The DAV Class 8 Maths Solutions and DAV Class 8 Maths Chapter 9 Brain Teasers Solutions of Linear Equations in One Variable offer comprehensive answers to textbook questions.
DAV Class 8 Maths Ch 9 Brain Teasers Solutions
Question 1A.
Tick (✓) the correct option.
(i) If \(\frac{5 x}{3}\) – 4 = \(\frac{2 x}{5}\), then the numerical value of (2x – 7) is
(a) \(\frac{19}{13}\)
(b) \(-\frac{13}{19}\)
(c) 0
(d) \(\frac{13}{19}\)
Solution:
(b) \(-\frac{13}{19}\)
\(\frac{5 x}{3}\) – 4 = \(\frac{2 x}{5}\)
\(\frac{5 x}{3}\) – \(\frac{2 x}{5}\) = 4
(Transposing – 4 to R.H.S. and \(\frac{2 x}{5}\) to L.H.S.)
\(\frac{25 x-6 x}{15}\) = 4
\(\frac{19 x}{15}\) = 4
19x = 60
x = \(\frac{60}{19}\)
∴ 2x – 7 = 2 × \(\frac{60}{19}\) – 7
= \(\frac{120-133}{19}=\frac{-13}{19}\)
Thus, option (b) is the correct answer.
(ii) A number 351 is divided into two parts in the ratio 2 : 7. The product of the numbers is
(a) 20,294
(b) 21,294
(c) 25,294
(d) 31,294
Solution:
(b) 21,294
Let the parts of the number 351 be 2x and lx.
So, 2x + 7x = 351
⇒ x = 39.
Thus, numbers are 2 × 39 and 7 × 39 i.e., 78 and 273.
Product of the numbers is 78 × 273 = 21,294.
Hence, option (b) is the correct answer.
(iii) Value of x for which \(\frac{2 x-3}{3 x+2}=\frac{-2}{3}\) is
(a) \(\frac{5}{10}\)
(b) \(\frac{5}{11}\)
(c) \(\frac{11}{5}\)
(d) \(\frac{5}{12}\)
Solution:
(d) \(\frac{5}{12}\)
\(\frac{2 x-3}{3 x+2}=\frac{-2}{3}\)
⇒ 3 (2x – 3) = – 2 (3x + 2) (Cross multiplication)
⇒ 6x – 9 = – 6x – 4
⇒ 6x + 6x = 9 – 4
(Transposing – 9 to R.H.S. and – 6x to L.H.S.)
⇒ 12x = 5
⇒ x = \(\frac{5}{12}\)
Thus, option (d) is the correct answer.
(iv) Solution of a linear equation in one variable is always
(a) a natural number
(b) a whole number
(c) a real number
(d) an integer
Solution:
(c) a real number
Solution of a linear equation in one variable is always a real number.
Thus, option (c) is the correct answer.
(v) The perimeter of a rectangle is numerically equal to its area. If the width of a rectangle is 2\(\frac{3}{4}\) cm, then its length is 4
(a) \(\frac{11}{3}\) cm
(b) \(\frac{22}{3}\) cm
(c) 11 cm
(d) 10 cm
Solution:
(b) \(\frac{22}{3}\) cm
Let the length of the rectangle be x cm.
and, area of the rectangle = 2 (x + 2\(\frac{3}{4}\))
= 2 (x + \(\frac{11}{4}\))
= 2x + \(\frac{11}{2}\)
and, area of the rectangle = x × 2\(\frac{3}{4}\) = \(\frac{11}{4}\) x
ATQ,
2x + \(\frac{11}{2}\) = \(\frac{11}{4}\) x
2x – \(\frac{11}{4}\)x = – \(\frac{11}{2}\)
(Transposing \(\frac{11}{4}\) x to L.H.S. and \(\frac{11}{2}\) to R.H.S.)
⇒ \(\frac{8 x-11 x}{4}=-\frac{11}{2}\)
⇒ \(\frac{-3 x}{4}=-\frac{11}{2}\)
⇒ – 6x = -44 (Cross multiplication)
⇒ x = \(\frac{-44}{-6}=\frac{22}{3}\) cm
Thus, option (b) is the correct answer.
Question 1B.
Answer the following questions.
