Class 10 equation MCQ Questions

Question 1/19 Expert Mathematics Arithmetic Progressions (AP) Finding the sum of the first $n$ terms of an AP Class 10 Level 67

समांतर श्रेढ़ी $11,20,29,\ldots$ के पहले कितने पदों का योग (3973) होगा?

How many first terms of the AP $11,20,29,\ldots$ have sum (3973)?

Explanation opens after your attempt

Correct Answer

B. (29)

Step 1

Concept

Solving (\frac{n}{2}[22+9(n-1)]=3973) gives (n=29). The number of terms must be a positive integer.

Step 2

Why this answer is correct

The correct answer is B. (29). Solving (\frac{n}{2}[22+9(n-1)]=3973) gives (n=29). The number of terms must be a positive integer.

Step 3

Exam Tip

(\frac{n}{2}[22+9(n-1)]=3973) हल करने पर (n=29) मिलता है। पदों की संख्या धनात्मक पूर्णांक होनी चाहिए।

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Question 2/19 Hard Mathematics Arithmetic Progressions (AP) Finding the sum of the first $n$ terms of an AP Class 10 Level 69

समांतर श्रेढ़ी $16,25,34,\ldots$ के पहले कितने पदों का योग (2030) होगा?

How many first terms of the AP $16,25,34,\ldots$ have sum (2030)?

Explanation opens after your attempt

Correct Answer

A. (20)

Step 1

Concept

Solving (\frac{n}{2}[32+9(n-1)]=2030) gives (n=20). The number of terms must be a positive integer.

Step 2

Why this answer is correct

The correct answer is A. (20). Solving (\frac{n}{2}[32+9(n-1)]=2030) gives (n=20). The number of terms must be a positive integer.

Step 3

Exam Tip

(\frac{n}{2}[32+9(n-1)]=2030) हल करने पर (n=20) मिलता है। पदों की संख्या धनात्मक पूर्णांक होनी चाहिए।

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Question 3/19 Hard Mathematics Arithmetic Progressions (AP) Finding the sum of the first $n$ terms of an AP Class 10 Level 68

समांतर श्रेढ़ी $11,18,25,\ldots$ के पहले कितने पदों का योग (1701) होगा?

How many first terms of the AP $11,18,25,\ldots$ have sum (1701)?

Explanation opens after your attempt

Correct Answer

C. (21)

Step 1

Concept

Solving (\frac{n}{2}[22+7(n-1)]=1701) gives (n=21). The number of terms must be a positive integer.

Step 2

Why this answer is correct

The correct answer is C. (21). Solving (\frac{n}{2}[22+7(n-1)]=1701) gives (n=21). The number of terms must be a positive integer.

Step 3

Exam Tip

(\frac{n}{2}[22+7(n-1)]=1701) हल करने पर (n=21) मिलता है। पदों की संख्या धनात्मक पूर्णांक होनी चाहिए।

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Question 4/19 Hard Mathematics Arithmetic Progressions (AP) Finding the sum of the first $n$ terms of an AP Class 10 Level 67

समांतर श्रेढ़ी $7,12,17,\ldots$ के पहले कितने पदों का योग (1090) होगा?

How many first terms of the AP $7,12,17,\ldots$ have sum (1090)?

Explanation opens after your attempt

Correct Answer

C. (20)

Step 1

Concept

Solving (\frac{n}{2}[14+5(n-1)]=1090) gives (n=20). The number of terms must always be a positive integer.

Step 2

Why this answer is correct

The correct answer is C. (20). Solving (\frac{n}{2}[14+5(n-1)]=1090) gives (n=20). The number of terms must always be a positive integer.

Step 3

Exam Tip

(\frac{n}{2}[14+5(n-1)]=1090) हल करने पर (n=20) मिलता है। पदों की संख्या हमेशा धनात्मक पूर्णांक होनी चाहिए।

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Question 5/19 Hard Mathematics Polynomials Representing real numbers on the number line Class 10 Level 49

संख्या रेखा पर (x) ऐसा है कि (|x-2|=3), तो (x) के मान क्या होंगे?

On the number line, (x) satisfies (|x-2|=3). What are the values of (x)?

