The probability of throwing a number greater than 2 with a fair die is
Question: The probability of throwing a number greater than 2 with a fair die is (a) $\frac{2}{5}$ (b) $\frac{5}{6}$ (C) $\frac{1}{3}$ (d) $\frac{2}{3}$ Solution: Total number of outcomes = 6.Out of the given numbers, numbers greater than 2 are 3, 4, 5 and 6.Numbers of favourable outcomes = 4. $\therefore \mathrm{P}($ getting a number greater than 2$)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\frac{4}{6}=\frac{2}{3}$ Thus, the probability of ge...
Read More →Find the volume in cubic metre (cu. m) of each of the cuboids whose dimensions are:
Question: Find the volume in cubic metre (cu. m) of each of the cuboids whose dimensions are: (i) length = 12 m, breadth = 10 m, height = 4.5 cm (ii) length = 4 m, breadth = 2.5 m, height = 50 cm. (iii) length = 10 m, breadth = 25 dm, height = 50 cm. Solution: (i) Length $=12 \mathrm{~m}$ Breadth $=10 \mathrm{~m}$ Height $=4.5 \mathrm{~m}$ $\therefore$ Volume of the cubo id $=$ length $\times$ breadth $\times$ height $=12 \times 10 \times 4.5$ $=540 \mathrm{~m}^{3}$ (ii) Length $=4 \mathrm{~m}$ ...
Read More →The probability of throwing a number greater than 2 with a fair die is
Question: The probability of throwing a number greater than 2 with a fair die is (a) $\frac{2}{5}$ (b) $\frac{5}{6}$ (C) $\frac{1}{3}$ (d) $\frac{2}{3}$ Solution: Total number of outcomes = 6.Out of the given numbers, numbers greater than 2 are 3, 4, 5 and 6.Numbers of favourable outcomes = 4. $\therefore \mathrm{P}($ getting a number greater than 2$)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\frac{4}{6}=\frac{2}{3}$ Thus, the probability of ge...
Read More →Solve this
Question: $x+y+z=0$ $x-y-5 z=0$ $x+2 y+4 z=0$ Solution: Here, $x+y+z=0$ ....(1) $x-y-5 z=0$ ....(2) $x+2 y+4 z=0$ ....(3) The given system of homogeneous equations can be written in matrix form as follows: $\left[\begin{array}{ccc}1 1 1 \\ 1 -1 -5 \\ 1 2 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ $A X=O$ Here, $A=\left[\begin{array}{ccc}1 1 1 \\ 1 -1 -5 \\ 1 2 4\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z...
Read More →A die is thrown once. The probability of getting an even number is
Question: A die is thrown once. The probability of getting an even number is (a) $\frac{1}{2}$ (b) $\frac{1}{3}$ (C) $\frac{1}{6}$ (d) $\frac{5}{6}$ Solution: Total number of outcomes = 6.Out of the given numbers, even numbers are 2, 4 and 6.Numbers of favourable outcomes = 3. $\therefore P$ (getting an even number) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\frac{3}{6}=\frac{1}{2}$ Thus, the probability of getting an even number is $\frac{1}{...
Read More →A die is thrown once. The probability of getting an even number is
Question: A die is thrown once. The probability of getting an even number is (a) $\frac{1}{2}$ (b) $\frac{1}{3}$ (C) $\frac{1}{6}$ (d) $\frac{5}{6}$ Solution: Total number of outcomes = 6.Out of the given numbers, even numbers are 2, 4 and 6.Numbers of favourable outcomes = 3. $\therefore P$ (getting an even number) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\frac{3}{6}=\frac{1}{2}$ Thus, the probability of getting an even number is $\frac{1}{...
Read More →If Zeba were younger by 5 yr than what she really is,
Question: If Zeba were younger by 5 yr than what she really is, then the square of, her age (in years) would have been 11, more than five times her actual age, what is her age now? Solution: Let the actual age of Zeba = x yr, Herage when she was 5 yr younger = (x 5)yr. Now, by given condition, Square of her age = 11 more than five times her actual age $(x-5)^{2}=5 \times$ actual age $+11$ $\Rightarrow \quad(x-5)^{2}=5 x+11$ $\Rightarrow \quad x^{2}+25-10 x=5 x+11$ $\Rightarrow \quad x^{2}-15 x+1...
Read More →Cards, each marked with one of the numbers 6, 7, 8,..., 15, are placed in a box and mixed thoroughly.
Question: Cards, each marked with one of the numbers 6, 7, 8,..., 15, are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with a number less than 10? (a) $\frac{3}{5}$ (b) $\frac{1}{3}$ (c) $\frac{1}{2}$ (d) $\frac{2}{5}$ Solution: (d) $\frac{2}{5}$ Explanation:All possible outcomes are 6, 7, 8................15.Number of all possible outcomes = 10 Out of these, the numbers that are less than 10 are 6, 7, 8 and 9.Number of...
Read More →Cards, each marked with one of the numbers 6, 7, 8,..., 15, are placed in a box and mixed thoroughly.
