There are 180 multiple choice questions in a test.
Question: There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly? Solution: Let the number of correctly answered questions be $x$. Therefore, the number of unattempted or wrongly answered questions will be $(180-\mathrm{x})$. According to the que...
Read More →If a line is drawn parallel to the base
Question: If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. Solution: Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E. Proof in $\triangle A B C$,[equal sides of an isosceles triangle] $\Rightarrow$ $\angle A C B=\angle A B C$ $\ldots(1)$ [angles opposite t...
Read More →At a party, colas, squash and fruit juice were offered to guests.
Question: At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two fifths drank fruit juice and just three did not drink any thing. How many guests were in all? Solution: Let the total number of guests be $\mathrm{x}$. Therefore, the number of guests, who drank colas, would be $\frac{1}{4} \mathrm{x}$. The number of guests, who drank squash, would be $\frac{1}{3} \mathrm{x}$. The number of guests, who drank fruit juice, would...
Read More →The area of a square is the same as the area of a square.
Question: The area of a square is the same as the area of a square. Their perimeters are in the ratio(a) 1 : 1(b) 2 : (c) : 2 (d) $\sqrt{\pi}: 2$ Solution: (d) $\sqrt{\pi}: 2$ Letabe the side of the square.We know: Area of a square $=a^{2}$ Letrbe the radius of the circle.We know: Area of a circle $=\pi r^{2}$ Because the area of the square is the same as the area of the circle, we have: $a^{2}=\pi r^{2}$ $\Rightarrow \frac{r^{2}}{a^{2}}=\frac{1}{\pi}$ $\Rightarrow \frac{r}{a}=\frac{1}{\sqrt{\pi...
Read More →The area of a square is the same as the area of a square.
Question: The area of a square is the same as the area of a square. Their perimeters are in the ratio(a) 1 : 1(b) 2 : (c) : 2 (d) $\sqrt{\pi}: 2$ Solution: (d) $\sqrt{\pi}: 2$ Letabe the side of the square.We know: Area of a square $=a^{2}$ Letrbe the radius of the circle.We know: Area of a circle $=\pi r^{2}$ Because the area of the square is the same as the area of the circle, we have: $a^{2}=\pi r^{2}$ $\Rightarrow \frac{r^{2}}{a^{2}}=\frac{1}{\pi}$ $\Rightarrow \frac{r}{a}=\frac{1}{\sqrt{\pi...
Read More →Solve this
Question: If $A=\left[\begin{array}{cc}3 1 \\ -1 2\end{array}\right]$, show that $A^{2}-5 A+7 I=O$. Hence, find $A^{-1}$. Solution: $A=\left[\begin{array}{cc}3 1 \\ -1 2\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}3 1 \\ -1 2\end{array}\right]\left[\begin{array}{cc}3 1 \\ -1 2\end{array}\right]=\left[\begin{array}{cc}9-1 3+2 \\ -3-2 -1+4\end{array}\right]=\left[\begin{array}{cc}8 5 \\ -5 3\end{array}\right]$ and $A^{2}-5 A+7 I=\left[\begin{array}{cc}8 5 \\ -5 3\end{array}\right]...
Read More →I have Rs 1000 in ten and five rupee notes.
Question: I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination? Solution: Let the number of five $-$ rupee notes be $\mathrm{x} .$ Therefore, the number of ten $-$ rupee notes will be $(\mathrm{x}+10)$. Now, Value of five - rupee notes $=$ Rs. $5 \mathrm{x}$ Value of ten $-$ rupee notes $=$ Rs. $10(\mathrm{x}+10)$ According to the question, $5 \mathrm{x}+10(\mathrm{x...
Read More →I am currently 5 times as old as my son.
Question: I am currently 5 times as old as my son. In 6 years time I will be three times as old as he will be then. What are our ages now? Solution: Let the age of my son be $\mathrm{x}$ years. Therefore, my age will be $5 \mathrm{x}$ years. After 6 years : Age of my son $=(\mathrm{x}+6)$ years My age $=(5 \mathrm{x}+6)$ years According to the question, $5 \mathrm{x}+6=3(\mathrm{x}+6)$ or $5 \mathrm{x}-3 \mathrm{x}=18-6$ or $2 \mathrm{x}=12$ or $\mathrm{x}=6$ $\therefore$ Age of my son $=6$ year...
Read More →If BM and CN are the perpendiculars
Question: If BM and CN are the perpendiculars drawn on the sides AC and AB of the ΔABC, prove that the points B, C, M and N are concyclic. Solution: Given In ΔABC, BM AC and CN AB. To prove Points B, C, M and N are con-cyclic. Construction Draw a circle passing through the points B, C, M and N. Proof Suppose, we consider SC as a diameter of the circle. Also, we know that SC subtends a 90 to the circle. So, the points M and N should be on a circle. Hence, BCMN form a con-cyclic quadrilateral. Hen...
Read More →Five years ago a man was seven times as old as his son.
