Test Yourself
Q1: Multiple Choice Type
i. For \(\frac{x}{2}+\frac{x}{3}=15\); the value of x is:
Step 1: LCM of 2 and 3 = 6
\(\frac{3x}{6} + \frac{2x}{6} = 15 → \frac{5x}{6} = 15\)
Step 2: Solve for x:
\(5x = 15 × 6 → 5x = 90 → x = 18\)
Answer: a. 18
ii. The sum of three consecutive odd whole numbers is 87. The largest of these is:
Step 1: Let the numbers be \(x, x+2, x+4\)
Equation: \(x + (x+2) + (x+4) = 87 → 3x + 6 = 87\)
Step 2: Solve for x:
\(3x = 81 → x = 27\)
Largest number = \(x + 4 = 27 + 4 = 31\)
Answer: b. 31
iii. If \(\frac{x+3}{\left(x+4\right)+3}=\frac{4}{5}\), the value of x is:
Step 1: Simplify denominator:
\(\frac{x+3}{x+7} = \frac{4}{5}\)
Cross multiply: \(5(x+3) = 4(x+7)\)
Step 2: Solve for x:
\(5x + 15 = 4x + 28 → 5x – 4x = 28 – 15 → x = 13\)
Answer: d. 13
iv. When four consecutive integers are added, their sum is -46. The smallest integer is:
Step 1: Let the numbers be \(x, x+1, x+2, x+3\)
Equation: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = -46\)
Step 2: Solve for x:
\(4x = -46 – 6 → 4x = -52 → x = -13\)
Answer: b. -13
v. The perimeter of a rectangular plot is 560 m and the length of the plot is three times its width; the length of the plot is:
Step 1: Let the width = \(x\), length = \(3x\)
Perimeter: \(2(length + width) = 2(3x + x) = 8x = 560\)
Step 2: Solve for x:
\(x = \frac{560}{8} = 70\)
Length = \(3x = 3 × 70 = 210\) m
Answer: b. 210 m
vi. Statement 1: \(\frac{ax+b}{cx+d}=\frac{p}{q}\ \Rightarrow q\left(cx+d\right)=p\left(ax+b\right)\). This process is cross-multiplication.
Statement 2: \(ax+b=c\) becomes \(ax=c-b\) after transposition.
Which of the following options is correct?
Step 1: Check statement 1:
Cross multiplication is not correct → False
Step 2: Check statement 2:
Transposing b to RHS: \(ax = c – b\) → Correct → True
Answer: d. Statement 1 is false, and statement 2 is true.
vii. Assertion (A): \(\left(2x+5\right)^2-\left(2x-5\right)^2=40\) is a linear equation in terms of one variable.
Reason (R): An equation in which the greatest exponent of the variable after simplification is one is called a linear equation.
Step 1: Simplify the Assertion using difference of squares:
\((2x+5)^2 – (2x-5)^2 = [(2x+5) – (2x-5)] \cdot [(2x+5) + (2x-5)] = (10) \cdot (4x) = 40x\)
Step 2: Equation becomes \(40x = 40 → x = 1\)
This is a linear equation (highest power of x is 1)
Step 3: Reason statement is correct definition of linear equation
Answer: a. Both A and R are correct, and R is the correct explanation for A
viii. Assertion (A) : The solution of : \(\frac{x}{2}+\frac{x}{3}=15\) is 18.
Reason (R): In the process of solving a linear equation, only the method of transposition is applied.
Step 1: Solve the equation:
LCM of 2 and 3 = 6 → \(\frac{3x}{6} + \frac{2x}{6} = 15 → \frac{5x}{6} = 15 → 5x = 90 → x = 18\)
Assertion is correct
Step 2: Reason is not fully correct, because solving linear equations may involve cross-multiplication or LCM, not only transposition
Answer: c. A is true, but R is false
ix. Assertion (A): The solution of: \(\frac{x-3}{x+4}=\frac{x+1}{x-2}\) is 5.
Reason (R): \(\frac{ax+b}{cx+d}=\frac{p}{q}\ \Rightarrow q\left(ax+b\right)=p\left(cx+d\right)\). This process is cross-multiplication.
