CBSE Class X Mathematics Standard (041) Sample Question Paper Review

Comprehensive practice flashcards covering multiple-choice, short-answer, and case-study problems from the CBSE Class 10 Mathematics Standard sample paper.

What is the greatest number which divides both $134$ and $188$, leaving remainders $4$ and $6$ respectively?

$26$. To find this, compute the greatest common divisor of $(134 - 4) = 130$ and $(188 - 6) = 182$. Since $\text{HCF}(130, 182) = 26$, the greatest number is $26$.

If $f(x) = px^2 + qx + r$, $p \neq 0$, and $p + r = q$, what is one of the zeroes of $f(x)$?

$-\frac{r}{p}$. Substituting $x = -1$ gives $f(-1) = p(-1)^2 + q(-1) + r = p - q + r = 0$. Since $-1$ is one root and the product of zeroes is $\frac{r}{p}$, the other zero is $\frac{r/p}{-1} = -\frac{r}{p}$.

Tarun solved a pair of linear equations and found their unique intersection to be $(5, -1)$. If one line was $x - y = 6$, which of the following lines could have been the other: I: $3x - 3y = 18$, II: $2x - 3y = 13$, III: $2x - 3y = 16$?

II only. Line I ($3x - 3y = 18$) is coincident with $x - y = 6$ (infinitely many solutions). Line III does not satisfy $(5, -1)$, as $2(5) - 3(-1) = 13 \neq 16$. Line II satisfies $2(5) - 3(-1) = 13$ and has a different slope, yielding a unique intersection point.

If $(1 - p)$ is a root of the quadratic equation $x^2 + px + 1 - p = 0$, what are the roots of the equation?

$0$ and $-1$. Substituting $x = (1 - p)$ gives $(1 - p)^2 + p(1 - p) + (1 - p) = 0 \implies (1 - p)(2) = 0 \implies p = 1$. For $p = 1$, the equation becomes $x^2 + x = 0$, which yields roots $x = 0$ and $x = -1$.

What is the middle term of the arithmetic progression $10, 7, 4, \dots, -62$?

$-26$. The A.P. has first term $a = 10$ and common difference $d = -3$. Solving $10 + (n - 1)(-3) = -62$ gives $n = 25$. The middle term is the $13^{\text{th}}$ term: $a_{13} = 10 + 12(-3) = -26$.

The perimeters of two similar triangles are $56\,cm$ and $70\,cm$ respectively. If one side of the first triangle is $14\,cm$, what is the corresponding side of the second triangle?

$17.5\,cm$. The ratio of the perimeters equals the ratio of corresponding sides: $\frac{56}{70} = \frac{14}{x} \implies \frac{4}{5} = \frac{14}{x} \implies x = 17.5\,cm$.

Which point lies on the perpendicular bisector of the line segment joining the points $A(-3, -4)$ and $B(3, 4)$?

$(0, 0)$. The midpoint of line segment $AB$ is $\left(\frac{-3+3}{2}, \frac{-4+4}{2}\right) = (0, 0)$, which lies on its perpendicular bisector.

If $\tan(A + B) = \sqrt{3}$ and $\tan(A - B) = \frac{1}{\sqrt{3}}$, where $0^\circ < A + B < 90^\circ$ and $A > B$, what are the values of $A$ and $B$?

$A = 45^\circ$ and $B = 15^\circ$. Since $A + B = 60^\circ$ and $A - B = 30^\circ$, adding them gives $2A = 90^\circ \implies A = 45^\circ$, and subtracting gives $B = 15^\circ$.

T-shirts marked with numbers $4$ to $99$ are placed in a box. What is the probability of randomly choosing a number that is either a perfect square or a perfect cube?

$\frac{5}{48}$. Total outcomes from $4$ to $99$ are $96$. Perfect squares are $\{4, 9, 16, 25, 36, 49, 64, 81\}$ ($8$ numbers) and perfect cubes are $\{8, 27, 64\}$ ($3$ numbers). Since $64$ is common, there are $8 + 3 - 1 = 10$ favorable numbers. The probability is $\frac{10}{96} = \frac{5}{48}$.

The diameter of a car wheel is $21\,cm$. How many complete revolutions will it make in traveling $66\,km$?

$10^5$. The circumference is $\pi d = \frac{22}{7} \times 21\,cm = 66\,cm$. The total distance is $66\,km = 6,600,000\,cm$. The number of revolutions is $\frac{6,600,000}{66} = 100,000 = 10^5$.

Two cubes each of volume $64\,cm^3$ are joined end to end to form a solid. What is the total surface area of the resulting cuboid?

$160\,cm^2$. Each cube has an edge length of $\sqrt[3]{64} = 4\,cm$. Joining two cubes creates a cuboid with dimensions $8\,cm \times 4\,cm \times 4\,cm$. Its surface area is $2(8 \times 4 + 4 \times 4 + 8 \times 4) = 2(32 + 16 + 32) = 160\,cm^2$.

The mean age of a combined group of men and women is $35$ years. If the mean ages of the men and women are $38$ years and $30$ years respectively, what is the percentage of women in the group?

