Exercise-15B
Q1: 17 less than four times a number is 11. Find the number.
Step 1:Let the number be \(x\).
Step 2:Translate the given statement into an equation:
\[
4x – 17 = 11
\]Step 3:Add 17 to both sides:
\[
4x = 11 + 17 \\
4x = 28
\]Step 4:Divide both sides by 4:
\[
x = \frac{28}{4} = 7
\]Answer:x = 7
Q2: If 10 be added to four times a certain number, the result is 5 less than five times the number. Find the number.
Step 1:Let the number be \(x\).
Step 2:Translate the statement into an equation:
\[
4x + 10 = 5x – 5
\]Step 3:Bring variables to one side and constants to the other:
\[
4x – 5x = -5 – 10 \\
– x = -15
\]Step 4:Multiply both sides by -1:
\[
x = 15
\]Answer:x = 15
Q3: \(\frac{2}{3}\) of a number is 20 less than the original number. Find the original number.
Step 1:Let the number be \(x\).
Step 2:Translate the statement into an equation:
\[
\frac{2}{3}x = x – 20
\]Step 3:Bring terms to one side:
\[
\frac{2}{3}x – x = -20 \\
-\frac{1}{3}x = -20
\]Step 4:Multiply both sides by -3:
\[
x = 60
\]Answer:x = 60
Q4: A number is 25 more than its \(\frac{5}{6}\)th part. Find the number.
Step 1:Let the number be \(x\).
Step 2:Translate the statement into an equation:
\[
x = \frac{5}{6}x + 25
\]Step 3:Bring terms involving \(x\) together:
\[
x – \frac{5}{6}x = 25 \\
\frac{1}{6}x = 25
\]Step 4:Multiply both sides by 6:
\[
x = 150
\]Answer:x = 150
Q5: A number is as much greater than 21 as is less than 71. Find the number.
Step 1:Let the number be \(x\).
Step 2:Translate the statement into an equation:
\[
x – 21 = 71 – x
\]Step 3:Add \(x\) to both sides:
\[
x + x – 21 = 71 \\
2x – 21 = 71
\]Step 4:Add 21 to both sides:
\[
2x = 92
\]Step 5:Divide both sides by 2:
\[
x = 46
\]Answer:x = 46
Q6: 6 more than one-fourth of a number is two-fifths of the number. Find the number.
Step 1:Let the number be \(x\).
Step 2:Translate the statement into an equation:
\[
\frac{1}{4}x + 6 = \frac{2}{5}x
\]Step 3:Bring \(\frac{1}{4}x\) to the right side:
\[
6 = \frac{2}{5}x – \frac{1}{4}x
\]Step 4:Find the LCM of 5 and 4 to subtract fractions:
\[
\frac{2}{5}x – \frac{1}{4}x = \frac{8 – 5}{20}x = \frac{3}{20}x \\
6 = \frac{3}{20}x
\]Step 5:Multiply both sides by \(\frac{20}{3}\):
\[
x = 6 \times \frac{20}{3} = 40
\]Answer:x = 40
Q7: One-third of a number exceeds one-fourth of the number by 15. Find the number.
Step 1:Let the number be \(x\).
Step 2:Translate the statement into an equation:
\[
\frac{1}{3}x – \frac{1}{4}x = 15
\]Step 3:Find LCM of 3 and 4 to subtract fractions:
\[
\frac{4 – 3}{12}x = \frac{1}{12}x \\
\frac{1}{12}x = 15
\]Step 4:Multiply both sides by 12:
\[
x = 15 \times 12 = 180
\]Answer:x = 180
Q8: If one-fifth of a number decreased by 5 is 16, find the number.
Step 1:Let the number be \(x\).
Step 2:Translate the statement into an equation:
\[
\frac{1}{5}x – 5 = 16
\]Step 3:Add 5 to both sides:
\[
\frac{1}{5}x = 21
\]Step 4:Multiply both sides by 5:
\[
x = 105
\]Answer:x = 105
Q9: A number when divided by 6 is diminished by 40. Find the number.
Step 1:Let the number be \(x\).
