If a – b = 6 and ab = 20, find the value of
Question: If $a-b=6$ and $a b=20$, find the value of $a^{3}-b^{3}$ Solution: Given, a b = 6 and ab = 20 We know that, $a^{3}-b^{3}=(a-b)^{3}+3 a b(a-b)$ $\Rightarrow a^{3}-b^{3}=(a-b)^{3}+3 a b(a-b)$ $\Rightarrow a^{3}-b^{3}=(6)^{3}+3(20)(6)$ $\Rightarrow a^{3}-b^{3}=216+360$ $\Rightarrow a^{3}-b^{3}=576$ Hence, the value of $a^{3}-b^{3}$ is 576...
Read More →If a + b = 8 and ab = 6, find the value of
Question: If $a+b=8$ and $a b=6$, find the value of $a^{3}+b^{3}$ Solution: Given, a + b = 8 and ab = 6 We know that, $a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)$ $\Rightarrow a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)$ $\Rightarrow a^{3}+b^{3}=(8)^{3}-3(6)(8)$ $\Rightarrow a^{3}+b^{3}=512-144$ $\Rightarrow a^{3}+b^{3}=368$ Hence, the value of $a^{3}+b^{3}$ is 368...
Read More →Write the value of k for which the system of equations has infinitely many solutions.
Question: Write the value ofkfor which the system of equations has infinitely many solutions. $2 x-y=5$ $6 x+k y=15$ Solution: The given systems of equations are $2 x-y=5$ $6 x+k y=15$ $a_{1}=2, a_{2}=6, b_{1}=1, b_{2}=k, c_{1}=5, c_{2}=15$ $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$ For the equations to have infinite number of solutions, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ $\frac{2}{6}=\frac{-1}{k}$ By cross Multiplication we get, $2 k=-6$ $k=\frac{-6}{2}$ $k=-3$ Hence t...
Read More →If a + b = 10 and ab = 16, find the value of
Question: If $a+b=10$ and $a b=16$, find the value of $a^{2}-a b+b^{2}$ and $a^{2}+a b+b^{2}$ Solution: Given, a + b = 10, ab = 16 We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ $\Rightarrow a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)$ $\Rightarrow a^{3}+b^{3}=(10)^{3}-3(16)(10)$ $\Rightarrow a^{3}+b^{3}=1000-480$ $\Rightarrow a^{3}+b^{3}=520$ Substitute, $a^{3}+b^{3}=520$ $a+b=10 i n a^{3}+b^{3}$ $=(a+b)\left(a^{2}+b^{2}-a b\right) a^{3}+b^{3}$ $=(a+b)\left(a^{2}+b^{2}-a b\right) 520$ $=10\left(a^{2}+b^...
Read More →Write the value of k for which the system of equations x + y − 4 = 0
Question: Write the value of $k$ for which the system of equations $x+y-4=0$ and $2 x+k y-3=0$ has no solution. Solution: The given system of equations is $x+y-4=0$ $2 x+k y-3=0$ $a_{1}=1, a_{2}=2, b_{1}=1, b_{2}=k, c_{1}=4, c_{2}=3$ For the equations to have no solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ $\frac{1}{2}=\frac{1}{k}$ By cross multiplication we get, $1 \times k=1 \times 2$ $k=2$ Hence, the value of $k$ is 2 when system equations has no solution....
Read More →One says. "give me hundred, friend! I shall then become twice as rich as you" The other replies,
Question: One says. "give me hundred, friend! I shall then become twice as rich as you" The other replies, "If you give me ten, I shall be six times as rich as you". Tell me what is the amount of their respective capital? Solution: 21. Let the money with first person be Rsx and the money with the second person be Rsy. Then, $(x+100)=2(y-100)$ $(y+10)=6(x-10)$ If first person gives $R s 100$ to second person then the second person will become twice as rich as first person, According to the given ...
Read More →The students of a class are made to stand in rows.
Question: The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find the number of student in the class. Solution: Let the number of students be $x$ and the number of row be $y$.then, Number of students in each row $=\frac{x}{y}$ Where three students is extra in each row, there are one row less that is when each row has $\left(\frac{x}{y}+3\right)$ students the number of rows is ...
