**LCM and HCF Aptitude Questions With Solutions:** Want to practice questions from HCF and LCM, so here we are with an exercise of **LCM and HCF Aptitude Questions With Solutions**. These questions are really important from your examination point of view. So, if you are a candidate who is preparing for some government examination like **banking, railways, SSC,** and many more, then you must scroll down the page and solve these questions, so that you will have the idea about the type of questions that are asked, and the art of solving them.

## LCM and HCF Aptitude Questions With Solutions

Ques 1: Find the HCF of 2^{3}x3^{2}x5x7^{4}, 2^{2}x3^{5}x5^{2}x7^{3}, 2^{3}x5^{3}x7^{2}

- 980
- 935
- 940
- 945

980

Sol. The prime numbers common to given numbers are 2, 5, and 7.

HCF = 2^{2}x5x7^{2} = 980

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

- 24
- 45
- 15
- 36

36

Sol. 108 = 2^{2} x 3^{3}, 288 = 2^{5} x 3^{2} and 360 = 2^{3} x 5 x 3^{2}.

H.C.F. = 2^{2} x 3^{2} = 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 2^{2}x3^{3}x5x7^{2}, 2^{3}x3^{2}x5^{2}x7^{4}, 2x3x5^{3}x7x11

- 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 = 2^{3}x3^{3}x5^{3}x7^{4}x11

= 713,097,000

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

- 42300
- 37800
- 43200
- 35000

37800

Sol. 72 = 2^{3} x 3^{2} ,. 108 = 3^{3} x 2^{2} ,. 2100 = 2^{2} x 5^{2} x 3 x 7. L.C.M. = 2^{3} x 3^{3} x 5^{2} 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 = 2^{2}×3

15 = 3×5

20 = 2×2×5 = 2^{2}×5

27 = 3×3×3 = 3^{3}

LCM = 2^{2}×3^{3}×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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