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QUESTION: 1

How many pairs of integers (x, y) exist such that the product of x, y and HCF (x, y) = 1080?

Solution:

We need to find ordered pairs (x, y) such that xy * HCF(x, y) = 1080.

Let x = ha and y = hb where h = HCF(x, y)

⇒ HCF(a, b) = 1.

So h^{3}(ab) = 1080 = (2^{3})(3^{3})(5).

We need to write 1080 as a product of a perfect cube and another number.

Four cases:

1. h = 1, ab = 1080 and b are co-prime. We gave 4 pairs of 8 ordered pairs (1, 1080), (8, 135), (27, 40) and (5, 216). (Essentially we are finding co-prime a,b such that a*b = 1080).

2. h = 2, We need to find number of ways of writing (3^{3}) * (5) as a product of two co-prime numbers. This can be done in two ways - 1 and (3^{3}) * (5) , (3^{3}) and (5)

number of pairs = 2, number of ordered pairs = 4

3. h = 3, number of pairs = 2, number of ordered pairs = 4

4. h = 6, number of pairs = 1, number of ordered pairs = 2

Hence total pairs of (x, y) = 9, total number of ordered pairs = 18.

The pairs are (1, 1080), (8, 135), (27, 40), (5, 216), (2, 270), (10, 54), (3, 120), (24, 15) and (6, 30).

QUESTION: 2

Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

Solution:

**Note:** When a number is divided by 8, a remainder of 7 can be thought of as a remainder of -1.

This idea is very useful in a bunch of questions. So, N = 5a - 1 or N + 1 = 5a

N = 6b - 1 or N + 1 = 6b

N = 7c - 1 or N + 1 = 7c

N = 8d - 1 or N + 1 = 8d

N = 9e - 1 or N + 1 = 9e

N + 1 can be expressed as a multiple of (5, 6, 7, 8, 9)

N + 1 = 5a*6b*7c*8d*9e

Or N = (5a*6b*7c*8d*9e) - 1

Smallest value of N will be when we find the smallest common multiple of (5, 6, 7, 8, 9) or LCM of (5, 6, 7, 8, 9)

N = LCM (5, 6, 7, 8, 9) - 1 = 2520 - 1 = 2519.

The question is "Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?"

Hence the answer is "2519"

QUESTION: 3

How many pairs of positive integers x, y exist such that HCF of x, y = 35 and sum of x and y = 1085?

Solution:

Let HCF of (x, y) be h. Then we can write x = h * a and y = h * b.

Furthermore, note that HCF (a, b) = 1. This is a very important property. One that seems obvious when it is mentioned but a property a number of people overlook.

So, we can write x = 35a; y = 35b

x + y = 1085 => 35(a + b) = 1085. => (a + b) = 31. We need to find pairs of co-prime integers that add up to 31. (Another way of looking at it is to find out integers less than 31 those are co-prime with it or phi(31) as had mentioned. More on this wonderful function in another post).

Since 31 is prime. All pairs of integers that add up to 31 will be co-prime to each other. Or, there are totally 15 pairs that satisfy this condition.

The question is "How many pairs of positive integers x, y exist such that HCF of x, y = 35 and sum of x and y = 1085?"

Hence the answer is "15"

QUESTION: 4

How many pairs of positive integers x, y exist such that HCF (x, y) + LCM (x, y) = 91?

Solution:

Let us x = h * a; y = h * b

a and b are co-prime. So, LCM of (x, y) = h * a * b

So, in essence h + h * a * b = 91. Or h(ab + 1) = 91

Now, 91 can be written as 1 * 91 or 7 * 13

Or, we can have HCF as 1, LCM as 90 -

There are 4 pairs of numbers like this (2, 45), (9, 10), (1, 90) and (5, 18)

We can have HCF as 7, ab + 1 = 13 => ab = 12 => 1 * 12 or 4 * 3

Or, the pairs of numbers are (7, 84) or (21, 28)

The third option is when HCF = 13, ab + 1 = 7 => ab = 6

Or (a, b) can be either (1, 6) or (2, 3)

The pairs possible are (13, 78) and (26, 39)

There are totally 8 options possible - (2, 45), (9, 10), (1, 90), (5, 18), (7, 84), (21, 28), (13, 78) and (26, 39).

