RBSE Maths Chapter 13 – Comparing quantities Class 8 Important questions and solutions is available here. The important, additional and text exercise questions and solutions of Chapter 13, available at BYJU’S, contain step by step answers. All these important questions are based on the new pattern prescribed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 8 solutions.
Chapter 13 of the RBSE Class 8 Maths will help the students to solve problems related to the percentage, profit-loss, simple interest, compound interest, understanding direct and inverse relationship.
RBSE Maths Chapter 13: Exercise 13.1 Textbook Important Questions and Solutions
Question 1: Convert the following ratios into percentages.
[i] 1:4
[ii] 3:4
Solution:
[i] 1:4 = [1 / 4] = [1 * 25] / [4 * 25] = [25 / 100] = 25% [ii] 3:4 = [3 / 4] = [3 * 25] / [4 * 25] = [75 / 100] = 75%Question 2: Himi travelled 240 km by bus and 360 km by train then determine,
(i) The ratio of travelling by train and travelling by bus.
(ii) The ratio of travelling by bus and by train.
(iii) The ratio of travelling by train and total travelling.
(iv) The ratio of travelling by bus and total travelling.
Solution:
(i) The distance travelled by train = 360 km.
The distance travelled by bus = 240 km
The ratio of the distance travelled travelled by train to the distance travelled by bus = 360 : 240 = 3 : 2 km
(ii) The ratio of the distance travelled by bus to the distance travelled by train = 240 : 360 = 2 : 3
(iii) The ratio of distance travelled by train to total distance = 360: (240 + 360) = 360 : 600 = 3 : 5
(iv) The ratio of the distance travelled by bus to total distance = 240 : 600 = 2 : 5
Question 3: In grade VIII, 68% of students obtained grade A out of 75 students. How many students obtained grade A?
Solution:
Percentage of the students who obtained grade A
= 68% of 75
= 75 x [68 / 100 ]
= 75 * 0.68
= 51
So, 51 students obtained grade A.
Question 4: A Kabaddi team of school has won 15 matches out of the total matches played this year. If the winning percentage was 75 then how many matches were played by the team?
Solution:
Let the number of matches played by the team be x matches.
The number of matches won = 15
Wins of the matches in percentage = [15 / x] * 100%
According to the question,
[15 / x] * 100% = 75 [15 / x] = [75 / 100] [15 / x] = 0.7515 = 0.75 * x
15 / 0.75 = x
x = 20
The total number of matches played by the team is 20.
Question 5: There are 1275 trees in total in the field of Mohan. There are 36% of trees that have fruits. Determine the number of trees having fruits in the field.
Solution:
The total number of trees = 1275
Number of trees with fruits = 36% of 1275
= [36 / 100] * 1275
= [0.36] * 1275
= 459
The number of trees having fruits in the field is 459.
Question 6: Kamli spent 75% of the amount deposited in her account under Pradhan Mantri Jan Dhan Yojana. Now the amount left in her account is Rs, 600. Determine the amount deposited in her account.
Solution:
Let the amount deposited in Kamli’s account be Rs x.
The percentage of amount spent by Kamli = 75% of x
= Rs. [x] * [75 / 100]
= Rs. [x] * [3 / 4]
= Rs. [3x / 4]
Balance Amount = [x – [3x / 4]] = [4x – 3x] / [4] = [x / 4]
According to the question,
[x / 4] = 600x = 600 x 4 = 2400
The total amount deposited in account was Rs 2400.
Question 7: According to a survey of 50,000 students under mid-day meal in five districts of Rajasthan, 60% of students like pulses-chapati, 25% like vegetables and chapati and rest like khichdi. Determine the percentage of students who like khichdi.