Explanation opens after your attempt

Correct Answer

A. (-1) और (5)(-1) and (5)

Step 1

Concept

(|x-2|=3) means (x) is (3) units away from (2), so (x=-1,5). Place the distance on both sides of the center.

Step 2

Why this answer is correct

The correct answer is A. (-1) और (5) / (-1) and (5). (|x-2|=3) means (x) is (3) units away from (2), so (x=-1,5). Place the distance on both sides of the center.

Step 3

Exam Tip

(|x-2|=3) का अर्थ है (x), (2) से (3) इकाई दूर है, इसलिए (x=-1,5)। केंद्र से दोनों ओर दूरी लगाएँ।

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Question 6/19 Medium Mathematics Polynomials Representing real numbers on the number line Class 10 Level 51

यदि $\sqrt{a}$ संख्या रेखा पर ठीक (4.5) पर है तो (a) का मान क्या होगा?

If $\sqrt{a}$ is exactly at (4.5) on the number line, what will be the value of (a)?

Explanation opens after your attempt

Correct Answer

C. (20.25)

Step 1

Concept

If $\sqrt{a}=4.5$, then (a=(4.5)²=20.25). Square both sides to remove the square root.

Step 2

Why this answer is correct

The correct answer is C. (20.25). If $\sqrt{a}=4.5$, then (a=(4.5)²=20.25). Square both sides to remove the square root.

Step 3

Exam Tip

$\sqrt{a}=4.5$ होने पर (a=(4.5)²=20.25) है। वर्गमूल हटाने के लिए वर्ग करें।

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Question 7/19 Medium Mathematics Polynomials Polynomials in one variable Class 10 Level 48

(p(x)=9x-27) का शून्य क्या है?

What is the zero of (p(x)=9x-27)?

Explanation opens after your attempt

Correct Answer

B. (3)

Step 1

Concept

From (9x-27=0), we get (x=3). Find the zero of a linear polynomial by solving the equation.

Step 2

Why this answer is correct

The correct answer is B. (3). From (9x-27=0), we get (x=3). Find the zero of a linear polynomial by solving the equation.

Step 3

Exam Tip

(9x-27=0) से (x=3) मिलता है। रैखिक बहुपद का शून्य समीकरण हल करके निकालें।

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Question 8/19 Medium Mathematics Polynomials Polynomials in one variable Class 10 Level 47

यदि (p(x)=4x-12), तो किस मान पर (p(x)=0) होगा?

If (p(x)=4x-12), for which value will (p(x)=0)?

Explanation opens after your attempt

Correct Answer

B. (3)

Step 1

Concept

From (4x-12=0), we get (4x=12) and (x=3). This is the zero of the polynomial.

Step 2

Why this answer is correct

The correct answer is B. (3). From (4x-12=0), we get (4x=12) and (x=3). This is the zero of the polynomial.

Step 3

Exam Tip

(4x-12=0) से (4x=12) और (x=3) मिलता है। यह बहुपद का शून्य है।

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Question 9/19 Medium Mathematics Polynomials Polynomials in one variable Class 10 Level 46

(p(x)=7x-14) का शून्य क्या है?

What is the zero of (p(x)=7x-14)?

Explanation opens after your attempt

Correct Answer

B. (2)

Step 1

Concept

From (7x-14=0), we get (x=2). Solve the equation to find the zero of a linear polynomial.

Step 2

Why this answer is correct

The correct answer is B. (2). From (7x-14=0), we get (x=2). Solve the equation to find the zero of a linear polynomial.

Step 3

Exam Tip

(7x-14=0) से (x=2) मिलता है। रैखिक बहुपद का शून्य जल्दी निकालने के लिए समीकरण हल करें।

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Question 10/19 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि $x^2-16x+48=0$ के मूल $\alpha,\beta$ हैं, तो नए मूल $\alpha+6,\beta+6$ वाला समीकरण कौनसा है?

If roots of $x^2-16x+48=0$ are $\alpha,\beta$, which equation has roots $\alpha+6,\beta+6$?