Step 2:Translate the statement into an equation:
\[
\frac{x}{6} = x – 40
\]Step 3: To solve for \(x\), first, multiply both sides of the equation by 6 to eliminate the fraction. :
\[
6\left(\frac{x}{6}\right)=6(x-40) \\
x=6x-240
\]Step 4: Subtract \(6x\) from both sides:
\[
x-6x=-240 \\
-5x=-240
\]
Step 5: Finally, divide both sides by -5:
\[
x=\frac{-240}{-5} \\
x=48
\]Answer:x = 48
Q10: Four-fifths of a number exceeds two-thirds of the number by 10. Find the number.
Step 1:Let the number be \(x\).
Step 2:Translate the statement into an equation:
\[
\frac{4}{5}x – \frac{2}{3}x = 10
\]Step 3:Find LCM of 5 and 3 to subtract fractions:
\[
\frac{12 – 10}{15}x = \frac{2}{15}x \\
\frac{2}{15}x = 10
\]Step 4:Multiply both sides by \(\frac{15}{2}\):
\[
x = 10 \times \frac{15}{2} = 75
\]Answer:x = 75
Q11: Two numbers are in the ratio 3 : 4 and their sum is 84. Find the numbers.
Step 1:Let the numbers be \(3x\) and \(4x\).
Step 2:Sum of the numbers is 84:
\[
3x + 4x = 84
\]Step 3:Combine like terms:
\[
7x = 84
\]Step 4:Divide both sides by 7:
\[
x = 12
\]Step 5:Find the numbers:
\[
3x = 3 \times 12 = 36 \\
4x = 4 \times 12 = 48
\]Answer:The numbers are 36 and 48
Q12: Three numbers are in the ratio 4 : 5 : 6 and their sum is 135. Find the numbers.
Step 1:Let the numbers be \(4x\), \(5x\), and \(6x\).
Step 2:Sum of the numbers is 135:
\[
4x + 5x + 6x = 135
\]Step 3:Combine like terms:
\[
15x = 135
\]Step 4:Divide both sides by 15:
\[
x = 9
\]Step 5:Find the numbers:
\[
4x = 4 \times 9 = 36 \\
5x = 5 \times 9 = 45 \\
6x = 6 \times 9 = 54
\]Answer:The numbers are 36, 45, and 54
Q13: Two numbers are in the ratio 3 : 5. If each is increased by 10, the ratio between the new numbers formed is 5 : 7. Find the original numbers.
Step 1:Let the original numbers be \(3x\) and \(5x\).
Step 2:After increasing each by 10, the numbers become \(3x + 10\) and \(5x + 10\).
Step 3:Form the equation using the new ratio:
\[
\frac{3x + 10}{5x + 10} = \frac{5}{7}
\]Step 4:Cross-multiply:
\[
7(3x + 10) = 5(5x + 10) \\
21x + 70 = 25x + 50
\]Step 5:Simplify:
\[
25x – 21x = 70 – 50 \\
4x = 20 \\
x = 5
\]Step 6:Find the original numbers:
\[
3x = 3 \times 5 = 15 \\
5x = 5 \times 5 = 25
\]Answer:The original numbers are 15 and 25
Q14: The sum of three consecutive odd numbers is 75. Find the numbers.
Step 1:Let the three consecutive odd numbers be \(x\), \(x+2\), and \(x+4\).
Step 2:Sum of the numbers is 75:
\[
x + (x+2) + (x+4) = 75
\]Step 3:Combine like terms:
\[
3x + 6 = 75
\]Step 4:Subtract 6 from both sides:
\[
3x = 69
\]Step 5:Divide both sides by 3:
\[
x = 23
\]Step 6:Find the numbers:
\[
x = 23, \quad x+2 = 25, \quad x+4 = 27
\]Answer:The numbers are 23, 25, and 27
Q15: Divide 25 into two parts such that 7 times the first part added to 5 times the second part makes 139.
Step 1:Let the two parts be \(x\) and \(25 – x\).
Step 2:According to the problem:
\[
7 \cdot x + 5 \cdot (25 – x) = 139
\]Step 3:Expand the terms:
\[
7x + 125 – 5x = 139
\]Step 4:Combine like terms:
\[
2x + 125 = 139
\]Step 5:Subtract 125 from both sides:
\[
2x = 14
\]Step 6:Divide both sides by 2:
\[
x = 7
\]Step 7:The second part is:
\[
25 – x = 25 – 7 = 18
\]Answer:The two parts are 7 and 18
Q16: Divide 180 into two parts such that the first part is 12 less than twice the second part.
