A game consists of spinning an
Question: A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (see figure). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons Solution: No, the outcomes are not equally likely, because 3 contains half part of the total region, so it is more likely than 1 and 2, since 1 and 2, each contains half part of the remaining part of the region....
Read More →Factorise:
Question: Factorise:(x 2y)2+ 4x 8y Solution: We have: $(x-2 y)^{2}+4 x-8 y=(x-2 y)^{2}+4(x-2 y)$ $=(x-2 y)(x-2 y)+4(x-2 y)$ $=(x-2 y)\{(x-2 y)+4\}$ $=(x-2 y)(x-2 y+4)$ $\therefore(x-2 y)^{2}+4 x-8 y=(x-2 y)(x-2 y+4)$...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\log \sqrt{\frac{1-\cos x}{1+\cos x}}$ Solution: Let $y=\log \sqrt{\frac{1-\cos x}{1+\cos x}}$ On differentiating $y$ with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left(\log \sqrt{\frac{1-\cos x}{1+\cos x}}\right)$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\log \left(\frac{1-\cos \mathrm{x}}{1+\cos \mathrm{x}}\right)^{\frac{1}{2}}\right]$ We know $\frac{\mathrm{d}}{\mathrm{d...
Read More →In a family having three children,
Question: In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is $\frac{1}{4}$. Is this correct? Justify your answer. Solution: No, the probability of each is not $\frac{1}{4}$ because the probability of no girl in three children is zero and probability of three girls in three children is one. Justification So, these events are not equally likely as outcome one girl, means gbb, bgb, bbg 'three girls' means 'ggg' and so on....
Read More →Factorise
Question: Factorise:6abb2+ 12ac 2bc Solution: By suitably arranging the terms: $6 a b-b^{2}+12 a c-2 b c=6 a b+12 a c-b^{2}-2 b c$ $=(6 a b+12 a c)-\left(b^{2}+2 b c\right)$ $=6 a(b+2 c)-b(b+2 c)$ $=(b+2 c)(6 a-b)$ $\therefore 6 a b-b^{2}+12 a c-2 b c=(b+2 c)(6 a-b)$...
Read More →Will the median class and modal
Question: Will the median class and modal class of grouped data always be different? Justify your answer. Solution: Not always, It depends on the given data....
Read More →Is it true to say that the mean,
Question: Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer Solution: the value of these three measures can be the same, it depends on the type of data....
Read More →Factorise:
Question: Factorise:x2xz+xyyz Solution: By suitably arranging the terms: $x^{2}-x z+x y-y z=x^{2}+x y-x z-y z$ $=\left(x^{2}+x y\right)-(x z+y z)$ $=x(x+y)-z(x+y)$ $=(x+y)(x-z)$ $\therefore x^{2}-x z+x y-y z=(x+y)(x-z)$...
Read More →In calculating the mean of grouped data,
Question: In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula, $\bar{x}=a+\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$ Where, a is the assumed mean, a must be one of the mid-point of the classes. Is the last statement correct? Justify your answer. Solution: No, it is not necessary that assumed mean consider as the mid-point of the class interval. It is considered as any value which is easy to simplify it....
Read More →Factorise:
Question: Factorise:ab2bc2ab+c2 Solution: By suitably arranging the terms: $a b^{2}-b c^{2}-a b+c^{2}=a b^{2}-a b-b c^{2}+c^{2}$ $=\left(a b^{2}-a b\right)-\left(b c^{2}-c^{2}\right)$ $=a b(b-1)-c^{2}(b-1)$ $=(b-1)\left(a b-c^{2}\right)$ $\therefore a b^{2}-b c^{2}-a b+c^{2}=(b-1)\left(a b-c^{2}\right)$...
Read More →The median of an ungrouped data and the median
Question: The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason. Solution: Not always, because for calculating median of a grouped data, the formula used is based on the assumption that the observations in the classes are uniformal distributed (or equally spaced)....
Read More →Factorise:
Question: Factorise:x2axbx+ab Solution: By suitably arranging the terms: $x^{2}-a x-b x+a b=x^{2}-b x-a x+a b$ $=\left(x^{2}-b x\right)-(a x-a b)$ $=x(x-b)-a(x-b)$ $=(x-b)(x-a)$ $\therefore x^{2}-a x-b x+a b=(x-b)(x-a)$...
Read More →A school has five houses A, B, C, D and E.
Question: A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A, B and C is (a) $\frac{4}{23}$ (b) $\frac{6}{23}$ (c) $\frac{8}{23}$ (d) $\frac{17}{23}$ Solution: (b)Total number of students = 23 Number of students in house A, B and C = 4+ 8 + 5 = 17 $\therefore \quad$ Remains st...
Read More →Someone is asked to take a number from 1 to 100.
Question: Someone is asked to take a number from 1 to 100. The probability that it is a prime,is (a) $\frac{1}{5}$ (b) $\frac{6}{25}$ (c) $\frac{1}{4}$ (d) $\frac{13}{50}$ Solution: (c) Total numbers of outcomes $=100$ So, the prime numbers between 1 to 100 are $2,3,5,7,11,13,17,19,23,29,31,37,41.43,47,53,56,61,67,71$, $73,79,83,89$ and 97 . $\therefore$ Total number of possible outcomes $=25$ $\therefore$ Required probability $=\frac{25}{100}=\frac{1}{4}$...
