Lakshya Education MCQs

Question:

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

Options:
A.123
B.127
C.235
D.305
Answer: Option B

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

  = H.C.F. of 1651 and 2032 = 127.

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More Questions on This Topic :

Question 1.

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

Options:
  1.    504
  2.    536
  3.    544
  4.    548
Answer: Option D

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

   = 540 + 8

   = 548.

Question 2.

Which of the following fraction is the largest ?

Options:
  1.    \(\frac{7}{8}\)
  2.    \(\frac{13}{16}\)
  3.    \(\frac{31}{40}\)
  4.    \(\frac{43}{80}\)
Answer: Option A

L.C.M. of 8, 16, 40 and 80 = 80.

\(\frac{7}{8}=\frac{70}{80}; \frac{13}{16}=\frac{65}{80}; \frac{31}{40}=\frac{62}{80}\)  

   since,\(\frac{70}{80}>\frac{65}{80}>\frac{63}{80} so, \frac{7}{8}>\frac{13}{16}>\frac{69}{80}>\frac{31}{40}\)

 

 \(so,\frac{7}{8}is the largest\)

 

 

Question 3.

Find the highest common factor of 36 and 84.

Options:
  1.    4
  2.    6
  3.    12
  4.    18
Answer: Option D

36 = 22 x 32

84 = 22 x 3 x 7

Therefore H.C.F. = 22 x 3 = 12.

Question 4.

Which of the following has the most number of divisors?

Options:
  1.    99
  2.    101
  3.    176
  4.    182
Answer: Option C

99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

Question 5.

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

Options:
  1.    28
  2.    32
  3.    40
  4.    64
Answer: Option C

Let the numbers be 2x and 3x.

Then, their L.C.M. = 6x.

So, 6x = 48 or x = 8.

Therefore The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.

Question 6.

The H.C.F. of \(\frac{9}{10}, \frac{12}{25}, \frac{18}{35}and\frac{21}{40} is:\) 

Options:
  1.    \(\frac{3}{5}\)
  2.    \(\frac{252}{5}\)
  3.    \(\frac{3}{1400}\)
  4.    \(\frac{63}{700}\)
Answer: Option C

\( Required H.F.C = \frac{H.F.C of 9,12,18,21}{L.C.M.of 10,25,35,40}= \frac{3}{1400}\)