Step 1:Let the first part be \(x\) and the second part be \(180 – x\).
Step 2:According to the problem:
\[
x = 2 \cdot (180 – x) – 12
\]Step 3:Expand the right-hand side:
\[
x = 360 – 2x – 12 \\
x = 348 – 2x
\]Step 4:Add \(2x\) to both sides:
\[
x + 2x = 348 \\
3x = 348
\]Step 5:Divide both sides by 3:
\[
x = 116
\]Step 6:The second part is:
\[
180 – x = 180 – 116 = 64
\]Answer:The two parts are 116 and 64
Q17: The denominator of a fraction is 4 more than its numerator. On subtracting 1 from each numerator and the denominator, the fraction becomes \(\frac{1}{2}\). Find the original fraction.
Step 1:Let the numerator be \(x\). Then the denominator is \(x + 4\).
Step 2:The original fraction is:
\[
\frac{x}{x + 4}
\]Step 3:After subtracting 1 from numerator and denominator, the fraction becomes:
\[
\frac{x – 1}{(x + 4) – 1} = \frac{x – 1}{x + 3} = \frac{1}{2}
\]Step 4:Cross-multiply:
\[
2(x – 1) = 1 \cdot (x + 3) \\
2x – 2 = x + 3
\]Step 5:Simplify:
\[
2x – x = 3 + 2 \\
x = 5
\]Step 6:Original fraction:
\[
\frac{x}{x + 4} = \frac{5}{5 + 4} = \frac{5}{9}
\]Answer:The original fraction is 5/9
Q18: The denominator of a fraction is 1 more than double the numerator. On adding 2 to the numerator and subtracting 3 from the denominator, we obtain 1. Find the original fraction.
Step 1:Let the numerator be \(x\). Then the denominator is \(2x + 1\).
Step 2:The original fraction is:
\[
\frac{x}{2x + 1}
\]Step 3:After adding 2 to numerator and subtracting 3 from denominator, the fraction becomes:
\[
\frac{x + 2}{(2x + 1) – 3} = \frac{x + 2}{2x – 2} = 1
\]Step 4:Equate numerator and denominator since the fraction equals 1:
\[
x + 2 = 2x – 2
\]Step 5:Simplify:
\[
2 + 2 = 2x – x \\
x = 4
\]Step 6:Original fraction:
\[
\frac{x}{2x + 1} = \frac{4}{2 \cdot 4 + 1} = \frac{4}{9}
\]Answer:The original fraction is 4/9
Q19: The sum of the digits of a two-digit number is 5. On adding 27 to the number, its digits are reversed. Find the original number.
Step 1:Let the tens digit be \(x\) and the units digit be \(y\).
Step 2:The number is \(10x + y\).
Step 3:According to the problem, the sum of the digits is 5:
\[
x + y = 5
\]Step 4:After adding 27, the digits are reversed:
\[
10x + y + 27 = 10y + x
\]Step 5:Simplify the equation:
\[
10x + y + 27 = 10y + x \\
10x – x + y – 10y = -27 \\
9x – 9y = -27 \\
x – y = -3
\]Step 6:Solve the two equations:
1) \(x + y = 5\)
2) \(x – y = -3\)
Add the equations:
\[
(x + y) + (x – y) = 5 + (-3) \\
2x = 2 \\
x = 1
\]Step 7:Substitute \(x = 1\) in \(x + y = 5\):
\[
1 + y = 5 \\
y = 4
\]Step 8:Original number:
\[
10x + y = 10 \cdot 1 + 4 = 14
\]Answer:The original number is 14
Q20: What same number should be added to each one of the numbers 15, 23, 29, 44 to obtain numbers which are in proportion?
Step 1:Let the same number to be added be \(x\).
Step 2:After adding \(x\), the numbers become:
\[
15 + x,\ 23 + x,\ 29 + x,\ 44 + x
\]Step 3:Numbers are in proportion if:
\[
\frac{15 + x}{23 + x} = \frac{29 + x}{44 + x}
\]Step 4:Cross-multiply:
\[
(15 + x)(44 + x) = (23 + x)(29 + x)
\]Step 5:Expand both sides:
\[
15 \cdot 44 + 15x + 44x + x^2 = 23 \cdot 29 + 23x + 29x + x^2 \\
660 + 59x + x^2 = 667 + 52x + x^2
\]Step 6:Subtract \(x^2\) from both sides:
\[
660 + 59x = 667 + 52x
\]Step 7:Simplify:
\[
59x – 52x = 667 – 660 \\
7x = 7 \\
x = 1
\]Answer:The same number to be added is 1
Q21: The sum of two numbers is 110. One-fifth of the larger number is 8 more than one-ninths of the smaller number. Find the numbers.
