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The least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case, but when divided by 9 leaves no remainder, is:

A.

1677

B.

1677

C.

2523

D.

3363

Answer with explanation

Answer: Option BExplanation

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

Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 X 2 + 3) = 1683

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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.

16/41

B.

14/57

C.

13/125

D.

11/120

Answer with explanation

Answer: Option DExplanation

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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The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A.

127

B.

236

C.

123

D.

305

Answer with explanation

Answer: Option AExplanation

Required number = H.C.F. of (1657 – 6) and (2037 – 5)

= H.C.F. of 1651 and 2032 = 127.

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There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :

A.

87

B.

85

C.

83

D.

81

Answer with explanation

Answer: Option BExplanation

As given the questions these numbers are co-primes, so there is only 1 as their common factor.

It is also given that two products have the middle number in common.

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

So first number is : 551/29 = 19

Third number = 1073/29 = 37

So sum of these numbers is = (19 + 29 + 37) = 85

Workspace

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

A.

56 lts

B.

28 lts

C.

14 lts

D.

7 lts

Answer with explanation

Answer: Option CExplanation

To know 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 liters is the largest cylinder that can fill all the given cylinders.

(or)

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

Workspace

A rectangular courtyard 3.78 meters long 5.25 meters wide is to be paved exactly with square tiles, all of the same size. what is the largest size of the tile which could be used for the purpose?

A.

11 cm

B.

48 cm

C.

42 cm

D.

21 cm

Answer with explanation

Answer: Option DExplanation

3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7

5.25 meters=525 cm = 5 × 5 × 3 × 7

Hence common factors are 3 and 7

Hence LCM = 3 × 7 = 21

Hence largest size of square tiles that can be paved exactly with square tiles is 21 cm.

Workspace

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

A.

45 minutes

B.

46 minutes and 12 seconds

C.

26 minutes and 18 seconds

D.

42 minutes and 36 seconds

Answer with explanation

Answer: Option BExplanation

C.M. of 252, 308 and 198 = 2772.

So, A, B, and C will again meet at the starting point in 2772 sec. i.e., 46 minutes and 12 seconds.

Workspace

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A.

364

B.

104

C.

84

D.

74

Answer with explanation

Answer: Option AExplanation

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.

Required number = (90 x 4) + 4 = 364.

Workspace

The HCF and LCM of two numbers is 70 and 1050 respectively. If the first number is 210, find the second one.

A.

250

B.

350

C.

520

D.

480

Answer with explanation

Answer: Option BExplanation

Product of two numbers = HCF * LCM

So 2nd number = 70*1050/210 =====>>> 350

Workspace

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:

A.

3363

B.

2523

C.

1677

D.

1683

Answer with explanation

Answer: Option DExplanation

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

∴ 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.

∴ Required number = (840 x 2 + 3) = 1683.

Workspace

Find the largest number of four digits which is exactly divisible by 27,18,12,15

A.

9710

B.

9720

C.

9700

D.

9730

Answer with explanation

Answer: Option BExplanation

LCM of 27-18-12-15 is 540.

After dividing 9999 by 540 we get 279 remainder.

So answer will be 9999-279 = 9720

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Find the greatest number which on dividing 1661 and 2045, leaves a remainder of 10 and 13 respectively

A.

129

B.

131

C.

127

D.

125

Answer with explanation

Answer: Option CExplanation

In this type of question, its obvious we need to calculate the HCF, trick is

HCF of (1661 – 10) and (2045 -13)

= HCF (1651, 2032) = 127

Workspace

The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above condition is

A.

12

B.

8

C.

6

D.

4

Answer with explanation

Answer: Option DExplanation

Let the required numbers be 33a and 33b.

Then 33a +33b= 528 => a+b = 16.

Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).

Therefore, Required numbers are ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)

The number of such pairs is 4

Workspace

Find the lowest common multiple of 24, 36 and 40.

A.

360

B.

120

C.

480

D.

240

Answer with explanation

Answer: Option AExplanation

2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.

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