The sum of the two numbers is 528 and their hcf is 33. how many pairs of such numbers can be?

136

Answer: A) 4

Explanation:

Let the required numbers be 33a and 33b.

Then 33a +33b= 528 => a+b = 16.

Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).

Therefore, Required numbers are ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)

The number of such pairs is 4

Subject: HCF and LCM - Quantitative Aptitude - Arithmetic Ability

Free

100 Qs. 200 Marks 60 Mins

Given

The sum of two numbers = 528

HCF = 33

Concept

1) HCF: The greatest common divisor of two or more positive integers happens to be the largest positive integer that divides the numbers without leaving a remainder.

2) Coprime numbers: Two numbers are said to be coprime when they have no common factors other than 1. For example : (3, 5), (14, 25), etc.

Calculation

33 is a factor of both the numbers so, let the numbers be 33a and 33b, where 'a' and 'b' are coprime numbers

33a + 33b = 528

⇒ a + b = 528/33 = 16

For a + b = 16, a and b can have the following possible values :

(a = 1, b = 15)

(a = 3, b = 13)

(a = 5, b = 11)

(a = 7, b = 9)

∴ Number of possible pairs is 4

India’s #1 Learning Platform

Start Complete Exam Preparation

Video Lessons & PDF Notes

Get Started for Free Download App

Trusted by 2,74,81,895+ Students

Answer

Verified

Hint: By reading the questions you get that the sum of the two numbers is 528 and the HCF is 33 which can be used in the sum of two different variables. secondly, the no having sum 33, should also have 1 only as an HCF

Complete step by step solution: We can stepwise get through the information provided as you can see that the first information is the sum of two numbers and second both have the highest common factor as 33.

From this we can conclude, both are divisible by 33 and we can assume both no as a multiple of 33.So let’s Assume one no. as $33 \times x$ and the second number as $33 \times y$.As we get back to our first information. Sum of both this number is equal to 524Therefore, $33x + 33y = 528$ $33\left( {x + y} \right) = 528$ $x + y = \dfrac{{528}}{{33}} = 16$ Now we need a pair of x and y which can have a sum equal to 16.So pairs can be identified by assuming x as one variable then $16 - x$ as the second.Therefore at \[\begin{array}{l}x = 1,\;\;y = 16 - x = 15\\x = 2,\;\;y = 16 - 2 = 14\\x = 3,\;\;y = 16 - 3 = 13\\x = 4,\;\;y = 16 - 4 = 12\\x = 5,\;\;y = 16 - 5 = 11\\x = 6,\;\;y = 16 - 6 = 10\\x = 7,\;\;y = 16 - 7 = 9\\x = 8,\;\;y = 16 - 8 = 8\end{array}\] Further, for x it will repeat the same changing the values of x and y.The 8 pairs (1, 15) (2, 14) (3, 13) (4, 12) (5, 11) (6, 10) (7, 9) and (8, 8) satisfies the condition appropriately, but the condition of being a highest common factor as 33 will be satisfied by few above pairs.So now let’s eliminate the Pairs having HCF between them.The Pairs which do not have HCF other than 1 are (1, 15) (3, 13) (5, 11) (7, 9).Therefore, there are 4 Pairs. Option C is the correct answer.

Note: As we proceed with the situation given in the question. Generally, we square or Square root both the side, which leads to the wrong solution turning into right for that particular stage. we need to check the final solution obtained with our initial conditions to get the final solution. Similarly, we got 8 pairs at some stage. but the condition of HCF reduced it to 4.

Read Less