RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2

RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2

Other Exercises

Question 1.
Find :
(i) 22% of 120
(ii) 25% of Rs. 1000
(iii) 25% of 10 kg
(iv) 16.5% of 5000 metres
(v) 135% of 80 cm
(vi) 2.5% of 10000 ml
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 1

Question 2.
Find the number ‘a’, if
(i) 8.4% of a is 42
(ii) 0.5% of a is 3
(iii) \(\frac { 1 }{ 2 }\) % of a is 50
(iv) 100% of a is 100
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 2
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 3

Question 3.
x is 5% of y, y is 24% of z. If x = 480, find the values of y and z.
Solution:
x = 5% of y, y = 24% of z.
x = 480
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 4

Question 4.
A coolie deposits Rs. 150 per month in his post office Saving Bank account. If this is 15% of his monthly income, find his monthly income.
Solution:
Let his monthly income = Rs. x
15% of x = Rs. 150
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 5

Question 5.
Asha got 86.875% marks in the annual examination. If she got 695 marks, find the number of marks of the Examination.
Solution:
Let total marks of the examination = x
86.875% of x = 695
=> 86.875 x \(\frac { 1 }{ 100 }\) x x = 695
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 6

Question 6.
Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened ?
Solution:
Let the school opened for = x days = 90% of x = 216
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 7

Question 7.
A garden has 2000 trees. 12% of these are mango trees, 18% lemon and the rest are orange trees. Find the number of orange trees.
Solution:
Number of total trees = 2000
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 8
Rest trees = 2000 – (240 + 360) = 2000 – 600 = 1400
Number of orange trees = 1400

Question 8.
Balanced diet should contain 12% of protein, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food in take.
Solution:
Balance diet contains
Protein = 12%
Fats = 25%
Carbohydrates = 63%
Number of total calories = 2600
Number of calories of proteins = 12% of 2600 = \(\frac { 12 }{ 100 }\) x 2600 = 312
Number of calories of fats = 25% of 2600 = \(\frac { 25 }{ 100 }\) x 2600 = 650
Number of calories of carbohydrates = 63% of 2600 = \(\frac { 63 }{ 100 }\) x 2600 = 1638

Question 9.
A cricketer scored a total of 62 runs in 96 balls. He hits 3 sixes, 8 fours, 2 twos and 8 singles. What percentage of the total runs came in :
(i) Sixes
(ii) Fours
(iii) Twos
(iv) Singles
Solution:
Total score of a cricketer = 62 runs
(z) Number of sixes = 3
Run from 3 sixes = 3 x 6 = 18
Percentage = \(\frac { 18 }{ 62 }\) x 100 = 29.03%
(ii) Number of fours = 8
Total run from 8 fours = 4 x 8 = 32
Percentage = \(\frac { 32 }{ 62 }\) x 100 = 51.61%
(iii) Number of twos = 2
Total score from 2 twos = 2 x 2 = 4
Percentage = \(\frac { 4 }{ 62 }\) x 100 = \(\frac { 400 }{ 62 }\) = 6.45%
(iv) Number of single run = 8
Percentage = \(\frac { 8 }{ 62 }\) x 100 = \(\frac { 800 }{ 62 }\) = 12.9%

Question 10.
A cricketer hits 120 runs in 150 balls during a test match. 20% of the runs came in 6’s, 30% in 4’s, 25% in 2’s and the rest in 1’s. How many runs did he score in :
(i) 6’s
(ii) 4’s
(iii) 2’s
(iv) singles
What % of his shots were scoring ones ?
Solution:
Total runs scored by a cricketer =120
(i) Number of runs from sixes (6’s) = 20% of 120
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 9

Question 11.
Radha earns 22% of her investment. If she earns Rs. 187, then how much did she invest ?
Solution:
Total earning from investment = Rs. 187
Percent earning = 22%
Let his investment = x
Then 22% of x = Rs. 187
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 10

Question 12.
Rohit deposits 12% his income in a bank. He deposited Rs. 1440 in the bank during 1997. What was his total income for the year 1997 ?
Solution:
Deposit in the bank = Rs. 1440
Percentage = 12% of his total income
Let his total income = Rs. x
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 11

Question 13.
Gunpowder contains 75% nitre and 10% sulphur. Find the amount of the gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.3 kg sulphur ?
Solution:
(i) In gunpowder,
Nitre = 75%
Sulphur = 10%
Let total amount of gunpowder = x kg
Nitre = 9 kg
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 12

Question 14.
An alloy of tin and copper consists of 15 parts of tin and 105 parts of copper. Find the percentage of copper in the alloy ?
Solution:
In an alloy,
Number of parts of tin = 15
and number of parts of copper = 105
Total parts = 15 + 105 = 120
Percentage of copper in the alloy = \(\frac { 105 }{ 120 }\) x 100 = 87.5%

Question 15.
An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy.
Solution:
In an alloy,
Copper = 32%
Nickel = 40%
Rest is zinc = 100 – (32 + 40) = 100 – 72 = 28%
Mass of zinc in 1 kg = 28% of 1 kg = \(\frac { 28 }{ 100 }\) x 100 gm = 280 gm.

Question 16.
A motorist travelled 122 kilometres before his first stop. If he had 10% of his journey to complete at this point, how long was the total ride ?
Solution:
Distance travelled before first stop = 122 km
Let total journey = x km
10% of x = 122
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 13

Question 17.
A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female ?
Solution:
Total numbers of teachers = 30
Number of male teachers = 12
Number of female teacher = 30 – 12 = 18
Percentage of female teachers = \(\frac { 18 x 100 }{ 30 }\) = 60%

Question 18.
Aman’s income is 20% less than that of Anil. How much percent is Anil’s income more than Aman’s income ?
Solution:
Let Anil’s income = Rs. 100
Then Aman’s income = Rs, 100 – 20 = Rs. 80
Now, difference of both’s incomes = 100 – 80 = Rs. 20
Anil income is Rs. 20 more than that of Aman’s
Percentage = \(\frac { 20 x 100 }{ 80 }\) = 25%

Question 19.
The value of a machine depreciates every year by 5%. If the present value of the machine be Rs. 100000, what will be its value after 2 years ?
Solution:
Present value of machine = Rs. 100000
Rate of depreciation per year = 5%
Period = 2 years
Value of machine after 2 years
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 14

Question 20.
The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years ?
Solution:
Present population of the town = 60000
Increase annually = 10%
Period = 2 years
Population after 2 years will be
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 15

