CDC SEE Model Set A (2080)

Subject Code: 1031 Set:A

NATIONAL EXAMINATION BOARD
SEE MODEL QUESTION -2080 (SET A)

Subject: C. Maths
Time: 3 hours Full Marks:75

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दिइएका निर्देशनका आधारमा आफ्नै शैलीमा सिर्जनात्मक उत्तर दिनुहोस् ।
सबै प्रश्नहरुको उत्तर दिनुहोस् (Answer all the questions)

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1. 300 जना मानिसहरूको समूहमा गरिएको सर्वेक्षणमा 150 जनाले आइफोन र 200 जनाले एन्ड्रोइडफोन मन पराएको पाइयो । तर 25 जनाले यी दुई मध्ये कुनैपनि फोनहरू मन नपराएको पाइयो ।
   In a survey of 300 people, it was found that 150 people liked I-phone and 200 people like Android phone. But 25 people did not like any of these two phones

   1. यदि I र A ले क्रमशः आइफोन र एन्ड्रोइड फोन मन पराउने मानिसहरूको समूहलाई जनाउँदछ भने $n(\overline{I\cup A})$ को गणनात्मकता लेख्नुहोस ।
      If I and A denote the sets of people who like I-phone and Android phone respectively, write the cardinality of $n(\overline{I\cup A})$ [1]
   2. माथिको जानकारीलाई भेनचित्रमा प्रस्तुत गर्नुहोस्।
      Present the above information in a Venn-diagram. [1]
   3. आइफोनमात्र मन पराउने मानिसहरूको संख्या पत्ता लगाउनुहोस्।
      Find the number of people who liked I-phone only.[3]
   4. आइफोन र एन्ड्रोइड फोन दुवै मन पराउनेको संख्या र यी दुई मध्ये कुनै पनि फोन मन नपराउने मानिसहरूको संख्याबीच तुलना गर्नुहोस्।
      Compare the number of people who like both I-phone and Android phone and who do not like any of these two phones. [1]

   Solution 👉 Click Here

   Solution
   Let
   U is universal set of all survayed people
   Given that
   I is set of people who like I-phone
   A is set of people who like Android phone
   Thus
   Cardinality of all surveyed people as Universal Set U is
   $n(U)=300$
   Cardinality of set I, people who liked I-phone is
   $n(I)=150$
   Cardinality of set A, people who liked Android phone is
   $n(A)=200$
   Cardinality of set, people who did not like any of these two phones is
   $n(\overline{I\cup A})=25$
   Now the itemwise solution are :

   1. If I and A denote the sets of people who like I-phone and Android Phone respectively, write the cardinality of $n(\overline{I\cup A})$
      According the question,
      $n(\overline{I\cup A})=25$
      
   2. Present the above information in a Venn-diagram.
      According the question,
      Let us suppose that
      $n(I\cap A)=x$
      Then, the Venn-diagram of given information are as follows.
      0,0

      U

      I

      A

      150-x

      x

      200-x

      25
   3. Find the number of people who liked I-phone only.
      According to the Venn-diagram given above
      $(150-x)+(x)+(200-x)+(25)=300$
      or$x=75$
      Therefore,
      The number of people who liked I-phone only is
      $n(I_o)=150-x=150-75=75$
      
   4. Compare the number of people who like both I-phone and Android phone and who do not like any of these two phones.
      According to the solution
      The number of people who like both I-phone and Android phone is
      $n(I\cap A)=x=75$
      The number of people who do not like any of these two phones is
      $n(\overline{I\cup A})=25$
      It shows that
      Number of people who like both I-phone and Android phone is THREE TIMES greater than people who do not like any of these two phones

   This completes the solution

2. एक जना सेवानिवृत्त शिक्षकले रु. 80,000 दुई वर्षका लागि प्रतिवर्ष 10% को दरमा अर्धवार्षिक चक्रिय व्याज पाउने गरी एउटा विकास बैङ्कमा रहेको आफ्नो खातामा जम्मा गरेछ ।
   A retired teacher deposited Rs.80,000 in own account of development bank for two years to get the half yearly compound interest at the rate of 10% per annum

   1. अर्धवार्षिक चक्रिय व्याज अनुसार २ वर्षमा कति पटक व्याजको गणना गरिन्छ ? लेख्नुहोस्।
      How many times the interest is calculated according to the semi-annual compound interest in 2 years? Write it [1]
   2. उक्त शिक्षकले 2 वर्षको अन्त्यमा अर्धवार्षिक चक्रीय व्याज अनुसार कति चक्रीय व्याज प्राप्त गर्दछ ? पत्ता लगाउनुहोस्।
      According to the half yearly compound interest, what would be the compound interest received by teacher at the end of 2 years? Find it [2]
   3. सोही व्याजदरमा वार्षिक चक्रीय व्याजअनुसार कति समयमा रु. 80,000 को मिश्रधन रु.1,06,480 हुन्छ ? पत्ता लगाउनुहोस्।
      According to the same rate of yearly compound interest, in how many years will the compound amount of Rs. 80,000 be Rs. 1,06,480 ? Find it.[2]

   Solution 👉 Click Here

   Solution

   1. अर्धवार्षिक चक्रिय व्याज अनुसार २ वर्षमा कति पटक व्याजको गणना गरिन्छ ? लेख्नुहोस्।
      How many times the interest is calculated according to the semi-annual compound interest in 2 years? Write it.
      अर्धवार्षिक चक्रिय व्याज अनुसार बर्षमा 2 पटक (प्रत्येक 6 महिनामा ) व्याजको गणना हुने हुनाले, 2 वर्षमा 4 पटक व्याजको गणना गरिन्छ ।
   2. उक्त शिक्षकले 2 वर्षको अन्त्यमा अर्धवार्षिक चक्रीय व्याज अनुसार कति चक्रीय व्याज प्राप्त गर्दछ ? पत्ता लगाउनुहोस्।
      According to the half yearly compound interest, what would be the compound interest received by teacher at the end of 2 years? Find it.
      According to the question
      Principal (P)=80000
      Rate of semi-annual compound interest (R)=10%
      Time in Years (T)=2 years
      Therefore,semi-annual compound interest in 2 years is
      $CI=P\{{(1+\frac{R}{200})}^{2T}-1\}$
      or $CI=80000\{{(1+\frac{10}{200})}^{2\times 2}-1\}$
      or $CI=80000\{{\left(\frac{210}{200}\right)}^4-1\}$
      or $CI=80000\{{\left(\frac{21}{20}\right)}^4-1\}$
      or $CI=80000\left\{\frac{34481}{160000}\right\}$
      or $CI=\left\{\frac{34481}{2}\right\}$
      or $CI=17240.5$
      So, the semi-annually compound interest received by teacher at the end of 2 years is Rs 17240.5.
   3. सोही व्याजदरमा वार्षिक चक्रीय व्याजअनुसार कति समयमा रु. 80,000 को मिश्रधन रु.1,06,480 हुन्छ ? पत्ता लगाउनुहोस्।
      According to the same rate of yearly compound interest, in how many years will the compound amount of Rs. 80,000 be Rs. 1,06,480 ? Find it.
      According to the question
      Principal (P)=80000
      Amount (A)=106480
      Rate of annual compound interest (R)=10%
      Time in Years (T)=?
      Therefore, using formula of compound amount, we get
      $A=P{(1+\frac{R}{100})}^T$
      or $106480=80000{(1+\frac{10}{100})}^T$
      or $\frac{106480}{80000}={\left(\frac{110}{100}\right)}^T$
      or $\frac{1331}{1000}={\left(\frac{11}{10}\right)}^T$
      or ${\left(\frac{11}{10}\right)}^3={\left(\frac{11}{10}\right)}^T$
      or $T=3$
      So, in 3 years, the annual compound amount of Rs. 80,000 will be Rs. 1,06,480.

