LCM and HCF Test

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LCM and HCF Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

The smallest positive number which leaves a remainder of 1 when it is divided by 3, 4, 5 or 7 is

A
85

B
106

C
141

D
420

E
421

Solution

The smallest positive number which leaves a remainder of 1 when it is divided by 3, 4, 5 or 7 is
LCM of (3,4,5,7) + 1
Required no. = 420 + 1 = 421
Question 2
1 / -0

What is the least number by which 825 must be multiplied in order to produce a multiple of 715?

A
13

B
11

C
9

D
8

E
7

Solution

825 = 3 × 5 × 5 × 11
715 = 5 × 11 × 13.
In 825, 13 is missing. Rest all the factors of 715 are present in 825.
Question 3
1 / -0

HCF of 4 × 27 × 3125; 8 × 9 × 25 × 7 and 16 × 81 × 5 × 11 × 49 is

A
180

B
360

C
540

D
260

E
240

Solution

4 × 27 × 3125 = 4 × 9 × 3 × 5 × 5 × 125
8 × 9 × 25 × 7 = 4 × 2 × 9 × 5 × 5 × 7
16 × 81 × 5 × 11 × 49 = 4 × 4 × 9 × 9 × 5 × 11 × 7 × 7
Common factors with least powers = 4 × 9 × 5 = 180.
Question 4
1 / -0

What is the smallest sum of money which contains an integral number of $2.50, $20, $1.20 and $7.50 each?

A
$60

B
$80

C
$90

D
$100

E
$120

Solution

The smallest sum of money that contains an integral number of $2.50, $20, $1.20 and $7.50 is the money the numerical value of which is equivalent to the LCM of 2.50, 20, 1.20 and 7.50
2.50 = 25/ 10 = 5/2
1.20 = 12/10 = 6/5
7.50 = 75/10 = 15/2
Thus, we need to evaluate the LCM of 5/2, 20/1, 6/5 and 15/2 = LCM of (5, 20, 6, 15)/(HCF of 2, 1 , 5) = 60/1
Thus the smallest sum of money is $ 60
Question 5
1 / -0

Find the least square number which is exactly divisible by 4, 5, 6, 15 and 18.

A
900

B
1200

C
1500

D
1600

E
1800

Solution

LCM of 4, 5, 6, 15, and 18 = 180
So, Required answer must be the multiple of 180.
180 can be written as = 2 × 2 × 3 × 3 × 5 Only 5 is not in pair,
We need to multiply by 5 for perfect square.
180 × 5 = 900
So, the least number be 900.
Therefore, Option 1 is correct.
Question 6
1 / -0

From 3 drums of milk, 27 L, 33 L and 45 L are to be drawn. To do it in the minimum number of repetitions, the capacity of the measuring can should be

A
27 L

B
11 L

C
9 L

D
3 L

E
2 L

Solution

To find the required answer, we need to find the HCF of these numbers (27, 33 and 45). The HCF is the required answer.
27 = 3³
33 = 3 × 11
45 = 3² × 5
HCF = 3
Therefore, option 4 is the correct answer.
Question 7
1 / -0

Two cyclists are preparing for the Olympics. The first cyclist takes 10 minutes to cover one full round, whereas the second cyclist takes 9 minutes. Find the time when they will both be together at the starting point, if they both start simultaneously.

A
90 minutes

B
95 minutes

C
98 minutes

D
99 minutes

E
109 minutes

Solution

This might deceptively seem to be a problem on Time, Speed and Distance. But the logic is based on the concept of LCM.
The first cyclist will be at the starting point at every multiple of 10 minutes. The second will be at the starting point at every multiple of 9 minutes. So, they will be together at the starting point for the first time, after the LCM of 10 and 9 = 90 minutes.
In this problem, in 90 minutes, the first cyclist will complete 9 rounds and the second will complete 10 rounds. So, the second cyclist has taken a lead of 1 full round over the first. It coincidentally happens that they will be meeting for the first time at the starting point after their start.
Question 8
1 / -0

A heap of coconuts is divided into groups of 2, 3 or 5 and each time, no coconut is left over. Find the least number of coconuts in the heap.

