Class 11 Mathematics - Trigonometric Functions - Angles Easy Quiz

Mathematics Trigonometric Functions Angles Easy Class 11 Level 54

Level 54 • 50/50 questions • 40 seconds per question.

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Question 1/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक पहचान कोड में पहला स्थान (4) अक्षरों से दूसरा स्थान (5) अंकों से और तीसरा स्थान (2) प्रतीकों से भरा जाता है। कुल कोड कितने बनेंगे?

In an identification code the first place is filled from (4) letters the second from (5) digits and the third from (2) symbols. How many codes can be formed?

Explanation opens after your attempt
Correct Answer

A. (40) कोड(40) codes

Step 1

Concept

For three independent places \(4 \times 5 \times 2=40\). Use the multiplication principle when stages occur together.

Step 2

Why this answer is correct

The correct answer is A. (40) कोड / (40) codes. For three independent places \(4 \times 5 \times 2=40\). Use the multiplication principle when stages occur together.

Step 3

Exam Tip

तीनों स्वतंत्र स्थानों के लिए \(4 \times 5 \times 2=40\)। चरण एक साथ हों तो गुणन सिद्धांत लगाएं।

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Question 2/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक भोजन पैक में (3) मुख्य व्यंजन (4) पेय और (2) मिठाई विकल्प हैं। एक-एक वस्तु चुनकर कितने पैक बनेंगे?

A meal pack has (3) main dishes (4) drinks and (2) dessert options. How many packs can be made by choosing one item from each?

Explanation opens after your attempt
Correct Answer

B. (24) पैक(24) packs

Step 1

Concept

Total packs are \(3 \times 4 \times 2=24\). Multiply the choices when one item is selected from each category.

Step 2

Why this answer is correct

The correct answer is B. (24) पैक / (24) packs. Total packs are \(3 \times 4 \times 2=24\). Multiply the choices when one item is selected from each category.

Step 3

Exam Tip

कुल पैक \(3 \times 4 \times 2=24\) होंगे। हर श्रेणी से एक चयन हो तो विकल्पों को गुणा करें।

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Question 3/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक झंडे में (5) रंगों में से ऊपरी पट्टी (3) रंगों में से बीच की पट्टी और (4) रंगों में से निचली पट्टी चुनी जाती है। कुल झंडे कितने बनेंगे?

In a flag the top stripe is chosen from (5) colors the middle stripe from (3) colors and the bottom stripe from (4) colors. How many flags can be made?

Explanation opens after your attempt
Correct Answer

C. (60) झंडे(60) flags

Step 1

Concept

Each stripe choice is independent so \(5 \times 3 \times 4=60\). Multiply for different positions.

Step 2

Why this answer is correct

The correct answer is C. (60) झंडे / (60) flags. Each stripe choice is independent so \(5 \times 3 \times 4=60\). Multiply for different positions.

Step 3

Exam Tip

हर पट्टी का चुनाव स्वतंत्र है इसलिए \(5 \times 3 \times 4=60\)। अलग-अलग स्थानों के लिए गुणा करें।

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Question 4/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक छात्र को (2) बैग (6) पेन और (3) कॉपियों में से एक-एक चुनना है। कुल चयन कितने हैं?

A student has to choose one each from (2) bags (6) pens and (3) notebooks. How many total selections are possible?

Explanation opens after your attempt
Correct Answer

D. (36) चयन(36) selections

Step 1

Concept

Total selections are \(2 \times 6 \times 3=36\). For three independent choices total ways are found by multiplication.

Step 2

Why this answer is correct

The correct answer is D. (36) चयन / (36) selections. Total selections are \(2 \times 6 \times 3=36\). For three independent choices total ways are found by multiplication.

Step 3

Exam Tip

कुल चयन \(2 \times 6 \times 3=36\) हैं। तीन स्वतंत्र विकल्पों में कुल तरीके गुणा से मिलते हैं।

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Question 5/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक यात्रा योजना में (4) शहर (5) होटल और (2) वाहन विकल्प हैं। एक शहर एक होटल और एक वाहन चुनकर कुल योजनाएं कितनी बनेंगी?

A travel plan has (4) cities (5) hotels and (2) vehicle options. How many plans can be made by choosing one city one hotel and one vehicle?

Explanation opens after your attempt
Correct Answer

B. (40) योजनाएं(40) plans

Step 1

Concept

City hotel and vehicle all must be chosen so \(4 \times 5 \times 2=40\). Do not add independent stages multiply them.

Step 2

Why this answer is correct

The correct answer is B. (40) योजनाएं / (40) plans. City hotel and vehicle all must be chosen so \(4 \times 5 \times 2=40\). Do not add independent stages multiply them.

Step 3

Exam Tip

शहर होटल और वाहन तीनों चुनने हैं इसलिए \(4 \times 5 \times 2=40\)। स्वतंत्र चरणों को जोड़ें नहीं गुणा करें।

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Question 6/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक वर्दी में (3) कमीज (2) पतलून और (4) जूते विकल्प हैं। पूरी वर्दी कितने तरीकों से बनाई जा सकती है?

A uniform has (3) shirt options (2) trouser options and (4) shoe options. In how many ways can a complete uniform be made?

Explanation opens after your attempt
Correct Answer

A. (24) तरीके(24) ways

Step 1

Concept

For the complete uniform \(3 \times 2 \times 4=24\). The multiplication principle is correct when one choice is made for each part.

Step 2

Why this answer is correct

The correct answer is A. (24) तरीके / (24) ways. For the complete uniform \(3 \times 2 \times 4=24\). The multiplication principle is correct when one choice is made for each part.

