Quantitative Ability Tech Test L.C.M and H.C.F Test 1

Let the numbers be 3$ x $ and 4$ x $. Then, their H.C.F. = $ x $. So, $ x $ = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

Let the numbers be $2x$, $3x$ and $4x$

LCM of $2x$, $3x$ and $4x$ = $12x$

12x=240

$\Rightarrow x=\dfrac{240}{12}=20$

H.C.F of $2x$, $3x$ and $4x$ $=x=20$

L.C.M. of 8, 16, 40 and 80 = 80.

$ \dfrac{7}{8} $=$ \dfrac{70}{80} $;  $ \dfrac{13}{16} $=$ \dfrac{65}{80} $;  $ \dfrac{31}{40} $=$ \dfrac{62}{80} $

Since,$ \dfrac{70}{80} $>$ \dfrac{65}{80} $>$ \dfrac{63}{80} $>$ \dfrac{62}{80} $, so$ \dfrac{7}{8} $>$ \dfrac{13}{16} $>$ \dfrac{63}{80} $>$ \dfrac{31}{40} $

So,$ \dfrac{7}{8} $is the largest.

N = H.C.F. of $\left(4665 - 1305\right)$, $\left(6905 - 4665\right)$ and $\left(6905 - 1305\right)$

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = $\left( 1 + 1 + 2 + 0 \right)$ = 4

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.

L.C.M. of 5, 6, 4 and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

$\therefore$ Number to be added = $\left(60 - 37\right)$ = 23.

LCM of 8, 12, 15 and 20 = 120

Required Number = 120 + 5 = 125