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Q:

# Product of two co-prime numbers is 117. Their L.C.M  should be

 A) 1 B) 117 C) Equal to their H.C.F D) cannot be calculated

Answer:   B) 117

Explanation:

H.C.F of co-prime numbers is 1. So, L.C.M = $\inline&space;\fn_jvn&space;\frac{117}{1}$ =117

Q:

The difference of two numbers is 14. Their LCM and HCF are 441 and 7. Find the two numbers ?

 A) 63 and 49 B) 64 and 48 C) 62 and 46 D) 64 and 49

Answer & Explanation Answer: A) 63 and 49

Explanation:

Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y

Difference = 14
=> 7x - 7y = 14
=> x - y = 2

product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63

Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7

The numbers are 7x and 7y
=> 63 and 49

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3 116
Q:

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 ?

 A) 13/125 B) 14/57 C) 11/120 D) 16/41

Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120

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1 136
Q:

The largest measuring cylinder that can accurately fill 3 tanks of capacity 98, 182 and 266 litres each, is of capacity ?

 A) 7 lts B) 14 lts C) 98 lts D) 42 lts

Answer & Explanation Answer: B) 14 lts

Explanation:

To know the the measuring cylinder that can fill all the given capacities , they must be divisible by the required number.

98,182,266 all are divisible by 14
So  14 litres  is the largest cylinder that can fill all the given cylinders.

(or)

The other method is take HCF of all given capacities i.e 98, 182 and 266.

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2 194
Q:

If N is the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. What is the sum of the digits of N ?

 A) 6 B) 8 C) 4 D) 9

Answer & Explanation Answer: C) 4

Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

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

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

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4 100
Q:

Find the greatest number that will divide 964, 1238 and 1400 leaving remainder of 41,31 and 51 respectively ?

 A) 64 B) 69 C) 71 D) 58

Answer & Explanation Answer: C) 71

Explanation:

To find the greatest number which divides the numbers 964, 1238 and 1400 leaving the remainders 41, 31 and 51 is nothing but the HCF of (964 - 41), (1238 - 31), (1400 - 51).

Therefore, HCF of 923, 1207 and 1349 is 71.

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2 223