Maths plz solve

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1 the numbr of parallelograms that can be formed from a set of 4 parallel lines intersecting another set of 3 parallel lines is

a 6  b 18  c 12  d 9

2 the numbr of ways in wich 12 students cn be equally divided into 3 groups is

a5775 b7575 c7755 d none

3 a court hs givn a 6 to 3 decision upholding a lower court the numbr of ways it cn give a majority decision reversing the lower court is

a256 b276 c245 d226

4 a qstn paper has 6 qstns each having an alternative. the number of ways an examine cn answr one or more qstns is

a 720  b 728  c 729  d none

5 the numbr of words that cn b made by  rearranging the letters of the word APURNA so that vowels and consonents appear alternate is

a 18  b 35  c 36  d none

6 the result of 8 matches(win loss or draw) are 2 b predicted. the numbr of different forecasts containing exactly 6 correct result is

a 316  b 214  c 112  d none

7 the number of different factors the number 75600 has is

a 120  b 121  c 119  d none

 

 

Replies (4)

7. 120 is the answer.

75600 can be factored into 24×33×52×7

Let X be a factor now

X is of the form 2a×3b×5c×7d

a can take values from 0 to 4 i.e. 5 values

b can take values from 0 to 3 i.e. 4 values

c can take values from 0 to 2 i.e. 3 values

d can take values from 0 to 1 i.e. 2 values

so number of distinct factors = 5×4×3×2 = 120

 

6. 112 is the answer

 Let the 6 correctly predicted places be chosen from the 8 matches. This can be done in 8C6 ways, i.e. is 28 ways. These results are already unique. Now for the other 2 games they can any result other than the correct one (as there should be exactly 6 correct results) so we can choose them in 2×2 ways i.e. 4 ways therefore 28×4 i.e. is 112 is the correct answer.

5. 36 is the answer.

This can be done easily if we divide the problem into 2 cases

  • Starting with a vowel and ending with a consonant.
  • Starting with a consonant and ending with a vowel.

Case (i): the three vowels A,A,U can be arranged in 3 ways.(A_A_U_, A_U_A_, U_A_A_). Now the three consonants can be placed in 6 different ways 3×6 ways i.e. 18 ways.

Case (ii): as in the above case there will be 18 ways in which this can be done(just swap the places of the words generated above alternately)

Therefore there are 18+18 i.e. 36 ways.

 

4. 729 is the answer

There are 3 ways in which each question can be answered 1.left blank 2. 1st option 3. 2nd option. There are 6 questions as such number of ways is 3 multiplied 6 times i.e. 36 = 729

 

2. 5775 is the answer

First we give the three groups separate identities A,B,C.

For A we have to choose 4 students from 12. This can be done in 12C4 ways i.e. 495 ways. Now we choose 4 students for group B from 8 remaining students. This can be done in 8C4 ways i.e. 70 ways. Therefore total number of ways is 495×70 ways i.e. 34650 ways. But in the original question groups are indistinguishable. Hence we should divide 34650 by 6 (the number of ways 3 things can be placed in 3 places) 34650/6 = 5775.

 

1. 6 is the answer.

Each line divides the plane into two halves. Two parallel lines divide the plane into 3 different divisions.(one of them between the two and two to the either side of the lines ) 3 lines divide the plane into 4 divisions (two in between and two at either side) 4 lines divide the plane into 5 divisions (3 in between and 2 at either side). Only the divisions which are between lines can form parallelograms. i.e. 3×2 =6

3. 256 is the answer There are 9 Judges in total. We need a majority decision to reverse the decision of a lower court. Majority decision can happen by 5-4 majority,6-3 majority,7-2 majority, 8-1 majority or be unanimous(9-0) Number of ways 5-4 can happen – choose 5 judges who consent to the reversal from 9 judges i.e. 9C5 ways, i.e. 126 ways. Similarly 6-3 – 9C6 i.e. 84 ways 7-2 – 9C7 i.e. 36 ways 8-1 – 9C8 i.e. 9 ways 9-0 –1 way Total number of ways = 126+84+36+9+1 ways i.e. 256 ways

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