CDC SEE Model Set B (2080)

Subject Code: 1031 Set:B

NATIONAL EXAMINATION BOARD
SEE MODEL QUESTION -2080 (SET B)[RE-1031]

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. 200 जना मानिसहरूमा गरिएको सर्वेक्षणमा 120 जनालई फूटवल खेल्न, 85 जनालई भलिवल खेल्न मन पर्दो रहेछ र 30 जनालई दुवै खेलमध्ये कुनै पनि खेल्न मन पर्दो रहेनछ ।
   In a survey of 200 people, 120 like to play football, 85 like to play volleyball and 30 like to play none of these two games

   1. यदि F र V ले क्रमशः फूटबल र भलिवल खेल्न मन पराउने मानिसहरूको समूहहरू जनाउँछ भने $n(\overline{F\cup V})$ को गणनात्मकता लेख्नुहोस्।
      If F and V denote the sets of people who like to play football and volleyball respectively then write the cardinality of $n(\overline{F\cup V})$ [1]
   2. माथिको जानकारीलाई भेनचित्रमा प्रस्तुत गर्नुहोस्।
      Present the above information in a Venn-diagram. [1]
   3. कति जना मानिसलाई भलिबल मात्र खेल्न मनपर्दो रहेछ ? पत्ता लगाउनुहोस्।
      How many people like to play volleyball only? Find it. [3]
   4. दुवै खेल मन पराउने मानिसको सङ्ख्या र यी मध्ये कुनै पनि खेल मन नपराउने मानिसको सङ्ख्याहरू बिच तुलना गर्नुहोस्।
      Compare the number of people who like to play both the games and the number of people who don't like any of these two games [1]

   Solution 👉 Click Here

   Solution
   Let
   U is universal set of all survayed people
   Given that
   F denote the set of people who like to play football
   V denote the set of people who like to play volleyball
   Thus
   Cardinality of all surveyed people as Universal Set U is
   $n(U)=200$
   Cardinality of set F, people who like to play football is
   $n(F)=120$
   Cardinality of set V, people who like to play volleyball is
   $n(V)=85$
   Cardinality of set, people who like to play none of these two games is
   $n(\overline{F\cup V})=30$
   Now the itemwise solution are :

   1. If F and V denote the sets of people who like to play football and volleyball respectively then write the cardinality of $n(\overline{F\cup V})$
      According the question,
      $n(\overline{F\cup V})=30$
      
   2. Present the above information in a Venn-diagram.
      According the question,
      Let us suppose that
      $n(F\cap V)=x$
      Then, the Venn-diagram of given information are as follows.
      0,0

      A

      B

      C

      D

      U

      F

      V

      120-x

      x

      85-x

      30
   3. How many people like to play volleyball only? Find it.
      According to the Venn-diagram given above
      $(120-x)+(x)+(85-x)+(30)=200$
      or$x=35$
      Therefore,
      The number of people like to play volleyball only is
      $n(V_o)=85-x=85-35=50$
      
   4. Compare the number of people who like to play both the games and the number of people who don't like any of these two games
      According to the solution
      The number of people who like to play both the games is
      $n(F\cap V)=x=35$
      The number of people who don't like any of these two games is
      $n(\overline{F\cup V})=30$
      It shows that
      Number of people who like to play both the games is Greater by FIVE than the number of people who don't like any of these two games

   This completes the solution

2. कक्षा १० मा अध्ययनरत छात्रा रीतालाई आफ्नो आमाले पढाई खर्चका लागि रु.50000 वार्षिक चक्रिय व्याजदरमा 2 वर्षको अवधिका लागि एउटा बैंकको मुद्दती निक्षेपमा राख्नु भएछ ,जसअनुसार 1 वर्षको अन्त्यमा चक्रिय मिश्रधन रु. 56000 हुन्छ ।
   Rita is a student studying in class 10. Her mother deposited Rs.50,000 for 2 years in fixed deposit of a bank at compound interest compounded annually for her study expenses and the compound amount at the end of one year is Rs. 56,000.

   1. मूलधन रु. P समय T वर्ष र व्याजदर R% प्रतिवर्ष हुँदा वार्षिक चव्रिmय मिश्रधन CA पत्ता लगाउने सूत्र लेख्नुहोस्।
      For principal Rs. 'P', time T years and rate of interest R% per year, write the formula to find yearly compound amount 'CA'. [1]
   2. वैँकले प्रदान गरेको वार्षिक चक्रिय व्याजदर पत्ता लगाउनुहोस्।
      Find the annual rate of compound interest offered by the bank. [2]
   3. दोस्रो वर्षको अन्त्यमा रीताले प्राप्त गर्ने चक्रिय मिश्रधन कति हुन्छ ? पत्ता लगाउनुहोस्।
      What will be the compound amount that Rita get at the end of 2 years? Find it. [2]

   Solution 👉 Click Here

   Solution
   

   1. मूलधन रु. P समय T वर्ष र व्याजदर R% प्रतिवर्ष हुँदा वार्षिक चक्रिय मिश्रधन CA पत्ता लगाउने सूत्र लेख्नुहोस्।
      For principal Rs. 'P', time T years and rate of interest R% per year, write the formula to find yearly compound amount 'CA'.
      If 'P' is principal, time is T years and rate of interest is R% per year, then the formula to find yearly compound amount 'CA' is.
      $CA=P{(1+\frac{R}{100})}^T$
   2. वैँकले प्रदान गरेको वार्षिक चक्रिय व्याजदर पत्ता लगाउनुहोस्।
      Find the annual rate of compound interest offered by the bank.
      According to the question
      Principal (P)=50000
      Amount (A)=56000
      Rate of annual compound interest (R)=?
      Time in Years (T)=1
      Therefore, using formula of compound amount, we get
      $CA=P{(1+\frac{R}{100})}^T$
      or $56000=50000{(1+\frac{R}{100})}^1$
      or $56=50(1+\frac{R}{100})$
      or $\frac{56}{50}=(1+\frac{R}{100})$
      or $\frac{56}{50}-1=\frac{R}{100}$
      or $\frac{6}{50}=\frac{R}{100}$
      or $R=12$
      Therefore, the annual rate of compound interest offered by the bank is 12%.
   3. दोस्रो वर्षको अन्त्यमा रीताले प्राप्त गर्ने चक्रिय मिश्रधन कति हुन्छ ? पत्ता लगाउनुहोस्।
      What will be the compound amount that Rita get at the end of 2 years? Find it.
      According to the question
      Principal (P)=50000
      Rate of annual compound interest (R)=12%
      Time in Years (T)=2
      Therefore, using formula of compound amount, we get
      $CA=P{(1+\frac{R}{100})}^T$
      or $CA=50000{(1+\frac{12}{100})}^2$
      or $CA=50000{\left(\frac{112}{100}\right)}^2$
      or $CA=50000\times \frac{{112}^2}{10000}$
      or $CA=5\times {112}^2$
      or $CA=62720$
      Therefore, the compound amount that Rita get at the end of 2 years is Rs 62720.