3. एउटा मिनीबस रु.40,00,000 मा खरिद गरियो । तीन वर्षसम्म प्रयोग गर्दा रु.15,00,000 आम्दानी भयो । बसको मूल्यमा प्रतिवर्ष 15% को दरले ह्रास आउँछ र तीनवर्षपछि सो मिनीबस बिक्री गरियो ।
   A minibus is purchased for Rs.40,00,000. After using bus for three years. Rs.15,00,000 is earned. The value of the bus depreciates by the rate of 15% per annum and the minibus is sold after three years.

   1. यदि बसको खरिद मूल्य रु. $V_o$, वार्षिक मिश्रह्रासदर R% र बसको T वर्षपछिको मूल्य रु.$V_T$ भए $V_T$ लाई $V_o$, R% र T को रूपमा व्यक्त गर्नुहोस्।
      If the purchasing price of the bus is Rs.$V_o$, the annual rate of compound depreciation is R% and price of the bus after T years is Rs.$V_T$, then express $V_T$ in terms of $V_o$, R% and T. [1]
   2. तीन वर्ष पछिको सो बसको बिक्रीमूल्य पत्ता लगाउनुहोस्।
      Find the selling price of the bus after three years. [2]
   3. उक्त मिनीबसको कुल कारोवारबाट भएको कूल नाफा वा नोक्सान प्रतिशतमा पत्ता लगाउनुहोस्।
      Find the total profit or loss in percent through the total transaction of that minibus. [1]

   Solution 👉 Click Here

   Solution

   1. यदि बसको खरिद मूल्य रु. $V_o$, वार्षिक मिश्रह्रासदर R% र बसको T वर्षपछिको मूल्य रु.$V_T$ भए $V_T$ लाई $V_o$, R% र T को रूपमा व्यक्त गर्नुहोस्।
      If the purchasing price of the bus is Rs.$V_o$, the annual rate of compound depreciation is R% and price of the bus after T years is Rs.$V_T$, then express $V_T$ in terms of $V_o$, R% and T.
      Using formula of compound depreciation, the expression is
      $V_T=V_0{(1-\frac{R}{100})}^T$
      
   2. तीन वर्ष पछिको सो बसको बिक्रीमूल्य पत्ता लगाउनुहोस्।
      Find the selling price of the bus after three years.
      According to the question
      बसको खरिद मूल्य रु. $V_o=4000000$
      Rate of compound depreciation (R)=15%
      Time in Years (T)=3 years
      Therefore, using formula of compound depreciation, the selling price of the bus after three years is
      $V_T=V_0{(1-\frac{R}{100})}^T$
      or $V_T=4000000{(1-\frac{15}{100})}^3$
      or $V_T=2456500$
      So, the selling price of the bus after three years is 2456500.
   3. उक्त मिनीबसको कुल कारोवारबाट भएको कूल नाफा वा नोक्सान प्रतिशतमा पत्ता लगाउनुहोस्।
      Find the total profit or loss in percent through the total transaction of that minibus.
      According to the question
      बसको खरिद मूल्य रु. $V_o=4000000$
      After using bus for three years, the earning $=1500000$
      तीन वर्ष पछिको सो बसको बिक्रीमूल्य $V_T=2456500$
      मिनीबसको कुल कारोवारबाट भएको,
      खरिद मूल्य रु. $CP=4000000$
      तीन वर्षको आम्दानी र बिक्रीमूल्य सहित जम्मा रकम $SP=1500000+2456500=3956500$
      Therefore, Loss is
      Loss=CP-SP
      or Loss=4000000-3956500=43500
      Now, Loss percentage is
      $L\mathrm{\%}=\frac{43500}{4000000}\times 100\mathrm{\%}$
      or $L\mathrm{\%}=1.0875\mathrm{\%}$
      or $L\mathrm{\%}=1.0875\mathrm{\%}$
      So, the total loss in percent through the total transaction of that minibus is 1.0875 %.

4. एक जना मानिसले विदेश जानको लागि अमेरिकी डलर साट्न बैङ्क गएछ । उक्त दिनको मुद्रा विनिमय दर अनुसार अमेरिकी डलर 1 को खरिद दर रु.132 र बिक्रीदर रु.133 छन्।
   A man went to the Bank to exchange American dollars to visit abroad. In that day, according to the money exchange rate, the buying rate of American Dollar is Rs.132 and selling rate is Rs.133.

   1. सो मानिसले रु.3,32,500 सँग अमेरिकी डलर साट्दा कति डलर प्राप्त गर्दछ ? पत्ता लगाउनुहोस्।
      How many dollars does the man receive while exchanging American dollar with Rs. 3,32,500? Find it. [2]
   2. सोही दिन उसको साथीले अमेरिकी डलर 2800 साट्दा कति नेपाली रुपैयाँ पाउँछ ? पत्ता लगाउनुहोस्।
      How much Nepali rupees does his friend receive while exchanging American dollar 2800 in the same day? Find it [1]
   3. 10 दिनपछि अमेरिकी डलर 1 को बिक्रीदर रु.138.32 हुन्छ भने नेपाली मुद्रा कति प्रतिशतले अवमूल्यन भएछ ? पत्ता लगाउनुहोस्।
      After 10 days, the selling rate for American dollar 1 becomes Rs.138.32 then by what percent the Nepali currency was devaluated? Find it. [1]

