं दिया गया, तो numbers के pairwise differences का HCF लें।
उदाहरण:
391,527 एवं731 को divide करने पर same remainder छोड़ने वाली greatest number ज्ञात करें।
Differences:
136,204,340।
HCF=68।
Different Remainders:
यदि a,b,c से क्रमशः r₁,r₂,r₃ remainders प्राप्त हों, तो:
HCF(a−r₁, b−r₂, c−r₃)
Necessary Check:
Required divisor प्रत्येक given remainder से बड़ा होना चाहिए।
Common Mistake:
Non-zero remainder दिया होने पर original numbers का direct HCF न लें। पहले remainder subtract करें।
9. Remainder-Based LCM Applications
General Form:
यदि N को a,b,c से divide करने पर same remainder r प्राप्त होता है, तो:
N−r सभी divisors से divisible होगा।
अतः:
N = k × LCM(a,b,c) + r
महत्वपूर्ण Mathematical Note:
यदि कोई lower-bound या extra condition नहीं दी हो और r सभी divisors से छोटा हो, तो mathematically r स्वयं भी congruence satisfy कर सकता है। Competitive-exam questions में सामान्यतः wording या additional condition intended larger value निर्धारित करती है।
उदाहरण:
यदि कोई number 8,16,18,20 एवं25 से divide होने पर remainder3 छोड़े, तो:
LCM=3600।
ऐसे numbers का form:
3600k+3।
Exam Strategy:
यदि additional condition हो कि required number किसी अन्य number से भी divisible है, तो suitable k ज्ञात करें।
10. HCF-LCM Applications में Unit Conversion
HCF या LCM निकालने से पहले सभी measurements को same unit में convert करना आवश्यक है।
Common Conversions:
metre → centimetre
hour → minute
minute → second
kilogram → gram
उदाहरण:
4 m 90 cm=490 cm
3 m 85 cm=385 cm।
इसके बाद ही HCF calculate करें।
Exam Trap:
Different units में दी गई numerical values पर direct HCF या LCM calculate न करें।
11. Fast Decision Strategy
| Question Pattern | Method |
|---|
| Greatest equal measure या piece | HCF |
| Largest square tile | HCF |
| Maximum identical groups | HCF |
| Greatest divisor leaving remainders | Remainder adjust करके HCF |
| Least exactly divisible number | LCM |
| Events फिर together होना | Intervals का LCM |
| Common divisibility के लिए least addition/subtraction | LCM + remainder method |
| Several divisors पर same remainder | k×LCM+r form |
Exam Tip:
Arithmetic शुरू करने से पहले word problem को divisibility statement में बदलें।
12. Common Exam Traps
Trap 1:
Largest equal measure के लिए LCM लेना।
Trap 2:
Repeated events के earliest common time के लिए HCF लेना।
Trap 3:
Measurements को same unit में convert न करना।
Trap 4:
Times-together question में initial occurrence को गलत तरीके से include या exclude करना।
Trap 5:
Same known remainder को सभी numbers से subtract न करना।
Trap 6:
Different remainder question में प्रत्येक number से उसका corresponding remainder subtract न करना।
Trap 7:
ऐसी divisor स्वीकार करना जो किसी given remainder से छोटी या बराबर हो।
Trap 8:
Question की wording check किए बिना LCM+r को automatically least mathematical solution मान लेना।
13. Quick Revision
एक नज़र में:
• Greatest equal measure → HCF।
• Maximum equal piece length → HCF।
• Largest square tile side → Dimensions का HCF।
• Maximum identical groups → Quantities का HCF।
• Events together again → Intervals का LCM।
• Least exactly divisible number → LCM।
• Starting occurrence include हो तो times-together question में उसे count करें।
• Divisibility adjustment के लिए पहले LCM निकालें।
• Least subtraction = N÷LCM का remainder।
• Least addition = LCM−remainder, जब remainder non-zero हो।
• Known same remainder r → Numbers से r subtract करके HCF।
• Same unknown remainder → Pairwise differences का HCF।
• Different remainders → Corresponding remainders subtract करके HCF।
• Same-remainder LCM problem का form k×LCM+r होता है।
• सभी measurements को same unit में convert करें।
14. Verified Previous-Year Questions
SSC GD PYQ
Greatest Measurement
24 नवंबर 2021 · Shift I
प्रश्न 1. 4 m 90 cm, 3 m 85 cm एवं12 m 95 cm की lengths को exactly measure करने वाली greatest possible length ज्ञात करें।
A. 35 cm
B. 25 cm
C. 20 cm
D. 45 cm
सही उत्तर: A. 35 cm
Lengths:
490 cm, 385 cm, 1295 cm।
Required length=HCF(490,385,1295)
=35 cm।
SSC GD PYQ
Repeated Events
31 जनवरी 2023 · Shift I
प्रश्न 2. तीन bells together start करती हैं और 8 sec, 10 sec एवं12 sec के intervals पर बजती हैं। Initial toll सहित30 minutes में वे कितनी बार together बजेंगी?
