Two coins are tossed simultaneously. What is the probability of getting at most one head?
Question: Two coins are tossed simultaneously. What is the probability of getting at most one head? (a) $\frac{1}{4}$ (b) $\frac{1}{2}$ (C) $\frac{2}{3}$ (d) $\frac{3}{4}$ Solution: (d) $\frac{3}{4}$ Explanation:When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT.Total number of possible outcomes = 4 Let E be the event of getting at most one head. Then, the favourable outcomes are HT, TH and TT.Number of favourable outcomes = 3 $\therefore P($ getting at most 1...
Read More →Two coins are tossed simultaneously. What is the probability of getting at most one head?
Question: Two coins are tossed simultaneously. What is the probability of getting at most one head? (a) $\frac{1}{4}$ (b) $\frac{1}{2}$ (C) $\frac{2}{3}$ (d) $\frac{3}{4}$ Solution: (d) $\frac{3}{4}$...
Read More →In an AP, if a = 3.5, d = 0
Question: In an AP, if a = 3.5, d = 0 and n = 101, then anwill be (a) 0 (b) 3.5 (c) 103.5 (d) 104.5 Solution: (b) For an AP an= a + (n 1)d= 3.5+ (101 1 )x 0 [by given conditions] = 3.5...
Read More →Solve this
Question: $2 x+3 y-z=0$ $x-y-2 z=0$ $3 x+y+3 z=0$ Solution: The given system of homogeneous equations can be written in matrix form as follows: $\left[\begin{array}{ccc}2 3 -1 \\ 1 -1 -2 \\ 3 1 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ $A X=O$ Here, $A=\left[\begin{array}{ccc}2 3 -1 \\ 1 -1 -2 \\ 3 1 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $O=\left[\begin{array}{l}0 \\ 0 \\ 0...
Read More →In an AP, if d = – 4, n = 7
Question: In an AP, if d = 4, n = 7 and an= 4, then a is equal to (a) 6 (b) 7 (c) 20 (d) 28 Solution: (d) In an AP $a_{n}=a+(n-1) d$ $\Rightarrow \quad 4=a+(7-1)(-4) \quad$ [by given conditions] $\Rightarrow \quad 4=a+6(-4)$ $\Rightarrow \quad 4+24=a$ $\therefore \quad a=28$...
Read More →A water tank is 3 m long, 2 m broad and 1 m deep.
Question: A water tank is 3 m long, 2 m broad and 1 m deep. How many litres of water can it hold? Solution: Length of the water $\operatorname{tank}=3 \mathrm{~m}$ Breadth $=2 \mathrm{~m}$ Height $=1 \mathrm{~m}$ Volume of the water $\operatorname{tank}=3 \times 2 \times 1=6 \mathrm{~m}^{3}$ We $\mathrm{know}$ that $1 \mathrm{~m}^{3}=1000 \mathrm{~L}$ i.e., $6 \mathrm{~m}^{3}=6 \times 1000 \mathrm{~L}=6000 \mathrm{~L}$ $\therefore$ The water tank can hold $6000 \mathrm{~L}$ of water in it....
Read More →Two dice are thrown together. The probability of getting a doublet is
Question: Two dice are thrown together. The probability of getting a doublet is (a) $\frac{1}{3}$ (b) $\frac{1}{6}$ (c) $\frac{1}{4}$ (d) $\frac{2}{3}$ Solution: (2) $\frac{1}{6}$ Explanation:When two dice are thrown simultaneously, all possible outcomes are:(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)N...
Read More →The length , breadth and height of a room are 5 m, 4.5 m and 3 m, respectively.
Question: The length , breadth and height of a room are 5 m, 4.5 m and 3 m, respectively. Find the volume of the air it contains. Solution: Length of the room $=5 \mathrm{~m}$ Breadth $=4.5 \mathrm{~m}$ Height $=3 \mathrm{~m}$ Now, volume $=$ length $\times$ breadth $\times$ height $=5 \times 4.5 \times 3$ $=67.5 \mathrm{~m}^{3}$ $\therefore$ The volume of air in the room is $67.5 \mathrm{~m}^{3}$....
Read More →The probability of getting 2 heads, when two coins are tossed, is
Question: The probability of getting 2 heads, when two coins are tossed, is (a) 1 (b) $\frac{3}{4}$ (c) $\frac{1}{2}$ (d) $\frac{1}{4}$ Solution: All possible outcomes are HH, HT, TH and TT.Total number of outcomes = 4.Getting 2 heads means getting HH.Numbers of favourable outcomes = 1 $\therefore \mathrm{P}($ getting 2 heads $)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\frac{1}{4}$ Thus, the probability of getting 2 heads is $\frac{1}{4}$. Hen...