Explanation opens after your attempt

Correct Answer

A. $x^2-28x+180=0$

Step 1

Concept

The roots are (4,12), so new roots are (10,18), and the equation is ((x-10)(x-18)=0). In exams, form the new roots and then the new equation.

Step 2

Why this answer is correct

The correct answer is A. $x^2-28x+180=0$. The roots are (4,12), so new roots are (10,18), and the equation is ((x-10)(x-18)=0). In exams, form the new roots and then the new equation.

Step 3

Exam Tip

मूल (4,12) हैं, इसलिए नए मूल (10,18) होंगे और समीकरण ((x-10)(x-18)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।

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Question 11/19 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि $x^2-14x+33=0$ के मूल $\alpha,\beta$ हैं, तो नए मूल $\alpha+5,\beta+5$ वाला समीकरण कौनसा है?

If roots of $x^2-14x+33=0$ are $\alpha,\beta$, which equation has roots $\alpha+5,\beta+5$?

Explanation opens after your attempt

Correct Answer

A. $x^2-24x+128=0$

Step 1

Concept

The roots are (3,11), so new roots are (8,16), and the equation is ((x-8)(x-16)=0). In exams, form the new roots and then the new equation.

Step 2

Why this answer is correct

The correct answer is A. $x^2-24x+128=0$. The roots are (3,11), so new roots are (8,16), and the equation is ((x-8)(x-16)=0). In exams, form the new roots and then the new equation.

Step 3

Exam Tip

मूल (3,11) हैं, इसलिए नए मूल (8,16) होंगे और समीकरण ((x-8)(x-16)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।

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Question 12/19 Expert Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि $x^2-12x+20=0$ के मूल $\alpha,\beta$ हैं, तो नए मूल $\alpha+4,\beta+4$ वाला समीकरण कौनसा है?

If roots of $x^2-12x+20=0$ are $\alpha,\beta$, which equation has roots $\alpha+4,\beta+4$?

Explanation opens after your attempt

Correct Answer

A. $x^2-20x+84=0$

Step 1

Concept

The roots are (2,10), so new roots are (6,14), and the equation is ((x-6)(x-14)=0). In exams, form the new roots and then the new equation.

Step 2

Why this answer is correct

The correct answer is A. $x^2-20x+84=0$. The roots are (2,10), so new roots are (6,14), and the equation is ((x-6)(x-14)=0). In exams, form the new roots and then the new equation.

Step 3

Exam Tip

मूल (2,10) हैं, इसलिए नए मूल (6,14) होंगे और समीकरण ((x-6)(x-14)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।

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Question 13/19 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36

यदि $x^2-10x+16=0$ के मूल $\alpha,\beta$ हैं, तो नए मूल $\alpha+3,\beta+3$ वाला समीकरण कौनसा है?

If roots of $x^2-10x+16=0$ are $\alpha,\beta$, which equation has roots $\alpha+3,\beta+3$?

Explanation opens after your attempt

Correct Answer

A. $x^2-16x+55=0$

Step 1

Concept

The roots are (2,8), so new roots are (5,11), and the equation is ((x-5)(x-11)=0). In exams, form the new roots and then the new equation.

Step 2

Why this answer is correct

The correct answer is A. $x^2-16x+55=0$. The roots are (2,8), so new roots are (5,11), and the equation is ((x-5)(x-11)=0). In exams, form the new roots and then the new equation.

Step 3

Exam Tip

मूल (2,8) हैं, इसलिए नए मूल (5,11) होंगे और समीकरण ((x-5)(x-11)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।

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Question 14/19 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35

यदि $x^2-8x+12=0$ के मूल $\alpha,\beta$ हैं, तो नए मूल $\alpha+2,\beta+2$ वाला समीकरण कौनसा है?

If roots of $x^2-8x+12=0$ are $\alpha,\beta$, which equation has roots $\alpha+2,\beta+2$?

Explanation opens after your attempt

Correct Answer

A. $x^2-12x+32=0$

Step 1

Concept

The roots are (2,6), so new roots are (4,8), and the equation is ((x-4)(x-8)=0). In exams, form the new roots and then the new equation.

