A number whose fifth part increased by 5 is equal to its fourth part diminished by 5.
Question: A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number. Solution: Let the number be $\mathrm{x}$. According to the question, $\frac{\mathrm{x}}{5}+5=\frac{\mathrm{x}}{4}-5$ or $\frac{\mathrm{x}}{5}-\frac{\mathrm{x}}{4}=-5-5$ or $\frac{4 \mathrm{x}-5 \mathrm{x}}{20}=-10$ or $-\mathrm{x}=-200[$ After cross multiplication $]$ or $x=200$ Thus, the number is 200 ....
Read More →Find a number such that when 5 is subtracted from 5 times the number,
Question: Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number. Solution: Let the number be $\mathrm{x}$. According to the question, $5 \mathrm{x}-5=2 \mathrm{x}+4$ or $5 \mathrm{x}-2 \mathrm{x}=4+5$ or $3 \mathrm{x}=9$ or $\mathrm{x}=\frac{9}{3}$ or $\mathrm{x}=3$ Thus, the number is 3 ....
Read More →If a line segment joining mid-points
Question: If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. Solution: Given $A B$ and $C D$ are two chords of a circle whose centre is $O$ and $P Q$ is a diameter bisecting the chord $A B$ and $C D$ at $L$ and $M$, respectively and the diameter $P Q$ passes through the centre $O$ of the circle. To prove $A B \| C D$ Proof Since, $L$ is the mid-point of $A B$. $\therefore \quad O L \perp A B$ [since, the...
Read More →Find a number whose double is 45 greater than its half.
Question: Find a number whose double is 45 greater than its half. Solution: Let the number be $\mathrm{x}$. According to the question, $2 \mathrm{x}=\frac{1}{2} \mathrm{x}+45$ or $2 \mathrm{x}-\frac{1}{2} \mathrm{x}=45$ or $\frac{4 \mathrm{x}-\mathrm{x}}{2}=45$ or $3 \mathrm{x}=90[$ After c ross multiplication $]$ or $\mathrm{x}=\frac{90}{3}$ or $\mathrm{x}=30$ Thus, the number is 30 ....
Read More →The difference between the squares of two consecutive numbers is 31.
Question: The difference between the squares of two consecutive numbers is 31. Find the numbers. Solution: Let the numbers be $x$ and $x+1$. According to the question, $(\mathrm{x}+1)^{2}-\mathrm{x}^{2}=31$ or $\mathrm{x}^{2}+2 \mathrm{x}+1-\mathrm{x}^{2}=31$ or $2 \mathrm{x}=31-1$ or $\mathrm{x}=\frac{30}{2}$ or $\mathrm{x}=15$ Thus, the numbers are 15 and 16 ....
Read More →Four-fifth of a number is more than three-fourth of the number by 4.
Question: Four-fifth of a number is more than three-fourth of the number by 4. Find the number. Solution: Let the number be $\mathrm{x}$. According to the question, $\frac{4}{5} \mathrm{x}-\frac{3}{4} \mathrm{x}=4$ or $\frac{16 \mathrm{x}-15 \mathrm{x}}{20}=4$ or $\mathrm{x}=80[$ After cross multiplication $]$ Thus, the required number is 80 ....
Read More →AB and AC are two equal chords of a circle.
Question: AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle. Solution: Given AS and AC are two equal chords whose centre is O. To prove Centre O lies on the bisector of BAC. Construction Join SC, draw bisector AD of BAC. Proof In ΔSAM and ΔCAM, AS = AC [given] BAM = CAM [given]...
Read More →Find a positive value of x for which the given equation is satisfied:
Question: Find a positive value ofxfor which the given equation is satisfied: (i) $\frac{x^{2}-9}{5+x^{2}}=-\frac{5}{9}$ (ii) $\frac{y^{2}+4}{3 y^{2}+7}=\frac{1}{2}$ Solution: $($ i $) \frac{x^{2}-9}{5+x^{2}}=\frac{-5}{9}$ or $9 \mathrm{x}^{2}-81=-25-5 \mathrm{x}^{2}[$ After c ross multiplication $]$ or $9 \mathrm{x}^{2}+5 \mathrm{x}^{2}=-25+81$ or $14 \mathrm{x}^{2}=56$ or $\mathrm{x}^{2}=\frac{56}{14}$ or $x^{2}=4=2^{2}$ or $\mathrm{x}=2$ Thus, $x=2$ is the solution of the given equation. Chec...