   Solution 👉 Click Here

   Solution

   1. सो मानिसले रु.3,32,500 सँग अमेरिकी डलर साट्दा कति डलर प्राप्त गर्दछ ? पत्ता लगाउनुहोस्।
      How many dollars does the man receive while exchanging American dollar with Rs. 3,32,500? Find it.
      According to the question
      The selling [Bank will sell] rate of American Dollar is Rs.133
      So, when exchanging Rs.3,32,500 to get $, he will receive
      $\frac{332500}{133}=2500$
      So, man will receive $\mathrm{\$}2500$ while exchanging with Rs. 3,32,500.
   2. सोही दिन उसको साथीले अमेरिकी डलर 2800 साट्दा कति नेपाली रुपैयाँ पाउँछ ? पत्ता लगाउनुहोस्।
      How much Nepali rupees does his friend receive while exchanging American dollar 2800 in the same day? Find it
      According to the question
      The buying [Bank will buy] rate of American Dollar is Rs.132
      So, when exchanging $\mathrm{\$}2800$ to get Rs, he will receive
      $2800\times 132=369600$
      So, his friend receive Rs $369600$ while exchanging with $\mathrm{\$}2800$ .
   3. 10 दिनपछि अमेरिकी डलर 1 को बिक्रीदर रु.138.32 हुन्छ भने नेपाली मुद्रा कति प्रतिशतले अवमूल्यन भएछ ? पत्ता लगाउनुहोस्।
      After 10 days, the selling rate for American dollar 1 becomes Rs.138.32 then by what percent the Nepali currency was devaluated? Find it.
      According to the question
      Before 10 days
      The selling rate of American Dollar is $S_0$=Rs.133
      After 10 days,
      The selling rate of American Dollar is $S_1$=Rs.138.32
      The difference in serring rate is
      $S_1-S_0$=138.32-133=5.32
      So, the rate of devaluation of Nepali currency is
      $D=\frac{5.32}{133}\times 100\mathrm{\%}$
      or $D=4\mathrm{\%}$

5. वर्ग आधार भएको पिरामिडको उचाइ 21cm र आधार भुजाको लम्बाई 20cm छन्।
   The height of a square based pyramid is 21cm and the length of the base is 20cm.

   1. वर्ग आधार भएको पिरामिडमा कति ओटा त्रिभुजाकार सतहहरू हुन्छन् ?लेख्नुहोस्।
      How many triangular surfaces are there in a square based pyramid?Write it. [1]
   2. उक्त पिरामिडको छड्के उचाइ पत्ता लगाउनुहोस्।
      Find the slant height of the pyramid. [1]
   3. उक्त पिरामिडको पुरा सतहमा प्रति वर्ग से.मी. रु.5 का दरले रङ लगाउँदा जम्मा कति खर्च लाग्छ ? पत्ता लगाउनुहोस्।
      What is the total cost of painting the total surface area of the pyramid at the rate of Rs.5 per square cm? Find it [2]

   Solution 👉 Click Here

   Solution

   1. वर्ग आधार भएको पिरामिडमा कति ओटा त्रिभुजाकार सतहहरू हुन्छन् ?लेख्नुहोस्।
      How many triangular surfaces are there in a square based pyramid? Write it.
      There are 4 triangular surfaces in a square based pyramid.
      0,0
   2. उक्त पिरामिडको छड्के उचाइ पत्ता लगाउनुहोस्।
      Find the slant height of the pyramid.
      According to the question
      Height of the pyramid
      $h=21$
      Length of base of the pyramid
      $a=20$
      For slant height,
      $l^2=h^2+(a/2{)}^2$
      or $l^2={21}^2+{10}^2$
      or $l^2=541$
      or $l=23.26$
      So, the slant height of the pyramid is $l=23.26$cm.
      0,0

      h=21cm

      a=20cm

      h=21cm

      10cm
   3. उक्त पिरामिडको पुरा सतहमा प्रति वर्ग से.मी. रु.5 का दरले रङ लगाउँदा जम्मा कति खर्च लाग्छ ? पत्ता लगाउनुहोस्।
      What is the total cost of painting the total surface area of the pyramid at the rate of Rs.5 per square cm? Find it
      According to the question
      Base area of pyramid is
      BA$=a^2$
      or BA$={20}^2$
      or BA$=400$
      Lateral surface area of pyramid is
      LSA$=2al$
      or LSA$=2\times 20\times 23.26$
      or LSA$=930.4$
      Now, total surface area of pyramid is
      TSA=BA+LSA
      or TSA=400+930.4
      or TSA=1330.4 square cm
      The unit cost for the painting in the TSA of pyramids is
      C=Rs.5 per square cm
      So, total cost of painting is
      Total Cost=TSA x C
      or Total Cost=1330.4 x 5
      or Total Cost=6652
      So, the total cost of painting the total surface area of the pyramid at the rate of Rs.5 per square cm is Rs 6652.

6. चित्रमा सोली र वेलना मिलि बनेको धातुको ठोस वस्तु दिइएको छ । बेलना र सोलीको आधारको अर्धव्यासहरू बराबर छन् । बेलनाको उचाई 40से.मी., सोलीको उचाई 24से.मी. र सोलीकोआधारको अर्धव्यास 7से.मी. छन्।
   A metallic solid made up of a cone and a cylinder is given in the figure. The radii of the base of the cone and cylinder are equal. The height of the cylinder is 40cm, height of the cone is 24cm and radius of the base of cone is 7cm.

   0,0

   40cm

   24cm

   7cm
   
   
   1. सोलीको आधारको अर्धव्यास र छड्के उचाई दिइएको छ भने सोलीको बक्रसतहको क्षेत्रफल पत्ता लगाउने सूत्र लेख्नुहोस्।
      If radius of the base and slant height of the cone are given then write the formula for finding the curved surface area of the cone। [1]
   2. उक्त ठोस वस्तुको आयतन पत्ता लगाउनुहोस्।
      Find the volume of the solid object [2]
   3. वेलनाकार भागको आयतन र सोली भागको आयतन तुलना गर्नुहोस्।
      Compare the volume of the cylindrical part and the volume of conical part? [1]

   Solution 👉 Click Here

   Solution

   1. सोलीको आधारको अर्धव्यास र छड्के उचाई दिइएको छ भने सोलीको बक्रसतहको क्षेत्रफल पत्ता लगाउने सूत्र लेख्नुहोस्।
      If radius of the base and slant height of the cone are given then write the formula for finding the curved surface area of the cone.
      If radius of the base $(r)$ and slant height of the cone $(l)$ are given then, formula for finding the curved surface area of the cone is
      $CSA=\pi rl$
   2. उक्त ठोस वस्तुको आयतन पत्ता लगाउनुहोस्।
      Find the volume of the solid object
      According to the question
      The radius of the base of conical object is
      $r=7$
      The height of conical object is
      $h_1=24$
      0,0

      24cm

      7cm
      The volume of conical object is
      $V_1=\frac{1}{3}\pi r^2{h}_1$
      or $V_1=\frac{1}{3}\times \pi \times 7^2\times 24$
      or $V_1=\pi \times 7^2\times 8$
      or $V_1=392\pi$
      Next
      The radius of the base of cylinderical object is
      $r=7$
      The height of cylinderical object is
      $h_2=40$
      0,0

      40cm

      7cm
      The volume of cylinderical object is
      $V_2=\pi r^2{h}_2$
      or $V_2=\pi \times 7^2\times 40$
      or $V_2=1960\pi$
      Hence,the volume of the solid object is
      $V=V_1+V_2$
      or $V=392\pi +1960\pi$
      or $V=2352\pi$
      or $V=7389$
      So, the volume of the solid object is 7389 cubic cm.
   3. वेलनाकार भागको आयतन र सोली भागको आयतन तुलना गर्नुहोस्।
      Compare the volume of the cylindrical part and the volume of conical part?
      According to the question
      The volume of conical object is
      $V_1=392\pi$
      The volume of cylinderical object is
      $V_2=1960\pi$
      It shows that,
      $\frac{V_2}{V_1}=\frac{\pi r^2{h}_2}{\frac{1}{3}\pi r^2{h}_1}=\frac{3h_2}{h_1}=\frac{3\times 40}{24}=5$
      Therefore, the volume of the cylindrical part is FIVE TIMES larger than the volume of conical part.