Solution:
Total number of students = 50,000
The number of students who like pulses and chapati
= 60% of 50,000
= [60 / 100] * [50,000]
= [0.6] * [50,000]
= 30,000
The number of students who like vegetable and chapati
= 25% of 50,000
= [25 / 100] * [50,000]
= [0.25] * [50,000]
= 12,500
Remaining students
= 50,000 – (30,000 + 12,500)
= 50,000 – 42,500
= 7,500
The number of students who like khichdi = 7,500
The percentage of students who like khichdi
= [7500 / 50000] * [100]
= [0.15] * [100]
= 15
15% of students like Khichdi.
RBSE Maths Chapter 13: Exercise 13.2 Textbook Important Questions and Solutions
Question 1: Mohan purchases some mattresses for Rs. 7250. After some time he sold them for Rs. 6090. Find the loss percentage.
Solution:
The cost price of mattresses = Rs. 7250
The selling price of mattresses = Rs 6090
Loss = Cost price – Selling price
= 7250 – 6090
= Rs. 1160
The loss incurred on Rs. 7250 = Rs. 1160
Loss on Re. 1 = Rs. [1160 / 7250] = 0.16
Loss on Rs. 100 = Rs. [1160 / 7250] * 100
= Rs 16
= 16% [when converted to percentage]
The percentage loss of Mohan = 16%.
Question 2: Due to an increase in salary of Ajit Singh by 12%, the new salary becomes Rs. 25760. Find his previous salary.
Solution:
Let Rs. x be the previous salary of Ajit Singh.
The percentage increase in his salary = 12%
Increase = Rs. [x] * [12 / 100]
= Rs. [3x] / 25
Total new salary = Rs. [x + (3x / 25]
= Rs. [(25x + 3x) / 25]
= Rs. [(28x) / 25]
According to the question,
Rs. [(28x) / 25] = Rs. 25760
28x = Rs. [25760 * 25]
28x = 644000
x = [644000] / 28
x = Rs. 23000
So, the previous salary was Rs 23,000.
Question 3: Manjeet markup price on a pump is increasing by 40%. If he wishes to sell it after providing a subsidy of 40%, then find the profit or loss percentage.
Solution:
Let Rs. x be the cost price of the pump.
Markup price = x + [40% of x]
= [x] + [(40 / 100) * x]
= [x] + [2x / 5]
= [5x + 2x] / 5
= Rs. [7x] / 5
Discount = 40% of ([7x] / 5)
= Rs. [7x / 5] * [40 / 100]
= Rs. [280x / 500]
= Rs. [14x / 25]
Selling Price = Rs. [7x / 5] – Rs. [14x / 25]
= Rs. [35x – 14x] / 25
= Rs. [21x / 25]
Loss = cost price – selling price
= x – Rs. [21x / 25]
= [25x – 21x] / 25
= [4x] / 25
Loss percentage = ([4x / 25] / x) * 100 = 16%
So, the loss percentage is 16%.
Question 4: Cost of a moped is Rs. 54000. Now the price increased by 14%, then what is the price to be paid for the moped?
Solution:
Price of Moped = Rs 54,000
Percentage increase in cost = 14%
Increase = 14% of 54000
= [14 / 100] * 54000
= [7 / 50] * 54000
= [378000 / 50]
= Rs. 7560
New cost of Moped = 54,000 + 7,560 = Rs 61,560
The required price of the moped is Rs 61,560.
Question 5: A businessman purchased goods for Rs.14000. He paid Rs. 350 as auto rent and Rs. 150 as wages. For earning a 5% profit, at what price should he sell goods?
Solution:
Cost price of products = Rs 14,000
Transportation = Rs 350
Labor = Rs 150
Total cost price of products = 14000 + 350 + 150 = Rs 14,500
Profit percentage = 5%
Profit = 5% of 14,500
= [5 / 100] * 14500
= [1 / 20] * 14500
= 725
= Rs 725
Selling Price = Total Cost Price + Profit = 14,500 + 725 = Rs 15,225
So, he will sell the products for Rs 15,225.