A
1

B
10

C
15

D
25

E
30

Solution

Let X be the number of coconuts desired. X must be a common multiple of 2, 3 and 5, otherwise some coconuts will be left over. Further, the number X must be the least possible number. So, X has to be the LCM of 2, 3, 5, which is 30. So, the least number of coconuts in the heap is 30.
Question 9
1 / -0

Four bells ring at intervals of 6, 8, 12 and 18 seconds. They start ringing simultaneously at 12 o'clock. How many times will they ring simultaneously in 6 minutes?

A
7

B
5

C
6

D
4

E
8

Solution

LCM of 6, 8, 12, 18 is 72.
The bells will ring again simultaneously after 72 seconds, i.e. 1 min 12 seconds.
In six minutes, they will ring simultaneously for + 1 = 5 + 1 = 6 times.
Question 10
1 / -0

An electric wire is sold only in multiples of 1 m and a customer requires several lengths of wire, each 85 cm long. To avoid any wastage and to minimise labour, he should purchase a minimum length of

A
1.7 m

B
8.5 m

C
17 m

D
85 m

E
90 m

Solution

Take the LCM of 100 cm and 85 cm = 1700 cm = 17 m.
Question 11
1 / -0

Three men start together to walk along a road at the same rate. The lengths of their strides are 68 cm, 51 cm and 85 cm. How far will they go before they are "IN STEP" again?

A
10.2 m

B
102 m

C
120 m

D
150 m

E
1020 m

Solution

Let the three men be A, B and C. A takes steps in multiples of 68; B takes steps in multiples of 51; and C takes steps in multiples of 85. They start in sync. They will again be in step when they are at a distance which is an integral multiple of 68, 51 and 85, which is the LCM, i.e., 2 × 2 × 3 × 5 × 17 = 1020 cm = 10.2 m.
Question 12
1 / -0

Find the smallest weight which can be divided into exact number of parcels weighing 9 kg, 12 kg, 16 kg and 18 kg.

A
128 kg

B
134 kg

C
144 kg

D
125 kg

E
162 kg

Solution

LCM of 9, 12, 16 and 18 can be calculated as:
9 = 3 × 3
12 = 2 × 3 × 2
16 = 2 × 2 × 2 × 2
18 = 2 × 3 × 3
LCM = 3 × 3 × 2 × 2 × 2 × 2 = 144 (Taking the highest power out of the factorisation of all the numbers)
So, the smallest weight that can be divided into the exact number of parcels weighing 9 kg, 12 kg, 16 kg and 18 kg is 144 kg.
OR
Among options, only 144 is exactly divisible by 9, 12, 16 and 18.
Question 13
1 / -0

The HCF and LCM of two numbers are 13 and 455 respectively. If one of the numbers lies between 75 and 125, find that number.

A
78

B
91

C
104

D
110

E
117

Solution

Let b the number be n
As the HF is 13, so n = 13 a
Now, 455 = 5 × 7 × 13
So, n = 13 , or 5 × 13 or 7 × 13 or 5 × 7 × 13
i.e., n = 13, or 65, or 91 or 455
As n lies between 75 and 125, n = 91
Thus, answer option 2 is correct.
Question 14
1 / -0

There are 408 boys and 312 girls in a school. They are to be divided into equal sections of either boys or girls alone. Find the total number of sections that can be formed.

A
36

B
28

C
30

D
25

E
35

Solution

HCF of 408 and 312 = 24
The number of boys or girls that can be placed in a section = 24.
Total sections formed = = 30
Question 15
1 / -0

Find the greatest number which will divide 1625, 2281 and 4218 leaving 8, 4 and 5 as remainders respectively.

A
9

B
11

C
23

D
33

E
45

Solution

Numbers without remainder are (1625 – 8), (1281 – 4) and (4218 – 5) = (1617), (1277), (4213).
These are exactly divisible by 11.
Hence, 11 is the required answer.
Question 16
1 / -0

There are 408 boys and 312 girls in a school who are to be divided into groups of either boys or girls alone with each group having the same strength. Find the largest number of boys or girls that can be placed in a group.

A
12

B
16

C
18

D
24

E
26

Solution

Largest number of boys and girls that can be placed in a group = HCF of 408 and 312 = 24.
Question 17
1 / -0

The LCM of two numbers is (a + b) and their HCF is k(a - b). If one of the numbers is 'k', then the other number is

A
k(a² - b²)

B
kab

C
a² - b²

D
k²(a² - b²)

E
ka²b²

Solution

LCM = a + b; HCF = k (a - b);
Let the number to be found out be n.
Now, product of numbers = HCF × LCM
So, nk = k(a + b)(a - b) = k(a² - b²)
Or n = (a² - b²)
Thus, answer option 3 is correct.
Question 18
1 / -0

Six bells commence tolling together and toll at the intervals of 2, 4, 6, 8, 10 and 12 seconds, respectively. How many times will they toll together in 30 minutes?