Step 1: Solve the equation using cross-multiplication:
\((x-3)(x-2) = (x+1)(x+4) → x^2 -5x +6 = x^2 +5x +4 → -10x = -2 → x = \frac{1}{5}\)
So Assertion is false (x ≠ 5)
Step 2: Reason is correct about cross-multiplication
Answer: d. A is false, but R is true
x. Assertion (A): \(7x=-21\ \Rightarrow\frac{7x}{7}=-\frac{21}{7}\ \Rightarrow x=-3\).
Reason (R): An equation remains unchanged on dividing both sides by the same (non-zero) number.
Step 1: Assertion is correct: \(7x = -21 → x = -3\)
Step 2: Reason is correct definition of operation on equations
Answer: a. Both A and R are correct, and R is the correct explanation for A
Q2: Solve:
i. \(\frac{1}{3}x – 6 = \frac{5}{2}\)
Step 1: Add 6 to both sides:
\(\frac{1}{3}x = \frac{5}{2} + 6 = \frac{5}{2} + \frac{12}{2} = \frac{17}{2}\)
Step 2: Multiply both sides by 3:
\(x = 3 × \frac{17}{2} = \frac{51}{2}\)
Answer: x = \(\frac{51}{2}\)
ii. \(\frac{2x}{3} – \frac{3x}{8} = \frac{7}{12}\)
Step 1: LCM of 3 and 8 = 24
\(\frac{16x}{24} – \frac{9x}{24} = \frac{7}{12} → \frac{7x}{24} = \frac{7}{12}\)
Step 2: Multiply both sides by 24:
\(7x = 14 → x = 2\)
Answer: x = 2
iii. \((x+2)(x+3) + (x-3)(x-2) – 2x(x+1) = 0\)
Step 1: Expand each term:
\((x^2 + 5x + 6) + (x^2 -5x +6) – 2x^2 -2x = 0 → x^2 + 5x +6 + x^2 -5x +6 – 2x^2 -2x = 0\)
Combine like terms:
\((x^2 + x^2 – 2x^2) + (5x -5x -2x) + (6+6) = 0 → -2x + 12 = 0\)
Step 2: Solve for x:
\(-2x + 12 = 0 → -2x = -12 → x = 6\)
Answer: x = 6
iv. \(\frac{1}{10} – \frac{7}{x} = 35\)
Step 1: Move \(-\frac{7}{x}\) to RHS:
\(\frac{1}{10} – 35 = \frac{7}{x} → -\frac{349}{10} = \frac{7}{x}\)
Step 2: Solve for x:
\(x = \frac{7}{-349/10} = -\frac{70}{349}\)
Answer: x = -\(\frac{70}{349}\)
v. \(13(x-4) – 3(x-9) – 5(x+4) = 0\)
Step 1: Expand:
\(13x -52 -3x +27 -5x -20 = 0 → (13x -3x -5x) + (-52+27-20) = 0 → 5x -45 = 0\)
Step 2: Solve for x:
\(5x = 45 → x = 9\)
Answer: x = 9
vi. \(x + 7 – \frac{8x}{3} = \frac{17x}{6} – \frac{5x}{8}\)
Step 1: Take LCM 24:
\(24(x+7) – 64x = 68x -15x → 24x +168 -64x = 53x → -40x +168 = 53x\)
Step 2: Solve for x:
\(-40x -53x +168 = 0 → -93x +168 = 0 → x = \frac{168}{93} = \frac{56}{31}\)
Answer: x = \(\frac{56}{31}\)
vii. \(\frac{3x-2}{4} – \frac{2x+3}{3} = \frac{2}{3} – x\)
Step 1: Take LCM of 4 and 3 = 12:
\(\frac{9x-6}{12} – \frac{8x+12}{12} = \frac{8-12x}{12} → \frac{9x-6-(8x+12)}{12} = \frac{8-12x}{12} → \frac{x-18}{12} = \frac{8-12x}{12}\)
Step 2: Multiply both sides by 12:
\(x -18 = 8 -12x → x +12x = 8 +18 → 13x = 26 → x = 2\)
Answer: x = 2
viii. \(\frac{x+2}{6} – \left(\frac{11-x}{3} – \frac{1}{4}\right) = \frac{3x-4}{12}\)
Step 1: Simplify brackets:
\(\frac{x+2}{6} – \frac{11-x}{3} + \frac{1}{4} = \frac{3x-4}{12}\)
Step 2: Take LCM 12:
\(\frac{2(x+2) -4(11-x) +3}{12} = \frac{3x-4}{12} → \frac{2x+4 -44+4x+3}{12} = \frac{3x-4}{12} → \frac{6x -37}{12} = \frac{3x-4}{12}\)
Step 3: Multiply both sides by 12:
\(6x -37 = 3x -4 → 6x -3x = -4 +37 → 3x = 33 → x = 11\)
Answer: x = 11
ix. \(\frac{2}{5x} – \frac{5}{3x} = \frac{1}{15}\)
Step 1: LCM of denominators 5x and 3x = 15x:
\(\frac{6 – 25}{15x} = \frac{-19}{15x} = \frac{1}{15} → -19 = x → x = -19\)
Answer: x = -19
x. \(\frac{x+2}{3} – \frac{x+1}{5} = \frac{x-3}{4} -1\)
Step 1: Take LCM of denominators 3,5,4 = 60
\(\frac{20(x+2) -12(x+1)}{60} = \frac{15(x-3) -60}{60} → \frac{20x+40 -12x -12}{60} = \frac{15x-45-60}{60} → \frac{8x +28}{60} = \frac{15x -105}{60}\)
Step 2: Multiply both sides by 60:
\(8x +28 = 15x -105 → 7x = 133 → x = 19\)
Answer: x = 19
xi. \(\frac{3x-2}{3} + \frac{2x+3}{2} = x + \frac{7}{6}\)
Step 1: LCM of denominators 3,2,6 = 6
\(\frac{2(3x-2) + 3(2x+3)}{6} = \frac{6x +7}{6} → \frac{6x-4 +6x+9}{6} = \frac{6x+7}{6} → \frac{12x+5}{6} = \frac{6x+7}{6}\)
Step 2: Multiply both sides by 6:
\(12x +5 = 6x +7 → 6x = 2 → x = \frac{1}{3}\)
Answer: x = \(\frac{1}{3}\)
xii. \(x – \frac{x-1}{2} = 1 – \frac{x-2}{3}\)
Step 1: Take LCM of 2 and 3 = 6
\(\frac{6x -3(x-1)}{6} = \frac{6 -2(x-2)}{6} → \frac{6x -3x +3}{6} = \frac{6 -2x +4}{6} → \frac{3x +3}{6} = \frac{10-2x}{6}\)
Step 2: Multiply both sides by 6:
\(3x +3 = 10 -2x → 5x = 7 → x = \frac{7}{5}\)
Answer: x = \(\frac{7}{5}\)
xiii. \(\frac{9x+7}{2} – \left(x – \frac{x-2}{7}\right) = 36\)
Step 1: Simplify brackets:
\(\frac{9x+7}{2} – x + \frac{x-2}{7} = 36 → \frac{9x+7}{2} – x + \frac{x-2}{7} = 36\)
Step 2: LCM of 2 and 7 = 14:
\(\frac{7(9x+7) + 2(x-2)}{14} – x = 36 → \frac{63x+49 +2x-4}{14} – x = 36 → \frac{65x +45}{14} – x =36\)
Step 3: Combine terms:
\(\frac{65x +45 -14x}{14} = 36 → \frac{51x +45}{14} =36 → 51x +45 =504 → 51x =459 → x = 9\)
Answer: x = 9
xiv. \(\frac{6x+1}{2} +1 = \frac{7x-3}{3}\)
Step 1: Combine LHS:
\(\frac{6x+1}{2} +1 = \frac{6x+1+2}{2} = \frac{6x+3}{2}\)
Step 2: Equation becomes:
\(\frac{6x+3}{2} = \frac{7x-3}{3}\)
Step 3: Cross multiply:
\(3(6x+3) = 2(7x-3) → 18x +9 =14x -6 → 4x = -15 → x = -\frac{15}{4}\)
Answer: x = -\(\frac{15}{4}\)
Q3: After 12 years, I shall be 3 times as old as I was 4 years ago. Find my present age.
Step 1: Let my present age be \(x\) years.
Step 2: Age after 12 years = \(x + 12\)
Age 4 years ago = \(x – 4\)
According to the question:
\(x + 12 = 3(x – 4)\)
Step 3: Simplify the equation:
\(x + 12 = 3x – 12 → 12 +12 = 3x – x → 24 = 2x → x = 12\)
Answer: My present age is 12 years.
