Drill # 3:  HCF and LCM

Drill # 3: HCF and LCM

Assessment

Flashcard

Mathematics

4th - 5th Grade

Hard

Created by

Ma Irene Adrales

Used 1+ times

FREE Resource

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6 questions

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1.

FLASHCARD QUESTION

Front

Find the LCM of 63 and 45.

Back

315

Answer explanation

  • 63 = 3 × 3 × 7 = 3² × 7

  • 45 = 3 × 3 × 5 = 3² × 5

LCM=3×3×5×7=315​

2.

FLASHCARD QUESTION

Front

Find the HCF of 152 and 114.

Back

38

Answer explanation

  • 152 = 2³ × 19

  • 114 = 2 × 3 × 19

HCF=2×19

=38​

3.

FLASHCARD QUESTION

Front

Find the smallest number that leaves a remainder of 7 when divided by 12, 16, and 125.

Back

6007

Answer explanation

This means:

  • N − 7 is divisible by 12

  • N − 7 is divisible by 16

  • N − 7 is divisible by 125

Steps

  • 1. Subtract the remainder → N − 7 must be divisible by 12, 16, and 125.

  • 2. Find the LCM of 12, 16, and 125 → 6000

  • 3. N - 7 = 6000, therefore add the remainder back → 6000 + 7 = 6007

4.

FLASHCARD QUESTION

Front

How long, in minutes, will it take for Car X, Car Y, and Car Z to be back at the starting point at the same time if Car X makes a lap every 60 seconds, Car Y every 80 seconds, and Car Z every 90 seconds?

Back

720 seconds

Answer explanation

This is a sample of a perfect timing: synchronizing events problem. To solve, find the LCM of 60, 80 and 90 seconds.

5.

FLASHCARD QUESTION

Front

A basket company uses 1260 boxes of chocolates, 420 bottles of wine, and 630 tins of biscuits to make as many gift baskets as possible for an event. Each basket has the same number of each item. Find the greatest number of baskets that can be packed.

Back

210

Answer explanation

This is an example of a perfect split problem using the greatest common factor. So find the HCF of 1260, 420 and 630.

6.

FLASHCARD QUESTION

Front

Find two numbers greater than 20 that have an LCM of 72 and an HCF of 12. Write the numbers in ascending order using comma as separator without space in between.

Back

24,36

Answer explanation

Prime factorization:

  • 12 = 2² × 3

  • 72 = × 3²

Let a and b be the two numbers.

a = 2² × 3

b = 2² × 3

  • 72 = × 3² so 2 and 3 should be included as factors

Possible combinations

a = 2² × 3 × 2 = 24

b = 2² × 3 × 3 = 36

or

a = 2² × 3 × 2 x 3 = 72

b = 2² × 3 = 12

The condition states that the numbers must be greater than 20. Therefore, combination 1 is the answer.

24 and 36.