Step 3

Exam Tip

पूरी वर्दी के लिए \(3 \times 2 \times 4=24\)। हर भाग का एक चुनाव होने पर गुणन सिद्धांत सही है।

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Question 7/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक मोबाइल पासकोड में (7) अंकों में से पहला स्थान (4) अक्षरों में से दूसरा स्थान और (3) प्रतीकों में से तीसरा स्थान चुना जाता है। कुल पासकोड कितने हैं?

In a mobile passcode the first place is chosen from (7) digits the second from (4) letters and the third from (3) symbols. How many passcodes are possible?

Explanation opens after your attempt
Correct Answer

C. (84) पासकोड(84) passcodes

Step 1

Concept

The positions are independent so \(7 \times 4 \times 3=84\). In passcodes multiply the choices for each place.

Step 2

Why this answer is correct

The correct answer is C. (84) पासकोड / (84) passcodes. The positions are independent so \(7 \times 4 \times 3=84\). In passcodes multiply the choices for each place.

Step 3

Exam Tip

स्थान स्वतंत्र हैं इसलिए \(7 \times 4 \times 3=84\)। पासकोड में प्रत्येक स्थान के विकल्प गुणा करें।

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Question 8/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक पुस्तक सेट में (6) विषय (2) कवर प्रकार और (5) आकार हैं। एक विषय एक कवर और एक आकार चुनने के कितने तरीके हैं?

A book set has (6) subjects (2) cover types and (5) sizes. How many ways are there to choose one subject one cover and one size?

Explanation opens after your attempt
Correct Answer

D. (60) तरीके(60) ways

Step 1

Concept

Total ways are \(6 \times 2 \times 5=60\). Multiply when selecting one from each of three lists.

Step 2

Why this answer is correct

The correct answer is D. (60) तरीके / (60) ways. Total ways are \(6 \times 2 \times 5=60\). Multiply when selecting one from each of three lists.

Step 3

Exam Tip

कुल तरीके \(6 \times 2 \times 5=60\) हैं। तीन सूचियों से एक-एक चयन में गुणा करें।

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Question 9/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक विद्यालय ऐप में (4) कक्षाएं (3) विषय और (6) टेस्ट प्रकार हैं। एक कक्षा एक विषय और एक टेस्ट प्रकार कितने तरीकों से चुने जा सकते हैं?

A school app has (4) classes (3) subjects and (6) test types. In how many ways can one class one subject and one test type be chosen?

Explanation opens after your attempt
Correct Answer

C. (72) तरीके(72) ways

Step 1

Concept

With each class there are (3) subjects and (6) test types so \(4 \times 3 \times 6=72\). The product principle gives total options.

Step 2

Why this answer is correct

The correct answer is C. (72) तरीके / (72) ways. With each class there are (3) subjects and (6) test types so \(4 \times 3 \times 6=72\). The product principle gives total options.

Step 3

Exam Tip

हर कक्षा के साथ (3) विषय और (6) टेस्ट प्रकार हैं इसलिए \(4 \times 3 \times 6=72\)। उत्पाद सिद्धांत से कुल विकल्प मिलते हैं।

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Question 10/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक कमरे की सजावट में (5) दीवार रंग (2) पर्दे और (4) लैंप विकल्प हैं। कुल सजावटें कितनी बनेंगी?

A room decoration has (5) wall colors (2) curtain options and (4) lamp options. How many decorations can be made?

Explanation opens after your attempt
Correct Answer

A. (40) सजावटें(40) decorations

Step 1

Concept

For wall curtain and lamp \(5 \times 2 \times 4=40\). Multiply all independent choices.

Step 2

Why this answer is correct

The correct answer is A. (40) सजावटें / (40) decorations. For wall curtain and lamp \(5 \times 2 \times 4=40\). Multiply all independent choices.

Step 3

Exam Tip

दीवार पर्दे और लैंप के लिए \(5 \times 2 \times 4=40\)। सभी स्वतंत्र विकल्पों को गुणा करें।

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Question 11/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक खेल किट में (3) बल्ले (5) गेंदें और (4) दस्ताने विकल्प हैं। एक-एक चुनने के कुल तरीके कितने हैं?

A sports kit has (3) bats (5) balls and (4) glove options. How many total ways are there to choose one each?

Explanation opens after your attempt
Correct Answer

D. (60) तरीके(60) ways

Step 1

Concept

Total ways are \(3 \times 5 \times 4=60\). While forming a kit the choices of each component are multiplied.

Step 2

Why this answer is correct

The correct answer is D. (60) तरीके / (60) ways. Total ways are \(3 \times 5 \times 4=60\). While forming a kit the choices of each component are multiplied.

Step 3

Exam Tip

कुल तरीके \(3 \times 5 \times 4=60\) हैं। किट बनाने में हर घटक का चयन गुणा होता है।

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Question 12/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक मंच प्रस्तुति में (6) गीत (3) पोशाकें और (2) प्रकाश प्रभाव हैं। कुल प्रस्तुति विकल्प कितने हैं?

A stage performance has (6) songs (3) costumes and (2) lighting effects. How many performance options are possible?

Explanation opens after your attempt
Correct Answer

B. (36) विकल्प(36) options

Step 1

Concept

For each song there are (3) costumes and (2) effects so \(6 \times 3 \times 2=36\). Multiply independent creative choices.

Step 2

Why this answer is correct

The correct answer is B. (36) विकल्प / (36) options. For each song there are (3) costumes and (2) effects so \(6 \times 3 \times 2=36\). Multiply independent creative choices.

