1 Mark
1
The greatest number which divides both 134 and 188, leaving remainders 4 and 6 respectively, is:
A. 13
B. 26
C. 39
D. 65
Concept:
1
When a number divides two other numbers leaving specific remainders, the exact divisible numbers are found by subtracting the remainders from the original numbers. The "greatest" divisor is then the Highest Common Factor (HCF) of those resulting numbers.
Hint:
1
Subtract the remainders: $134 - 4 = 130$. $188 - 6 = 182$.
2
Find the HCF of 130 and 182.
Mistakes to Avoid:
!
Finding LCM instead of HCF: The keyword "greatest number which divides" signals HCF. LCM is for "smallest number divisible by".
Official CBSE Solution:
Subtract the remainders from the respective numbers:
$134 - 4 = 130$
$188 - 6 = 182$
Now, find the HCF of $130$ and $182$:
$\text{HCF}(130, 182) = 26$
Ans: Option (B) 26
🎯 CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
2
If $f(x)=px^{2}+qx+r$, $p \neq 0$ and $p+r=q$, then one of the zeroes of $f(x)$ is:
A. $\frac{q}{p}$
B. $\frac{r}{p}$
C. $-\frac{r}{p}$
D. $-\frac{q}{p}$
Concept:
1
Substitute the given condition ($q = p + r$) directly into the polynomial. Then, factorize the quadratic equation by splitting the middle term to find the explicit zeroes.
Hint:
1
$f(x) = px^2 + (p+r)x + r$.
2
Expand: $px^2 + px + rx + r$. Take common factors to form binomial brackets.
Mistakes to Avoid:
!
Using sum/product formulas blindly: While $\alpha\beta = r/p$, without definitively isolating one root, you can't easily find the other just by the product formula. Substituting the $q$ value is the intended algebraic path.
Official CBSE Solution:
Substituting $q = p+r$ in $f(x)$:
$f(x) = px^2 + (p+r)x + r$
Factorizing:
$= px^2 + px + rx + r$
$= px(x+1) + r(x+1)$
$= (x+1)(px+r)$
To find zeroes, set factors to 0:
$x+1 = 0 \implies x = -1$
$px+r = 0 \implies x = -\frac{r}{p}$
Ans: Option (C) $-\frac{r}{p}$
🎯 CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
3
Tarun correctly solved a pair of linear equations in two variables and found their only point of intersection as (5,1). One of the lines was $x-y=6$. Which of the following could have been the other line?
I: $3x-3y=18$
II: $2x-3y=13$
III: $2x-3y=16$
I: $3x-3y=18$
II: $2x-3y=13$
III: $2x-3y=16$
A. I only
B. II only
C. I and II
D. II and III
Concept:
1
A unique intersection point means two things: First, the point $(5,1)$ must satisfy the equation of the second line. Second, the two lines must be intersecting, meaning their coefficients must not be proportional ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$). Coincident lines yield infinite solutions, not an "only point".
Hint:
1
Substitute $x=5, y=1$ into equation I. $3(5)-3(1) = 15-3=12 \neq 18$. Also note $3x-3y=18$ simplifies to $x-y=6$, making it coincident.
2
Note: The official CBSE marking scheme uses $2x+3y=13$ as the intended line II, which perfectly satisfies $2(5)+3(1)=13$.
Mistakes to Avoid:
!
Typo Alert: The original CBSE Question Paper has a typo in statement II ($2x-3y=13$). If you calculate $2(5)-3(1)$ it equals 7. However, the official marking scheme dictates the correct answer is B, assuming line II was intended to be $2x+3y=13$ or expecting students to intuitively select the only non-parallel option that closely resembles a valid solution.
Official CBSE Solution:
Check line I: $3x-3y=18$. This simplifies to $x-y=6$.
This makes the system of equations consistent and dependent (coincident lines). This means infinite solutions, not an "only point".
Only II fulfills the given condition of having only one point of intersection (intersecting lines).
(Note: The CBSE Marking Scheme explicitly writes II as $2x+3y=13$ in their solution text, correcting the typo in the question paper.)
Ans: Option (B) II only
🎯 CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
4
If $(1-p)$ is a root of the quadratic equation $x^{2}+px+1-p=0$, then its roots are:
A. 0, 1
B. -1, 1
C. 0, -1
D. -1, 2
Concept:
1
If a value is a root of an equation, substituting that value for $x$ will satisfy the equation (make it equal zero). Substitute $x = (1-p)$ to solve for $p$. Once $p$ is known, rewrite the original equation and find its actual numerical roots.
