Class 10: Question Bank & Solutions (ICSE)

$\displaystyle \text{MULTIPLE CHOICE QUESTIONS TYPE (1 Mark Each)}$

$\displaystyle \textbf{Question 1: }\text{A retailer buys an article at its listed price from a}$
$\displaystyle \text{wholesaler and sells it to a consumer in the same state after marking up the}$
$\displaystyle \text{price by }20\%.\text{ The list price of the article is Rs. }2500\text{ and the rate of GST is}$
$\displaystyle 12\%.\text{ What is the tax liability of the retailer to the central government?}$
$\displaystyle \text{(a) Rs. }0\qquad\text{(b) Rs. }15\qquad\text{(c) Rs. }30\qquad\text{(d) Rs. }60$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the sale is within the same state, CGST rate}=\frac{12\%}{2}=6\%.$
$\displaystyle \text{Purchase price of the retailer}=\text{Rs. }2500.$
$\displaystyle \text{Input CGST}=6\%\text{ of Rs. }2500=\text{Rs. }150.$
$\displaystyle \text{Selling price before GST}=2500+\frac{20}{100}\times2500=\text{Rs. }3000.$
$\displaystyle \text{Output CGST}=6\%\text{ of Rs. }3000=\text{Rs. }180.$
$\displaystyle \text{Tax liability to the central government}=\text{Output CGST}-\text{Input CGST}.$
$\displaystyle =180-150=\text{Rs. }30.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Dev bought an electrical fan which has a marked price of Rs. }800.$
$\displaystyle \text{ If the GST on the} \ \text{goods is }7\%,\text{ then the SGST is:}$
$\displaystyle \text{(a) Rs. }24\qquad\text{(b) Rs. }28\qquad\text{(c) Rs. }56\qquad\text{(d) Rs. }80$
$\displaystyle \text{Answer:}$
$\displaystyle \text{GST}=7\%.$
$\displaystyle \therefore \text{SGST}=\frac{7\%}{2}=3.5\%.$
$\displaystyle \text{SGST on Rs. }800=\frac{3.5}{100}\times800=\text{Rs. }28.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Rs. }P\text{ is deposited for }n\text{ number of months in a recurring}$
$\displaystyle \text{deposit account which pays interest at the rate of }r\%\text{ per annum. The nature and}$
$\displaystyle \text{time of interest calculated is:}$
$\displaystyle \text{(a) compound interest for }n\text{ number of months.}$
$\displaystyle \text{(b) simple interest for }n\text{ number of months.}$
$\displaystyle \text{(c) compound interest for one month.}$
$\displaystyle \text{(d) simple interest for one month.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a recurring deposit, the sum of monthly products is}$
$\displaystyle P[n+(n-1)+\cdots+2+1]=P\frac{n(n+1)}{2}.$
$\displaystyle \text{Interest}=\frac{P\,n(n+1)}{2}\times\frac{r}{100}\times\frac1{12}.$
$\displaystyle \text{Thus, simple interest for one month is calculated on the sum of the monthly products.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Anwesha intended to open a Recurring Deposit account of Rs. }1000$
$\displaystyle \text{per month for one year in a bank paying }5\%\text{ per annum simple interest. The bank}$
$\displaystyle \text{reduced the rate to }4\%\text{ per annum. How much must Anwesha deposit monthly for}$
$\displaystyle \text{one year so that her interest remains the same?}$
$\displaystyle \text{(a) Rs. }12325\qquad\text{(b) Rs. }1250\qquad\text{(c) Rs. }1200\qquad\text{(d) Rs. }1000$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the same period, interest on an RD is proportional to the monthly deposit and rate.}$
$\displaystyle \therefore 1000\times5=P\times4.$
$\displaystyle P=\frac{1000\times5}{4}=\text{Rs. }1250.$
$\displaystyle \therefore \text{Anwesha must deposit Rs. }1250\text{ per month.}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Mr. Das invests in Rs. }100,\ 12\%\text{ shares of Company A}$
$\displaystyle \text{available at Rs. }60\text{ each. Mr. Singh invests in Rs. }50,\ 16\%\text{ shares of Company B}$
$\displaystyle \text{available at Rs. }40\text{ each. Which of the following statements is true?}$
$\displaystyle \text{(a) The rate of return for Mr. Das is }12\%$
$\displaystyle \text{(b) The rate of return for Mr. Singh is }10\%$
$\displaystyle \text{(c) Both Mr. Das and Mr. Singh have the same rate of return of }10\%$
$\displaystyle \text{(d) Both Mr. Das and Mr. Singh have the same rate of return of }20\%$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For Mr. Das, dividend per share}=12\%\text{ of Rs. }100=\text{Rs. }12.$
$\displaystyle \text{Rate of return}=\frac{12}{60}\times100=20\%.$
$\displaystyle \text{For Mr. Singh, dividend per share}=16\%\text{ of Rs. }50=\text{Rs. }8.$
$\displaystyle \text{Rate of return}=\frac{8}{40}\times100=20\%.$
$\displaystyle \therefore \text{Both have the same rate of return of }20\%.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Amit invested a certain sum in Rs. }100\text{ shares paying a }7.5\%$
$\displaystyle \text{dividend. The rate of return on his investment is }10\%.\text{ The money invested by Amit}$
$\displaystyle \text{to purchase }10\text{ shares is:}$
$\displaystyle \text{(a) Rs. }250\qquad\text{(b) Rs. }750\qquad\text{(c) Rs. }900\qquad\text{(d) Rs. }1100$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Dividend per share}=7.5\%\text{ of Rs. }100=\text{Rs. }7.50.$