Read More →Find x = 3 and y = -1, Find the values of each of the following using in identity:
Question: Find x = 3 and y = -1, Find the values of each of the following using in identity: (a) $\left(9 x^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right)$ (b) $(3 / x-5 / y)\left(9 / x^{2}+25 / y^{2}+15 / x y\right)$ (c) $(x / 7+y / 3)\left(x^{2} / 49+y^{2} / 9-x y / 21\right)$ (d) $(x / 4-y / 3)\left(x^{2} / 16+y^{2} / 9+x y / 21\right)$ (e) $(5 / x+5 x)\left(25 / x^{2}-25+25 x^{2}\right)$ Solution: (a) We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$ $\left(9 x^{...
Read More →Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer.
Question: Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awanded for each correct answer and 2 marks been deducted for each incorrect answer, the Yash would have scored 50 marks. How many question were there in the test? Solution: Let take right answer will be $x$ and wrong answer will be $y$. Hence total number of questions will be $x+y \cdots(i)$ If yash scored 40 marks in atleast getting 3 marks for each right an...
Read More →Meena went to a bank to withdraw Rs 2000.
Question: Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes Rs 50 and Rs 100 she received. Solution: Let $R s . x$ be the notes of $R s .50$ and $R s .100$ notes will be $R s . y$ If Meena ask for $R s .50$ and $R s .100$ notes only, then the equation will be, $50 x+100 y=2000$ Divide both sides by 50 then we get, $x+2 y=40 \cdots(i)$ If Meena got 25 notes in all then the equation will be, $x+y=...
Read More →A wizard having powers of mystic in candations and magical medicines seeing a cock,
Question: A wizard having powers of mystic in candations and magical medicines seeing a cock, fight going on, spoke privately to both the owners of cocks. To one he said; if your bird wins, than you give me your stake-money, but if you do not win, I shall give you two third of that'. Going to the other, he promised in the same way to give three fourths. From both of them his gain would be only 12 gold coins. Find the the stake of money each of the cock-owners have. Solution: Let the strike money...
Read More →2 women and 5 men can together finish a piece of embroidery in 4 days,
Question: 2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery, and that taken by 1 man alone. Solution: 1 women alone can finish the work in $x$ days and 1 man alone can finish it in $y$ days then One woman one day work $=\frac{1}{x}$ One man one days work $=\frac{1}{y}$ 2 women's one days work $=\frac{2}{x}$ 5 man's one days work $=\frac{5}{y}$ Since 2 women and 5 me...
Read More →Find the following products
Question: Find the following products (a) $(3 x+2 y)\left(9 x^{2}-6 x y+4 y^{2}\right)$ (b) $(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$ (c) $\left(7 p^{4}+q\right)\left(49 p^{8}-7 p^{4} q+q^{2}\right)$ (d) $(x / 2+2 y)\left(x^{2} / 4-x y+4 y^{2}\right)$ (e) $(3 / x-5 / y)\left(9 / x^{2}+25 / y^{2}+15 / x y\right)$ (f) $(3+5 / x)\left(9-15 / x+25 / x^{2}\right)$ (g) $(2 / x+3 x)\left(4 / x^{2}+9 x^{2}-6\right)$ (h) $\left(3 / x-2 x^{2}\right)\left(9 / x^{2}+4 x^{4}-6 x\right)$ (i) $(1-x)\lef...
Read More →The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Question: The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. Solution: We know that the sum of supplementary angles will be $180^{\circ}$. Let the longer supplementary angles will be' $y^{\prime}$. Then, $x+y=180^{\circ} \cdots(i)$ If larger of supplementary angles exceeds the smaller by 18 degree, According to the given condition. We have, $x=y+18 \cdots(i i)$ Substitute $x=y+18$ in equation $(i)$, we get, $x+y=180^{\circ}$ $y+18+y=180^{\circ}$ $2 y+18=180^{\ci...
Read More →Half the perimeter of a garden, whose length is 4 more than its width is 36 m.
