Exercise-15C
Q1: If \(\frac{x}{2}-\frac{3x}{4}+\frac{5x}{6}=21\), then, x =?
Step 1:Given equation:
\[
\frac{x}{2} – \frac{3x}{4} + \frac{5x}{6} = 21
\]Step 2:Take LCM of denominators \(2, 4, 6\).
LCM = 12.
Step 3:Convert each term with denominator 12:
\[
\frac{x}{2} = \frac{6x}{12}, \quad \frac{3x}{4} = \frac{9x}{12}, \quad \frac{5x}{6} = \frac{10x}{12}
\]Step 4:Substitute back:
\[
\frac{6x}{12} – \frac{9x}{12} + \frac{10x}{12} = 21 \\
\frac{(6x – 9x + 10x)}{12} = 21 \\
\frac{7x}{12} = 21
\]Step 5:Multiply both sides by 12:
\[
7x = 21 \times 12 \\
7x = 252 \\
x = \frac{252}{7} = 36
\]Answer: c. 36
Q2: \(\frac{x+5}{2}+\frac{x-5}{3}=\frac{25}{6}\) gives:
Step 1:Given equation:
\[
\frac{x+5}{2} + \frac{x-5}{3} = \frac{25}{6}
\]Step 2:Take LCM of denominators \(2, 3, 6\).
LCM = 6.
Step 3:Convert terms with denominator 6:
\[
\frac{x+5}{2} = \frac{3(x+5)}{6}, \quad \frac{x-5}{3} = \frac{2(x-5)}{6}
\]Step 4:Substitute back:
\[
\frac{3(x+5)}{6} + \frac{2(x-5)}{6} = \frac{25}{6} \\
\frac{3x+15 + 2x-10}{6} = \frac{25}{6} \\
\frac{5x+5}{6} = \frac{25}{6}
\]Step 5:Multiply both sides by 6:
\[
5x + 5 = 25 \\
5x = 20 \\
x = 4
\]Answer: c. \(x = 4\)
Q3: \(\frac{4}{5}\) of a number exceeds its \(\frac{2}{3}\) by 8. The number is
Step 1:Let the number be \(x\).
Step 2:According to the question:
\[
\frac{4}{5}x – \frac{2}{3}x = 8
\]Step 3:Take LCM of denominators \(5\) and \(3\).
LCM = 15.
\[
\frac{12x}{15} – \frac{10x}{15} = 8 \\
\frac{2x}{15} = 8
\]Step 4:Multiply both sides by 15:
\[
2x = 120 \\
x = 60
\]Answer: b. 60
Q4: A number whose fifth part increased by 5 is equal to its fourth part diminished by 5, is
Step 1:Let the number be \(x\).
Step 2:According to the question:
\[
\frac{x}{5} + 5 = \frac{x}{4} – 5
\]Step 3:Take LCM of denominators \(5\) and \(4\).
LCM = 20.
\[
\frac{4x}{20} + 5 = \frac{5x}{20} – 5
\]Step 4:Multiply through by 20:
\[
4x + 100 = 5x – 100
\]Step 5:Rearrange terms:
\[
5x – 4x = 100 + 100 \\
x = 200
\]Answer: c. 200
Q5: A student was asked to find \(\frac{7}{8}\) of a positive number. He found \(\frac{7}{18}\) of the same by mistake. If his was 770 less than the correct one, then the original number was
Step 1:Let the number be \(x\).
Step 2:Correct value = \(\frac{7}{8}x\).
Wrong value = \(\frac{7}{18}x\).
Step 3:Difference between correct and wrong values = 770.
\[
\frac{7}{8}x – \frac{7}{18}x = 770
\]Step 4:Take LCM of 8 and 18 → LCM = 72.
\[
\frac{63x}{72} – \frac{28x}{72} = 770 \\
\frac{35x}{72} = 770
\]Step 5:Multiply both sides by 72:
\[
35x = 770 \times 72 \\
35x = 55440 \\
x = \frac{55440}{35} = 1584
\]Step 6 (Check options):The answer \(x = 1584\) is not in the given options.
Answer: d. None of these (Correct number = 1584)
Q6: A drum of kerosene is \(\frac{3}{4}\) full. When 80 litres of kerosene is drawn from it, it remains \(\frac{7}{2}\) full. The capacity of the drum is
Step 1:Let the total capacity of the drum be \(x\) litres.
