Factorize:
Question: Factorize: $125 a^{3}+b^{3}+64 c^{3}-60 a b c$ Solution: $125 a^{3}+b^{3}+64 c^{3}-60 a b c=(5 a)^{3}+(b)^{3}+(4 c)^{3}-3 \times 5 a \times b \times 4 c$ $=(5 a+b+4 c)\left[(5 a)^{2}+(b)^{2}+(4 c)^{2}-5 a \times b-b \times 4 c-5 a \times 4 c\right]$ $=(5 a+b+4 c)\left(25 a^{2}+b^{2}+16 c^{2}-5 a b-4 b c-20 a c\right)$...
Read More →A letter lock consists of three rings each marked with 10 different letters.
Question: A letter lock consists of three rings each marked with 10 different letters. In how many ways it is possible to make an unsuccessful attempt to open the lock? Solution: Number of ways of marking each of the ring = 10 different letters $\therefore$ Total number of ways of marking any letter on these three rings $=10 \times 10 \times 10=1000$ Out of these 1000 combinations of the lock, 1 combination will be successful. $\therefore$ Total number of unsuccessful attempts $=1000-1=999$...
Read More →A pole has to be erected at a point on the boundary of
Question: A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected? Solution: Let $P$ be the required location on the boundary of a circular park such that its distance from gate $B$ is $=x$ metres that is $B P=x$ metres Then, $A P...
Read More →Find the product:
Question: Find the product: $(3 x-5 y+4)\left(9 x^{2}+25 y^{2}+15 x y-20 y+12 x+16\right)$ Solution: $(3 x-5 y+4)\left(9 x^{2}+25 y^{2}+15 x y-20 y+12 x+16\right)$ $=(3 x+(-5 y)+4)\left(9 x^{2}+25 y^{2}+16+15 x y-20 y+12 x\right)$ $(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=a^{3}+b^{3}+c^{3}-3 a b c$ Here, $a=3 x, b=-5 y, c=4$ $(3 x+(-5 y)+4)\left(9 x^{2}+25 y^{2}+16+15 x y-20 y+12 x\right)$ $=(3 x)^{3}+(-5 y)^{3}+4^{3}-3 \times 3 x(-5 y)(4)$ $=27 x^{3}-125 y^{3}+64+180 x y$...
Read More →In how many ways can an examinee answer a set of ten true/false type questions?
Question: In how many ways can an examinee answer a set of ten true/false type questions? Solution: Number of ways of answering the first question = 2 (either true or false) Similarly, each question can be answered in 2 ways. $\therefore$ Total number of ways of answering all the 10 questions $=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=2^{10}=1024$...
Read More →A coin is tossed five times and outcomes are recorded.
Question: A coin is tossed five times and outcomes are recorded. How many possible outcomes are there? Solution: Number of outcomes when the coin is tossed for the first time = 2 Number of outcomes when the coin is tossed for the second time = 2 Thus, there would be 2 outcomes, each time the coin is tossed. Total number of possible outcomes on tossing the coin five times $=2 \times 2 \times 2 \times 2 \times 2=32$...
Read More →There are four parcels and five post-offices.
Question: There are four parcels and five post-offices. In how many different ways can the parcels be sent by registered post? Solution: Number of ways of sending 1 parcel via registered post = 5 Number of ways of sending 4 parcels via registered post through 5 post offices $=5 \times 5 \times 5 \times 5=625$...
Read More →Find the product:
Question: Find the product: $(x-2 y+3)\left(x^{2}+4 y^{2}+2 x y-3 x+6 y+9\right)$ Solution: $(\mathrm{x}-2 \mathrm{y}+3)\left(x^{2}+4 y^{2}+2 x y-3 x+6 y+9\right)$ $=(\mathrm{x}-2 \mathrm{y}+3)\left(x^{2}+4 y^{2}+9+2 x y+6 y-3 x\right)$ $=[x+(-2 y)+3]\left[x^{2}+(-2 y)^{2}+(3)^{2}-x \times(-2 y)-(-2 y) \times 3-3 \times x\right]$ $=(x)^{3}+(-2 y)^{3}+(3)^{3}-3(x)(-2 y)(3)$ $=x^{3}-8 y^{3}+27+18 x y$...
Read More →A mint prepares metallic calendars specifying months,
Question: A mint prepares metallic calendars specifying months, dates and days in the form of monthly sheets (one plate for each month). How many types of calendars should it prepare to serve for all the possibilities in future years? Solution: The first day of the year can be any one of the days of the week, i.e the first day can be selected in 7 ways. But, the year could also be a leap year. So, the mint should prepare 7 calendars for the non-leap year and 7 calendars for the leap year. So, to...
Read More →From Goa to Bombay there are two roots; air, and sea. From Bombay to Delhi there are three routes;
Question: From Goa to Bombay there are two roots; air, and sea. From Bombay to Delhi there are three routes; air, rail and road. From Goa to Delhi via Bombay, how many kinds of routes are there? Solution: Number of routes from Goa to Bombay = 2 Number of routes from Bombay to Delhi = 3 Using fundamental principle of multiplication: Number of routes from Goa to Delhi via Bombay = 23 = 6...
