RD Sharma solution class 8 chapter 9 Linear Equation In One Variable Exercise 9.4

Exercise 9.4

Page-9.29

Question 1:

Four-fifth of a number is more than three-fourth of the number by 4. Find the number.

Answer 1:

Let the number be x.According to the question, 45x-34x=4or 16x-15x20=4or x=80  [After cross multiplication]Thus, the required number is 80.

Question 2:

The difference between the squares of two consecutive numbers is 31. Find the numbers.

Answer 2:

Let the numbers be x and x+1.According to the question,(x + 1)2 - x2 = 31or x2 + 2x + 1 - x2 = 31or 2x = 31 - 1or x = 302or x = 15Thus, the numbers are 15 and 16.

Question 3:

Find a number whose double is 45 greater than its half.

Answer 3:

Let the number be x.According to the question,2x = 12x + 45or 2x - 12x = 45or 4x-x2 = 45or 3x = 90 [After cross multiplication]or x = 903or x = 30Thus, the number is 30.

Question 4:

Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.

Answer 4:

Let the number be x.According to the question,5x - 5 = 2x + 4or 5x - 2x = 4 + 5or 3x = 9or x = 93or x =  3Thus, the number is 3.

Question 5:

A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number.

Answer 5:

Let the number be x.According to the question,x5 + 5 = x4 - 5or x5 - x4 = -5 - 5or 4x-5x20 = -10or -x = -200 [After cross multiplication]or x = 200Thus, the number is 200.

Question 6:

A number consists of two digits whose sum is 9. If 27 is subtracted from the number, its digits are reversed. Find the number.

Answer 6:

Let the units digit be x. Sum of two digits = 9 Tens digit = (9-x) Original number =10×(9-x)+x     Reversed number = 10x+(9-x)According to the question,10×(9-x)+x-27 = 10x+(9-x)or 90-10x+x-27 = 10x+9-xor 9x+9x = 90-27-9or 18x = 54or x = 5418 = 3 The number =10×(9-3)+3 = 63

Question 7:

Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8.

Answer 7:

Let the first part of 184 be x.Therefore, the other part will be (184-x).According to the question,13x-17(184-x) = 8or 7x-552+3x21 = 8or 10x-552 = 168 [After cross multiplication]or 10x = 168+552or x = 72010 = 72Thus, the parts of 184 are 72 and 112 (184-72 = 112).

Question 8:

The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to 23. What is the original fraction equal to?

Answer 8:

Let the denominator of the fraction be x.Therefore, the numerator will be ( x-6). Fraction = x-6xAccording to the question,x-6+3x = 23or x-3x = 23or 3x-9 = 2x [After cross multiplication]or 3x-2x = 9or x = 9Thus, the original fraction = 9-69=13

Question 9:

A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50, find the number of notes of each type.

Answer 9:

Let the number of Rs. 10 notes be x.Therefore, the number of Rs. 20 notes will be (50-x).Value of Rs. 10 notes = 10xValue of Rs. 20 notes = 20(50-x)According to the question,10x+20(50-x) = 800or 10x+1000-20x = 800or 10x = 1000-800or x = 20010 = 20 Number of Rs. 10 notes = 20     Number of Rs. 20 notes = (50-20) = 30. 

Question 10:

Seeta Devi has Rs 9 in fifty-paise and twenty five-paise coins. She has twice as many twenty-five paise coins as she has fifty-paise coins. How many coins of each kind does she have?

Answer 10:

Let the number of 50 paise coins be x.Therefore, the number of 25 paise coins will be 2x.Value of 50 paise coins = Rs. 0.5xValue of 25 paise coins = Rs. 0.25×2xAccording to the question,0.5x+0.25×2x = 9or x = 9 Number of fifty paise coins = 9     Number of twenty five paise coins = 2×9 = 18     Total number of coins = 9+18 = 27.

Page-9.30

Question 11:

Sunita is twice as old as Ashima. If six years is subtracted from Ashima's age and four years added to Sunita's age, then Sunita will be four times Ashima's age. How old were they two years ago?

Answer 11:

Let the age of Ashima be x years.Therefore, the age of Sunita will be 2x years.According to the question,4(x-6) = 2x+4or 4x-24 = 2x+4or 4x-2x = 4+24or 2x = 28or x = 14 Age of Ashima = 14 years.     Age of Sunita = 2×14 = 28 years.

Question 12:

The ages of sonu and Monu are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.