Read More →Factorise: ar + br + at + bt
Question: Factorise:ar+br+at+bt Solution: By grouping the terms: $a r+b r+a t+b t=(a r+b r)+(a t+b t)$ $=r(a+b)+t(a+b)$ $=(a+b)(r+t)$ $\therefore a r+b r+a t+b t=(a+b)(r+t)$...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\log \left(\frac{\sin x}{1+\cos x}\right)$ Solution: Let $y=\log \left(\frac{\sin x}{1+\cos x}\right)$ $\Rightarrow y=\log \left(\frac{\sin 2 \times \frac{x}{2}}{1+\cos 2 \times \frac{x}{2}}\right)$ We have $\sin 2 \theta=2 \sin \theta \cos \theta$ and $1+\cos 2 \theta=2 \cos ^{2} \theta$ $\Rightarrow y=\log \left(\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right)$ $\Rightarrow y=\log \left(\frac{\sin ...
Read More →One ticket is drawn at random from
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 5 is (a) $\frac{1}{5}$ (b) $\frac{3}{5}$ (c) $\frac{4}{5}$ (d) $\frac{1}{3}$ Solution: (a)Number of total outcomes = 40 Multiples of 5 between 1 to 40 = 5,10,15,20,25. 30 35, 40 $\therefore$ Total number of possible outcomes $=8$ $\therefore$Required probability $=\frac{8}{40}=\frac{1}{5}$...
Read More →Factorise: (x + y)(2x + 5) − (x + y)(x + 3)
Question: Factorise:(x+y)(2x+ 5) (x+y)(x+ 3) Solution: We have: $(x+y)(2 x+5)-(x+y)(x+3)=(x+y)\{(2 x+5)-(x+3)\}$ $=(x+y)(2 x+5-x-3)$ $=(x+y)(x+2)$...
Read More →A girl calculates that the probability
Question: A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, then how many tickets has she bought? (a) 40 (b) 240 (c) 480 (d) 750 Solution: (c) Given, total number of sold tickets = 6000 Let she bought x tickets. Then, probability of her winning the first prize $=\frac{x}{6000}=0.08$ [given] $\Rightarrow$ $x=0.08 \times 6000$ $\therefore$ $x=480$ Hence, she bought 480 tickets....
Read More →Factorise:
Question: Factorise:12(2x 3y)2 16(3y 2x) Solution: We have: $12(2 x-3 y)^{2}-16(3 y-2 x)=12(2 x-3 y)^{2}+16(2 x-3 y)$ $=(2 x-3 y)\{12(2 x-3 y)+16\}$ $=(2 x-3 y)(24 x-36 y+16)$ $\therefore 12(2 x-3 y)^{2}-16(3 y-2 x)=(2 x-3 y)(24 x-36 y+16)$...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\log \left(\frac{\sin x}{1+\cos x}\right)$ Solution: Let $y=\log \left(\frac{\sin x}{1+\cos x}\right)$ $\Rightarrow y=\log \left(\frac{\sin 2 \times \frac{x}{2}}{1+\cos 2 \times \frac{x}{2}}\right)$ We have $\sin 2 \theta=2 \sin \theta \cos \theta$ and $1+\cos 2 \theta=2 \cos ^{2} \theta$ $\Rightarrow y=\log \left(\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right)$ $\Rightarrow y=\log \left(\frac{\sin ...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\log \left(\frac{\sin x}{1+\cos x}\right)$ Solution: Let $y=\log \left(\frac{\sin x}{1+\cos x}\right)$ $\Rightarrow y=\log \left(\frac{\sin 2 \times \frac{x}{2}}{1+\cos 2 \times \frac{x}{2}}\right)$ We have $\sin 2 \theta=2 \sin \theta \cos \theta$ and $1+\cos 2 \theta=2 \cos ^{2} \theta$ $\Rightarrow y=\log \left(\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right)$ $\Rightarrow y=\log \left(\frac{\sin ...
Read More →Factorise: x(a − 3) + y(3 − a)
Question: Factorise:x(a 3) +y(3 a) Solution: We have: $x(a-3)+y(3-a)=x(a-3)-y(a-3)$ $=(a-3)(x-y)$ $\therefore x(a-3)+y(3-a)=(a-3)(x-y)$...
Read More →The probability of getting a bad egg
Question: The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is (a) 7 (b) 14 (c) 21 (d) 28 Solution: (b)Here, total number of eggs = 400 Probability of getting a bad egg = 0.035 $\Rightarrow \quad \frac{\text { Number of bad eggs }}{\text { Total number of eggs }}=0.035$ $\Rightarrow \quad \frac{\text { Number of bad eggs }}{400}=0.035$ $\therefore \quad$ Number of bad eggs $=0.035 \times 400=14$...
Read More →Factorise:
Question: Factorise:16(2p 3q)2 4(2p 3q) Solution: We have: $16(2 p-3 q)^{2}-4(2 p-3 q)=(2 p-3 q)\{16(2 p-3 q)-4\}$ $=(2 p-3 q)(32 p-48 q-4)$ $\therefore 16(2 p-3 q)^{2}-4(2 p-3 q)=(2 p-3 q)(32 p-48 q-4)$...
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