Question 21.
The population of a town increases 10% annually. If the present population is 22000, find its population a year ago.
Solution:
Let the population of the town a year ago was = x
Increase in population = 10%
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 16

Question 22.
Ankit was given an increment of 10% on his salary. His new salary is Rs. 3575. What was his salary before increment ?
Solution:
Let the salary of Ankit before increment = x
Increment given = 10% of the salary
Salary after increment will be
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 17

Question 23.
In the new budget, the price of petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase ?
Solution:
Let price of petrol before budged = Rs. 100
Increase = 10%
Price after budget = Rs. 100 + 10 = Rs. 110
Let the consumption of petrol before budget = 100 l
Price pf 100 l = Rs. 110
Now of new price is Rs. 110, consumption = 100 l
are of new price will be 100, then
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 18

Question 24.
Mohan’s income is Rs. 15500 per month. He saves 11% of his income. If his income increases by 10% then he reduces his saving by 1%, how much does he save now ?
Solution:
Mohan’s income = Rs. 15500
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 19
We see that the savings is same
There is no change in savings.

Question 25.
Shikha’s income is 60% more than that of Shalu. What percent is Shalu’s income less than Shikha’s ?
Solution:
Let Shalu’s income = Rs. 100
Then Shikha’s income will be = Rs. 100 + 60 = Rs. 160
Now difference in their incomes = Rs. 160 – 100 = Rs. 60
Shalu’s income is less than Shikha’s income by Rs. 60
Percentage less = \(\frac { 60 x 100 }{ 160 }\) = \(\frac { 75 }{ 2 }\) % = 37.5%

Question 26.
Rs. 3500 is to be shared among three people so that the first person gets 50% of the second who in turn gets 50% of the third. How much will each of them get ?
Solution:
Let the third person gets = Rs. x
Then second person will get
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 20
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 21

Question 27.
After a 20% hike, the cost of Chinese Vase is Rs. 2000. What was the original price of the object ?
Solution:
Let the original price of the vase = Rs. x
Hike in price = 20%
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 22
Original price of the vase = Rs. 1666.66

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RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1

RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1

Other Exercises

Question 1.
Write each of the following as percent: Solution—
(i) \(\frac { 7 }{ 25 }\)
(ii) \(\frac { 16 }{ 625 }\)
(iii) \(\frac { 5 }{ 8 }\)
(iv) 0.8
(v) 0.005
(vi) 3 : 25
(vii) 11 : 80
(viii) 111 : 125
(ix) 13 : 75
(x) 15 : 16
(xi) 0.18
(xii) \(\frac { 7 }{ 125 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 1
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 2

Question 2.
Convert the following percentages to fractions and ratios :
(i) 25%
(ii) 2.5%
(iii) 0.25%
(iv) 0.3%
(v) 125%
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 3
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 4

Question 3.
Express the following as decimal fractions :
(i) 27%
(ii) 6.3%
(iii) 32%
(iv) 0.25%
(v) 7.5%
(vi) \(\frac { 1 }{ 8 }\) %
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 5

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RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1

RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1

Question 1.
Rakesh can do a piece of work in 20 days. How much work can he do in 4 days ?
Solution:
Rakesh can do it in 20 days = 1
his 1 day’s work = \(\frac { 1 }{ 20 }\)
and his 4 days work = \(\frac { 1 }{ 20 }\) x 4 = \(\frac { 1 }{ 5 }\) th work

Question 2.
Rohan can paint \(\frac { 1 }{ 3 }\) of a painting in 6 days. How many days will he take to complete the painting ?
Solution:
Rohan can paint \(\frac { 1 }{ 3 }\) of painting in = 6 days
he will complete the painting in = \(\frac { 6 x 3 }{ 1 }\) = 18 days

Question 3.
Anil can do a piece of work in 5 days and Ankur in 4 days. How long will they take to do the same work, if they work together ?
Solution:
Anil’s 1 day’s work = \(\frac { 1 }{ 5 }\)
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 1

Question 4.
Mohan takes 9 hours to mow a large lawn. He and Sohan together can mow it in 4 hours. How long will Sohan take to mow the lawn if he works alone ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 2

Question 5.
Sita can finish typing a 100 page document in 9 hours, Mita in 6 hours and Rita in 12 hours. How long will they take to type a 100 page document if they work together?
Solution:
Sita can do a work in 1 hour = \(\frac { 1 }{ 9 }\)
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 3

Question 6.
A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 4

Question 7.
A and B can do a piece of work in 18 days; B and C in 24 days and A and C in 36 days. In what time can they do it, all working together ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 5

Question 8.
A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. How much time will A alone take to finish the work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 6
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 7

Question 9.
A, B and C can reap a field in 15\(\frac { 3 }{ 4 }\) days; B, C and D in 14 days; C, D and A in 18 days; D, A and B in 21 days. In what time can A, B, C and D together reap it ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 8
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 9

Question 10.
A and B can polish the floors of a building in 10 days A alone can do \(\frac { 1 }{ 4 }\) th of it in 12 days. In how many days can B alone polish the floor ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 10

Question 11.
A and B can finish a work in 20 days. A alone can do \(\frac { 1 }{ 5 }\) th of the work in 12 days. In how many days can B alone do it ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 11
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 12

Question 12.
A and B can do a piece of work in 20 days and B in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the remaining work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 13

Question 13.
A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days will A finish the remaining work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 14
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 15

Question 14.
Aasheesh can paint his doll in 20 minutes and his sister Chinki can do so in 25 minutes. They paint the doll together for five minutes. At this juncture they have a quarrel and Chinki withdraws from painting. In how many minutes will Aasheesh finish the painting of the remaining doll ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 16
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 17

Question 15.
A and B can do a piece of work in 6 days and 4 days respectively. A started the work; worked at it for 2 days and then was joined by B. Find the total time taken to complete the work.
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 18
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 19

Question 16.
6 men can complete the electric fitting in a building in 7 days. How many days will it take if 21 men do the job ?
Solution:
6 men can complete the work in = 7 days
1 man will complete the same work in = 7 x 6 days (Less men, more days)
21 men will finish the work in = \(\frac { 7 x 6 }{ 21 }\) days (More men, less days) = 2 days

Question 17.
8 men can do a piece of work in 9 days. In how many days will 6 men do it ?
Solution:
8 men can do a work in = 9 days
1 men will do the work in = 9 x 8 days (Less men, more days)
6 men will do the work in = \(\frac { 9 x 8 }{ 6 }\) days (More men, less days)
= \(\frac { 72 }{ 6 }\) = 12 days