Hint:
1
Substitute $x = (1-p)$: $(1-p)^2 + p(1-p) + 1-p = 0$.
2
Factor out $(1-p)$: $(1-p) [ (1-p) + p + 1 ] = 0$.
Mistakes to Avoid:
!
Expanding unnecessarily: While expanding $(1-p)^2 = 1 - 2p + p^2$ works, it creates a messy quadratic in terms of $p$. Factoring out the common term $(1-p)$ is much faster and cleaner.
Official CBSE Solution:
As $(1-p)$ is a root of the quadratic equation $x^2+px+1-p=0$:
$\therefore (1-p)^2 + p(1-p) + 1-p = 0$
Factor out $(1-p)$:
$(1-p) [ (1-p) + p + 1 ] = 0$
$(1-p) [ 1 - p + p + 1 ] = 0$
$(1-p) [ 2 ] = 0 \implies 1-p = 0 \implies p = 1$
Now quadratic equation will be:
$x^2 + 1x + 1-1 = 0 \implies x^2 + x = 0$
$x(x+1) = 0$
So the roots are $0, -1$.
Ans: Option (C) 0, -1
🎯 CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
5
The middle term of the A.P.: 10, 7, 4, ..., -62 is:
A. -26
B. -29
C. -32
D. -35
Concept:
1
To find the middle term, first find the total number of terms ($n$) using the $n^{th}$ term formula: $a_n = a + (n-1)d$. If $n$ is odd, the middle term is the $(\frac{n+1}{2})^{th}$ term.
Hint:
1
Here, $a = 10$ and $d = 7 - 10 = -3$. Last term $a_n = -62$.
2
Solve $-62 = 10 + (n-1)(-3)$ for $n$.
Mistakes to Avoid:
!
Averaging the first and last term: Calculating $(10 + (-62)) / 2 = -26$ works as a shortcut *only* for arithmetic progressions. However, the standard method proves it. Be careful, averaging terms is finding the arithmetic mean, which happens to equal the middle term perfectly in a linear AP!
Official CBSE Solution:
Find total number of terms ($n$):
$-62 = 10 + (n-1)(-3)$
$-72 = -3(n-1) \implies n-1 = 24 \implies n = 25$
Find the middle term index:
Middle term = $\frac{25+1}{2} = 13^{th}$ term.
Calculate the 13th term:
$a_{13} = 10 + 12(-3)$
$a_{13} = 10 - 36 = -26$
Ans: Option (A) -26
🎯 CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
6
The perimeters of two similar triangles are 56 cm and 70 cm respectively. If one side of the first triangle is 14 cm, then the corresponding side of the second triangle (in cm) is:
A. 5
B. 7.5
C. 10
D. 17.5
Concept:
1
For similar triangles, the ratio of their perimeters is exactly equal to the ratio of their corresponding sides.
Hint:
1
Set up the proportion: $\frac{56}{70} = \frac{14}{\text{side}}$.
Mistakes to Avoid:
!
Squaring the ratio: Do not confuse perimeter ratios with area ratios. Area ratio = $(\text{Side Ratio})^2$. Perimeter ratio = Side Ratio.
Official CBSE Solution:
Ratio of perimeters of similar triangles = ratio of their corresponding sides.
$\frac{56}{70} = \frac{14}{\text{corresponding side}}$
$\text{corresponding side} = \frac{14 \times 70}{56}$
$\text{corresponding side} = \frac{70}{4} = 17.5$
Ans: Option (D) 17.5
🎯 CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
7
The point which lies on the perpendicular bisector of the line segment joining the points A (-3,-4) and B (3, 4) is:
A. (0, 0)
B. (0, 3)
C. (3, 0)
D. (0, 4)
Concept:
1
A perpendicular bisector of a line segment passes exactly through its mid-point. Therefore, the mid-point of the segment AB will inherently lie on its perpendicular bisector.
Hint:
1
Use the Mid-point formula: $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$.
2
Substitute A(-3, -4) and B(3, 4).
Mistakes to Avoid:
!
Calculating the full equation of the line: You don't need to find slopes and write the equation of the bisector. Just find the midpoint, as it is guaranteed to lie on the bisector and happens to exactly match one of the simple options.
Official CBSE Solution:
Mid-point of AB will lie on the perpendicular bisector of the line segment joining the points A(-3, -4) and B(3, 4).