(i) Find a number whose fifth part increased by 30 is equal to its fourth part decreased by 30.
(ii) Raman had 30 flowers. He offered some flowers in a temple and found that the ratio of the number of remaining flowers to that of flowers in the beginning is 2 : 5. Find the number of flowers offered by him in the temple.
(iii) The tens and ones digits of a two-digit number are the same. When the number is added to its reverse, the sum is 88. What is the number?
(iv) Solve for x : \(\frac{x^2-9}{x+3}=\frac{4}{7}\)
(v) Rohan brought a pack of 100 chocolates on his birthday to distribute in his class. After giving two chocolates to each student and his class teacher, he is left with ten chocolates. How many students are there in the class?
Solution:
(ii) Let the number be x.
According to question,
\(\frac{x}{5}\) + 30 = \(\frac{x}{4}\) – 30
⇒ \(\frac{x}{5}\) – \(\frac{x}{4}\) = – 30 – 30
(Transposing 30 to R.H.S. and \(\frac{x}{4}\) to L.H.S.)
⇒ \(\frac{4 x-5 x}{20}\) = – 60
⇒ \(\frac{-x}{20}\) = – 60
⇒ – x = – 60 × 20 = – 1200
⇒ x = 1200 (Multiplying both sides by – 1)
(ii) Let the number of flowers offered in a temple be x.
Then, the number of remaining flowers = 30 – x.
ATQ,
\(\frac{30-x}{30}=\frac{2}{5}\)
⇒ 5 (30 – x) = 60 (Cross Multiplication)
⇒ 150 – 5x = 60
⇒ 150 – 60 = 5x
(Transposing – 5x to R.H.S. and 60 to L.H.S.)
⇒ 5x = 90
⇒ x = 18
Thus, the number of flowers offered in the temple is 18.
(iii) Let the digits at both tens and ones places be x.
Then, the 2-digits number formed = 10x + x = 11x.
Reverse of the number = 10x + x = 11x.
ATQ,
11x + 11x = 88
⇒ 22x = 88
⇒ x = 4
Thus, the required number is 4 × 10 + 4, i.e. 44.
(iv) We have,
\(\frac{x^2-9}{x+3}=\frac{4}{7}\)
⇒ \(\frac{x^2-3^2}{x+3}=\frac{4}{7}\)
⇒ \(\frac{(x+3)(x-3)}{x+3}=\frac{4}{7}\)
⇒ x – 3 = \(\frac{4}{7}\)
⇒ 7 (x – 3) = 4
⇒ 7x – 21 = 4
⇒ 7x = 25
⇒ x = \(\frac{25}{7}\)
(v) Let the number of persons having chocolates be x.
According to question,
2x + 10 = 100
⇒ 2x = 100 – 10 = 90 (Transposing 10 to R.H.S.)
⇒ 2x = 90
⇒ x = 45 (Dividing both sides by 2)
So, the number of persons having chocolates is 45.
Thus, number of students = No. of persons having chocolates – No. of class teacher
= 45 – 1 = 44
Hence, there are 44 students in the class.
Question 2.