Question: Half the perimeter of a garden, whose length is 4 more than its width is 36 m. Find the dimension of the garden. Solution: Let perimeter of rectangular garden will be $2(l+b)$.if half the perimeter of a garden will be $36 \mathrm{~m}$ $(l+b)=36 \cdots(i)$ When the length is four more than its width then $(b+4)$ Substituting $l=b+4$ in equation $(i)$ we get $l+b=36$ $b+4+b=36$ $2 b=36-4$ $2 b=32$ $b=\frac{32}{2}$ $b=16$ Putting $b=16$ in equation $(i)$ we get $(l+b)=36$ $l+16=36$ $l=36-...
Read More →A part of monthly hostel charges in a college are fixed and the remaining depend on the number of days one has taken food in the mess.
Question: A part of monthly hostel charges in a college are fixed and the remaining depend on the number of days one has taken food in the mess. When a student A takes foods for 20 days, he has to pay Rs 1000 as hostel charges whereas a students B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charge and the cost of food per day. Solution: Let the fixed charges of hostel be $R s . x$ and the cost of food charges be $R s . y$ per day According to the given condition w...
Read More →The car hire charges in a city comprise of a fixed charges together with the charge for the distance covered.
Question: The car hire charges in a city comprise of a fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is Rs 89 and the journey of 20 km, the charge paid is Rs 145. What will a person have to pay for travelling a distance of 30 km? Solution: Let the fixed charges of car be Rs.x per km and the running charges be $R s . y \mathrm{~km} / \mathrm{hr}$ According to the given condition we have $x+12 y=89 \cdots(i)$ $x+20 y=145 \cdots(i i)$ $x+12...
Read More →The value of
Question: If $x-\frac{1}{x}=3+2 \sqrt{2}$, Find the value of $x^{3}-\frac{1}{x^{3}}$ Solution: Given, $x-\frac{1}{x}=3+2 \sqrt{2}$ Cubing $x-\frac{1}{x}=3+2 \sqrt{2}$ on both sides We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ $\left(x-\frac{1}{x}\right)^{3}=(3+2 \sqrt{2})^{3}$ $x^{3}-\frac{1}{x^{3}}-3 * x * \frac{1}{x}\left(x-\frac{1}{x}\right)=3^{2}+(2 \sqrt{2})^{3}+3 * 3 * 2 \sqrt{2}(3+2 \sqrt{2})$ $x^{3}-\frac{1}{x^{3}}-3(3+2 \sqrt{2})=27+16 \sqrt{2}+18 \sqrt{2}(3+2 \sqrt{2})$ $x^{3}-\fra...
Read More →In a ∆ABC, ∠A = x°, ∠B = 3x° and ∠C = y°. If 3y − 5x = 30,
Question: In a $\triangle A B C, \angle A=x^{\circ}, \angle B=3 x^{\circ}$ and $\angle C=y^{\circ}$. If $3 y-5 x=30$, prove that the triangle is right angled. Solution: We have to prove that the triangle is right Given $\angle A=x^{\circ}, \angle B=3 x^{\circ}$ and $\angle C=y^{\circ}$ Sum of three angles in triangle are $\angle A+\angle B+\angle c=180^{\circ}$ $\angle A+\angle B+\angle c=180^{\circ}$ $x+3 x+y=180$ $4 x+y=180 \cdots(i)$ By solving $4 x+y=180$ with $3 y-5 x=30$ we get, $4 x+y=180...
Read More →Find the value of 64x3
Question: Find the value of $64 x^{3}-125 z^{3}$, if $4 x-5 z=16$ and $x z=12$ Solution: Given, $64 x^{3}-125 z^{3}$ Here, $4 x-5 z=16$ and $x z=12$ Cubing $4 x-5 z=16$ on both sides $(4 x-5 z)^{3}=16^{3}$ We know that, $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$ $(4 x)^{3}-(5 z)^{3}-3(4 x)(5 z)(4 x-5 z)=16^{3}$ $64 x^{3}-125 z^{3}-60(x z)(16)=4096$ $64 x^{3}-125 z^{3}-60(12)(16)=4096$ $64 x^{3}-125 z^{3}-11520=4096$ $64 x^{3}-125 z^{3}=4096+11520$ $64 x^{3}-125 z^{3}=15616$ The value of $64 x^{3}-125 z^...
