If 70% of a no. is subtracted from itself, it reduces to 81. What is two fifth of that Number?

Problems on Percentage - Bent examination, questions, shortcuts, solved example videos

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Video on Problems on Percentage - shortcuts, tips and tricks Percentage Points to Retrieve one) Y % is expressed as Y 100 2) To find percent of x = x × 100 % y y Quick Tips and Tricks 1. Prices of Goods (i) If the price of appurtenances increases by R %, then the reduction in consumption so equally not to increment the expenditure can be calculated using the formula: R × 100 % (100 + R) (2) If the price of goods decreases past R %, then the increment in consumption so every bit not to decrease the expenditure can exist calculated using the formula: R × 100 % (100 – R) 2. Numerical on Population: Population of a city at present is P and it increases at the charge per unit of R% per annum. (one) To find population after n years = P i + R north 100 -------- (Later on due north years population increases, thus 1 + R is used) 100 3. Numerical on Depreciation: Present value of machine is M, If it depreciates at the rate of R% per annum. (ane) To find value of auto afterward due north years = P one – R n 100 -------- (Later on due north years the value of machine decreases, thus ane – R is used) 100 Question Variety Mostly five types of questions are asked from this chapter. Understanding these concepts will help in solving the issues related to this chapter. Type 1: Numerical on numbers Q 1. The difference between 2 numbers is 1550. If viii % of one number is ten % of the other number, then notice the ii numbers a. 4973, 6523 b. 5450, 7000 c. 6200, 7750 d. 6500, 4950 View solution Right Option: (c) Let 2 numbers be x and y. It is given that, viii % of x = 10 % of y Therefore, x = 10 y = five y 8 4 Divergence between two numbers (x – y)= 1550 Substituting the value of 10, we get y=1550×4=6200 The ii numbers whose departure is 1550 are 6200 and 7750. Q two. Two numbers P and Q are such that, the sum of 2 % of P and Sum of 2 % of Q is two-third of the sum of 2 % of P and 6 % of Q. Find the ratio of P and Q. a. 2 : 5 b. 3 : one c. one : 4 d. 5 : one View solution Correct Option: (b) The sum of 2 % of P and Sum of 2% of Q is two-third of the sum of 2 % of P and vi % of Q. This sentence ways that, ii% of P + 2% of Q = 2 (2% of P + 6% of Q) 3 2 P + ii Q = two 2 P + half dozen Q 100 100 three 100 100 1 P + 1 Q = 1 P + ane Q 50 50 75 25 ane – 1 P = 1 – 1 fifty 75 25 50 P = 150 = iii Q 50 1 Alternate method ii P + ii Q = 2 ii P + vi Q 100 100 3 100 100 Only eliminate 100 from both the sides, nosotros get three(2P + 2Q) = ii(2P + 6Q) 6P + 6Q = 4P + 12Q 6P – 4P =12Q – 6Q 2P = 6Q P = 6 = 3 Q 2 1 Q three. 50 % of a number is 18 less than 2-third of that number. Notice the number. a. 123 b. 119 c. 115 d. 108 View solution Correct Option: (d) Let the number be 10. It is given that, fifty % of a number is 18, less than two-tertiary of that number. This means that, 50x=5400 x=108 The number is 108 Q 4. When 35 is subtracted from a number; information technology reduces to its 80 %. Detect the four-fifth of that number. a. 140 b. 125 c. 137 d. 129 View solution Right Option: (a) We are given, 35 when subtracted from a number, reduces to its 80 %. Therefore, Let the number exist ten. Solving, nosotros get the value of 10 Now, we are asked to find the iv-fifth of that number i.east 10 Hence, the number is 140.

Type 2: Numerical on Depreciation Q v. The value of lathe motorcar depreciates at the rate of 10 % per annum. If the cost of machine at nowadays is Rs. 160,000, then what will exist its worth subsequently 2 years? a. Rs. 122,365 b. Rs. 153,680 c. Rs. 129,600 d. Rs. 119,900 View solution Correct Choice: (c) Hint: Nowadays value of machine is M. If the cost depreciates at the rate of R% per annum, then The value of automobile after north years = P 1 – R n 100 In this numerical, we are given the present cost of the car i.e 160,000 and the cost decreases by 10 % per annum. We take to find the toll of this machine later on 2 years. Nosotros tin can solve this numerical in a minute, if we know the trick used to solve such numerical related to depreciation of cost. Given: Present amount = Rs. one,60,000 Rate of depreciation = 10 % Substituting the given values, we go The value of auto afterwards n years = P 1 – R n 100 = 1,60,000 × 1 – 10 two 100 = Rs. 129,600 Subsequently 2 years, the cost of automobile = Rs. 129,600 Q 6. The value of Xerox car depreciates at the charge per unit of 10 % per annum. If the cost of machine at present is Rs. 75,000 and so what was the value of auto before 2 years? a. Rs. ninety,000 b. Rs. 92,600 c. Rs. 93,800 d. Rs. 95,000 View solution Correct Selection: (b) Hint: Present value of machine is M. If the toll depreciates at the rate of R% per annum, then Given: Cost of Xerox automobile at present = Rs. 75,000 Charge per unit of depreciation = ten % Substituting the given values, we get =Rs.92592.60 Therefore, the value of machine before 2 years = Rs. 92592.60 Merely this value is non available in the given options. Hence, select the nearby value to the amount of Rs. 92592.60. Among the listed options option (b) is the correct answer.