Step 1:Let the larger number be \(x\) and the smaller number be \(y\).
Step 2:According to the problem:
1) Sum of numbers: \(x + y = 110\)
2) One-fifth of the larger number is 8 more than one-ninth of the smaller number:
\[
\frac{x}{5} = \frac{y}{9} + 8
\]Step 3:From the first equation, express \(y\) in terms of \(x\):
\[
y = 110 – x
\]Step 4:Substitute \(y = 110 – x\) into the second equation:
\[
\frac{x}{5} = \frac{110 – x}{9} + 8
\]Step 5:Eliminate denominators by multiplying both sides by 45 (LCM of 5 and 9):
\[
9x = 5(110 – x) + 360
\]Step 6:Simplify:
\[
9x = 550 – 5x + 360 \\
9x + 5x = 910 \\
14x = 910 \\
x = 65
\]Step 7:Find \(y\):
\[
y = 110 – x = 110 – 65 = 45
\]Answer:The two numbers are 65 and 45
Q22: A number is subtracted from the numerator of the fraction \(\frac{12}{13}\) and six times that number is added to the denominator. If the new fraction is \(\frac{1}{11}\) , then find the number.
Step 1:Let the number to be found be \(x\).
Step 2:According to the problem, the new fraction is:
\[
\frac{12 – x}{13 + 6x} = \frac{1}{11}
\]Step 3:Cross-multiply:
\[
11(12 – x) = 1(13 + 6x)
\]Step 4:Expand both sides:
\[
132 – 11x = 13 + 6x
\]Step 5:Bring variables to one side and constants to the other:
\[
-11x – 6x = 13 – 132 \\
-17x = -119
\]Step 6:Solve for \(x\):
\[
x = \frac{-119}{-17} = 7
\]Answer:The number is 7
Q23: A right angled triangle having perimeter 120 cm has its two perpendicular sides in the ratio 5 : 12. Find the lengths of its sides.
Step 1:Let the perpendicular sides be \(5x\) and \(12x\).
Step 2:Let the hypotenuse be \(c\). By the Pythagoras theorem:
\[
c^2 = (5x)^2 + (12x)^2
\]Step 3:Calculate \(c^2\):
\[
c^2 = 25x^2 + 144x^2 = 169x^2 \\
c = \sqrt{169x^2} = 13x
\]Step 4:Perimeter = sum of all sides = 120 cm:
\[
5x + 12x + 13x = 120 \\
30x = 120 \\
x = 4
\]Step 5:Lengths of the sides:
\[
5x = 5 \cdot 4 = 20\text{ cm},\quad 12x = 12 \cdot 4 = 48\text{ cm},\quad 13x = 13 \cdot 4 = 52\text{ cm}
\]Answer:The sides of the triangle are 20 cm, 48 cm, 52 cm
Q24: The sum of the digits of a two-digit number is 9. If 9 is added to the number formed by reversing the digits, then the result is thrice the original number. Find the original number.
Step 1:Let the tens digit be \(x\) and the units digit be \(y\).
Step 2:Original number = \(10x + y\)
Reversed number = \(10y + x\)
Step 3:Given: sum of digits = 9:
\[
x + y = 9
\]Step 4:Also, 9 added to reversed number = thrice original number:
\[
10y + x + 9 = 3(10x + y)
\]Step 5:Simplify the equation:
\[
10y + x + 9 = 30x + 3y \\
10y – 3y + x – 30x + 9 = 0 \\
7y – 29x + 9 = 0 \\
7y = 29x – 9
\]Step 6:From Step 3, \(y = 9 – x\). Substitute into above:
\[
7(9 – x) = 29x – 9 \\
63 – 7x = 29x – 9 \\
63 + 9 = 29x + 7x \\
72 = 36x \\
x = 2
\]Step 7:Find \(y\):
\[
y = 9 – x = 9 – 2 = 7
\]Answer:The original number is 27
Q25: The length of a rectangular plot of land exceeds its breadth by 23 m. If the length is decreased by 15 m and the breadth is increased by 7 m, the area is reduced by 360 m². Find the length and breadth of the plot.
