LCM AND HCF TCS Questions: Asked in SSC MTS 2019 For 2021
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In this article, we provide LCM and HCF questions with their solutions asked by TCS in SSC MTS exams 2019 of MTS. These questions are useful for upcoming exams of SSC, Railways, and other competitive exams conducted by TCS.
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SSC LCM and HCF Maths Questions
Q1. The product of two numbers is 6845, if the HCF of the number is 37, then the greater number is:
दो संख्याओं का गुणनफल 6845 है | यदि संख्याओं का H.C.F 37 है, तो बड़ी संख्या ज्ञात करें |
SSC MTS 9 August 2019 (Morning)
(a). 111
(b). 37
(c). 148
(d). 185
Q2. If x×y denotes HCF of x and y and x @ denotes LCM of x and y, then the value of (72×84) @ 144 is: यदि x×y, x तथा y का महत्तम समापवर्तक है और x @ y, x तथा y का लघुत्तम समापवर्त्य है, तो (72×84)@144 का मान होगा :
SSC MTS 9 August 2019 (Afternoon)
(a). 144
(b). 504
(c). 210
(d). 420
Q3. The Highest Common Factor and Lowest Common Multiple of two numbers p and q are A and B respectively. IF A+B = p+q, then the value of A3+B3 is :
दो संख्याओं p तथा q का महत्तम समापवर्तक (HCF) एवं लघुत्तम समापवर्तक (LCM) क्रमसः A तथा B है, यदि A+B=p+q है, तो A3+B3का मान है:
SSC MTS 9 August 2019 (Evening)
(a) p3
(b) q3
(c) p3+q3
(d) p3-q3
Q4. If LCM of two numbers is 231, HCF of these two numbers is 11 and the first number is 77, then find the second number. यदि दो संख्याओं का लघुतम समापवर्तक 231 है, दोनों संख्याओं का महत्तम समापवर्तक 11 है तथा पहली संख्या 77 है, तो दूसरी संख्या है:
SSC MTS 13 August 2019 (Morning)
(a). 47
(b). 37
(c). 57
(d). 33
Q5. What is the least number of square tiles required to pave the floor of a room 15m 17 cm long and 9 m 43 cm broad? 15 मी 17 सेमी लंबे तथा 9 मी 43 सेमी चौड़े कमरे की फर्श पर बिछाने के लिए आवश्यक वर्गाकार टाइलों की न्यूनतम संख्या ज्ञात करें |
SSC MTS 13 August 2019 (Evening)
(a). 851
(b). 841
(c). 840
(d). 830
Q6. The Highest Common Factor and Lowest Common Factor of two numbers are 20 and 120. If one number is 50% more than the other number, then what is the smaller number?
दो संख्याओं का महत्तम समापवर्तक और लघुतम समापवर्तक क्रमसः 20 और 120 है | यदि एक संख्या दूसरी संख्या से 50 % अधिक है, तो छोटी संख्या कौन सी है |
SSC MTS 14 August 2019 (Morning)
(a). 2
(b). 60
(c). 40
(d). 80
Q7. If the LCM of two numbers 390 and 420 is 5460, then the HCF of two numbers is:
यदि दो संख्याओं 390 तथा 420 का लघुत्तम समापवर्त्य 5460 है, तो इन दोनों संख्याओं का महत्तम समापवर्तक क्या होगा?
SSC MTS 14 August 2019 (Evening)
(a). 35
(b). 45
(c). 30
(d). 42
Q8. A temple has five bells which ring at intervals of 12, 15, 16, 20, and 25 minutes respectively. If they ring together at midnight, then at what time next will they ring together?
.एक मंदिर में 5 घंटियाँ हैं जो क्रमशः 12, 15, 16, 20 और 25 मिनट के अंतराल पर बजती हैं | यदि वे आधी रात में एक साथ बजती हैं, तो अगली बार वे किस समय एक साथ बजेंगी ?
SSC MTS 16 August 2019 (Morning)
(a). 8:00 PM
(b). 7:00 PM
(c). 7:30 PM
(d). 8:30 PM
Q9. What is the HCF of 20, 250, and 120?
20, 250 तथा 120 का महत्तम समापवर्तक (HCF) कितना है?
SSC MTS 19 August 2019 (Afternoon)
(a). 15
(b). 10
(c). 25
(d). 20
Q10. What is the LCM of the HCF of 2/3, 3/4, and LCM of 5/6, 7/8?
2/3, 3/4 के म. स. तथा 5/6, 7/8 के ल. स. का ल. स. क्या है?
SSC MTS 20 August 2019 (Evening)
(a) 1/2
(b) 35/2
(c) 1/12
(d) 35/12
Q11. If two numbers are in the ratio of 7: 11, and their HCF is 13 then find their LCM.
