Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M

31.

Which of the following has most number of divisors?

• (A) 99
• (B) 101
• (C) 176
• (D) 182

Show Answer

 99 = 1 × 3 × 3 × 11 101 = 1 × 101 176 = 1 × 2 × 2 × 2 × 2 × 11 182 = 1 × 2 × 7 × 13 So, divisor of 99 are 1, 3, 9, 11, 33 and 99 Divisor of 101 are 1 and 101 Divisor of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176 Divisor of 182 are 1, 2, 7, 13, 14, 26, 91 and 182 Hence, 176 has the most number of divisors.

32.

The greatest 4-digit number exactly divisible by 10, 15, 20 is = ?

• (A) 9990
• (B) 9960
• (C) 9980
• (D) 9995

Show Answer

 LCM of (10, 15, 20) ⇒ 5 × 2 × 3 × 2 = 60 ⇒ Largest 4-digit number = 9999 divide 9999 by LCM of given number ⇒ We get remainder = 39 ⇒ So, to divide completely subtract it from (9999 - 39) = 9960 ∴ 9960 is the largest four digit number which is completely divided by the given numbers.

33.

Three sets of English , Mathematics and Science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the height of each stack is the same. Total number of stakes will be ?

• (A) 14
• (B) 21
• (C) 22
• (D) 48

Show Answer

 Number of books in each stack = HCF of 336, 240 and 96 = 48. Hence, total number of stacks $$\eqalign{ & {\text{ = }}\left( {\frac{{336 + 240 + 96}}{{48}}} \right) \cr & = \frac{{672}}{{48}} \cr & = 14 \cr} $$

34.

An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 am. The next time, when they would beep together at the earliest is = ?

• (A) 10.30 am
• (B) 10.31 am
• (C) 10.59 am
• (D) 11 am

Show Answer

 Interval of change = (LCM of 60 and 62) sec = 1860 sec = 31 min So, the devices would beep together 31 min after 10 am i,e; at 10.31 am

35.

Three numbers which 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. The sum of the three numbers is:

• (A) 75
• (B) 81
• (C) 85
• (D) 89

Show Answer

 Since the numbers are co-prime, they contain only 1 as the common factor. Also, the given two products have the middle number in common. So, middle number = H.C.F. of 551 and 1073 = 29 $$\eqalign{ & {\text{First}}\,{\text{number}} \cr & = {\frac{{551}}{{29}}} \cr & = 19 \cr & {\text{Third}}\,{\text{number}} \cr & = {\frac{{1073}}{{29}}} \cr & = 37 \cr & \therefore {\text{Required}}\,{\text{sum}} \cr & = {19 + 29 + 37} \cr & = 85 \cr} $$

36.

Calculate the HCF of $$\frac{{12}}{5},\,\frac{{14}}{{15}}$$  and $$\frac{{16}}{{17}}.$$

• (A) $$\frac{4}{{255}}$$
• (B) $$\frac{3}{{255}}$$
• (C) $$\frac{2}{{255}}$$
• (D) $$\frac{1}{{255}}$$

Show Answer

 $$\eqalign{ & {\text{HCF of }}\frac{{12}}{5},\,\frac{{14}}{{15}}{\text{ and}}\frac{{16}}{{17}} \cr & \frac{{{\text{HCF of}}\,12,\,14,\,16}}{{{\text{LCM of}}\,5,\,15,\,17}} = \frac{2}{{255}} \cr} $$

37.

Find the least multiple of 23, which when divided by 18, 21 and 24 leaves the remainder 7, 10 and 13 respectively.

• (A) 3013
• (B) 3024
• (C) 3002
• (D) 3036

Show Answer

 18 - 7 = 11 21 - 10 = 11 24 - 13 = 11 Take LCM of (18, 21, 24) ⇒ 9 × 2 × 7 × 4 = 504 ⇒ Required number = (504k - 11) which is divided by 23 ∴ For $$\frac{{504k - 11}}{{23}},$$   remainder should be zero Put minimum value of k so that it completely divides 23 ⇒ at k = 6, 504k - 11 = 3013 completely divisible by 23 ∴ Required number is = 3013

38.

A number when divided by 361 gives remainder 47. When the same number is divided by 19 then find the remainder ?

• (A) 9
• (B) 1
• (C) 8
• (D) 3

Show Answer

 $$\eqalign{ & \frac{{{\text{Remainder of number}}}}{{19}} \cr & = \frac{{47}}{{19}} \cr & \left[ {{\text{Remainder = 9}}} \right] \cr} $$

39.

What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7 ?

• (A) 37
• (B) 36
• (C) 39
• (D) 30

Show Answer

 $$\eqalign{ & {\text{LCM of }}\left( {9,10,15} \right) \cr & = 3 \times 3 \times 10 \cr & = 90 \cr & \frac{{1936}}{{90}},{\text{remainder = 46}} \cr} $$ Least number when is subtracted from 1936 which gives remainder 7 when divided by (9, 10, 15) is = (46 - 7) = 39

40.

The number between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is = ?

• (A) 4203
• (B) 4023
• (C) 4032
• (D) 4302

Show Answer

 LCM of 12, 18, 21, 32 = 2016 = 2016 × K = 2016 × 2 = 4032 (K = 2) "4032" is the number which is completely divided by 12, 18, 21, 32