Step 3

Exam Tip

प्रत्येक गीत के साथ (3) पोशाकें और (2) प्रभाव हैं इसलिए \(6 \times 3 \times 2=36\)। स्वतंत्र कलात्मक विकल्पों में गुणा करें।

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Question 13/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक (3)-स्थान कोड में हर स्थान पर (5) विकल्प हैं और पुनरावृत्ति की अनुमति है। कुल कोड कितने होंगे?

In a (3)-place code each place has (5) choices and repetition is allowed. How many codes are possible?

Explanation opens after your attempt
Correct Answer

A. (125) कोड(125) codes

Step 1

Concept

Each place has (5) choices so \(5 \times 5 \times 5=125\). With repetition allowed the choices remain the same at every place.

Step 2

Why this answer is correct

The correct answer is A. (125) कोड / (125) codes. Each place has (5) choices so \(5 \times 5 \times 5=125\). With repetition allowed the choices remain the same at every place.

Step 3

Exam Tip

हर स्थान पर (5) विकल्प हैं इसलिए \(5 \times 5 \times 5=125\)। पुनरावृत्ति हो तो हर स्थान के विकल्प समान रहते हैं।

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Question 14/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक लाइसेंस प्लेट में (2) अक्षर और (2) अंक हैं। अक्षरों के लिए (4) विकल्प और अंकों के लिए (5) विकल्प हैं तथा पुनरावृत्ति मान्य है। कुल प्लेटें कितनी बनेंगी?

A license plate has (2) letters and (2) digits. There are (4) letter choices and (5) digit choices and repetition is allowed. How many plates can be formed?

Explanation opens after your attempt
Correct Answer

C. (400) प्लेटें(400) plates

Step 1

Concept

Total plates are \(4 \times 4 \times 5 \times 5=400\). Multiply the choices for every position separately.

Step 2

Why this answer is correct

The correct answer is C. (400) प्लेटें / (400) plates. Total plates are \(4 \times 4 \times 5 \times 5=400\). Multiply the choices for every position separately.

Step 3

Exam Tip

कुल प्लेटें \(4 \times 4 \times 5 \times 5=400\) हैं। हर स्थान के विकल्प अलग से गुणा करें।

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Question 15/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक ऑनलाइन पंजीकरण में (3) पाठ्यक्रम (4) बैच और (5) भुगतान विकल्प हैं। कुल पंजीकरण प्रकार कितने हैं?

In an online registration there are (3) courses (4) batches and (5) payment options. How many registration types are possible?

Explanation opens after your attempt
Correct Answer

D. (60) प्रकार(60) types

Step 1

Concept

Total types are \(3 \times 4 \times 5=60\). Course batch and payment are independent choices.

Step 2

Why this answer is correct

The correct answer is D. (60) प्रकार / (60) types. Total types are \(3 \times 4 \times 5=60\). Course batch and payment are independent choices.

Step 3

Exam Tip

कुल प्रकार \(3 \times 4 \times 5=60\) होंगे। पाठ्यक्रम बैच और भुगतान स्वतंत्र चुनाव हैं।

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Question 16/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक उपहार टोकरी में (2) फल प्रकार (5) चॉकलेट प्रकार और (3) पैकिंग विकल्प हैं। एक-एक चुनने के कितने तरीके हैं?

A gift basket has (2) fruit types (5) chocolate types and (3) packing options. How many ways are there to choose one each?

Explanation opens after your attempt
Correct Answer

B. (30) तरीके(30) ways

Step 1

Concept

Total ways are \(2 \times 5 \times 3=30\). Treat every component of the basket as a separate stage.

Step 2

Why this answer is correct

The correct answer is B. (30) तरीके / (30) ways. Total ways are \(2 \times 5 \times 3=30\). Treat every component of the basket as a separate stage.

Step 3

Exam Tip

कुल तरीके \(2 \times 5 \times 3=30\) हैं। टोकरी के हर घटक को अलग चरण मानें।

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Question 17/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक मार्ग में (4) रास्ते घर से बाजार तक और (6) रास्ते बाजार से विद्यालय तक हैं। घर से विद्यालय तक बाजार होकर कुल रास्ते कितने हैं?

There are (4) routes from home to market and (6) routes from market to school. How many routes are there from home to school via market?

Explanation opens after your attempt
Correct Answer

A. (24) रास्ते(24) routes

Step 1

Concept

After each first route there are (6) second routes so \(4 \times 6=24\). Multiply for successive travel stages.

Step 2

Why this answer is correct

The correct answer is A. (24) रास्ते / (24) routes. After each first route there are (6) second routes so \(4 \times 6=24\). Multiply for successive travel stages.

Step 3

Exam Tip

हर पहले रास्ते के बाद (6) दूसरे रास्ते मिलते हैं इसलिए \(4 \times 6=24\)। क्रमिक यात्रा चरणों में गुणा करें।

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Question 18/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक विद्यार्थी (7) अध्यायों में से एक अध्याय और (4) प्रश्न प्रकारों में से एक प्रकार चुनता है। कुल अभ्यास विकल्प कितने हैं?

A student chooses one chapter from (7) chapters and one type from (4) question types. How many practice options are possible?

Explanation opens after your attempt
Correct Answer

C. (28) विकल्प(28) options

Step 1

Concept

For chapter and question type \(7 \times 4=28\). The multiplication principle applies to two independent choices.

Step 2

Why this answer is correct

The correct answer is C. (28) विकल्प / (28) options. For chapter and question type \(7 \times 4=28\). The multiplication principle applies to two independent choices.