3. राजिवले आफूसँग भएको रु. 1200000 बाट रु. 200000 को मोटरसाइकल र रु. 1000000 को एउटा जग्गा किनेछ । 2 वर्षसम्म मोटरसाइकलको मूल्य वार्षिक 10% का दरले चक्रिय ह्रास हुँदै गएछ भने जग्गाको मूल्य वार्षिक १२% को दरले चक्रिय वृद्धि हुँदै गएको छ ।
   Rajiv has got Rs. 12,00,000. He purchased a motorcycle for Rs. 2,00,000 and a land for Rs. 10,00,000. The price of motorcycle has been depreciating at a compound rate of 10% for 2 years, while the price of land has been increasing at the compound rate of 12% .

   1. चक्रिय वृद्धि निकाल्ने सूत्र लेख्नुहोस्।
      Write the formula to calculate compound growth. [1]
   2. 2 वर्षपछि जग्गाको मूल्य कति पुग्ला ? पत्ता लगाउनुहोस्।
      What will be the price of land after 2 years? Find it [1]
   3. के 2 वर्षपछि मोटरसाइकल र जग्गाको जम्मा मूल्य रु. 1500000 हुन्छ ? गणना गरी लेख्नुहोस्।
      Will the total price of motorcycle and land after 2 years be Rs. 15,00,000 ? Write with calculation. [2]

   Solution 👉 Click Here

   Solution
   

   1. चक्रिय वृद्धि निकाल्ने सूत्र लेख्नुहोस्।
      Write the formula to calculate compound growth.
      If 'P' is initial amount, time is T years and rate of compound interest is R% per year, then the formula to find compound growth CA is.
      $CA=P{(1+\frac{R}{100})}^T$
   2. 2 वर्षपछि जग्गाको मूल्य कति पुग्ला ? पत्ता लगाउनुहोस्।
      What will be the price of land after 2 years? Find it
      According to the question
      Principal (P)=1000000
      Rate of annual compound growth (R)=12%
      Time in Years (T)=2
      Therefore, using formula of compound amount after 2 years, is
      $CA=P{(1+\frac{R}{100})}^T$
      or $CA=1000000{(1+\frac{12}{100})}^2$
      or $CA=1000000{\left(\frac{112}{100}\right)}^2$
      or $CA=1000000\times (\frac{{112}^2}{10000}$
      or $CA=100\times {112}^2$
      or $CA=1254400$
      
   3. के 2 वर्षपछि मोटरसाइकल र जग्गाको जम्मा मूल्य रु. 1500000 हुन्छ? गणना गरी लेख्नुहोस्।
      Will the total price of motorcycle and land after 2 years be Rs. 15,00,000 ? Write with calculation.
      According to the question, for the case of motercycle
      Principal (P)=200000
      Rate of annual compound depreciation (R)=10%
      Time in Years (T)=2
      Therefore, using formula of amount of motercycle after 2 years, is
      $CA=P{(1-\frac{R}{100})}^T$
      or $CA=200000{(1-\frac{10}{100})}^2$
      or $CA=200000{\left(\frac{90}{100}\right)}^2$
      or $CA=200000{\left(\frac{9}{10}\right)}^2$
      or $CA=200000\frac{81}{100}$
      or $CA=2000\times 81$
      or $CA=162000$
      Therefore, total price of motorcycle and land after 2 years be
      $1254400+162000=1416400$
      Hence, the total price of motorcycle and land after 2 years will NOT be. 15,00,000., because ut will be 1416400.

4. एक जना व्यापारीले पाउण्ड स्टर्लिङ्ग (£)1= ने.रु.168.11 को दरमा रु.7,06,062 को पाउण्ड स्टर्लिङ्ग सटही गरेछ । एक हप्तापछि नेपाली रुपियाँ 2% ले अवमूल्यन भएछ ।
   A merchant exchanged Rs.7,06,062 with pound sterling (£) at the rate of pound sterling (£)1 = NRs. 168.11. After one week, Nepali rupees is devaluated by 2%.

   1. आफुसँग भएको नेपाली रुपैयाँ साट्दा व्यापारीले पाउण्ड स्टर्लिङ्ग कति प्राप्त गर्छन्? पत्ता लगाउनुहोस्।
      What amount of pound sterling (£) does the merchant exchange with the Nepali rupees he had? Find it. [1]
   2. नेपाली रुपियाँ २% अवमूल्यन हुँदा नयाँ सटही दर कति कायम हुन्छ? पत्ता लगाउनुहोस्।
      What would be the new exchange rate after 2% devaluation of Nepali rupees? Find it. [1]
   3. माथिको भनाइअनुसार नेपाली रुपियाँ अवमूल्यन भएको दिन पाउण्ड स्टलिङ्गसँग साट्दा व्यापारीलाई कति रुपियाँ नाफा वा नोक्सान हुन्छ ? पत्ता लगाउनुहोस्।
      How much rupees will the merchant gain or lose when he exchanged Nepali rupees with the sterling pound at the time of devaluation? Find it [2]

