1 MarkQ52. The length of a rectangular hall is $5\text{ m}$ more than its breadth. If the area of the hall is $84\text{ m}^2$, then the perimeter of the hall is:
1 MarkQ55. A fast train takes $3\text{ hours}$ less than a slow train for a journey of $600\text{ km}$. If the speed of the slow train is $10\text{ km/h}$ less than that of the fast train, the speed of the fast train is:
1 MarkQ60. A piece of cloth costs ₹ $200$. If the piece were $5\text{ m}$ longer and each metre of cloth cost ₹ $2$ less, the cost of the piece would remain unchanged. The original length of the piece is:
(a) $20\text{ m}$
(b) $25\text{ m}$
(c) $15\text{ m}$
(d) $30\text{ m}$
Solution:
Let original length be $x\text{ m}$. Original price per metre = $\frac{200}{x}$.
$(x + 5)\left(\frac{200}{x} - 2\right) = 200 \implies 200 - 2x + \frac{1000}{x} - 10 = 200$
$-2x + \frac{1000}{x} - 10 = 0 \implies 2x^2 + 10x - 1000 = 0 \implies x^2 + 5x - 500 = 0$
$(x + 25)(x - 20) = 0 \implies x = 20\text{ m}$. Answer: (a) $20\text{ m}$
1 MarkQ61. If $\alpha$ and $\beta$ are the roots of $x^2 - p(x + 1) - c = 0$, then the value of $(\alpha + 1)(\beta + 1)$ is:
(a) $1 - c$
(b) $1 + c$
(c) $c - 1$
(d) $p + c$
Solution:
Equation: $x^2 - px - p - c = 0$. Sum of roots $\alpha + \beta = p$, Product of roots $\alpha\beta = -(p + c)$.
$(\alpha + 1)(\beta + 1) = \alpha\beta + (\alpha + \beta) + 1 = -(p + c) + p + 1 = -p - c + p + 1 = 1 - c$. Answer: (a) $1 - c$
1 MarkQ62. The value of $\sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}}$ is:
(a) $3$
(b) $2$
(c) $-2$
(d) $6$
Solution:
Let $x = \sqrt{6 + \sqrt{6 + \dots}} \implies x = \sqrt{6 + x}$
$x^2 = 6 + x \implies x^2 - x - 6 = 0 \implies (x - 3)(x + 2) = 0 \implies x = 3$ (since value must be positive). Answer: (a) $3$
1 MarkQ63. If the quadratic equation $p x^2 + 4x + 1 = 0$ has real and distinct roots, then:
(a) $p < 4$
(b) $p > 4$
(c) $p \le 4$
(d) $p = 4$
Solution:
For real and distinct roots, $D > 0$:
$4^2 - 4(p)(1) > 0 \implies 16 - 4p > 0 \implies 4p < 16 \implies p < 4$. Answer: (a) $p < 4$
1 MarkQ64. If one root of $a x^2 + b x + c = 0$ is $k$ times the other, then:
1 MarkQ65. Two water taps together can fill a tank in $9\frac{3}{8}\text{ hours}$. The tap of larger diameter takes $10\text{ hours}$ less than the smaller one to fill the tank separately. The time taken by the smaller tap alone is:
(a) $25\text{ hours}$
(b) $15\text{ hours}$
(c) $20\text{ hours}$
(d) $30\text{ hours}$
Solution:
Let smaller tap take $x\text{ hours}$. Larger tap takes $(x - 10)\text{ hours}$. Total time = $\frac{75}{8}\text{ hours}$.
$\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75} \implies \frac{2x - 10}{x(x - 10)} = \frac{8}{75} \implies 75(2x - 10) = 8(x^2 - 10x)$
$150x - 750 = 8x^2 - 80x \implies 8x^2 - 230x + 750 = 0 \implies 4x^2 - 115x + 375 = 0$
$(4x - 15)(x - 25) = 0 \implies x = 25\text{ hours}$ ($x = \frac{15}{4}$ is rejected as larger tap would take negative time). Answer: (a) $25\text{ hours}$
1 MarkQ66. The discriminant of the quadratic equation $(x - 1)(2x - 1) = 0$ is:
1 MarkQ67. If the sum of a number and its reciprocal is $\frac{10}{3}$, then the number is:
(a) $3$
(b) $4$
(c) $2$
(d) $5$
Solution:
Let number be $x$. $x + \frac{1}{x} = \frac{10}{3} \implies \frac{x^2 + 1}{x} = \frac{10}{3}$
$3x^2 + 3 = 10x \implies 3x^2 - 10x + 3 = 0 \implies (3x - 1)(x - 3) = 0 \implies x = 3 \text{ or } \frac{1}{3}$. Answer: (a) $3$
1 MarkQ68. The nature of the roots of the quadratic equation $9x^2 - 6x - 2 = 0$ is:
(a) Real and distinct
(b) Real and equal
(c) No real roots
(d) Purely imaginary
Solution:
Discriminant $D = b^2 - 4ac = (-6)^2 - 4(9)(-2) = 36 + 72 = 108$.
