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.
   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.
      \(CA=P \left ( 1+\frac{R}{100} \right )^T \)
   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
      \(CA=P \left ( 1+\frac{R}{100} \right )^T \)
      or \(56000=50000 \left ( 1+\frac{R}{100} \right )^1 \)
      or \(56=50 \left ( 1+\frac{R}{100} \right ) \)
      or \(\frac{56}{50}= \left ( 1+\frac{R}{100} \right ) \)
      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
      \(CA=P \left ( 1+\frac{R}{100} \right )^T \)
      or \(CA=50000 \left ( 1+\frac{12}{100} \right )^2 \)
      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.
      \(CA=P \left ( 1+\frac{R}{100} \right )^T \)
   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
      \(CA=P \left ( 1+\frac{R}{100} \right )^T \)
      or \(CA=1000000 \left ( 1+\frac{12}{100} \right )^2 \)
      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
      \(CA=P \left ( 1-\frac{R}{100} \right )^T \)
      or \(CA=200000 \left ( 1-\frac{10}{100} \right )^2 \)
      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.
      
      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.

   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
      \(V_2=\frac{1}{3} \pi r^2h\)
      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
      \(\frac{V_2}{V_1}=\frac{\frac{1}{3}\pi r^2h}{\frac{2}{3}\pi r^3}= \frac{h}{2r}=\frac{24}{14}=\frac{12}{7}\)
      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
   

   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> \(S_n=\frac{n}{2}\left [ 2a+(n-1)d\right ]\)
      
   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
      \(S_n=\frac{n}{2}\left [ a+l\right ]\)
      or\(220=\frac{n}{2}\left [ 4+40\right ]\)
      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 \(ax^2+bx+c=0,a\ne 0\) मा \(x\) का मूलहरू कतिओटा हुन्छन्? लेख्नुहोस्।
      How many roots does quadratic equation \(ax^2+bx+c=0,a\ne 0\) 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 \(ax^2+bx+c=0,a\ne 0\) मा \(x\) का मूलहरू कतिओटा हुन्छन्? लेख्नुहोस्।
      How many roots does quadratic equation \(ax^2+bx+c=0,a\ne 0\) have? Write it.
      There are two roots in the quadratic equation \(ax^2+bx+c=0,a\ne 0\). 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.
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.

    1. वराबर क्षेत्रफल भएका र्दुओटा त्रिभुजहरूको नाम लेख्नुहोस्।
       Write the name of two triangles having equal area. [1]
    2. प्रमाणित गर्नुहोस् (Prove that): \( \triangle PAM=\frac{1}{2} \square AMKN\)
       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 \(\triangle QRT\) 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 \(\triangle QRT\) so formed ? Find it. [2]

    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
       \(\triangle PAM\) and \(\triangle SAM\)
       
       
    2. प्रमाणित गर्नुहोस् (Prove that): \( \triangle PAM=\frac{1}{2} \square AMKN\)
       Given that
       NS||AM
       To prove
       Area of \( \triangle PAM=\frac{1}{2} \square AMKN\)
       Construction
       draw altitude between two parallel lines NS||AM, say the height is \(h\).
       Now, the proof are as follows.
       

       SN	Statements	Reasons
       1	Area of \(\triangle PAM=\frac{1}{2} AM \times h\)	Formula for area of triangle
       2	Area of \(\square AMKN=PQ \times h\)	Formula for area of rectangle
       3	Area of \( \triangle PAM=\frac{1}{2} \square AMKN\)	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 \(\triangle QRT\) 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 \(\triangle QRT\) so formed ? Find it.
       Given that
       PQRS is a square
       So
       Area of \(\square PQRS=\frac{1}{2}d^2\)
       orArea of \(\square PQRS=\frac{1}{2}10^2\)
       orArea of \(\square PQRS=50\)
       Next
       Area of \( \triangle QRT=\frac{1}{2} \square PQRS\)
       orArea of \(\triangle QRT=\frac{1}{2} \times 50\)
       orArea of \(\triangle QRT=25\)
       Therefore, the area of \(\triangle QRT=25cm^2\).

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.

    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.
       

       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.
       2. At the point A, draw an angle of \(60^o\), and also draw a ray AC making AC=5cm
       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.
       4. Required quadrilateral ABCD with ABCD in which AB=4.5cm,AC=CD=5cm, AD=6cm and \(\measuredangle BAC=60^o\) is constructed.
       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.
       6. The required triangle PBC whose area is equal to the area of the quadrilateral ABCD is constructed.
          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, \(\triangle ACP=\triangle ACD\) in area, both are added to \(\triangle ACB\).

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.

    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 ।
    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
       \(M_d=L+\left ( \frac{n}{2}-cf \right ) \times \frac{i}{f}\)
    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 \(= 25^th\) item
       Here, \(25^th\) 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, \(25^th\) 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=\(\frac{1}{2}\)
       
    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.