$\displaystyle \text{Dividend on }10\text{ shares}=10\times7.50=\text{Rs. }75.$
$\displaystyle \text{Let the total investment be Rs. }P.$
$\displaystyle 10=\frac{75}{P}\times100.$
$\displaystyle P=\frac{75\times100}{10}=\text{Rs. }750.$
$\displaystyle \therefore \text{Amit invested Rs. }750.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }-3\leq -4x+5\text{ and }x\in W,\text{ then the solution set is:}$
$\displaystyle \text{(a) }\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}\qquad \text{(b) }\{1,2\}$
$\displaystyle \text{(c) }\{0,1,2\}\qquad \text{(d) }\{2,3,4,5\}$
$\displaystyle \text{Answer:}$
$\displaystyle -3\leq -4x+5.$
$\displaystyle -8\leq -4x.$
$\displaystyle \text{Dividing by }-4\text{ and reversing the inequality sign, we get}$
$\displaystyle x\leq 2.$
$\displaystyle \text{Since }x\in W,\quad x=0,1,2.$
$\displaystyle \therefore \text{Solution set}=\{0,1,2\}.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If }-4x>8y,\text{ then:}$
$\displaystyle \text{(a) }x>2y\qquad \text{(b) }x>-2y\qquad \text{(c) }x<-2y\qquad \text{(d) }x<2y$
$\displaystyle \text{Answer:}$
$\displaystyle -4x>8y.$
$\displaystyle \text{Dividing both sides by }-4\text{ and reversing the inequality sign, we get}$
$\displaystyle x<-2y.$
$\displaystyle \therefore \text{Option (c) is correct.}$

$\displaystyle \textbf{Question 9: }\text{The value(s) of }k\text{ for which the quadratic equation }$
$\displaystyle 2x^2-kx+k=0 \ \text{has equal roots is (are):}$
$\displaystyle \text{(a) }0\text{ only}\qquad \text{(b) }4,0\qquad \text{(c) }8\text{ only}\qquad \text{(d) }0,8$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For equal roots, the discriminant must be zero.}$
$\displaystyle b^2-4ac=0.$
$\displaystyle (-k)^2-4(2)(k)=0.$
$\displaystyle k^2-8k=0.$
$\displaystyle k(k-8)=0.$
$\displaystyle \therefore k=0\text{ or }k=8.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }x=-2\text{ is one of the solutions of the quadratic equation }$
$\displaystyle x^2+3a-x=0, \ \text{then the value of }a\text{ is:}$
$\displaystyle \text{(a) }-8\qquad \text{(b) }-2\qquad \text{(c) }-\frac{1}{3}\qquad \text{(d) }\frac{1}{3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }x=-2\text{ is a solution, substitute }x=-2.$
$\displaystyle (-2)^2+3a-(-2)=0.$
$\displaystyle 4+3a+2=0.$
$\displaystyle 3a=-6.$
$\displaystyle \therefore a=-2.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{In solving a quadratic equation, one of the values of the variable }x\text{ is }233.356.$
$\displaystyle \text{The solution rounded to two significant figures is:}$
$\displaystyle \text{(a) }233.36\qquad \text{(b) }233.35\qquad \text{(c) }233.3\qquad \text{(d) }230$
$\displaystyle \text{Answer:}$
$\displaystyle 233.356\text{ rounded to two significant figures is }230.$
$\displaystyle \text{The first two significant digits are }2\text{ and }3.$
$\displaystyle \text{Since the next digit is }3<5,\text{ the second significant digit remains unchanged.}$
$\displaystyle \therefore 233.356\approx230\text{ (to two significant figures).}$
$\displaystyle \therefore \text{Option (d) is correct.}$

$\displaystyle \textbf{Question 12: }\text{In the adjoining diagram, }AB=x\text{ cm, }BC=y\text{ cm and }$
$\displaystyle x-y=7\text{ cm.} \ \text{Area of }\triangle ABC=30\text{ cm}^2.\text{ The length of }AC\text{ is:}$
$\displaystyle \text{(a) }10\text{ cm}\qquad \text{(b) }12\text{ cm}\qquad \text{(c) }13\text{ cm}\qquad \text{(d) }15\text{ cm}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times AB\times BC.$
$\displaystyle 30=\frac{1}{2}xy.$
$\displaystyle xy=60.$
$\displaystyle \text{Also, }x-y=7\quad\Rightarrow\quad x=y+7.$
$\displaystyle y(y+7)=60.$
$\displaystyle y^2+7y-60=0.$
$\displaystyle (y+12)(y-5)=0.$
$\displaystyle \text{Since }y>0,\quad y=5,\quad x=12.$
$\displaystyle AC=\sqrt{AB^2+BC^2}=\sqrt{12^2+5^2}.$
$\displaystyle AC=\sqrt{169}=13\text{ cm}.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{If }p,\ q,\text{ and }r\text{ are in continued proportion, then:}$
$\displaystyle \text{(a) }p:q=p:r\qquad \text{(b) }q:r=p^2:q^2$
$\displaystyle \text{(c) }p:q^2=r:p^2\qquad \text{(d) }p:r=p^2:q^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }p,\ q,\ r\text{ are in continued proportion,}$
$\displaystyle p:q=q:r.$
$\displaystyle q^2=pr.$
$\displaystyle \text{For option (c),}\quad p:q^2=r:p^2.$
$\displaystyle \frac{p}{q^2}=\frac{r}{p^2}.$
$\displaystyle p^3=q^2r.$
$\displaystyle \text{But }q^2=pr\quad\Rightarrow\quad q^2r=pr^2,$
$\displaystyle \text{which does not imply }p^3=pr^2\text{ in general.}$
$\displaystyle \therefore \text{Option (c) is not generally true.}$

$\displaystyle \textbf{Question 14: }\text{The ratio of diameter to height of a Borosil cylindrical glass is}$
$\displaystyle 3:5.\text{ If the actual diameter of the glass is }6\text{ cm, then the curved surface area}$
$\displaystyle \text{of the glass, in cm}^2,\text{ is:}$