A
10 times

B
12 times

C
16 times

D
15 times

E
18 times

Solution

LCM of 2, 4, 6, 8, 10 and 12 = 120 seconds = 2 minutes
So, in 30 minutes, no. of times they toll together = 30/2 = 15
The bells also toll in the starting, that is, at 0 minute.
Therefore, they will toll together 16 times.
Question 19
1 / -0

How many numbers less than 10,000 are divisible by each of the numbers 21, 36, 66?

A
2

B
3

C
4

D
5

E
6

Solution

21 = 3 × 7 ; 36 = 2²× 3²66 = 2 × 3 × 11
LCM of 21, 36, 66 = 2 × 2 × 3 × 3 × 7 × 11 = 2772
Multiples of 2772 less than 10,000 are : 2772, 5544, 8316.
Question 20
1 / -0

There are 1323 peaches, 2457 apricots and 3024 apples are to be piled up into heaps of equal number of fruits, without mixing the different types. Find the greatest possible number of fruits in a heap.

A
171

B
183

C
187

D
189

E
195

Solution

Greatest possible number of fruits in a heap = H.C.F. of 1323, 2457 and 3024 = 189
Question 21
1 / -0

The smallest number by which 72 should be multiplied in order that the resulting number is exactly divisible by both 27 and 48, is

A
3

B
6

C
9

D
12

E
15

Solution

Let the number be x.
72 × x = LCM of 27 and 48
⇒ 72x = 16 × 27
⇒ x = 6
Question 22
1 / -0

How many times is the HCF of 48, 36, 72 and 24 contained in their LCM?

A
12

B
24

C
36

D
48

E
72

Solution

LCM of 48, 36, 72, 24 = 144
HCF of 48, 36, 72, 24 = 12
So, number of times HCF is contained in LCM = LCM/HCF = 144/12 = 12.
Question 23
1 / -0

Find the least number which when divided by 8, 12 and 16, leaves 3 as remainder in each case, but when divided by 7 leaves no remainder.

A
147

B
63

C
51

D
39

E
36

Solution

N = L.C.M of 8, 12, 16 + Remainder = 48K + 3 = 42K + (6K + 3)
First part of the above equation is an multiple of 7, for second part to be multiple of 7, K has to be 3.
So, N = 147
Question 24
1 / -0

What least number should be added to 3,500 to make it exactly divisible by 42, 49, 56 and 63?

A
42

B
28

C
25

D
24

E
20

Solution

Divisor = L.C.M of 42, 49, 56 and 63 = 3528. So the required no. is 28, which should be added in 3500 to make it divisible by the given numbers.
Question 25
1 / -0

Find the least number which, when divided by 72, 80 and 88, leaves remainders 52, 60 and 68, respectively.

A
8000

B
7900

C
4900

D
4500

E
3900

Solution

As 72 - 52 = 20,
80 - 60 = 20,
88 - 68 = 20,
So, the least number which when divided by 72, 80 and 88, respectively leaves remainders 52, 60 and 68 is given by
LCM of (72, 80, 88) - 20 = 7920 - 20 = 7900.

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Re-Attempt Weekly Quiz Competition

LCM and HCF Test - 3

Result

LCM and HCF Test - 3

Score

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Rank

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SHARING IS CARING

Question 1

85

106

141

420

421

Solution

Question 2

13

11

9

8

7

Solution

Question 3

180

360

540

260

240

Solution

Question 4

$60

$80

$90

$100

$120

Solution

Question 5

900

1200

1500

1600

1800

Solution

Question 6

27 L

11 L

9 L

3 L

2 L

Solution

Question 7

90 minutes

95 minutes

98 minutes

99 minutes

109 minutes

Solution

Question 8

1

10

15

25

30

Solution

Question 9

7

5

6

4

8

Solution

Question 10

1.7 m

8.5 m

17 m

85 m

90 m

Solution

Question 11

10.2 m

k(a² - b²)

a² - b²

k²(a² - b²)

ka²b²