Step 1:Let the breadth of the rectangle be \(x\) m.
Then, the length = \(x + 23\) m.
Step 2:Original area = length × breadth = \(x(x + 23)\) m²
Step 3:After changes: new length = \(x + 23 – 15 = x + 8\)
New breadth = \(x + 7\)
New area = \((x + 8)(x + 7)\) m²
Step 4:The area is reduced by 360 m²:
\[
\text{Original area} – \text{New area} = 360 \\
x(x + 23) – (x + 8)(x + 7) = 360
\]Step 5:Expand both sides:
\[
x^2 + 23x – (x^2 + 15x + 56) = 360 \\
x^2 + 23x – x^2 – 15x – 56 = 360 \\
8x – 56 = 360
\]Step 6:\[
8x = 416 \\
x = 52
\]Step 7:Length = \(x + 23 = 52 + 23 = 75\) m
Breadth = \(x = 52\) m
Answer:Length = 75 m, Breadth = 52 m
Q26: The length of a rectangular park is twice its breadth. If the perimeter of the park is 186 m, find its length and breadth.
Step 1:Let the breadth of the park be \(x\) m.
Then, the length = \(2x\) m.
Step 2:Perimeter of a rectangle = 2 × (length + breadth)
\[
2 \left( \text{length} + \text{breadth} \right) = 186
\]Step 3:Substitute length and breadth:
\[
2(2x + x) = 186 \\
2(3x) = 186 \\
6x = 186
\]Step 4:\[
x = \frac{186}{6} = 31
\]Step 5:Length = \(2x = 2 \times 31 = 62\) m
Breadth = \(x = 31\) m
Answer:Length = 62 m, Breadth = 31 m
Q27: The length of a rectangle is 7 cm more than its breadth. If the perimeter of the rectangle is 90 cm, find its length and breadth.
Step 1:Let the breadth of the rectangle be \(x\) cm.
Then, the length = \(x + 7\) cm.
Step 2:Perimeter of a rectangle = 2 × (length + breadth)
\[
2 \left( \text{length} + \text{breadth} \right) = 90
\]Step 3:Substitute length and breadth:
\[
2 \big( (x + 7) + x \big) = 90 \\
2 (2x + 7) = 90 \\
4x + 14 = 90
\]Step 4:\[
4x = 90 – 14 = 76 \\
x = \frac{76}{4} = 19
\]Step 5:Length = \(x + 7 = 19 + 7 = 26\) cm
Breadth = \(x = 19\) cm
Answer:Length = 26 cm, Breadth = 19 cm
Q28: The length of a rectangle is 7 cm less than twice its breadth. If the length is decreased by 2 cm and breadth increased by 3 cm, the perimeter of the resulting rectangle is 66 cm. Find the length and breadth of the original rectangle.
Step 1:Let the breadth of the original rectangle be \(x\) cm.
Then, length = \(2x – 7\) cm.
Step 2:After changes: new length = \(2x – 7 – 2 = 2x – 9\) cm
New breadth = \(x + 3\) cm
Step 3:Perimeter of rectangle = 2 × (length + breadth)
\[
2 \big( (2x – 9) + (x + 3) \big) = 66
\]Step 4:\[
2 (3x – 6) = 66 \\
6x – 12 = 66
\]Step 5:\[
6x = 66 + 12 = 78 \\
x = \frac{78}{6} = 13
\]Step 6:Original length = \(2x – 7 = 2 \times 13 – 7 = 19\) cm
Original breadth = \(x = 13\) cm
Answer:Length = 19 cm, Breadth = 13 cm
Q29: A man is five times as old as his son. In two years time, he will be four times as old as his son. Find their present ages.
Step 1:Let the present age of the son = \(x\) years.
Then, present age of the father = \(5x\) years.
Step 2:In 2 years: Son’s age = \(x + 2\)
Father’s age = \(5x + 2\)
Step 3:According to the question:
\[
5x + 2 = 4(x + 2)
\]Step 4:\[
5x + 2 = 4x + 8 \\
5x – 4x = 8 – 2 \\
x = 6
\]Step 5:Present age of the son = \(x = 6\) years
Present age of the father = \(5x = 5 \times 6 = 30\) years
Answer:Son = 6 years, Father = 30 years
Q30: A man is twice as old as his son. Twelve years ago, the man was thrice as old as his son. Find their present ages.