Question: Cards, each marked with one of the numbers 6, 7, 8,..., 15, are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with a number less than 10? (a) $\frac{3}{5}$ (b) $\frac{1}{3}$ (c) $\frac{1}{2}$ (d) $\frac{2}{5}$ Solution: (d) $\frac{2}{5}$ Explanation:All possible outcomes are 6, 7, 8................15.Number of all possible outcomes = 10 Out of these, the numbers that are less than 10 are 6, 7, 8 and 9.Number of...
Read More →There are 25 tickets numbered 1, 2, 3, 4,..., 25 respectively. One ticket is draw at random.
Question: There are 25 tickets numbered 1, 2, 3, 4,..., 25 respectively. One ticket is draw at random. What is the probability that the number on the ticket is a multiple of 3 or 5? (a) $\frac{2}{5}$ (b) $\frac{11}{25}$ (C) $\frac{12}{25}$ (d) $\frac{13}{25}$ Solution: (c) $\frac{12}{25}$ Explanation:Total number of tickets = 25Let E be the event of getting a multiple of 3 or 5.Then,Multiples of 3 are 3, 6, 9, 12, 15, 18, 21 and 24.Multiples of 5 are 5, 10, 15, 20 and 25.Number of favourable ou...
Read More →There are 25 tickets numbered 1, 2, 3, 4,..., 25 respectively. One ticket is draw at random.
Question: There are 25 tickets numbered 1, 2, 3, 4,..., 25 respectively. One ticket is draw at random. What is the probability that the number on the ticket is a multiple of 3 or 5? (a) $\frac{2}{5}$ (b) $\frac{11}{25}$ (C) $\frac{12}{25}$ (d) $\frac{13}{25}$ Solution: (c) $\frac{12}{25}$ Explanation:Total number of tickets = 25Let E be the event of getting a multiple of 3 or 5.Then,Multiples of 3 are 3, 6, 9, 12, 15, 18, 21 and 24.Multiples of 5 are 5, 10, 15, 20 and 25.Number of favourable ou...
Read More →A train, travelling at a uniform speed for 360 km,
Question: A train, travelling at a uniform speed for 360 km, would have taken 48 min less to travel the same distance, if its speed were 5 km/h more. Find the original speed of the train. Solution: Let the original speed of the train = x km/h Then, the increased speed of the train = (x + 5) km/h [by given condition] and distance = 360 km According to the question, $\frac{360}{x}-\frac{360}{x+5}=\frac{4}{5}$$\left[\begin{array}{l}\because \text { time }=\frac{\text { Distance }}{\text { Speed }} ...
Read More →Find the area of the following regular hexagon.
Question: Find the area of the following regular hexagon. Solution: The given figure is: It is given that the hexagon is regular. So, all its sides must be equal to $13 \mathrm{~cm}$. Also, $\mathrm{AN}=\mathrm{BQ}$ $\mathrm{QB}+\mathrm{BA}+\mathrm{AN}=\mathrm{QN}$ $\mathrm{AN}+13+\mathrm{AN}=23$ $2 \mathrm{AN}=23-13=10$ $\mathrm{AN}=\frac{10}{2}=5 \mathrm{~cm}$ Hence, $\mathrm{AN}=\mathrm{BQ}=5 \mathrm{~cm}$ Now, in the right angle triangle MAN: $\mathrm{MN}^{2}=\mathrm{AN}^{2}+\mathrm{AM}^{2}$...
Read More →There are 20 tickets numbered as 1, 2, 3,..., 20 respectively.
Question: There are 20 tickets numbered as 1, 2, 3,..., 20 respectively. One ticket is drawn at random. What is the probability that the number on the ticket drawn is a multiple of 5? (a) $\frac{1}{4}$ (b) $\frac{1}{5}$ (C) $\frac{2}{5}$ (d) $\frac{3}{10}$ Solution: (b) $\frac{1}{5}$ Explanation:Total number of tickets = 20Out of the given ticket numbers, multiples of 5 are 5, 10, 15 and 20.Number of favourable outcomes = 4 $\therefore P($ getting a multiple of 5$)=\frac{4}{20}=\frac{1}{5}$...
Read More →Solve this
Question: $x+y-6 z=0$ $x-y+2 z=0$ $-3 x+y+2 z=0$ Solution: Here, $x+y-6 z=0$ ....(1) $x-y+2 z=0$ .....(2) $-3 x+y+2 z=0$ ...(3) The given system of homogeneous equations can be written in matrix form as follows: $\left[\begin{array}{ccc}1 1 -6 \\ 1 -1 2 \\ -3 1 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ $A X=O$ Here, $A=\left[\begin{array}{ccc}1 1 -6 \\ 1 -1 2 \\ -3 1 2\end{array}\right], X=\left[\begin{array}{l}x \...
Read More →There are 20 tickets numbered as 1, 2, 3,..., 20 respectively.