Step 2

Why this answer is correct

The correct answer is A. $x^2-12x+32=0$. The roots are (2,6), so new roots are (4,8), and the equation is ((x-4)(x-8)=0). In exams, form the new roots and then the new equation.

Step 3

Exam Tip

मूल (2,6) हैं, इसलिए नए मूल (4,8) होंगे और समीकरण ((x-4)(x-8)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।

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Question 15/19 Hard Mathematics Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34

यदि $x^2-6x+5=0$ के मूल $\alpha,\beta$ हैं, तो नए मूल $\alpha+1,\beta+1$ वाला समीकरण कौनसा है?

If roots of $x^2-6x+5=0$ are $\alpha,\beta$, which equation has roots $\alpha+1,\beta+1$?

Explanation opens after your attempt

Correct Answer

A. $x^2-8x+12=0$

Step 1

Concept

The roots are (1,5), so new roots are (2,6), and the equation is ((x-2)(x-6)=0). In exams, form the new roots and then the new equation.

Step 2

Why this answer is correct

The correct answer is A. $x^2-8x+12=0$. The roots are (1,5), so new roots are (2,6), and the equation is ((x-2)(x-6)=0). In exams, form the new roots and then the new equation.

Step 3

Exam Tip

मूल (1,5) हैं, इसलिए नए मूल (2,6) होंगे और समीकरण ((x-2)(x-6)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।

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Question 16/19 Hard Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32

यदि $x^2-4x+3=0$ के मूल $\alpha$ और $\beta$ हैं तो $3\alpha$ और $3\beta$ को मूल मानकर समीकरण कौन सा होगा?

If $\alpha$ and $\beta$ are roots of $x^2-4x+3=0$, which equation has $3\alpha$ and $3\beta$ as roots?

Explanation opens after your attempt

Correct Answer

A. $x^2-12x+27=0$

Step 1

Concept

The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is $x^2-12x+27=0$.

Step 2

Why this answer is correct

The correct answer is A. $x^2-12x+27=0$. The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is $x^2-12x+27=0$.

Step 3

Exam Tip

पुराने योग (4) और गुणनफल (3) हैं। नए योग (12) और गुणनफल (27) होंगे इसलिए $x^2-12x+27=0$ है।

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Question 17/19 Hard Mathematics Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31

यदि $x^2-3x+2=0$ के मूल $\alpha$ और $\beta$ हैं तो $2\alpha$ और $2\beta$ को मूल मानकर समीकरण कौन सा होगा?

If $\alpha$ and $\beta$ are roots of $x^2-3x+2=0$, which equation has $2\alpha$ and $2\beta$ as roots?

Explanation opens after your attempt

Correct Answer

A. $x^2-6x+8=0$

Step 1

Concept

The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is $x^2-6x+8=0$.

Step 2

Why this answer is correct

The correct answer is A. $x^2-6x+8=0$. The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is $x^2-6x+8=0$.

Step 3

Exam Tip

पुराने योग (3) और गुणनफल (2) हैं। नए योग (6) और गुणनफल (8) होंगे इसलिए $x^2-6x+8=0$ है।

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Question 18/19 Hard Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 26

यदि $x=\sqrt{3}+\sqrt{2}$, तो (x) किस द्विघात समीकरण को संतुष्ट करता है?

If $x=\sqrt{3}+\sqrt{2}$, which quadratic equation does (x) satisfy?

Explanation opens after your attempt

Correct Answer

C. $x^4-10x^2+1=0$

Step 1

Concept

Here $x^2=5+2\sqrt{6}$ and then ($x^2-5$²=24), so $x^4-10x^2+1=0$. In exams a sum of two radicals may lead to a fourth-degree relation.

Step 2

Why this answer is correct

The correct answer is C. $x^4-10x^2+1=0$. Here $x^2=5+2\sqrt{6}$ and then ($x^2-5$²=24), so $x^4-10x^2+1=0$. In exams a sum of two radicals may lead to a fourth-degree relation.