Step 3

Exam Tip

अध्याय और प्रश्न प्रकार के लिए \(7 \times 4=28\)। दो स्वतंत्र चयन में गुणा सिद्धांत लागू होता है।

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Question 19/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक परीक्षा में (5) खंड हैं और हर खंड में (3) वैकल्पिक प्रश्न हैं। हर खंड से एक प्रश्न चुनने के कितने तरीके हैं?

An exam has (5) sections and each section has (3) optional questions. In how many ways can one question be chosen from each section?

Explanation opens after your attempt
Correct Answer

B. (243) तरीके(243) ways

Step 1

Concept

There are (3) choices from each section so \(3^5=243\). Use powers when the same number of choices repeats across stages.

Step 2

Why this answer is correct

The correct answer is B. (243) तरीके / (243) ways. There are (3) choices from each section so \(3^5=243\). Use powers when the same number of choices repeats across stages.

Step 3

Exam Tip

हर खंड से (3) विकल्प हैं इसलिए \(3^5=243\)। समान विकल्प कई चरणों में हों तो घात का उपयोग करें।

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Question 20/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक (4)-अंकीय कोड में हर स्थान पर (2) विकल्प हैं। पुनरावृत्ति मान्य है तो कुल कोड कितने हैं?

In a (4)-digit code each place has (2) choices. If repetition is allowed how many codes are possible?

Explanation opens after your attempt
Correct Answer

D. (16) कोड(16) codes

Step 1

Concept

Total codes are \(2^4=16\). If each place is independent use the power according to the number of places.

Step 2

Why this answer is correct

The correct answer is D. (16) कोड / (16) codes. Total codes are \(2^4=16\). If each place is independent use the power according to the number of places.

Step 3

Exam Tip

कुल कोड \(2^4=16\) होंगे। हर स्थान स्वतंत्र हो तो स्थानों की संख्या के अनुसार घात लें।

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Question 21/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक क्विज सेटिंग में (4) भाषा विकल्प (5) समय सीमा विकल्प और (3) अंक विकल्प हैं। कुल सेटिंग संयोजन कितने हैं?

In a quiz setting there are (4) language options (5) time-limit options and (3) mark options. How many setting combinations are possible?

Explanation opens after your attempt
Correct Answer

C. (60) संयोजन(60) combinations

Step 1

Concept

Total combinations are \(4 \times 5 \times 3=60\). Multiply the separate setting decisions.

Step 2

Why this answer is correct

The correct answer is C. (60) संयोजन / (60) combinations. Total combinations are \(4 \times 5 \times 3=60\). Multiply the separate setting decisions.

Step 3

Exam Tip

कुल संयोजन \(4 \times 5 \times 3=60\) हैं। सेटिंग के अलग-अलग निर्णयों को गुणा करें।

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Question 22/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक पेंटिंग में (3) पृष्ठभूमि रंग (4) सीमा रंग और (2) हस्ताक्षर रंग चुने जाते हैं। कुल पेंटिंग रूप कितने हैं?

In a painting (3) background colors (4) border colors and (2) signature colors are chosen. How many painting forms are possible?

Explanation opens after your attempt
Correct Answer

A. (24) रूप(24) forms

Step 1

Concept

For three color decisions \(3 \times 4 \times 2=24\). Color choices for different places are multiplied.

Step 2

Why this answer is correct

The correct answer is A. (24) रूप / (24) forms. For three color decisions \(3 \times 4 \times 2=24\). Color choices for different places are multiplied.

Step 3

Exam Tip

तीन रंग निर्णयों के लिए \(3 \times 4 \times 2=24\)। अलग-अलग स्थानों के रंग विकल्प गुणा होते हैं।

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Question 23/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक प्रमाणपत्र में (6) शीर्षक शैली (2) मुहर विकल्प और (5) किनारा डिज़ाइन हैं। कुल प्रमाणपत्र डिज़ाइन कितने हैं?

A certificate has (6) title styles (2) seal options and (5) border designs. How many certificate designs are possible?

Explanation opens after your attempt
Correct Answer

B. (60) डिज़ाइन(60) designs

Step 1

Concept

Total designs are \(6 \times 2 \times 5=60\). Treat each design part as an independent stage and multiply.

Step 2

Why this answer is correct

The correct answer is B. (60) डिज़ाइन / (60) designs. Total designs are \(6 \times 2 \times 5=60\). Treat each design part as an independent stage and multiply.

Step 3

Exam Tip

कुल डिज़ाइन \(6 \times 2 \times 5=60\) हैं। हर डिज़ाइन भाग को स्वतंत्र चरण मानकर गुणा करें।

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Question 24/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक पुस्तकालय में (8) अलमारियां हैं और हर अलमारी में (4) पंक्तियां हैं। एक अलमारी और एक पंक्ति चुनने के कितने तरीके हैं?

A library has (8) shelves and each shelf has (4) rows. How many ways are there to choose one shelf and one row?

Explanation opens after your attempt
Correct Answer

D. (32) तरीके(32) ways

Step 1

Concept

One shelf is chosen in (8) ways and one row in (4) ways so \(8 \times 4=32\). In structure-based selection multiply the levels.

Step 2

Why this answer is correct

The correct answer is D. (32) तरीके / (32) ways. One shelf is chosen in (8) ways and one row in (4) ways so \(8 \times 4=32\). In structure-based selection multiply the levels.

Step 3

Exam Tip

एक अलमारी (8) तरीकों से और पंक्ति (4) तरीकों से चुनी जाती है इसलिए \(8 \times 4=32\)। संरचना आधारित चयन में स्तरों को गुणा करें।

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Question 25/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक बस यात्रा में (3) सीट श्रेणियां (5) बस समय और (2) भुगतान माध्यम हैं। कुल बुकिंग विकल्प कितने हैं?