Question 6: A furniture seller sold two dressing tables at the rate of Rs. 7200. Out of them, 20% profit obtained on one table and 20% loss on another. How much profit or gain percentage is obtained in the whole transaction?
Solution:
The selling price of the first dressing table = Rs 7,200
Let cost price be Rs x
Gain percentage = 20%
Gain = 20% of x
= [x] * [20 / 100]
= Rs. [x / 5]
Selling price = cost price + profit
= [x] + [x / 5]
= [5x + x] / 5
= 6x / 5
According to the question,
6x / 5 = 7200
6x = 7200 * 5
6x = 36000
x = 36000 / 6
x = 6000
The cost price of one dressing table is Rs. 6000.
The selling price of the second dressing table = 7200
Let cost Price = y
Loss % = 20%
Loss = 20% of y
= y * [20 / 100]
= [y / 5]
Selling price = Cost price – Loss
= y – [y / 5]
= [5y – y] / 5
= 4y / 5
According to the question,
4y / 5 = 7200
4y = 7200 * 5
4y = 36000
y = 36000 / 4
y = 9000
The cost price of second dressing table is Rs. 9000
Total Cost Price = 6000 + 9000 = Rs. 15,000
Total Selling Price = 7,200 + 7,200 = Rs. 14,400
Total selling price < Total cost price
Hence, a loss is incurred.
Loss = 15,000 – 14,400 = Rs. 600
Loss on Rs. 15000 = Rs. 600
Loss on Re.1 = 600 / 15000 = 1 / 25
Loss on Rs. 100 = [600 * 100] / 15000 = Rs. 4
The loss percentage of the two tables is 4%.
Question 7: Manoj paid Rs 6500 as interest on a loan of Rs 52000 after 2 years. Find the percentage of interest paid by Manoj.
Solution:
Interest paid for 2 years Rs 6,500.
Interest of 1 year = Rs. 6500 / 2 = Rs. 3250.
Interest on Rs. 52000 = Rs. 3250.
Interest on Re. 1 = Rs. [3250 / 52000].
Interest on Rs. 100 = Rs. [3250 * 100] / [52000] = 6.25%.
The required rate of interest is 6.25%.
Question 8: In what time will the principal of Rs. 3200 at a rate of 8% become Rs. 3840.
Solution:
Principal amount Rs 3,200
Amount = Rs 3,840
Simple Interest = [Principal * Time * Rate of interest] / [100]
Simple interest = Amount – Principal = 3840 – 3200 = Rs. 640
640 = [3200 * T * 8] / [100]
640 * 100 = 25600 * T
64000 = 25600 * T
T = 64000 / 25600
T = 2.5 years = 2 years 6 months
The principal of Rs. 3200 at a rate of 8% will become Rs. 3840 in 2 years 6 months.
Question 9: Bhupendra took a loan of Rs. 6300 at a rate of 7% for 2 years and 8 months. Find the amount that will be paid by him.
Solution:
Principal= Rs 6,300
Time = 2 years 8 months
= 2 years + 8 months
= 2 years + [8 / 12] year
= 2 year + [2 / 3] year
= (2 + [2 / 3]) year
= [8 / 3] years
Rate of interest = 7%
Simple Interest = [Principal * Time * Rate of interest] / [100]
= [6300 * [8 / 3] * 7] / 100
= Rs. 1176
Amount = Principal + Simple interest
= 6300 + 1176
= Rs. 7476
Hence, the amount that will be paid by him will be Rs 7,476.
RBSE Maths Chapter 13: Exercise 13.3 Textbook Important Questions and Solutions
Question 1: Number of visitors on the first day of the book fair in the city was 3000 which increased to 3600 on the next day. Determine the increase in fair visitors.
Solution:
Number of visitors of the first day = 3000
Number of visitors on the second day = 3600
Increase in number of visitors = 3600 – 3000 = 600
Percentage increase in visitors
= [increase / starting number of visitors] * 100
= [600 / 3000] * 100
= [1 / 5] * 100
= 20%
The increase in visitors is by 20% on the next day.
