Quantitative Ability Tech Test L.C.M and H.C.F Test 2

Required H.C.F. =$ \dfrac{H.C.F. of 9, 12, 18, 21}{L.C.M. of 10, 25, 35, 40} $=$ \dfrac{3}{1400} $

Let the numbers be $ a $ and $ b $.

Then, $ a $ + $ b $ = 55 and ab = 5 x 120 = 600.

$\therefore$ The required sum =$ \dfrac{1}{a} $+$ \dfrac{1}{b} $=$ \dfrac{a + b}{ab} $=$ \dfrac{55}{600} $=$ \dfrac{11}{120} $

99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

L.C.M. of 5, 6, 7, 8 = 840.

$\therefore$ Required number is of the form 840 k + 3

Least value of $ k $ for which [840$ k $ + 3] is divisible by 9 is $ k $ = 2.

$\therefore$ Required number = $\left(840 \times 2 + 3\right)$ = 1683.

Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29;

First number =$ \left(\dfrac{551}{29} \right) $= 19;    Third number =$ \left(\dfrac{1073}{29} \right) $= 37.

$\therefore$ Required sum = $\left(19 + 29 + 37\right)$ = 85.

Solution 1

LCM of 5, 6, 7 and 8 = 840

Hence the number can be written in the form (840k + 3) which is divisible by 9.

If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9.

If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9.

Hence 1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder

3, but when divided by 9 leaves no remainder.

Solution 2 - Hit and Trial Method

Just see which of the given choices satisfy the given condtions.

Take 3363. This is not even divisible by 9. Hence this is not the answer.

Take 1108. This is not even divisible by 9. Hence this is not the answer.

Take 2007. This is divisible by 9.

2007 ÷ 5 = 401, remainder = 2 . Hence this is not the answer

Take 1683. This is divisible by 9.

1683 ÷ 5 = 336, remainder = 3

1683 ÷ 6 = 280, remainder = 3

1683 ÷ 7 = 240, remainder = 3

1683 ÷ 8 = 210, remainder = 3

Hence 1683 is the answer

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90$ k $ + 4, which is multiple of 7.

Least value of $ k $ for which [90$ k $ + 4] is divisible by 7 is $ k $ = 4.

$\therefore$ Required number = $\left(90 \times 4\right)$ + 4   = 364.

[Refer " How to calculate LCM and HCF of Decimals " here]

Step 1: Make the same number of decimal places in all the given numbers by suffixing

zero(s) in required numbers as needed.

=> 2.04, 0.24 and 0.80

Step 2: Now find the HCF of these numbers without decimal.

=> HCF of 204, 24 and 80 = 4

Step 3: Put the decimal point in the result obtained in step 2 leaving as many digits on its

right as there are in each of the numbers.

i.e., here we need to put decimal point in the result obtained in step 2 leaving two digits

on its right.

=> HCF of 2.04, 0.24 and 0.8 = 0.04

Given numbers are 1.08, 0.36 and 0.90.   H.C.F. of 108, 36 and 90 is 18,

$\therefore$ H.C.F. of given numbers = 0.18.

Let the numbers 13$ a $ and 13$ b $.

Then, 13$ a $ x 13$ b $ = 2028

$\Rightarrow$ ab = 12.

Now, the co-primes with product 12 are $\left(1, 12\right)$ and $\left(3, 4\right)$.

[Note: Two integers $ a $ and $ b $ are said to be co-prime or relatively prime if they have no

common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are $\left(13 \times 1, 13 \times 12\right)$ and $\left(13 \times 3, 13 \times 4\right)$.

Clearly, there are 2 such pairs.