Question 18.
Reema weaves 35 baskets in 25 days. In how many days will she weave 55 baskets?
Solution:
Reema can weave 35 baskets in = 25 days
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 20

Question 19.
Neha types 75 pages in 14 hours. How many pages will she type in 20 hours ?
Solution:
Neha types pages in 14 hours = 75 pages
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 21

Question 20.
If 12 boys earn Rs. 840 in 7 days, what will 15 boys earn in 6 days ?
Solution:
12 boys in 7 days earn an amount of = Rs. 840
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 22

Question 21.
If 25 men earn Rs. 1000 in 10 days, how much will 15 men earn in 15 days ?
Solution:
25 men can earn in 10 days = Rs. 1000
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 23

Question 22.
Working 8 hours a day, Ashu can copy a book in 18 days. How many hours a day should he work so as to finish the work in 12 days ?
Solution:
Ashu can copy a book in 18 days working in a day = 8 hours
He will copy the book in 1 day working = 8 x 18 hours a day (Less days, more hours a day)
He will copy the book in 12 days working in a day = \(\frac { 8 x 18 }{ 12 }\) hours
(More days, less hours a day)
= \(\frac { 144 }{ 12 }\) = 12 hours a day

Question 23.
If 9 girls can prepare 135 garlands in 3 hours, how many girls are needed to prepare 270 garlands in 1 hour.
Solution:
135 garlands in 3 hours are prepared by = 9 girls
1 garland in 3 hours will be prepared by
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 24

Question 24.
A cistern can be filled by one tap in 8 hours, and by another in 4 hours. How long will it take to fill the cistern if both taps are opened together ?
Solution:
First tap’s 1 hour work to fill the cistern = \(\frac { 1 }{ 8 }\)
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 25

Question 25.
Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and then B is turned off. How much time will A take to fill the remaining tank ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 26

Question 26.
A pipe can fill a cistern in 10 hours. Due to a leak in the bottom, it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 27

Question 27.
A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 28
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 29

Question 28.
A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern of both the tap and the pipe are opened together ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 30

Hope given RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 are helpful to complete your math homework.

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1

More Exercises

Question 1.
A manufacturing company sells a T.V. to a trader A for ₹ 18000. Trader A sells it to a trader B at a point of ₹ 750 and trader B sells it to a consumer at a profit of ₹ 900. If the rate of sales tax (under VAT) is 10%, find
(i) the amount of tax received by the Government.
(ii) the amount paid by the consumer for the T.V.
Solution:
Sale price of a T.V. to trader A = ₹ 18000
Rate of VAT tax = 10%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q1.1
(i) Total tax paid to Govt. = ₹ (I800 + 75 + 90) = ₹ 1965
(ii) Amount paid by the consumer to trader B = ₹ 18000 + 750 + 900 + Tax 1965 = ₹ 21615

Question 2.
A manufacturer sells a washing machine to a wholesaler for ₹ 15000. The wholesaler sells it to a trader at a profit of ₹ 1200 and the trader sells it to a consumer at a profit of ₹ 1800. If the rate of VAT is 8%, find :
(i) The amount of VAT received by the State Government on the sale of this machine from the manufacturer and the wholesaler.
(ii) The amount that the consumer pays for the machine.
Solution:
Total amount under VAT = ₹ 15000 + ₹ 1200 + ₹ 1800 = ₹ 18000
(i) VAT = 8% of ₹ 18000
= \(\frac { 8 }{ 100 }\) x 18000 = ₹ 1440
(ii) Consumer pays for the machine = ₹ 18000 + ₹ 1440 = ₹ 19440

Question 3.
A manufacturer buys raw material for ₹ 40000 and pays sales tax at the rate of 4%. He sells the ready stock for ₹ 78000 and charges sales tax at the rate of 7.5%. Find the VAT paid by the manufacturer.
Solution:
Cost price of raw material = ₹ 40000
Rate of sales tax = 4%
Total tax = ₹ \(\frac { 40000 x 4 }{ 100 }\) = ₹ 1600
Selling price = ₹ 78000
Rate of sales tax = 7.5%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q3.1
VAT paid by the manufacturer = ₹ 5850 – ₹ 1600 = ₹ 4250

Question 4.
A shopkeeper buys a camera at a discount of 20% from the wholesaler, the printed price of the camera being ₹ 1600 and the rate of sales tax is 6%. The shopkeeper sells it to the buyer at the printed price and charges sales tax at the same rate. Find
(i) the price at which the camera can be bought.
(ii) the VAT (Value Added Tax) paid by the shopkeeper.
Solution:
Printed price of the camera (MP) = ₹ 1600
Rate of discount = 20%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q4.1
(i) Price of camera = ₹ 1600 + ₹ 96 = ₹ 1696
(ii) VAT paid by the shopkeeper= ₹ 96 – ₹ 76.80 = ₹ 19.20

Question 5.
The printed price of an article is ₹ 60000. The wholesaler allows a discount of 20% to the shopkeeper. The shopkeeper sells the article to the customer at the printed price. Sales tax (under VAT) is charged at the rate of 6% at every stage. Find :
(i) the cost to the shopkeeper inclusive of tax.
(ii) VAT paid by the shopkeeper to the Government.
(iii) the cost to the customer inclusive of tax.
Solution:
Printed price of an article = ₹ 60000
Rate of discount allowed = 20%
Total discount = ₹ 60000 x \(\frac { 20 }{ 100 }\) = ₹ 12000
S.P. after discount = ₹ 60000 – ₹ 12000 = ₹ 48000
Rate of VAT = 6%
(i) Amount paid by the shopkeeper
= ₹ 48000 + ₹ 48000 x \(\frac { 6 }{ 100 }\)
= ₹ 48000 + ₹ 2880 = ₹ 50880
(ii) The price at which the shopkeeper sold to the customer = ₹ 60000
Profit = ₹ 60000 – ₹ 48000 = ₹ 12000
VAT paid by the customer to the Govt.
= ₹ 12000 x \(\frac { 6 }{ 100 }\) = ₹ 720
(iii) Total cost to the customer = ₹ 60000 + VAT inclusive of tax
= ₹ 60000 + \(\frac { 60000 x 6 }{ 100 }\)
= ₹ 60000 + ₹ 3600 = ₹ 63600

Question 6.
A shopkeeper bought a TV at a discount of 30% of the listed price of ₹ 24000. The shopkeeper offers a discount of 10% of the listed price to his customer. If the VAT (Value Added Tax) is 10%, find : the amount paid by the customer, the VAT to be paid by the shopkeeper.
Solution:
List price = ₹ 24000
Discount = 30%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q6.1
(i) Amount paid by customer = ₹ 21600 + ₹ 2160 = ₹ 23760
(ii) Total VAT to be paid by shopkeeper = ₹ 2160 – ₹ 1680 = ₹ 480