$\displaystyle \text{(a) }120\pi\qquad\text{(b) }60\pi\qquad\text{(c) }30\pi\qquad\text{(d) }18\pi$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Diameter : Height}=3:5.$
$\displaystyle \text{Given diameter}=6\text{ cm}.$
$\displaystyle \therefore \text{Height}=\frac{5}{3}\times6=10\text{ cm}.$
$\displaystyle \text{Radius}=\frac{6}{2}=3\text{ cm}.$
$\displaystyle \text{Curved surface area of a cylinder}=2\pi rh.$
$\displaystyle =2\pi(3)(10)=60\pi\text{ cm}^2.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{If the polynomial }2x^3+3x^2-2x-3\text{ is completely divisible by }$
$\displaystyle (2x+a), \ \text{and the quotient is equal to }(x^2-1),\text{ then one of the values of }a\text{ is:}$
$\displaystyle \text{(a) }-3\qquad\text{(b) }-1\qquad\text{(c) }1\qquad\text{(d) }3$
$\displaystyle \text{Answer:}$
$\displaystyle 2x^3+3x^2-2x-3=(2x+a)(x^2-1).$
$\displaystyle (2x+a)(x^2-1)=2x^3+ax^2-2x-a.$
$\displaystyle \text{Comparing with }2x^3+3x^2-2x-3,$
$\displaystyle a=3.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{A polynomial in }x\text{ is }x^3+5x^2-kx-24.\text{ Which of the following is a factor}$
$\displaystyle \text{of the given polynomial so that the value of }k\text{ is }2\text{?}$
$\displaystyle \text{(a) }(x+2)\qquad\text{(b) }(x-3)\qquad\text{(c) }(x+4)\qquad\text{(d) }(x-4)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }k=2,\text{ the polynomial is }f(x)=x^3+5x^2-2x-24.$
$\displaystyle \text{For }(x+4)\text{ to be a factor, }f(-4)=0.$
$\displaystyle f(-4)=(-4)^3+5(-4)^2-2(-4)-24.$
$\displaystyle =-64+80+8-24=0.$
$\displaystyle \therefore (x+4)\text{ is a factor of the polynomial.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{If }A=[\,a\quad b\,]\text{ and }B= \begin{bmatrix}c\\d\end{bmatrix},\text{ then:}$
$\displaystyle \text{(a) only matrix }AB\text{ is possible}\qquad \text{(b) only matrix }BA\text{ is possible}$
$\displaystyle \text{(c) both matrices }AB\text{ and }BA\text{ are possible}\qquad \text{(d) both are possible, }AB=BA$
$\displaystyle \text{Answer:}$
$\displaystyle A\text{ is of order }1\times2\text{ and }B\text{ is of order }2\times1.$
$\displaystyle AB\text{ is possible since }(1\times2)(2\times1)\text{ gives a }1\times1\text{ matrix.}$
$\displaystyle BA\text{ is also possible since }(2\times1)(1\times2)\text{ gives a }2\times2\text{ matrix.}$
$\displaystyle \therefore \text{Both }AB\text{ and }BA\text{ are possible.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Matrix }A= \begin{bmatrix}6&9\\-4&k\end{bmatrix}\text{ such that }A^2= \begin{bmatrix}0&0\\0&0\end{bmatrix}.\text{ Then }k\text{ is:}$
$\displaystyle \text{(a) }6\qquad\text{(b) }-6\qquad \text{(c) }36\qquad\text{(d) }\pm6$
$\displaystyle \text{Answer:}$
$\displaystyle A^2= \begin{bmatrix}6&9\\-4&k\end{bmatrix} \begin{bmatrix}6&9\\-4&k\end{bmatrix}.$
$\displaystyle = \begin{bmatrix} 36-36&54+9k\\ -24-4k&-36+k^2 \end{bmatrix}.$
$\displaystyle \text{Since }A^2= \begin{bmatrix}0&0\\0&0\end{bmatrix},\quad54+9k=0.$
$\displaystyle 9k=-54.$
$\displaystyle k=-6.$
$\displaystyle \text{Also, }-36+k^2=-36+(-6)^2=0,\text{ as required.}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{If the sum of }n\text{ terms of an arithmetic progression is }$
$\displaystyle S_n=n^2-n,\text{ then the} \ \text{third term of the series is:}$
$\displaystyle \text{(a) }2\qquad\text{(b) }4\qquad\text{(c) }6\qquad\text{(d) }9$
$\displaystyle \text{Answer:}$
$\displaystyle S_n=n^2-n.$
$\displaystyle S_3=3^2-3=6.$
$\displaystyle S_2=2^2-2=2.$
$\displaystyle \text{Third term}=S_3-S_2.$
$\displaystyle =6-2=4.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{Which of the following is not a geometric progression?}$
$\displaystyle \text{(a) }\frac13,\ 1,\ 3,\ 9\qquad\text{(b) }\frac15,\ \frac15,\ \frac15,\ \frac15$
$\displaystyle \text{(c) }-2,\ 4,\ -8,\ 16\qquad\text{(d) }2,\ 0,\ 4,\ 0,\ 8,\ 0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For option (a), the common ratio is }3.$
$\displaystyle \text{For option (b), the common ratio is }1.$
$\displaystyle \text{For option (c), the common ratio is }-2.$
$\displaystyle \text{In option (d), there is no constant ratio between consecutive terms.}$
$\displaystyle \therefore \text{Option (d) is not a geometric progression.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{In the adjoining diagram, }G\text{ is the centroid of }\triangle ABC.\ A(3,-3),\ B(2,-6),$
$\displaystyle C(x,y)\text{ and }G(5,-5).\text{ The coordinates of point }D\text{ are:}$
$\displaystyle \text{(a) }(2,-6)\qquad\text{(b) }(3,-6)\qquad\text{(c) }(6,-6)\qquad\text{(d) }(10,-6)$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }G\text{ is the centroid, it divides the median }AD\text{ in the ratio }2:1.$
$\displaystyle AG:GD=2:1.$
$\displaystyle \text{Let }D=(x,y).$
$\displaystyle 5=\frac{1(3)+2x}{3}.$
$\displaystyle 15=3+2x.$
$\displaystyle \therefore x=6.$
$\displaystyle -5=\frac{1(-3)+2y}{3}.$
$\displaystyle -15=-3+2y.$
$\displaystyle \therefore y=-6.$
