Find whether the following equations
Question: Find whether the following equations have real roots. If real roots exist, find them (i) 8x2+2x -3=0 (ii) 2x2+ 3x + 2 = 0 (iii) 5x2 2x- 10 = 0 (iv) $\frac{1}{2 x-3}+\frac{1}{x-5}=1, x \neq \frac{3}{2}, 5$ (v) $x^{2}+5 \sqrt{5} x-70=0$ Solution: (i) Given equation is $8 x^{2}+2 x-3=0$. On comparing with $a x^{2}+b x+c=0$, we get $a=8, b=2$ and $c=-3$ $\therefore \quad$ Discriminant, $D=b^{2}-4 a c$ $=(2)^{2}-4(8)(-3)$ $=4+96=1000$ Therefore, the equation $8 x^{2}+2 x-3=0$ has two distin...
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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Question: $3 x-y+2 z=0$ $4 x+3 y+3 z=0$ $5 x+7 y+4 z=0$ Solution: Here, $3 x-y+2 z=0$ ....(1) $4 x+3 y+3 z=0$ .....(2) $5 x+7 y+4 z=0$ ....(3) The given system of homogeneous equations can be written in matrix form as follows: $\left[\begin{array}{ccc}3 -1 2 \\ 4 3 3 \\ 5 7 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ $A X=O$ Here, $A=\left[\begin{array}{ccc}3 -1 2 \\ 4 3 3 \\ 5 7 4\end{array}\right], X=\left[\begin{a...
Read More →One ticket is drawn at random from a bag containing tickets numbered 1 to 40.
Question: One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number, which is a multiple of 7, is (a) $\frac{1}{7}$ (b) $\frac{1}{8}$ (c) $\frac{1}{5}$ (d) $\frac{7}{40}$ Solution: Total number of tickets = 40.Out of the given numbers, multiples of 7 are 7, 14, 21, 28 and 35.Numbers of favourable outcomes = 5. $\therefore P($ getting a number which is a multiple of 7$)=\frac{\text { Number of favourable outcomes }}{\text ...
Read More →One ticket is drawn at random from a bag containing tickets numbered 1 to 40.
Question: One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number, which is a multiple of 7, is (a) $\frac{1}{7}$ (b) $\frac{1}{8}$ (c) $\frac{1}{5}$ (d) $\frac{7}{40}$ Solution: Total number of tickets = 40.Out of the given numbers, multiples of 7 are 7, 14, 21, 28 and 35.Numbers of favourable outcomes = 5. $\therefore P($ getting a number which is a multiple of 7$)=\frac{\text { Number of favourable outcomes }}{\text ...
Read More →Find the area of the pentagon shown in fig. 20.48,
Question: Find the area of the pentagon shown in fig. 20.48, ifAD= 10 cm,AG= 8 cm,AH= 6 cm,AF= 5 cm,BF= 5 cm,CG= 7 cm andEH= 3 cm. Solution: The given figure is: Given: $\mathrm{AD}=10 \mathrm{~cm}, \mathrm{AG}=8 \mathrm{~cm}, \mathrm{AH}=6 \mathrm{~cm}, \mathrm{AF}=5 \mathrm{~cm}$ $\mathrm{BF}=5 \mathrm{~cm}, \mathrm{CG}=7 \mathrm{~cm}, \mathrm{EH}=3 \mathrm{~cm}$ $\therefore \mathrm{FG}=\mathrm{AG}-\mathrm{AF}=8-5=3 \mathrm{~cm}$ And, $\mathrm{GD}=\mathrm{AD}-\mathrm{AG}=10-8=2 \mathrm{~cm}$ F...
Read More →Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at random from the bag.
Question: Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is (a) $\frac{1}{2}$ (b) $\frac{2}{5}$ (C) $\frac{3}{10}$ (d) $\frac{5}{9}$ Solution: Total number of cards = 10.Out of the given numbers, prime numbers are 2, 3, 5, 7 and 11.Numbers of favourable outcomes = 5. $\therefore \mathrm{P}$ (getting a prime number) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possib...
Read More →Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at random from the bag.
Question: Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is (a) $\frac{1}{2}$ (b) $\frac{2}{5}$ (C) $\frac{3}{10}$ (d) $\frac{5}{9}$ Solution: Total number of cards = 10.Out of the given numbers, prime numbers are 2, 3, 5, 7 and 11.Numbers of favourable outcomes = 5. $\therefore \mathrm{P}$ (getting a prime number) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possib...
Read More →Find the area of the field shown in Fig. 20.39 by dividing it into a square,
Question: Find the area of the field shown in Fig. 20.39 by dividing it into a square, a rectangle and a trapezium. Solution: The given figure can be divided into a square, a parallelogram and a trapezium as shown in following figure: From the above figure: Area of the figure $=$ (Area of square AGFM with sides $4 \mathrm{~cm}$ ) $+$ (Area of rectangle ME DN with length $8 \mathrm{~cm}$ and width $4 \mathrm{~cm}$ )+(Area of trapezium NDCB with parallel sid es $8 \mathrm{~cm}$ and $3 \mathrm{~cm}...