Question 7.
A shopkeeper sells an article at the listed price of ₹ 1500 and the rate of VAT is 12% at each stage of sale. If the shopkeeper pays a VAT of ₹ 36 to the Government, what was the amount inclusive of tax at which the shopkeeper purchased the articles from the wholesaler?
Solution:
List price (M.P.) of an article = ₹ 1500
Rate of VAT = 12%
Total VAT = ₹ \(\frac { 1500 x 12 }{ 100 }\) = ₹ 180
But VAT paid by the shopkeeper = ₹ 36
Total VAT paid by wholeseller = ₹ 180 – ₹ 36 = ₹ 144
Rate of VAT = 12%
S.P. of the whole seller = \(\frac { 144 x 100 }{ 12 }\) = ₹ 1200
Total amount paid by the wholeseller including VAT = ₹ 1200 + ₹ 144 = ₹ 1344

Question 8.
A shopkeeper buys an article whose list price is ₹ 800 at some rate of discount from a wholesaler. He sells the article to a consumer at the list price and charges sales tax at the prescribed rate of 7.5%. If the shopkeeper has to pay a VAT of ₹ 6, find the rate of discount at which he bought the article from the wholesaler.
Solution:
List price (MP) of an article = ₹ 800
S.P. of the shopkeeper = ₹ 800
Rate of VAT = 7.5%
Total VAT = ₹ \(\frac { 800 x 7.5 }{ 100 }\) = ₹ 60
VAT paid by the shopkeeper = ₹ 6
VAT paid by the wholeseller = ₹ 60 – ₹ 6 = ₹ 54
Rate of VAT = 7.5%
S.P. of wholeseller
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q8.1

Question 9.
A manufacturing company ‘P’ sells a Desert cooler to a dealer A for ₹ 8100 including sales tax (under VAT). The dealer A sells it to a dealer B for ₹ 8500 plus sales tax and the dealer B sells it to a consumer at a profit of ₹ 600. If the rate of sales tax (under VAT) is 8%, find
(i) the cost price of the cooler for the dealer A.
(ii) the amount of tax received by the Government.
(iii) the amount which the consumer pays for the cooler.
Solution:
Manufactures ‘P’ selling price for Desert cooler including sales tax (VAT) = ₹ 8100
Rate of sales tax (VAT) = 8%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q9.1
Cost price of dealer A = ₹ 7500
and sale price of dealer A = ₹ 8500
Gain = ₹ 8500 – ₹ 7500 = ₹ 1000
or cost price of dealer B = ₹ 8500
Gain = ₹ 600
S.P. of dealer B = ₹ 8500 + ₹ 600 = ₹ 9100
Consumers cost price = ₹ 8500 + ₹ 600 = ₹ 9100
(ii) Tax paid to the Govt.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q9.2
= ₹ 600 + ₹ 80 + ₹ 48 = ₹ 728
The amount which the consumer pays = ₹ 7500 + ₹ 1000 + ₹ 600 + ₹ 728 = ₹ 9828

Question 10.
A manufacturer marks an article for ₹ 5000. He sells it to a wholesaler at a discount of 25% on the marked price and the wholseller sells it to a retailer at a discount of 15% on the marked price. The retailer sells it to a consumer at the marked price and at each stage the VAT is 8%.
Calculate the amount of VAT received by the Government from :
(i) the wholesaler.
(ii) the retailer.
Solution:
Marked price (M.P.) of an article = ₹ 5000
Discount given to the wholesaler = 25%
Cost price of wholesaler or S.P. of the manufacturer
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q10.2
(i) VAT received from the wholesaler = ₹ 340 – ₹ 300 = ₹ 40
(ii) and VAT received by the retailer = ₹ 400 – ₹ 340 = ₹ 60

Question 11.
A manufacturer listed the price of his goods at ₹ 160 per article. He allowed a discount of 25% to a wholesaler who in his turn allowed a discount of 20% on the listed price to a retailer. The rate of sales tax on the goods is 10%. If the retailer sells one article to a consumer at a discount of 5% on the listed price, then find
(i) the VAT paid by the wholesaler.
(ii) the VAT paid by the retailer.
(iii) the VAT received by the Government.
Solution:
List price (MP) of the goods = ₹ 160 per article
Rate of discount = 25%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q11.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q11.2
(i) VAT paid by the wholesaler = ₹ 12.80 – ₹ 12 = ₹ 0.80
(ii) VAT paid by the retailer = 15.20 – 12.80 = ₹ 2.40
(iii) Total VAT paid to the Govt. = ₹ 15.20

Question 12.
Kiran purchases an article for ₹ 5, 400 which includes 10% rebate on the marked price and 20% sales tax (under VAT) on the remaining price. Find the marked price of the article.
Solution:
Let market price of the article be ₹ x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q12.1
Hence, market price of the article is ₹ 5000

Question 13.
A shopkeeper buys an article for ₹ 12000 and marks up its price by 25%. The shopkeeper gives a discount of 10% on the marked up price. He gives a further off-season discount of 5% on the balance. But the sales tax (under VAT) is charged at 8% on the remaining price. Find :
(i) the amount of VAT which a customer has to pay.
(ii) the final price he has to pay for the article.
Solution:
Cost price of an article = ₹ 12000
Rate of mark up in price = 25%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q13.1
(i) Amount of sales tax = ₹ 1026
(ii) Price to be paid = ₹ 12825 + ₹ 1026 = ₹ 13851

Question 14.
In a particular tax period, Mr. Sunder Dass, a shopkeeper pruchased goods worth ₹ 960000 and paid a total tax of ₹ 62750 (under VAT). During this period, his sales consisted of taxable turnover of ₹ 400000 of goods taxable at 6% and ₹ 480000 for goods taxable at 12.5%. He also sold tax exempted goods worth ₹ 95640 in the same period. Calculate his tax liability (under VAT) for this period.
Solution:
Cost price of good purchased by Sunder Dass = ₹ 960000
Tax paid (VAT) = ₹ 62750
Sale of goods worth = ₹ 400000
Rate of VAT = 6%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q14.1
Tax paid to Govt. (VAT) = ₹ 62750
Tax liability = ₹ 84000 – ₹ 62750 = ₹ 21250

Question 15.
In the tax period ended March 2015, M/S Hari Singh & Sons purchased floor tiles worth ₹ 800000 taxable at 7.5% and sanitary fittings worth ₹ 750000 taxable at 10%. During this period, the sales turnover for floor tiles and sanitary fittings are worth ₹ 840000 and ₹ 920000 respectively. However, the floor tiles worth ₹ 60000 were returned by the firm during the same period. Calculate the tax liability (under VAT) of the firm for this tax period.
Solution:
Cost of floor tiles = ₹ 800000
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q15.2
Liability of tax of the firm = 155000 – (135000 + 4500) = ₹ 155000 – ₹ 139500 = ₹ 15500

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax EX 1 are helpful to complete your math homework.