   Solution 👉 Click Here

   Solution
   

   1. आफुसँग भएको नेपाली रुपैयाँ साट्दा व्यापारीले पाउण्ड स्टर्लिङ्ग कति प्राप्त गर्छन्? पत्ता लगाउनुहोस्।
      What amount of pound sterling (£) does the merchant exchange with the Nepali rupees he had? Find it.
      According to the question
      The exchange rate of pound sterling (£)1 = NRs. 168.11
      So, when exchanging Rs.7,06,062 to get £, he will receive
      $\frac{706062}{168.11}=4200$
      So, merchant will receive £4200 while exchanging with Rs. 7,06,062.
   2. नेपाली रुपियाँ २% अवमूल्यन हुँदा नयाँ सटही दर कति कायम हुन्छ? पत्ता लगाउनुहोस्।
      What would be the new exchange rate after 2% devaluation of Nepali rupees? Find it.
      According to the question
      The old exchange rate of pound sterling (£)1 = NRs. 168.11
      The rate of devaluation is
      2%
      The new exchange rate of pound sterling is
      (£)1 = NRs. 168.11 x (100-2)%
      or (£)1 = NRs. 164.7478
      So, the new exchange rate after 2% devaluation of Nepali rupees is (£)1 = NRs. 164.7478.
   3. माथिको भनाइअनुसार नेपाली रुपियाँ अवमूल्यन भएको दिन पाउण्ड स्टलिङ्गसँग साट्दा व्यापारीलाई कति रुपियाँ नाफा वा नोक्सान हुन्छ ? पत्ता लगाउनुहोस्।
      How much rupees will the merchant gain or lose when he exchanged Nepali rupees with the sterling pound at the time of devaluation? Find it
      At the time of devaluation [2%], the merchant will loss 2% amount [because the devaluation rate is 2%], so the loss amount is
      Loss=Rs.7,06,062 x 2%=14121.24
      

5. एउटा वर्गाकार आधार भएको पिरामिडको छड्के उचाइ र आधार भुजाको अनुपात 5:6 छ र पूरा सतहको क्षत्रेफल 1536 वर्ग से.मि.छन्।
   The ratio of slant height and a side of base of square based pyramid is 5:6 and its total surface area is 1536 sq.cm

   1. आधारको क्षेत्रफल A, उचाइ h र आयतन V को सम्बन्ध लेख्नुहोस्।
      Write the relation among base area (A), height(h) and volume(v) of the pyramid [1]
   2. आधारको क्षेत्रफल र त्रिभुजाकार सतहहरूको क्षेत्रफल तुलना गर्नुहोस्।
      Compare the base area and the area of triangular surfaces. [2]
   3. उक्त पिरामिडको आयतन पत्ता लगाउनुहोस्।
      Find the volume of the pyramid. [2]

   Solution 👉 Click Here

   Solution
   

   1. आधारको क्षेत्रफल A, उचाइ h र आयतन V को सम्बन्ध लेख्नुहोस्।
      Write the relation among base area (A), height(h) and volume(v) of the pyramid
      The relation among base area (A), height(h) and volume(v) of the pyramid is
      $V=\frac{1}{3}A\times h$
      
   2. आधारको क्षेत्रफल र त्रिभुजाकार सतहहरूको क्षेत्रफल तुलना गर्नुहोस्।
      Compare the base area and the area of triangular surfaces.
      
      0,0

      A

      B

      C

      D

      E

      l=5k

      a=6k

      F

      G

      H

      I

      J

      h=?

      a/2=3k

      l=5k
      According to the question, let
      slant height of the pyramid $l=5k$
      side of base of pyramid$a=6k$
      Now, base area of the square pyramid is
      BA$=a^2$
      or BA$=(6k{)}^2$
      or BA$=36k^2$
      Next, lateral surface area (area of triangular surfaces) of pyramid is
      LSA$=2al$
      or LSA$=2\times 6k\times 5k$
      or LSA$=60k^2$
      It shows that, The base area and the area of triangular surfaces of the pyramid are in the ratio $\frac{36k^2}{60k^2}=3:5$
   3. उक्त पिरामिडको आयतन पत्ता लगाउनुहोस्।
      Find the volume of the pyramid.
      According to the question, total area of the pyramid is 1536 sq.cm, therefore
      BA+LSA=TSA
      or $36k^2+60k^2=1536$
      or $96k^2=1536$
      or $k^2=16$
      or $k=4$
      Now, vertical height (h) of the pyramis is
      $h^2=l^2-(a/2{)}^2$
      or$h^2=(5k{)}^2-(3k{)}^2$
      or$h^2=(5\times 4{)}^2-(3\times 4{)}^2$
      or$h^2=400-144$
      or$h^2=256$
      or$h=16$
      Hence, the volume of the pyramid is
      $V=\frac{1}{3}A\times h$
      or $V=\frac{1}{3}36k^2\times 16$
      or $V=12\times 4^2\times 16$
      or $V=12\times 16\times 16$
      or $V=3072$
      So, the volume of the pyramid is 3072 cubic cm.

6. चित्रमा माथि अर्धगोलाकार हुने गरि एउटा सोलीमा आइसक्रिम भरिएको छ । उक्त सोलीको छड्के उचाइ 25 से.मि. र अर्धव्यास 7 से.मि. छन्।
   In the figure, a cone is filled with ice-cream whose upper part is hemispherical. The slant height of the cone is 25cm and its radius is 7 cm.

   0,0

   A

   B

   C

   D

   25cm

   7cm
   1. उक्त सोलिको उचाइ h, अर्धव्यास r र छड्के उचाइ $l$ बिचको सम्बन्ध लेख्नुहोस्।
      Write the relation among the height (h), radius (r) and slant height (l) of the cone [1]
   2. सोली र अर्धगोलाकार भागमा रहेको आइसक्रिमको जम्मा आयतन पत्ता लगाउनु्होस्।
      Find the total volume of ice-cream in conical and hemi spherical parts. [2]
   3. सोली आकार र अर्धगोलाकार भागमा रहेको आइसक्रिमको परिमाणहरू तुलना गर्नुहोस्।
      Compare the quantities of ice-cream in the conical and hemi-spherical parts [1]