Q4: A man sold an article for ₹396 and gained 10% on it. Find the cost price of the article.
Step 1: Let the cost price (C.P.) of the article be ₹x.
Step 2: Selling Price (S.P.) = C.P. + Gain = \(x + \frac{10}{100}x = \frac{110}{100}x = \frac{11}{10}x\)
Step 3: According to the question:
\(\frac{11}{10}x = 396\)
Step 4: Solve for x:
\(x = 396 \times \frac{10}{11} → x = 360\)
Answer: The cost price of the article is ₹360.
Q5: The sum of two numbers is 4500. If 10% of one number is 12.5% of the other, find the numbers.
Step 1: Let the two numbers be \(x\) and \(y\).
Step 2: According to the question:
1) Sum of numbers: \(x + y = 4500\) …(i)
2) 10% of one = 12.5% of other: \(\frac{10}{100}x = \frac{12.5}{100}y → \frac{1}{10}x = \frac{1}{8}y\)
Step 3: Simplify the second equation:
\(\frac{x}{10} = \frac{y}{8} → 8x = 10y → 4x = 5y → y = \frac{4}{5}x\) …(ii)
Step 4: Substitute \(y = \frac{4}{5}x\) in equation (i):
\(x + \frac{4}{5}x = 4500 → \frac{9}{5}x = 4500 → x = 4500 \times \frac{5}{9} = 2500\)
Step 5: Find \(y\):
\(y = 4500 – 2500 = 2000\)
Answer: The two numbers are 2500 and 2000.
Q6: The sum of two numbers is 405 and ratio is 8 : 7. Find numbers.
Step 1: Let the two numbers be \(8x\) and \(7x\) (since their ratio is 8:7).
Step 2: According to the question, their sum is 405:
\(8x + 7x = 405 → 15x = 405\)
Step 3: Solve for \(x\):
\(x = \frac{405}{15} = 27\)
Step 4: Find the numbers:
First number = \(8x = 8 × 27 = 216\)
Second number = \(7x = 7 × 27 = 189\)
Answer: The two numbers are 216 and 189.
Q7: The ages of A and B are in the ratio 7 : 5. Ten years hence, ratio of their ages will be 9 : 7. Find their present ages.
Step 1: Let the present ages of A and B be \(7x\) and \(5x\) respectively.
Step 2: After 10 years, their ages will be:
A = \(7x + 10\)
B = \(5x + 10\)
According to the question:
\(\frac{7x + 10}{5x + 10} = \frac{9}{7}\)
Step 3: Cross multiply:
\(7(7x + 10) = 9(5x + 10)\)
\(49x + 70 = 45x + 90\)
Step 4: Simplify:
\(49x – 45x = 90 – 70 → 4x = 20 → x = 5\)
Step 5: Find present ages:
A = \(7x = 7 × 5 = 35\) years
B = \(5x = 5 × 5 = 25\) years
Answer: The present ages of A and B are 35 years and 25 years respectively.
Q8: Find the number whose double is 45 greater than its half.
Step 1: Let the number be \(x\).
Step 2: According to the question:
Double of the number = \(2x\)
Half of the number = \(\frac{x}{2}\)
Equation: \(2x = \frac{x}{2} + 45\)
Step 3: Solve the equation:
Multiply both sides by 2 to remove fraction:
\(4x = x + 90\)
\(4x – x = 90 → 3x = 90 → x = 30\)
Answer: The number is 30.
Q9: The difference between the squares of two consecutive numbers is 31. Find numbers.
Step 1: Let the smaller number be \(x\). Then the next consecutive number = \(x + 1\).
Step 2: According to the question:
\((x+1)^2 – x^2 = 31\)
Step 3: Expand and simplify:
\((x^2 + 2x + 1) – x^2 = 31 → 2x + 1 = 31\)
Step 4: Solve for \(x\):
\(2x = 31 – 1 → 2x = 30 → x = 15\)
Step 5: Find the next consecutive number:
\(x + 1 = 15 + 1 = 16\)
Answer: The two consecutive numbers are 15 and 16.
Q10: Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.
Step 1: Let the number be \(x\).