$37.5\%$. Using the combined mean equation $38m + 30w = 35(m + w) \implies 3m = 5w \implies \frac{w}{m} = \frac{3}{5}$. The fraction of women is $\frac{w}{m + w} = \frac{3}{8}$, which corresponds to $\frac{3}{8} \times 100\% = 37.5\%$.

What is the length of median $PM$ in $\Delta PQR$ having vertices $P(5, -6)$, $Q(6, 4)$, and $R(0, 0)$?

$2\sqrt{17}\,\text{units}$. The midpoint $M$ of $QR$ has coordinates $\left(\frac{6+0}{2}, \frac{4+0}{2}\right) = (3, 2)$. The length of $PM$ is $\sqrt{(5 - 3)^2 + (-6 - 2)^2} = \sqrt{2^2 + (-8)^2} = \sqrt{68} = 2\sqrt{17}\,\text{units}$.

If the arc of a circle has a length of $6\pi\,cm$ and the sector it bounds has an area of $24\pi\,cm^2$, what is the radius of the circle?

$8\,cm$. The area of a sector is given by $\text{Area} = \frac{1}{2} l r$. Substituting the given values gives $24\pi = \frac{1}{2}(6\pi)r \implies 24\pi = 3\pi r \implies r = 8\,cm$.

If the mean and median of a dataset are $35.5$ and $32$ respectively, what is the mode of the data?

$25$. Using the empirical formula $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$, we calculate $\text{Mode} = 3(32) - 2(35.5) = 96 - 71 = 25$.

A kite is tied to a ground point with a string of length $52\,m$. If the string makes an angle $\theta$ with the ground such that $\tan(\theta) = \frac{12}{5}$, what is the height of the kite above the ground?

$48\,m$. Since $\tan(\theta) = \frac{12}{5}$, $\sin(\theta) = \frac{12}{\sqrt{12^2 + 5^2}} = \frac{12}{13}$. The height of the kite is $h = 52 \times \sin(\theta) = 52 \times \frac{12}{13} = 48\,m$.

Two cards of diamonds and four cards of spades are missing from a pack of $52$ cards. What is the probability of drawing a heart at random from the remaining cards?

$\frac{13}{46}$. The total number of remaining cards is $52 - 6 = 46$. None of the hearts were removed, so there are still $13$ hearts. The probability is $\frac{13}{46}$.

What condition must be satisfied for the zeroes of the quadratic polynomial $ax^2 + bx + c$, $a \neq 0$, to be reciprocal of each other?

$a = c$. If the zeroes are $\alpha$ and $\frac{1}{\alpha}$, their product is $\alpha \times \frac{1}{\alpha} = 1$. By Vieta's relations, the product of zeroes is $\frac{c}{a} = 1 \implies a = c$.

Three bells toll at intervals of $5$, $6$, and $8$ minutes respectively. They toll together at $10:00\,\text{a.m.}$. How many more times will they toll together until the park closes at $7:00\,\text{p.m.}$?

$4$ more times. The LCM of $5$, $6$, and $8$ is $120\,\text{minutes} = 2\,\text{hours}$. The total elapsed time between $10:00\,\text{a.m.}$ and $7:00\,\text{p.m.}$ is $9\,\text{hours}$. Tolls occur at $+2\,\text{h}$, $+4\,\text{h}$, $+6\,\text{h}$, and $+8\,\text{h}$ (at $12:00\,\text{p.m.}$, $2:00\,\text{p.m.}$, $4:00\,\text{p.m.}$, and $6:00\,\text{p.m.}$), totaling $4$ times.

In the triangle shown, if $\angle PQR = \angle QSP$, $PQ = 6\,cm$, and $PS = 3\,cm$, what is the length of $PR$?

$12\,cm$. Triangles $\Delta PQR$ and $\Delta PSQ$ are similar because $\angle P$ is common and $\angle PQR = \angle QSP$. Therefore, $\frac{PQ}{PS} = \frac{PR}{PQ} \implies PQ^2 = PS \times PR \implies 6^2 = 3 \times PR \implies PR = 12\,cm$.

If $\tan(\theta) = \frac{1}{\sqrt{5}}$, what is the value of $\frac{\csc^2(\theta) - \sec^2(\theta)}{\csc^2(\theta) + \sec^2(\theta)}$?

$\frac{2}{3}$. Using identities, $\csc^2(\theta) = 1 + \cot^2(\theta) = 1 + 5 = 6$, and $\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \frac{1}{5} = \frac{6}{5}$. The expression simplifies to $\frac{6 - 6/5}{6 + 6/5} = \frac{24/5}{36/5} = \frac{2}{3}$.

Two dice are rolled simultaneously. What is the probability that the product of the numbers appearing on them is a prime number?

$\frac{1}{6}$. A product is prime only when one die shows $1$ and the other shows a prime number ($2, 3,$ or $5$). The favorable outcomes are $(1, 2), (2, 1), (1, 3), (3, 1), (1, 5), (5, 1)$ ($6$ outcomes). The probability is $\frac{6}{36} = \frac{1}{6}$.