Read More →A box contains 90 discs, numbered from 1 to 90.
Question: A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears prime number less than 23 is (a) $\frac{7}{90}$ (b) $\frac{1}{9}$ (c) $\frac{4}{15}$ (d) $\frac{8}{89}$ Solution: Total number of discs = 90.Out of the given numbers, prime numbers less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19.Numbers of favourable outcomes = 8. $\therefore P$ (getting a prime number which is less than 23 ) $=\frac{\text { Number of favourable out...
Read More →Find the roots of the following quadratic
Question: Find the roots of the following quadratic equations by the factorisation method. (i) $2 x^{2}+\frac{5}{3} x-2=0$ (ii) $\frac{2}{5} x^{2}-x-\frac{3}{5}=0$ (iii) $3 \sqrt{2} x^{2}-5 x-\sqrt{2}=0$ (iv) $3 x^{2}+5 \sqrt{5} x-10=0$ (v) $21 x^{2}-2 x+\frac{1}{21}=0$ Solution: (i) Given equation is $2 x^{2}+\frac{5}{3} x-2=0$ On multiplying by 3 on both sides, we get $6 x^{2}+5 x-6=0$ $\Rightarrow \quad 6 x^{2}+(9 x-4 x)-6=0 \quad$ [by splitting the middle term] $\Rightarrow \quad 6 x^{2}+9 x...
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Question: $2 x-y+2 z=0$ $5 x+3 y-z=0$ $x+5 y-5 z=0$ Solution: Here, $2 x-y+2 z=0$ ....(1) $5 x+3 y-z=0$ ....(2) $x+5 y-5 z=0$ .....(3) The given system of homogeneous equations can be written in matrix form as follows: $\left[\begin{array}{ccc}2 -1 2 \\ 5 3 -1 \\ 1 5 -5\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ $A X=O$ Here, $A=\left[\begin{array}{ccc}2 -1 2 \\ 5 3 -1 \\ 1 5 -5\end{array}\right], X=\left[\begin{array...
Read More →A box contains 90 discs, numbered from 1 to 90.
Question: A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears prime number less than 23 is (a) $\frac{7}{90}$ (b) $\frac{1}{9}$ (c) $\frac{4}{15}$ (d) $\frac{8}{89}$ Solution: Total number of discs = 90.Out of the given numbers, prime numbers less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19.Numbers of favourable outcomes = 8. $\therefore P$ (getting a prime number which is less than 23 ) $=\frac{\text { Number of favourable out...
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Question: $2 x-y+2 z=0$ $5 x+3 y-z=0$ $x+5 y-5 z=0$ Solution: Here, $2 x-y+2 z=0$ ....(1) $5 x+3 y-z=0$ ....(2) $x+5 y-5 z=0$ .....(3) The given system of homogeneous equations can be written in matrix form as follows: $\left[\begin{array}{ccc}2 -1 2 \\ 5 3 -1 \\ 1 5 -5\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ $A X=O$ Here, $A=\left[\begin{array}{ccc}2 -1 2 \\ 5 3 -1 \\ 1 5 -5\end{array}\right], X=\left[\begin{array...
Read More →In Fig. 20.38, a parallelogram is drawn in a trapezium,
Question: In Fig. 20.38, a parallelogram is drawn in a trapezium, the area of the parallelogram is 80 cm2, find the area of the trapezium. Solution: The given figure is: From above figure, it is clear that the length of the parallel sides of the trapezium are $22 \mathrm{~cm}$ and $10 \mathrm{~cm}$. Also, it is given that the area of the parallelogram is $80 \mathrm{~cm}^{2}$ and its base is $10 \mathrm{~cm}$. We know: Area of parallelogram $=$ Base $\times$ Height $\therefore 80=10 \times$ Heig...
Read More →A box conatins cards numbered 6 to 50. A card is drawn at random from the box.
Question: A box conatins cards numbered 6 to 50. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square is (a) $\frac{1}{45}$ (b) $\frac{2}{15}$ (c) $\frac{4}{45}$ (d) $\frac{1}{9}$ Solution: Total number of outcomes = 45.Out of the given numbers, perfect square numbers are 9, 16, 25, 36 and 49.Numbers of favourable outcomes = 5. $\therefore \mathrm{P}$ (getting a number which is a perfect square) $=\frac{\text { Number of favourable ou...