Solve the equations:
(i) \(\frac{2 z+7}{3 z+8}=\frac{1}{4}\)
(ii) \(\frac{3+2 y}{2+5 y}=\frac{7}{12}\)
(iii) \(\frac{2 x-1}{1+5 x}=\frac{1}{2}\)
(iv) \(\frac{5 k-7}{3 k-9}=\frac{-6}{7}\)
(v) \(\frac{\frac{3}{7} z-\frac{1}{2}}{z-\frac{1}{4}}-\frac{3}{14}=\frac{1}{7}\)
(vi) \(\frac{p-\frac{2}{5}}{p+\frac{2}{5}}\) = 5
(vii) \(\frac{(x+1)(4 x-3)-4 x^2+5}{4 x+1}\) = 2
(viii) \(\frac{x^2-(x+2)(x-2)}{x+3}=\frac{1}{2}\)
Solution:
(i) \(\frac{2 z+7}{3 z+8}=\frac{1}{4}\)
⇒ 4 (2z + 7) = 1 (3z + 8)
⇒ 8z + 28 = 3z + 8
⇒ 8z – 3z = 8 – 28
⇒ 5x = – 20
∴ x = – 4
(ii) \(\frac{3+2 y}{2+5 y}=\frac{7}{12}\)
⇒ 12 (3 + 2y) = 7 (2 + 5y)
⇒ 36 + 24y = 14 + 35y
⇒ 24y – 35y = 14 – 36
⇒ – 11y = – 22
∴ y = 2
(iii) \(\frac{2 x-1}{1+5 x}=\frac{1}{2}\)
⇒ 2 (2x – 1) = 1 + 5x
⇒ 4x – 2 = 1 + 5x
⇒ 4x – 5x = 1 + 2
⇒ – x = 3
∴ x = – 3
(iv) \(\frac{5 k-7}{3 k-9}=\frac{-6}{7}\)
⇒ 7 (5k – 7) = – 6 (3k – 9)
⇒ 35k – 49 = – 18k + 54
⇒ 35k + 18k = 49 + 54
⇒ 53k = 103
∴ k = \(\frac{103}{53}\)
(v) \(\frac{\frac{3}{7} z-\frac{1}{2}}{z-\frac{1}{4}}-\frac{3}{14}=\frac{1}{7}\)
⇒ \(\frac{\frac{3}{7} z-\frac{1}{2}}{z-\frac{1}{4}}=\frac{1}{7}+\frac{3}{14}\)
⇒ \(\frac{\frac{3}{7} z-\frac{1}{2}}{z-\frac{1}{4}}=\frac{5}{14}\)
⇒ \(14\left(\frac{3}{7} z-\frac{1}{2}\right)=5\left(z-\frac{1}{4}\right)\)
⇒ 6z – 7 = 5z – \(\frac{5}{4}\)
6z – 5z = 7 – \(\frac{5}{4}\)
z = \(\frac{33}{4}\)
∴ z = \(\frac{33}{4}\) or 5\(\frac{3}{4}\)
(vi) \(\frac{p-\frac{2}{5}}{p+\frac{2}{5}}\) = 5
⇒ p – \(\frac{2}{5}\) = 5 (p + \(\frac{2}{5}\))
⇒ p – \(\frac{2}{5}\) = 5p + 2
⇒ p – 5p = 2 + p – \(\frac{2}{5}\)
⇒ – 4p = \(\frac{12}{5}\)
⇒ p = \(\frac{12}{5(-4)}=\frac{-3}{5}\)
∴ p = \(\frac{-3}{5}\)
(vii) \(\frac{(x+1)(4 x-3)-4 x^2+5}{4 x+1}\) = 2
⇒ \(\frac{4 x^2-3 x+4 x-3-4 x^2+5}{4 x+1}\) = 2
⇒ \(\frac{x+2}{4 x+1}\) = 2
⇒ x + 2 = 2 (4x + 1)
⇒ x + 2 = 8x + 2
⇒ x – 8x = 0
⇒ – 7x = 0
⇒ x = 0.
(viii) \(\frac{x^2-(x+2)(x-2)}{x+3}=\frac{1}{2}\)
⇒ \(\frac{x^2-\left(x^2-4\right)}{x+3}=\frac{1}{2}\)
⇒ \(\frac{x^2-x^2+4}{x+3}=\frac{1}{2}\)
⇒ \(\frac{4}{x+3}=\frac{1}{2}\)
⇒ x + 3 = 8
⇒ x = 8 – 3 = 5
⇒ x = 5.
Question 3.
Find the positive value of the variable for which the given equation is satisfied.
(i) \(\frac{1-x^2}{1+x^2}=\frac{-4}{5}\)
(ii) \(\frac{3 z^2+7}{4+z^2}\) = 2.
Solution:
(i) \(\frac{1-x^2}{1+x^2}=\frac{-4}{5}\)
⇒ 5 (1 – x2) = – 4 (1 + x2)
⇒ 5 – 5x2 = – 4 – 4x2
⇒ – 5x2 + 4x2 = – 4 – 5
⇒ – x2 = – 9
⇒ x2 = 9
⇒ x = ± 3.
Hence, x = 3.
(ii) \(\frac{3 z^2+7}{4+z^2}\) = 2
⇒ 3z2 + 7 = 8 + 2z2
⇒ 3z2 – 2z2 = 8 – 7
⇒ z2 = 1
⇒ z = ± 1
Hence, z = 1.