Step 2:Initially, the drum has \(\frac{3}{4}x\) litres.
Step 3:After removing 80 litres, remaining quantity = \(\frac{3}{4}x – 80\).
Step 4:According to the problem:
\[
\frac{3}{4}x – 80 = \frac{7}{20}x
\]Step 5:Take LCM of 4 and 20 → 20.
\[
\frac{15x}{20} – 80 = \frac{7x}{20} \\
\frac{15x – 7x}{20} = 80 \\
\frac{8x}{20} = 80
\]Step 6:Simplify:
\[
\frac{2x}{5} = 80 \\
2x = 400 \\
x = 200
\]Answer: 200 litres (Not listed in options)
Q7: The sum of seven consecutive integers is 175. What the difference between twice the largest number and thrice the smallest number?
Step 1:
Let the seven consecutive integers be:
\(x,\ (x+1),\ (x+2),\ (x+3),\ (x+4),\ (x+5),\ (x+6)\).
Step 2:
Sum of these integers = \(175\).
\[
x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) = 175
\]Step 3:
Simplify:
\[
7x + (1+2+3+4+5+6) = 175 \\
7x + 21 = 175
\]Step 4:
\[
7x = 175 – 21 \\
7x = 154 \\
x = 22
\]Step 5:
So, the seven numbers are:
\(22, 23, 24, 25, 26, 27, 28\).
– Smallest number = \(22\)
– Largest number = \(28\)
Step 6:
Twice the largest = \(2 \times 28 = 56\)
Thrice the smallest = \(3 \times 22 = 66\)
Step 7:
Difference = \(56 – 66 = -10\)
Answer: c. -10
Q8: The sum of five consecutive odd numbers is 575 What is the sum of the next set of five consecutive odd numbers?
Step 1:
Let the five consecutive odd numbers be:
\(x,\ (x+2),\ (x+4),\ (x+6),\ (x+8)\).
Step 2:
Their sum = \(575\).
\[
x + (x+2) + (x+4) + (x+6) + (x+8) = 575
\]Step 3:
Simplify:
\[
5x + (2+4+6+8) = 575 \\
5x + 20 = 575
\]Step 4:
\[
5x = 575 – 20 \\
5x = 555 \\
x = 111
\]Step 5: So, the numbers are: \(111, 113, 115, 117, 119\).
Their sum = \(575\) (verified ✅).
Step 6:
Next set of 5 consecutive odd numbers will be:
\(121, 123, 125, 127, 129\).
Step 7:
Sum of these =
\[
121 + 123 + 125 + 127 + 129
\]Step 8:
\[
= (121+129) + (123+127) + 125 \\
= 250 + 250 + 125 \\
= 625
\]Answer: c. 625
Q9: The sum of four consecutive even integers is 1284. The greatest of them is
Step 1:
Let the four consecutive even integers be:
\(x,\ (x+2),\ (x+4),\ (x+6)\).
Step 2:
Their sum is given as 1284.
\[
x + (x+2) + (x+4) + (x+6) = 1284
\]Step 3:
Simplify:
\[
4x + (2+4+6) = 1284 \\
4x + 12 = 1284
\]Step 4:
\[
4x = 1284 – 12 \\
4x = 1272 \\
x = \frac{1272}{4} = 318
\]Step 5:
The four consecutive even integers are:
\(318,\ 320,\ 322,\ 324\).
Step 6:
The greatest number = **324**.
Answer: c. 324
Q10: The ratio of the ages of a father and his is 17 : 7. Six years ago, the ratio of their ages was 3 : 1. The father’s present age is
Step 1:Let the present ages of father and son be \(17x\) and \(7x\) respectively.
Step 2:Six years ago, their ages were:
Father = \(17x – 6\)
Son = \(7x – 6\)
Step 3:Given ratio six years ago was 3 : 1.
\[
\frac{17x – 6}{7x – 6} = \frac{3}{1}
\]Step 4:Cross multiply:
\[
17x – 6 = 3(7x – 6) \\
17x – 6 = 21x – 18
\]Step 5:
\[
-6 + 18 = 21x – 17x \\
12 = 4x \\
x = 3
\]Step 6:Father’s present age = \(17x = 17 \times 3 = 51\).