Read More →Solve this
Question: $3 x+y-2 z=0$ $x+y+z=0$ $x-2 y+z=0$ Solution: The given system of homogeneous equations can be written in matrix form as follows: $\left[\begin{array}{ccc}3 1 -2 \\ 1 1 1 \\ 1 -2 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ or, $A X=O$ where, $A=\left[\begin{array}{ccc}3 1 -2 \\ 1 1 1 \\ 1 -2 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $O=\left[\begin{array}{l}0 \\ 0 \\ 0\...
Read More →A rectangular diesel tanker is 2 m long, 2 m wide and 40 cm deep.
Question: A rectangular diesel tanker is 2 m long, 2 m wide and 40 cm deep. How many litres of diesel can it hold? Solution: Lenght of the rectangular diesel tanker $=2 \mathrm{~m}$ Breadth $=2 \mathrm{~m}$ Height $=40 \mathrm{~cm}$ $=40 \times \frac{1}{100} \mathrm{~m} \quad(\because 1 \mathrm{~m}=100 \mathrm{~cm})$ $=0.4 \mathrm{~m}$ So, volume of the tanker $=$ lenght $\times$ breadth $\times$ height $=2 \times 2 \times 0.4$ $=1.6 \mathrm{~m}^{3}$ We konw that $1 \mathrm{~m}^{3}=1000 \mathrm{...
Read More →At t min past 2 pm, the time needed by the
Question: At t min past $2 \mathrm{pm}$, the time needed by the minute hand of a clock to show $3 \mathrm{pm}$ was found to be $3 \mathrm{~min}$ less than $\frac{t^{2}}{4}$ min. Find $t$. Solution: We know that, the time between 2 pm to 3 pm = 1 h = 60 min Given that, at f min past 2pm, the time needed by the min hand of a clock to show 3 pm was found to be 3 min less than $\frac{t^{2}}{4}$ min $i, e$ $t+\left(\frac{t^{2}}{4}-3\right)=60$ $\Rightarrow \quad 4 t+t^{2}-12=240$ $\Rightarrow \quad t...
Read More →Two dice are thrown together. The probability of getting the same number on the both dice is
Question: Two dice are thrown together. The probability of getting the same number on the both dice is . (a) $\frac{1}{2}$ (b) $\frac{1}{3}$ (C) $\frac{1}{6}$ (d) $\frac{1}{12}$ Solution: Total number of outcomes = 36.Getting the same number on both the dice means (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).Numbers of favourable outcomes = 6. $\therefore \mathrm{P}($ getting the same number on both dice $)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outco...
Read More →The capacity of a certain cuboidal tank is 50000 litres of water.
Question: The capacity of a certain cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its height and length are 10 m and 2.5 m respectively. Solution: Capacity of the cuboidal tank $=50000 \mathrm{~L}$ $1000 \mathrm{~L}=1 \mathrm{~m}^{3}$ i. e., $50000 \mathrm{~L}=50 \times 1000$ litres $=50 \mathrm{~m}^{3}$ $\therefore$ The volume of the tank is $50 \mathrm{~m}^{3}$. Also, it is given that the length of the tank is $10 \mathrm{~m}$. Height $=2.5 \mathrm{~m}$ Suppose that ...
Read More →Solve this
Question: $x+y-z=0$ $x-2 y+z=0$ $3 x+6 y-5 z=0$ Solution: $x+y-z=0$ ....(1) $x-2 y+z=0$ .....(2) $3 x+6 y-5 z=0$ ....(3) The given system of homogeneous equations can be written in matrix form as follows: $\left[\begin{array}{ccc}1 1 -1 \\ 1 -2 1 \\ 3 6 -5\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ $A X=O$ Here, $A=\left[\begin{array}{ccc}1 1 -1 \\ 1 -2 1 \\ 3 6 -5\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ ...
Read More →A tank is 8 m long, 6 m broad and 2 m high.
Question: A tank is 8 m long, 6 m broad and 2 m high. How much water can it contain? Solution: Length of the tank $=8 \mathrm{~m}$ Breadth $=6 \mathrm{~m}$ Height $=2 \mathrm{~m}$ $\therefore$ Its volume $=$ length $\times$ breadth $\times$ height $=(8 \times 6 \times 2) \mathrm{m}^{3}=96 \mathrm{~m}^{3}$ We know that $1 \mathrm{~m}^{3}=1000 \mathrm{~L}$ Now, $96 \mathrm{~m}^{3}=96 \times 1000 \mathrm{~L}=96000 \mathrm{~L}$ $\therefore$ The tank can store $96000 \mathrm{~L}$ of water....
Read More →In the centre of a rectangular lawn
Question: In the centre of a rectangular lawn of dimensions 50m x 40m, a rectangular pond has to be constructed, so that the area of the grass surrounding the pond would be 1184 m2[see figure]. Find the length and breadth of the pond. Solution: Given that a rectangular pond has to be constructed in the centre of a rectangular lawn of dimensions 50 mx 40 m So, the distance between pond and lawn would be same around the pond. Say x m. Now, length of rectangular lawn (l1) = 50 m and breadth of rect...