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Karnataka SSLC Maths Model Question Paper 4 Kannada Medium

Karnataka SSLC Maths Model Question Paper 4 Kannada Medium is part of Karnataka SSLC Maths Model Question Papers. Here we have given Karnataka SSLC Maths Model Question Paper 4 Kannada Medium.

BoardKSEEB, Karnataka Board
TextbookKTBS, Karnataka
ClassSSLC Class 10
SubjectMaths
Paper SetModel Paper 4
CategoryKarnataka Board Model Papers

Karnataka SSLC Maths Model Question Paper 4 Kannada Medium

ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80

I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)

Question 1.
ಕೆಳಗಿನ ಸಂಖ್ಯೆಗಳ ಗಣಗಳಲ್ಲಿ ಸಮರೂಪ ತ್ರಿಭುಜಗಳಾಗುವಂತಹ ಜೋಡಿಯು
(A) (3, 4, 6) (9, 12, 24)
(B) (3, 4, 6) (9, 12, 18)
(C) (2,4, 6) (2,3, 14)
(D) (5, 10, 15) (10, 30, 45)

Question 2.
ಚತುರ್ಥಕದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರ
(A) πr2
(B) \(\frac { 1 }{ 2 }\) πr2
(C) \(\frac { 1 }{ 3 }\) πr2
(D) \(\frac { 1 }{ 4 }\) πr2

Question 3.
(2, 3) ಮತ್ತು (4, 5) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡದ ಮಧ್ಯಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳು
(A) (3, 4)
(B) (4, 5)
(C) (5, 6)
(D) (6, 7)

Question 4.
ಧನ ಪೂಣಾಂಕ ‘a’ ಅನ್ನು ಪೂಣಾಂಕ ‘b’ ನಿಂದ ಭಾಗಿಸಿದಾಗ ಅನನ್ಯ ಪೂರ್ಣಾಂಕಗಳಾದ ‘q’ ಮತ್ತು ‘r’ ಗಳು ಇರುತ್ತವೆ. ಇದಕ್ಕೆ ಸರಿಹೊಂದುವ ಹೇಳಿಕೆ.
(A) b = a × q + r
(B) q = a × b + r
(C) a = b × q – r
(D) a = b × q + r

Question 5.
tan260° ಯ ಬೆಲೆ
(A) \(\frac { 1 }{ 3 }\)
(B) \(\frac { 1 }{ \surd 3 }\)
(C) 3
(D) √3

Question 6.
m ನ ಯಾವ ಧನಾತ್ಮಕ ಬೆಲೆಗೆ 3x2 + ka + 3 = 0 ಸಮೀಕರಣದ ಮೂಲಗಳು ಸಮವಾಗಿರುತ್ತವೆ.
(A) 2
(B) 3
(C) 5
(D) 6

Question 7.
ಒಂದು ಪ್ರಯೋಗದ ಎಲ್ಲಾ ಪ್ರಾಥಮಿಕ ಘಟನೆಗಳ ಸಂಭವನೀಯತೆಗಳ ಮೊತ್ತವು
(A) 0
(B) 1
(C) 2
(D) 3

Question 8.
ಒಂದು ಸಿಲಿಂಡರ್‌ನ ಪಾದದ ವಿಸ್ತೀರ್ಣ 24 cm2 ಮತ್ತು ಎತ್ತರ 10 cm ಆದರೆ ಸಿಲಿಂಡರ್‌ನ ಘನಫಲ
(A) 24 cm3
(B) 48 cm3
(C) 240 cm3
(D) 480 cm3

II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)

Question 9.
ಥೇಲ್ಸ್‌ನ ಪ್ರಮೇಯವನ್ನು ನಿರೂಪಿಸಿ,

Question 10.
135 ಮತ್ತು 225 ರ ಮ.ಸಾ.ಅ. ವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 11.
ಬಹುಪದೋಕ್ತಿ x2 – 3x +5 ರ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 12.
ಬಾಹ್ಯ ಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕದ ಉದ್ದವು 8 cm ಮತ್ತು ವೃತ್ತಕೇಂದ್ರ ಹಾಗು ಬಾಹ್ಯಬಿಂದುವಿನ ದೂರ 10 cm ಆದರೆ ವೃತ್ತದ ತ್ರಿಜ್ಯವೆಷ್ಟು?

Question 13.
sin A = \(\frac { 3 }{ 4 }\) ಆದರೆ cosec A ನ ಬೆಲೆಯೇನು?

Question 14.
ಹಂತ ವಿಚಲನಾ ವಿಧಾನದಿಂದ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರವನ್ನು ಬರೆಯಿರಿ.

III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)

Question 15.
ವಾರ್ಷಿಕ ಸಂಬಳ ₹ 5000 ಮತ್ತು ಪ್ರತಿವರ್ಷಕ್ಕೆ ಹೆಚ್ಚುವರಿ ಬತ್ಯ ₹ 200 ಇರುವ ಕೆಲಸಕ್ಕೆ ಸುಬ್ಬರಾವ್ 1995 ರಲ್ಲಿ ಸೇರಿದರು. ಯಾವ ವರ್ಷದಲ್ಲಿ ಅವರ ಸಂಬಳ ₹ 7000 ಆಗುತ್ತದೆ?

Question 16.
ಚಿತ್ರದಲ್ಲಿ LM || CB ಮತ್ತು LN || CD ಆದರೆ \(\frac { AM }{ AB }\) = \(\frac { AN }{ AD }\) ಎಂದು ಸಾಧಿಸಿ.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 1

Question 17.
ಈ ಜೋಡಿ ಸಮೀಕರಣಗಳನ್ನು ವರ್ಜಿಸುವ ವಿಧಾನದಿಂದ ಬಿಡಿಸಿ, 3x + 4y = 10 & 2x – 2y = 2.