In a bus journey there are (3) seat categories (5) bus times and (2) payment modes. How many booking options are possible?

Explanation opens after your attempt
Correct Answer

A. (30) विकल्प(30) options

Step 1

Concept

Booking options are \(3 \times 5 \times 2=30\). Category time and payment are separate choices.

Step 2

Why this answer is correct

The correct answer is A. (30) विकल्प / (30) options. Booking options are \(3 \times 5 \times 2=30\). Category time and payment are separate choices.

Step 3

Exam Tip

बुकिंग विकल्प \(3 \times 5 \times 2=30\) हैं। श्रेणी समय और भुगतान अलग चयन हैं।

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Question 26/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक (2)-स्थान संकेत में पहले स्थान पर (9) रंग और दूसरे स्थान पर (7) आकार संभव हैं। कुल संकेत कितने हैं?

In a (2)-place signal the first place can have (9) colors and the second place can have (7) shapes. How many signals are possible?

Explanation opens after your attempt
Correct Answer

C. (63) संकेत(63) signals

Step 1

Concept

Total signals are \(9 \times 7=63\). The first and second places are independent so multiply.

Step 2

Why this answer is correct

The correct answer is C. (63) संकेत / (63) signals. Total signals are \(9 \times 7=63\). The first and second places are independent so multiply.

Step 3

Exam Tip

कुल संकेत \(9 \times 7=63\) हैं। प्रथम और द्वितीय स्थान स्वतंत्र हैं इसलिए गुणा करें।

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Question 27/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक स्कूल टीम में (4) कप्तान विकल्प (3) उपकप्तान विकल्प और (2) मैनेजर विकल्प हैं। एक-एक पद भरने के कितने तरीके हैं?

A school team has (4) captain options (3) vice-captain options and (2) manager options. How many ways are there to fill one of each post?

Explanation opens after your attempt
Correct Answer

D. (24) तरीके(24) ways

Step 1

Concept

For the three posts \(4 \times 3 \times 2=24\). Multiply the choices for different posts.

Step 2

Why this answer is correct

The correct answer is D. (24) तरीके / (24) ways. For the three posts \(4 \times 3 \times 2=24\). Multiply the choices for different posts.

Step 3

Exam Tip

तीनों पदों के लिए \(4 \times 3 \times 2=24\)। अलग-अलग पदों के विकल्पों को गुणा करें।

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Question 28/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक ऑनलाइन स्टोर में (5) टी-शर्ट रंग (4) आकार और (3) प्रिंट हैं। कुल टी-शर्ट प्रकार कितने हैं?

An online store has (5) T-shirt colors (4) sizes and (3) prints. How many T-shirt types are possible?

Explanation opens after your attempt
Correct Answer

B. (60) प्रकार(60) types

Step 1

Concept

Total types are \(5 \times 4 \times 3=60\). Color size and print are three independent features.

Step 2

Why this answer is correct

The correct answer is B. (60) प्रकार / (60) types. Total types are \(5 \times 4 \times 3=60\). Color size and print are three independent features.

Step 3

Exam Tip

कुल प्रकार \(5 \times 4 \times 3=60\) हैं। रंग आकार और प्रिंट तीन स्वतंत्र विशेषताएं हैं।

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Question 29/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक शिक्षक (6) अध्यायों में से एक अध्याय और (5) कठिनाई स्तरों में से एक स्तर चुनकर प्रश्न बनाता है। कुल प्रश्न सेटिंग कितनी हैं?

A teacher creates a question by choosing one chapter from (6) chapters and one level from (5) difficulty levels. How many question settings are possible?

Explanation opens after your attempt
Correct Answer

A. (30) सेटिंग(30) settings

Step 1

Concept

For chapter and difficulty level \(6 \times 5=30\). Multiply when there are two independent choices.

Step 2

Why this answer is correct

The correct answer is A. (30) सेटिंग / (30) settings. For chapter and difficulty level \(6 \times 5=30\). Multiply when there are two independent choices.

Step 3

Exam Tip

अध्याय और कठिनाई स्तर के लिए \(6 \times 5=30\)। दो स्वतंत्र चयन हों तो गुणा करें।

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Question 30/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक डिजिटल पोस्टर में (4) फॉन्ट (6) रंग और (2) पृष्ठभूमि बनावट हैं। कुल पोस्टर शैली कितनी होंगी?

A digital poster has (4) fonts (6) colors and (2) background textures. How many poster styles are possible?

Explanation opens after your attempt
Correct Answer

C. (48) शैली(48) styles

Step 1

Concept

Total styles are \(4 \times 6 \times 2=48\). Each design property is a separate stage.

Step 2

Why this answer is correct

The correct answer is C. (48) शैली / (48) styles. Total styles are \(4 \times 6 \times 2=48\). Each design property is a separate stage.

Step 3

Exam Tip

कुल शैली \(4 \times 6 \times 2=48\) होंगी। हर डिज़ाइन गुण अलग चरण है।

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Question 31/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक कॉलोनी गेट पास में (2) पास प्रकार (7) रंग और (3) वैधता अवधि विकल्प हैं। कुल गेट पास कितने बनेंगे?

A colony gate pass has (2) pass types (7) colors and (3) validity-period options. How many gate passes can be made?

Explanation opens after your attempt
Correct Answer

B. (42) पास(42) passes

Step 1

Concept

Total passes are \(2 \times 7 \times 3=42\). Multiply pass type color and period.