Step 2: According to the question:
5 times the number minus 5 = \(5x – 5\)
Twice the number plus 4 = \(2x + 4\)
Equation: \(5x – 5 = 2x + 4\)
Step 3: Solve for \(x\):
\(5x – 2x = 4 + 5 → 3x = 9 → x = 3\)
Answer: The number is 3.
Q11: The numerator of a fraction is 5 less than its denominator. If 3 is added to numerator and denominator both, the fraction becomes \(\frac{4}{5}\). Find the original fraction.
Step 1: Let the denominator be \(x\).
Then, numerator = \(x – 5\).
So, the fraction = \(\frac{x-5}{x}\).
Step 2: According to the question:
\[
\frac{(x-5)+3}{x+3} = \frac{4}{5}\\
\frac{x-2}{x+3} = \frac{4}{5}
\]Step 3: Cross multiply:
\[
5(x-2) = 4(x+3)\\
5x – 10 = 4x + 12\\
5x – 4x = 12 + 10\\
x = 22
\]Step 4: Denominator = \(22\)
Numerator = \(22 – 5 = 17\)
Original fraction = \(\frac{17}{22}\).
Answer: The original fraction is \(\frac{17}{22}\).
Q12: The difference between two numbers is 36. The quotient when one number is divided by other is 4. Find the two numbers.
Step 1: Let the smaller number be \(x\).
Then, the larger number = \(4x\) (since quotient = 4).
Step 2: According to the question:
\[
4x – x = 36\\
3x = 36\\
x = 12
\]Step 3: Smaller number = \(x = 12\)
Larger number = \(4x = 48\)
Answer: The two numbers are 12 and 48.
Q13: Five years ago, Mohit was thrice of Manish 10 years later. Mohit will be twice as old as Manish. Find their present ages.
Step 1: Let the present age of Mohit = \(M\), and Manish = \(N\).
Step 2: Condition 1 (Five years ago):
\[
M – 5 = 3(N – 5)\\
M – 5 = 3N – 15\\
M = 3N – 10 \quad …(1)
\]Step 3: Condition 2 (Ten years later):
\[
M + 10 = 2(N + 10)\\
M + 10 = 2N + 20\\
M = 2N + 10 \quad …(2)
\]Step 4: From (1) and (2):
\[
3N – 10 = 2N + 10\\
N = 20
\]
Substitute in (1):
\[
M = 3(20) – 10 = 60 – 10 = 50
\]Answer: Mohit’s present age = 50 years, Manish’s present age = 20 years.
Q14: The denominator of a fraction is 3 more its denominator. If the numerator is increased by 7 and the denominator is decreased by 2, the fraction is 2. Find the sum of the numerator and the denominator of the fraction.
Step 1: Let the numerator = \(x\).
Then the denominator = \(x + 3\).
So, the fraction = \(\frac{x}{x+3}\).
Step 2: According to the condition:
\[
\frac{x+7}{(x+3)-2} = 2\\
\frac{x+7}{x+1} = 2
\]Step 3: Cross-multiplying:
\[
x + 7 = 2(x + 1)\\
x + 7 = 2x + 2\\
7 – 2 = 2x – x\\
x = 5
\]Step 4: Denominator = \(x + 3 = 5 + 3 = 8\).
So, fraction = \(\frac{5}{8}\).
Answer: The numerator = 5, denominator = 8, and their sum = 13.
Q15: 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, the result is 143. Find the original number.
Step 1: Let the tens digit of the number be x and the units digit be y.
So,
Original number = 10x + y
Interchanged number = 10y + x
Step 2: According to the question, the digits differ by 3.
Since tens digit is greater than units digit,
x − y = 3 …… (1)
Step 3: The sum of the original number and the interchanged number is 143.
(10x + y) + (10y + x) = 143
Step 4: Simplify the equation.
11x + 11y = 143
Divide both sides by 11,
x + y = 13 …… (2)
Step 5: Solve equations (1) and (2).
x − y = 3
x + y = 13
Adding both equations,
2x = 16
x = 8
Substitute x = 8 in equation (2),
8 + y = 13
y = 5
Step 6: Find the original number.
Original number = 10x + y
= 10(8) + 5
= 85
Answer:The original number is 85.
Verification: Original number = 85 Interchanged number = 58 85 + 58 = 143 ✔
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