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

2928.The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
12
16
24
48
Explanation:

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.

2931.Three numbers are in the ratio of 2 : 3 : 4 and their L.C.M. is 240. Their H.C.F. is:
40
30
20
10
Explanation:

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 $4x$ $=x=20$

2936.Which of the following fraction is the largest ?
$ \dfrac{7}{8} $
$ \dfrac{13}{16} $
$ \dfrac{31}{40} $
$ \dfrac{63}{80} $
Explanation:

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.

2939.Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
4
5
6
8
Explanation:

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

2942.The H.C.F of two 9/10,12/25,18/35 and 21/40 is:
$ \dfrac{3}{5} $
$ \dfrac{252}{5} $
$ \dfrac{3}{1400} $
$ \dfrac{63}{700} $
Explanation:

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

2943.If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
$ \dfrac{55}{601} $
$ \dfrac{601}{55} $
$ \dfrac{11}{120} $
$ \dfrac{120}{11} $
Explanation:

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} $

2944.Which of the following has the most number of divisors?
99
101
176
182
Explanation:

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.

2945.The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
1677
1683
2523
3363
Explanation:

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.

2946.The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
3
13
23
33
Explanation:

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.

2947.What is the least number which when divided by 8, 12, 15 and 20 leaves in each case a remainder of 5 ?
125
117
132
112
Explanation:

LCM of 8, 12, 15 and 20 = 120

Required Number = 120 + 5 = 125


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