7. एउटा आयताकार कोठाको लम्बाइ, चौडाइ र उचाइ क्रमशः 14ft, 13ft र 10ft छन् । उक्त कोठमा 3ft किनारा भएका दुईओटा वर्गाकार झ्याल छन् र दुईओटा 6ft × 3ft का ढोका छन्।
   The length, breadth and height of a rectangular room are 14 ft, 13 ft, and 10 ft respectively. There are two square windows with 3 feet edges and two doors of size 6ft × 3ft in the room.

   1. ढोका र झ्याल बाहेक उक्त कोठको चार भित्ता र सिलिङमा प्रति वर्ग फिटको रु.36 दरले रङ लगाउन कति खर्च लाग्छ ? पत्ता लगाउनुहोस्।
      How much does it cost to paint four walls and ceiling of the room excluding doors and windows at the rate of Rs.36 per square feet? Find it [3]
   2. बजारमा भएको महङ्गीले गर्दा सोही भागमा रङ लगाउन प्रति वर्ग मिटरमा पहिलेको भन्दा एक चौथाइले बढ्दा जम्मा खर्च कतिले वृद्धि हुन्छ ? पत्ता लगाउनुहोस्।
      How much the total cost will increase to paint on same part if the cost of painting per square meter is increased by one third of what it was before due to the increase in the market price? Find it [2]

   Solution 👉 Click Here

   Solution

   0,0

   14ft

   13ft

   10ft

   3x6

   3x3

   3x6

   3x3

   1. ढोका र झ्याल बाहेक उक्त कोठको चार भित्ता र सिलिङमा प्रति वर्ग फिटको रु.36 दरले रङ लगाउन कति खर्च लाग्छ ? पत्ता लगाउनुहोस्।
      How much does it cost to paint four walls and ceiling of the room excluding doors and windows at the rate of Rs.36 per square feet? Find it
      In the rectangular room
      Length (l) = 14 ft
      Breadth (b) = 13 ft
      Height (h) = 10 ft
      The formula to find out the area of 4 walls and ceiling is
      $A_1=2h(l+b)+lb$
      or $A_1=2\times 10(14+13)+14\times 13$
      or $A_1=722$
      Area of two square windows with 3 feet edges is
      $A_2=2l^2$
      or $A_2={2.3}^3$
      or $A_2=18$
      Area of two two doors of size 6ft × 3ft is
      $A_3=2lb$
      or $A_3=2.3\times 6$
      or $A_3=36$
      Area of four walls and ceiling of the room excluding doors and windows is
      $A=A_1-A_2-A_3$
      or $A=722-18-36$
      or $A=668$
      Now, rate of unit cost to paint four walls and ceiling of the room excluding doors and windows is
      C=Rs.36 per square feet
      Hence, the total cost is
      Total Cost=A x C
      or Total Cost=668 x 36
      or Total Cost=24048
      So, cost to paint four walls and ceiling of the room excluding doors and windows at the rate of Rs.36 per square feet is Rs 24048
   2. बजारमा भएको महङ्गीले गर्दा सोही भागमा रङ लगाउन प्रति वर्ग मिटरमा पहिलेको भन्दा एक चौथाइले बढ्दा जम्मा खर्च कतिले वृद्धि हुन्छ ? पत्ता लगाउनुहोस्।
      How much the total cost will increase to paint on same part if the cost of painting per square meter is increased by one third of what it was before due to the increase in the market price? Find it
      According to the question
      The cost of painting per square meter is increased by one third of what it was before due to the increase in the market price, so increased rate is
      C1=$\frac{1}{4}$ of 36
      or C1=9
      Therefore, the total increased cost is
      Total Cost=A x C1
      or Total Cost=668 x 9
      or Total Cost=6012
      So, Rs 6012 will increase to paint on same part if the cost of painting per square meter is increased by one third of what it was before.

8. 3 र 243 का बिचमा 3 ओटा गुणोत्तरीय मध्यमाहरू छन्।
   There are 4 geometric means between 3 and 243

   1. पहिलो पद a, अन्तिम पद b र गुणोत्तर मध्यमाको सङ्ख्या n दिइएको अवस्थामा समान अनुपात r निकाल्ने सूत्र लेख्नुहोस्।
      First term 'a', last term 'b' and number of geometric means 'n' are given. Write the formula for the calculation of common ratio in the given condition. [1]
   2. दिइएको श्रेणीको तेस्रो मध्यमा कति होला ? पत्ता लगाउनुहोस्।
      What is the third mean of the given series? Find it. [2]
   3. 3 र 243 कोसमानान्तरीय मध्यमा र गुणोत्तरीय मध्यमामा कुन कतिले ठूलो छ ? तुलना गर्नुहोस्।
      In arithmetic mean and geometric mean between 3 and 243, which one is greater and by how much? Compare it. [1]

   Solution 👉 Click Here

   Solution

   1. पहिलो पद a, अन्तिम पद b र गुणोत्तर मध्यमाको सङ्ख्या n दिइएको अवस्थामा समान अनुपात r निकाल्ने सूत्र लेख्नुहोस्।
      First term 'a', last term 'b' and number of geometric means 'n' are given. Write the formula for the calculation of common ratio in the given condition.
      According to the question,
      First term =a
      Last term =b
      Number of terms =n
      Then, formula for the calculation of common ratio r is
      $r={\left(\frac{b}{a}\right)}^{\frac{1}{n+1}}$
      
   2. दिइएको श्रेणीको तेस्रो मध्यमा कति होला ? पत्ता लगाउनुहोस्।
      What is the third mean of the given series? Find it.
      According to the question, let $m_1,m_2,m_3$ are 3 geometric means between 3 and 243, then
      $r={\left(\frac{b}{a}\right)}^{\frac{1}{n+1}}={\left(\frac{243}{3}\right)}^{\frac{1}{4}}=3$
      Now, third mean of the given series is
      $m_3=a{r}^3=3\times 3^3=81$
      So, the third mean of the given series is 81.
   3. 3 र 243 कोसमानान्तरीय मध्यमा र गुणोत्तरीय मध्यमामा कुन कतिले ठूलो छ ? तुलना गर्नुहोस्।
      In arithmetic mean and geometric mean between 3 and 243, which one is greater and by how much? Compare it.
      According to the question,
      The arithmetic mean between 3 and 243 is
      $AM=\frac{a+b}{2}=\frac{3+243}{2}=123$
      The geometric mean between 3 and 243 is
      $AM=\sqrt{ab}=\sqrt{3\times 243}=\sqrt{729}=27$
      The difference between AM and GM is
      $AM-GM=123-27=96$
      It shows that, AM is GREATER than 96 then GM.

