HCF and LCM ( 11 to 15 )

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Questions:


11. Four bells ring at intervals 4, 6, 8 and 14 s. They start ringing simultaneously at 12 ‘o’ clock. At what time will they again ring simultaneously?

A)    12h 2min 48s

B)    12h 3min

C)    12h 3min 20s

D)    12h 3min 44s

E)     None of these

View Answer

Answer: option (A)

Explanation:

LCM of 4, 6, 8 and 14 s = 168s => 2min 48s

Thus, they will ring again simultaneously at 12h 2min 48s.

12. Let the least number of six digits which when divided by 4, 6, 10, 15 leaves in each case same remainder 2 be N. The sum of digit in N is:

A)    3

B)    5

C)    4

D)    6

E)    None of the these

View Answer

Answer: option (B)

Explanation:

LCM of 4, 6, 10, 15 = 60

∴ N = 60x + 2, where x has the value which makes N the least six digit number.

∴ N = 60 * 1667 + 2 = 100022

∴ Sum of digits of N = 1 + 2 + 2 = 5

13. Two numbers, both greater than 29, have HCF 29 and LCM 4147. The sum of the numbers is:

A)    666

B)    669

C)    696

D)    966

E)    None of these

View Answer

Answer: option (C)

Explanation:

Product of two numbers = 29 * 4147.

Let the numbers be 29a and 29b. Then, 29a * 29b = (24 * 447) => ab = 143

The product of two numbers is 143 and its co-prime numbers are (1, 143) and (11, 13)


So, the numbers are (29 * 1, 29 * 143), (29 * 11, 29 * 13)

Since both numbers are greater than 29, the suitable pair is (29 * 11, 29 * 13) i.e. (319, 377).

14. Find the maximum number of students among which 1200 pens and 900 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is:

A)    300

B)    910

C)    1001

D)    1911

E)    None of these

View Answer

Answer: option (A)

Explanation:

Total Pens = 1200

Total Pencils = 900

We need to find maximum no. of students among 1200 pens and 900 pencils that can be distributed in such a way that each students get same no. of pens and pencils.

Then, we need to find HCF of 1200 and 900.

Prime Factorization of

1200 = 2 * 2 * 2 * 2 * 5 * 5 * 3

900 = 2 * 2 * 5 * 5 * 3 * 3
HCF = product of common prime factor of least power

HCF = 2 * 2* 5 * 5 * 3 = 300

Required number of students = HCF of 1200 and 900 = 300.

15. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

A)    7

B)    4

C)    9

D)    13

E)    None of these

View Answer

Answer: option (B)

Explanation:

Required Number = (91 – 43), (183 – 91) and (183 – 43)

  • HCF of 48 = 2 * 2 * 2 * 2 * 3
  • HCF of 92 = 2 * 2 * 23
  • HCF of 140 = 2 * 2 * 5 * 7

HCF is = 2 * 2 = 4

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