Step 2

Why this answer is correct

The correct answer is B. (42) पास / (42) passes. Total passes are \(2 \times 7 \times 3=42\). Multiply pass type color and period.

Step 3

Exam Tip

कुल पास \(2 \times 7 \times 3=42\) बनेंगे। पास प्रकार रंग और अवधि को गुणा करें।

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Question 32/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक कक्षा में (5) पंक्तियां और (8) स्तंभ हैं। एक सीट कितने तरीकों से चुनी जा सकती है?

In a classroom there are (5) rows and (8) columns. In how many ways can one seat be chosen?

Explanation opens after your attempt
Correct Answer

D. (40) तरीके(40) ways

Step 1

Concept

A seat needs both row and column so \(5 \times 8=40\). In a grid total places are rows multiplied by columns.

Step 2

Why this answer is correct

The correct answer is D. (40) तरीके / (40) ways. A seat needs both row and column so \(5 \times 8=40\). In a grid total places are rows multiplied by columns.

Step 3

Exam Tip

सीट के लिए पंक्ति और स्तंभ दोनों चाहिए इसलिए \(5 \times 8=40\)। ग्रिड में कुल स्थान पंक्ति गुणा स्तंभ होते हैं।

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Question 33/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक फोटो फिल्टर ऐप में (3) फिल्टर (4) फ्रेम और (5) स्टिकर विकल्प हैं। एक-एक चुनने पर कुल रूप कितने हैं?

A photo filter app has (3) filters (4) frames and (5) sticker options. How many total looks are possible by choosing one each?

Explanation opens after your attempt
Correct Answer

A. (60) रूप(60) looks

Step 1

Concept

Total looks are \(3 \times 4 \times 5=60\). Each photo element is an independent choice that is multiplied.

Step 2

Why this answer is correct

The correct answer is A. (60) रूप / (60) looks. Total looks are \(3 \times 4 \times 5=60\). Each photo element is an independent choice that is multiplied.

Step 3

Exam Tip

कुल रूप \(3 \times 4 \times 5=60\) हैं। प्रत्येक फोटो तत्व का स्वतंत्र चयन गुणा होता है।

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Question 34/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक छात्र (4) वैकल्पिक विषयों और (9) परियोजना विषयों में से एक-एक चुनता है। कुल चयन कितने हैं?

A student chooses one each from (4) optional subjects and (9) project topics. How many selections are possible?

Explanation opens after your attempt
Correct Answer

B. (36) चयन(36) selections

Step 1

Concept

Total selections are \(4 \times 9=36\). Choosing a subject and a project topic is independent.

Step 2

Why this answer is correct

The correct answer is B. (36) चयन / (36) selections. Total selections are \(4 \times 9=36\). Choosing a subject and a project topic is independent.

Step 3

Exam Tip

कुल चयन \(4 \times 9=36\) हैं। एक विषय और एक परियोजना विषय का चयन स्वतंत्र है।

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Question 35/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक परीक्षा ऐप में (2) मोड (5) प्रश्न संख्या विकल्प और (6) समय विकल्प हैं। कुल टेस्ट सेटअप कितने हैं?

An exam app has (2) modes (5) question-count options and (6) time options. How many test setups are possible?

Explanation opens after your attempt
Correct Answer

C. (60) सेटअप(60) setups

Step 1

Concept

Total setups are \(2 \times 5 \times 6=60\). Mode count and time are three independent settings.

Step 2

Why this answer is correct

The correct answer is C. (60) सेटअप / (60) setups. Total setups are \(2 \times 5 \times 6=60\). Mode count and time are three independent settings.

Step 3

Exam Tip

कुल सेटअप \(2 \times 5 \times 6=60\) हैं। मोड संख्या और समय तीन स्वतंत्र सेटिंग हैं।

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Question 36/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक पिन में पहले (2) स्थानों पर (3) विकल्प और अगले (2) स्थानों पर (4) विकल्प हैं। पुनरावृत्ति मान्य है तो कुल पिन कितने हैं?

In a PIN the first (2) places have (3) choices each and the next (2) places have (4) choices each. If repetition is allowed how many PINs are possible?

Explanation opens after your attempt
Correct Answer

D. (144) पिन(144) PINs

Step 1

Concept

Total PINs are \(3 \times 3 \times 4 \times 4=144\). Multiply the choices for each place separately.

Step 2

Why this answer is correct

The correct answer is D. (144) पिन / (144) PINs. Total PINs are \(3 \times 3 \times 4 \times 4=144\). Multiply the choices for each place separately.

Step 3

Exam Tip

कुल पिन \(3 \times 3 \times 4 \times 4=144\) हैं। हर स्थान के विकल्पों को अलग-अलग गुणा करें।

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Question 37/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक कैफे में (6) कॉफी (3) कप आकार और (2) चीनी स्तर हैं। एक कॉफी एक आकार और एक चीनी स्तर कितने तरीकों से चुने जा सकते हैं?

A cafe has (6) coffees (3) cup sizes and (2) sugar levels. In how many ways can one coffee one size and one sugar level be chosen?

Explanation opens after your attempt
Correct Answer

A. (36) तरीके(36) ways

Step 1

Concept

Total ways are \(6 \times 3 \times 2=36\). Multiply every independent choice of the drink.

Step 2

Why this answer is correct

The correct answer is A. (36) तरीके / (36) ways. Total ways are \(6 \times 3 \times 2=36\). Multiply every independent choice of the drink.