Read More →On increasing the diameter of a circle by 40%, its area will be increased by
Question: On increasing the diameter of a circle by 40%, its area will be increased by(a) 40%(b) 80%(c) 96%(d) 82% Solution: (c) 96%Letdbe the original diameter. Radius $=\frac{d}{2}$ Thus, we have: Original area $=\pi \times\left(\frac{d}{2}\right)^{2}$ $=\frac{\pi d^{2}}{4}$ New diameter $=140 \%$ of $d$ $=\left(\frac{140}{100} \times d\right)$ $=\frac{7 d}{5}$ Now, New radius $=\frac{7 d}{5 \times 2}$ $=\frac{7 d}{10}$ New area $=\pi \times\left(\frac{7 d}{10}\right)^{2}$ $=\frac{49 \pi d^{2}...
Read More →A, B and C are three points on a circle.
Question: A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent.Thinking Process Firstly, inscribe a ΔABC in a circle, then draw the perpendicular bisecters of any two sides of a triangle. Secondly, prove that ΔOEA and ΔOEB are congruent by SAS rule and also ΔOMB and ΔOMC are congruent by RHS rule. Further, prove the required result. Solution: Given $A$ circle passing through three points $A, B$ and $C$. Construction Draw perpendicular bi...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{(2 x+3)-(5 x-7)}{6 x+11}=-\frac{8}{3}$ Solution: $\frac{(2 x+3)-(5 x-7)}{6 x+11}=\frac{-8}{3}$ or $\frac{-3 x+10}{6 x+11}=\frac{-8}{3}$ or $-9 \mathrm{x}+30=-48 \mathrm{x}-88[$ After c ross multipl ication $]$ or $-9 \mathrm{x}+48 \mathrm{x}=-88-30$ or $39 \mathrm{x}=-118$ or $\mathrm{x}=\frac{-118}{39}$ Thus, $\mathrm{x}=\frac{-118}{39}$ is the solution of the given equation. Check: Substituting $\mathrm{x}=\frac{-118}{39}$ i...
Read More →Solve this
Question: Given $A=\left[\begin{array}{lll}5 0 4 \\ 2 3 2 \\ 1 2 1\end{array}\right], B^{-1}=\left[\begin{array}{lll}1 3 3 \\ 1 4 3 \\ 1 3 4\end{array}\right]$. Compute $(A B)^{-1}$. Solution: We have, $A=\left[\begin{array}{lll}5 0 4 \\ 2 3 2 \\ 1 2 1\end{array}\right]$ $B^{-1}=\left[\begin{array}{lll}1 3 3 \\ 1 4 3 \\ 1 3 4\end{array}\right]$ We know $(A B)^{-1}=B^{-1} A^{-1}$ For matrix $A$, $C_{11}=\left|\begin{array}{ll}3 2 \\ 2 1\end{array}\right|=-1, C_{12}=-\left|\begin{array}{ll}2 2 \\ ...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{x^{2}-(x+1)(x+2)}{5 x+1}=6$ Solution: $\frac{x^{2}-(x+1)(x+2)}{5 x+1}=6$ or $\frac{x^{2}-x^{2}-2 x-x-2}{5 x+1}=6$ or $\frac{-3 x-2}{5 x+1}=6$ or $30 \mathrm{x}+6=-3 \mathrm{x}-2[$ After c ross multipl ication $]$ or $30 \mathrm{x}+3 \mathrm{x}=-2-6$ or $33 \mathrm{x}=-8$ or $\mathrm{x}=\frac{-8}{33}$ Thus, $\mathrm{x}=\frac{-8}{33}$ is the solution of the given equation. Check : Substituting $\mathrm{x}=\frac{-8}{33}$ in the g...