Question 4.
5 years ago, a mother was 7 times as old as her daughter. 5 years hence, she will be 3 times as old as her daughter. Find their present ages.
Solution:
Let the daughter’s age before 5 years be x years.
∴ Mother’s age before 5 years = 7x years
After 10 years (5 + 5), the daughter’s age = (x + 10) years
After 10 years, the mother’s age = (7x + 10) years
As per the condition,
7x + 10 = 3(x + 10)
⇒ 7x + 10 = 3x + 30
⇒ 7x – 3x = 30-10
⇒ 4x = 20
∴ x = 5
Hence, the present age of the daughter = 5 + 5 = 10 years
and the present age of mother = 7 × 5 + 5 = 40 years.
Question 5.
The digit in the one’s place of a 2-digit number is 4 times the digit in the ten’s place. The number obtained by reversing the digits exceeds the given number by 54. Find the given number.
Solution:
Let the ten’s place be x.
∴ One’s place = 4x
Number = 10x + 4x = 14x
Number obtained by reversing the digits = x + 10 × 4x = 41x
As per the condition,
41x – 14x = 54
⇒ 27x = 54
⇒ x = 2
Ten’s place = 2 and one’s place = 2 × 4 = 8
Hence, the required number = 8 + 2 × 10 = 28.
Question 6.
The sum of the digits of a 2-digit number is 10. The number obtained by interchanging the digits exceeds the original number by 36. Find the original number.
Solution:
Let the unit place of the number be x.
∴ Ten’s place = (10 – x)
∴ Original number = x + 10(10 – x)
= x + 100 – 10x
= – 9x + 100
New number obtained by interchanging the digits = 1x + 10 – x = 9x + 10
As per the condition,
(9x + 10) – (- 9x + 100) = 36
⇒ 9x + 10 + 9x – 100 = 36
⇒ 18x – 90 = 36
⇒ 18x = 36 + 90
⇒ 18x = 126
⇒ x = \(\frac{126}{18}\) = 7
∴ Unit place = 7
and Ten’s place = 10 – 7 = 3
Hence, the required number = 7 + 10 × 3 = 37.
Question 7.
The sum of two consecutive multiples of 6 is 66. Find these multiples.
Solution:
Let the two consecutive multiples of 6 be x and x + 6.
As per the condition,
x + x + 6 = 66
⇒ 2x = 66 – 6
⇒ 2x = 60
∴ x = 30
Hence, the required multiples of 6 are 30 and 30 + 6 = 36.
Question 8.
The numerator of a rational number is 3 less than its denominator. If the numerator is increased by 1 and the denominator is increased by 3, the number becomes \(\frac{1}{2}\). Find the rational number.
Solution:
Let the denominator be x.
∴ Numerator = x – 3
Rational number = \(\frac{x-3}{x}\)
As per the condition,
\(\frac{x-3+1}{x+3}=\frac{1}{2}\)
\(\frac{x-2}{x+3}=\frac{1}{2}\)
⇒ 2x – 4 = x + 3
⇒ 2x – x = 3 + 4
⇒ x = 7
∴ Denominator = 7 and
Numerator = 7 – 3 = 4
Hence, the required number = \(\frac{4}{7}\).
Question 9.
A race boat covers a distance of 60 km downstream in one and a half hour. It covers this distance upstream in 2 hours. The speed of the race-boat in still water is 35 km/hr. Find the speed of the stream.
Solution:
Let the speed of the stream be x km/hr.
∴ Speed of race boat in downstream = (35 + x) kmJhr
Speed of race boat in upstream = (35 – x) km/hr
Distance travelled by the boat going downstream = \(\frac{3}{2}\) (35 + x) km
and the distance travelled by this boat going upstream = 2 (35 – x)
∴ Total distance covered = [\(\frac{3}{2}\) (35 + x) + 2 (35 – x)] km
As per the condition,
⇒ \(\frac{3}{2}\) (35 + x) + 2 (35 – x) = 60 + 60
⇒ 3 (35 + x) + 4(35 – x) = 2 × 120
⇒ 105 + 3x + 104 – 4x = 240
⇒ – x + 245 = 240
⇒ – x = 240 – 245
⇒ – x = – 5
∴ x = 5
Hence, the speed of the stream = 5 km/hr.