Since $D > 0$, the roots are real and distinct. Answer: (a) Real and distinct
1 MarkQ69. If $\alpha, \beta$ are the roots of $x^2 - kx + 6 = 0$ such that $\alpha - \beta = 1$, then $k$ equals:
1 MarkQ73. The difference of two natural numbers is $3$ and the sum of their reciprocals is $\frac{7}{10}$. The numbers are:
(a) $5\text{ and } 2$
(b) $6\text{ and } 3$
(c) $7\text{ and } 4$
(d) $8\text{ and } 5$
Solution:
Let numbers be $x$ and $x - 3$.
$\frac{1}{x - 3} + \frac{1}{x} = \frac{7}{10} \implies \frac{2x - 3}{x(x - 3)} = \frac{7}{10} \implies 10(2x - 3) = 7(x^2 - 3x)$
$20x - 30 = 7x^2 - 21x \implies 7x^2 - 41x + 30 = 0 \implies (7x - 6)(x - 5) = 0 \implies x = 5$ (since $x$ must be a natural number).
The numbers are $5$ and $5 - 3 = 2$. Answer: (a) $5\text{ and } 2$
1 MarkQ74. If the quadratic equation $x^2 - b x + c = 0$ has two consecutive integers as roots, then $b^2 - 4c$ is equal to:
(a) $1$
(b) $2$
(c) $4$
(d) $0$
Solution:
Let roots be $n$ and $n + 1$.
$b^2 - 4c$ is the discriminant of the equation, which is equal to $(\alpha - \beta)^2$.
$[n - (n + 1)]^2 = (-1)^2 = 1$. Answer: (a) $1$
1 MarkQ75. The sum of the areas of two squares is $468\text{ m}^2$. If the difference of their perimeters is $24\text{ m}$, then the side of the larger square is:
(a).$18\text{ m}$
(b).$12\text{ m}$
(c).$15\text{ m}$
(d).$24\text{ m}$
Solution:
Let the sides of the two squares be $x$ and $y$ ($x > y$).
Difference of perimeters: $4x - 4y = 24 \implies x - y = 6 \implies x = y + 6$.
Sum of areas: $x^2 + y^2 = 468 \implies (y + 6)^2 + y^2 = 468$.
$2y^2 + 12y + 36 = 468 \implies y^2 + 6y - 216 = 0 \implies (y + 18)(y - 12) = 0$.
Since side cannot be negative, $y = 12\text{ m}$, so $x = 12 + 6 = 18\text{ m}$. Answer: (a) $18\text{ m}$
1 MarkQ76. If $x = 1$ is a common root of $a x^2 + a x + 3 = 0$ and $x^2 + x + b = 0$, then $a b$ equals:
(a).$3$
(b).$-3$
(c).$6$
(d).$-6$
Solution:
Substitute $x = 1$ in $a(1)^2 + a(1) + 3 = 0 \implies 2a + 3 = 0 \implies a = -\frac{3}{2}$.
Substitute $x = 1$ in $(1)^2 + 1 + b = 0 \implies 2 + b = 0 \implies b = -2$.
$ab = \left(-\frac{3}{2}\right)(-2) = 3$. Answer: (a) $3$
1 MarkQ77. If the equation $x^2 - k x + 9 = 0$ has no real roots, then $k$ lies in the interval:
(a).$-6 < k < 6$
(b).$k > 6$ or $k < -6$
(c).$k = 6$
(d).$0 < k < 6$
Solution:
For no real roots, discriminant $b^2 - 4ac < 0$.
$(-k)^2 - 4(1)(9) < 0 \implies k^2 - 36 < 0 \implies -6 < k < 6$. Answer: (a) $-6 < k < 6$
1 MarkQ78. The hypotenuse of a right triangle is $3\sqrt{5}\text{ cm}$. If the smaller side is tripled and the larger side is doubled, the new hypotenuse becomes $15\text{ cm}$. The length of the smaller side is:
(a).$3\text{ cm}$
(b).$6\text{ cm}$
(c).$5\text{ cm}$
(d).$4\text{ cm}$
Solution:
Let smaller side be $a$ and larger side be $b$.
$a^2 + b^2 = (3\sqrt{5})^2 = 45$.
$(3a)^2 + (2b)^2 = 15^2 \implies 9a^2 + 4b^2 = 225$.
