Q 65. LCM AND HCF

What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3?

a) 1063
b) 1073
c) 1083
d) 1183
Answer: c
Practice This Question in Exam Mode

  •   LCM of 6, 9, 12, 15, 18 = 180.
  •   Means this is the smallest number that will be divided by every number.
  •   Such number above 1000 will also be a multiplication of 180.
  •   180 × 5 = 900 < 1000
  •   180 × 6 = 1080. > 1000 however this is the smallest multiple of 180 above 1000.
  •   Now, we need the remainder as 3. So, desired number = 1083.
  •   This question can also be solved backward by trying every options.
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Q 12. LCM AND HCF

If you have two straight sticks of length 7.5 feet and 3.25 feet, what is the minimum length can you measure?

a) 0.05 foot
b) 0.25 foot
c) 1 foot
d) 3.25 feet
Answer: b
Practice This Question in Exam Mode

Length of sticks are 7.5 feet ad 3.25 feet

Formula used:

HCF of (A/B) and (C/D) = (HCF of A and C)/(LCM of B and D)

Calculation:

First convert decimal into fraction

7.5 = 75/10 = 15/2

3.25 = 325/100 = 13/4

Now, minimum length will be measured by HCF of {15/2, 13/4}

HCF = 1/4 = 0.25

The minimum length can be measured by two stick is 0.25 foot.

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Q 51. LCM AND HCF

What is the greatest length x such that 3 1/2 m and 8 3/4 m are integral multiples of x?

a) 1 1/2 m
b) 1 1/3 m
c) 1 1/4 m
d) 1 3/4 m
Answer: d
Practice This Question in Exam Mode

3 ½ = 7/2 and 8 ¾= 35/4

HCF (Fractions) = HCF (Numerator) / LCM (Denominator)

x = H. C. F. of (7/2) and (35/4) = H.C.F. of 7 and 35 / L.C.M. of 2 and 4 = 7/4 Hence, x = 7/4 m or 1 3/4 m

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Q 54. LCM AND HCF

What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case?

a) 1012
b) 1022
c) 1122
d) 1222
Answer: b
Practice This Question in Exam Mode

  L.C.M. of 3, 4, 5 and 6 = 60

Let the required number be 60x + 2

If x = 17, then the number = 60 × 17 + 2 = 1020 + 2 = 1022

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Q 15. LCM AND HCF

In a school every student is assigned a unique identification number. A student is a football player if and only if the identification number is divisible by 4, whereas a student is a cricketer if and only if the identification number is divisible by 6. If every number from 1 to 100 is assigned to a student, then how many of them play cricket as well as football?

a) 4
b) 8
c) 10
d) 12
Answer: b
Practice This Question in Exam Mode

  The required number should be completely divisible by both 4 and 6.

  That means it should be divisible by LCM of 4 and 6, which is 12.

  Such 8 numbers are possible which are completely divisible by 12. They are 12, 24, 36, 48, 60, 72, 84 and 96.

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Q 6. LCM AND HCF

Five persons fire bullets at a target at an interval of 6, 7, 8, 9 and 12 seconds respectively. The number of times they would fire the bullets together at the target in an hour is

a) 6
b) 7
c) 8
d) 9
Answer: b
Practice This Question in Exam Mode

The five persons will together fire at the target at an interval of 'r second where

t = LCM (6, 7, 8, 9. 12)

LCM (6, 7, 8, 9, 12) = 23 x 32 x 7 = 504

⸫ t = 504 s

Number of seconds in an hour = 3600

⸫ Number of times the five persons would fire the bullets together at the target in an hour =

(where  denotes the greatest integer less than or equal to t) =  = 7

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Q 46. LCM AND HCF

A bell rings every 18 minutes. A second bell rings every 24 minutes. A third bell rings every 32 minutes. If all the three bells ring at the same time at 8 o'clock in the morning, at what other time will they all ring together?

a) 12:40 hrs
b) 12:48 hrs
c) 12:56 hrs
d) 13:04 hrs
Answer: b
Practice This Question in Exam Mode

Interval of first bell ring = 18 min

Interval of second bell ring = 24 min

Interval of third bell ring = 32 min

Now interval after which they will ring together = LCM (18 min, 24 min and 32 min)

= 4 x 3 x 3 x 2 x 2 x 2 = 288 min

⸫ Bells will ring after 288 min = = 4 h 48 min

Now, if they ring at 8 O’ clock in the morning, then they will ring again at = 8 + 4 h 48 min = 12 : 48 h

Hence. option (b) is correct.

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Q 40. LCM AND HCF

Three persons start walking together and their steps measure 40 cm, 42 cm and 45 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

a) 25 m 20 cm
b) 50 m 40 cm
c) 75 m 60 cm
d) 100 m 80 cm
Answer: a
Practice This Question in Exam Mode

It’s a simple LCM problem. Take the LCM of 40, 42 and 45 to know at what point will they find a common multiple. It is 2520 cm i.e. 25 m 20 cm.

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