   Solution 👉 Click Here

   Solution
   

   1. उक्त सोलिको उचाइ h, अर्धव्यास r र छड्के उचाइ $l$ बिचको सम्बन्ध लेख्नुहोस्।
      Write the relation among the height (h), radius (r) and slant height (l) of the cone
      The relation among the height (h), radius (r) and slant height (l) of the cone is
      $l^2=h^2+r^2$
   2. सोली र अर्धगोलाकार भागमा रहेको आइसक्रिमको जम्मा आयतन पत्ता लगाउनु्होस्।
      Find the total volume of ice-cream in conical and hemi spherical parts.
      According to the question
      The volume of spherical is
      radius $r=7$
      The volume of speherical part is
      $V_1=\frac{2}{3}\pi r^3$
      or $V_1=\frac{2}{3}\pi 7^3$
      or $V_1=\frac{686}{3}\pi$
      or $V_1=718.37$
      Again
      The volume of conical part is
      radius $r=7$
      slant height $l=25$
      vertical height $h^2=l^2-r^2={25}^2-7^2$
      or vertical height $h=24$
      The volume of conical part is
      $V_2=\frac{1}{3}\pi r^{2}h$
      or $V_2=\frac{1}{3}\pi 7^2\times 24$
      or $V_2=\pi 7^2\times 8$
      or $V_2=392\pi$
      or $V_2=1231.50$
      It shows that,
      The total volume of conical and hemi spherical parts is
      $V=V_1+V_2$
      $V=718.37+1231.50$
      $V=1949.87$
      Hence, The total volume of conical and hemi spherical parts is 1949.87 cubic cm.
   3. सोली आकार र अर्धगोलाकार भागमा रहेको आइसक्रिमको परिमाणहरू तुलना गर्नुहोस्।
      Compare the quantities of ice-cream in the conical and hemi-spherical parts
      According to solution
      $\frac{V_2}{V_1}=\frac{\frac{1}{3}\pi r^{2}h}{\frac{2}{3}\pi r^3}=\frac{h}{2r}=\frac{24}{14}=\frac{12}{7}$
      Therefore, the ratio of volume of conical and hemi-spherical part is 12:7.

7. एउटा आयतकार कोठाको लम्बाइ, चौडाइ र उचाइ क्रमशः 13ft, 12ft र 10ft छन्। उक्त कोठामा 3ftx4ft का दुइओटा झ्याल र 3ftx6ft को एउटा ढोका छन्।
   The length, breadth and height of a rectangular room are 13ft, 12ft and 10ft respectively. There are 2 windows of size 3ft×4ft and a door of size 3ft×6ft.

   1. ढोका र झ्याल बाहेक चार भित्ता र सिलिङमा रु. 40 प्रति वर्ग फिटको दरले रङ लगाउन कति खर्च लाग्छ ? पत्ता लगाउनुहोस्।
      How much does it cost to paint four walls and celling excluding door and windows at the rate of Rs. 40 per square ft? Find it. [3]
   2. यदि प्रति वर्ग फिटको दर 25% ले बृद्धि भयो भने कोठाको उही भागहरूमा रङ लगाउन जम्मा खर्च कतिले बढ्न जान्छ ? पत्ता लगाउनुहोस्।
      By how much will the total cost of painting the same parts of the room be increased if the rate of cost per square feet is increased by 25%? Find it. [2]

   Solution 👉 Click Here

   Solution
   

   0,0

   A

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   13ft

   12ft

   10ft

   3x6

   3x4

   3x4

   1. ढोका र झ्याल बाहेक चार भित्ता र सिलिङमा रु. 40 प्रति वर्ग फिटको दरले रङ लगाउन कति खर्च लाग्छ ? पत्ता लगाउनुहोस्।
      How much does it cost to paint four walls and celling excluding door and windows at the rate of Rs. 40 per square ft? Find it.
      In the rectangular room
      Length (l) = 13 ft
      Breadth (b) = 12 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 $A_1=2\times 10(13+12)+13\times 12$
      or $A_1=656$
      Area of two windows of size 3ft x 4ft is
      $A_2=2\times (3\times 4)$
      or $A_2=2.12$
      or $A_2=24$
      Area of a doors of size 6ft × 3ft is
      $A_3=lb$
      or $A_3=3\times 6$
      or $A_3=18$
      Area of four walls and ceiling of the room excluding door and windows is
      $A=A_1-A_2-A_3$
      or $A=656-24-16$
      or $A=614$
      Now, rate of unit cost to paint four walls and ceiling of the room excluding door and windows is
      C=Rs. 40 per square feet
      Hence, the total cost is
      Total Cost=A x C
      or Total Cost=614 x 40
      or Total Cost=24560
      So, cost to paint four walls and ceiling of the room excluding door and windows at the rate of Rs.36 per square feet is Rs 24560
   2. यदि प्रति वर्ग फिटको दर 25% ले बृद्धि भयो भने कोठाको उही भागहरूमा रङ लगाउन जम्मा खर्च कतिले बढ्न जान्छ ? पत्ता लगाउनुहोस्।
      By how much will the total cost of painting the same parts of the room be increased if the rate of cost per square feet is increased by 25%? Find it.
      According to the question
      The cost of painting per square meter is increased by 25%, so increased rate is
      C1=25% of 40
      or C1=10
      Therefore, the total increased cost is
      Total Cost=A x C1
      or Total Cost=614 x 10
      or Total Cost=6140
      So, Rs 6140 will increase to paint on same part if the cost of painting per square meter is increased by 25%.

8. केही पदहरू भएको एउटा समानान्तरीय श्रेणीको पहिलो पद र अन्तिम पद क्रमशः 4 र 40 छन्। सवै पदहरूको योगफल 220 छ।
   The first and last term of an arithmetic series having some terms are 4 and 40 respectively. The sum of all terms is 220.

   1. सो श्रेणीको पहिलो n ओटा पदहरूको योगफल निकाल्ने सूत्र लेख्नुहोस्?
      Write the formula to calculate sum of the first n terms of the series. [1]
   2. उक्त श्रेणीको जम्मा पदहरूको सङ्ख्या पत्ता लगाउनुहोस्।
      Find the total number of terms in the series. [2]
   3. उक्त श्रेणीको तेस्रो पदमा कति थप्दा पहिला तीनओटा पदहरू गुणोत्तर श्रेणीका हुन्छन्? पत्ता लगाउनुहोस्।
      What should be added to the third term of the series so that the first three terms form a geometric series? Find it. [2]

   Solution 👉 Click Here

   Solution
   

   1. सो श्रेणीको पहिलो n ओटा पदहरूको योगफल निकाल्ने सूत्र लेख्नुहोस्?
      Write the formula to calculate sum of the first n terms of the series.
      The formula to calculate sum of the first n terms of the series is>br> $S_n=\frac{n}{2}[2a+(n-1)d]$
      