Solving $4(a^2 + b^2) = 180$ and $9a^2 + 4b^2 = 225$ gives $5a^2 = 45 \implies a = 3\text{ cm}$. Answer: (a) $3\text{ cm}$
1 MarkQ79. If one root of the equation $4x^2 - 2x + (p - 4) = 0$ is the reciprocal of the other, then the value of $p$ is:
(a).$8$
(b).$4$
(c).$-4$
(d).$2$
Solution:
Product of roots for $ax^2 + bx + c = 0$ is $\frac{c}{a}$.
Since roots are reciprocals, product $= 1 \implies \frac{p - 4}{4} = 1 \implies p - 4 = 4 \implies p = 8$. Answer: (a) $8$
1 MarkQ80. The roots of the quadratic equation $\frac{x+1}{x-1} + \frac{x-2}{x+2} = 3$ are:
1 MarkQ81. If the roots of $(a - b)x^2 + (b - c)x + (c - a) = 0$ are equal, then:
(a).$2a = b + c$
(b).$2b = a + c$
(c).$2c = a + b$
(d).$a + b + c = 0$
Solution:
Sum of coefficients $(a-b) + (b-c) + (c-a) = 0$, so $x = 1$ is a root. Since roots are equal, both roots are $1$.
Product of roots $= \frac{c-a}{a-b} = 1 \times 1 = 1 \implies c - a = a - b \implies 2a = b + c$. Answer: (a) $2a = b + c$
1 MarkQ82. If $\alpha$ and $\beta$ are the roots of $x^2 - p(x + 1) - c = 0$, then the value of $\frac{\alpha^2 + 2\alpha + 1}{\alpha^2 + 2\alpha + c} + \frac{\beta^2 + 2\beta + 1}{\beta^2 + 2\beta + c}$ is:
(a).$1$
(b).$0$
(c).$-1$
(d).$2$
Solution:
Rewrite equation as $x^2 - px - p - c = 0$. Since $\alpha$ is a root, $\alpha^2 - p\alpha - p - c = 0 \implies \alpha^2 + c = p\alpha + p$.
Simplifying terms yields the expression evaluating to $1$. Answer: (a) $1$
1 MarkQ83. If the ratio of the roots of $x^2 + px + q = 0$ is equal to the ratio of the roots of $x^2 + lx + m = 0$, then:
(a).$p^2 m = l^2 q$
(b).$p m^2 = l q^2$
(c).$p^2 q = l^2 m$
(d).$p m = l q$
Solution:
Equating the square of the sum-to-product ratio $\frac{(\text{Sum})^2}{\text{Product}}$ for both equations gives $\frac{p^2}{q} = \frac{l^2}{m} \implies p^2 m = l^2 q$. Answer: (c) $p^2 q = l^2 m$
1 MarkQ84. An express train takes $1\text{ hour}$ less than a passenger train to travel $132\text{ km}$ between Mysore and Bangalore. If the average speed of the express train is $11\text{ km/h}$ more than that of the passenger train, the speed of the express train is:
1 MarkQ85. The value of $k$ for which $x^2 - (k + 6)x + 2(2k - 1) = 0$ has sum of roots equal to half of their product is:
(a).$7$
(b).$5$
(c).$3$
(d).$1$
Solution:
Sum $= k + 6$, Product $= 2(2k - 1)$.
Given sum = $\frac{1}{2} \times \text{Product} \implies k + 6 = \frac{1}{2}[2(2k - 1)] \implies k + 6 = 2k - 1 \implies k = 7$. Answer: (a) $7$
1 MarkQ86. If the roots of $a x^2 + b x + c = 0$ are in the ratio $2 : 3$, then:
(a).$6b^2 = 25ac$
(b).$5b^2 = 6ac$
(c).$2b^2 = 3ac$
(d).$3b^2 = 2ac$
Solution:
Let roots be $2\alpha$ and $3\alpha$. Sum $= 5\alpha = -\frac{b}{a}$, Product $= 6\alpha^2 = \frac{c}{a}$.
Eliminating $\alpha$: $\left(-\frac{b}{5a}\right)^2 = \frac{c}{6a} \implies \frac{b^2}{25a^2} = \frac{c}{6a} \implies 6b^2 = 25ac$. Answer: (a) $6b^2 = 25ac$
1 MarkQ87. A plane left $30\text{ minutes}$ later than its scheduled time and in order to reach the destination $1500\text{ km}$ away in time, it had to increase its speed by $250\text{ km/h}$ from its usual speed. The usual speed of the plane is:
1 MarkQ89. If $\alpha, \beta$ are the roots of $2x^2 + 2(p + q)x + p^2 + q^2 = 0$, then the roots are:
(a).Real and distinct if $p \neq q$
(b).Real and equal if $p = q$
(c).Non-real if $p \neq q$
(d).Equal for all values of $p$ and $q$
Solution:
Discriminant $D = [2(p+q)]^2 - 4(2)(p^2 + q^2) = 4(p^2 + 2pq + q^2 - 2p^2 - 2q^2) = -4(p - q)^2$.
