# LCM and HCF Aptitude Questions With Solutions [Practice now]

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## LCM and HCF Aptitude Questions With Solutions

Ques 1: Find the HCF of 23x32x5x74, 22x35x52x73, 23x53x72

• 980
• 935
• 940
• 945

980
Sol. The prime numbers common to given numbers are 2, 5, and 7.
HCF = 22x5x72 = 980

Ques 2: Find the HCF of 108, 288, and 360.

• 24
• 45
• 15
• 36

36
Sol. 108 = 22 x 33, 288 = 25 x 32 and 360 = 23 x 5 x 32.
H.C.F. = 22 x 32 = 36.

Ques 3: Find the HCF of 513, 1134, and 1215.

• 25
• 27
• 34
• 43

27
Sol. 513 = 3 x 3 x 3 x 19
1134 = 2 x 3 x 3 x 3 x 3 x 7
1215 = 3 x 3 x 3 x 3 x 3 x 5
Therefore, HCF = 27

Ques 4: Reduce 391/667 to lowest terms.

• 21/29
• 15/29
• 17/29
• 17/31

17/29
Sol. HCF of 391 and 667 is 23.
On dividing the numerator and denominator by 23, we get :
391/667 = (391+23)/(667+23) = 17/29

Ques 5: Find the LCM Of 22x33x5x72, 23x32x52x74, 2x3x53x7x11

• 750,213,000
• 713,097,000
• 615,243,000
• 771,089,000

713,097,000
Sol. LCM = Product of highest powers of 2, 3, ,5, 7, and 11 = 23x33x53x74x11

= 713,097,000

Ques 6: Find the LCM of 72, 108, and 2100.

• 42300
• 37800
• 43200
• 35000

37800
Sol. 72 = 23 x 32 ,. 108 = 33 x 22 ,. 2100 = 22 x 52 x 3 x 7. L.C.M. = 23 x 33 x 52 x 7 = 37800

Ques 7: Find the LCM of 16, 24, 36, and 54.

• 350
• 442
• 432
• 435

432
Sol. LCM of 16, 24, 36, 54
16 = 2x2x2x2
24 = 2x2x2x3
36 = 2x2x3x3
54 = 2x3x3x3
LCM of 16, 24, 36, 54 is = 2x2x2x2x3x3x3 = 432

Ques 8: Find the HCF and LCM of 2/3, 8/9, 16/81, and 10/27.

• 2/61, 79/5
• 2/81, 80/3
• 5/21, 92/6
• 3/25, 71/4

2/81, 80/3
Sol. HCF of given fractions = HCF of 2, 8, 16, 10/ LCM of 3, 9, 81, 27 = 2/81
LCM of given fractions = LCM of 2, 8, 16, 10/ HCF of 3, 9, 81, 27 = 80/3

Ques 9: Find the HCF and LCM of 0.63, 1.05, and 2.1.

• 1.21, 5.74
• 0.21, 6.30
• 0.52, 4.63
• 0.12, 5.30

0.21, 6.30
Sol. Making the same number of decimal places, the given numbers are 0.63, 1.05 and 2.10.
Without decimal places, these numbers are 63, 105 and 210.
Now, H.C.F. of 63, 105 and 210 is 21.
H.C.F. of 0.63, 1.05 and 2.1 is 0.21.
L.C.M. of 63, 105 and 210 is 630.
L.C.M. of 0.63, 1.05 and 2.1 is 6.30.

Ques 10: Two number are in the ratio of 15:11. If their HCF is 13, find the numbers.

• 190, 157
• 195, 150
• 195, 143
• 185, 157

195, 143
Sol. From given,
let the numbers be,
15x,11x
since given,
HCF=13
we have,
15×13=195
11×13=143
195,143 are the numbers

Ques 11: The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other.

• 96
• 99
• 95
• 93

99
Sol. Other number = ((11×693)/77) = 99

Ques 12: Find the greatest possible length which can be used to measure exactly the lengths 4m 95 cm, 9m and 16m 65cm.