Read More →The perimeter of a circular field is 242 m. The area of the field is
Question: The perimeter of a circular field is 242 m. The area of the field is(a) 9317 m2(b) 18634 m2(c) 4658.5 m2(d) none of these Solution: (c) 4658.5 m2Let the radius berm.We know: Perimeter of a circle $=2 \pi r$ Thus, we have: $2 \pi r=242$ $\Rightarrow 2 \times \frac{22}{7} \times r=242$ $\Rightarrow \frac{44}{7} \times r=242$ $\Rightarrow r=\left(242 \times \frac{7}{44}\right)$ $\Rightarrow r=\frac{77}{2}$ $\therefore$ Area of the circle $=\pi r^{2}$ $=\left(\frac{22}{7} \times \frac{77}{...
Read More →The perimeter of a circular field is 242 m. The area of the field is
Question: The perimeter of a circular field is 242 m. The area of the field is(a) 9317 m2(b) 18634 m2(c) 4658.5 m2(d) none of these Solution: (c) 4658.5 m2Let the radius berm.We know: Perimeter of a circle $=2 \pi r$ Thus, we have: $2 \pi r=242$ $\Rightarrow 2 \times \frac{22}{7} \times r=242$ $\Rightarrow \frac{44}{7} \times r=242$ $\Rightarrow r=\left(242 \times \frac{7}{44}\right)$ $\Rightarrow r=\frac{77}{2}$ $\therefore$ Area of the circle $=\pi r^{2}$ $=\left(\frac{22}{7} \times \frac{77}{...
Read More →Find the inverse of the matrix
Question: Find the inverse of the matrix $A=\left[\begin{array}{cc}a b \\ c \frac{1+b c}{a}\end{array}\right]$ and show that $a A^{-1}=\left(a^{2}+b c+1\right) I-a A$. Solution: We have, $A=\left[\begin{array}{cc}a b \\ c \frac{1+b c}{a}\end{array}\right]$ So, $\operatorname{adj}(A)=\left[\begin{array}{cc}\frac{1+b c}{a} -b \\ -c a\end{array}\right]$ and $|A|=1$ $\therefore A^{-1}=\left[\begin{array}{cc}\frac{1+b c}{a} -b \\ -c a\end{array}\right]$ Now, $a A^{-1}=\left(a^{2}+b c+1\right) I-a A$ ...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{(x+2)(2 x-3)-2 x^{2}+6}{x-5}=2$ Solution: $\frac{(x+2)(2 x-3)-2 x^{2}+6}{x-5}=2$ or $\frac{2 x^{2}+x-6-2 x^{2}+6}{x-5}=2$ or $\frac{\mathrm{x}}{\mathrm{x}-5}=2$ or $2 \mathrm{x}-10=\mathrm{x}[$ After c ross multiplication $]$ or $2 \mathrm{x}-\mathrm{x}=10$ or $\mathrm{x}=10$ Thus, $x=10$ is the solution of the given equation. Check : Substituting $x=10$ in the given equation, we get: L. H.S. $=\frac{(10+2)(2 \times 10-3)-2 \t...
Read More →The difference between the circumference and radius of a circle is 37 cm.
Question: The difference between the circumference and radius of a circle is 37 cm. The area of the circle is(a) 111 cm2(b) 184 cm2(c) 154 cm2(d) 259 cm2 Solution: (c) 154 cm2Let the radius bercm.We know: Circumference of the circle $=2 \pi \mathrm{r}$ Thus, we have: $2 \pi r-r=37$ $\Rightarrow r(2 \pi-1)=37$ $\Rightarrow r\left(2 \times \frac{22}{7}-1\right)=37$ $\Rightarrow r\left(\frac{37}{7}\right)=37$ $\Rightarrow r=\left(37 \times \frac{7}{37}\right)$ $\Rightarrow r=7 \mathrm{~cm}$ Radius ...
Read More →The difference between the circumference and radius of a circle is 37 cm.