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.
   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
      \(CI=P \left \{ \left ( 1+\frac{R}{200} \right )^{2T} -1 \right \} \)
      or \(CI=80000 \left \{ \left ( 1+\frac{10}{200} \right )^{2 \times 2} -1 \right \} \)
      or \(CI=80000 \left \{ \left ( \frac{210}{200} \right )^{4} -1 \right \} \)
      or \(CI=80000 \left \{ \left ( \frac{21}{20} \right )^{4} -1 \right \} \)
      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
      \(A=P \left ( 1+\frac{R}{100} \right )^T \)
      or \(106480=80000 \left ( 1+\frac{10}{100} \right )^T \)
      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
      \(V_T=V_0 \left ( 1-\frac{R}{100} \right )^T \)
      
   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
      \(V_T=V_0 \left ( 1-\frac{R}{100} \right )^T \)
      or \(V_T=4000000 \left ( 1-\frac{15}{100} \right )^3 \)
      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
      \(L\%=\frac{43500}{4000000} \times 100 \%\)
      or \(L\%=1.0875 \%\)
      or \(L\%=1.0875 \%\)
      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 \(\$2500\) 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 \(\$2800\) to get Rs, he will receive
      \(2800 \times 132=369600\)
      So, his friend receive Rs \(369600\) while exchanging with \(\$2800\) .
   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
      \(D=\frac{5.32}{133} \times 100\%\)
      or \(D=4\%\)

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.
   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.
   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.

   
   
   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
      \(CSA=\pi r l\)
   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
      \(h_1=24\)
      The volume of conical object is
      \(V_1=\frac{1}{3} \pi r^2h_1\)
      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
      \(h_2=40\)
      The volume of cylinderical object is
      \(V_2=\pi r^2h_2\)
      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,
      \(\frac{V_2}{V_1}=\frac{\pi r^2h_2}{\frac{1}{3}\pi r^2h_1}= \frac{3h_2}{h_1}=\frac{3 \times 40}{24}=5\)
      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

   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
      \(A_2 =2 l^2\)
      or \(A_2 = 2.3^3\)
      or \(A_2 = 18\)
      Area of two two doors of size 6ft × 3ft is
      \(A_3 =2 lb\)
      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
      \( m_3= ar^3=3 \times 3^3=81 \)
      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 \(ax^2+bx+c=0,a\ne 0\) मा \(x\) का मूलहरू उल्लेख गर्नुहोस्।
      Illustrate the roots of x in the quadratic equation \(ax^2+bx+c=0,a\ne 0\) [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 \(ax^2+bx+c=0,a\ne 0\) मा \(x\) का मूलहरू उल्लेख गर्नुहोस्।
      Illustrate the roots of x in the quadratic equation \(ax^2+bx+c=0,a\ne 0\)
      The roots of x in the quadratic equation \(ax^2+bx+c=0,a\ne 0\) 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.
      
      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.

    1. दिइएकोदुई त्रिभुजको क्षेत्रफल बीचको सम्बन्ध लेख्नुहोस्।
       Write the relation between the areas of given two triangles [1]
    2. प्रमाणित गर्नुहोस्ः \(\triangle PQR\) को क्षेत्रफल = \(\triangle PQS\) को क्षेत्रफल
       Prove that : Area of \(\triangle PQR\) = Area of \(\triangle PQS\) [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]

    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 \(\triangle PQR\) को क्षेत्रफल = \(\triangle PQS\) को क्षेत्रफल
       Prove that : Area of \(\triangle PQR\) = Area of \(\triangle PQS\)
       Given that
       PQ||RS
       To prove
       Area of \(\triangle PQR=\triangle PQS\)
       Construction
       draw altitude between two parallel lines PQ||RS, say the height is \(h\).
       Now, the proof are as follows.
       

       SN	Statements	Reasons
       1	Area of \(\triangle PQR=\frac{1}{2} PQ \times h\)	Formula for area of triangle
       2	Area of \(\triangle PQS=\frac{1}{2} PQ \times h\)	Formula for area of triangle
       3	Area of \(\triangle PQR=\triangle PQS\)	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
       
       Given that
       AC||DR
       To prove
       Area of \(\square ABCD=\triangle ABE\)
       Construction
       Intersect AE and CD at F
       Now, the proof are as follows.

       SN	Statements	Reasons
       1	Area of \(\triangle ACD=\triangle ACE\)	Being AC||DE, trangles on same base and between same parallel lines AC||DE
       2	Area of \(\triangle ACD+\triangle ABC=\triangle ACE+\triangle ABC\)	Adding equal quantity \(\triangle ABC\) in (1)
       3	Area of \(\square ABCD=\triangle ABE\)	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.

    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
       

       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.
       2. At the point A, draw an angle of \(60^o\), and also draw a ray AD making AD=5.1cm
       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.
       4. Required quadrilateral ABCD with AB = BC = 6cm, AD = CD = 5.1 cm and \( \measuredangle DAB=60^o\) is constructed.
       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).
       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.
          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

    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 ।
    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.
    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.
       
       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 \(Q_1=L+\left ( \frac{N}{4}-cf \right ) \times \frac{i}{f}\) मा \(i\) ले के जनाउँदछ ? लेख्नुहोस्।
       What does i denote in the formula \(Q_1=L+\left ( \frac{N}{4}-cf \right ) \times \frac{i}{f}\) 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 \(Q_1=L+\left ( \frac{N}{4}-cf \right ) \times \frac{i}{f}\) मा \(i\) ले के जनाउँदछ ? लेख्नुहोस्।
       What does i denote in the formula \(Q_1=L+\left ( \frac{N}{4}-cf \right ) \times \frac{i}{f}\) for the calculation of first quartile of the continuous series? Write it.
       i denote in the length of "Class-Interval" in the formula \(Q_1=L+\left ( \frac{N}{4}-cf \right ) \times \frac{i}{f}\) 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 \(= 5^th\) item
       Here, \(5^th\) 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
       \(\frac{15}{20} \times 100\%=75\%\)

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.
       
       
    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.