Question 10.
The sides (other than the hypotenuse) of a right triangle are in the ratio 3 :4. A rectangle is described on its hypotenuse, the hypotenuse being the longer side of the rectangle. The breadth of the rectangle is \(\frac{4}{5}\) of its length. Find the shortest side of the right triangle, if the perimeter of the rectangle is 180 cm.
Solution:
Let the lengths of the two sides (other than hypotenuse) be 3x cm and 4x cm respectively.
∴ Hypotenuse AC = \(\sqrt{\mathrm{AB}^2+\mathrm{BC}^2}\)
= \(\sqrt{(3 x)^2+(4 x)^2}\)
= \(\sqrt{9 x^2+16 x^2}\)
= \(\sqrt{25 x^2}\)
= 5x cm
∴ Breadth of the rectangle = \(\frac{4}{5}\) × 5x = 4x cm.
Perimeter of rectangle = 2 [l + b]
= 2 [5x + 4x] = 18x cm
∴ 18x=180
⇒ x = 10
∴ Shortest side of the rectangle = 3 × 10 = 30 cm.
Question 11.
Divide 243 into 3 parts such that half of the first part, one-third of second part and one- fourth of the third part are all equal.
Solution:
Let the first part be x
∴ Second part = \(\frac{3 x}{2}\)
and third part = 2x
As per the condition, x + \(\frac{3 x}{2}\) + 2x = 243
⇒ \(\frac{9 x}{2}\) = 243
⇒ x = 243 × \(\frac{2}{9}\)
⇒ x = 
⇒ x = 54
Hence first part = 54
second part = \(\frac{3}{2}\) × 54 = 81
and third part = 2 × 54 = 108.
Question 12.
A purse has only two-rupee and five-rupee coins. The sum of the coins is 36 and the total value of the coins is ₹ 84. Find the number of five-rupee coins.
Solution:
Let the number of five-rupee coins be x.
Number of two-rupee coins = (36 – x)
As per the condition,
5x + 2 (36 – x) = 84
⇒ 5x + 72 – 2x – 84
⇒ 3x + 72 = 84
⇒ 3x = 84 – 72
⇒ 3x = 12
∴ x = 4
Hence, the number of five-rupee coins is 4.
DAV Class 8 Maths Chapter 9 HOTS
Question 1.
A number is 6 more than the average of, its half, its one-third and its one-sixth. Find the number.
Solution:
Let the number be x.
Then, half of the number = \(\frac{x}{2}\)
one-third of the number = \(\frac{x}{3}\) and
one-sixth of the number = \(\frac{x}{6}\)
According to question,
\(\frac{\frac{x}{2}+\frac{x}{3}+\frac{x}{6}}{3}\) + 6 = x
⇒ \(\frac{\frac{3 x+2 x+x}{6}}{3}\) + 6 = x
⇒ \(\frac{6 x}{6 \times 3}\) + 6 = x
⇒ \(\frac{x}{3}\) + 6 = x
⇒ \(\frac{x+18}{3}\) = x
⇒ x + 18 = 3x
⇒ 3x – x = 18
⇒ 2x = 18
⇒ x = 9
Thus, the required number is 9.
Question 2.
Solve for x. \(\frac{x^2-3 x-28}{x^2-49}=\frac{3}{17}\), x ≠ ± 7
[Hint: x2 – 49 = (x + 7) (x – 7) Factorise the numerator.]
Solution:
We have,
\(\frac{x^2-3 x-28}{x^2-49}=\frac{3}{17}\)
⇒ \(\frac{x^2-7 x+4 x-28}{x^2-7^2}=\frac{3}{17}\) (Factorising the numerator)
⇒ \(\frac{x(x-7)+4(x-7)}{(x+7)(x-7)}=\frac{3}{17}\)
(Applying a2 – b2 identity in the numerator)
⇒ \(\frac{(x-7)(x+4)}{(x+7)(x-7)}=\frac{3}{17}\)
⇒ \(\frac{x+4}{x+7}=\frac{3}{17}\)
⇒ 17 (x + 4) = 3 (x + 7) (Cross Multiplication)
⇒ 17x + 68 = 3x + 21
⇒ 17x – 3x = 21 – 68
⇒ 14x = – 47
⇒ x = \(-\frac{47}{14}\)
Additional Questions:
Question 1.