$\displaystyle \therefore D=(6,-6).$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{In the given diagram, }O\text{ is the origin, and }P\text{ is the midpoint}$
$\displaystyle \text{of }AB.\text{ The equation of }OP\text{ is:}$
$latex \displaystyle \text{(a) }y=x\qquad\text{(b) }2y=x\qquad\text{(c) }y=2x\qquad\text{(d) }y=-x &fg=0000ff&s=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{From the graph, }A=(4,0)\text{ and }B=(0,2).$
$\displaystyle \text{Since }P\text{ is the midpoint of }AB,$
$\displaystyle P=\left(\frac{4+0}{2},\frac{0+2}{2}\right)=(2,1).$
$\displaystyle \text{The line }OP\text{ passes through }O(0,0)\text{ and }P(2,1).$
$\displaystyle \text{Slope of }OP=\frac{1-0}{2-0}=\frac{1}{2}.$
$\displaystyle \therefore y=\frac{1}{2}x.$
$\displaystyle \therefore 2y=x.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{In the given figure, line }l_1\text{ is parallel to line }l_2.\text{ If line }l_3\text{ is perpendicular}$
$\displaystyle \text{to line }l_1,\text{ then the slopes of lines }l_2\text{ and }l_3\text{ respectively are:}$
$latex \displaystyle \text{(a) }1,1\qquad\text{(b) }-1,-1\qquad\text{(c) }1,-1\qquad\text{(d) }-1,1 &fg=0000ff&s=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Line }l_1\text{ makes an angle of }45^\circ\text{ with the positive }x\text{-axis.}$
$\displaystyle \therefore \text{Slope of }l_1=\tan45^\circ=1.$
$\displaystyle \text{Since }l_1\parallel l_2,\text{ their slopes are equal.}$
$\displaystyle \therefore \text{Slope of }l_2=1.$
$\displaystyle \text{Since }l_3\perp l_1,\text{ the product of their slopes is }-1.$
$\displaystyle \therefore \text{Slope of }l_3=-1.$
$\displaystyle \therefore \text{The slopes of }l_2\text{ and }l_3\text{ are }1\text{ and }-1\text{ respectively.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{Which of the following lines cut the positive }x\text{-axis and positive }$
$\displaystyle y\text{-axis at equal} \ \text{distances from the origin?}$
$\displaystyle \text{(a) }3x+3y=6\qquad\text{(b) }5x+10y=10$
$\displaystyle \text{(c) }-x+y=1\qquad\text{(d) }10x+5y=5$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a line to cut equal positive intercepts on both axes, its }x\text{- and }y\text{-intercepts must be equal.}$
$\displaystyle \text{For option (a),}\quad3x+3y=6.$
$\displaystyle x+y=2.$
$\displaystyle \text{Putting }y=0,\quad x=2.$
$\displaystyle \text{Putting }x=0,\quad y=2.$
$\displaystyle \therefore \text{Both intercepts are equal and positive.}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{In the given diagram, railway stations }A,\ B,\ C,\ P\text{ and }Q$
$\displaystyle \text{are connected by straight tracks. Track }PQ\text{ is parallel to }BC.\text{ The time taken by a}$
$\displaystyle \text{train travelling at }90\text{ km/hr to reach }B\text{ from }A\text{ by the shortest route is:}$
$\displaystyle \text{(a) }8\text{ minutes}\qquad\text{(b) }12\text{ minutes}\qquad \text{(c) }16.8\text{ minutes}\qquad\text{(d) }20\text{ minutes}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }PQ\parallel BC,\quad\triangle APQ\sim\triangle ABC.$
$\displaystyle \frac{AP}{AB}=\frac{PQ}{BC}=\frac{20}{50}=\frac25.$
$\displaystyle \therefore AP=\frac25AB.$
$\displaystyle PB=AB-AP.$
$\displaystyle 18=AB-\frac25AB=\frac35AB.$
$\displaystyle \therefore AB=18\times\frac53=30\text{ km}.$
$\displaystyle \text{Time taken}=\frac{\text{Distance}}{\text{Speed}}=\frac{30}{90}\text{ hour}.$
$\displaystyle =\frac13\text{ hour}=20\text{ minutes}.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{In the given diagram, }\triangle ABC\text{ and }\triangle DEF\text{ are such that}$
$\displaystyle \angle C=\angle F\text{ and }\frac{AB}{DE}=\frac{BC}{EF},\text{ then:}$ $\displaystyle \text{(a) }\triangle ABC\sim\triangle DEF\qquad \text{(b) }\triangle BCA\sim\triangle DEF$
$\displaystyle \text{(c) }\triangle CAB\sim\triangle DEF\qquad \text{(d) The similarity of the given triangles cannot be determined.}$
$\displaystyle \text{Answer:}$
$\displaystyle \angle C=\angle F\quad\text{and}\quad\frac{AB}{DE}=\frac{BC}{EF}.$
$\displaystyle \text{The given equal angle is not included between the two pairs of proportional sides.}$
$\displaystyle \text{Hence, the SAS similarity criterion cannot be applied.}$
$\displaystyle \text{Also, no second pair of equal angles is given for the AA similarity criterion.}$
$\displaystyle \therefore \text{The similarity of the triangles cannot be determined from the given information.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{In the adjoining diagram, }ST\text{ is not parallel to }PQ.\text{ The necessary}$
$\displaystyle \text{and sufficient condition for }\triangle PQR\sim\triangle TSR\text{ is:}$ $\displaystyle \text{(a) }\angle PQR=\angle STR\qquad \text{(b) }\angle QPR=\angle TSR$
$\displaystyle \text{(c) }\angle PQR=\angle TSR\qquad \text{(d) }\angle PRQ=\angle RST$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }\triangle PQR\sim\triangle TSR,\text{ the correspondence is }P\leftrightarrow T,\ Q\leftrightarrow S,\ R\leftrightarrow R.$