Question 2: Price of television is Rs. 30,000. If the value of an object decreases (devaluates) by 20%, then determine its value after 2 years.
Solution:
Cost after 2 years
= 30,000 * (1 – [20 / 100])^{2}
= 30,000 * (1 – [⅕])^{2}
= 30,000 * ([5-1] / 5)^{2}
= 30,000 * [4 / 5]^{2}
= 30,000 * [4 / 5] * [4 / 5]
= 19200
= Rs 19,200
The value of the television after two years is Rs. 19200.
Question 3: Kapil took a loan of Rs. 52800 at an annual rate of 12% from a bank for purchasing a scooter when the accumulated rate is annual. After one year and 6 months, what amount is to be paid for repaying the loan?
Solution:
Principal = Rs 52,800
Rate = 12% yearly
Amount = P * (1 + [R / 100])^{n}
Amount after 1 year
= 52800 * [1 + (12 / 100)]^{1}
= 52800 * [1 + (3 / 25)]
= 52800 * [(25 + 3) / 25]
= 52800 * [28 / 25]
= 1478400 / 25
= Rs. 59136
Interest after 6 months
= [59136 * [1 / 2] * 12] / 100
= [59136 * 6] / 100
= Rs. 3548.16
Amount to be paid = 59136 + 3548.16 = 62,684.16 after one year and 6 months.
Question 4: In the year 2013, the number of road accidents was 10,000. By traffic police awareness programs which were conducted for avoiding road accidents, it decreased by 20%. Find the number of road accidents in 2015.
Solution:
Initial number = 10,000
Time = 2015 – 2013 = 2 years
Rate of decrease = 20%
Number of road accidents in 2015
= 10,000 * [1 – (20 / 100)]^{2}
= 10,000 * [1 – [1 / 5]]^{2}
= 10,000 * [(5 – 1) / (5)]^{2}
= 10,000 * [4 / 5]^{2}
= 10,000 * [16 / 25]
= 160000 / 25
= 6400
The number of road accidents in 2015 was 6400.
Question 5: Determine compound interest on Rs. 10,000 for 2 years at 8% annual rate. The interest is calculated at an annual rate.
Solution:
Amount = P * (1 + [R / 100])^{n}
Amount = 10000 * [1 + (8 / 100)]^{2}
= 10000 * [1 + (2 / 25)]^{2}
= 10000 * [(25 + 2) / 25]^{2}
= 10000 * [27 / 25]^{2}
= 10000 * 1.1664
= Rs. 11664
Interest = Amount – Principal = 11,664 – 10,000 = Rs. 1664
Question 6: Payal took a loan of Rs. 12,000 for parlour from a nationalized bank. How much will she repay after 2 years 6 months at an annual rate of 8% when the interest Is accumulated annually?
Solution:
Time = 2 years 6 months
= 2 years + 6 months
= 2 years + [6 / 12] year
= 2 years + [1 / 2] year
= (2 + [1 / 2]) year
= [5 / 2] years
Amount = P * (1 + [R / 100])^{n}
Amount to be paid after 2 years
= 12000 * [1 + [8 / 100]]^{2}
= 12000 * [1 + [2 / 25]]^{2}
= 12000 * [27 / 25]^{2}
= 12000 * 1.1664
= Rs. 13996.80
Compound interest = Amount – Principal = 13996.80 – 12000 = 1996.8
The amount for half a year = [P * T * R] / 100
= [12000 * 0.5 * 8] / 100
= Rs. 480
Therefore, total interest = 1996.8 + 480 = 2476.8.
Question 7: Calculate the compound interest on Rs. 18000 for 1.5 years at a rate of 10%. The interest is calculated half-yearly.