   2. उक्त श्रेणीको जम्मा पदहरूको सङ्ख्या पत्ता लगाउनुहोस्।
      Find the total number of terms in the series.
      According to the question,
      First term a=4
      Last term l=40
      Sum of all terms $S_n$=220
      Then, formula for the calculation of sum is
      $S_n=\frac{n}{2}[a+l]$
      or$220=\frac{n}{2}[4+40]$
      or$220=\frac{n}{2}\times 44$
      or$220=n\times 22$
      or$n=10$
      So, the total number of terms in the series is n=10.
   3. उक्त श्रेणीको तेस्रो पदमा कति थप्दा पहिला तीनओटा पदहरू गुणोत्तर श्रेणीका हुन्छन्? पत्ता लगाउनुहोस्।
      What should be added to the third term of the series so that the first three terms form a geometric series? Find it.
      According to the question,
      First term a=4
      Last term l=40
      Number of terms n=10
      Then, formula for the calculation of common difference is
      $l=a+(n-1)d$
      or $40=4+9d$
      or $36=9d$
      or $d=4$
      Now, The firdt three terms of the series are
      a,a+d,a+2d
      or4,8,12
      As per the question, let x be addes to the third, then the three terms form a geometric series
      Hence, three terms are
      4,8,12+x for GP
      So
      $8^2=4\times (12+x)$
      or$16=12+x$
      or$x=4$
      Thus, 4 should be added to the third term of the series so that the first three terms form a geometric series

9. एउटा आयतकार खेतको लम्बाइ चौडाइभन्दा 8 मि. ले बढी छ । उक्त खेतको क्षेत्रफल 384 वर्ग मि. छ ।
   The length of a rectangular plot is 8m more than its breadth. The area of the plot is 384 sq.m.

   1. वर्ग समिकरण $a{x}^2+bx+c=0,a\ne 0$ मा $x$ का मूलहरू कतिओटा हुन्छन्? लेख्नुहोस्।
      How many roots does quadratic equation $a{x}^2+bx+c=0,a\ne 0$ have? Write it. [1]
   2. उक्त खेतको लम्बाइ र चौडाइ कति हुन्छ ? पत्ता लगाउनुहोस्।
      What are length and breadth of the plot? Find it. [3]
   3. उक्त खेतलाई वर्गाकार बनाउन लम्बाइबाट कति घटाउनुपर्छ? गणना गर्नुहोस्।
      How long the plot should be decreased from its length to form it is a square plot? Calculate it [1]

   Solution 👉 Click Here

   Solution
   

   1. वर्ग समिकरण $a{x}^2+bx+c=0,a\ne 0$ मा $x$ का मूलहरू कतिओटा हुन्छन्? लेख्नुहोस्।
      How many roots does quadratic equation $a{x}^2+bx+c=0,a\ne 0$ have? Write it.
      There are two roots in the quadratic equation $a{x}^2+bx+c=0,a\ne 0$. They are
      $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
   2. उक्त खेतको लम्बाइ र चौडाइ कति हुन्छ ? पत्ता लगाउनुहोस्।
      What are length and breadth of the plot? Find it.
      According to the question, let $l$ and $b$ are the length and breadth of the given ground, then
      $l-b=8$
      $lb=384$
      Using the formula, we get
      $l+b=\sqrt{(l-b{)}^2+4lb}=\sqrt{(8{)}^2+4\times 384}=40$
      Since, $l-b=8,l+b=40$, the values of $l,b$ are
      $l=24,b=16$
      So, length and breadth of the plot are l=24m, b=16m.
   3. उक्त खेतलाई वर्गाकार बनाउन लम्बाइबाट कति घटाउनुपर्छ? गणना गर्नुहोस्।
      How long the plot should be decreased from its length to form it is a square plot? Calculate it
      Since $l-b=8$, 8m should be decreased from its length to form it is a square plot.
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      A

      B

      C

      D

      E

      F

      8m

      24m

      16m
10. 
    
    1. हल गर्नुहोस (Solve): $2^{x-2}+3^{3-x}=3$ [2]
    2. सरल गर्नुहोस (Simplify): $\frac{1}{x^2-5x+6}+\frac{2}{4x-x^2-3}$ [3]

    Solution 👉 Click Here

    Solution
    

    1. हल गर्नुहोस (Solve): $2^{x-2}+3^{3-x}=3$
       The solution is
       $2^{x-2}+3^{3-x}=3$
       
       or $2^{x-2}+3^{3-x}=2^1+3^0$
       
       Comparing the corresponding terms, we get
       $x-2=1,3-x=0$
       
       or $x=3$
       
       So, the solution is $x=3$.
    2. सरल गर्नुहोस (Simplify): $\frac{1}{x^2-5x+6}+\frac{2}{4x-x^2-3}$
       The simplification is
       $\frac{1}{x^2-5x+6}+\frac{2}{4x-x^2-3}$
       
       or $\frac{1}{(x-2)(x-3)}-\frac{2}{x^2-4x+3}$
       
       or $\frac{1}{(x-2)(x-3)}-\frac{2}{(x-3)(x-1)}$
       
       or $\frac{1(x-1)-2(x-2)}{(x-1)(x-2)(x-3)}$
       
       or $\frac{x-1-2x+4}{(x-1)(x-2)(x-3)}$
       
       or $\frac{-x+3}{(x-1)(x-2)(x-3)}$
       
       or $\frac{-1(x-3)}{(x-1)(x-2)(x-3)}$
       
       or $\frac{-1}{(x-1)(x-2)}$
       
       So, the simplified term is $\frac{-1}{(x-1)(x-2)}$



11. सँगैको चित्रमा NS||AM र NA||KM छन्। NK लाई विन्दु S सम्म लम्ब्याइएको छ ।
    In the adjoining figure, NS//AM and NA//KM. NK is extended to the point S.