Blazon iii: Numerical on Population Q 7. The electric current birth charge per unit per k is 30, whereas respective death charge per unit is 10 per thousand. Find the net growth charge per unit in terms of population increment in percent. a. i.v % b. 2 % c. 2.5 % d. 3 % View solution Correct Option: (b) We are given that, 1) Current nascency rate per thousand is 30 2) Corresponding death rate is 10 per thousand Hence, net growth on 1000 = Current birth rate - death rate = 30 – 10 =20 We are asked to find, internet growth rate in terms of population increase in percent (which ways cyberspace growth on 100) Net worth on 100 = Net worth on grand × 100 1000 Cyberspace worth on 100 = 20 × 100 = 2% thou Q 8. The full population of a metropolis is 6500.The number of males and females increases by 5 % and 10 % respectively and consequently the population becomes 7000. Notice the number of males in the hamlet. a. 4000 b. 3000 c. 3500 d. 2950 View solution Correct Option: (b) We are given that, ane) Total population of city = 6500 2) Increment in male and female person population = 5 % & x% respectively. 3) Last population of city = 7000 Hence, Let's assume that number of males = x Number of female = 6500 – x Therefore, subsequently increase in five % male and ten % female, the population becomes 7000 v % male +ten % female person = Difference betwixt new and original population 5 ten + 10 (6500 – 10) = 7000 – 6500 100 100 5x+65000-10x=50000 5x= 15000 10=3000 Number of males = 3000 Number of females = 3500 Q 9. The nowadays population of a state is 10 crores. If it rises to 17.28 crores during next 3 years, and so find uniform rate of growth in population. a. 20 % b. xxx % c. 40 % d. sixty % View solution Q ten. The population of dissimilar trees in a field increased by 10 % in first year, increased by viii % in second year and decreased by 10 % in 3rd year. If now the number of trees is 26730, then find the number of tress in the offset. a. 30000 b. 25000 c. 27000 d. 27865 View solution We are given, number of trees increased in 1) First twelvemonth = increased past 10 % 2) Second year = increased by 8 % iii) Third year = decreased past 10 % Therefore, to find number of copse in the start employ the trick. = 26730 × 10 × 25 × 10 11 27 9 = 25000

Type four: Numerical on Prices of Appurtenances Q xi. The price of diesel fuel increases by 50 %. Detect past how much percent a truck owner must reduce his consumption in club to maintain the aforementioned budget? a. eleven.11 % b. 22.22 % c. 33.33 % d. 44.44 % View solution Correct Option: (c) Hint: If the price of goods increases by R %, and so the reduction in consumption so as not to increase the expenditure can be calculated using the formula: R × 100 % (100 + R) = 50 × 100 % (100 + fifty) = 33.33 % The truck owner must reduce its consumption in club to maintain the aforementioned upkeep by 33.33 % Q 12. The price of rice falls past 15 %. By what per centum a person tin increase the consumption of rice so that his overall budget does not modify? a. 10.74 % b. 17.64 % c. twenty.46 % d. 21.90 % View solution Correct Option: (b) Hint: If the price of goods decreases past R %, then the increase in consumption then as non to decrease the expenditure tin be calculated using the formula: R × 100 % (100 – R) Using this pull a fast one on, nosotros can hands solve such type of numerical. The cost of rice falls past 15 %, therefore substituting this value, we get fifteen × 100 % = 17.64 % (100 – 15) Therefore, the person can increase his consumption by 17.64 %

Type 5: Numericals based on Marks of students Q 13. In an exam, P scored 30 % marks and failed by 15 marks. Q scored 40 % marks and obtained 35 marks more those required to pass. Find the pass per centum. a. 30 % b. 33 % c. 35 % d. 40 % View solution Correct Option: (b) We have to calculate the laissez passer percent. In example of P: He scores 30 % out of full marks, but fails by fifteen marks. Hence, the unproblematic equation formed is (xxx % of 10) + fifteen In case of Q: He scores 40 % out of total marks, just gets 35 marks more than required to pass. Hence, the unproblematic equation formed is (40 % of x) – 35 1) First summate total marks. Let total marks be x. (30 % of x) + fifteen = (40 % of ten) – 35 30 × x + 15 = 40 × 10 – 35 100 100 35 + 15 = xl × x – 30 × 10 100 100 10=500 Full marks = 500 2) As we know the full marks, we tin calculate the passing. Passing Marks = xxx × ten + 15 100 Passing Marks = 30 × 500 + 15 = 165 100 Therefore, Passing Percentage = 165 × 100 = 33% 500 Required laissez passer percentage = 33 % Q 14. In a scientific discipline examination, the average obtained by entire class was 80 marks. If 10 % of students scored 92 marks and 20 % of students scored 90 marks, and so what was the boilerplate of remaining students? a. 65.32 b. seventy.56 c. 75.43 d. 77.96 View solution Right Pick: (c) Here, we do not know the number of students in the class. So let the number of students be 100 and the required boilerplate is y. 1) 10 % of students scored 92 marks 2) 20 % of students scored 90 marks iii) Therefore, from 100 students, the remaining students are 70 iv) Average obtained past 100 students = 80 marks Considering the given parameters, form the equation. (10 ten 92) + (20 10 90) + (seventy 10 y) = (100 x 80) 70 y = 8000 – (1800 + 920) y = 75.43 The average of remaining students = 75.43 Q fifteen. A student attempts 10 number of questions. He answers 15 correctly out of first xx questions and of the remaining questions, he answers 1/three correctly. If all questions take same credit and the student gets l % marks, and so find the value of ten. a. thirty b. 35 c. 45 d. 50 View solution Right Option: (d) Given: 1) Student attempts x questions. 2) Out of twenty questions he answers 15 correctly and of (10 – 20) questions he answered one/3 correctly. iii) The student gets fifty % marks. Therefore, 15 + 1 (x – 20) = 50% of ten 3 15 + 1 (x – 20) = 50 × x iii 100 90+2 (10-twenty)=3x Solving this equation, we get x=50 Hence, the number of question attempted by the students = 50

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