Question: Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages. Solution: Five years ago: Let the age of the son be $x$ years. Therefore, the age of the father will be $7 \mathrm{x}$ years. $\therefore$ Present age of the son $=(\mathrm{x}+5)$ years Present age of the father $=(7 \mathrm{x}+5)$ years After five years: Age of the son $=(\mathrm{x}+5+5)=(\mathrm{x}+10)$ years Age of the father $=(7 \math...
Read More →Show that
Question: Show that $A=\left[\begin{array}{cc}-8 5 \\ 2 4\end{array}\right]$ satisfies the equation $A^{2}+4 A-42 I=O$. Hence, find $A^{-1}$. Solution: $A=\left[\begin{array}{cc}-8 5 \\ 2 4\end{array}\right]$ $\therefore A^{2}=\left[\begin{array}{cc}74 -20 \\ -8 26\end{array}\right]$ and $A^{2}+4 A-42 I=\left[\begin{array}{cc}74 -20 \\ -8 26\end{array}\right]+\left[\begin{array}{cc}-32 20 \\ 8 16\end{array}\right]-\left[\begin{array}{cc}42 0 \\ 0 42\end{array}\right]$ $\Rightarrow A^{2}+4 A-42 I...
Read More →On decreasing the radius of a circle by 30%, its area is decreased by
Question: On decreasing the radius of a circle by 30%, its area is decreased by(a) 30%(b) 60%(c) 45%(d) none of these Solution: (d) None of theseLetrbe the original radius.Thus, we have: Original area $=\pi r^{2}$ Also, New radius $=70 \%$ of $r$ $=\left(\frac{70}{100} \times r\right)$ $=\frac{7 r}{10}$ New area $=\pi \times\left(\frac{7 r}{10}\right)^{2}$ $=\frac{49 \pi r^{2}}{100}$ Decrease in the area $=\left(\pi r^{2}-\frac{49 \pi r^{2}}{100}\right)$ $=\frac{59 \pi r^{2}}{100}$ Thus, we have...
Read More →Two chords AB and AC of a circle subtends
Question: Two chords AB and AC of a circle subtends angles equal to 90 and 150, respectively at the centre. Find BAC, if AB and AC lie on the opposite sides of the centre. Solution: In $\triangle B O A$, $O B=O A$ [both are the radius of circle] $\therefore$ $\angle O A B=\angle O B A$ ...(i) [angles opposite to equal sides are equal] $\begin{array}{lll}\ln \triangle O A B, \angle O B A+\angle O A B+\angle A O B=180^{\circ} \text { [by angle sum property of a triangle] }\end{array}$ $\Rightarrow...
Read More →The ages of sonu and Monu are in the ratio 7 : 5. Ten years hence,
Question: The ages of sonu and Monu are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages. Solution: It is given that the ratio of the ages of Sonu and Monu is $7: 5$. Let the present ages of Sonu and Monu be $7 x$ and $5 x$ years. After ten years: Age of Sonu $=7 \mathrm{x}+10$ years Age of Monu $=5 \mathrm{x}+10$ years According to the question, $\frac{7 x+10}{5 x+10}=\frac{9}{7}$ or $49 x+70=45 x+90$ or $49 x-45 x=90-70$ or $4 x=20$ or $x=5$ $...
Read More →Sunita is twice as old as Ashima.
Question: Sunita is twice as old as Ashima. If six years is subtracted from Ashima's age and four years added to Sunita's age, then Sunita will be four times Ashima's age. How old were they two years ago? Solution: Let the age of Ashima be x years. Therefore, the age of Sunita will be $2 \mathrm{x}$ years. According to the question, $4(x-6)=2 x+4$ or $4 x-24=2 x+4$ or $4 x-2 x=4+24$ or $2 x=28$ or $x=14$ $\therefore$ Age of Ashima $=14$ years. Age of Sunita $=2 \times 14=28$ years....
Read More →On a common hypotenuse AB,
Question: On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. Prove that BAC = BDC. Solution: Given, $\triangle A C B$ and $\triangle A D B$ are two right angled triangles with common hypotenuse $A B$. To prove $\angle B A C=\angle B D C$ Construction Join $C D$. Proof Let $O$ be the mid-point of $A B$. Then, $O A=O B=O C=O D$. Since, mid-point of the hypotenuse of a right triangle is equidistant from its vertices. Now, draw a circle to pass through...
Read More →Seeta Devi has Rs 9 in fifty-paise and twenty five-paise coins.
Question: Seeta Devi has Rs 9 in fifty-paise and twenty five-paise coins. She has twice as many twenty-five paise coins as she has fifty-paise coins. How many coins of each kind does she have? Solution: Let the number of 50 paise coins be $\mathrm{x}$. Therefore, the number of 25 paise coins will be $2 \mathrm{x}$. Value of 50 paise coins $=$ Rs. $0.5 \mathrm{x}$ Value of 25 paise coins $=$ Rs. $0.25 \times 2 \mathrm{x}$ According to the question, $0.5 \mathrm{x}+0.25 \times 2 \mathrm{x}=9$ or $...
Read More →A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20.