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    M
    1. वराबर क्षेत्रफल भएका र्दुओटा त्रिभुजहरूको नाम लेख्नुहोस्।
       Write the name of two triangles having equal area. [1]
    2. प्रमाणित गर्नुहोस् (Prove that): $\mathrm{△}PAM=\frac{1}{2}◻AMKN$
       What are length and breadth of the plot? Find it. [2]
    3. दइएको चित्रमा PQRS एउटा वर्ग हो जसमा PR=10 से.मि. छ । PS लाई T सम्म लम्व्याउँदा बन्ने $\mathrm{△}QRT$ को क्षेत्रफल कति हुन्छ ? पत्ता लगाउनुहोस्।
       In the given figure, PQRS is a square in which PR=10cm. PS is produced to T. What is the area of $\mathrm{△}QRT$ so formed ? Find it. [2]
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    Solution 👉 Click Here

    Solution
    

    1. वराबर क्षेत्रफल भएका र्दुओटा त्रिभुजहरूको नाम लेख्नुहोस्।
       Write the name of two triangles having equal area.
       The name of two triangles having equal area is
       $\mathrm{△}PAM$ and $\mathrm{△}SAM$
       
       
    2. प्रमाणित गर्नुहोस् (Prove that): $\mathrm{△}PAM=\frac{1}{2}◻AMKN$
       Given that
       NS||AM
       To prove
       Area of $\mathrm{△}PAM=\frac{1}{2}◻AMKN$
       Construction
       draw altitude between two parallel lines NS||AM, say the height is $h$.
       Now, the proof are as follows.
       
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       h

       SN	Statements	Reasons
       1	Area of $\mathrm{△}PAM=\frac{1}{2}AM\times h$	Formula for area of triangle
       2	Area of $◻AMKN=PQ\times h$	Formula for area of rectangle
       3	Area of $\mathrm{△}PAM=\frac{1}{2}◻AMKN$	From 1 and 2

       This completes the proof.
    3. दिइएको चित्रमा PQRS एउटा वर्ग हो जसमा PR=10 से.मि. छ । PS लाई T सम्म लम्व्याउँदा बन्ने $\mathrm{△}QRT$ को क्षेत्रफल कति हुन्छ ? पत्ता लगाउनुहोस्।
       In the given figure, PQRS is a square in which PR=10cm. PS is produced to T. What is the area of $\mathrm{△}QRT$ so formed ? Find it.
       Given that
       PQRS is a square
       So
       Area of $◻PQRS=\frac{1}{2}{d}^2$
       orArea of $◻PQRS=\frac{1}{2}{10}^2$
       orArea of $◻PQRS=50$
       Next
       Area of $\mathrm{△}QRT=\frac{1}{2}◻PQRS$
       orArea of $\mathrm{△}QRT=\frac{1}{2}\times 50$
       orArea of $\mathrm{△}QRT=25$
       Therefore, the area of $\mathrm{△}QRT=25c{m}^2$.

12. केन्द्रविन्दु O भएको वृत्तमा एउटै चाप PR मा दुईओटा परिधि कोणहरू ∡RMP र ∡RNP खिचिएका छन्। $\measuredangle ROP$ केन्द्रीय कोण हो ।
    In a circle with centre O, circumference angles RMP and RNP are drawn on the same arc RP. $\measuredangle ROP$ is the central angle.

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    1. ∡RMP र ∡RNP बिचको सम्बन्ध लेख्नुहोस्।
       Write the relation between ∡RMP and ∡RNP. [1]
    2. यदि $\measuredangle MRN=(7x-2{)}^o$ र $\measuredangle MPN=(3x+10{)}^o$ भए $\measuredangle MRN$ को मान पत्ता लगाउनुहोस्।
       If $\measuredangle MRN=(7x-2{)}^o$ and $\measuredangle MPN=(3x+10{)}^o$, find the value of $\measuredangle MRN$. [1]
    3. कम्तिमा 3 से.मि. अर्धव्यास भएका दुइओटा वृत्तहरू खिचि ∡RMP र ∡ROP बिचको सम्बन्ध प्रयोगात्मक रुपमा सिद्ध गर्नुहोस्।
       Verify experimentally that the relation between ∡RMP and ∡ROP after drawing two circles having radii at least 3 cm. [2]

    Solution 👉 Click Here

    Solution
    

    1. ∡RMP र ∡RNP बिचको सम्बन्ध लेख्नुहोस्।
       Write the relation between ∡RMP and ∡RNP.
       The relation between ∡RMP and ∡RNP is that
       $\measuredangle RMP=\measuredangle RNP$
       
    2. यदि $\measuredangle MRN=(7x-2{)}^o$ र $\measuredangle MPN=(3x+10{)}^o$ भए $\measuredangle MRN$ को मान पत्ता लगाउनुहोस्।
       If $\measuredangle MRN=(7x-2{)}^o$ and $\measuredangle MPN=(3x+10{)}^o$, find the value of $\measuredangle MRN$.
       According to the question
       $\measuredangle MRN=\measuredangle MPN$
       or $7x-2=3x+10$
       or $4x=12$
       or $x=3$
       So, the value of $\measuredangle MRN$ is
       $\measuredangle MRN=7x-2={26}^o$
    3. कम्तिमा 3 से.मि. अर्धव्यास भएका दुइओटा वृत्तहरू खिचि ∡RMP र ∡ROP बिचको सम्बन्ध प्रयोगात्मक रुपमा सिद्ध गर्नुहोस्।
       Verify experimentally that the relation between ∡RMP and ∡ROP after drawing two circles having radii at least 3 cm.
       
       0,0

       O

       M

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       Fig No.:2

       Fig No.:1

       According to the question
       Two circles of radii 3cm and 4 cm respectively are drawn.
       To verify: relation between
       $\measuredangle RMP$ and $\measuredangle ROP$
       Experimental table:
       Angles are measured by protractor and written the values on the following table:

       Fig No.	Measurements	Results
       1	$\measuredangle RMP=x={66}^o,\measuredangle ROP=y={132}^o$	$\measuredangle ROP=2\measuredangle RMP$
       2	$\measuredangle RMP=x={52}^o,\measuredangle ROP=y={104}^o$	$\measuredangle ROP=2\measuredangle RMP$

       This shows that, angle substended by same arc RP at the center O is twice than the angle at the circumference at M
13. 
    
