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The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

A.

216

B.

389

C.

443

D.

548

Answer with explanation

Answer: Option DExplanation

Required number = (L.C.M. of 12, 15, 20, 54) + 8

= 540 + 8

= 548.

**(OR)**

Given numbers are 12, 15, 20, 54

To Find: least no. which is divided by given nos. and leaves remainder 8

Least no which is divisible by all given no is LCM of all nos.

LCM means the least common multiple.

First, we find LCM of 12, 15, 20, 54 by prime factorization method

12 = 2 × 2 × 3

15 = 3 × 5

20 = 2 × 2 × 5

54 = 2 × 3 × 3 × 3

LCM ( 12, 15 , 20 , 54 ) = 2 × 2 × 3 × 3 × 3 × 5 = 540

To find the required no. we add 8 to LCM

⇒ Required No. = 540 + 8 = 548

Thus, 548 is the least no which when divided by 12 15 20 and 54 leaves in each case a remainder of 8.

Workspace

The LCM of two numbers is 520 and their HCF is 4. If one of the number is 52, then the other number is

A.

42

B.

32

C.

50

D.

40

Answer with explanation

Answer: Option DExplanation

First number * second number = LCM * HCF

52*Second number = 520 * 4

Second number = (520*4)/52

= 40

Then the other number is 40.

Workspace

What is the number nearest to 10000 which is exactly divisible by 3, 4, 5, 6, 7, and 8?

A.

9956

B.

10080

C.

10096

D.

9240

Answer with explanation

Answer: Option BExplanation

Take LCM of 3, 4, 5, 6, 7, and 8. It will be = 840.

Now, 10000/840 = 760 (remainder).

840 – 760 = 80. If we add 80 in the given number, the number is exactly divisible by 840.

So, the required number is 10080.

Workspace

If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other

A.

34

B.

45

C.

77

D.

12

Answer with explanation

Answer: Option CExplanation

For any this type of question, remember

Product of two numbers = Product of their HCF and LCM

So Other number = 693×11/99 = 77

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.

1677

B.

1683

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 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

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

A.

135

B.

127

C.

235

D.

307

Answer with explanation

Answer: Option BExplanation

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

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

Workspace

The LCM of two numbers is 30 and their HCF is 5. One of the numbers is 10. The other is

A.

20

B.

15

C.

32

D.

23

Answer with explanation

Answer: Option BExplanation

First number * Second number = LCM * HCF

Let the second number be x.

10x = 30*5

x = 150/10 = 15

Workspace

Which is the greatest three-digit number which when divided by 6, 9 and 12 leaves a remainder of 3 in each case?

A.

975

B.

996

C.

939

D.

903

Answer with explanation

Answer: Option AExplanation

Greatest three digit number = 999

LCM of 6, 9 and 12

LCM = 2 x 2 x 3 x 3 = 36

On dividing 999 by 36,

Remainder = 27.

∴ The greatest three digit number divisible by 6, 9, and 12

= (999 – 27) = 972

As per the question, the required number is (972 + 3) = 975.

Workspace

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

A.

253

B.

356

C.

384

D.

364

Answer with explanation

Answer: Option DExplanation

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

Find the probability of getting a numbered card when a card is drawn from the pack of 52 cards.

A.

9/13

B.

2/13

C.

4/13

D.

5/13

Answer with explanation

Answer: Option AExplanation

Total Cards = 52. Numbered Cards = (2, 3, 4, 5, 6, 7, 8, 9, 10) 9 from each suit 4 × 9 = 36

P (E) = 36/52 = 9/13

Or

If you count an ace as a number card there are 40 in a pack so p(get a number card) = 40/52 = 10/13.

If aces don’t count then there are 9 x 4 = 36 in a pack so p(get a number card) = 36/52 = 9/13.

Workspace

A.

35

B.

36

C.

37

D.

38

Answer with explanation

Answer: Option CExplanation

If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.

So, the number of liters in each can = HCF of 80,144 and 368 = 16 liters.

Now, number of cans of Maaza = 80/16 = 5

Number of cans of Pepsi = 144/16 = 9

Number of cans of Sprite = 368/16 = 23

Thus, the total number of cans required = 5 + 9 + 23 = 37

Workspace

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