Step 3

Exam Tip

कुल तरीके \(6 \times 3 \times 2=36\) हैं। पेय के हर स्वतंत्र चुनाव को गुणा करें।

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Question 38/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक छात्रावास में (4) ब्लॉक (5) मंजिल और (6) कमरे प्रति मंजिल हैं। कमरा चुनने के कुल तरीके कितने हैं?

A hostel has (4) blocks (5) floors and (6) rooms on each floor. How many ways are there to choose a room?

Explanation opens after your attempt
Correct Answer

B. (120) तरीके(120) ways

Step 1

Concept

To choose a room there are \(4 \times 5 \times 6=120\) ways. Block floor and room are successive stages.

Step 2

Why this answer is correct

The correct answer is B. (120) तरीके / (120) ways. To choose a room there are \(4 \times 5 \times 6=120\) ways. Block floor and room are successive stages.

Step 3

Exam Tip

कमरा चुनने के लिए \(4 \times 5 \times 6=120\) तरीके हैं। ब्लॉक मंजिल और कमरा क्रमिक चरण हैं।

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Question 39/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक ट्रेन टिकट में (3) श्रेणियां (4) प्रस्थान समय और (2) सीट पसंद हैं। कुल टिकट विकल्प कितने हैं?

A train ticket has (3) classes (4) departure times and (2) seat preferences. How many ticket options are possible?

Explanation opens after your attempt
Correct Answer

C. (24) विकल्प(24) options

Step 1

Concept

Ticket options are \(3 \times 4 \times 2=24\). Multiply class time and seat preference.

Step 2

Why this answer is correct

The correct answer is C. (24) विकल्प / (24) options. Ticket options are \(3 \times 4 \times 2=24\). Multiply class time and seat preference.

Step 3

Exam Tip

टिकट विकल्प \(3 \times 4 \times 2=24\) हैं। श्रेणी समय और सीट पसंद को गुणा करें।

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Question 40/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक संदेश टेम्पलेट में (5) अभिवादन (3) मुख्य वाक्य और (4) समाप्ति वाक्य हैं। कुल संदेश कितने बनेंगे?

A message template has (5) greetings (3) main sentences and (4) closing sentences. How many messages can be formed?

Explanation opens after your attempt
Correct Answer

D. (60) संदेश(60) messages

Step 1

Concept

Total messages are \(5 \times 3 \times 4=60\). Multiply when one option is taken from each part.

Step 2

Why this answer is correct

The correct answer is D. (60) संदेश / (60) messages. Total messages are \(5 \times 3 \times 4=60\). Multiply when one option is taken from each part.

Step 3

Exam Tip

कुल संदेश \(5 \times 3 \times 4=60\) बनेंगे। हर भाग से एक विकल्प लेने पर गुणा करें।

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Question 41/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक सिक्का खेल में (2) सिक्के और (6) पासे के परिणाम माने जाते हैं। एक सिक्के का परिणाम और एक पासे का परिणाम मिलाकर कुल परिणाम कितने हैं?

In a coin-die game (2) coin outcomes and (6) die outcomes are considered. How many total outcomes are there by combining one coin outcome and one die outcome?

Explanation opens after your attempt
Correct Answer

A. (12) परिणाम(12) outcomes

Step 1

Concept

Total outcomes are \(2 \times 6=12\). In combined experiments the number of outcomes is multiplied.

Step 2

Why this answer is correct

The correct answer is A. (12) परिणाम / (12) outcomes. Total outcomes are \(2 \times 6=12\). In combined experiments the number of outcomes is multiplied.

Step 3

Exam Tip

कुल परिणाम \(2 \times 6=12\) हैं। संयुक्त प्रयोगों में परिणामों की संख्या गुणा होती है।

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Question 42/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक छात्र आईडी में (4) विभाग कोड और (25) क्रमांक विकल्प हैं। एक विभाग कोड और एक क्रमांक चुनने पर कुल आईडी कितनी बनेंगी?

A student ID has (4) department codes and (25) serial-number options. How many IDs can be made by choosing one department code and one serial number?

Explanation opens after your attempt
Correct Answer

B. (100) आईडी(100) IDs

Step 1

Concept

Total IDs are \(4 \times 25=100\). Code and serial number are independent parts.

Step 2

Why this answer is correct

The correct answer is B. (100) आईडी / (100) IDs. Total IDs are \(4 \times 25=100\). Code and serial number are independent parts.

Step 3

Exam Tip

कुल आईडी \(4 \times 25=100\) बनेंगी। कोड और क्रमांक स्वतंत्र भाग हैं।

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Question 43/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक खिलौना रोबोट में (3) सिर प्रकार (4) शरीर प्रकार और (5) पहिया प्रकार हैं। कुल रोबोट मॉडल कितने हैं?

A toy robot has (3) head types (4) body types and (5) wheel types. How many robot models are possible?

Explanation opens after your attempt
Correct Answer

C. (60) मॉडल(60) models

Step 1

Concept

Total models are \(3 \times 4 \times 5=60\). Multiply the choices for different parts.

Step 2

Why this answer is correct

The correct answer is C. (60) मॉडल / (60) models. Total models are \(3 \times 4 \times 5=60\). Multiply the choices for different parts.

Step 3

Exam Tip

कुल मॉडल \(3 \times 4 \times 5=60\) हैं। अलग-अलग भागों के विकल्पों को गुणा करें।

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Question 44/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक घर नंबर में (2) अक्षर विकल्प और (10) अंक विकल्प हैं। एक अक्षर और एक अंक चुनकर कुल नंबर कितने बनेंगे?

A house number has (2) letter options and (10) digit options. How many numbers can be formed by choosing one letter and one digit?