$\displaystyle \text{Since }R,Q,T\text{ are collinear and }R,P,S\text{ are collinear,}$
$\displaystyle \angle PRQ=\angle TRS.$
$\displaystyle \text{If }\angle PQR=\angle TSR,\text{ then two corresponding angles are equal.}$
$\displaystyle \therefore \triangle PQR\sim\triangle TSR\text{ by the AA similarity criterion.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{The scale factor of a picture and the actual height of Sonia is }$
$\displaystyle 20\text{ cm}:1.6\text{ m. If her} \ \text{height in the picture is }18\text{ cm, then her actual height is:}$
$\displaystyle \text{(a) }14.4\text{ m}\qquad \text{(b) }2.25\text{ m}\qquad \text{(c) }1.78\text{ m}\qquad \text{(d) }1.44\text{ m}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Picture height}:\text{Actual height}=20\text{ cm}:1.6\text{ m}.$
$\displaystyle \therefore \frac{20}{18}=\frac{1.6}{x}$
$\displaystyle 20x=18\times1.6$
$\displaystyle x=\frac{18\times1.6}{20}=1.44\text{ m}.$
$\displaystyle \therefore \text{The actual height of Sonia is }1.44\text{ m}.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{In the adjoining figure, }O\text{ is the centre of the circle, and a semicircle is}$
$\displaystyle \text{drawn on } OA\text{ as the diameter. If }\angle APQ=20^\circ,\text{ then the degree measure of }\angle OAQ\text{ is:}$
$\displaystyle \text{(a) }25^\circ\qquad\text{(b) }40^\circ\qquad\text{(c) }50^\circ\qquad\text{(d) }65^\circ$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }P,\ O\text{ and }Q\text{ are collinear, }\angle APO=\angle APQ=20^\circ.$
$\displaystyle OA=OP,\text{ being radii of the circle.}$
$\displaystyle \therefore \angle OAP=\angle APO=20^\circ.$
$\displaystyle \angle AOP=180^\circ-20^\circ-20^\circ=140^\circ.$
$\displaystyle \text{Since }OP\text{ and }OQ\text{ are opposite rays,}$
$\displaystyle \angle AOQ=180^\circ-140^\circ=40^\circ.$
$\displaystyle \text{Since }OA\text{ is the diameter of the semicircle, }\angle OQA=90^\circ.$
$\displaystyle \therefore \angle OAQ=180^\circ-90^\circ-40^\circ=50^\circ.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{In the given diagram, }O\text{ is the centre of the circle, and }PQ\text{ is a tangent at }A.$
$\displaystyle \text{If }\angle ABC=50^\circ,\text{ then the values of }x,\ y\text{ and }z\text{ respectively are:}$ $\displaystyle \text{(a) }50^\circ,100^\circ,40^\circ\qquad\text{(b) }50^\circ,50^\circ,65^\circ$
$\displaystyle \text{(c) }40^\circ,80^\circ,50^\circ\qquad\text{(d) }50^\circ,25^\circ,78^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \angle ABC=50^\circ.$
$\displaystyle \text{By the alternate segment theorem, the angle between tangent }AQ\text{ and chord }AC$
$\displaystyle \text{is equal to the angle in the alternate segment.}$
$\displaystyle \therefore x=\angle ABC=50^\circ.$
$\displaystyle \text{The angle subtended by chord }AC\text{ at the centre is twice the angle subtended at the circle.}$
$\displaystyle \therefore y=\angle AOC=2\angle ABC=2(50^\circ)=100^\circ.$
$\displaystyle OA=OC,\text{ being radii of the circle.}$
$\displaystyle \therefore \triangle AOC\text{ is isosceles.}$
$\displaystyle z=\angle OCA=\frac{180^\circ-100^\circ}{2}=40^\circ.$
$\displaystyle \therefore x=50^\circ,\quad y=100^\circ,\quad z=40^\circ.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{In the given figure, }PT\text{ and }QT\text{ are tangents to a circle such that}$
$\displaystyle \angle TPS=45^\circ\text{ and }\angle TQS=30^\circ.\text{ Then the value of }x\text{ is:}$ $\displaystyle \text{(a) }30^\circ\qquad\text{(b) }45^\circ\qquad\text{(c) }75^\circ\qquad\text{(d) }105^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{By the alternate segment theorem,}$
$\displaystyle \angle TPS=\angle PQS=45^\circ.$
$\displaystyle \text{Also, }\angle TQS=\angle QPS=30^\circ.$
$\displaystyle \text{In }\triangle PQS,$
$\displaystyle x+\angle PQS+\angle QPS=180^\circ.$
$\displaystyle x+45^\circ+30^\circ=180^\circ.$
$\displaystyle x=180^\circ-75^\circ=105^\circ.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{A cylindrical metallic wire is stretched to double its length. Which of the}$
$\displaystyle \text{following will NOT change for the wire after stretching?}$
$\displaystyle \text{(a) Its curved surface area}\qquad\text{(b) Its total surface area}$
$\displaystyle \text{(c) Its volume}\qquad\text{(d) Its radius}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When a metallic wire is stretched, the amount of material remains unchanged.}$
$\displaystyle \therefore \text{The volume of the wire remains constant.}$
$\displaystyle \text{If }r_1,h_1\text{ and }r_2,h_2\text{ are its dimensions before and after stretching,}$
$\displaystyle \pi r_1^2h_1=\pi r_2^2h_2.$
$\displaystyle \therefore \text{Its radius and surface areas change, but its volume does not change.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{A right circular cone has the radius of the base equal to the height of the cone.}$