Question: There are 20 tickets numbered as 1, 2, 3,..., 20 respectively. One ticket is drawn at random. What is the probability that the number on the ticket drawn is a multiple of 5? (a) $\frac{1}{4}$ (b) $\frac{1}{5}$ (C) $\frac{2}{5}$ (d) $\frac{3}{10}$ Solution: (b) $\frac{1}{5}$ Explanation:Total number of tickets = 20Out of the given ticket numbers, multiples of 5 are 5, 10, 15 and 20.Number of favourable outcomes = 4 $\therefore P($ getting a multiple of 5$)=\frac{4}{20}=\frac{1}{5}$...
Read More →Find the area of the following polygon,
Question: Find the area of the following polygon, ifAL= 10 cm,AM= 20 cm,AN= 50 cm,AO= 60 cm andAD= 90 cm. Solution: The given polygon is: Given: $\mathrm{AL}=10 \mathrm{~cm}, \mathrm{AM}=20 \mathrm{~cm}, \mathrm{AN}=50 \mathrm{~cm}$ $\mathrm{AO}=60 \mathrm{~cm}, \mathrm{AD}=90 \mathrm{~cm}$ Hence, we have the following: $\mathrm{MO}=\mathrm{AO}-\mathrm{AM}=60-20=40 \mathrm{~cm}$ $\mathrm{OD}=\mathrm{AD}-\mathrm{AO}=90-60=30 \mathrm{~cm}$ $\mathrm{ND}=\mathrm{AD}-\mathrm{AN}=90-50=40 \mathrm{~cm}...
Read More →A natural number, when increased by 12,
Question: A natural number, when increased by 12, equals 160 times its reciprocal.Find the number. Solution: Let the natural number be x. According to the question, $x+12=\frac{160}{x}$ On multiplying by $x$ on both sides, we get $\Rightarrow \quad x^{2}+12 x-160=0$ $x^{2}+(20 x-8 x)-160=0$ $\Rightarrow \quad x^{2}+20 x-8 x-160=0$[by factorisation method] $\Rightarrow \quad x(x+20)-8(x+20)=0$ $\Rightarrow \quad ; \quad(x+20)(x-8)=0$ Now, x + 20 = 0 ⇒ x = 20 which is not possible because natural ...
Read More →There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita divided it in two different ways.
Question: There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita divided it in two different ways.Find the area of this park using both ways. Can you suggest some another way of finding its areas? Solution: A pentagonal park is given below: Jyoti and Kavita divided it in two different ways. (i) Jyoti divided is into two trapeziums as shown below: It is clear that the park is divided in two equal trapeziums whose parallel sides are $30 \mathrm{~m}$ and $15 \mathrm{~m}$. And, t...
Read More →If an event cannot occur then its probability is
Question: If an event cannot occur then its probability is (a) 1 (b) $\frac{1}{2}$ (c) $\frac{3}{4}$ (d) 0 Solution: (d) 0If an event cannot occur, its probability is 0....
Read More →Find a natural number whose square
Question: Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. Solution: Let n be a required natural number. Square of a natural number diminished by 84 = n2 84 and thrice of 8 more than the natural number = 3 (n + 8) Now, by given condition, $n^{2}-84=3(n+8)$ $\Rightarrow \quad n^{2}-84=3 n+24$ $\Rightarrow \quad n^{2}-3 n-108=0$ $\Rightarrow \quad n^{2}-12 n+9 n-108=0 \quad$ [by splitting the middle term] $\Rightarrow \quad n(n-12)+9(n-12)=0$ ...
Read More →If an event cannot occur then its probability is
Question: If an event cannot occur then its probability is (a) 1 (b) $\frac{1}{2}$ (c) $\frac{3}{4}$ (d) 0 Solution: (d) 0If an event cannot occur, its probability is 0....
Read More →Find the area enclosed by each of the following
Question: Find the area enclosed by each of the following figures [Fig. 20.49 (i)-(iii)] as the sum of the areas of a rectangle and a trapezium: Solution: (i) The given figure can be divided into a rectangle and a trapezium as shown below: From the above firgure: Area of the complete figure $=($ Area of square $\mathrm{ABCF})+($ Area of trapezium $\mathrm{CDEF})$ $=(\mathrm{AB} \times \mathrm{BC})+\left[\frac{1}{2} \times(\mathrm{FC}+\mathrm{ED}) \times(\right.$ Distance between $\mathrm{FC}$ an...
Read More →If the probability of winning a game is 0.4, the probability of losing it is
Question: If the probability of winning a game is 0.4, the probability of losing it is(a) 0.96 (b) $\frac{1}{0.4}$ (c) 0.6(d) none of these Solution: (c) 0.6Explanation:LetEbe the event of winning a game.Then, ( notE) is the event of not winning the game or of losing the game.Then,P(E) = 0.4Now,P(E) +P(notE) = 1⇒ 0.4 +P(notE) = 1⇒P(notE) = 1 0.4 = 0.6P(losing the game) =P(notE) = 0.6...
Read More →Which of the following cannot be the probability of an event?
Question: Which of the following cannot be the probability of an event? (a) $\frac{1}{3}$ (b) $0.3$ (c) $33 \%$ (d) $\frac{7}{6}$ Solution: (d) $\frac{7}{6}$ Explanation:Probability of an event can't be more than 1....
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