Since $D \le 0$ for all real $p, q$, roots are non-real when $p \neq q$, and real/equal when $p = q$. Answer: (b) Real and equal if $p = q$
1 MarkQ90. A two-digit number is four times the sum of its digits and twice the product of its digits. The number is:
(a).$36$
(b).$24$
(c).$48$
(d).$12$
Solution:
Test option (a): Number $= 36$. Sum $= 3 + 6 = 9$. Four times sum $= 4 \times 9 = 36$. Product $= 3 \times 6 = 18$, twice product $= 36$. Both conditions satisfied. Answer: (a) $36$
1 MarkQ91. If the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$ are equal, then:
(a).$a = 0$ or $a^3 + b^3 + c^3 = 3abc$
(b).$b = 0$ or $a^3 + b^3 + c^3 = 3abc$
(c).$c = 0$ or $a^3 + b^3 + c^3 = 0$
(d).$a + b + c = 0$
Solution:
Setting discriminant $D = 0$ for equal roots leads to the standard symmetric condition $a = 0$ or $a^3 + b^3 + c^3 = 3abc$. Answer: (a) $a = 0$ or $a^3 + b^3 + c^3 = 3abc$
1 MarkQ92. A motor boat whose speed is $18\text{ km/h}$ in still water takes $1\text{ hour}$ more to go $24\text{ km}$ upstream than to return downstream to the same spot. The speed of the stream is:
1 MarkQ93. If $\alpha$ and $\beta$ are roots of $x^2 - 5x + 6 = 0$, the quadratic equation whose roots are $(\alpha + 1)$ and $(\beta + 1)$ is:
(a).$x^2 - 7x + 12 = 0$
(b).$x^2 - 7x + 10 = 0$
(c).$x^2 - 5x + 12 = 0$
(d).$x^2 - 6x + 8 = 0$
Solution:
Roots are $2$ and $3$ (since $x^2 - 5x + 6 = (x-2)(x-3) = 0$). New roots are $3$ and $4$.
Sum $= 7$, Product $= 12$. Equation: $x^2 - 7x + 12 = 0$. Wait, new roots are $\alpha+1 = 3$ and $\beta+1 = 4$. Sum $= 7$, product $= 12$. Option (a)? Wait, let's check: alternative method replacing $x$ with $x-1$: $(x-1)^2 - 5(x-1) + 6 = x^2 - 2x + 1 - 5x + 5 + 6 = x^2 - 7x + 12 = 0$. Answer: (a) $x^2 - 7x + 12 = 0$
1 MarkQ94. If the roots of $(b - c)x^2 + (c - a)x + (a - b) = 0$ are equal, then $a, b, c$ satisfy:
(a).$2b = a + c$
(b).$2c = a + b$
(c).$2a = b + c$
(d).$a + b + c = 0$
Solution:
Sum of coefficients is zero, so $x = 1$ is a root. Since roots are equal, both are $1$. Product $= \frac{a-b}{b-c} = 1 \implies a - b = b - c \implies 2b = a + c$. Wait, let's check option (a). Answer: (a) $2b = a + c$
1 MarkQ95. The sum of two numbers is $15$. If the sum of their reciprocals is $\frac{3}{10}$, the numbers are:
1 MarkQ99. The quadratic equation whose roots are the reciprocals of the roots of $a x^2 + b x + c = 0$ (where $a, c \neq 0$) is:
(a).$c x^2 + b x + a = 0$
(b).$c x^2 - b x + a = 0$
(c).$a x^2 - b x + c = 0$
(d).$b x^2 + a x + c = 0$
Solution:
Substituting $x = \frac{1}{y}$ in $a\left(\frac{1}{y}\right)^2 + b\left(\frac{1}{y}\right) + c = 0$ and clearing denominators gives $c y^2 + b y + a = 0$. Answer: (a) $c x^2 + b x + a = 0$
1 MarkQ100. A cottage industry produces a certain number of pottery articles in a day. The cost of production of each article (in ₹) was $3$ more than twice the number of articles produced on that day. If the total cost was ₹ $90$, the number of articles produced is:
(a).$6$
(b).$10$
(c).$12$
(d).$15$
Solution:
Let articles produced be $x$. Cost per article $= 2x + 3$.
Total cost $= x(2x + 3) = 90 \implies 2x^2 + 3x - 90 = 0 \implies (2x + 15)(x - 6) = 0 \implies x = 6$. Answer: (a) $6$
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English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.
