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Question 1 of 5
1. Question
In how many ways a selection can be made of atleast one fruit out of 5 bananas, 4 mangoes and 4 almonds?
Correct
Ans-149
From 5 bananas we have 6 choices available (0, 1 ,2, 3, 4 or 5). Si milarly 4 mangoes and 4 almond can be chosen in 5 ways each. So total ways = 6 x 5 x 5 = 150 possible selection But in this 150, there is one selection where no fruit chosen So required no. of ways 150-1=149 Hence Option (b) is correct.
Incorrect
Ans-149
From 5 bananas we have 6 choices available (0, 1 ,2, 3, 4 or 5). Si milarly 4 mangoes and 4 almond can be chosen in 5 ways each. So total ways = 6 x 5 x 5 = 150 possible selection But in this 150, there is one selection where no fruit chosen So required no. of ways 150-1=149 Hence Option (b) is correct.
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Question 2 of 5
2. Question
There is a question paper consisting of 15 questions Each question has an internal choice of 2 options In how many ways can a student attempt one more questions of the given fifteen questions in the paper?
Correct
Ans-d
For each question we have 3 choices of answering the question (2 internal choices + 1 non-attempt). Thus, there are a total of 315 ways of answering the question paper. Out of this there is exactly one way in which the student does not answer any question. Thus there are a total of 315 -1 ways in which atleast one question is answered .Hence, Option (d) is correct.
Incorrect
Ans-d
For each question we have 3 choices of answering the question (2 internal choices + 1 non-attempt). Thus, there are a total of 315 ways of answering the question paper. Out of this there is exactly one way in which the student does not answer any question. Thus there are a total of 315 -1 ways in which atleast one question is answered .Hence, Option (d) is correct.
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Question 3 of 5
3. Question
How many numbers can be formed with the digits 1,6.7.8, 7, 6, 1 so that the odd digits always occupy the odd places?
Correct
Ans-c
The digits are 1, 6, 7, 8, 7, 6, 1. In this seven-digit no. there are four odd places and three even places OEOEOEO. The four odd digits 1, 7, 7, 1 can be arranged in four odd places in [4!/2! x 2] = 6 ways [as 1 and 7 are both occurring twice] The even digits 6, 8, 6 can be arranged in three even places in 3!/2! = 3 ways. Total no. of ways = 6 x 3 = 18. Hence, Option (c) is correct.
Incorrect
Ans-c
The digits are 1, 6, 7, 8, 7, 6, 1. In this seven-digit no. there are four odd places and three even places OEOEOEO. The four odd digits 1, 7, 7, 1 can be arranged in four odd places in [4!/2! x 2] = 6 ways [as 1 and 7 are both occurring twice] The even digits 6, 8, 6 can be arranged in three even places in 3!/2! = 3 ways. Total no. of ways = 6 x 3 = 18. Hence, Option (c) is correct.
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Question 4 of 5
4. Question
From a group of 12 dancers, five have to be taken for a stage show. Among them Radha and Mohan decide either both of them would join or none of them would join. In how many ways can the 5 dancers be chosen?
Correct
Ans-d
We have to select 5 out of 12.
If Radha and Mohan join- then we have to select only 5 – 2 = 3 dancers out of 12 – 2 = 10 which can be done in ‘°C3 = 120 ways.If Radha and Mohan do not join, then we have to
select 5 out of 12 – 2 = 10-> 10C5, = 252 ways.
Total number of ways = 120 + 252 = 372.
Hence, Option (d) is correct.
Incorrect
Ans-d
We have to select 5 out of 12.
If Radha and Mohan join- then we have to select only 5 – 2 = 3 dancers out of 12 – 2 = 10 which can be done in ‘°C3 = 120 ways.If Radha and Mohan do not join, then we have to
select 5 out of 12 – 2 = 10-> 10C5, = 252 ways.
Total number of ways = 120 + 252 = 372.
Hence, Option (d) is correct.
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Question 5 of 5
5. Question
Find the number of 6-digit numbers that can be found using the digits 1, 2, 3, 4, 5, 6 once such that the 6-digit number is divisible by its unit digit. (The unitdigit is not 1.)
Correct
Ans-d
The unit digit can either be 2, 3, 4, 5 or 6. When the unit digit is 2, the number would be even and hence will be divisible by 2. Hence all numbers with unit digit 2 will be included which is equal to 5! Or 120.When the unit digit is 3, then in every case the sum of the digits of the number would be 21 ,which is a multiple of 3 ,hence all numbers with unit digit 3 will be divisible by 3 and hence will be included ,total no of such numbers is !5 or 120.
Similarly for unit dogit is 4 ,then the number would be divisible by 4 only if the tens digit is 2 or 6,total no of such numbers is 2*!4 or 48.
Hence total no of required nos is (4*120)+48=528
Hence, Option (d) is correct.
Incorrect
Ans-d
The unit digit can either be 2, 3, 4, 5 or 6. When the unit digit is 2, the number would be even and hence will be divisible by 2. Hence all numbers with unit digit 2 will be included which is equal to 5! Or 120.When the unit digit is 3, then in every case the sum of the digits of the number would be 21 ,which is a multiple of 3 ,hence all numbers with unit digit 3 will be divisible by 3 and hence will be included ,total no of such numbers is !5 or 120.
Similarly for unit dogit is 4 ,then the number would be divisible by 4 only if the tens digit is 2 or 6,total no of such numbers is 2*!4 or 48.
Hence total no of required nos is (4*120)+48=528
Hence, Option (d) is correct.
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