Explanation opens after your attempt
Correct Answer

D. (20) नंबर(20) numbers

Step 1

Concept

Total numbers are \(2 \times 10=20\). Multiply when making a pair of one letter and one digit.

Step 2

Why this answer is correct

The correct answer is D. (20) नंबर / (20) numbers. Total numbers are \(2 \times 10=20\). Multiply when making a pair of one letter and one digit.

Step 3

Exam Tip

कुल नंबर \(2 \times 10=20\) हैं। एक अक्षर और एक अंक की जोड़ी बनाने में गुणा करें।

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Question 45/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक परीक्षा हॉल में (6) पंक्तियां और प्रत्येक पंक्ति में (7) सीटें हैं। कुल सीटें कितनी हैं?

An exam hall has (6) rows and (7) seats in each row. How many total seats are there?

Explanation opens after your attempt
Correct Answer

A. (42) सीटें(42) seats

Step 1

Concept

Total seats are \(6 \times 7=42\). Multiply rows and seats per row.

Step 2

Why this answer is correct

The correct answer is A. (42) सीटें / (42) seats. Total seats are \(6 \times 7=42\). Multiply rows and seats per row.

Step 3

Exam Tip

कुल सीटें \(6 \times 7=42\) हैं। पंक्तियों और प्रति पंक्ति सीटों को गुणा करें।

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Question 46/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक भाषा प्रयोगशाला में (5) भाषा विकल्प (3) अभ्यास स्तर और (2) रिकॉर्डिंग मोड हैं। कुल अभ्यास सेट कितने बनेंगे?

A language lab has (5) language options (3) practice levels and (2) recording modes. How many practice sets can be formed?

Explanation opens after your attempt
Correct Answer

B. (30) सेट(30) sets

Step 1

Concept

Practice sets are \(5 \times 3 \times 2=30\). Language level and mode choices are independent.

Step 2

Why this answer is correct

The correct answer is B. (30) सेट / (30) sets. Practice sets are \(5 \times 3 \times 2=30\). Language level and mode choices are independent.

Step 3

Exam Tip

अभ्यास सेट \(5 \times 3 \times 2=30\) हैं। भाषा स्तर और मोड के चुनाव स्वतंत्र हैं।

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Question 47/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक संगीत अभ्यास में (4) राग (5) ताल और (2) गति विकल्प हैं। एक-एक चुनने के कुल तरीके कितने हैं?

In music practice there are (4) ragas (5) rhythms and (2) speed options. How many total ways are there to choose one each?

Explanation opens after your attempt
Correct Answer

C. (40) तरीके(40) ways

Step 1

Concept

Total ways are \(4 \times 5 \times 2=40\). Multiply the three independent music choices.

Step 2

Why this answer is correct

The correct answer is C. (40) तरीके / (40) ways. Total ways are \(4 \times 5 \times 2=40\). Multiply the three independent music choices.

Step 3

Exam Tip

कुल तरीके \(4 \times 5 \times 2=40\) हैं। तीन स्वतंत्र संगीत विकल्पों को गुणा करें।

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Question 48/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक भोजनालय कूपन में (3) भोजन प्रकार (6) छूट विकल्प और (4) वैधता दिन विकल्प हैं। कुल कूपन प्रकार कितने हैं?

A restaurant coupon has (3) meal types (6) discount options and (4) validity-day options. How many coupon types are possible?

Explanation opens after your attempt
Correct Answer

D. (72) प्रकार(72) types

Step 1

Concept

Total types are \(3 \times 6 \times 4=72\). Each coupon feature is an independent choice.

Step 2

Why this answer is correct

The correct answer is D. (72) प्रकार / (72) types. Total types are \(3 \times 6 \times 4=72\). Each coupon feature is an independent choice.

Step 3

Exam Tip

कुल प्रकार \(3 \times 6 \times 4=72\) हैं। हर कूपन विशेषता का स्वतंत्र चयन है।

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Question 49/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक कंप्यूटर लैब में (4) कंप्यूटर पंक्तियां और हर पंक्ति में (9) कंप्यूटर हैं। एक कंप्यूटर चुनने के कितने तरीके हैं?

A computer lab has (4) rows of computers and (9) computers in each row. In how many ways can one computer be chosen?

Explanation opens after your attempt
Correct Answer

A. (36) तरीके(36) ways

Step 1

Concept

The ways to choose a computer are \(4 \times 9=36\). Both row and position need to be determined.

Step 2

Why this answer is correct

The correct answer is A. (36) तरीके / (36) ways. The ways to choose a computer are \(4 \times 9=36\). Both row and position need to be determined.

Step 3

Exam Tip

कंप्यूटर चुनने के तरीके \(4 \times 9=36\) हैं। पंक्ति और स्थान दोनों निर्धारित करने होते हैं।

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Question 50/50 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 54

एक (3)-अक्षर कोड में हर स्थान पर (6) अक्षर विकल्प हैं और पुनरावृत्ति की अनुमति है। कुल कोड कितने बनेंगे?

In a (3)-letter code each place has (6) letter choices and repetition is allowed. How many codes can be formed?

Explanation opens after your attempt
Correct Answer

B. (216) कोड(216) codes

Step 1

Concept

Total codes are \(6^3=216\). When repetition is allowed each place keeps (6) choices.

Step 2

Why this answer is correct

The correct answer is B. (216) कोड / (216) codes. Total codes are \(6^3=216\). When repetition is allowed each place keeps (6) choices.

Step 3

Exam Tip

कुल कोड \(6^3=216\) होंगे। पुनरावृत्ति मान्य होने पर हर स्थान पर (6) विकल्प रहते हैं।

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