Step 1:Let the present age of the son = \(x\) years.
Then, present age of the man = \(2x\) years.
Step 2:Twelve years ago:
Son’s age = \(x – 12\)
Man’s age = \(2x – 12\)
Step 3:According to the question:
\[
2x – 12 = 3(x – 12)
\]Step 4:\[
2x – 12 = 3x – 36 \\
2x – 3x = -36 + 12 \\
-x = -24 \\
x = 24
\]Step 5:Present age of the son = \(x = 24\) years
Present age of the man = \(2x = 48\) years
Answer:Son = 24 years, Man = 48 years
Q31: Seema is 10 years elder than Rekha. The ratio of their ages is 5 : 3. Find their ages.
Step 1:Let Rekha’s age = \(x\) years.
Then, Seema’s age = \(x + 10\) years.
Step 2:According to the question, the ratio of their ages:
\[
\frac{x + 10}{x} = \frac{5}{3}
\]Step 3:Cross-multiply:
\[
3(x + 10) = 5x \\
3x + 30 = 5x \\
5x – 3x = 30 \\
2x = 30 \\
x = 15
\]Step 4:Rekha’s age = \(x = 15\) years
Seema’s age = \(x + 10 = 25\) years
Answer:Rekha = 15 years, Seema = 25 years
Q32: 5 years ago, the age of Parvathi was 4 times the age of her son. The sum of their present ages is 55 years. Find Parvathi’s present age.
Step 1:Let the present age of Parvathi’s son = \(x\) years.
Then, present age of Parvathi = \(55 – x\) years.
Step 2:Five years ago:
Son’s age = \(x – 5\)
Parvathi’s age = \(55 – x – 5 = 50 – x\)
Step 3:According to the question, 5 years ago, Parvathi’s age was 4 times her son’s age:
\[
50 – x = 4(x – 5)
\]Step 4:\[
50 – x = 4x – 20 \\
50 + 20 = 4x + x \\
70 = 5x \\
x = 14
\]Step 5:Present age of Parvathi = \(55 – x = 55 – 14 = 41\) years
Answer:Parvathi = 41 years
Q33: A man is 56 years old and his son is 24 years old. In how many years, the father will be twice as old as his son at that time?
Step 1:Let the number of years after which the father will be twice as old as his son = \(x\) years.
Step 2:After \(x\) years:
Father’s age = \(56 + x\)
Son’s age = \(24 + x\)
Step 3:According to the question, father’s age will be twice son’s age:
\[
56 + x = 2(24 + x)
\]Step 4:\[
56 + x = 48 + 2x \\
56 – 48 = 2x – x \\
8 = x
\]Step 5:Number of years after which father will be twice as old as son = \(x = 8\) years
Answer:8 years
Q34: 9 years hence, a girl will be 3 times as old as she was 9 years ago. How old is she now?
Step 1:Let the present age of the girl = \(x\) years.
Step 2:Age 9 years hence = \(x + 9\)
Age 9 years ago = \(x – 9\)
Step 3:According to the question:
\[
x + 9 = 3(x – 9)
\]Step 4:\[
x + 9 = 3x – 27 \\
3x – x = 9 + 27 \\
2x = 36 \\
x = 18
\]Answer:Present age of the girl = 18 years
Q35: A man made a trip of 480 km in 9 hours. Some part of the trip was covered at 45 km/hr and the remaining at 60 km/hr. Find the part of the trip covered by him at 60 km/hr.
Step 1:Let the distance covered at 60 km/hr = \(x\) km.
Then, the distance covered at 45 km/hr = \(480 – x\) km.
Step 2:Time = Distance ÷ Speed
Time taken for 60 km/hr part = \(\frac{x}{60}\) hours
Time taken for 45 km/hr part = \(\frac{480 – x}{45}\) hours
Step 3:Total time = 9 hours
\[
\frac{x}{60} + \frac{480 – x}{45} = 9
\]Step 4:LCM of 60 and 45 = 180
\[
\frac{3x}{180} + \frac{4(480 – x)}{180} = 9 \\
\frac{3x + 1920 – 4x}{180} = 9 \\
\frac{1920 – x}{180} = 9 \\
1920 – x = 1620 \\
x = 1920 – 1620 = 300
\]Answer:Distance covered at 60 km/hr = 300 km
Q36: A motorist travelled from town A to town B at an average speed of 54 km/hr. On his return journey, his average speed was 60 kmph. If the total time taken is \(9\frac{1}{2}\) hours, find the distance between the two towns.
