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4 results found for "estimated quantity" in Class 10.

Question Easy Mathematics Quadratic Equations Nature of Roots Class 10 Level 38

द्विघात समीकरण \(ax^2+bx+c=0\) में मूलों की प्रकृति जानने के लिए किस राशि का उपयोग किया जाता है?

Which quantity is used to know the nature of roots of the quadratic equation \(ax^2+bx+c=0\)?

Explanation opens after your attempt
Correct Answer

A. विविक्तकर \(D=b^2-4ac\)Discriminant \(D=b^2-4ac\)

Step 1

Concept

The nature of roots is decided by \(D=b^2-4ac\). In exams first identify (a), (b), and (c).

Step 2

Why this answer is correct

The correct answer is A. विविक्तकर \(D=b^2-4ac\) / Discriminant \(D=b^2-4ac\). The nature of roots is decided by \(D=b^2-4ac\). In exams first identify (a), (b), and (c).

Step 3

Exam Tip

मूलों की प्रकृति \(D=b^2-4ac\) से तय होती है। परीक्षा में पहले (a), (b), (c) पहचानें।

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Question Medium Mathematics Quadratic Equations Word Problems and Applications Class 10 Level 42

एक वस्तु पर (1200) रुपये खर्च होते हैं। यदि प्रति वस्तु मूल्य (10) रुपये कम हो जाए तो उसी धन से (4) वस्तुएँ अधिक मिलती हैं। मूल प्रति वस्तु मूल्य क्या था?

A total of (1200) rupees is spent on some items. If the price per item decreases by (10) rupees, (4) more items can be bought for the same money. What was the original price per item?

Explanation opens after your attempt
Correct Answer

B. (60) रुपये(60) rupees

Step 1

Concept

If original price is (x), then \(\frac{1200}{x-10}-\frac{1200}{x}=4\). Solving gives (x=60).

Step 2

Why this answer is correct

The correct answer is B. (60) रुपये / (60) rupees. If original price is (x), then \(\frac{1200}{x-10}-\frac{1200}{x}=4\). Solving gives (x=60).

Step 3

Exam Tip

मूल मूल्य (x) हो तो \(\frac{1200}{x-10}-\frac{1200}{x}=4\)। हल करने पर (x=60) मिलता है।

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Question Expert Mathematics Quadratic Equations Word Problems and Applications Class 10 Level 41

एक वस्तु की कीमत (₹x) है। (₹5) कम कीमत पर खरीदी जाए तो (₹300) में (3) वस्तुएँ अधिक मिलती हैं। मूल कीमत क्या है?

The price of an article is (₹x). If it is bought at (₹5) less, (3) more articles can be bought for (₹300). What is the original price?

Explanation opens after your attempt
Correct Answer

B. (₹25)

Step 1

Concept

Let original price be (x), then \(\frac{300}{x-5}-\frac{300}{x}=3\). This gives \(x^2-5x-500=0\), so (x=25).

Step 2

Why this answer is correct

The correct answer is B. (₹25). Let original price be (x), then \(\frac{300}{x-5}-\frac{300}{x}=3\). This gives \(x^2-5x-500=0\), so (x=25).

Step 3

Exam Tip

मूल कीमत (x) हो, तो \(\frac{300}{x-5}-\frac{300}{x}=3\)। इससे \(x^2-5x-500=0\), इसलिए (x=25)।

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Question Expert Mathematics Quadratic Equations Word Problems and Applications Class 10 Level 41

एक दुकानदार (₹800) में कुछ पेन खरीदता है। यदि प्रत्येक पेन की कीमत (₹2) कम होती, तो वह (20) पेन अधिक खरीदता। वास्तविक प्रति पेन कीमत क्या थी?

A shopkeeper buys some pens for (₹800). If each pen cost (₹2) less, he would buy (20) more pens. What was the actual price per pen?

Explanation opens after your attempt
Correct Answer

B. (₹10)

Step 1

Concept

Let the actual price be (x), then \(\frac{800}{x-2}-\frac{800}{x}=20\). This gives \(x^2-2x-80=0\), so (x=10).

Step 2

Why this answer is correct

The correct answer is B. (₹10). Let the actual price be (x), then \(\frac{800}{x-2}-\frac{800}{x}=20\). This gives \(x^2-2x-80=0\), so (x=10).

Step 3

Exam Tip

वास्तविक कीमत (x) हो, तो \(\frac{800}{x-2}-\frac{800}{x}=20\)। इससे \(x^2-2x-80=0\), इसलिए (x=10)।

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