9. एउटा आयतकार खेल मैदानको परमिति र क्षेत्रफल क्रमशः 66 मि. र 260 बर्ग मिटर छन्।
   The perimeter and area of a rectangular ground are 66 m and 260 sq.m. respectively.

   1. वर्ग समिकरण $a{x}^2+bx+c=0,a\ne 0$ मा $x$ का मूलहरू उल्लेख गर्नुहोस्।
      Illustrate the roots of x in the quadratic equation $a{x}^2+bx+c=0,a\ne 0$ [1]
   2. दिइएको खेल मैदानको लम्बाइ र चौडाइ पत्ता लगाउनुहोस्।
      Find the length and breadth of the given ground [2]
   3. उक्त आयताकार खेल मैदानबाट (13 x 4) वर्ग मिटरका कतिवटा जग्गाका टुक्राहरू तयार गर्न सकिएला ? गणना गर्नुहोस्।
      How many pieces of land can be made with dimension (13×4) square meter from that rectangular field? Calculate it. [1]

   Solution 👉 Click Here

   Solution

   1. वर्ग समिकरण $a{x}^2+bx+c=0,a\ne 0$ मा $x$ का मूलहरू उल्लेख गर्नुहोस्।
      Illustrate the roots of x in the quadratic equation $a{x}^2+bx+c=0,a\ne 0$
      The roots of x in the quadratic equation $a{x}^2+bx+c=0,a\ne 0$ are
      $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
   2. दिइएको खेल मैदानको लम्बाइ र चौडाइ पत्ता लगाउनुहोस्।
      Find the length and breadth of the given ground
      According to the question, let $l$ and $b$ are the length and breadth of the given ground, then
      Perimeter=2(l+b)=66
      Area=lb=260
      It shows that
      $l+b=33,lb=260$
      Using the formula, we get
      $l-b=\sqrt{(l+b{)}^2-4lb}=\sqrt{(33{)}^2-4\times 260}=7$
      Since, $l+b=33,l-b=7$, the values of $l,b$ are
      $l=20,b=13$
      
   3. उक्त आयताकार खेल मैदानबाट (13 x 4) वर्ग मिटरका कतिवटा जग्गाका टुक्राहरू तयार गर्न सकिएला ? गणना गर्नुहोस्।
      How many pieces of land can be made with dimension (13×4) square meter from that rectangular field? Calculate it.
      
      0,0

      13m

      20m

      4m
      According to the side adjustment
      Area of rectangular field is
      A=lb=20x13
      Are of piece of pieces of land is
      a=lb=13x4
      Possible piece of land is
      $N=\frac{A}{a}=\frac{20\times 13}{13\times 4}=5$
      So, 5 pieces of land can be made with dimension (13×4) square meter from that rectangular field.
10. 
    
    1. सरल गर्नुहोस (Simplify): $\frac{x+y}{x-y}-\frac{x-y}{x+y}$ [2]
    2. हल गर्नुहोस (Solve): $4^x+\frac{1}{4^x}=16+\frac{1}{16}$ [3]

    Solution 👉 Click Here

    Solution

    1. सरल गर्नुहोस (Simplify): $\frac{x+y}{x-y}-\frac{x-y}{x+y}$
       The simplification is
       $\frac{x+y}{x-y}-\frac{x-y}{x+y}$
       
       or $\frac{(x+y{)}^2-(x-y{)}^2}{(x-y)(x+y)}$
       
       or $\frac{4xy}{x^2-y^2}$
       
    2. हल गर्नुहोस (Solve): $4^x+\frac{1}{4^x}=16+\frac{1}{16}$
       Let $4^x=a$, then the solution is
       $4^x+\frac{1}{4^x}=16+\frac{1}{16}$
       
       or $a+\frac{1}{a}=16+\frac{1}{16}$
       
       or $\frac{a^2+1}{a}=\frac{{16}^2+1}{16}$
       
       or $\frac{a^2+1}{a}=\frac{257}{16}$
       
       or $16a^2-257a+16=0$
       
       or $(16a-1)(a-16)=0$
       
       or $a=16$ or $a=\frac{1}{16}$
       
       If $a=16$, then
       $4^x=16$ or $x=2$
       If $a=\frac{1}{16}$, then
       $4^x=\frac{1}{16}$ or $x=-2$
       Therefore
       $x=\pm 2$



11. दइएको चित्रमा एउटैआधार PQ र उही समानान्तर रेखाहरू PQ र RS का बीचमा दुईओटा त्रिभुजहरू PQR र PQS बनेका छन्।
    In the given figure, there are two triangles PQR and PQS on the same base PQ and between same parallel lines PQ and RS.

    0,0

    P

    Q

    S

    R
    1. दिइएकोदुई त्रिभुजको क्षेत्रफल बीचको सम्बन्ध लेख्नुहोस्।
       Write the relation between the areas of given two triangles [1]
    2. प्रमाणित गर्नुहोस्ः $\mathrm{△}PQR$ को क्षेत्रफल = $\mathrm{△}PQS$ को क्षेत्रफल
       Prove that : Area of $\mathrm{△}PQR$ = Area of $\mathrm{△}PQS$ [2]
    3. दिइएको चित्रमा $AC||DE$ छन् भने चतुर्भुज ABCD को क्षेत्रफल र त्रिभुज ABE को क्षेत्रफल बराबर हुन्छ भनी प्रमाणित गर्नुहोस्।
       In the given figure, $AC||DE$. Prove that the area of quadrilateral ABCD and area of triangle ABE are equal [2]
       0,0

       A

       B

       C

       D

       E

    Solution 👉 Click Here

    Solution

    1. दिइएकोदुई त्रिभुजको क्षेत्रफल बीचको सम्बन्ध लेख्नुहोस्।
       Write the relation between the areas of given two triangles
       Two triangles PQR and PQS on the same base PQ and between same parallel lines PQ and RS, so
       areas of given two triangles are equal (Area of ∆PQR=Area of ∆PQS)
    2. प्रमाणित गर्नुहोस्ः $\mathrm{△}PQR$ को क्षेत्रफल = $\mathrm{△}PQS$ को क्षेत्रफल
       Prove that : Area of $\mathrm{△}PQR$ = Area of $\mathrm{△}PQS$
       Given that
       PQ||RS
       To prove
       Area of $\mathrm{△}PQR=\mathrm{△}PQS$
       Construction
       draw altitude between two parallel lines PQ||RS, say the height is $h$.
       Now, the proof are as follows.
       