Read More →A die is thrown once. The probability of getting a prime number is
Question: A die is thrown once. The probability of getting a prime number is (a) $\frac{2}{3}$ (b) $\frac{1}{3}$ (C) $\frac{1}{2}$ (d) $\frac{1}{6}$ Solution: (C) $\frac{1}{2}$ Explanation:In a single throw of a die, the possible outcomes are:1, 2, 3, 4, 5, 6Total number of possible outcomes = 6Let E be the event of getting a prime number.Then, the favourable outcomes are 2, 3 and 5. Number of favourable outcomes = 3 $\therefore$ Probability of getting a prime number $=P(E)=\frac{\text { Number ...
Read More →A die is thrown once. The probability of getting a prime number is
Question: A die is thrown once. The probability of getting a prime number is (a) $\frac{2}{3}$ (b) $\frac{1}{3}$ (C) $\frac{1}{2}$ (d) $\frac{1}{6}$ Solution: (C) $\frac{1}{2}$ Explanation:In a single throw of a die, the possible outcomes are:1, 2, 3, 4, 5, 6Total number of possible outcomes = 6Let E be the event of getting a prime number.Then, the favourable outcomes are 2, 3 and 5. Number of favourable outcomes = 3 $\therefore$ Probability of getting a prime number $=P(E)=\frac{\text { Number ...
Read More →What will be the height of a cuboid of volume 168 m
Question: What will be the height of a cuboid of volume 168 m3, if the area of its base is 28 m2? Solution: Volume of the cuboid $=168 \mathrm{~m}^{3}$ Area of its base $=28 \mathrm{~m}^{2}$ Let $h \mathrm{~m}$ be the height of the cuboid. Now, we have the following: Area of the rectangular base $=$ lenght $\times$ breadth Volume of the cuboid $=$ lenght $\times$ breadth $\times$ height $\Rightarrow$ Volume of the cuboid $=($ area of the base $) \times$ height $\Rightarrow 168=28 \times h$ $\Rig...
Read More →How much clay is dug out in digging a well measuring 3 m
Question: How much clay is dug out in digging a well measuring 3 m by 2 m by 5 m? Solution: The measure of well is $3 \mathrm{~m} \times 2 \mathrm{~m} \times 5 \mathrm{~m}$. $\therefore$ Volume of the clay dug out $=(3 \times 2 \times 5) \mathrm{m}^{3}=30 \mathrm{~m}^{3}$...
Read More →Find the volume in cubic decimetre of each of the cubes whose side is
Question: Find the volume in cubic decimetre of each of the cubes whose side is (i) 1.5 m (ii) 75 cm (iii) 2 dm 5 cm Solution: (i) Side of the cube $=1.5 \mathrm{~m}$ $=1.5 \times 10 \mathrm{dm}(\because 1 \mathrm{~m}=10 \mathrm{dm})$ $=15 \mathrm{dm}$ $\therefore$ Volume of the cube $=(\text { side })^{3}=(15)^{3}=3375 \mathrm{dm}^{3}$ (ii) Side of the cube $=75 \mathrm{~cm}$ $=75 \times \frac{1}{10} \mathrm{dm}(\because 1 \mathrm{dm}=10 \mathrm{~cm})$ $=7.5 \mathrm{dm}$ $\therefore$ Volume of ...
Read More →A die is thrown once. The probability of getting an odd number greater than 3 is
Question: A die is thrown once. The probability of getting an odd number greater than 3 is (a) $\frac{1}{3}$ (b) $\frac{1}{6}$ (c) $\frac{1}{2}$ (d) 0 Solution: Total number of outcomes = 6.Out of the given numbers, odd number greater than 3 is 5.Numbers of favourable outcomes = 1. $\therefore \mathrm{P}$ (getting an odd number greater than 3 ) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\frac{1}{6}$ Thus, the probability of getting an odd numb...
Read More →A die is thrown once. The probability of getting an odd number greater than 3 is
Question: A die is thrown once. The probability of getting an odd number greater than 3 is (a) $\frac{1}{3}$ (b) $\frac{1}{6}$ (c) $\frac{1}{2}$ (d) 0 Solution: Total number of outcomes = 6.Out of the given numbers, odd number greater than 3 is 5.Numbers of favourable outcomes = 1. $\therefore \mathrm{P}$ (getting an odd number greater than 3 ) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$ $=\frac{1}{6}$ Thus, the probability of getting an odd numb...
Read More →At present Asha’s age (in years) is 2 more
Question: At present Ashas age (in years) is 2 more than the square of her daughter Nishas age. When Nisha grows to her mothers present age. Ashas age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha. Solution: Let Nisha's present age be $x y r$. Then, Asha's present age $=x^{2}+2$ [by given condition] Now, when Nisha grows to her mothers present age i.e., after [(x2+2)- x] yr. Then, Ashas age also increased by [(x2+ 2) x]yr. Again by gi...
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