Question 18.
ಕೆಳಗಿನ ರೇಖಾತ್ಮಕ ಸಮೀಕರಣಗಳ ಜೋಡಿಗಳು ಪ್ರತಿನಿಧಿಸುವ ಸರಳರೇಖೆಗಳು ಒಂದು ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆಯೇ? ಸಮಾಂತರವಾಗಿವೆಯೆ? ಅಥವಾ ಐಕ್ಯಗೊಂಡಿವೆಯೆ? ಕಂಡುಹಿಡಿಯಿರಿ
9x + 3y + 12 = 0; 18x + 6y + 24 = 0.

Question 19.
ಪರಧಿಯು 22 cm ಇರುವ ಒಂದು ಅರ್ಧ ವೃತ್ತದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 20.
4 cm ಮತ್ತು 6 cm ತ್ರಿಜ್ಯಗಳಿರುವ ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳಿವೆ. 6 cm ತ್ರಿಜ್ಯದ ವೃತ್ತದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ 4 cm ತ್ರಿಜ್ಯದ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕವನ್ನು ರಚಿಸಿ.

Question 21.
(2, -5) ಮತ್ತು (-2, 9) ರಿಂದ ಸಮಾನ ದೂರದಲ್ಲಿರುವ x-ಅಕ್ಷದ ಮೇಲಿನ ಬಿಂದುವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 22.
ಶೃಂಗಬಿಂದುಗಳು (1, -3), (4, 1) ಮತ್ತು (2, 3) ಆಗಿರುವ ತ್ರಿಭುಜದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 23.
ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ \(\frac { -1 }{ 4 }\) ಮತ್ತು ಗುಣಲಬ್ಧ \(\frac { 1 }{ 4 }\) ಆಗಿರುವ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 24.
x2 + 6x – 7 = 0 ಸಮೀಕರಣವನ್ನು ವರ್ಗಪೂರ್ಣಗೊಳಿಸುವ ವಿಧಾನದಿಂದ ಬಿಡಿಸಿ.

Question 25.
ಗೋಪುರದ ಪಾದದಿಂದ 30m ದೂರದ ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ, ಗೋಪುರದ ತುದಿಯನ್ನು ನೋಡಿದಾಗ ಉಂಟಾಗುವ ಉನ್ನತ ಕೋನವು 30° ಆದರೆ, ಗೋಪುರದ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 26.
ಒಂದು ಕಟ್ಟಡದ ಮೇಲಿನಿಂದ ಹಾಗೂ ಕೆಳಗಿನಿಂದ ಬೆಟ್ಟದ ತುದಿಯನ್ನು ಗಮನಿಸಿದಾಗ ಉಂಟಾದ ಉನ್ನತ ಕೋನವು 45° ಮತ್ತು 60° ಆಗಿದೆ. ಕಟ್ಟಡದ ಎತ್ತರ 24 m ಆದರೆ ಬೆಟ್ಟದ ಎತ್ತರವೇನು?

Question 27.
ಈ ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 2
ಅಥವಾ
ಈ ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಬಹುಲಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 3

Question 28.
ಒಂದು ಚೀಲದಲ್ಲಿ 3 ಕೆಂಪು ಚೆಂಡುಗಳು ಮತ್ತು 5 ಕಪ್ಪು ಚೆಂಡುಗಳಿವೆ. ಚೀಲದಿಂದ ಯಾದೃಚ್ಛಿಕವಾಗಿ ಒಂದು ಚೆಂಡನ್ನು ತೆಗೆಯಲಾಗಿದೆ. ತೆಗೆದ ಚಂಡು ಕೆಂಪು ಆಗಿರುವ ಸಂಭವನೀಯತೆ ಎಷ್ಟು?

Question 29.
64 cm3 ಘನಫಲವನ್ನು ಹೊಂದಿರುವ 2 ವರ್ಗ ಘನಗಳ ಮುಖಗಳನ್ನು ಸೇರಿಸಿ ಒಂದು ಆಯತ ಘನಾಕೃತಿ ಮಾಡಿದೆ. ಈ ಘನಾಕೃತಿಯ ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
14 cm ಎತ್ತರವಿರುವ ಒಂದು ಕುಡಿಯುವ ನೀರಿನ ಗಾಜಿನ ಲೋಟವು ಶಂಕುವಿನ ಭಿನ್ನಕದ ರೂಪದಲ್ಲಿದೆ. ಅದರ ಎರಡು ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ವ್ಯಾಸಗಳು 4 cm ಮತ್ತು 2 cm ಗಳಾಗಿವೆ. ಗಾಜಿನ ಲೋಟ ಸಾಮರ್ಥ್ಯವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 30.
12 ಮತ್ತು 15 ರ ಲ.ಸಾ.ಅ. ಮತ್ತು ಮ.ಸಾ.ಅ. ವನ್ನು ಅವಿಭಾಜ್ಯ ಅಪವರ್ತನ ವಿಧಾನದಿಂದ ಕಂಡುಹಿಡಿಯಿರಿ.

IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)

Question 31.
ವೃತ್ತದ ಮೇಲಿನ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಒಂದು ಸಮಾಂತರ ಚತುರ್ಭುಜದಲ್ಲಿ ವೃತ್ತವು ಅಂತಸ್ಥವಾದಾಗ ಸಮಾಂತರ ಚತುರ್ಭುಜವು ವಜ್ರಾಕೃತಿಯಾಗುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.

Question 32.
ಪಾದ 8 cm ಮತ್ತು ಎತ್ತರ 4 cm ಇರುವ ಒಂದು ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು ಅದರ ಬಾಹುಗಳು ಮೊದಲು ರಚಿಸಿದ ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜದ ಅನುರೂಪ ಬಾಹುಗಳ 1\(\frac { 1 }{ 2 }\) ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,

Question 33.
ಎರಡು ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳ ವರ್ಗಗಳ ಮೊತ್ತವು 290 ಆದರೆ ಆ ಪೂಣಾರ್ಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಆಯತಾಕಾರದ ಹೊಲದ ಕರ್ಣವು ಅದರ ಚಿಕ್ಕ ಬಾಹುವಿಗಿಂತ 60 m ಹೆಚ್ಚಾಗಿದೆ. ಅದರ ದೊಡ್ಡ ಬಾಹುವು ಚಿಕ್ಕ ಬಾಹುವಿಗಿಂತ 30 m ಹೆಚ್ಚಾಗಿದ್ದರೆ, ಆ ಹೊಲದ ಬಾಹುಗಳ ಉದ್ದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 34.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 4
ಅಥವಾ
sec A (1 – sin A) (sec A + tan A) = 1 ಎಂದು ಸಾಧಿಸಿ.