Question: The difference between the circumference and radius of a circle is 37 cm. The area of the circle is(a) 111 cm2(b) 184 cm2(c) 154 cm2(d) 259 cm2 Solution: (c) 154 cm2Let the radius bercm.We know: Circumference of the circle $=2 \pi \mathrm{r}$ Thus, we have: $2 \pi r-r=37$ $\Rightarrow r(2 \pi-1)=37$ $\Rightarrow r\left(2 \times \frac{22}{7}-1\right)=37$ $\Rightarrow r\left(\frac{37}{7}\right)=37$ $\Rightarrow r=\left(37 \times \frac{7}{37}\right)$ $\Rightarrow r=7 \mathrm{~cm}$ Radius ...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{x+3}{x-3}+\frac{x+2}{x-2}=2$ Solution: $\frac{x+3}{x-3}+\frac{x+2}{x-2}=2$ or $\frac{x+3}{x-3}=2-\frac{x+2}{x-2}$ or $\frac{x+3}{x-3}=\frac{2 x-4-x-2}{x-2}$ or $\frac{x+3}{x-3}=\frac{x-6}{x-2}$ or $x^{2}-2 x+3 x-6=x^{2}-3 x-6 x+18$ [After c ross multiplication] or $\mathrm{x}^{2}-\mathrm{x}^{2}+\mathrm{x}+9 \mathrm{x}=18+6$ or $10 \mathrm{x}=24$ or $\mathrm{x}=\frac{24}{10}$ or $\mathrm{x}=\frac{12}{5}$ Thus, $\mathrm{x}=\frac...
Read More →If the perpendicular bisector of a chord
Question: If the perpendicular bisector of a chord $A B$ of a circle $P X A Q B Y$ intersects the circle at $P$ and $Q$, prove that arc $P X A=\operatorname{arc} P Y B$. Thinking Process Firstly, prove that $\triangle \mathrm{APM}$ is congruent to $\triangle \mathrm{BPM}$ by SAS rule, then further prove the required result by CPCT rule. Solution: Let $A B$ be a chord of a circle having centre at $O P Q$ be the perpendicular bisector of the chord $A B$, which intersects at $M$ and it always passe...
Read More →If arcs AXB and CYD of a circle are congruent,
Question: If $\operatorname{arcs} A X B$ and $C Y D$ of a circle are congruent, find the ratio of $A B$ and $C D$. Solution: Let $A X B$ and $C Y D$ are arcs of circle whose centre and radius are $O$ and $r$ units, respectively. Hence, the ratio of $A B$ and $C D$ is 1:1....
Read More →Solve this
Question: If $A=\left[\begin{array}{ll}4 5 \\ 2 1\end{array}\right]$, then show that $A-3 I=2\left(I+3 A^{-1}\right)$ Solution: We have, $A=\left[\begin{array}{ll}4 5 \\ 2 1\end{array}\right]$ Now, $\operatorname{adj}(A)=\left[\begin{array}{cc}1 -5 \\ -2 4\end{array}\right]$ and $|A|=-6$ $\therefore A^{-1}=-\frac{1}{6}\left[\begin{array}{cc}1 -5 \\ -2 4\end{array}\right]$ Now, $A-3 I=I+3 A^{-1}$ $\mathrm{LHS}=A-3 I=\left[\begin{array}{ll}4 5 \\ 2 1\end{array}\right]-3\left[\begin{array}{ll}1 0 \...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{15(2-x)-5(x+6)}{1-3 x}=10$ Solution: $\frac{15(2-\mathrm{x})-5(\mathrm{x}+6)}{1-3 \mathrm{x}}=10$ or $\frac{30-15 \mathrm{x}-5 \mathrm{x}-30}{1-3 \mathrm{x}}=10$ or $\frac{-20 \mathrm{x}}{1-3 \mathrm{x}}=10$ or $10-30 \mathrm{x}=-20 \mathrm{x}[$ After c ross multiplication $]$ or $-20 \mathrm{x}+30 \mathrm{x}=10$ or $10 \mathrm{x}=10$ or $\mathrm{x}=1$ Thus, $x=1$ is the solution of the given equation. Check : Substituting $\m...
Read More →The area of a circle is 49 π cm2. Its circumference is
Question: The area of a circle is 49 cm2. Its circumference is(a) 7 cm(b) 14 cm(c) 21 cm(d) 28 cm Solution: (b) 14 cmLet the radius bercm.We know: Area of a circle $=\pi r^{2}$ Thus, we have: $\pi r^{2}=49 \pi$ $\Rightarrow r^{2}=49$ $\Rightarrow r=\sqrt{49}$ $\Rightarrow r=7$ Now, Circumference of the circle $=2 \pi \mathrm{r}$ $=\left(2 \times \frac{22}{7} \times 7\right) \mathrm{cm}$ $=14 \pi \mathrm{cm}$...
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