Read More →A person wants to buy one fountain pen, one ball pen and one pencil from a stationery shop.
Question: A person wants to buy one fountain pen, one ball pen and one pencil from a stationery shop.If there are 10 fountain pen varieties, 12 ball pen varieties and 5 pencil varieties, in how many ways can he select these articles? Solution: Number of fountain pen varieties = 10 Number of ball pen varieties = 12 Number of pencil varieties = 5 Ways to select a fountain pen = 10 Ways to select a ball pen = 12 Ways to select a pencil = 5 Ways to select a fountain pen, a ball pen and a pencil $=10...
Read More →Find the product:
Question: Find the product: $(x-y-z)\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right)$ Solution: $(x-y-z)\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right)$ $=(x+(-y)+(-z))\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right)$ We know $(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=a^{3}+b^{3}+c^{3}-3 a b c$ Here, $a=x, b=-y, c=-z$ $(x+(-y)+(-z))\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right)=x^{3}-y^{3}-z^{3}-3 x y z$...
Read More →In a class there are 27 boys and 14 girls. The teacher wants to select 1 boy and 1 girl to represent the class in a function.
Question: In a class there are 27 boys and 14 girls. The teacher wants to select 1 boy and 1 girl to represent the class in a function. In how many ways can the teacher make this selection? Solution: No. of boys in the class = 27 No. of girls in the class = 14 Ways to select a boy = 27 Similarly, ways to select a girl = 14 $\therefore$ Number of ways to select 1 boy and $1 \mathrm{girl}=27 \times 14=378$...
Read More →The hypotenuse of a right triangle is 310−−√ cm.
Question: The hypotenuse of a right triangle is $3 \sqrt{10} \mathrm{~cm}$. If the smaller leg is tripled and the longer leg doubled, new hypotenuse wll be $9 \sqrt{5} \mathrm{~cm}$. How long are the legs of the triangle? Solution: Let the length of smaller side of right triangle be $=x \mathrm{~cm}$ then larger side be $=y \mathrm{~cm}$ Then, as we know that by Pythagoras theorem $x^{2}+y^{2}=(3 \sqrt{10})^{2}$ $x^{2}+y^{2}=90$ .......(1) If the smaller side is triple and the larger side be dou...
Read More →Find the product:
Question: Find the product: $(x+y-z)\left(x^{2}+y^{2}+z^{2}-x y+y z+z x\right)$ Solution: $(x+y-z)\left(x^{2}+y^{2}+z^{2}-x y+y z+z x\right)$ $=[x+y+(-z)]\left[x^{2}+y^{2}+(-z)^{2}-x y-y \times(-z)-[-z] \times x\right]$ $=x^{3}+y^{3}+(-z)^{3}-3 x \times y \times(-z)$ $=x^{3}+y^{3}-z^{3}+3 x y z$...
Read More →Prove that
Question: Prove that $\frac{59 \times 59 \times 59-9 \times 9 \times 9}{59 \times 59+59 \times 9+9 \times 9}=50$ Solution: $\frac{59 \times 59 \times 59-9 \times 9 \times 9}{59 \times 59+59 \times 9+9 \times 9}$ $=\frac{(59)^{3}-9^{3}}{59^{2}+59 \times 9+9^{2}}$ We know $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$ Here $a=59, b=9$ So $, \frac{(59-9)\left(59^{2}+9^{2}+59 \times 9\right)}{59^{2}+9^{2}+59 \times 9}=59-9=50: \mathrm{RHS}$ Thus, LHS=RHS...
Read More →The hypotenuse of a right triangle is 25 cm.
Question: The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides. Solution: Let the length of one side of right triangle be $=x \mathrm{~cm}$ then other side be $=(x+5) \mathrm{cm}$ And given that hypotenuse $=25 \mathrm{~cm}$ As we know that by Pythagoras theorem, $x^{2}+(x+5)^{2}=(25)^{2}$ $x^{2}+x^{2}+10 x+25=625$ $2 x^{2}+10 x+25-625=0$ $2 x^{2}+10 x-600=0$ $x^{2}+5 x-300=0$ $x^{2}-15 x+2...
Read More →Prove that
Question: Prove that $\frac{0.85 \times 0.85 \times 0.85+0.15 \times 0.15 \times 0.15}{0.85 \times 0.85-0.85 \times 0.15+0.15 \times 0.15}=1$ Solution: LHS: $\frac{0.85 \times 0.85 \times 0.85+0.15 \times 0.15 \times 0.15}{0.85 \times 0.85-0.85 \times 0.15+0.15 \times 0.15}$ $=\frac{(0.85)^{3}+(0.15)^{3}}{(0.85)^{2}-0.85 \times 0.15+(0.15)^{2}}$ We know $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$ Here $a=0.85, b=0.15$ $\frac{(0.85)^{3}+(0.15)^{3}}{(0.85)^{2}-0.85 \times 0.15+(0.15)^{2}}$ $=\...