Solve the given equation and check the answer: 2y + \(\frac{5}{3}\) = \(\frac{26}{3}\) – y.
Solution:
2y + \(\frac{5}{3}\) = \(\frac{26}{3}\) – y
⇒ 2y + y = \(\frac{26}{3}\) – \(\frac{5}{3}\)
⇒ 3y = \(\frac{26-5}{3}\)
⇒ 3y = 7
∴ y = \(\frac{7}{3}\)
Check:
2y + \(\frac{5}{3}\) = \(\frac{26}{3}\)
Put y = \(\frac{7}{3}\), bothsides, we get
⇒ 2 × \(\frac{7}{3}\) + \(\frac{5}{3}\) = \(\frac{26}{3}\) – \(\frac{7}{3}\)
⇒ \(\frac{14}{3}\) + \(\frac{5}{3}\) = \(\frac{26-7}{3}\)
⇒ \(\frac{19}{3}\) = \(\frac{19}{3}\)
LHS = RHS.
Hence verified.
Question 2.
The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What can be the original number?
Solution:
Case 1:
Let the unit place digit be x.
∴ Ten’s place digit may be x + 3 or x – 3
∴ Number = x + 10 (x + 3)
= x + 10x + 30
= 11x + 30
Number obtained by reversing the digits = 10x + x + 3 = 11x + 3
As per the condition,
(11x + 30) + (11x + 3) = 143
⇒ 11x + 30 + 11x + 3 = 143
⇒ 22x = 143 – 33
⇒ 22x = 110
∴ x = 5
Unit place = 5
∴ Ten’s place = 5 + 3 = 8
Hence, the original number = 5 + 8 × 10 = 85
Case II:
Unit place = x
Ten’s place = x – 3
Number = x + 10 (x – 3) = 11x – 30
Number obtained by reversing the digits = 10x + x – 3 = 11x – 3
As per the condition,
11x – 30 + 11x – 3 = 143
⇒ 22x – 33 = 143
⇒ 22x = 143 + 33
⇒ 22x = 176
⇒ x = \(\frac{176}{22}\) = 8
Unit place = 8
Ten’s place = 8 – 3 = 5
Hence, the required number = 8 + 5 × 10 = 58.
Question 3.
The present ages of Anu and Raj are in the ratio 4 : 5. Eight years from now, the ratio of their ages will be 5 : 6. Find their present ages.
Solution:
Let the present ages of Anu and Raj be 4x years and 5x years respectively.
After 8 years, Anu’s age = (4x + 8) years
After 8 years, Raj’s age = (5x + 8) years
As per the condition,
\(\frac{4 x+8}{5 x+8}=\frac{5}{6}\)
⇒ 6 (4x + 8) = 5 (5x + 8)
⇒ 24x + 48 = 25x + 40
⇒ 24x – 25x = 40 – 48
⇒ – x = – 8
∴ x = 8
Hence, Anu’s present age = 8 × 4 = 32 years and
Raj’s present age = 8 × 5 = 40 years.
Question 4.
The perimeter of a rectangle is 100 m. If the length is decreased by 2 m and the breadth is in creased by 3 m, the area increases by 44 m2. Find the length and the breadth of the rectangle.
Solution:
Perimeter = 100
⇒ 2 [length + breadth] = 100
⇒ length + breadth = 50
Let the length be x m.
∴ Breadth = (50 – x) m
∴ Area = x (50 – x) m2
Now new length= (x – 2) m
and new breadth = (50 – x) + 3 = (53 – x) m
∴ New area = (x – 2) (53 – x) m2
As per the condition,
∴ (x – 2) (53 – x) – x (50 – x) = 44
⇒ 53x – x2 – 106 + 2x – 50x + x2 = 44
⇒ 5x – 106 = 44
⇒ 5x = 106 + 44
∴ x = \(\frac{150}{4}\) = 150
Hence, the required length = 30 m
and breadth = 50 – 30 = 20 m.