Question 35.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಓಜೀವ್ ನಕ್ಷೆಯನ್ನು ರಚಿಸಿ, (ಕಡಿಮೆ ವಿಧಾನ)
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 5

Question 36.
3x2 – x – 4 ಎಂಬ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ ಹಾಗು ಶೂನ್ಯತೆಗಳು ಮತ್ತು ಸಹಗುಣಕಗಳ ನಡುವಿನ ಸಂಬಂಧವನ್ನು ತಾಳೆ ನೋಡಿ.
ಅಥವಾ
x4 – 3x2 + 4x + 5 ನ್ನು x2 + 1 – x ದಿಂದ ಭಾಗಿಸಿ ಭಾಗಲಬ್ದ ಮತ್ತು ಶೇಷವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)

Question 37.
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೂರು ಪದಗಳ ಮೊತ್ತವು 15 ಮತ್ತು ಅವುಗಳ ಅಂತ್ಯಪದಗಳ ವರ್ಗಗಳ ಮೊತ್ತವು 58 ಆಗಿದೆ. ಶ್ರೇಢಿಯ ಆ ಮೂರು ಪದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೊದಲ 5 ಪದಗಳ ಮೊತ್ತವು ಮುಂದಿನ 5 ಪದಗಳ ಮೊತ್ತದ ನಾಲ್ಕನೇ ಒಂದು ಭಾಗದಷ್ಟಿದೆ. ಮೊದಲ ಪದ 2 ಆದರೆ, a20 = -112 ಎಂದು ಸಾಧಿಸಿ ಮತ್ತು S20 ನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 38.
ಎರಡು ಸಮರೂಪ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳ ಅನುಪಾತವು ಅವುಗಳ ಅನುರೂಪ ಬಾಹುಗಳ ವರ್ಗಗಳ ಅನುಪಾತಕ್ಕೆ ಸಮನಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.

Question 39.
ಒಂದು ಬಾವಿಯ ವ್ಯಾಸ 3 m ಮತ್ತು ಆಳ 14 m ಇರುವಂತೆ ತೋಡಿದೆ. ಭೂಮಿಯಿಂದ ತೆಗೆದ ಮಣ್ಣನ್ನು ಬಾವಿಯ ಸುತ್ತಲು ಸಮವಾಗಿ ಹರಡಿ 4 m ಅಗಲವಿರುವ ವೃತ್ತಾಕಾರದ ಕಟ್ಟೆಯನ್ನು ಕಟ್ಟಿದೆ. ಕಟ್ಟೆಯ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 40.
ನಕ್ಷೆಯ ಮೂಲಕ ಸಮೀಕರಣಗಳನ್ನು ಬಿಡಿಸಿ.
2x + y = 8
x + 2y = 7

Solutions

I.
Solution 1.
(B) (3, 4, 6) (19, 12, 18)

Solution 2.
(D) \(\frac { 1 }{ 4 }\) πr2

Solution 3.
(A) (3, 4)

Solution 4.
(D) a = b × q + r

Solution 5.
(C) 3

Solution 6.
(D) 6

Solution 7.
(B) 1

Solution 8.
(C) 240 cm3

II.
Solution 9.
ತ್ರಿಭುಜದ ಎರಡು ಬಾಹುಗಳನ್ನು ಎರಡು ವಿಭಿನ್ನ ಬಿಂದುಗಳಲ್ಲಿ ಛೇದಿಸುವಂತೆ ಒಂದು ಬಾಹುವಿಗೆ ಸಮಾಂತರವಾಗಿ ಎಳೆದ ಸರಳರೇಖೆಯು ಉಳಿದೆರಡು ಬಾಹುಗಳನ್ನು ಸಮಾನುಪಾತದಲ್ಲಿ ವಿಭಾಗಿಸುತ್ತದೆ.

Solution 10.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 6

Solution 11.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 7

Solution 12.
OA2 = OP2 – AP2 =102 – 82
⇒ OA2 = 100 – 64
⇒ OA2 = 100 – 64
⇒ OA = √36
⇒ OA = 6 cm
ವೃತ್ತದ ತ್ರಿಜ್ಯ = 6 cm

Solution 13.
cosec A = \(\frac { 4 }{ 3 }\)

Solution 14.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 8

III.
Solution 15.
5000, 5200, 5400, ……… 7000.
a = 5000, d = 200 an = 7000, n = ?
an = a + (n – 1) d
⇒ 7000 = 5000 + (n – 1) 200
⇒ 7000 – 5000 = (n – 1) 200
⇒ 2000 = 200 (n – 1)
⇒ (n – 1) = 10
⇒ n = 10 + 1
⇒ n = 11
11 ನೇ ವರ್ಷದಲ್ಲಿ ಅವರ ಸಂಬಳ ₹ 7000 ಆಗುತ್ತದೆ.

Solution 16.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 9

Solution 17.
3x + 4y = 10 …. (1)
2x – 2y = 2 …… (2)
ಸಮೀಕರಣ 2ನ್ನು 2ರಿಂದ ಗುಣಿಸಿದಾಗ
3x + 4y = 10
4x – 4y = 4
…………………
7x = 14
x = 2
3x + 4y = 10
3(2) + 4y = 10
4y = 10 – 6
y = 1
∴ x = 2, y = 1

Solution 18.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 10

Solution 19.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 11
ಅರ್ಧ ವೃತ್ತದ ವಿಸ್ತೀರ್ಣ = \(\frac { 77 }{ 4 }\) cm2

Solution 20.
R = 6 cm, r = 4 cm
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 12

Solution 21.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 13

Solution 22.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 14

Solution 23.
ಬಹುಪದೋಕ್ತಿ = ax2+ bx + c ಆಗಿರಲಿ
ಶೂನ್ಯತೆಗಳು α ಮತ್ತು β ಆಗಿರಲಿ
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 15

Solution 24.
x2 + 6x – 7 = 0
⇒ x2 + 6x = 7
⇒ x2 + 6x + 32 – 32 = 7
⇒ (x + 3)2 – 9 = 7
⇒ (x + 3)2 = 16
⇒ x + 3 = ±√16
⇒ x + 3 = ±4
⇒ x = ±4 – 3
⇒ x = +4 – 3 ಅಥವಾ x = -4 – 3
⇒ x = 1 ಅಥವಾ x = -7

Solution 25.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 16

Solution 26.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 17
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 18

