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Q 4    Here is an incomplete net for making a cube. Complete it in at least two different
ways. Remember that a cube has six faces. How many are there in the net here?
(Give two separate diagrams. If you like, you may use a squared sheet for easy
manipulation.)

There are three faces in the given net. The two possibilities of forming nets are 

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Safeer PP

Q 2     Dice are cubes with dots on each face. Opposite faces of a die always have a total of seven dots on them. Here are two nets to make dice (cubes); the numbers inserted in each square indicate
            the number of dots in that box.

          

Insert suitable numbers in the blanks, remembering that the number on the
opposite faces should total to 7.

The sum of the opposite sides of a dice =7 

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Safeer PP

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Q. 5.     Match the nets with appropriate solids:

            

Thsese are the correct matches 

(a) \rightarrow (ii)
(b) \rightarrow (iii)
(c) \rightarrow (iv)
(d) \rightarrow (i)
 

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Gautam harsolia

Q. 3.     Can this be a net for a die? Explain your answer.

            

NO, 
This can not be a net for a die because the opposite faces of a die always have a total sum of 7.
Now, this case one pair is  4  opposite to 1
So, their sum is equal to  5  and not equal to 7
And also, the 2nd pair is 6  opposite to 3 .
So, their sum is equal to 9 and not equal to 7
 

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Gautam harsolia

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2.(i) A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

Probability -A die is numbered in such a way that its faces show
What is the probability that the total score is (i) even?

+ 1 2 2 3 3 6
1 2 3 3 4 4 7
2 3 4 4 5 5 8
2 3 4 4 5 5 8
3 4 5 5 6 6 9
3 4 5 5 6 6 9
6 7 8 8 9 9 12

Total possible outcomes when two dice are thrown = 6\times6=36

(1) Number of times when sum is even = 18

\therefore P(sum\ is\ even) = \frac{18}{36} = \frac{1}{2}

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HARSH KANKARIA

1.(iii) Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (iii) different days?

Total possible ways Shyam and Ekta can visit the shop = 5\times5 = 25

(1) Case that both will visit the same day.

Shyam can go on any day between Tuesday to saturday in 5 ways.

For any day that Shyam goes, Ekta will go on a different day in (5-1) = 4 ways.

Total ways that they both go in the same day = 5\times4 = 20 

\therefore P(both\ go\ on\ different\ days) = \frac{20}{25} = \frac{4}{5}

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HARSH KANKARIA

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1.(i)   Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day?

Total possible ways Shyam and Ekta can visit the shop = 5\times5 = 25

(1) Case that both will visit the same day.

Shyam can go on any day between Tuesday to saturday in 5 ways.

For any day that Shyam goes, Ekta will go on the same day in 1 way.

Total ways that they both go in the same day = 5\times1 = 5 

\therefore P(both\ go\ on\ same\ day) = \frac{5}{25} = \frac{1}{5}

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Posted by

HARSH KANKARIA

25.(ii) Which of the following arguments are correct and which are not correct? Give reasons for your answer. (ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2

The possible outcomes when a die is thrown= {1,2,3,4,5,6}

Total number of possible outcomes = 6

Number of odd number, {1,3,5} = 3

And, number of even numbers {2,4,6} = 3

Hence, both these events are equally likely

\therefore P(getting\ an\ odd) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{3}{6} = \frac{1}{2}

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HARSH KANKARIA

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25.(i)   Which of the following arguments are correct and which are not correct? Give reasons for your answer. (i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3 

The possible outcomes when two coins are tossed = {HH, HT, TH, TT}

Total number of possible outcomes = 4

\therefore P(getting\ two\ heads) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{1}{4}

Hence, the given statement is not correct. This is because one of each can occur in two different ways. Hence the mentioned events are not equally likely.

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HARSH KANKARIA

24.(ii)    A die is thrown twice. What is the probability that 5 will come up at least once?

When a die is thrown twice, the possible outcomes =

\{(x,y): x,y\in\{1,2,3,4,5,6\}\}

Total number of possible outcomes = 6\times6 = 36

The outcomes when 5 comes up at least once =

{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}

Number of such favourable outcomes = 11 

\therefore P(5\ comes\ up\ at\ least\ once)= \frac{11}{36}

Therefore, the probability that 5 comes at least once is \frac{11}{36}

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HARSH KANKARIA

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