Percentage Class 6 RS Aggarwal Exe-10C Goyal Brothers ICSE Maths Solutions 10. In this article you will learn about Application of Percentage. Visit official Website CISCE for detail information about ICSE Board Class-6 Mathematics.
Percentage Class 6 RS Aggarwal Exe-10C Goyal Brothers ICSE Maths Solutions 10
| Board | ICSE |
| Subject | Maths |
| Class | 6th |
| Ch-10 | Percentage |
| Exercise | 10C |
| Writer | RS Aggrawal |
| Book Name | Foundation |
| Topics | Application of Percentage |
| Academic Session | 2024 – 2025 |
Word Problems on Percentage
To solve Word Problems on Percentage you should have first practice previous exercise of percentage. There are variety of questions / word problems on application of percentage.
Exercise- 10C
( Percentage Class 6 RS Aggarwal Exe-10C Goyal Brothers ICSE Maths Solutions 10 )
Que-1: In a class test, the marks were awarded out of 80. Kamal obtained 65% marks. How many marks did he gets ?
Sol: Marks awarded out of 80
Kamal obtained = 65%
Marks he gets = 65% of 80
= (65/100) * 80
= 52
Que-2: In an examination, Renu obtained 480 marks out of 750. What percentage of marks did she get ?
Sol: Rennu obtained = 480 marks
Out of = 750 marks
480 / 750 * 100
= 48000 / 750
= 64 %
Que-3: In her annual examination Geeta scored 39 marks out of 60 in English and 51 out of 75 in Hindi. In which subject her performance was better ?
Sol: For English
percentage = (39/60)×100
= 13×5
= 65%
For Hindi
percentage = (51/75)×100
= 17×4
= 68%
In Hindi subject her performance was better.
Que-4: In an examination, 640 students appeared. Of them, 544 passed and the rest failed. What percentage of the students failed ?
Sol: Total number of students appeared= 640
Number of students passed= 544
So, number of students failed= 640-544
= 96.
Percentage of failed students=(96/640 x 100)%
= (0.15 x 100)%
= 15%
Que-5: In an examination 1360 students appeared. Of them, 85% passed. How many students failed ?
Sol: 100% – 85% = 15% of the students failed.
15% of 1360
15/100 times 1360 = 204
Therefore, 204 students failed.
Que-6: In an examination, a student has to secure 48% marks to pass. If Rahul gets 247 marks and fails by 5 marks, what are the maximum marks?
Sol: Let’s denote the maximum marks as “M”. According to the given information, Rahul got 247 marks and failed by 5 marks, which means he needed 252 marks to pass (247 + 5).
Since 252 marks represent 48% of the maximum marks, we can set up the following equation:
(252 / M) * 100 = 48
Solving for M:
M = (252 * 100) / 48
M = 525
Que-7: On a particular day, 96% of the students were present in a school. If the number of absentees a on that day was 82, find the total strength of a school.
Sol: Percentage of students present= 96%
Percentage of total students absent= (100% – 96%)
= 4%
Number of students absent= 82
Since 4%= 82 students, the total strength of school = 96/4= 24
= 24 x 82
= 1986 + 82
= 2050
Que-8: The population of a town was increased by 5% in a year. Last year its population was 20320. What is the present population of the town ?
Sol: Increase in population = 5%*20320
Increase in population = (5*20320)/100
Performing multiplication and division
Increase in population = 1016
Present population = 20320 + 1016
Present population = 21,336
Que-9: The price of a bat was Rs650. Now, its price has been increased by 6%. What is the increased price of that bat ?
Sol: Price of bat = Rs650
Increased = 6%
650+6% of 650
650+0.06×650
650+39
689
Que-10: The price of a cooler during last summer was quoted at Rs7250. In off season, the price has been reduced by 12%. What is the reduced price of a cooler ?
Sol: Given that the price of the cooler was 7250 during summer.
Since the price was reduced by 12 %, we need to find what is 12% of 7250. It is calculated as,
12% of 7250
= (12/100) * 7250
= 870
So the price is reduced by 870. Hence the reduced price of the cooler is,
7250 – 870
= 6380.
Que-11: The value of a car depreciates (decreases) by 10% every year. If the value of the new car is Rs225000, what will be its value after one year
Sol: Cost of car is : 225000
Depreciation(10% every year) : 22500 (225000*10/100)
New Value : 202500 (Cost – Depreciation)
Que-12: Ashish is 175 cm tall. His sister Annu is 8% shorter than him. What is Annu’s height ?
Sol: Ashish is 175 cm taller than his sister.
Her sister Annu is 8% shorter than him
So, 8% of 175 cm
= 8/100×175
= 14cm.
So,
175 – 14
= 161 cm.
Her sister height is 161 cm.
Que-13: An alloy consists of copper and zinc. It contains 60% copper and rest is zinc. Find the quantity of copper and zinc in 425g of his alloy.
Sol: copper = 60%
zinc = 40%
wt of alloy=425g
wt. of Cu = (425×60)/100 = 255g
wt. of Zinc = (425×40)/100 = I70g
Que-14: A tin contained 25 litres of oil. Due to leakage 2 litres of oil was lost. What percentage of oil is still there in the tin ?
Sol: total amount of oil = 25 L
amount of oil lost = 2 L
amount of oil in tin after leakage = 25- 2 = 23 L
percentage of oil left in tin = (23/25) * 100 = 92%
Que-15: An election was contested by two candidates A and B. In all there were 25000 voters. 80% of the votes were polled. If A got 60% of the total votes polled, how many votes did B get ?
Sol: Total voters = 25000
80% of the votes were polled.
= (80/100)×25000 = 20000
A got 60% votes
= (60/100)×20000 = 12000
So,
B will get votes
= 20000−12000
= 8000
Que-16: Mr. Mehta had Rs60000. Out of this money, he gave 20% to his daughter, 35% to his son, 30% to his wife and the rest he donated to a Charitable Trust. How much did he donate ?
Sol: Daughter’s share = 60000 × 20%
= 60000 × (20/100)
= 12000
Son’s share = 60000 × 35%
= 60000 × (35/100)
= 21000
Wife’s share = 60000 × 30%
= 60000 × (30/100)
= 18000
Family’s share = 12000 + 21000 + 18000
Family’s share = 51000
Donated amount = total amount – family’s share
Donated amount = 60000 – 51000
Donated amount = Rupees 9000
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