A. 14
B. 17
C. 16
D. 18
सही उत्तर: C. 16
LCM(8,10,12)=120 sec।
30 minutes=1800 sec।
1800÷120=15।
Initial toll सहित:
15+1=16 times।
SSC GD PYQ
Same Remainder
26 नवंबर 2021 · Shift III
प्रश्न 3. 265,580 एवं825 को divide करने पर प्रत्येक case में same remainder छोड़ने वाली greatest number क्या है?
A. 35
B. 55
C. 25
D. 45
सही उत्तर: A. 35
Same remainder unknown है, इसलिए differences लें:
580−265=315
825−580=245
825−265=560।
HCF(315,245,560)=35।
SSC GD PYQ
Least Number & LCM
3 मार्च 2019 · Shift III
प्रश्न 4. वह least positive number क्या है जिसमें8 add करने पर result 4,5,6 एवं7 से exactly divisible हो?
A. 322
B. 312
C. 412
D. 422
सही उत्तर: C. 412
LCM(4,5,6,7)=420।
Required number:
420−8=412।
15. Practice MCQs
Practice MCQ
प्रश्न 1. 4.2 m, 6.3 m एवं8.4 m को exactly measure करने वाली greatest length क्या है?
A. 1.4 m
B. 1.8 m
C. 2.1 m
D. 4.2 m
सही उत्तर: C. 2.1 m
420,630 एवं840 cm का HCF=210 cm=2.1 m। Practice MCQ
प्रश्न 2. 8.4 m × 6.3 m के rectangular floor को largest possible equal square tiles से cover करना है। Minimum कितनी tiles चाहिए?
A. 6
B. 8
C. 12
D. 16
सही उत्तर: C. 12
Largest tile side=210 cm। Tiles=(840/210)×(630/210)=4×3=12। Practice MCQ
प्रश्न 3. 48 red,72 blue एवं120 green balls से maximum identical packets बनाए जाएँ, तो packets की संख्या क्या होगी?
A. 12
B. 18
C. 24
D. 36
सही उत्तर: C. 24
HCF(48,72,120)=24। Practice MCQ
प्रश्न 4. तीन alarms 12 sec,18 sec एवं30 sec के intervals पर repeat होते हैं। वे next time together कब होंगे?
A. 90 sec
B. 120 sec
C. 150 sec
D. 180 sec
सही उत्तर: D. 180 sec
LCM(12,18,30)=180 sec। Practice MCQ
प्रश्न 5. 8 sec,10 sec एवं12 sec के intervals वाली bells together start करती हैं। Initial ringing सहित30 minutes में कितनी बार together बजेंगी?
A. 15
B. 16
C. 17
D. 18
सही उत्तर: B. 16
LCM=120 sec। 1800÷120=15 intervals। Initial ringing सहित=16। Practice MCQ
प्रश्न 6. 18,24 एवं30 से exactly divisible least positive number ज्ञात करें।
A. 180
B. 240
C. 360
D. 720
सही उत्तर: C. 360
LCM(18,24,30)=360। Practice MCQ
प्रश्न 7. 758 में least कितना add करें ताकि result 8,12 एवं18 से divisible हो?
A. 24
B. 30
C. 34
D. 38
सही उत्तर: C. 34
LCM=72 तथा remainder=38। Required addition=72−38=34। Practice MCQ
प्रश्न 8. 275,455 एवं635 को divide करने पर प्रत्येक case में remainder5 छोड़ने वाली greatest number क्या है?
A. 45
B. 60
C. 90
D. 135
सही उत्तर: C. 90
5 subtract करने पर270,450,630 मिलते हैं। HCF=90। Practice MCQ
प्रश्न 9. 391,527 एवं731 को divide करने पर same remainder छोड़ने वाली greatest number क्या है?
A. 34
B. 51
C. 68
D. 102
सही उत्तर: C. 68
Differences=136,204,340। इनका HCF=68। Practice MCQ
प्रश्न 10. 465,724 एवं983 को divide करने पर क्रमशः remainders3,4 एवं5 छोड़ने वाली greatest number क्या है?
A. 6
B. 8
C. 12
D. 18
सही उत्तर: A. 6
Corresponding remainders subtract करने पर462,720,978 मिलते हैं। इनका HCF=6 है, जो सभी remainders से बड़ा है।