       0,0

       P

       Q

       S

       R

       h

       SN	Statements	Reasons
       1	Area of $\mathrm{△}PQR=\frac{1}{2}PQ\times h$	Formula for area of triangle
       2	Area of $\mathrm{△}PQS=\frac{1}{2}PQ\times h$	Formula for area of triangle
       3	Area of $\mathrm{△}PQR=\mathrm{△}PQS$	Being PQ||RS, both height (h) in 1 and 2 are equal

       This completes the proof.
    3. दिइएको चित्रमा $AC||DE$ छन् भने चतुर्भुज ABCD को क्षेत्रफल र त्रिभुज ABE को क्षेत्रफल बराबर हुन्छ भनी प्रमाणित गर्नुहोस्।
       In the given figure, $AC||DE$. Prove that the area of quadrilateral ABCD and area of triangle ABE are equal
       
       0,0

       A

       B

       C

       D

       E

       F
       Given that
       AC||DR
       To prove
       Area of $◻ABCD=\mathrm{△}ABE$
       Construction
       Intersect AE and CD at F
       Now, the proof are as follows.

       SN	Statements	Reasons
       1	Area of $\mathrm{△}ACD=\mathrm{△}ACE$	Being AC||DE, trangles on same base and between same parallel lines AC||DE
       2	Area of $\mathrm{△}ACD+\mathrm{△}ABC=\mathrm{△}ACE+\mathrm{△}ABC$	Adding equal quantity $\mathrm{△}ABC$ in (1)
       3	Area of $◻ABCD=\mathrm{△}ABE$	Adding traingles from (2)

       This completes the proof.

12. चक्रीय चतुर्भुज PQRS मा $\measuredangle P$ र $\measuredangle R$ सम्मूख कोणहरू हुन्।
    In a cyclic quadrilateral, $\measuredangle P$ and $\measuredangle R$ are opposite angles.

    0,0

    P

    Q

    R

    S

    2x^o

    4x^o
    1. $\measuredangle P$ र $\measuredangle R$ बीचको सम्बन्ध लेख्नुहोस्।
       Write the relation between $\measuredangle P$ and $\measuredangle R$ [1]
    2. यदि $\measuredangle P=2x^o$ र $\measuredangle R=4x^o$ भए $x$ को मान पत्ता लगाउनुहोस्।
       Find the value of $x$ when $\measuredangle P=2x^o$ and $\measuredangle R=4x^o$ [1]
    3. $\measuredangle Q+\measuredangle S={180}^o$ ण् हुन्छ भनी प्रयोगबाट प्रमाणित गर्नुहोस् । कम्तीमा 3cm अर्धव्यास भएका 2 वटा वृत्तहरू जरुरी छन्।
       Prove experimentally that $\measuredangle Q+\measuredangle S={180}^o$. Two circles having at least 3 cm radii are necessary [2]

    Solution 👉 Click Here

    Solution

    1. $\measuredangle P$ र $\measuredangle R$ बीचको सम्बन्ध लेख्नुहोस्।
       Write the relation between $\measuredangle P$ and $\measuredangle R$
       Since "opposite angles of a cyclic quadrilateral are supplementry"
       The relation between $\measuredangle P$ and $\measuredangle R$ is $\measuredangle P+\measuredangle R={180}^o$
    2. यदि $\measuredangle P=2x^o$ र $\measuredangle R=4x^o$ भए $x$ को मान पत्ता लगाउनुहोस्।
       Find the value of $x$ when $\measuredangle P=2x^o$ and $\measuredangle R=4x^o$
       According to the relation
       $\measuredangle P+\measuredangle R={180}^o$
       or $2x+4x=180$
       or $6x=180$
       or $x=30$
       So, the value of $x$ is ${30}^o$.
    3. $\measuredangle Q+\measuredangle S={180}^o$ हुन्छ भनी प्रयोगबाट प्रमाणित गर्नुहोस् । कम्तीमा 3cm अर्धव्यास भएका 2 वटा वृत्तहरू जरुरी छन्।
       Prove experimentally that $\measuredangle Q+\measuredangle S={180}^o$. Two circles having at least 3 cm radii are necessary
       
       0,0

       Q

       R

       S

       P

       y

       x

       x

       y

       ∡Q+∡S=180^o

       P

       Q

       R

       S

       y

       x

       x

       y

       ∡Q+∡S=180^o

       Fig No.:2

       Fig No.:1

       According to the question
       Two circles of radii 3cm and 4 cm respectively are drawn.
       The cyclic quadrilateral PQRS is drawn.
       To verify:
       $\measuredangle Q+\measuredangle S={180}^o$
       Experimental table:
       Angles are measured by protractor and written the values on the following table:

       Fig No.	Measurements	Results
       1	$\measuredangle Q=80,\measuredangle S=100$	$\measuredangle Q+\measuredangle S={180}^o$
       2	$\measuredangle Q=111,\measuredangle S=69$	$\measuredangle Q+\measuredangle S={180}^o$

       This completes the experimental proof with the conclusion that "opposite angles of a cyclic quadrilateral are supplementry".
13. 
    
    1. एउटा चतुर्भुज ABCD को रचना गर्नुहोस जसमा AB=BC=6 से.मि., AD=CD=5.1 से.मि. र $\measuredangle DAB={60}^o$ छन् । उक्त चतुर्भुजको क्षेत्रफलसँग बराबर हुने गरी एउटा त्रिभुज ADM को पनि रचना गर्नुहोस्।
       Construct a quadrilateral ABCD in which AB = BC = 6cm, AD = CD = 5.1 cm and $\measuredangle DAB={60}^o$. Also, construct a triangle ADM whose area is equal to the area of the quadrilateral. [3]
    2. के $BD||MC$ छन्? कारण दिनुहोस्।
       Are $BD||MC$? Give reason. [1]

    Solution 👉 Click Here

    Solution

    1. एउटा चतुर्भुज ABCD को रचना गर्नुहोस जसमा AB=BC=6 से.मि., AD=CD=5.1 से.मि. र $\measuredangle DAB={60}^o$ छन् । उक्त चतुर्भुजको क्षेत्रफलसँग बराबर हुने गरी एउटा त्रिभुज ADM को पनि रचना गर्नुहोस्।
       Construct a quadrilateral ABCD in which AB = BC = 6cm, AD = CD = 5.1 cm and $\measuredangle DAB={60}^o$. Also, construct a triangle ADM whose area is equal to the area of the quadrilateral.
       The construction steps of the questions are as follows.
       1. Draw a line segment AB=6cm.
          0,0

          6cm

          A

          B
       2. At the point A, draw an angle of ${60}^o$, and also draw a ray AD making AD=5.1cm
          0,0

          6cm

          A

          B

          5.1cm

          D

          60^o
       3. Taking B as center, draw an arc of 6cm, and also taking D ac center, draw another arc of 5.1cm.
          Label the point of intersection as C.
          0,0

