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Quantitative Ability Tech Test L.C.M and H.C.F Test 3

2923.Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
4
7
9
13
Explanation:

Required number = H.C.F. of $\left(91 - 43\right)$, $\left(183 - 91\right)$ and $\left(183 - 43\right)$

     = H.C.F. of 48, 92 and 140 = 4.

2924.Three numbers which are co-prime to each other are such that the product of the first two is 119 and that of the last two is 391. What is the sum of the three numbers?
47
43
53
51
Explanation:

Since the numbers are co-prime, their HCF = 1

Product of first two numbers = 119

Product of last two numbers = 391

The middle number is common in both of these products. Hence, if we take HCF of 119

and 391, we get the common middle number.

HCF of 119 and 391 = 17

=> Middle Number = 17

First Number $=\dfrac{119}{17}=7$

Last Number $=\dfrac{391}{17}=23$

Sum of the three numbers = 7 + 17 + 23 = 47

2925.The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
101
107
111
185
Explanation:

Let the numbers be 37$ a $ and 37$ b $.

Then, 37$ a $ x 37$ b $ = 4107

$\Rightarrow$ ab = 3.

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

So, the required numbers are $\left(37 \times 1, 37 \times 3\right)$ i.e., $\left(37, 111\right)$.

$\therefore$ Greater number = 111.

2925.The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
101
107
111
185
Explanation:

Let the numbers be 37$ a $ and 37$ b $.

Then, 37$ a $ x 37$ b $ = 4107

$\Rightarrow$ ab = 3.

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

So, the required numbers are $\left(37 \times 1, 37 \times 3\right)$ i.e., $\left(37, 111\right)$.

$\therefore$ Greater number = 111.

2926.Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
40
80
120
200
Explanation:

Let the numbers be 3$ x $, 4$ x $ and 5$ x $.

Then, their L.C.M. = 60$ x $.

So, 60$ x $ = 2400 or x = 40.

$\therefore$ The numbers are $\left(3 \times 40\right)$, $\left(4 \times 40\right)$ and $\left(5 \times 40\right)$.

Hence, required H.C.F. = 40.

2927.The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
279
283
308
318
Explanation:

Other number =$ \left(\dfrac{11 \times 7700}{275} \right) $= 308.

2949.What is the greatest number which divides 24, 28 and 34 and leaves the same remainder in each case?
1
2
3
4
Explanation:

If the remainder is same in each case and remainder is not given, HCF of the

differences of the numbers is the required greatest number.

34 - 24 = 10

34 - 28 = 6

28 - 24 = 4

Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder

= HCF of 10, 6, 4

= 2

2950.The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
9000
9400
9600
9800
Explanation:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

$\therefore$ Required number $\left(9999 - 399\right)$ = 9600.

2951.What is the LCM of 2/3,5/6 and 4/9 ?
3/10
3/20
10/3
20/3
Explanation:

LCM of fractions $=\dfrac{\text{LCM of Numerators}}{\text{HCF of Denominators}}$

LCM of $\dfrac{2}{3}$, $\dfrac{5}{6}$ and $\dfrac{4}{9}$

$=\dfrac{\text{LCM (2, 5, 4)}}{\text{HCF (3, 6, 9)}}$

=$\dfrac{20}{3}$

2953.What is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22 ?
1286
1436
1216
1386
Explanation:

LCM of 12, 14, 18 and 22 = 2772

Hence the least number which will be exactly divisible by 12, 14, 18 and 22 = 2772

2772 ÷ 2 = 1386

1386 is the number which when doubled, we get 2772

Hence, 1386 is the least number which when doubled will be exactly divisible by 12, 14,

18 and 22.

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