Read More →Write the solution set of the inequation
Question: Write the solution set of the inequation |x 1| |x 3|. Solution: We have: $|x-1| \geq|x-3|$ $\Rightarrow|x-1|-|x-3| \geq 0$ The LHS of the inequation has two seperate modulus. Equating these to zero, we obtain $x=1,3$ as critical points. These points divide the real line in three regions, i.e $(-\infty, 1],[1,3],[3, \infty)$. CASE 1: When $-\inftyx \leq 1$, then $|x-1|=-(x-1)$ and $|x-3|=-(x-3)$ $\therefore|x-1|-|x-3| \geq 0$ $\Rightarrow-(x-1)-[-(x-3)] \geq 0$ $\Rightarrow-x+1+x-3 \geq...
Read More →The sum of the reciprocals of Rehman's ages
Question: The sum of the reciprocals of Rehman's ages (in years) 3 years ago and 5 years from now is 1/3. Find his present age. Solution: Let the present age of Rehman be $x$ years Then, 8 years later, age of her $=(x+5)$ years Five years ago, her age $=(x-3)$ years Then according to question, $\frac{1}{(x-3)}+\frac{1}{(x+5)}=\frac{1}{3}$ $\frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}$ $\frac{2 x+2}{x^{2}+5 x-3 x-15}=\frac{1}{3}$ $x^{2}+2 x-15=6 x+6$ $x^{2}+2 x-15-6 x-6=0$ $x^{2}-4 x-21=0$ $x^{2}-7 x+3...
Read More →Factorise
Question: Factorise $(x+2)^{3}-(x-2)^{3}$ Solution: $(x+2)^{3}-(x-2)^{3}$ $=(x+2-x+2)\left[(x+2)^{2}+(x-2)^{2}+\left(x^{2}-4\right)\right]$ $=4\left[x^{2}+4+4 x+x^{2}+4-4 x+x^{2}-4\right]$ $=4\left(3 x^{2}+4\right)$...
Read More →Write the solution of set of
Question: Write the solution of set of $\left|x+\frac{1}{x}\right|2$. Solution: We have: $\left|x+\frac{1}{x}\right|2$ $\Rightarrow\left|x+\frac{1}{x}\right|-20$ CASE $1:$ When $x+\frac{1}{x}0$, then $\left|x+\frac{1}{x}\right|=x+\frac{1}{x}$ Now, $\left|x+\frac{1}{x}\right|-20$ $\Rightarrow x+\frac{1}{x}-20$ $\Rightarrow \frac{x^{2}+1-2 x}{x}0$ $\Rightarrow \frac{(x-1)^{2}}{x}0$ $\Rightarrow x0$ and $x \neq 1$ $\Rightarrow x \in(0,1) \mathrm{U}(1, \infty) \quad \ldots$ (i) CASE $2:$ When $x+\fr...
Read More →Factorise
Question: Factorise $(x+2)^{3}+(x-2)^{3}$ Solution: $(x+2)^{3}+(x-2)^{3}$ $=(x+2+x-2)\left[(x+2)^{2}+(x-2)^{2}-\left(x^{2}-4\right)\right]$ $=2 x\left(x^{2}+4+4 x+x^{2}+4-4 x-x^{2}+4\right)$ $=2 x\left(x^{2}+12\right)$...
Read More →A girls is twice as old as her sister. Four years hence,
Question: A girls is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Find their present ages. Solution: Let the present age of girl be $x$ years then, age of her sister $\left(\frac{x}{2}\right)$ years Then, 4 years later, age of girl $=(x+4)$ years and her sister's age be $\left(\frac{x}{2}+4\right)$ years Then according to question, $(x+4)\left(\frac{x}{2}+4\right)=160$ $(x+4)(x+8)=160 \times 2$ $x^{2}+8 x+4 x+32=320$ $x^{2}+12 x+32-320=0$ $x^{2}...
Read More →Factorise
Question: Factorise $8(x+y)^{3}-27(x-y)^{3}$ Solution: $8(x+y)^{3}-27(x-y)^{3}$ $=[2(x+y)]^{3}-[3(x-y)]^{3}$ $=(2 x+2 y-3 x+3 y)\left[4(x+y)^{2}+9(x-y)^{2}+6\left(x^{2}-y^{2}\right)\right]$ $=(-x+5 y)\left[4\left(x^{2}+y^{2}+2 x y\right)+9\left(x^{2}+y^{2}-2 x y\right)+6\left(x^{2}-y^{2}\right)\right]$ $=(-x+5 y)\left[4 x^{2}+4 y^{2}+8 x y+9 x^{2}+9 y^{2}-18 x y+6 x^{2}-6 y^{2}\right]$ $=(-x+5 y)\left(19 x^{2}+7 y^{2}-10 x y\right)$...
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