    1. एउटा चतुर्भुज ABCD को रचना गर्नुहोस् जसमा AB=4.5 से.मि., AC=CD=5 से.मि. AD=6 से. मि. र $\measuredangle BAC={60}^o$ छन् । उक्त चतुर्भुजको क्षेत्रफलसँग बराबर हुने गरी एउटा त्रिभुज PBC को पनि रचना गर्नुहोस्।
       Construct a quadrilateral ABCD in which AB=4.5cm,AC=CD=5cm, AD=6cm and $\measuredangle BAC={60}^o$. Also, construct a triangle PBC whose area is equal to the area of the quadrilateral. [3]
    2. उक्त रचनाबाट प्राप्त चतुर्भुज ABCD र त्रिभुज PBC को क्षेत्रफल बराबर हुनाको कारण दिनुहोस्।
       Give the reason for being the area of the quadrilateral ABCD and the triangle PBC equal. [1]

    Solution 👉 Click Here

    Solution
    

    1. एउटा चतुर्भुज ABCD को रचना गर्नुहोस् जसमा AB=4.5 से.मि., AC=CD=5 से.मि. AD=6 से. मि. र $\measuredangle BAC={60}^o$ छन् । उक्त चतुर्भुजको क्षेत्रफलसँग बराबर हुने गरी एउटा त्रिभुज PBC को पनि रचना गर्नुहोस्।
       Construct a quadrilateral ABCD in which AB=4.5cm,AC=CD=5cm, AD=6cm and $\measuredangle BAC={60}^o$. Also, construct a triangle PBC 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=4.5cm.
          0,0

          A

          B

          4.5cm
       2. At the point A, draw an angle of ${60}^o$, and also draw a ray AC making AC=5cm
          0,0

          A

          B

          4.5cm

          C

          5cm

          60^o
       3. From the point A, draw an arc of radius 6cm, and from the point C, draw an arc of radius 5cm, both intersect at D giving AD=6cm and CD=5cm.
          0,0

          A

          B

          4.5cm

          C

          5cm

          D

          6cm

          5cm

          60^o
       4. Required quadrilateral ABCD with ABCD in which AB=4.5cm,AC=CD=5cm, AD=6cm and $\measuredangle BAC={60}^o$ is constructed.
          0,0

          A

          B

          4.5cm

          C

          5cm

          D

          6cm

          5cm

          60^o
       5. Now, to construct a triangle PBC whose area is equal to the area of the quadrilateral ABCD.
          Draw a line from D which is parallel to AC (use alternate angle to draw parallel lines). Then extend BA up to P.
          0,0

          A

          B

          4.5cm

          C

          5cm

          D

          6cm

          5cm

          60^o

          P
       6. The required triangle PBC whose area is equal to the area of the quadrilateral ABCD is constructed.
          0,0

          A

          B

          4.5cm

          C

          5cm

          D

          6cm

          5cm

          60^o

          P
          NOTE: In the exam, the last figure [Fig. 6] is sufficient.
    2. उक्त रचनाबाट प्राप्त चतुर्भुज ABCD र त्रिभुज PBC को क्षेत्रफल बराबर हुनाको कारण दिनुहोस्।
       Give the reason for being the area of the quadrilateral ABCD and the triangle PBC equal.
       The area of the quadrilateral ABCD and the triangle PBC equal, because, $\mathrm{△}ACP=\mathrm{△}ACD$ in area, both are added to $\mathrm{△}ACB$.

14. दिइएको चित्रमा AB एक जना बालकको उचाइ हो र विन्दु C आकाशमा उडिरहेको एउटा चङ्गाको स्थान हो । AC=80m चङ्गाको धागोको लम्बाइ हो ।
    In the given figure, AB is the height of a boy and a point C is the position of a flying kite in the sky. AC=80m is the length of the string of the kite.

    0,0

    A

    B

    C

    D

    H

    30^o

    1.3m
    1. उन्नतांश कोणलाई परिभाषित गर्नुहोस्।
       Define the angle of elevation. [1]
    2. यदि $\measuredangle CAH={30}^o$ भए CH को मान कति हुन्छ ? पत्ता लगाउनुहोस्।
       If $\measuredangle CAH={30}^o$, what is the value of CH? Find it [1]
    3. आकाशमा उडिरहेको चङ्गको उचाइ पत्ता लगाउनुहोस्।
       Find the height of the kite flying in the sky. [1]
    4. यदि $\measuredangle CAH$ को नाप ${30}^o$ बाट ${45}^o$ मा बदलियो भने चङ्गा पहिलो उचाइभन्दा कति उचाइमा उड्ने छ ? पत्ता लगाउनुहोस्।
       If the measure of $\measuredangle CAH$ be changed from ${30}^o$ to ${45}^o$, at what height the kite fly than the previous height ? 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

       Sight

       Object

       θ
    2. यदि $\measuredangle CAH={30}^o$ भए CH को मान कति हुन्छ ? पत्ता लगाउनुहोस्।
       If $\measuredangle CAH={30}^o$, what is the value of CH? Find it
       Given that
       Angle of elevation $\measuredangle CAH={30}^o$
       From right angled triangle CAH, we get
       $\sin {30}^o=\frac{CH}{AC}$
       or$\frac{1}{2}=\frac{CH}{80}$
       or$CH=40$
       Therefore, the value of CH=40m.
    3. आकाशमा उडिरहेको चङ्गको उचाइ पत्ता लगाउनुहोस्।
       Find the height of the kite flying in the sky.
       According to the question
       The height of the kite flying in the sky is
       $CD=CH+DH=40+1.3=41.3m$
       
    4. यदि $\measuredangle CAH$ को नाप ${30}^o$ बाट ${45}^o$ मा बदलियो भने चङ्गा पहिलो उचाइभन्दा कति उचाइमा उड्ने छ ? पत्ता लगाउनुहोस्।
       If the measure of $\measuredangle CAH$ be changed from ${30}^o$ to ${45}^o$, at what height the kite fly than the previous height ? Find it
       If the measure of $\measuredangle CAH$ be changed from ${30}^o$ to ${45}^o$, then
       Angle of elevation $\measuredangle CAH={45}^o$
       From right angled triangle CAH, we get
       $\sin {45}^o=\frac{CH}{AC}$
       or$\frac{1}{\sqrt{2}}=\frac{CH}{80}$
       or$CH=56.5$
       The difference in height of CH is
       $56.5-40=16.5$
       So, at the kite fly 16.6m high than the previous height when $\measuredangle CAH={45}^o$.

15. तलको तालिकामा 50 उपभोक्ताहरूको मासिक विद्युत खपत (युनिटमा) दिइएको छ ।
    The monthly consumption of electricity (in units) of 50 consumers is given in the table below.