$\displaystyle \text{If the volume of the cone is }9702\text{ cm}^3,\text{ then the diameter of the base of the cone is:}$
$\displaystyle \text{(a) }21\text{ cm}\qquad\text{(b) }42\text{ cm}\qquad\text{(c) }21\sqrt7\text{ cm}\qquad\text{(d) }2\sqrt7\text{ cm}$
$\displaystyle \text{[Use }\pi=\frac{22}{7}\text{]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the cone be }r\text{ cm.}$
$\displaystyle \text{Since the radius is equal to the height, }h=r.$
$\displaystyle \text{Volume of cone}=\frac13\pi r^2h.$
$\displaystyle 9702=\frac13\times\frac{22}{7}\times r^3.$
$\displaystyle r^3=\frac{9702\times3\times7}{22}=9261.$
$\displaystyle r=\sqrt[3]{9261}=21\text{ cm}.$
$\displaystyle \therefore \text{Diameter}=2r=2(21)=42\text{ cm}.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{A solid sphere with a radius of }4\text{ cm is cut into }4\text{ identical}$
$\displaystyle \text{pieces by two mutually perpendicular planes passing through its centre. Find the total}$
$\displaystyle \text{surface area of one-quarter piece.}$ $\displaystyle \text{(a) }24\pi\qquad\text{(b) }32\pi\qquad\text{(c) }48\pi\qquad\text{(d) }64\pi$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the sphere}=4\text{ cm}.$
$\displaystyle \text{Curved surface area of one-quarter piece}=\frac14(4\pi r^2)=\pi r^2.$
$\displaystyle =\pi(4)^2=16\pi\text{ cm}^2.$
$\displaystyle \text{Each cutting plane forms a semicircular face of radius }4\text{ cm}.$
$\displaystyle \text{Area of two semicircular faces}=2\left(\frac12\pi r^2\right)=\pi r^2.$
$\displaystyle =16\pi\text{ cm}^2.$
$\displaystyle \text{Total surface area}=16\pi+16\pi=32\pi\text{ cm}^2.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\text{Two identical solid hemispheres are kept in contact to form a sphere.}$
$\displaystyle \text{The ratio of the total surface areas of the two hemispheres to the surface area of the}$
$\displaystyle \text{sphere formed is:}$ $\displaystyle \text{(a) }1:1\qquad\text{(b) }3:2\qquad\text{(c) }2:3\qquad\text{(d) }2:1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Total surface area of one solid hemisphere}=3\pi r^2.$
$\displaystyle \text{Total surface area of two hemispheres}=2(3\pi r^2)=6\pi r^2.$
$\displaystyle \text{Surface area of the sphere formed}=4\pi r^2.$
$\displaystyle \text{Required ratio}=6\pi r^2:4\pi r^2.$
$\displaystyle =6:4=3:2.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\mathrm{cosec}^2\theta+\sec^2\theta\text{ is equal to:}$
$\displaystyle \text{(a) }\tan^2\theta+\cot^2\theta\qquad \text{(b) }\cot\theta+\tan\theta$
$\displaystyle \text{(c) }(\cot\theta+\tan\theta)^2\qquad \text{(d) }1$
$\displaystyle \text{Answer:}$
$\displaystyle \mathrm{cosec}^2\theta+\sec^2\theta =(1+\cot^2\theta)+(1+\tan^2\theta)$
$\displaystyle =\cot^2\theta+\tan^2\theta+2.$
$\displaystyle \text{Also, }\tan\theta\cot\theta=1.$
$\displaystyle \therefore 2\tan\theta\cot\theta=2.$
$\displaystyle \mathrm{cosec}^2\theta+\sec^2\theta =\cot^2\theta+\tan^2\theta+2\tan\theta\cot\theta.$
$\displaystyle =(\cot\theta+\tan\theta)^2.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: }\text{Given }a=3\sec^2\theta\text{ and } b=3\tan^2\theta-2. \\ \text{ The value of }(a-b)\text{ is:}$
$\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad \text{(c) }3\qquad\text{(d) }5$
$\displaystyle \text{Answer:}$
$\displaystyle a-b=3\sec^2\theta-(3\tan^2\theta-2).$
$\displaystyle =3\sec^2\theta-3\tan^2\theta+2.$
$\displaystyle =3(\sec^2\theta-\tan^2\theta)+2.$
$\displaystyle \text{Using }\sec^2\theta-\tan^2\theta=1,$
$\displaystyle a-b=3(1)+2=5.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: }\text{At a certain time of day, the ratio of the height of the pole to the}$
$\displaystyle \text{length of its shadow is }1:\sqrt3.\text{ Then the angle of elevation of the sun at that time}$
$\displaystyle \text{of the day is:}$
$\displaystyle \text{(a) }30^\circ\qquad\text{(b) }45^\circ\qquad \text{(c) }60^\circ\qquad\text{(d) }90^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the angle of elevation of the sun be }\theta.$
$\displaystyle \tan\theta=\frac{\text{Height of pole}}{\text{Length of shadow}}.$
$\displaystyle \tan\theta=\frac{1}{\sqrt3}.$
$\displaystyle \tan\theta=\tan30^\circ.$
$\displaystyle \therefore \theta=30^\circ.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 39: }\text{A man standing on a ship approaching the lighthouse is observing}$
$\displaystyle \text{the top of the lighthouse. In }10\text{ minutes, the angle of elevation of the top of the}$
$\displaystyle \text{lighthouse changes from }\alpha\text{ to }\beta.\text{ Then:}$
$\displaystyle \text{(a) }\alpha>\beta\qquad\text{(b) }\alpha<\beta\qquad \text{(c) }\alpha=\beta\qquad\text{(d) }\alpha\leq\beta$
$\displaystyle \text{Answer:}$
$\displaystyle \text{As the ship approaches the lighthouse, its horizontal distance from the lighthouse decreases.}$
$\displaystyle \tan\theta=\frac{\text{Height of lighthouse}}{\text{Horizontal distance from lighthouse}}.$