Question 5.
The sum of a two digit number is 9. If the digits are reversed, the new number exceeds the original number by 45. Find the original number.
Solution:
Let unit place digit be x.
∴ Ten’s place digit = (9 – x)
∴ Original number = x + 10 (9 – x)
= x + 90 – 10x
= – 9x + 90
Number obtained by reversing the digits = 10x + 9 – x = 9x + 9
As per the condition,
9x + 9 – (- 9x + 90) = 45
⇒ 9x + 9 + 9x – 90 = 45
⇒ 18x – 81 = 45
⇒ 18x = 81 + 45
⇒ 18x = 126
⇒ x = 7
Unit place = 7
∴ Ten’s place = 9 – 7 = 2
Hence, the required number = 7 + 2 × 10 = 27.
Question 6.
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \(\frac{3}{2}\). Find the rational number.
Solution:
Let the numerator be x.
∴ Denominator = x + 8
As per the condition,
\(\frac{x+17}{x+8-1}=\frac{3}{2}\)
⇒ \(\frac{x+17}{x+7}=\frac{3}{2}\)
⇒ 3x + 21 = 2x + 34
⇒ 3x – 2x = 34 – 21
⇒ x = 13
Numerator = 13
∴ Denominator = 13 + 8 = 21
∴ Required rationa1 number = \(\frac{13}{21}\).
Question 7.
A streamer goes downstream in 6 hours. It covers the same distance upstream in 7 hours. If the speed of the stream be 2 km/hr, find the speed of the streamer in still water.
Solution:
Let the speed of the streamer be x km/hr.
Speed of the streamer in downstream = (x + 2) km/hr
Speed of the streamer in upstream = (x – 2) km/hr
As per the condition,
Distance travelled going downstream = 6 (x + 2) km
Distance travelled going upstream = 7 (x – 2) km
∴ (7x – 14) = (6x + 12)
⇒ 7x – 6x = 14 + 12
⇒ x = 26
Hence, the speed of the streamer = 26 km/hr.
Question 8.
A bag contains 25 paise and 50 paise coins. If the total number of coins is 70 and their total value is ₹ 30, find the number of coins of each type.
Solution:
Let the number of 25 paise coins be x.
∴ 50 paise coins = (70 – x)
As per the condition,
\(\frac{25 x}{100}+\frac{50(70-x)}{100}\) = 30
\(\frac{25 x+50(70-x)}{100}\) = 30
⇒ 25x + 3500 50x = 3000
⇒ – 25x = – 3500 + 3000
⇒ – 25x = – 500
⇒ x = 20
Hence, the required number of 25 paise coins = 20
and the number of 50 paise coins = 70 – 20 = 50.
Question 9.
Solve for x and check your answer \(\frac{4}{5}\left(x+\frac{5}{6}\right)-\frac{2}{3}\left(x-\frac{1}{4}\right)=1 \frac{1}{9}\).
Solution:
Check:
Question 10.
Solve for x: \(\frac{3 x}{5 x-5}\) = 1.
Solution:
\(\frac{3 x}{5 x-5}\) = 1
⇒ 3x = 5x – 5
⇒ 3x – 5x = – 5
⇒ – 2x = – 5
∴ x = \(\frac{5}{2}\).
Question 11.
\(\frac{2}{3}\) of a number is 20 less than the number. Find the number.
Solution:
Let the number be x.
∴ x – \(\frac{2}{3}\)x = 20
⇒ \(\frac{x}{3}\) = 20
∴ x = 20 × 3 = 60.
Question 12.
Solve the equation: \(\frac{(x+1)(4 x-3)-4 x^2+5}{4 x+1}\) = 2.
Solution:
\(\frac{(x+1)(4 x-3)-4 x^2+5}{4 x+1}\) = 2
⇒ \(\frac{4 x^2-3 x+4 x-3-4 x^2+5}{4 x+1}\) = 2
⇒ x + 2 = 2 (4x + 1)
⇒ x + 2 = 8x + 2
⇒ x – 8x = 2 – 2
⇒ – 7x = 0
∴ x = 0.
Question 13.
The numerator and the denominator of a rational number are in the ratio 3 : 4. If the denominator is increased by 3, the ratio becomes 3 : 5. Find the rational number.