यदि दो संख्याओं का अनुपात 7:11 है और उनका म. स. 13 है, तो उसका ल. स. है:
SSC MTS 21 August 2019 (Morning)
(a) 101
(b) 1001
(c) 143
(d) 234
Q12. The H.C.F and L.C.M of two numbers are 37 and 444 respectively. If the first number is 111, then the other number will be :
दो संख्याओं का महत्तम और लघुतम समापवर्तक क्रमसः 37 और 444 है | यदि पहली संख्या 111 है तो दूसरी संख्या है:
SSC MTS 21 August 2019 (Afternoon)
(a) 333
(b) 74
(c) 148
(d) 111
Q13. What is the smallest number which when increased by 5 is divisible by 12, 18 and 30?
वह सबसे छोटी संख्या कौन सी है जिसमें 5 की वृद्धि करने पर वह 12, 18 और 30 से विभाजित हो जाती है ?
SSC MTS 2 August 2019 (Evening)
(a) 175
(b) 115
(c) 235
(d) 195
Q14. What is the largest 4-digit number that is divisible by 32, 40, 36 and 48 ?
4 अंकों की सबसे बड़ी संख्या कौन सी है, जो 32, 40, 36 तथा 48 से विभाज्य है?
SSC MTS 5 August 2019 (Evening)
(a) 9220
(b) 8820
(c) 8640
(d) 9120
Q15. What is the largest two digit number which when divided by 6 and 7 gives remainder 3 and 4 respectively?
दो अंकों की वह सबसे बड़ी संख्या कौन सी है जिसे 6 और 7 से विभाजित करने पर शेष-फल क्रमशः 3 और 4 आता है |
SSC MTS 8 August 2019 (Morning)
(a) 81
(b) 94
(c) 83
(d) 84
Q16. If A is the smallest three digit number divisible by both 6 and 7 and B is the largest four digit number divisible by both 6 and 7, then what is the value of B-A?
यदि A , 6 और 7 दोनों से विभाजित होने वाली तीन अंकों की सबसे छोटी संख्या है तथा B , 6 और 7 दोनों से विभाजित होने वाली चार अंकों की सबसे बड़ी संख्या है, तो B-A का मान क्या होगा ?
SSC MTS 8 August 2019 (Evening)
(a) 9912
(b) 9870
(c) 9996
(d) 9954
Q17. The largest three-digit number that is exactly divisible by 6, 7 and 8 is:
6,7 और 8 से विभाज्य तीन अंकों की सबसे बड़ी संख्या है :
SSC MTS 13 August 2019 (Afternoon)
(a) 999
(b) 168
(c) 358
(d) 840
Q18. What is the highest number which when divides the numbers 1026, 2052 and 4102, leaves remainders 2, 4 and 6 respectively.
वह सबसे बड़ी संख्या कौन सी है जो संख्याओं 1026, 2052 और 4102 को विभाजित करने पर शेषफल क्रमशः 2, 4 और 6 छोड़ती है ?
SSC MTS 14 August 2019 (Afternoon)
(a) 512
(b) 1024
(c) 128
(d) 256
Q19. Find the sum of digits of the largest 6-digit number that is divisible by 3, 4, 5 and 6.
3, 4, 5 और 6 द्वारा विभाजित होने वाली सबसे बड़ी 6-अंकीय संख्या के अंकों का योग कितना है?
SSC MTS 16 August 2019 (Afternoon)
(a) 45
(b) 39
(c) 48
(d) 42
Q20. Let x be the smallest number greater than 600 which gives the remainders 2, 3 and 4, when divided by 5, 6 and 7 respectively. The sum of digits of x is:
मान लीजिये कि x, 600 से बड़ी सबसे छोटी संख्या है जो 5, 6 और 7 से विभाजित होने पर शेषफल क्रमशः 2, 3 और 4 छोड़ती है | x के अंकों का जोड़ है -
SSC MTS 19 August 2019 (Evening)
(a) 14
(b) 15
(c) 13
(d) 16
Q21. When the smallest number x is divided by 5,6,8,9 and 12, it gives remainder 1 in each case. But x is divisible by 13. What will be the remainder when x will be divided by 31 ?
जब 5, 6, 8, 9 और 12 से सबसे छोटी संख्या x को विभाजित किया जाता है, तो प्रत्येक मामले में 1 शेष प्राप्त होता है, किन्तु x, 13 से विभाज्य है | जब x को 31 से विभाजित किया जाएगा तो प्राप्त शेष कितना होगा?