8 Pairs.

The question is "How many pairs of positive integers x, y exist such that HCF (x, y) + LCM (x, y) = 91?"

Hence the answer is "8 pairs"

QUESTION: 5

Sum of two numbers x, y = 1050. What is the maximum value of the HCF between x and y?

Solution:

x = 525 and y = 525 works best.

If the question states x, y have to be distinct, then the best solution would be x = 350, y = 700, HCF = 350.

So the HCF is 525.

The question is "What is the maximum value of the HCF between x and y?"

Hence the answer is "525"

QUESTION: 6

The sum of two non co–prime numbers added to their HCF gives us 91. How many such pairs are possible?

Solution:

Let HCF of the numbers be h. The numbers can be taken as ha + hb, where a, b are coprime.

h + ha + hb = 91

h(1 + a + b) = 91

h ≠ 1

h = 7

⇒ 1 + a + b = 13 a + b = 12

h = 13

⇒ 1 + a + b = 7

⇒ a + b = 6

Case 1: h = 7, a + b = 12

(1, 11), (5, 7) => Only 2 pairs are possible as a, b have to be coprime.

Case 2: h = 13, a + b = 6

(1, 5) only one pair is possible as a, b have to be coprime.

Overall, 3 pairs of numbers are possible – (7, 77) (35, 49) and (13, 65)

The question is "How many such pairs are possible?"

Hence the answer is "3 Pairs"

QUESTION: 7

There are 2 numbers such that a > b, HCF (a, b) = h and LCM (a, b) = l. What is the LCM of a – b and b?

Solution:

Product of 2 numbers = LCM * HCF.

Given a > b, HCF = h, LCM = l

From the above we can say, HCF of (a – b, b) = h

LCM x HCF = Product of 2 numbers

(a – b)b = h x LCM

LCM = (a - b) b / h

The question is "What is the LCM of a – b and b?"

Hence the answer is "(a - b) b / h"

QUESTION: 8

6 different sweet varieties of count 32, 216, 136, 88, 184, 120 were ordered for a particular occasion. They need to be packed in such a way that each box has the same variety of sweet and the number of sweets in each box is also the same. What is the minimum number of boxes required to pack?

Solution:

All sweets need to packed and each box has the same variety.

This implies the number of sweets in each box should be HCF of different count of sweets

HCF of 32, 216, 136, 88, 184, 120 = 23 = 8

Minimum number of boxes = (32 + 216 + 136 + 88 + 184 + 120) / 8 = 97

The question is "What is the minimum number of boxes required to pack?"

Hence the answer is "97 boxes"

QUESTION: 9

In a large school auditorium, the students are made to sit to watch the programmes. If the teachers make a row of students of 16 each, there will be 12 students left. If they make rows of 24 each, then there will be 20 students left, if they make rows of 25 each, there will be 21 students left and if they make rows of 30 each, there will be 26 students left. What is the minimum number of students present in the school?

Solution:

16 in a row --> 12 left

24 in a row --> 20 left

25 in a row --> 21 left

30 in a row --> 26 left

In all the 4 cases above, the remainder is 4.

(16 – 12) = (24 – 20) = (25 – 21) = (30 – 26)

Hence the required students = LCM (16, 24, 25, 30 ) – 4

= 1200 – 4

= 1196

The question is "What is the minimum number of students present in the school?"

Hence the answer is "1196 students minimum"

QUESTION: 10

LCM of 2 natural numbers p and q where p > q is 935. What is the maximum possible sum of the digits of q?

Solution:

We know, 935 = 5*11*17

Let p = hx

q = hy

Where h is the hcf of p and q.

Therefore, lcm of p and q= hxy

We have lcm = 935.

hxy = 935

If h = 1,

Then p = 935

q = 1

Sum of digits of q = 1

If h = 5,

p = 5*17 = 85

q = 5*11 = 55

Sum of digits of q = 10

If h = 11,

p = 11*17 = 187

q = 11*5 = 55

Sum of digits of q = 10

If h = 17,

p = 17*11 = 187

q = 17*5 = 85

Sum of digits of q = 13

Maximum possible sum of digits of q = 13.

The question is "What is the maximum possible sum of the digits of q?"

Hence the answer is "13"

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