Step 3

Exam Tip

$x^2=5+2\sqrt{6}$ और फिर ($x^2-5$²=24), इसलिए $x^4-10x^2+1=0$ है। परीक्षा में दो वर्गमूलों के योग से कभी द्विघात नहीं बल्कि चतुर्थ घात संबंध भी बन सकता है।

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Question 19/19 Expert Mathematics Real Numbers Proof of irrationality of √2, √3, √5 Class 10 Level 16

$\sqrt{3}$ की अपरिमेयता में $p^2=3q^2$ से $p^2$ किस कारण (3) से विभाज्य है?

In the irrationality proof of $\sqrt{3}$, why is $p^2$ divisible by (3) from $p^2=3q^2$?

Explanation opens after your attempt

Correct Answer

A. क्योंकि दाएँ पक्ष में (3) गुणक के रूप में हैBecause (3) appears as a factor on the right side

Step 1

Concept

In $p^2=3q^2$, the right side is a multiple of (3).

Step 2

Why this answer is correct

Since both sides are equal, $p^2$ is also a multiple of (3).

Step 3

Exam Tip

Understand divisibility of the square first, then of the original number. चरण 1: $p^2=3q^2$ में दायाँ पक्ष (3) का गुणज है। चरण 2: बराबरी के कारण $p^2$ भी (3) का गुणज होगा। चरण 3: पहले वर्ग की विभाज्यता समझें, फिर मूल संख्या की।

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equation MCQ Questions for Class 10

Practice Questions

समांतर श्रेढ़ी \(11,20,29,\ldots\) के पहले कितने पदों का योग (3973) होगा?

Concept

Why this answer is correct

Exam Tip

समांतर श्रेढ़ी \(16,25,34,\ldots\) के पहले कितने पदों का योग (2030) होगा?

Concept

Why this answer is correct

Exam Tip

समांतर श्रेढ़ी \(11,18,25,\ldots\) के पहले कितने पदों का योग (1701) होगा?

Concept

Why this answer is correct

Exam Tip

समांतर श्रेढ़ी \(7,12,17,\ldots\) के पहले कितने पदों का योग (1090) होगा?

Concept

Why this answer is correct

Exam Tip

संख्या रेखा पर (x) ऐसा है कि (|x-2|=3), तो (x) के मान क्या होंगे?

Concept

Why this answer is correct

Exam Tip

यदि \(\sqrt{a}\) संख्या रेखा पर ठीक (4.5) पर है तो (a) का मान क्या होगा?

Concept

Why this answer is correct

Exam Tip

(p(x)=9x-27) का शून्य क्या है?

Concept

Why this answer is correct

Exam Tip

यदि (p(x)=4x-12), तो किस मान पर (p(x)=0) होगा?

Concept

Why this answer is correct

Exam Tip

(p(x)=7x-14) का शून्य क्या है?

Concept

Why this answer is correct

Exam Tip

यदि \(x^2-16x+48=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+6,\beta+6\) वाला समीकरण कौनसा है?

Concept

Why this answer is correct

Exam Tip

यदि \(x^2-14x+33=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+5,\beta+5\) वाला समीकरण कौनसा है?

Concept

Why this answer is correct

Exam Tip

यदि \(x^2-12x+20=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+4,\beta+4\) वाला समीकरण कौनसा है?

Concept

Why this answer is correct

Exam Tip

यदि \(x^2-10x+16=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+3,\beta+3\) वाला समीकरण कौनसा है?

Concept

Why this answer is correct

Exam Tip

यदि \(x^2-8x+12=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+2,\beta+2\) वाला समीकरण कौनसा है?

Concept

Why this answer is correct

Exam Tip

यदि \(x^2-6x+5=0\) के मूल \(\alpha,\beta\) हैं, तो नए मूल \(\alpha+1,\beta+1\) वाला समीकरण कौनसा है?

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यदि \(x^2-4x+3=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(3\alpha\) और \(3\beta\) को मूल मानकर समीकरण कौन सा होगा?

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यदि \(x^2-3x+2=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(2\alpha\) और \(2\beta\) को मूल मानकर समीकरण कौन सा होगा?

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यदि \(x=\sqrt{3}+\sqrt{2}\), तो (x) किस द्विघात समीकरण को संतुष्ट करता है?

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\(\sqrt{3}\) की अपरिमेयता में \(p^2=3q^2\) से \(p^2\) किस कारण (3) से विभाज्य है?

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