Solution:
1.5 years = 1 (½) years = 3 half-yearly
Principal = Rs. 18000
Rate of interest = 10% yearly = [10% / 2] for half-yearly = 5%
Amount = P * (1 + [R / 100])^{n}
Amount = = 18000 * [1 + [5 / 100]]^{3}
= 18000 * [1 + [1 / 20]]^{3}
= 18000 * [21 / 20]^{3}
= 18000 * 1.1576
= Rs. 20837.25
Compound interest = Amount – Principal = 20837.25 – 18000 = Rs. 2837.25
Question 8: Vishnu invested Rs. 80,000 at an annual rate of 14%. If the interest accumulates half-yearly, then determine what amount he will receive if time is
(i) 6 months
(ii) 1 year
Solution:
(i) 6 months
Principal = Rs. 80,000
Rate of interest = 14% annually
So, for half-yearly the interest rate = [14% / 2] = 7%
Time = 6 months = 0.5 year = 1 half-yearly
Amount = P * (1 + [R / 100])^{n}
= 80000 * [1 + (7 / 100)]^{1}
= 80000 * [(100 + 7) / 100]
= 80000 * [107 / 100]
= 80000 * 1.07
= Rs. 85600
(ii) 1 year
Principal = Rs. 80,000
Rate of interest = 14% annually
So, for half-yearly the interest rate = [14% / 2] = 7%
Time = 1 year = 2 half-yearly
Amount = P * (1 + [R / 100])^{2}
= 80000 * [1 + (7 / 100)]^{2}
= 80000 * [(100 + 7) / 100]^{2}
= 80000 * [107 / 100]^{2}
= 80000 * 1.07 * 1.07
= 80000 * 1.1449
= Rs. 91592
Question 9: Kushwant borrowed Rs. 12,500 for 3 years at 5% annually on simple interest. If the same amount is borrowed 5% annually on compound interest, then what extra amount Khushwant has to pay?
Solution:
Simple interest = [P * T * R] / 100
= [12500 * 3 * 5] / 100
= [187500] / 100
= Rs. 1875
Compound interest = 12500 * [1 + (5 / 100)]^{3} – 12500
= 12500 * [1 + (1 / 20)]^{3}] – 12500
= 12500 * [(20 + 1) / 20]^{3} – 12500
= 12500 * 1.1576 – 12500
= 14470.31 – 12500
= 1970.31
The amount that Kushwant has to pay = 1970.31 – 1875 = Rs. 95.31 when the same amount is borrowed at compound interest when compared to simple interest.
RBSE Maths Chapter 13: Exercise 13.4 Textbook Important Questions and Solutions
Question 1: Vimla travelled 200 km by bus and she gave a fare of Rs.180. What is the fare that she has to pay if she travels a distance of 500 km?
Solution:
Let the fare that she pays for a distance of 500km be Rs x.
Distance |
Fare |
200km |
180 |
500km |
x |
Hence, it is a direct relation.
200:500 :: 180:x
Product of the outer terms = Product of middle terms
200 * [x] = 500 * 180
200 * x = 90000
x = 90000 / 200
x = 450
Hence, the required fare is Rs 450.
Question 2: Shadow of a 10-metre long tree is 18 m in the morning. What will be the height of shadow of 120 m high tower at the same time?
Solution:
Let x meter be the length of the shadow.
Height of tree |
Length of the shadow |
10m |
18m |
120m |
x |
It is a direct relation.
10:120 :: 18:x
Product of outer terms = Product of middle terms
10 * [x] = 120 * 18
10x = 2160
x = [2160] / 10
x = 216
Hence, the required shadow is 216 meter.
Question 3: If the weight of 5 books is 2.5 kg, then 30 kg will be the weight of how many books?
Solution:
Let x be the number of books weighing 30kg.
Number of books |
Weight |
5 |
2.5kg |
x |
30kg |
It is a direct relation.