SSC MTS 20 August 2019 (Afternoon)
(a) 1
(b) 5
(c) 3
(d) 0
Q22. Let x be the largest 4-digit number which when divided by 7, 8 and 11 leaves remainders 4, 5 and 8 respectively. When x is divided by (7+8+11), then the remainder will be :
मान लिजिएं की x, 4 अंकों की सबसे बड़ी संख्या है, जिसे 7, 8 और 11 से विभाजित करने पर क्रमशः 4, 5 और 8 शेष बचता है | जब x को (7+8+11) से विभाजित किया जाता है, तब शेष होगा:
SSC MTS 21 August 2019 (Evening)
(a) 23
(b) 25
(c) 21
(d) 19
Q23. The difference between the two numbers is 15 and their HCF and LCM are 3 and 108 respectively. Find the difference between their inverses?
दो संख्याओं के बीच का अंतर 15 है और उनका महत्तम समापवर्तक (एच सी एफ) और लघुतम समापवर्तक (एल सी एम) क्रमश : 3 और 108 है | उनके व्युत्कर्मो के बीच अंतर ज्ञात कीजिए |
SSC MTS 16 August 2019 (Evening)
(a) 5/54
(b) 5/108
(c) 5/81
(d) 5/112
Q24. What is the greatest number which can exactly divide 192, 1056 and 1584 ?
वह सबसे बड़ी संख्या कौन सी है जो 192, 1056 तथा 1584 को पूर्णतः विभाजित कर सकती है ?
SSC MTS 19 August 2019 (Morning)
(a) 48
(b) 56
(c) 44
(d) 36
Q25. Let x and y be the 3-digit numbers such that their difference is 729 and HCF is 81. Find the value of (x+y) ? / मान लीजिए एक x और y, 3-अंको की ऐसी संख्याएँ है कि उनका अंतर 729 और महत्तम समापवर्तक 81 है | (x+y) का मान ज्ञात कीजिए ?
SSC MTS 20 August 2019 (Morning)
(a) 1053
(b) 891
(c) 1539
(d) 1377
Here are the Solutions to the Above LCM and HCF Questions asked by TCS in MTS exams.
Sol 1.
Ans. (d)
Let the numbers are 37x and 37y
According to the question
6845 = 37x x 37y
⇒ x×y = 5
Only possible factors of x and y are 1 and 5.
⇒ Greatest number = 37 x 5 = 185
Sol 2.
Ans. (a)
72 = 2x2x2x3x3
84 = 2x2x3x7
HCF of 72 and 84 = 2x2x3 = 12
Now,
12 = 2x2x3
144 = 2x2x2x2x3x3
LCM of 12 and 144 =2x2x2x2x3x3 = 144
Sol 3. (c)
We know that
A3 +B3 =(A+B)(A2+B2-AB)
= (A+B){(A+B)2-2AB-AB}
= (A+B){(A+B)2-3AB} ……..(1)
We know that ,
HCF × LCM = Number1 × Number2
⇒A×B = p×q
Put the values of AB and (A+B) in equation (1)
A3 +B3 = (p+q){(p+q)2-3pq}
= (p+q)(p2+q2+2pq-3pq)
= (p+q)(p2+q2-pq)
= p3 +q3
Sol 4. (d)
Let the required number = k
According to the question
231 x 11 = 77 x k
⇒k = 
= 33
Sol 5.(a)
Length of the room = 1517 cm
Breadth of the room = 943 cm
HCF of 1517 and 943 = 41
Required number of tiles =
= 851
Sol 6.(c)
50% = 1/2
Let the smaller number = 2x and the bigger number = 3x
According to the question
20 x 120 = 2x3x
x=20
So, the smaller number = 2 x 20 = 40
Sol 7. (c)
Let HCF of the numbers = h
According to the question
5460 x h = 390 x 420
⇒ h=30
Sol 8. (a)
12 = 2x2x3
15 = 3x5
16 = 2x2x2x2
20 = 2x2x5
25 = 5x5
LCM of the numbers = 2x2x2x2x3x5x5 = 1200 minutes = 20 hours
Required time = 12:00 AM + 20 hours = 8:00 PM
Sol 9.(b)
20 = 2x2x5
250 = 2x5x5x5
120 = 2x2x2x3x5
Required HCF = 2x5 = 10
Sol 10. (b)
HCF of 2/3, 3/4 = HCF of 2 and 3 divided by LCM of 3 and 4.
HCF of 2 and 3 = 1
LCM of 3 and 4 = 12
HCF of 2/3, 3/4 = 1/12
LCM of 5/6, 7/8 = LCM of 5 and 7 divided by HCF of 6 and 8.