Answer 12:

It is given that the ratio of the ages of Sonu and Monu is 7:5.Let the present ages of Sonu and Monu be 7x and 5x years.After ten years:Age of Sonu = 7x + 10 years Age of Monu = 5x + 10 yearsAccording to the question,7x + 105x + 10 = 97or 49x + 70 = 45x + 90or 49x - 45x = 90 - 70or 4x = 20or x = 5 Present age of Sonu = 7 × 5 = 35 years.    Present age of Monu = 5 × 5 = 25 years.

Question 13:

Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Answer 13:

Five years ago:Let the age of the son be x years.Therefore, the age of the father will be 7x years. Present age of the son = (x + 5) years     Present age of the father = (7x + 5) yearsAfter five years:Age of the son = (x + 5 + 5) = (x + 10) yearsAge of the father = (7x + 5 + 5) = (7x + 10) yearsAccording to the question,7x + 10 = 3(x + 10)or 7x - 3x = 30 - 10or 4x = 20or x = 5 Present age of the son = (5 + 5) = 10 years.     Present age of the father = (7 × 5 + 5) = 40 years.

Question 14:

I am currently 5 times as old as my son. In 6 years time I will be three times as old as he will be then. What are our ages now?

Answer 14:

Let the age of my son be x years.Therefore, my age will be 5x years.After 6 years:Age of my son = (x + 6) yearsMy age = (5x + 6) yearsAccording to the question,5x + 6 = 3(x + 6)or 5x - 3x = 18 - 6or 2x = 12or x = 6 Age of my son = 6 years.     My age = 5 × 6 = 30 years.

Question 15:

I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?

Answer 15:

Let the number of five-rupee notes be x.Therefore, the number of ten-rupee notes will be (x+10).Now,Value of five-rupee notes = Rs. 5xValue of ten-rupee notes = Rs. 10(x + 10)According to the question,5x + 10(x + 10) = 1000or 15x = 1000 - 100or x = 90015 = 60 Number of five-rupee notes = 60.     Number of ten-rupee notes = 60 + 10 = 70. 

Question 16:

At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two fifths drank fruit juice and just three did not drink any thing. How many guests were in all?

Answer 16:

Let the total number of guests be x.Therefore, the number of guests, who drank colas, would be 14x.The number of guests, who drank squash, would be 13x.The number of guests, who drank fruit juice, would be 25x.The number of guests, who did not drink, would be 3.According to the question,x - (x4 + x3 + 2x5) = 3or 60x - 15x - 20x - 24x60 = 3or x = 180Thus, total number of guests =180.

Question 17:

There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?

Answer 17:

Let the number of correctly answered questions be x.Therefore, the number of unattempted or wrongly answered questions will be (180 - x).According to the question,4x - 1(180 - x) = 450or 5x = 450 + 180or x = 6305 = 126Thus, number of correctly answered questions =126.Number of unattempted or wrongly answered questions =180 - 126 = 54.

Question 18:

A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day, he works and he will be fined Rs 5 for each day, he is absent. If he receives Rs 745 in all, for how many days he remained absent?

Answer 18:

Let the number of days for which the labourer is absent be x.Therefore, the number of days for which he is present will be (20-x). Earnings = Rs. 60(20 - x)     Fine = Rs. 5xAccording to the question,60(20 - x) - 5x = 745or 1200 - 60x - 5x = 745or 65x = 1200 - 745or x = 45565 = 7Thus, the labourer was absent for 7 days.

Question 19:

Ravish has three boxes whose total weight is 6012 kg. Box B weighs 312 kg more than box A and box C weighs 513 kg more than box B. Find the weight of box A.

Answer 19:

Let the weight of box A be x kg.Therefore, the weights of box B and box C will be (x + 312) kg and (x + 312 + 513) kg, respectively.According to the question,x + (x + 312) + (x + 312 + 513) = 6012or 3x = 1212 - 72 - 72 - 163or 3x = 363 - 21 - 21 - 326or 3x = 2896or x = 28918Thus, weight of box A = 28918 kg

Question 20:

The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 1/2.  Find the rational number.

Answer 20:

Let, the denominator of the rational number be x. The numerator of the rational number will be x-3. The rational number =x-3xAccording to the question,x - 3 + 2x + 5 = 12or x - 1x + 5 = 12or 2x - 2 = x + 5or 2x - x = 5 + 2or x = 7 The rational number = 7 - 37 = 47

Question 21:

In a rational number, twice the numerator is 2 more than the denominator. If 3 is added to each, the numerator and the denominator, the new fraction is 2/3. Find the original number.