5:x :: 2.5:30
Product of outer terms = Product of middle terms
5 * 30 = x * 2.5
150 = 2.5x
150 / 2.5 = x
60 = x
Hence, 60 books weigh 30kg.
Question 4: A bus is moving with a uniform speed of 45 km per hour then what is the time taken by the bus to travel a distance of 225 km?
Solution:
Let the time taken by the bus to travel a distance of 225km be x hours.
Distance |
Time |
45km |
1 hour |
225km |
x hours |
It is a direct relation.
45:225 :: 1:x
Product of outer terms = Product of middle terms
45 * x = 225 * 1
45x = 225
x = 225 / 45
x = 5
The required time is 5 hours.
Question 5: Mamta can fill 30 parindahs with 15 litres of water. How many litres of water is required to fill 120 such parindahs?
Solution:
Let the number of litres of water required to fill 120 such parindahs be x.
Litres of Water |
Parindahs |
15 |
30 |
x |
120 |
It is a direct relation.
15:x :: 30:120
Product of outer terms = Product of middle terms.
15 * 120 = 30 * x
1800 = 30x
1800 / 30 = x
60 = x
Hence, 60 litres of water is required to fill 120 parindahs.
Question 6: 100 litres of water can be saved by washing 5 cars with jug and buckets instead of tap. In this way how many litres of water can be saved by washing 20 such cars?
Solution:
Let x be the number of litres of water that can be saved by washing 20 cars.
Litres of Water |
Number of cars washed |
100 |
5 |
x |
20 |
It is a direct relation.
100:x :: 5:20
Product of outer terms = Product of middle terms.
100 * 20 = 5 * x
2000 = 5x
2000 / 5 = x
400 = x
Hence, 400 litres of water can be saved while washing 20 cars.
Question 7: 9 workers took 16 days to complete the pucca boundary walls of the school. If the number of workers is 12, then in how many days can the wall be prepared?
Solution:
Let x be the number of days required to complete the wall.
Workers |
Days |
9 |
16 |
12 |
x |
It is an inverse relation.
12:9 :: 16:x
Product of outer terms = Product of middle terms
12 * x = 16 * 9
12x = 144
x = 144 / 12
Hence, 12 days are required to complete the wall.
Question 8: A camp has food for 40 soldiers for 20 days. After 5 days, 10 more soldiers joined. For how many days, will the rest of the food be sufficient?
Solution:
The remaining food was enough for 40 soldiers for 20 – 5 = 25 days.
Let 40 + 10 = 50 soldiers be there such that they have enough food for x days.
Soldiers |
Days |
40 |
20 – 5 = 15 |
40 + 10 = 50 |
x |
It is an inverse relation.
50:40 :: 15:x
Product of outer terms = Product of middle terms
50 * x = 40 * 15
50x = 600
x = 600 / 50
x = 12
The required number of days is 12.
Question 9: Under Swachh Bharat Mission 15 volunteers cleaned their village in 4 days. If the village needs to be cleaned in 3 days, then how many workers will be required?
Solution:
Let the number of volunteers required to clean the village be x.
Volunteers |
Days |
15 |
4 |
x |
3 |
It is an inverse relation.
3:4 :: 15:x
Product of outer terms = Product of middle terms
3 * x = 4 * 15
3x = 60
x = 60 / 3
x = 20
Hence, the required number of volunteers is 20.
Question 10: In a school under shramdan 12 students clean for 5 hours. If the same part needs to be cleaned in 3 days, then how many workers will be required?
Solution:
Let the number of workers required to clean the same part be x.
Students |
Time Taken |
12 |
5 hours |
x |
3 days |
It is an inverse relation.
3:5 :: 12:x
Product of outer terms = Product of middle terms
3 * x = 5 x 12
3x = 60
x = 60 / 3
x = 2
Hence, 20 students are required.
Question 11: Madhu prepares food for 12 days from a biogas plant by putting on 80 kilograms of cow dung. Find the amount of cow dung required to prepare food for 60 days?