Step 1:Let the distance between the towns = \(x\) km.
Step 2:Time from A to B = Distance ÷ Speed = \(\frac{x}{54}\) hours
Time from B to A = Distance ÷ Speed = \(\frac{x}{60}\) hours
Step 3:Total time = \(9\frac{1}{2} = \frac{19}{2}\) hours
\[
\frac{x}{54} + \frac{x}{60} = \frac{19}{2}
\]Step 4:LCM of 54 and 60 = 540
\[
\frac{10x}{540} + \frac{9x}{540} = \frac{19}{2} \\
\frac{19x}{540} = \frac{19}{2} \\
x = \frac{19}{2} \cdot \frac{540}{19} = 270
\]Answer:Distance between the two towns = 270 km
Q37: The distance between two stations is 300 km. Two motor-cyclists start simultaneously from these stations and move towards each other. The speed of one of them is 7 km/hr faster than that of the other. If the distance between them after 2 hours of their start is 34 km, find the speed of each motor-cyclist.
Step 1:Let the speed of the slower motor-cyclist = \(x\) km/hr.
Then, speed of the faster motor-cyclist = \(x + 7\) km/hr.
Step 2:Distance covered by both together in 2 hours = \(2 \cdot x + 2 \cdot (x + 7) = 2x + 2x + 14 = 4x + 14\) km
Step 3:The distance remaining between them after 2 hours = 34 km
So, distance covered by both = Total distance – Remaining distance = \(300 – 34 = 266\) km
Step 4:\[
4x + 14 = 266 \\
4x = 266 – 14 = 252 \\
x = \frac{252}{4} = 63
\]Step 5:Speed of slower motor-cyclist = 63 km/hr
Speed of faster motor-cyclist = 63 + 7 = 70 km/hr
Answer:Speeds are 63 km/hr and 70 km/hr
Q38: A boat travels 30 km upstream in a river in the speed period of time as it takes to travel 50 km downstream. If the rate of stream be 5 kmph, find the speed of the boat in still water.
Step 1:Let the speed of the boat in still water = \(x\) km/hr.
Speed of stream = 5 km/hr.
Step 2:Upstream speed = \(x – 5\) km/hr
Downstream speed = \(x + 5\) km/hr
Step 3:Time taken upstream = Time taken downstream
\[
\frac{30}{x – 5} = \frac{50}{x + 5}
\]Step 4:Cross multiply:
\[
30(x + 5) = 50(x – 5) \\
30x + 150 = 50x – 250
\]Step 5:\[
50x – 30x = 150 + 250 \\
20x = 400 \\
x = 20
\]Answer:Speed of the boat in still water = 20 km/hr
Q39: The length of each of the equal sides of an isosceles triangle is 4 cm longer than the base. If the perimeter of the triangle is 62 cm, find the lengths of the sides of the triangle.
Step 1:Let the length of the base of the triangle = \(x\) cm.
Then, length of each equal side = \(x + 4\) cm.
Step 2:Perimeter of triangle = sum of all sides
\[
x + (x + 4) + (x + 4) = 62
\]Step 3:\[
x + x + 4 + x + 4 = 62 \\
3x + 8 = 62
\]Step 4:\[
3x = 62 – 8 = 54 \\
x = \frac{54}{3} = 18
\]Step 5:Length of each equal side = \(x + 4 = 18 + 4 = 22\) cm
Answer:Base = 18 cm, Equal sides = 22 cm each
Q40: A certain number of candidates appeared for an examination in which one-fifth of the whole plus 16 secured first division, one-fourth plus 15 secured second division and one-fourth minus 25 secured third division. If the remaining 60 candidates failed, find the total number of candidates appeared.
Step 1:Let the total number of candidates = \(x\).