Question: A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50, find the number of notes of each type. Solution: Let the number of Rs. 10 notes be $\mathrm{x}$. Therefore, the number of Rs. 20 notes will be $(50-\mathrm{x})$. Value of Rs. 10 notes $=10 \mathrm{x}$ Value of Rs. 20 notes $=20(50-\mathrm{x})$ According to the question, $10 \mathrm{x}+20(50-\mathrm{x})=800$ or $10 \mathrm{x}+1000-20 \mathrm{x}=800$ or $10 \mathrm{x}=1000-800$ or $\ma...
Read More →O is the circumcentre of the ΔABC and D is the mid-point of the base BC.
Question: O is the circumcentre of the ΔABC and D is the mid-point of the base BC. Prove that BOD = A. Thinking Process Firstly, prove that ΔBOD and ΔCOD are congruent by SSS rule. Further, use the theorem that in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle and prove the required result. Solution: Given In a ΔABC a circle is circumscribed having centre O. Also, $D$ is the mid-point of $B C$. To prove $\angle B O D=\...
Read More →solving the problem
Question: If $A=\left[\begin{array}{ll}2 3 \\ 1 2\end{array}\right]$, verify that $A^{2}-4 A+I=O$, where $I=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]$ and $O=\left[\begin{array}{ll}0 0 \\ 0 0\end{array}\right]$. Hence, find $A^{-1}$. Solution: $A=\left[\begin{array}{ll}2 3 \\ 1 2\end{array}\right]$ $\therefore A^{2}=\left[\begin{array}{cc}7 12 \\ 4 7\end{array}\right]$ and $A^{2}-4 A+I=\left[\begin{array}{cc}7 12 \\ 4 7\end{array}\right]-\left[\begin{array}{cc}8 12 \\ 4 8\end{array}\ri...
Read More →The numerator of a fraction is 6 less than the denominator.
Question: The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to $\frac{2}{3}$. What is the original fraction equal to? Solution: Let the denominator of the fraction be $\mathrm{x}$. Therefore, the numerator will be $(x-6)$. $\therefore$ Fraction $=\frac{\mathrm{x}-6}{\mathrm{x}}$ According to the question, $\frac{\mathrm{x}-6+3}{\mathrm{x}}=\frac{2}{3}$ or $\frac{\mathrm{x}-3}{\mathrm{x}}=\frac{2}{3}$ or $3 \mathrm{x}-9=2 \mathrm{x}[...
Read More →Divide 184 into two parts such that one-third of
Question: Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8. Solution: Let the first part of 184 be $\mathrm{x}$. Therefore, the other part will be $(184-\mathrm{x})$. According to the question, $\frac{1}{3} \mathrm{x}-\frac{1}{7}(184-\mathrm{x})=8$ or $\frac{7 \mathrm{x}-552+3 \mathrm{x}}{21}=8$ or $10 \mathrm{x}-552=168[$ After c ross multiplication $]$ or $10 \mathrm{x}=168+552$ or $\mathrm{x}=\frac{720}{10}=72$ Thus, the parts of 184 are 72...
Read More →ABCD is such a quadrilateral that A is
Question: ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that CBD +CDB = BAD. Solution: Given in a circle, $A B C D$ is a quadrilateral having centre $A$. To prove$\angle C B D+\angle C D B=\frac{1}{2} \angle B A D$ Construction Join $A C$ and $B D$. Proof Since, arc $D C$ subtends $\angle D A C$ at the centre and $\angle C B D$ at a point $B$ in the remaining part of the circle. $\because$ $\angle D A C=2 \angle C B D \quad \ldots$ (i) In a cir...
Read More →Solve this
Question: Let $F(\alpha)=\left[\begin{array}{ccc}\cos \alpha -\sin \alpha 0 \\ \sin \alpha \cos \alpha 0 \\ 0 0 1\end{array}\right]$ and $G(\beta)=\left[\begin{array}{ccc}\cos \beta 0 \sin \beta \\ 0 1 0 \\ -\sin \beta 0 \cos \beta\end{array}\right]$ (i) $[F(\alpha)]^{-1}=F(-\alpha)$ (ii) $[G(\beta)]^{-1}=G(-\beta)$ (iii) $[F(\alpha) G(\beta)]^{-1}=G(-\beta) F(-\alpha)$. Solution: $(\mathrm{i}) F(\alpha)=\left[\begin{array}{ccc}\cos \alpha -\sin \alpha 0 \\ \sin \alpha \cos \alpha 0 \\ 0 0 1\end...
Read More →A number consists of two digits whose sum is 9.
Question: A number consists of two digits whose sum is 9. If 27 is subtracted from the number, its digits are reversed. Find the number. Solution: Let the units digit be $\mathrm{x}$. $\because$ Sum of two digits $=9$ $\therefore$ Tens digit $=(9-\mathrm{x})$ $\therefore$ Original number $=10 \times(9-\mathrm{x})+\mathrm{x}$ Reversed number $=10 \mathrm{x}+(9-\mathrm{x})$ According to the question, $10 \times(9-\mathrm{x})+\mathrm{x}-27=10 \mathrm{x}+(9-\mathrm{x})$ or $90-10 \mathrm{x}+\mathrm{...
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