Read More →If the area of a trapezium is 28 cm
Question: If the area of a trapezium is 28 cm2and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm. Solution: Given: Area of the trapezium $=28 \mathrm{~cm}^{2}$ Length of one of its parallel sides $=6 \mathrm{~cm}$ Altitude $=4 \mathrm{~cm}$ Let the other side be $\mathrm{x} \mathrm{cm} .$ Area of trapezium $=\frac{1}{2} \times($ Sum of the parallel sides $) \times($ Altitude $)$ $\Rightarrow 28=\frac{1}{2} \times(6+\mathrm{x}) \times(4)$ $\Rightarrow 28=2...
Read More →A box conatins cards numbered 6 to 50. A card is drawn at random from the box.
Question: A box conatins cards numbered 6 to 50. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square is (a) $\frac{1}{45}$ (b) $\frac{2}{15}$ (c) $\frac{4}{45}$ (d) $\frac{1}{9}$ Solution: Total number of outcomes = 45.Out of the given numbers, perfect square numbers are 9, 16, 25, 36 and 49.Numbers of favourable outcomes = 5. $\therefore \mathrm{P}$ (getting a number which is a perfect square) $=\frac{\text { Number of favourable ou...
Read More →Find the area of a trapezium whose parallel sides are 25 cm,
Question: Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and the other sides are 15 cm each. Solution: Given: Parallel sides of a trapezium are $25 \mathrm{~cm}$ and $13 \mathrm{~cm}$. Its nonparallel sides are equal in length and each is equal to $15 \mathrm{~cm}$. A rough skech of the trapezium is given below: In above figure, we observe that both the right angle triangles AMD and BNC are similar triangles. This is because both have two common sides as $15 \mathrm{~cm}$ and...
Read More →The probability that a number selected at random from the numbers 1, 2, 3, ...,
Question: The probability that a number selected at random from the numbers 1, 2, 3, ..., 15 is a multiple of 4, is (a) $\frac{4}{15}$ (b) $\frac{2}{15}$ (c) $\frac{1}{5}$ (d) $\frac{1}{3}$ Solution: Total number of outcomes = 15.Out of the given numbers, multiples of 4 are 4, 8 and 12.Numbers of favourable outcomes = 3. $\therefore \mathrm{P}($ getting a multiple of 4$)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\frac{3}{15}=\frac{1}{5}$ Thus, t...
Read More →The probability that a number selected at random from the numbers 1, 2, 3, ...,
Question: The probability that a number selected at random from the numbers 1, 2, 3, ..., 15 is a multiple of 4, is (a) $\frac{4}{15}$ (b) $\frac{2}{15}$ (c) $\frac{1}{5}$ (d) $\frac{1}{3}$ Solution: Total number of outcomes = 15.Out of the given numbers, multiples of 4 are 4, 8 and 12.Numbers of favourable outcomes = 3. $\therefore \mathrm{P}($ getting a multiple of 4$)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\frac{3}{15}=\frac{1}{5}$ Thus, t...
Read More →The parallel sides of a trapezium are 25 cm and 13 cm;
Question: The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm, find the area of the trapezium. Solution: Given: The parallel sides of a trapezium are $25 \mathrm{~cm}$ and $13 \mathrm{~cm}$. Its nonparallel sides are equal in length and each is equal to $10 \mathrm{~cm}$. A rough sketch for the given trapezium is given below: In above figure, we observe that both the right angle trangles AMD and BNC are congruent triangles. $\mathrm{AD}=\mathr...
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Question: $2 x-y+z=0$ $3 x+2 y-z=0$ $x+4 y+3 z=0$ Solution: (i) The given system of homogeneous equations can be written in matrix form as follows: $\left[\begin{array}{ccc}2 -1 1 \\ 3 2 -1 \\ 1 4 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ or, $A X=O$ where, $A=\left[\begin{array}{ccc}2 -1 1 \\ 3 2 -1 \\ 1 4 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $O=\left[\begin{array}{l}0 \\...
Read More →A number is selected at random from the numbers 1 to 30.
Question: A number is selected at random from the numbers 1 to 30. What is the probability that the selected number is a prime number? (a) $\frac{2}{3}$ (b) $\frac{1}{6}$ (C) $\frac{1}{3}$ (d) $\frac{11}{30}$ Solution: Total number of outcomes = 30.Prime numbers in 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.Number of favourable outcomes = 10. $\therefore \mathrm{P}$ (getting a prime number) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\fra...
Read More →A number is selected at random from the numbers 1 to 30.
Question: A number is selected at random from the numbers 1 to 30. What is the probability that the selected number is a prime number? (a) $\frac{2}{3}$ (b) $\frac{1}{6}$ (C) $\frac{1}{3}$ (d) $\frac{11}{30}$ Solution: Total number of outcomes = 30.Prime numbers in 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.Number of favourable outcomes = 10. $\therefore \mathrm{P}$ (getting a prime number) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\fra...
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