Candidates securing first division = \(\frac{x}{5} + 16\)
Candidates securing second division = \(\frac{x}{4} + 15\)
Candidates securing third division = \(\frac{x}{4} – 25\)
Candidates failing = 60
Step 2:Sum of all candidates = total candidates
\[
\frac{x}{5} + 16 + \frac{x}{4} + 15 + \frac{x}{4} – 25 + 60 = x
\]Step 3:Combine like terms:
\[
\frac{x}{5} + \frac{x}{4} + \frac{x}{4} + (16 + 15 – 25 + 60) = x \\
\frac{x}{5} + \frac{2x}{4} + 66 = x \\
\frac{x}{5} + \frac{x}{2} + 66 = x
\]Step 4:Find LCM of denominators 5 and 2 → 10
\[
\frac{2x}{10} + \frac{5x}{10} + 66 = x \\
\frac{7x}{10} + 66 = x
\]Step 5:\[
x – \frac{7x}{10} = 66 \\
\frac{3x}{10} = 66 \\
x = \frac{66 \cdot 10}{3} = 220
\]Answer:Total number of candidates = 220
Q41: Raman has three times as much money as Kamal. If Raman gives ₹750 to Kamal, then Kamal will have twice as much as left with Raman. How much had each originally?
Step 1:Let the amount Kamal originally had = ₹\(x\).
Then Raman originally had = ₹\(3x\).
Step 2:After giving ₹750 to Kamal:
Raman has = \(3x – 750\)
Kamal has = \(x + 750\)
Step 3:According to the question, Kamal now has twice as much as what Raman has:
\[
x + 750 = 2(3x – 750)
\]Step 4:Simplify the equation:
\[
x + 750 = 6x – 1500 \\
750 + 1500 = 6x – x \\
2250 = 5x \\
x = 450
\]Step 5:Raman originally had = \(3x = 3 \cdot 450 = 1350\)
Answer:Kamal originally had ₹450 and Raman originally had ₹1350
Q42: The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles.
Step 1:Let the angles of the triangle be \(2x, 3x\) and \(4x\).
Step 2:Sum of angles of a triangle = 180°
\[
2x + 3x + 4x = 180
\]Step 3:\[
9x = 180 \\
x = \frac{180}{9} = 20
\]Step 4:The angles are:
\[
2x = 2 \cdot 20 = 40^\circ \\
3x = 3 \cdot 20 = 60^\circ \\
4x = 4 \cdot 20 = 80^\circ
\]Answer:40°, 60°, 80°
Q43: A certain number of men can finish a piece of work in 50 days. If there are 7 more men, the work can be completed 10 days earlier. How many men were originally there?
Step 1:Let the original number of men = \(x\).
Then, the work can be done by \(x\) men in 50 days.
Step 2:If there are 7 more men, total men = \(x + 7\).
Time taken = 50 – 10 = 40 days.
Step 3:Work = Men × Days (assuming 1 unit of work)
Original work: \(x \times 50\)
New work: \((x+7) \times 40\)
Since total work is same:
\[
x \cdot 50 = (x + 7) \cdot 40
\]Step 4:Simplify the equation:
\[
50x = 40x + 280 \\
50x – 40x = 280 \\
10x = 280 \\
x = 28
\]Answer:Originally, there were 28 men
Q44: Divide 600 in two parts such that 40% of one exceed 60% of the other by 120.
Step 1:Let the two parts be \(x\) and \(600 – x\).
Step 2:According to the question:
\[
40\% \text{ of } x – 60\% \text{ of } (600 – x) = 120 \\
0.4x – 0.6(600 – x) = 120
\]Step 3:Expand the terms:
\[
0.4x – 360 + 0.6x = 120 \\
(0.4x + 0.6x) – 360 = 120 \\
x – 360 = 120
\]Step 4:\[
x = 120 + 360 = 480
\]
Other part = \(600 – 480 = 120\)
Answer:The two parts are 480 and 120
Q45: A workman is paid ₹150 for each day he works and is fined ₹50 for each day he is absent. In a month of 30 days he earned ₹2100, For how many days did he remain absent?
Step 1:Let the number of days he was absent be \(x\).
Then the number of days he worked = \(30 – x\).
Step 2:His earnings = (Payment for working days) – (Fine for absent days)
\[
150(30 – x) – 50(x) = 2100
\]Step 3:Expand the terms:
\[
4500 – 150x – 50x = 2100 \\
4500 – 200x = 2100
\]Step 4:\[
4500 – 2100 = 200x \\
2400 = 200x \\
x = \frac{2400}{200} = 12
\]Answer:He remained absent for 12 days.
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