$\displaystyle \text{Since the height is constant and the horizontal distance decreases, }\tan\theta\text{ increases.}$
$\displaystyle \therefore \text{The angle of elevation also increases.}$
$\displaystyle \therefore \beta>\alpha,\text{ or }\alpha<\beta.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 40: }\text{Assertion (A): The difference in class marks of the modal class and the}$
$\displaystyle \text{median class of the following frequency distribution is }0.$
$\displaystyle \begin{array}{c|ccccc} \text{Class interval}&20-30&30-40&40-50&50-60&60-70\\ \hline \text{Frequency}&1&3&2&6&4 \end{array}$
$\displaystyle \text{Reason (R): Modal class and median class are always the same for a given frequency distribution.}$
$\displaystyle \text{(a) Both A and R are correct, and R is the correct explanation for A.}$
$\displaystyle \text{(b) Both A and R are correct, and R is not the correct explanation for A.}$
$\displaystyle \text{(c) A is true, but R is false.}\qquad\text{(d) Both A and R are true.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The highest frequency is }6,\text{ so the modal class is }50-60.$
$\displaystyle \text{Total frequency}=1+3+2+6+4=16.$
$\displaystyle \frac{N}{2}=\frac{16}{2}=8.$
$\displaystyle \text{Cumulative frequencies are }1,\ 4,\ 6,\ 12,\ 16.$
$\displaystyle \text{The first cumulative frequency greater than }8\text{ is }12.$
$\displaystyle \therefore \text{The median class is }50-60.$
$\displaystyle \text{Class mark of }50-60=\frac{50+60}{2}=55.$
$\displaystyle \therefore \text{Difference in the class marks}=55-55=0.$
$\displaystyle \therefore \text{Assertion (A) is true.}$
$\displaystyle \text{Modal class and median class need not always be the same.}$
$\displaystyle \therefore \text{Reason (R) is false.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 41: }\text{Assertion (A): For a collection of }11\text{ arrayed data, the median is}$
$\displaystyle \text{the middle number.}$
$\displaystyle \text{Reason (R): For the data }5,\ 9,\ 7,\ 13,\ 10,\ 11,\ 10,\text{ the median is }13.$
$\displaystyle \text{(a) Both A and R are correct, and R is the correct explanation for A.}$
$\displaystyle \text{(b) Both A and R are correct, and R is not the correct explanation for A.}$
$\displaystyle \text{(c) A is true, but R is false.}\qquad\text{(d) Both A and R are true.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }11\text{ arrayed observations, the median is the }\frac{11+1}{2}=6\text{th observation.}$
$\displaystyle \therefore \text{Assertion (A) is true.}$
$\displaystyle \text{Arranging the given data in ascending order,}$
$\displaystyle 5,\ 7,\ 9,\ 10,\ 10,\ 11,\ 13.$
$\displaystyle \text{There are }7\text{ observations, so the median is the }\frac{7+1}{2}=4\text{th observation.}$
$\displaystyle \therefore \text{Median}=10,\text{ not }13.$
$\displaystyle \therefore \text{Reason (R) is false.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 42: }\text{Ankit had the option of investing in Company A, where }7\%,\text{ Rs. }100$
$\displaystyle \text{shares are available at Rs. }120,\text{ or in Company B, where }8\%,\text{ Rs. }1000\text{ shares}$
$\displaystyle \text{are available at Rs. }1620.$
$\displaystyle \text{Assertion (A): Investment in Company A is better than Company B.}$
$\displaystyle \text{Reason (R): The rate of income in Company A is better than in Company B.}$
$\displaystyle \text{(a) Both A and R are true, and R is the correct explanation for A.}$
$\displaystyle \text{(b) Both A and R are true, but R is not the correct explanation for A.}$
$\displaystyle \text{(c) A is false, but R is true.}\qquad\text{(d) Both A and R are false.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For Company A, dividend per share}=7\%\text{ of Rs. }100=\text{Rs. }7.$
$\displaystyle \text{Rate of income}=\frac{7}{120}\times100=5.83\%\text{ (approx.).}$
$\displaystyle \text{For Company B, dividend per share}=8\%\text{ of Rs. }1000=\text{Rs. }80.$
$\displaystyle \text{Rate of income}=\frac{80}{1620}\times100=4.94\%\text{ (approx.).}$
$\displaystyle \therefore \text{Company A gives a higher rate of income than Company B.}$
$\displaystyle \therefore \text{Both A and R are true, and R is the correct explanation for A.}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 43: }\text{Assertion (A): }x^3+2x^2-x-2\text{ is a polynomial of degree }3.$
$\displaystyle \text{Reason (R): }x+2\text{ is a factor of the polynomial.}$
$\displaystyle \text{(a) Both A and R are correct, and R is the correct explanation for A.}$
$\displaystyle \text{(b) Both A and R are incorrect.}$
$\displaystyle \text{(c) A is true, but R is false.}\qquad\text{(d) Both A and R are true.}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^3+2x^2-x-2.$
$\displaystyle \text{The highest power of }x\text{ is }3.$
$\displaystyle \therefore \text{Assertion (A) is true.}$
$\displaystyle \text{To check whether }x+2\text{ is a factor, put }x=-2.$
$\displaystyle f(-2)=(-2)^3+2(-2)^2-(-2)-2.$