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $2x^2 - 8x + 5$, what is the value of $\left(\alpha + \frac{1}{\beta}\right) \times \left(\beta + \frac{1}{\alpha}\right)$?

$\frac{49}{10}$ (or $4.9$). From Vieta's formulas, $\alpha + \beta = 4$ and $\alpha\beta = \frac{5}{2}$. Expanding the product gives $\alpha\beta + 2 + \frac{1}{\alpha\beta} = \frac{5}{2} + 2 + \frac{2}{5} = \frac{25 + 20 + 4}{10} = \frac{49}{10}$.

Ridhi drew a polygon with $n$ sides where the smallest exterior angle is $8^\circ$ and each subsequent exterior angle increases by $4^\circ$. How many sides does the polygon have?

$12$ sides. The sum of all exterior angles of any polygon is $360^\circ$. Using the sum of an A.P., $S_n = \frac{n}{2}[2(8) + (n - 1)4] = 360 \implies 2n^2 + 6n - 360 = 0 \implies n^2 + 3n - 180 = 0$. Factoring gives $(n + 15)(n - 12) = 0$, so $n = 12$.

In what ratio does the $x$-axis divide the line segment joining $(-4, -6)$ and $(-1, 7)$, and what are the coordinates of the point of division?

Ratio $6 : 7$; coordinates $\left(-\frac{34}{13}, 0\right)$. Let the ratio be $k : 1$. On the $x$-axis, $y = 0 \implies \frac{7k - 6}{k + 1} = 0 \implies k = \frac{6}{7}$. The $x$-coordinate is $\frac{6(-1) + 7(-4)}{6 + 7} = -\frac{34}{13}$.

In Jyoti's simplification of $\frac{\cos(\theta)}{1 - \sin(\theta)} + \frac{1 - \sin(\theta)}{\cos(\theta)}$, in which step did she make an error, and what is the correct final expression?

She made an error in Step 2 by writing $(1 - \sin(\theta))^2 = \cos^2(\theta)$ instead of $1 - 2\sin(\theta) + \sin^2(\theta)$. Expanding correctly gives $\frac{\cos^2(\theta) + 1 - 2\sin(\theta) + \sin^2(\theta)}{(1 - \sin(\theta))\cos(\theta)} = \frac{2(1 - \sin(\theta))}{(1 - \sin(\theta))\cos(\theta)} = \frac{2}{\cos(\theta)} = 2\sec(\theta)$.

A rectangular park measures $50\,m \times 40\,m$. A central pool is surrounded by a uniform grass strip of area $1184\,m^2$. What are the dimensions of the pool?

$34\,m \times 24\,m$. Total area is $50 \times 40 = 2000\,m^2$. The pool area is $2000 - 1184 = 816\,m^2$. For a uniform strip of width $w$, $(50 - 2w)(40 - 2w) = 816 \implies 4w^2 - 180w + 1184 = 0 \implies w^2 - 45w + 296 = 0$, yielding $w = 8\,m$. The dimensions are $(50 - 16) = 34\,m$ by $(40 - 16) = 24\,m$.

Manish observes a kite $200\,m$ away at an elevation of $30^\circ$. Mahesh stands on a $50\,m$ high building on the opposite side and observes the same kite at an elevation of $45^\circ$. What is the distance between the kite and Mahesh?

$50\sqrt{2}\,m$. Kite height above ground is $200\sin(30^\circ) = 100\,m$. Kite height above Mahesh is $100 - 50 = 50\,m$. The distance from the kite to Mahesh is $\frac{50}{\sin(45^\circ)} = 50\sqrt{2}\,m$.

For the circular meditation garden shown, the radius is $9\,m$ and the distance $OP = 18\,m$. What are the values of $\angle OAP$, the tangent length $PA$, and $\angle AOB$?

$\angle OAP = 90^\circ$, $PA = 9\sqrt{3}\,m$, and $\angle AOB = 120^\circ$. Radius is perpendicular to the tangent at contact, so $\angle OAP = 90^\circ$. By the Pythagorean theorem, $PA = \sqrt{18^2 - 9^2} = \sqrt{243} = 9\sqrt{3}\,m$. Since $\sin(\angle APO) = \frac{9}{18} = \frac{1}{2}$, $\angle APO = 30^\circ$, making $\angle APB = 60^\circ$ and $\angle AOB = 180^\circ - 60^\circ = 120^\circ$.

For the grouped score data: $0-10$ ($3$), $10-20$ ($6$), $20-30$ ($12$), $30-40$ ($15$), $40-50$ ($14$), identify the modal class, median class, and mean score.

Modal class: $30 - 40$; Median class: $30 - 40$; Mean: $31.2$. The highest frequency is $15$ (in $30 - 40$). For $N = 50$, $N/2 = 25$ falls in cumulative frequency class $30 - 40$. The mean is $\frac{\sum f_i x_i}{N} = \frac{3(5) + 6(15) + 12(25) + 15(35) + 14(45)}{50} = \frac{1560}{50} = 31.2$.