LCM and HCF Questions with Solutions

We are providing LCM and HCF questions with their solutions asked by TCS in SSC exams 2018, 2019, and 2020 of CGL, CHSL, CPO, MTS. These questions are useful for upcoming exams of SSC, Railways, and other competitive exams conducted by TCS.

SSC Previous Year LCM and HCF Questions

Q1. Two numbers are in ratio 6:11, If there HCF is 28, then the sum of these two numbers is: 

दो संख्याएं 6 : 11 के अनुपात में हैं | यदि उनका महत्तम समापवर्तक 28 है, तो इन दो संख्याओं का योग क्या है ?

SSC CPO 13 March 2019 (Evening)

(a). 476

(b). 448   

(c). 392

(d). 420

Q2. What is the sum of digits of the least number, which when divided by 15, 18, and 42 leaves the same remainder 8 in each case and is also divisible by 13?

उस न्यूनतम संख्या के अंकों का योग क्या होगा, जो 15, 18 और 42 से विभाजित होने पर प्रत्येक स्थिति में  एक ही शेष 8 रहता है और 13 से विभाज्य भी है?

SSC CPO 13 March 2019 (Evening)

(a). 25

(b). 24      

(c). 22

(d). 26

Q3. What is the sum of digits of the least number, which when divided by 15, 18, and 24 leaves the remainder 8 in each case and is also divisible by 13?

उस सबसे छोटी संख्या के अंकों का योग ज्ञात करें जिसे 15, 18 तथा 24 से भाग देने पर हर बार शेषफल 8 आता है तथा यह 13 से भी विभाजित है |

SSC CPO 12 March 2019 (Morning)

(a). 17

(b). 16       

(c). 15

(d). 18 

Q4. Two numbers are in the ratio 4:5. If their HCF is 16, then the sum of these two numbers is:

दो संख्याएं 4 : 5 के अनुपात में हैं | यदि उनका महत्तम समापवर्तक 16 है, तो इन दो संख्याओं का योग क्या होगा ?

SSC CPO 12 March 2019 (Morning)

(a). 144

(b). 124        

(c). 160

(d). 150

Q5. Two numbers are in the ratio 5:11. If their HCF is 24, then the sum of two of these numbers is:

दो संख्याएं 5 : 11 के अनुपात में हैं | यदि उनका महत्तम समापवर्तक 24 है, तो इन दो संख्याओं का योग क्या होगा ?

SSC CPO 13 March 2019 (Morning)

(a). 384

(b). 408         

(c). 120

(d). 264

Q6. What is the sum of the digits of the least number, which when divided by 15, 25, and 27 leaves the same remainder 9 in each case and is also completely divisible by 11?

उस सबसे छोटी संख्या के अंकों का योग ज्ञात करें जिसे 15, 25 और 27 से भाग देने पर हर बार शेषफल 9 आता है तथा यह 11 से भी पूर्णतया विभाजित है |

SSC CPO 13 March 2019 (Morning)

(a). 20

(b). 17   

(c). 18

(d). 19

Q7. The largest number of four digits that is exactly divisible by 15, 21, and 30 is:

चार अंकों की सबसे बड़ी संख्या ज्ञात करें जो 15, 21 एवं 30 से पूर्णतः विभाजित है |

SSC CPO 16 March 2019 (Afternoon)

(a). 9840

(b). 9910   

(c). 9830

(d). 9870

Q8. Four bells ring together at a certain time. After this, they rang at intervals of 6, 8, 10, and 12 seconds, respectively. After how many minutes will they ring together for the first time? 

चार घंटियाँ एक निश्चित समय पर एक साथ बजती हैं | इसके बाद वे क्रमश : 6, 8, 10 और 12 सेकंड के अंतराल पर बजती हैं | कितने मिनटों के बाद वे फिर से पहली बार एक साथ बजेगी ?

SSC CPO 15 March 2019 (Evening)

(a). 1 Minute 

(b). 112 Minute  

(c). 214 Minute 

(d). 2 Minute 

Q9. The greatest number of 5 digits that is exactly divisible by each of 8,12,15 and 20 is:

5 अंकों की सबसे बड़ी संख्या कौन सी है जो 8, 12, 15 और 20 में से प्रत्येक से पूर्णत : विभाज्य है :

SSC CPO 15 March 2019 (Evening)

(a). 99960

(b). 99940    

(c). 99980

(d). 99950

Here are the Solutions to above Questions.

Sol 1.

Ans. (a)

Sum of the numbers = 28x(6+11) = 28x17 = 476

Sol 2.

Ans. (d)

LCM of (15, 18 and 42) = 630

Let the number be (630k+8).

For (630k+8) to be divisible by 13, (6k+8) should be divisible by 13.

Therefore, for k=3, (6k+8) is divisible by 13.

Hence the number is 1898.

Sum of digits = 1+8+9+8 = 26

Sol 3. (a)

LCM of (15, 18 and 24) = 360

Required number  = 360k+8

For it to be divisible by 13. (9k+8) should also be divisible by 13.

Therefore, for k=2, (9k+8) is divisible by 13.

Therefore, the required number = 728

Sum of digits = 7+2+8 = 17 

Solutions 4. (a)

Sum of the numbers = 16x(4+5) = 16x9 =144

Solutions 5. (a)

Sum of number  = 24x(5+11) = 24x16 = 384

Solutions 6.(c)

LCM of (15, 25 and 27) = 675

Required number  = 675k+9

For (675k+9) to be divisible by 11. (4k+9) should also be divisible by 11.

At k=6, (4k+9) is divisible by 11.

Therefore, the Required number = 4059

Sum of digits = 4+0+5+9 = 18

Sol 7. (d) 

Largest number of 4 digits = 9999

15 = 3 × 5

21 = 3 × 7

30 = 2 × 3 × 5

LCM = 2 × 3 × 5 × 7 = 210

Required four digit number = 9999-129 = 9870

Solutions 8. (d)

LCM of (6, 8, 10, 12) = 120

Therefore, Required time = 120 s = 2 min

Sol 9. (a)

LCM of (8, 12, 15 & 20) = 120

Largest number of 5 digits = 99999

Clearly, On dividing 99999 by 120 we get remainder 39.

Therefore, Desired number = 99999-39 = 99960

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