$\displaystyle =-8+8+2-2=0.$
$\displaystyle \therefore x+2\text{ is a factor, so Reason (R) is true.}$
$\displaystyle \text{However, R does not explain why the polynomial is of degree }3.$
$\displaystyle \therefore \text{Both A and R are true.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 44: }\text{Assertion (A): The point }(-2,8)\text{ is invariant under reflection in the line } \\ x=-2.$
$\displaystyle \text{Reason (R): If a point has its }x\text{-coordinate }0,\text{ it is invariant under reflection in both axes.}$
$\displaystyle \text{(a) Both A and R are correct, and R is the correct explanation for A.}$
$\displaystyle \text{(b) Both A and R are correct, and R is not the correct explanation for A.}$
$\displaystyle \text{(c) A is true, but R is false.}\qquad\text{(d) Both A and R are true.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The point }(-2,8)\text{ lies on the line }x=-2.$
$\displaystyle \text{A point lying on the mirror line remains unchanged after reflection in that line.}$
$\displaystyle \therefore (-2,8)\text{ is invariant under reflection in }x=-2.$
$\displaystyle \therefore \text{Assertion (A) is true.}$
$\displaystyle \text{A point with }x\text{-coordinate }0\text{ lies on the }y\text{-axis.}$
$\displaystyle \text{It is invariant under reflection in the }y\text{-axis, but not necessarily in the }x\text{-axis.}$
$\displaystyle \therefore \text{Reason (R) is false.}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 45: }\text{When a die is cast with numbering on its faces, as shown, the ratio}$
$\displaystyle \text{of the probability of getting a composite number to the probability of getting a prime}$
$\displaystyle \text{number is:}$ $\displaystyle \text{(a) }2:3\qquad\text{(b) }3:2\qquad\text{(c) }1:3\qquad\text{(d) }1:2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The possible outcomes are }1,2,3,4,5,6.$
$\displaystyle \text{Composite numbers}=4,6.$
$\displaystyle \therefore P(\text{composite number})=\frac{2}{6}=\frac13.$
$\displaystyle \text{Prime numbers}=2,3,5.$
$\displaystyle \therefore P(\text{prime number})=\frac{3}{6}=\frac12.$
$\displaystyle \text{Required ratio}=\frac13:\frac12=2:3.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 46: }\text{The product of }A= \begin{bmatrix}1&-2\\-3&4\end{bmatrix}\text{ and matrix }M\text{ is }AM=B,\text{ where }$
$\displaystyle B= \begin{bmatrix}2\\24\end{bmatrix}. \ \text{Then the order of matrix }M\text{ is:}$
$\displaystyle \text{(a) }2\times2\qquad\text{(b) }2\times1\qquad \text{(c) }1\times2\qquad\text{(d) }4\times1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Order of }A=2\times2\text{ and order of }B=2\times1.$
$\displaystyle \text{Let the order of }M\text{ be }m\times n.$
$\displaystyle AM=(2\times2)(m\times n).$
$\displaystyle \text{For the product }AM\text{ to exist, }m=2.$
$\displaystyle \text{Also, the order of }AM\text{ is }2\times n.$
$\displaystyle \text{Since }AM=B\text{ and }B\text{ is of order }2\times1,\text{ we get }n=1.$
$\displaystyle \therefore \text{The order of matrix }M\text{ is }2\times1.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 47: }\text{Given }a_1,a_2,a_3,\ldots\text{ and }b_1,b_2,b_3,\ldots\text{ are real numbers such that}$
$\displaystyle a_1-b_1=a_2-b_2=a_3-b_3=\cdots\text{ are all equal.}$
$\displaystyle a_1-b_1,\ a_2-b_2,\ a_3-b_3,\ldots\text{ forms a }\underline{\hspace{1.5cm}}\text{ progression.}$
$\displaystyle \text{(a) Geometric }(r=1)\qquad\text{(b) Arithmetic }(d=1)$
$\displaystyle \text{(c) Geometric }(r<1)\qquad\text{(d) Arithmetic }(d=0)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a_1-b_1=a_2-b_2=a_3-b_3=\cdots=k.$
$\displaystyle \therefore \text{The sequence is }k,\ k,\ k,\ldots$
$\displaystyle \text{The difference between any two consecutive terms is }k-k=0.$
$\displaystyle \therefore \text{It forms an arithmetic progression with common difference }d=0.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 48: }\text{Locus of a moving point is }\underline{\hspace{1.5cm}}\text{ if it moves such that it keeps a}$
$\displaystyle \text{fixed distance from a fixed point.}$
$\displaystyle \text{(a) Circle}\qquad\text{(b) Line}\qquad\text{(c) Angle}\qquad\text{(d) Line segment}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The locus of a point moving at a fixed distance from a fixed point is a circle.}$
$\displaystyle \text{The fixed point is the centre and the fixed distance is the radius of the circle.}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 49: }\text{The point of concurrence of the angle bisectors of a triangle is called the}$
$\displaystyle \underline{\hspace{1.5cm}}\text{ of the triangle.}$
$\displaystyle \text{(a) centroid}\qquad\text{(b) incentre}\qquad\text{(c) circumcentre}\qquad\text{(d) orthocentre}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The three internal angle bisectors of a triangle meet at a common point called the incentre.}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$