Answer: b. 51 years
Q11: The age of a mother is thrice that of her daughter. After 12 years, the age of the mother will be twice that of her daughter. The present age the daughter is
Step 1:Let the present age of the daughter be \(x\) years.
Then the present age of the mother = \(3x\) years.
Step 2:After 12 years:
Daughter’s age = \(x + 12\)
Mother’s age = \(3x + 12\)
Step 3:According to the problem, after 12 years mother’s age = 2 × daughter’s age:
\[
3x + 12 = 2(x + 12)
\]Step 4:Simplify:
\[
3x + 12 = 2x + 24 \\
3x – 2x = 24 – 12 \\
x = 12
\]Answer: a. 12 years
Q12: The present ages of three persons are in the proportion 4 : 7 : 9. Eight years ago, the sum of their ages was 56 years. The present age of the eldest person is
Step 1:Let the present ages of the three persons be \(4x, 7x, 9x\) years.
Step 2:Eight years ago, their ages were:
\[
4x – 8, \quad 7x – 8, \quad 9x – 8
\]Step 3:Sum of ages eight years ago = 56:
\[
(4x – 8) + (7x – 8) + (9x – 8) = 56
\]Step 4:Simplify:
\[
4x + 7x + 9x – 24 = 56 \\
20x – 24 = 56
\]Step 5:
\[
20x = 56 + 24 \\
20x = 80 \\
x = 4
\]Step 6:Present ages:
\[
4x = 16, \quad 7x = 28, \quad 9x = 36
\]Step 7:The eldest person’s present age = \(36\) years.
Answer: b. 36 years
Q13: In a classroom, there are certain number of benches. If 6 students are made to sit on a bench, then to accommodate all of them, one more bench is needed. If 7 students are made to sit on a bench, then after accommodating all space for 5 students is left. What is the total number of students in the school?
Step 1:Let the number of benches be \(x\) and the total number of students be \(y\).
Step 2:If 6 students sit per bench, one more bench is needed:
\[
6x < y \le 6(x+1)
\]Step 3:If 7 students sit per bench, space for 5 students is left:
\[
y + 5 = 7x
\]Step 4:From step 2, we approximate \(y = 6(x+1) – k\) where \(k < 6\).
Better approach: Use equation from step 3:
\[
y = 7x – 5
\]Step 5:Also, one more bench needed when 6 students sit:
\[
y > 6x \quad \text{and} \quad y \le 6(x+1)
\]
Substitute \(y = 7x – 5\):
\[
7x – 5 > 6x \quad \text{and} \quad 7x – 5 \le 6(x+1)
\]Step 6:Simplify the inequalities:
\[
7x – 5 > 6x \\
x > 5 \\
7x – 5 \le 6x + 6 \\
x \le 11
\]Step 7:So \(x\) can be integers: \(6,7,8,9,10,11\)
Check which satisfies \(y = 7x – 5\) divisible by 6?
– \(x = 6 \\
y = 7*6 -5 = 37\) ❌ not divisible by 6
– \(x = 7 \\
y = 49 – 5 = 44\) ❌
– \(x = 8 \\
y = 56 – 5 = 51\) ❌
– \(x = 9 \\
y = 63 – 5 = 58\) ❌
– \(x = 10 \\
y = 70 – 5 = 65\) ❌
– \(x = 11 \\
y = 77 – 5 = 72\) ✅ worksStep 8:Thus, total number of students = 72
Answer: c. 72
Q14: 180 oranges are distributed among 70 boys and girls such that each boy gets 2 and each girl gets 3 oranges. The number of boys is
Step 1:Let the number of boys = \(x\) and the number of girls = \(y\).
Step 2:Total number of children:
\[
x + y = 70
\]Step 3:Total oranges:
\[
2x + 3y = 180
\]Step 4:From step 2, \(y = 70 – x\). Substitute in step 3:
\[
2x + 3(70 – x) = 180
\]Step 5:Simplify:
\[
2x + 210 – 3x = 180 \\
– x + 210 = 180 \\
-x = -30 \\
x = 30
\]Answer: b. 30 boys
Q15: A sum of ₹1000 is to be divided among three boys- Rahul, Sumit and Gaurav such that Rahul receive twice as much as Sumit, who receives one-fifth as much as Gaurav. How much money does Gaurav receive?