Solution:
Let x kg the amount of cow dung required to prepare food for 60 days.
Days |
Cow dung |
12 |
80 |
60 |
x |
It is a direct relation.
12:60 :: 80:x
Product of outer terms = Product of middle terms
12 * x = 60 * 80
12x = 4800
x = 4800 / 12
x = 400
Hence, the required amount of cow dung is 400 kg.
RBSE Maths Chapter 13: Additional Questions and Solutions
Question 1: Match the following.
Question |
Answer choices |
1] The formula for calculating simple interest is |
A] Cost price + profit |
2] Selling price [if profit exist] |
B] Constant |
3] If x and y are in an inverse relationship, then xy = |
C] 1:10 |
4] Ratio between 50 paise to Rs. 5 |
D] [P * T * R] / 100 |
Solution:
1] – D]
2] – A]
3] – B]
4] – C]
Question 2: The meaning of twice a number is to increase by 100%. What is the percentage decrease if the number reduces to its half?
Solution:
Percentage loss = [New number / Original number] * 100%
= ([0.5] * original number) / (original number) * 100%
= 50%
Question 3: An amount is borrowed for a year at the rate of 15% per annum. If the interest is compounded quarterly, then how many times will the interest be paid?
Solution:
1 year = 4 quarters
Interest will be paid 4 times a year.
Question 4: After selling a plot for Rs 61,200, the gain is 2%. What is the cost price of the plot?
Solution:
Let the cost price of the plot be Rs x
Profit = x * [2 / 100] = 2x / 100 = Rs. x / 50
Selling price = x + [x / 50]
= [(50x + x) / 50]
= 51x / 50
According to the question,
51x / 50 = 61200
51x = 61200 * 50
51x = 3060000
x = 3060000 / 51
x = 60000
Hence, the cost price of the plot is Rs. 60000.
Question 5: The cost price of a scooter is Rs 42,000. If it depreciates at a rate of 8%, what will be the price of the scooter after a year?
Solution:
Price of the scooter after a year = 42,000 * (1 – [8 / 100])^{1}
= 42,000 * [(100 – 8) / 100]
= 42,000 * – [92 / 100]
= 42,000 * [23 / 25]
= 42,000 * 0.92
= Rs. 38640
The price of the scooter after a year is Rs 38,640.
Question 6: In a government school 25% of neem plants were shown during environment pakwara (fortnight). If there are 180 plants in total, then how many neem plants are there?
Solution:
Total number of plants =180
Percentage of neem plants = 25%
The required number of Neem plants = 25% of 180
= [25 / 100] * 180
= [1 / 4] * 180
= 0.25 * 180
= 45
There are 45 neem plants.
Question 7: Convert 2:5 ratio into a percentage.
Solution:
= 2:5
= 2/ 5
= [2 * 20] / [5 * 20]
= 40 / 100
Question 8: The population of a town before 3 years was 25,000. If it grows at 10%, 15% and 8% in three consecutive years. Find the present population.
Solution:
Present Population = 25,000 * [1 + (10 / 100)] * [1 + (15 / 100)] * [1 + (8 / 100)]
= 25,000 * [110 / 100] * [115 / 100] * [108 / 100]
= 25,000 * [11 / 10] * [23 / 20] * [27 / 25]
= 25,000 * [1.1] * [1.15] * [1.08]
= 25,000 * 1.3662
= 34,155
The present population is 34,155.
Question 9: A farmer has enough food for 20 animals for 6 days. If he increases 10 more animals, then for how many days will this food be enough?
Solution:
Let the number of days be x.
Number of animals |
Days |
20 |
6 |
20 + 10 = 30 |
x days |
It is an inverse relation.
30:20 :: 6:x
Product of outer terms = Product of middle terms
30 * x = 20 * 6
30x = 120
x = 120 / 30
x = 4
The required number of days is 4.