          6cm

          A

          B

          5.1cm

          D

          60^o

          C

          5.1cm

          6cm
       4. Required quadrilateral ABCD with AB = BC = 6cm, AD = CD = 5.1 cm and $\measuredangle DAB={60}^o$ is constructed.
          0,0

          6cm

          A

          B

          5.1cm

          D

          60^o

          C

          5.1cm

          6cm
       5. Now, to construct a triangle ADM whose area is equal to the area of the quadrilateral ABCD.
          Join BD, and draw a line from C which is parallel to BD (use alternate angle to draw parallel lines).
          0,0

          6cm

          A

          B

          5.1cm

          D

          60^o

          C

          5.1cm

          6cm
       6. Now, take a point M , which is intersection of the parallel line and AB
          The required triangle ADM whose area is equal to the area of the quadrilateral ABCD is constructed.
          0,0

          6cm

          A

          B

          5.1cm

          D

          60^o

          C

          5.1cm

          6cm

          M
          NOTE: In the exam, the last figure [Fig. 6] is sufficient.
    2. के $BD||MC$ छन्? कारण दिनुहोस्।
       Are $BD||MC$? Give reason.
       Yes $BD||MC$, because, in the construction, $BD||MC$ is drawn, in which the alternate angles $\measuredangle DBC$ and $\measuredangle BCM$ are equal.

14. चित्रमा देखाइए अनुसार बत्ती सहित खम्बाको उचाइ AB=20.5 मि. र मानिसको उचाइ CD=1.5 मि. छन्।
    In the given figure, height of pole with bulb (AB) = 20.5m and the height of man (CD) = 1.5 m

    0,0

    D

    B

    E

    C

    A

    20.5m

    1.5m

    45^o
    1. उन्नतांश कोणलाई परिभाषित गर्नुहोस्।
       Define the angle of elevation. [1]
    2. मानिसको उचाई बत्ती सहितको खम्बाको उचाई भन्दा कति कम छ ? लेख्नुहोस्।
       How much is the height of man less than the height of the pole with bulb? Write it [1]
    3. यदि उन्नतांश कोण $\measuredangle EAC={45}^o$ भए मानिस र खम्बा बिचको दूरी पत्ता लगाउनुहोस्।
       If the angle of elevation $\measuredangle EAC={45}^o$ , then find the distance between the man and pole. [1]
    4. मानिसले बत्ती सहितको खम्बाको टुप्पो हेर्दा ${30}^o$ को उन्नतांश कोण हुने गरी उभिन अहिलेको स्थानबाट कति दुरी अगाडि वा पछाडि हिड्नु पर्दछ ? पत्ता लगाउनुहोस्।
       When the man looks at the top of the pole with bulb, how far does he move forward or backward from the current position so that the angle of elevation may be ${30}^o$? Find it. [1]

    Solution 👉 Click Here

    Solution

    1. उन्नतांश कोणलाई परिभाषित गर्नुहोस्।
       Define the angle of elevation.
       The angle of elevation is an angle formed between the horizontal line and the line of sight. If the line of sight is upward from the horizontal line, then the angle formed at the point of sight to the object is an angle of elevation.
       In the figure below, θ is angle of elevation.
       उन्नतांश कोण एक कोण हो जुन तेर्सो रेखा र दृष्टि रेखाको बीचमा बनेको हुन्छ। यदि दृष्टि रेखा तेर्सो रेखाबाट माथितिर छ भने, हेर्ने बिन्दु मा बनेको कोणलाई उन्नतांश कोण भनिन्छ।
       तलको चित्रमा θ भनेको उन्नतांश कोण हो ।
       0,0

       B

       Point of sight

       Object

       θ
    2. मानिसको उचाई बत्ती सहितको खम्बाको उचाई भन्दा कति कम छ ? लेख्नुहोस्।
       How much is the height of man less than the height of the pole with bulb? Write it
       According to question
       height of man= 1.5m
       height of pole with bulb= 20.5m
       So,
       difference= 20.5-1.5=19m
       Therefore, the height of man is 19m less than the height of the pole with bulb.
       0,0

       D

       B

       E

       C

       A

       19m

       1.5m

       1.5m

       45^o
    3. यदि उन्नतांश कोण $\measuredangle EAC={45}^o$ भए मानिस र खम्बा बिचको दूरी पत्ता लगाउनुहोस्।
       If the angle of elevation $\measuredangle EAC={45}^o$ , then find the distance between the man and pole.
       Given that
       Angle of elevation $\measuredangle EAC={45}^o$
       From right angled triangle AEC, we get
       $\tan {45}^o=\frac{AE}{EC}$
       or$1=\frac{19}{EC}$
       or$EC=19$
       Therefore, the distance between the man and pole is 19m.
    4. मानिसले बत्ती सहितको खम्बाको टुप्पो हेर्दा ${30}^o$ को उन्नतांश कोण हुने गरी उभिन अहिलेको स्थानबाट कति दुरी अगाडि वा पछाडि हिड्नु पर्दछ ? पत्ता लगाउनुहोस्।
       When the man looks at the top of the pole with bulb, how far does he move forward or backward from the current position so that the angle of elevation may be ${30}^o$? Find it.
       
       0,0

       G

       B

       E

       F

       A

       19m

       1.5m

       1.5m

       30^o

       C

       D

       45^o
       According to the question
       Let the man move backward (far from the pole) from the current position and reach at F so that angle of elevation will be ${30}^o$, then
       Angle of elevation $\measuredangle EAC={30}^o$
       From right angled triangle AEC, we get
       $\tan {30}^o=\frac{AE}{EF}$
       or$\frac{1}{\sqrt{3}}=\frac{19}{EF}$
       or$EF=19\sqrt{3}$
       Here, the distance between man and pole when angle of elevation is ${45}^o$ is
       $EC=19$
       Also, the distance between man and pole when angle of elevation is ${30}^o$ is
       $EF=19\sqrt{3}$
       The difference in distance is
       $EF-EC=19\sqrt{3}-19=19(\sqrt{3}-1)$
       Therefore, the man must move $19(\sqrt{3}-1)m$ backward (far from the pole) distance from the current position so that the angle of elevation will be ${30}^o$.

15. तल दिइएको तथ्याङ्क 20 जना विद्यार्थीहरूले गणितको परीक्षामा प्राप्त गरका अङ्कहरू हुन्।
    The data given below is the marks obtained by 20 students in an exam of Mathematics.