Read More →Find the values of
Question: Find the values of $27 x^{3}+8 y^{3}$, if (a) 3x + 2y = 14 and xy = 8 (b) 3x + 2y = 20 and xy = 14/9 Solution: (a) Given,3x + 2y = 14 and xy = 8 cubing on both sides $(3 x+2 y)^{3}=14^{3}$ We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ $27 x^{3}+8 y^{3}+3(3 x)(2 y)(3 x+2 y)=2744$ $27 x^{3}+8 y^{3}+18 x y(3 x+2 y)=2744$ $27 x^{3}+8 y^{3}+18(8)(14)=2744$ $27 x^{3}+8 y^{3}+2016=2744$ $27 x^{3}+8 y^{3}=2744-2016$ $27 x^{3}+8 y^{3}=728$ Hence, the value of $27 x^{3}+8 y^{3}=728$ (b) Given...
Read More →A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket.
Question: A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs Rs 216 and one full and one half reserved first class tickets cost Rs 327. What is the basic first class full fare and what is the reservation charge? Solution: Let take first class full of fare is Rs $x$ and reservation charge is Rs $y$ per ticket Then half of the ticket as on full ticket $=\frac{x}{2}$ A...
Read More →In a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°,
Question: In a cyclic quadrilateral $A B C D, \angle A=(2 x+4)^{\circ}, \angle B=(y+3)^{\circ}, \angle C=(2 y+10)^{\circ}, \angle D=(4 x-5)^{\circ}$. Find the four angles. Solution: We know that the sum of the opposite angles of cyclic quadrilateral is $180^{\circ}$ in the cyclic quadrilateral $A B C D$, angles $A$ and $C$ and angles $B$ and $D$ pairs of opposite angles Therefore $\angle A+\angle C=180^{\circ}$ and $\angle B+\angle D=180^{\circ}$ Taking $\angle A+\angle C=180^{\circ}$ By substit...
Read More →If x4 + 1/x4 = 194, calculate
Question: If $x^{4}+1 / x^{4}=194$, calculate $x^{2}+1 / x^{2}, x^{3}+1 / x^{3}, x+1 / x$ Solution: Given, $x^{4}+1 / x^{4}=194$ add and subtract $\left(2^{*} x^{2} * 1 / x^{2}\right)$ on left side in above given equation $x^{4}+1 / x^{4}+\left(2^{*} x^{2} * 1 / x^{2}\right)-2\left(2^{*} x^{2} * 1 / x^{2}\right)=194$ $x^{4}+1 / x^{4}+\left(2^{*} x^{2} * 1 / x^{2}\right)-2=194$ $\left(x^{2}+1 / x^{2}\right)^{2}-2=194$ $\left(x^{2}+1 / x^{2}\right)^{2}=194+2$ $\left(x^{2}+1 / x^{2}\right)^{2}=196$...
Read More →If x + 1/x = 3, calculate x2
Question: If $x+1 / x=3$, calculate $x^{2}+1 / x^{2}, x^{3}+1 / x^{3}, x^{4}+1 / x^{4}$ Solution: Given, $x+1 / x=3$ We know that $(x+y)^{2}=x^{2}+y^{2}+2 x y$ $(x+1 / x)^{2}=x^{2}+1 / x^{2}+(2 * x * 1 / x)$ $3^{2}=x^{2}+1 / x^{2}+2$ $9-2=x^{2}+1 / x^{2}$ $x^{2}+1 / x^{2}=7$ Squaring on both sides $\left(x^{2}+1 / x^{2}\right)^{2}=7^{2}$ $x^{4}+1 / x^{4}+2^{*} x^{2} * 1 / x^{2}=49$ $x^{4}+1 / x^{4}+2=49$ $x^{4}+1 / x^{4}=49-2$ $x^{4}+1 / x^{4}=47$ Again, cubing on both sides $(x+1 / x)^{3}=3^{3}...
Read More →