Step 1:Let the amount received by Gaurav = \(x\) ₹.
Step 2:Sumit receives one-fifth of Gaurav:
\[
\text{Sumit} = \frac{x}{5}
\]Step 3:Rahul receives twice as much as Sumit:
\[
\text{Rahul} = 2 \times \frac{x}{5} = \frac{2x}{5}
\]Step 4:Total sum = ₹1000:
\[
\text{Rahul} + \text{Sumit} + \text{Gaurav} = 1000 \\
\frac{2x}{5} + \frac{x}{5} + x = 1000
\]Step 5:Combine terms:
\[
\frac{2x + x}{5} + x = \frac{3x}{5} + x = \frac{3x}{5} + \frac{5x}{5} = \frac{8x}{5} \\
\frac{8x}{5} = 1000
\]Step 6:
\[
x = \frac{1000 \times 5}{8} = 625
\]Answer: d. ₹625
Q16: On a school’s Annual Day sweets were to be equally distributed amongst 112 children. But on that particular day, 32 children were absent. Thus the remaining children got 6 extra sweets each. How many sweets each child originally supposed to get?
Step 1:Let the number of sweets each child was supposed to get = \(x\).
Step 2:Total number of sweets = \(112 \cdot x = 112x\).
Step 3:Number of children present = \(112 – 32 = 80\).
Step 4:Each present child got 6 extra sweets:
\[
\frac{112x}{80} = x + 6
\]Step 5:Simplify:
\[
\frac{112x}{80} = x + 6 \\
\frac{112}{80}x = x + 6 \\
\frac{7}{5}x = x + 6
\]Step 6:
\[
\frac{7x – 5x}{5} = 6 \\
\frac{2x}{5} = 6 \\
2x = 30 \\
x = 15
\]Answer: a. 15 sweets
Q17: An employee claims ₹7 for each km when he travels by taxi and ₹6 for each km if he drives his own car. If in one week he claimed ₹675 for travelling 90 km, how many km did he travel by taxi?
Step 1:Let the distance travelled by taxi = \(x\) km.
Then the distance travelled by own car = \(90 – x\) km.
Step 2:Total claim = ₹675.
\[
7x + 6(90 – x) = 675
\]Step 3:Simplify the equation:
\[
7x + 540 – 6x = 675 \\
x + 540 = 675
\]Step 4:
\[
x = 675 – 540 \\
x = 135
\]Answer: a. 135 km
Q18: In an examination, a Student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. A student attempted all the 200 questions and in all 200 marks. The number of questions he answered correctly was
Step 1:Let the number of correct answers = \(x\).
Then the number of wrong answers = \(200 – x\).
Step 2:Total marks = (marks for correct answers) – (marks lost for wrong answers)
\[
4x – 1(200 – x) = 200
\]Step 3:Simplify the equation:
\[
4x – 200 + x = 200 \\
5x – 200 = 200 \\
5x = 400
\]Step 4:
\[
x = \frac{400}{5} = 80
\]Answer: c. 80 questions
Q19: For aiming at a target, a person gets one rupee each time when he hits it and loses one rupee when he misses it. he gets ₹30 after aiming at the target one hundred times, then how many times did he miss the target?
Step 1:Let the number of hits = \(x\).
Then the number of misses = \(100 – x\).
Step 2:Total gain = (gain for hits) – (loss for misses) = ₹30
\[
1 \cdot x – 1 \cdot (100 – x) = 30
\]Step 3:Simplify the equation:
\[
x – 100 + x = 30 \\
2x – 100 = 30 \\
2x = 130 \\
x = 65
\]Step 4:Number of misses = \(100 – x = 100 – 65 = 35\)
Answer: b. 35 times
Q20: A total of 324 coins of 20 paise and 25 paise make a sum of ₹71. The number of 25 paise coins is
Step 1:Let the number of 25 paise coins = \(x\).
Then the number of 20 paise coins = \(324 – x\).
Step 2:Convert ₹71 into paise: ₹71 = 7100 paise.
Total value of coins = 25x + 20(324 – x) = 7100
Step 3:Simplify the equation:
\[
25x + 6480 – 20x = 7100 \\
5x + 6480 = 7100
\]Step 4:
\[
5x = 7100 – 6480 \\
5x = 620 \\
x = \frac{620}{5} = 124
\]Answer: b. 124 coins
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