    प्राप्ताङ्क (Marks obtained)	0-10	10-20	20-30	30-40	40-50
    विद्यार्थीहरूको संख्या (No. of students)	2	3	6	5	4

    
    
    1. अविछिन्न श्रेणीको पहिलो चतुर्थांश निकाल्ने सूत्र $Q_1=L+(\frac{N}{4}-cf)\times \frac{i}{f}$ मा $i$ ले के जनाउँदछ ? लेख्नुहोस्।
       What does i denote in the formula $Q_1=L+(\frac{N}{4}-cf)\times \frac{i}{f}$ for the calculation of first quartile of the continuous series? Write it. [1]
    2. दिइएको तथ्याङ्कको पहिलो चतुर्थांश पत्ता लगाउनुहोस्।
       Find the first quartile of the given data [2]
    3. माथिको तथ्याङ्कबाट रीत पत्ता लगाउनुहोस्।
       Find the mode from the above data. [2]
    4. 20 वा 20 भन्दा बढी अङ्क प्राप्त गर्ने कति प्रतिशत विद्यार्थीहरू रहेछन्? पत्ता लगाउनुहोस्।
       How many percentage of students are there who obtained 20 or more than 20 marks? Find it. [1]
       
    
    

    Solution 👉 Click Here

    Solution

    1. अविछिन्न श्रेणीको पहिलो चतुर्थांश निकाल्ने सूत्र $Q_1=L+(\frac{N}{4}-cf)\times \frac{i}{f}$ मा $i$ ले के जनाउँदछ ? लेख्नुहोस्।
       What does i denote in the formula $Q_1=L+(\frac{N}{4}-cf)\times \frac{i}{f}$ for the calculation of first quartile of the continuous series? Write it.
       i denote in the length of "Class-Interval" in the formula $Q_1=L+(\frac{N}{4}-cf)\times \frac{i}{f}$ for the calculation of first quartile of the continuous series
    2. दिइएको तथ्याङ्कको पहिलो चतुर्थांश पत्ता लगाउनुहोस्।
       Find the first quartile of the given data
       Based on the data given above, the frequency table is prepared as below.

       Marks obtained $X$	Number of students $f$	Cumulative frequency $cf$
       0-10	2	2
       10-20	3	5
       20-30	6	11
       30-40	5	16
       40-50	4	20

       Here, the number of data are $n=20$, thus, based on the formula, the $Q_1$ class is
       $Q_1$ Class$={\left(\frac{n}{4}\right)}^{th}$ item
       or $Q_1$ Class $={\left(\frac{20}{4}\right)}^{th}$ item
       or $Q_1$ Class $=5^{t}h$ item
       Here, $5^{t}h$ item lies in the $cf$ of 5, thus
       $L=10,f=3,cf=5,i=10$
       Hence, the $Q_1$ is
       $Q_1=L+\frac{\frac{n}{4}-cf}{f}\times i$
       or $Q_1=10+\frac{\frac{20}{4}-2}{3}\times 10=20$
       Hence, the first quartile of the given data is
       $Q_1=20$
       

    3. माथिको तथ्याङ्कबाट रीत पत्ता लगाउनुहोस्।
       Find the mode from the above data.
       

       Marks obtained $X$	Number of students $f$
       0-10	2
       10-20	3
       20-30	6
       30-40	5
       40-50	4

       Being highest frequency 6, the model class is $20-30$. Thus,
       $L=20,f_0=3,f_1=6,f_2=5,i=10$
       Hence, using formula, the Mode is
       $M_0=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i$
       or $M_0=20+\frac{6-3}{2\times 6-3-5}\times 10=26.75$
       

    4. 20 वा 20 भन्दा बढी अङ्क प्राप्त गर्ने कति प्रतिशत विद्यार्थीहरू रहेछन्? पत्ता लगाउनुहोस्।
       How many percentage of students are there who obtained 20 or more than 20 marks? Find it.
       According to the question
       Number of students who obtained 20 or more than 20 marks is=6+5+4=15
       Therefore, percentage of students who obtained 20 or more than 20 marks is
       $\frac{15}{20}\times 100\mathrm{\%}=75\mathrm{\%}$

16. हिमानीले विवाह गरीसकेपछि 4 वर्षको अन्तरालमा दुई बच्चाहरू जन्माउने योजना बनाइछिन्।
    Himani planned to have two children at an interval of 4 years after married.

    1. कुनै घट्ना E को सम्भाव्यता मापन कति हुन्छ ? लेख्नुहोस्।
       What is the probability scale of any event 'E'? Write it. [1]
    2. दुवै बच्चाहरू छोरी नै हुने सम्भाव्यता पत्ता लगाउनुहोस्।
       Find the probability of having both children are daughter. [1]
    3. छोरा र छोरी जन्मिने सम्भावित परिणामहरूको सम्भाव्यताहरूलाई एउटा वृक्ष चित्रमा देखाउनुहोस्।
       Show the probabilities of possible outcomes of getting son and daughter in a tree-diagram [2]
    4. कमसेकम एउटा छोरी हुन सक्ने सम्भाव्यता निकाल्नुहोस्।
       Find the probability of having at least one daughter [1]
    
    

    Solution 👉 Click Here

    Solution

    1. कुनै घट्ना E को सम्भाव्यता मापन कति हुन्छ? लेख्नुहोस्।
       What is the probability scale of any event 'E'? Write it.
       The probability of an event E, denoted as P(E), is a measure of the likelihood that the event will occur. The probability scale ranges from 0 to 1, where:
       $P(E)=0$ indicates that the event E is impossible and will not occur.
       $P(E)=1$ indicates that the event E is certain and will occur.
       $0\le P(E)\le 1$ indicates the probability scale of event E
       
    2. दुवै बच्चाहरू छोरी नै हुने सम्भाव्यता पत्ता लगाउनुहोस्।
       Find the probability of having both children are daughter.
       According to the question, we assumed that probability of having son or daughter is equally likely. So
       P(B) = probability of having son/boy=$\frac{1}{2}$
       P(G) = probability of having daughter/girl=$\frac{1}{2}$
       Now, the probability of having both children are daughter is
       P(GG) = P(G) x P(G)=$\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$
       
    3. छोरा र छोरी जन्मिने सम्भावित परिणामहरूको सम्भाव्यताहरूलाई एउटा वृक्ष चित्रमा देखाउनुहोस्।
       Show the probabilities of possible outcomes of getting son and daughter in a tree-diagram
       The probabilities of all possible outcomes of getting son and daughter in a tree-diagram is given below.
       0,0

       P(B)=½ 👨

       P(G)=½ 👧🏽

       P(B)=½

       P(G)=½

       B 👨

       G 👧🏽

       P(B)=½

       P(G)=½

       B 👨

       G 👧🏽

       P(BB)=¼

       P(BG)=¼

       P(GB)=¼

       P(GG)=¼

       1st child

       2nd child
       
       
    4. कमसेकम एउटा छोरी हुन सक्ने सम्भाव्यता निकाल्नुहोस्।
       Find the probability of having at least one daughter
       The probability of having at least one daughter is given as
       P(at least one daughter) = 1-P(no daughter)
       orP(at least one daughter) = 1-P(BB)
       orP(at least one daughter) = 1-¼
       orP(at least one daughter) = ¾
       So, the probability of having at least one daughter is ¾.