    विद्युत खपत (Consumption of electricity)	50-60	60-70	70-80	80-90	90-100
    उपभोक्ताको सङ्ख्या (Number. of consumers)	6	8	15	12	9

    
    
    1. निरन्तर श्रेणीको मध्यिका पत्ता लगाउने सूत्र लेख्नुहोस्।
       Write the formula to find the median of a continuous series. [1]
    2. दइएको तथ्याङ्कको रीत पर्ने श्रेणी र मध्यिका पर्ने श्रेणी पत्ता लगाउनुहोस्।
       Find the modal class and median class of the given data. [1]
    3. दिइएको तथ्याङ्कको मध्यिका गणना गर्नुहोस्।
       Calculate the median of the given data. [2]
    4. के माथिको गणनाबाट रीत श्रेणी र मध्यिका श्रेणी एउटै पाउनु भयो ? के यी सधैँ उही हुन्छन्?कारण दिनुहोस्।
       Did you find the modal class and median class same from the above computation? Are they always same? Give reason [1]
       

    Solution 👉 Click Here

    Solution
    

    1. निरन्तर श्रेणीको मध्यिका पत्ता लगाउने सूत्र लेख्नुहोस्।
       Write the formula to find the median of a continuous series.
       The formula to find the median of a continuous series is
       $M_d=L+(\frac{n}{2}-cf)\times \frac{i}{f}$
    2. दिइएको तथ्याङ्कको रीत पर्ने श्रेणी र मध्यिका पर्ने श्रेणी पत्ता लगाउनुहोस्।
       Find the modal class and median class 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 $cf$
       50-60	6	6
       60-70	8	14
       70-80	15	29
       80-90	12	41
       90-100	9	50

       Here, highest frequency is 15, so
       The model class is 70-80
       Next, the median class is
       $M_d$ Class$={\left(\frac{n}{2}\right)}^{th}$ item
       or $M_d$ Class $={\left(\frac{50}{2}\right)}^{th}$ item
       or $M_d$ Class $={25}^{t}h$ item
       Here, ${25}^{t}h$ item lies in the $cf$ of 25, thus
       The median class is 70-80
    3. दिइएको तथ्याङ्कको मध्यिका गणना गर्नुहोस्।
       Calculate the median of the given data.
       Here, ${25}^{t}h$ item lies in the $cf$ of 29, thus
       $L=70,f=15,cf=14,i=10$
       Hence, the $M_d$ is
       $M_d=L+\frac{\frac{n}{2}-cf}{f}\times i$
       or $M_d=70+\frac{\frac{50}{2}-14}{15}\times 10=77.3$
       Hence, the Median of the given data is
       $M_d=77.33$
    4. के माथिको गणनाबाट रीत श्रेणी र मध्यिका श्रेणी एउटै पाउनु भयो ? के यी सधैँ उही हुन्छन्? कारण दिनुहोस्।
       Did you find the modal class and median class same from the above computation? Are they always same? Give reason
       Yes, I found the modal class and median class same from the above computation.
       They are NOT always same class. Because the model class lies in highest frequency class, the Median class lies in $\frac{n}{2}$ the item class.

16. 12 जना केटा र 18 जना केटीहरू भएको एउटा कक्षाबाट पहिलोलाई पुनःनपठाई दुइ जना विद्यार्थीहरू गोला प्रथाबाट छानेका छन्।
    From a class having 12 boys and 18 girls, two students are selected randomly without sending the first student back to the class.

    
    
    1. पारस्परिक निषेधित घट्नाहरूलाई परिभाषित गर्नुहोस्।
       Define mutually exclusive events. [1]
    2. केटा र केटी पर्ने सम्भावित परिणामहरूका सम्म्भाव्यताहरूलाई एउटा वृक्ष चित्रमा देखाउनुहोस्।
       Show the probabilities of possible outcomes of selecting boys and girls in a tree diagram. [2]
    3. दुई जना केटी पर्ने सम्भाव्यता पत्ता लगाउनुहोस्।
       Find the probability of selecting both girls. [2]
    4. कम्तिमा एकजना केटा पर्ने सम्भाव्यता पूर्ण सम्भाव्यताभन्दा कतिले कम हुन्छ ?गणना गर्नुहोस्।
       By how much the probability of getting at least one boy is less than the total probability? Calculate it. [1]
       

    Solution 👉 Click Here

    Solution
    

    1. पारस्परिक निषेधित घट्नाहरूलाई परिभाषित गर्नुहोस्।
       Define mutually exclusive events.
       Mutually exclusive events are events that cannot occur at the same time. In probability theory, two events are considered mutually exclusive if the occurrence of one event means that the other event cannot occur simultaneously, and vice versa. In other words, the events have no outcomes in common.
       पारस्परिक निषेधित घट्नाहरू भनेका एकै समयमा आउन नसक्ने घटनाहरू हो । सम्भाव्यतामा दुइ घटनाहरूलाई एकै समयमा आउन सक्दैन भने तिनिहरुलाई पारस्परिक निषेधित घट्नाहरू भनिन्छ। अथवा, कुनै पनि साझा घटनहरु नभएका अवस्थालाई पारस्परिक निषेधित घट्नाहरू भनिन्छ।
    2. केटा र केटी पर्ने सम्भावित परिणामहरूका सम्म्भाव्यताहरूलाई एउटा वृक्ष चित्रमा देखाउनुहोस्।
       Show the probabilities of possible outcomes of selecting boys and girls in a tree diagram.
       According to the question, we assumed that probability of having boys or girls is equally likely. So
       P(B) = probability of having boy=$\frac{1}{2}$
       P(G) = probability of having girl=$\frac{1}{2}$
       
       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
    3. दुई जना केटी पर्ने सम्भाव्यता पत्ता लगाउनुहोस्।
       Find the probability of selecting both girls.
       According to the question, we assumed that probability of having boy or girl is equally likely. So
       P(B) = probability of having boy=$\frac{1}{2}$
       P(G) = probability of having girl=$\frac{1}{2}$
       Now, the probability of having both children girl is
       P(GG) = P(G) x P(G)=$\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$
    4. कम्तिमा एकजना केटा पर्ने सम्भाव्यता पूर्ण सम्भाव्यताभन्दा कतिले कम हुन्छ ?गणना गर्नुहोस्।
       By how much the probability of getting at least one boy is less than the total probability? Calculate it.
       The probability of having at least one boy is given as
       P(at least one boy) = 1-P(no boy)
       orP(at least one boy) = 1-P(GG)
       orP(at least one boy) = 1-¼
       orP(at least one boy) = ¾
       So, the probability of getting at least one boy is ¼ less than the total probability.