Coordinates of mid-point of A and B:
$= \left(\frac{-3+3}{2}, \frac{-4+4}{2}\right)$
$= \left(\frac{0}{2}, \frac{0}{2}\right) = (0, 0)$
Ans: Option (A) (0, 0)
🎯 CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
8
If $\tan(A+B)=\sqrt{3}$ and $\tan(A-B)=\frac{1}{\sqrt{3}}$, $0^\circ < A+B < 90^\circ$, $A>B$, then the value of A and B respectively are:
A. $60^\circ, 30^\circ$
B. $60^\circ, 45^\circ$
C. $45^\circ, 15^\circ$
D. $60^\circ, 15^\circ$
Concept:
1
Match the trigonometric values to standard angles. $\tan(60^\circ) = \sqrt{3}$ and $\tan(30^\circ) = 1/\sqrt{3}$. Create a system of linear equations for $(A+B)$ and $(A-B)$ and solve simultaneously.
Hint:
1
$A + B = 60^\circ$.
2
$A - B = 30^\circ$. Add the two equations together.
Mistakes to Avoid:
!
Assuming A and B are the standard angles: Do not choose Option A ($60^\circ, 30^\circ$). $60^\circ$ and $30^\circ$ are the sums and differences, not the individual angles themselves.
Official CBSE Solution:
$\tan(A+B) = \sqrt{3} \implies (A+B) = 60^\circ$ ---- (i)
$\tan(A-B) = \frac{1}{\sqrt{3}} \implies (A-B) = 30^\circ$ ---- (ii)
Solving equation (i) and (ii) by adding them:
$(A+B) + (A-B) = 60^\circ + 30^\circ$
$2A = 90^\circ \implies A = 45^\circ$
Substitute A back into (i):
$45^\circ + B = 60^\circ \implies B = 15^\circ$
Ans: Option (C) $45^\circ, 15^\circ$
🎯 CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
9
T-shirts marked with numbers 4 to 99 are placed in a box. Gunika is fond of numbers. She randomly takes out a T-shirt from this box. The probability that she gets a T-shirt marked with a number that is either a perfect square or a perfect cube is:
A. $\frac{1}{12}$
B. $\frac{1}{32}$
C. $\frac{11}{96}$
D. $\frac{5}{48}$
Concept:
1
First, calculate the exact total number of outcomes (T-shirts from 4 to 99). Next, list all perfect squares in that range, and all perfect cubes in that range. Be careful to subtract any duplicates (numbers that are BOTH squares and cubes) to avoid double counting.
Hint:
1
Total outcomes = $99 - 4 + 1 = 96$.
2
Perfect squares from 4 to 99: {4, 9, 16, 25, 36, 49, 64, 81} (8 items).
3
Perfect cubes from 4 to 99: {8, 27, 64} (3 items). Is there an overlap?
Mistakes to Avoid:
!
Double counting 64: 64 is both $8^2$ and $4^3$. If you just add $8 + 3 = 11$, you get Option C ($\frac{11}{96}$). You must count 64 only once!
Official CBSE Solution:
Total outcomes = 96 (Numbers 4 to 99).
Perfect square numbers in range:
4, 9, 16, 25, 36, 49, 64, 81 (Count = 8)
Perfect cube numbers in range:
8, 27, 64 (Count = 3)
Common number is 64.
Number of favourable outcomes = $8 + 3 - 1 = 10$.
$P(\text{perfect square or cube}) = \frac{10}{96} = \frac{5}{48}$
Ans: Option (D) $\frac{5}{48}$
🎯 CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
10
The diameter of a car wheel is 21 cm. The number of complete revolutions it will make in moving 66 km is:
A. $10^4$
B. $10^5$
C. $10^6$
D. $10^7$
Concept:
1
The distance covered in 1 complete revolution equals the circumference of the wheel ($\pi \times d$). To find the total revolutions, divide the total distance by the distance of one revolution. Remember to convert all units to be identical (km to cm).
Hint:
1
Circumference = $\frac{22}{7} \times 21 = 66 \text{ cm}$.
2
Convert 66 km to cm: $66 \times 1000 \times 100 = 66,000,000 \text{ cm}$.
Mistakes to Avoid:
!
Unit conversion failure: If you forget to convert km to cm, or if you only convert km to m (missing the extra two zeroes for cm), you will get the wrong exponent ($10^4$ or $10^6$).
Official CBSE Solution:
Total distance = $66 \text{ km} = 66 \times 10^5 \text{ cm}$.
Distance per revolution = Circumference = $\pi d$.
Circumference = $\frac{22}{7} \times 21 = 66 \text{ cm}$.
Total revolutions:
$= \frac{\text{Total Distance}}{\text{Circumference}}$
$= \frac{66 \times 10^5}{66}$
$= 10^5$
Ans: Option (B) $10^5$
🎯 CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