• 45 cm
• 50 cm
• 44 cm
• 40 cm

45cm
Sol. Required length = H.C.F. of 495 cm, 900 cm and 1665 cm.
495 = 32 x 5 x 11, 900 = 22 x 32 x 52, 1665 = 32 x 5 x 37.
H.C.F. = 32 x 5 = 45.
Hence, required length = 45 cm

Ques 13: Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

• 124
• 126
• 128
• 127

127
Sol. Required number
= H.C.F. of (1657−6) and (2037−5)
= H.C.F. of 1651 and 2032
1657=13×127
2037=2×2×2×2×127
Common = 127
Required number = 127

Ques 14: Find the largest number which divides 62, 132, and 237 to leave the same remainder in each case.

• 38
• 45
• 35
• 42

35
Sol. Required number = HCF (132-62), (237-132), and (237-62)
HCF of 70, 105, and 175 = 35

Ques 15: Find the least number exactly divisible by 12, 15, 20, and 27.

• 540
• 530
• 542
• 545

540
Sol. Least number exactly divisible by 12, 15, 20 and 27 is the LCM of these 4 numbers.
12 = 2×2×3 = 22×3
15 = 3×5
20 = 2×2×5 = 22×5
27 = 3×3×3 = 33

LCM = 22×33×5 = 540

Thus the least number exactly divisible by 12, 15, 20 and 27 is 540.

Ques 16: Find the least number which when divided by 6, 7, 8, 9, and 12 leaves the same remainder 1 in each case.

• 501
• 506
• 505
• 510

505
Sol. Writing the prime factorization:
6=2×3
7=1×7
8=2×2×2
9=3×3
12=2×2×3
the least common multiple therefore is:
2×7×2×2×3×3=504
So numbers which on being divided by 6,7,8,9 and 12 leave a remainder 1 are : 504+1=505

Ques 17: Find the largest number of four digits exactly divisible by 12, 15, 18, and 27.

• 9900
• 9500
• 9720
• 9828

9720
Sol. The largest number of four digits is 9999.
Required number must be divisible by LCM of 12, 15, 18, 27 i.e., 540
On dividing 9999 by 540, we get 279 as remainder
Required number = (9999-279) = 9720

Ques 18: Find the smallest number of five digits exactly divisible by 16, 24, 36, and 54.

• 12000
• 10400
• 11500
• 10368

10368
Sol. Smallest number of 5 digits is 10000.
Required number must be divisible by LCM of 16, 24, 36, 54 i.e., 432
On dividing 10000 by 432, we get 64 as remainder
Required number = 10000 +(432-64) = 10368

Ques 19: Find the least number which when divided by 20, 25, 35, and 40 leaves remainders 14, 19, 29, and 34 respectively.

• 1548
• 1335
• 1428
• 1394

1394
Sol. Here, (20-14)= 6, (25-19) = 6, (35-29) = 6 and (40-34)= 6
Required number = (LCM of 20, 25, 35, 40) – 6 = 1394

Ques 20: Find the least number which when divided 5, 6, 7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.

• 1552
• 1683
• 1697
• 1600

1683
Sol. LCM of 5, 6, 7, 8 = 840
Required number is of the form 840k+3
Least value of k for which (840k+3) is divisible by 9 is k = 2
Required number = (840×2+3) = 1683

Ques 21: The traffic lights at three different road crossings change after every 48 sec, 72 sec, and 108 sec, respectively. If they all change simultaneously at 8 : 20 : 00 hours, then at what time will they again change simultaneously?

• 8 : 30 : 00 hours
• 8 : 35 : 08 hours
• 8 : 27 : 12 hours
• 8 : 25 : 26 hours

8 : 27 : 12 hours
Sol. Interval of change = (LCM of 48, 72, 108) sec = 432 sec
So, the light will again change simultaneously after every 432 sec, i.e., 7 min 12 sec.
Hence, the next simultaneous change will take place at 8:27:12 hrs.

Ques 22: Arrange the fractions 17/18, 31/36, 43/45, 59/60 in the ascending order.

• 31/36, 59/60, 17/18, 43/45
• 31/36, 17/18, 43/45, 59/60
• 17/18, 43/45, 31/36, 59/60
• 43/45, 59/60, 31/36, 17/18

31/36, 17/18, 43/45, 59/60
Sol. Step-by-step explanation:

Value of fractions:

17/18=0.9444

31/36=0.8611

43/45=0.9555

59/60=0.9833

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