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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

Workspace

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.

Workspace

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:

A.

8

B.

5

C.

4

D.

6

Answer with explanation

Answer: Option CExplanation

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

Workspace

HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers.

A.

32/221

B.

20/268

C.

15/275

D.

16/275

Answer with explanation

Answer: Option DExplanation

x + y = 80, xy = 5*275 = 1375

sum of reciprocals = 1/x + 1/y

= (x+y)/xy

= 80/1375 = 16/275

Workspace

Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.

A.

31

B.

30

C.

16

D.

14

Answer with explanation

Answer: Option AExplanation

LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.

Now 60/2 = 30

Adding one bell at the starting it will 30+1 = 31

Workspace

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

A.

3

B.

1

C.

2

D.

4

Answer with explanation

Answer: Option CExplanation

Let the numbers 13*a* and 13*b*.

Then, 13*a* x 13*b* = 2028

ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers *a* and *b* are said to be *coprime* or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

Workspace

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

A.

180

B.

120

C.

80

D.

40

Answer with explanation

Answer: Option DExplanation

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

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

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

The numbers are (3 x 40), (4 x 40) and (5 x 40).

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

Workspace

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