Answer 21:

Let the denominator be x. The numerator = x + 22 The rational number = x + 22xAccording to the question,x + 22 + 3x + 3 = 23or x + 2 +  62(x + 3) = 23or x + 82x + 6 = 23or 3x +  24 = 4x + 12or x = 24 - 12or x = 12The rational number=12 + 22 × 12 = 1424 = 712

Question 22:

The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/hr. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.

Answer 22:

Let, the speed of the first train be x km/h.Then, the speed of the other train will be (x + 5) km/h.2 hours after they started:Distance of the first train from the starting point = 2x kmDistance of the other train from the starting point = 2(x + 5) kmNow,2(x + 5) + 2x + 30 = 340or 4x + 10 + 30 = 340or 4x = 340 - 40or x = 3004 = 75 Speed of the first train = 75 km/h.     Speed of the other train = (75 + 5)=80 km/h.

Question 23:

A steamer goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream be 1 km/hr, find the speed of the steamer in still water and the distance between the ports.

Answer 23:

It is given that the speed of the stream is 1 km/h.Let the speed of the steamer in still water be x km/h. Downstream speed = (x + 1) km/h     Upstream speed = (x - 1) km/hThe downstream and upstream distances are same; therefore, we have:9(x + 1) = 10(x - 1)or 9x + 9 = 10x - 10or x = 19 Speed of the steamer in still water = 19 km/h.     Distance between the ports = 9(19 + 1) = 180 km.

Question 24:

Bhagwanti inherited Rs 12000.00. She invested part of it as 10% and the rest at 12%. Her annual income from these investments is Rs 1280.00. How much did she invest at each rate?

Answer 24:

At the rate of 10%, let the investment by Bhagwanti be Rs. x.Therefore, at the rate of 12%, the investment will be Rs. (12000-x).At the rate of 10%, her annual income = x × 10%At the rate of 12%, her annual income = (12000 - x) × 12%So,x × 0.1 + 0.12(12000 - x) = 1280or 0.1x - 0.12x = 1280 - 1440or 0.02x = 160or x = 8000Thus, at the rate of 10%, she invested Rs. 8000 and at the rate of 12%, she invested Rs. 4000 (12000-8000). 

Question 25:

The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 84 cm2 more than that of the given rectangle. Find the length and breath of the given rectangle.

Answer 25:

Let the breadth of the rectangle be x cm.Therefore, the length of the rectangle will be (x + 9) cm. Area of the rectangle = x(x + 9) cm2.If the length and breadth are increased by 3 cm each, area = (x + 3)(x + 9 + 3) cm2.Now,(x + 3)(x + 12) - x(x + 9) = 84or x2 + 15x + 36 - x2 - 9x = 84or 6x = 84 - 36or x = 486 = 8.Thus, breadth of the rectangle = 8 cm.          Length of the rectangle =(8+9)=17 cm.

Page-9.31

Question 26:

The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now?

Answer 26:

Let Anup's age be x years.Therefore, his father's age will be (100 - x) years.When Anup is as old as his father after (100 - 2x) years, Anuj's age = 100 - x5 + 100 - 2x years=600 - 11x5 years.Again, when Anup is as old as his father, Anuj's age = x + 8.Now,600 - 11x5 = x + 8or 600 - 11x = 5x + 40or 16x = 560or x = 35.Thus, Anup's age = 35 years          Anup's father's age=100 - x = 100 - 35 = 65 years          Anuj's age = x + 8 = 35 + 8 = 43 years

Question 27:

A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on a lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?

Answer 27:

Suppose, the lady started with x rupees.Money spent on shopping=x2 rupeesRemaining amount=x-x2=x2 rupeesAfter giving a rupee she had=(x2-1) rupeesMoney spent on lunch=12(x2-1) rupeesAfter giving a two-rupee tip she had=12(x2-1)-2=x-2-84=x-104 rupeesMoney spent on a book=12(x-104) rupeesAfter spending three rupees on bus fare she had=12(x-104)-3=x-10-248=x-348 rupeesNow,x-348=1or x-34=8or x=42Therefore, she started with 42 rupees.

No comments:

Post a Comment

Contact Form

Name

Email *

Message *