LCM of 5 and 7 = 35
HCF of 6 and 8 = 2
LCM of 5/6, 7/8 = 35/2
Now,
LCM of 1/12, 35/2 = LCM of 1 and 35 divided by HCF of 12 and 2.
LCM of 1 and 35 = 35
HCF of 12 and 2 = 2
⇒LCM of 1/12, 35/2 = 35/2
Sol 11.(b)
Ratio of two numbers = 7:11
HCF = 13
Required LCM = 7 x 11 x 13 = 1001
Sol 12. (c)
Let the second number = y
According to the question
37 x 444 = 111 x y
⇒y=148
Sol 13. (a)
12 = 2x2x3
18 = 2x3x3
30 = 2x3x5
LCM of 12, 18 and 30 =2x2x3x3x5 = 180
Required number = 180-5 = 175
Sol 14. (c)
Largest number of four digits = 9999
32 = 2x2x2x2x2
40 = 2x2x2x5
36 = 2x2x3x3
48 = 2x2x2x2x3
LCM of 32, 40, 36 and 48 = 2x2x2x2x2x3x3x5 = 1440
Required number = 9999-1359 = 8640
Sol 15. (a)
6-3 = 3 and 7-4 = 3
Largest number of two digit = 99
LCM of 6 and 7 = 6x7 = 42
Required number = 99-15-3 = 81
Sol 16. (b)
LCM of 6 and 7 = 7 x 6 = 42
Smallest number of three digits = 100
Smallest number of three digits divisible by 6 and 7 = A = 100 + (42-16) = 126
LCM of 6 and 7 = 7 x 6 = 42
Largest number of four digits = 9999
Largest number of four digits divisible by 6 and 7 = A = 9999-3 = 9996
Required difference = B-A = 9996-126 = 9870
Sol 17.(d)
Largest three digit number = 999
6 = 2x3
7 = 1x7
8 = 2x2x2
LCM of 6, 7 and 8 = 2x2x2x3x7 = 168
The largest three digit number divisible by 6,7 and 8 = 999-159 = 840
Sol 18. (b)
Let the required number = k
According to the question
kx + 2 = 1026
⇒kx = 1024
ky + 4 = 2052
⇒ky = 2048
kz + 6 = 4102
⇒kz = 4096
So, k will be the HCF of 1024, 2048 and 4096
1024 = 210
2048 = 211
4096 = 212
⇒k = 210=1024
Sol 19. (d)
Biggest number of 6 digits = 999999
LCM of ,3,4,5,6 = 60
Required number = 999999-39 = 999960
Desired sum = 9+9+9+9+6+0 = 42
Sol 20. (b)
5-2 = 3, 6-3 = 3 and 7-4=3
LCM of 5, 6 and 7 = 210
Smallest number greater than 600 which is divisible by 5,6 and 7 = 210 x 3 = 630
Desired number = 630-3 = 627
Desired sum = 6+2+7 = 15
Sol 21. (b)
LCM of 5,6,8,9,12 = 360
⇒x = 360k+1
360k+1 will be divisible by 31 for the minimum value of k = 10
Desired number = 3601
When 3601 is divided by 31 the remainder will be 5.
Sol 22.(b)
7-4 = 3, 8-5 = 3 and 11-8 = 3
Largest number of 4 digit = 9999
LCM of 7,8 and 11 = 616
Desired number = x = 9999-143-3 = 9853
Divisor = 7+8+11 = 26
Desired remainder = 985326= 25
Sol 23. (b)
Let the numbers are 3k and 3m
According to the question
3k-3m = 15
⇒k-m=5or k = 5+m
And
3 x 108 = 3k x 3m
⇒k.m=36
⇒m(5+m)=36
⇒5m+m2 -36=0
⇒m2+9m-4m-36=0
⇒m= 4 …….(m≠-9)
So, k=5+4 = 9
The two numbers will be 3(4) = 12 and 3(9) = 27
Desired difference = 1/12-1/27= 5/108
Sol 24. (a)
192 = 2x2x2x2x2x2x3
1056 = 2x2x2x2x2x3x11
1584 = 2x2x2x2x3x3x11
Required HCF = 2x2x2x3 = 48
Sol 25. (a)
According to the question
x-y = 729
HCF= 81
Let the numbers be 81a and 81b.
Now, 81a - 81b = 729
⇒a - b = 9
Since, 81a < 1000
⇒ a < 12.34
a could be 10, 11 and 12
So, b could be 1, 2 and 3
If 81b > 100
⇒ b > 1.23
Hence, b could be 2 and 3.
and a could be 11 and 12.
Possible pairs of x and y = (891, 162) or (972, 243)
Since, HCF of (972, 243) = 243
Therefore, the required pair is (891, 162)
Hence, x + y = 891+162 = 1053
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