Solution:
Let the numerator be 3x.
and the denominator = 4x
As per the question,
\(\frac{3 x}{4 x+3}=\frac{3}{5}\)
⇒ 15x = 12x + 9
⇒ 15x – 12x = 9
⇒ 3x = 9
∴ x = 3
∴ Numerator = 3 × 3 = 9
Denominator = 4 × 3 = 12
Hence, the rational number = \(\frac{9}{12}\).
Question 14.
One side of a parallelogram is \(\frac{3}{4}\) times its adjacent side. If the perimeter of the parallelogram is 70 cm, find the sides of the parallelogram.
Solution:
Let one side of the parallelogram be x cm.
∴ Other side = \(\frac{3}{4}\) x cm
∴ Perimeter = 2 [x + \(\frac{3}{4}\) x]
= 2 \(\frac{7 x}{4}\)
= \(\frac{7 x}{2}\) cm
∴ \(\frac{7 x}{2}\) = 70
⇒ 7x = 140
⇒ x = 20
Hence, the sides of the parallelogram are 20 cm and 20 × \(\frac{3}{4}\) = 15 cm.
Question 15.
The distance between two towns is 300 km. Two cars start simultaneously from these towns and move towards each other. The speed of one car is more than the other by 7 km/hr. If the distance between the cars still facing each other, after 2 hours is 34 km, find the speed of each car.
Solution:
Let the speed of 1st car be x km/hr.
∴ Speed of 2nd car = (x + 7) km/hr
Distance covered by one car in 2 hours = 2x km
Distance covered by the other car in 2 hours = 2 (x + 7)
As per the condition,
2x + 2 (x + 7) = 300 – 34
⇒ 2x + 2x + 14 = 266
⇒ 4x = 266 – 14
⇒ 4x = 252
⇒ x = \(\frac{252}{4}\) = 63
Hence the speed of 1st car = 63 km/hr
and the speed of 2nd car = 63 + 7 = 70 km/hr.
Question 16.
The sum of the digits of a two digit number is 10. The number obtained by interchanging the digits exceeds the original number by 36. Find the original number.
Solution:
Let unit place digit be x.
∴ Ten’s place digit = (10 – x)
∴ Original number = x + 10 (10 – x)
= x + 100 – 10x
= – 9x + 100
Number obtained by interchanging the digits = 10x + 10 – x = 9x + 10
As per the question,
(9x + 10) – (- 9x + 100) = 36
⇒ 9x + 10 + 9x – 100 = 36
⇒ 18x – 90 = 36
⇒ 18x = 90 + 36 = 126
∴ x = \(\frac{126}{18}\) = 7
Unit place = 7
Ten’s place = 10 – 7 = 3
∴ Required number = 7 + 3 × 10 = 37.
Multiple Choice Questions:
Question 1.
Given, p = – \(\frac{2}{3 q}\), the value of q if p = – \(\frac{1}{5}\) is
(a) \(\frac{10}{3}\)
(b) 3
(e) \(\frac{8}{3}\)
(d) \(\frac{11}{3}\)
Solution:
(a) \(\frac{10}{3}\)
Question 2.
If \(\frac{x}{4}\) = 5 + \(\frac{x}{3}\), then the value of x is
(a) 60
(b) – 60
(c) 0
(d) 160
Solution:
(b) – 60
Question 3.
If \(\sqrt{1+x \sqrt{x^2+24}}\) = x + 1, then is value of x is
(a) – 5
(b) 15
(c) 3
(d) 5
Solution:
(d) 5
Question 4.
If 2 + \(\frac{2 x-3}{2 x+3}\) = \(\frac{3 x-4}{x+2}\), then the value of x is
(a) \(\frac{9}{7}\)
(b) \(\frac{-9}{7}\)
(c) \(\frac{7}{9}\)
(d) \(-\frac{7}{9}\)
Solution:
(b) \(\frac{-9}{7}\)
Question 5.
If p = \(\frac{2 x-1}{3}\), q = \(\frac{8+3 x}{4}\) and \(\frac{p-q}{5}\) = 3, then x is equal to
(a) – 208
(b) 208
(c) 108
(d) – 108
Solution:
(a) – 208