Solution 27.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 19

Solution 28.
ಚೀಲದಲ್ಲಿರುವ ಒಟ್ಟು ಚೆಂಡುಗಳು = n(S) = 3 + 5 = 8
ಕೆಂಪು ಚೆಂಡುಗಳು ಸಂಖ್ಯೆ = 3 = n(A)
ಕೆಂಪು ಚೆಂಡು ತೆಗೆಯುವ ಸಂಭವನೀಯತೆ = \(\frac { n(A) }{ n(S) }\)
ಕೆಂಪು ಚೆಂಡು ತೆಗೆಯುವ ಸಂಭವನೀಯತೆ = \(\frac { 3 }{ 8 }\)

Solution 29.
ವರ್ಗಗಳ ಘನಫಲ = 64cm3
ಬಾಹುವಿನ ಅಳತೆ = \(\sqrt [ 3 ]{ 64 }\) = 4 cm
ಆಯತ ಘನದ ಅಗಲ b = 4 cm
ಆಯತ ಘನದ ಉದ್ದ l = 8 cm (4 + 4)
ಆಯತ ಘನದ ಎತ್ತರ h = 4 cm
ಆಯತ ಘನದ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ = 2 (lb + bh + hl)
= 2(8 × 4 + 4 × 4 + 4 × 8)
= 2(32 + 16 + 32)
= 160 cm2
ಆಯತ ಘನದ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ = 160 cm2
ಅಥವಾ
14 cm ಎತ್ತರವಿರುವ ಒಂದು ಕುಡಿಯುವ ನೀರಿನ ಗಾಜಿನ ಲೋಟವು ಶಂಕುವಿನ ಭಿನ್ನಕದ ರೂಪದಲ್ಲಿದೆ. ಅದರ ಎರಡು ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ವ್ಯಾಸಗಳು 4 cm ಮತ್ತು 2 cm ಗಳಾಗಿವೆ. ಗಾಜಿನ ಲೋಟ ಸಾಮರ್ಥ್ಯವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Solution 30.
12 = 2 × 2 × 5
15 = 3 × 5
ಮ.ಸಾ.ಅ. = 5
ಲ.ಸಾ.ಅ. = 2 × 2 × 3 × 5
ಲ.ಸಾ.ಅ. = 60

IV.
Solution 31.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 20
ದತ್ತ: ‘O’ ವೃತ್ತಕೇಂದ್ರ, XY ಸ್ಪರ್ಶಕ, Pಸ್ಪರ್ಶಬಿಂದು
ಸಾಧನೀಯ: OP ⊥ XY
ರಚನೆ: P ಯನ್ನು ಹೊರತುಪಡಿಸಿ XY ಮೇಲೆ ಮತ್ತೊಂದು ಬಿಂದು Q ಆಗಿರಲಿ, OQ ಸೇರಿಸಿ.
ಸಾಧನೆ: OQ ವೃತ್ತವನ್ನು ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸಿದೆ.
OP = OR (ಒಂದೇ ವೃತ್ತದ ತ್ರಿಜ್ಯಗಳು)
OQ = OR + RQ
OQ > OR
OQ > OP (OP = OR)
OP ಯು O ನಿಂದ ಸ್ಪರ್ಶಕಕ್ಕೆ ಕನಿಷ್ಟ ದೂರವಾಗಿದೆ
OP ⊥ XY
ಅಥವಾ
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 22
ಸಾಧನೀಯ: ABCD ಒಂದು ವಜ್ರಾಕೃತಿ. (AB = BC = CD = AD)
ಸಾಧನೆ: AB = CD ಮತ್ತು AD = BC …… (1)
AP = AS, BP = BQ, CQ = CR, DS = DR
(ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು)
AB + CD = AP + PB + DR + CR
⇒ AB + CD = AS + BQ + DS + CQ
⇒ AB + CD = AS + DS + BQ + CQ
⇒ AB + CD = AD + BC
⇒ AB + AB = AD + AD [AB = CD, AD = BC]
⇒ 2AB = 2AD
⇒ AB = AD ….. (2)
∴ AB = BC = CD = AD (ಸಮೀಕರಣ (1) & (2) ರಿಂದ)

Solution 32.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 23

Solution 33.
ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳು x & x + 2
x2 + (x + 2)2 = 290
⇒ x2 + x2 + 4x + 4 – 290 = 0
⇒ 2x2 + 4x – 286 = 0
⇒ x2 + 2x – 143 = 0
⇒ x2 + 13x – 11x – 143 = 0
⇒ x (x + 13) – 11(x + 13) = 0
⇒ (x + 13) (x – 11) = 0
⇒ x + 13 = 0 ಅಥವಾ x – 11 = 0
⇒ x = -13 ಅಥವಾ x = 11
ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳು 11 & 13
[∴ x = 11 & x + 2 = 11 + 2 = 13]
ಅಥವಾ
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 24
ಚಿಕ್ಕ ಬಾಹು x ಆಗಿರಲಿ
AC2 = AB2 + BC2
⇒ (x + 60)2 = x2 + (x + 30)2
⇒ x2 + 3600 + 120x = x2 + x2 + 900 + 60x
⇒ x2 + 3600 + 120x – 2x2 – 900 – 60x = 0
⇒ -x2 + 2700 + 60x = 0
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x (x – 90) + 30 (x – 90) = 0
⇒ x – 90 = 0 ಅಥವಾ x + 30 = 0
⇒ x = 90 ಅಥವಾ x = -30
ಚಿಕ್ಕ ಬಾಹುವಿನ ಉದ್ದ = x = 90 m
ದೊಡ್ಡ ಬಾಹುವಿನ ಉದ್ದ = x + 30 = 90 + 30 = 120 m

Solution 34.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 25
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 26

Solution 35.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 27

Solution 36.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 28
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 29

V.
Solution 37.
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೂರು ಪದಗಳು
a – d, a, a + d
a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
(a – d)2 + (a + d)2 = 58
⇒ a2 + d2 – 2ad + a2 + d2 + 2ad = 58
⇒ 2a2 + 2d2 = 58
⇒ 2(a2 + d2) = 58
⇒ a2 + d2 = 29
⇒ 52 + d2 = 29
⇒ d2 = 29 – 25
⇒ d2 = 4
⇒ d = ± 2
⇒ d = 2 ಅಥವಾ d = -2
ಮೂರು ಪದಗಳು
∴ a – d = 5 – 2 = 3
∴ a = 5
∴ a + d = 5 + 2 = 7
ಅಥವಾ
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 30
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 31

Solution 38.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 32
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 33

Solution 39.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 34
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 35

Solution 40.
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 36
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 37
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium 38

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