Direct and Inverse Variations

Q1: Multiple Choice Type

i. The value of a when

x → 2   a   9
y → 6  18  27

Step 1: Check ratio of x and y: \[ \frac{y}{x} = \frac{6}{2} = 3,\quad \frac{18}{a} = 3 \Rightarrow a = \frac{18}{3} = 6 \]Answer: d. 6

ii. The value of b when

x →	3	b	12
y →	12	9	3

Step 1: Since x and y vary inversely, \[ x \times y = \text{constant} \]Step 2: Use known pair (x = 3, y = 12): \[ 3 \times 12 = 36 \Rightarrow \text{Constant} = 36 \]Step 3: Use inverse variation to find \( b \): \[ x \times y = 36 \Rightarrow b \times 9 = 36 \Rightarrow b = \frac{36}{9} = 4 \]Answer: b = 4 → Option (a)

iii. 15 note books can be bought for ₹240. The number of note books that can be bought for ₹60 is:

Step 1: Let the number of notebooks that can be bought for ₹160 be \( x \)
Since more money buys more notebooks, this is a case of direct variation.
So, \[ \frac{15}{240} = \frac{x}{160} \]Step 2: Cross-multiply: \[ 240x = 15 \times 160 = 2400 \\ \Rightarrow x = \frac{2400}{240} = 10 \]Answer: 10 notebooks → Option (b)

iv. 6 men can do a certain piece of work in 15 days. The number of men required to complete the same work in 10 days:

Step 1: Work is constant, men × days = constant \[ 6 \times 15 = m \times 10 \Rightarrow m = \frac{90}{10} = 9 \]Answer: d. 9

v. If x is in inverse variation with y and x = 4 when y = 6, value of x when y = 12 is:

Step 1: In inverse variation, \( xy = k \) \[ 4 \times 6 = 24 = x \times 12 \Rightarrow x = \frac{24}{12} = 2 \]Answer: a. 2

vi. Statement 1: Work done varies inversely to the number of persons at work.
Statement 2: Time taken to finish a work varies directly to the number of persons at work.

Step 1: Let’s understand Statement 1:
More persons at work means each does less individually, but total work gets done faster. However, work done in fixed time increases with more persons, so work is directly proportional to number of persons, not inversely.
Statement 1 is false
Step 2: Check Statement 2:
If number of persons increases, the time taken decreases. So time varies inversely with number of persons. Statement 2 is also false
Answer: b. Both the statements are false

vii. Assertion (A): In the following table, p and q are in direct variation.

p → 3   2  4
q → 8  12  6

Reason (R): If two quantities p and q are in direct variation, then \(\frac{p}{q}\) is always constant.

Step 1: In direct variation, \( \frac{p}{q} = k \), a constant.
Then \(\frac{3}{8} = \frac{2}{12} = \frac{4}{6}\), which is not true . So assertion is False. Step 2: In direct variation, \( \frac{p}{q} = k \), a constant.
So Reason is True.
Answer: d. A is false, but R is true.

viii. Assertion (A): If 3 pipes can fill a tank in 6 hours, then 6 pipes will fill the same tank in half the time. Reason (R): \(\text{Time required to complete a certain work} = \frac{\text{Work to be completed}} {\text{One day^’ s work}}\)

Step 1: More pipes ⇒ Less time ⇒ Inverse variation. So 6 pipes will take \( \frac{6}{2} = 3 \) hours ✅
Step 2: R gives the correct formula for time — so it also holds ✅
And R explains A correctly.
Answer: a. Both A and R are correct, and R is the correct explanation for A.

ix. Assertion (A): In the following table, p and q are in inverse variation.

p →  5   2   4
q → 10  25  12.5

Reason (R): If two quantities p and q are in inverse variation then \(\frac{p}{q}\) is always constant.

Step 1: In inverse variation: \( p \times q = \text{constant} \), not \( \frac{p}{q} \)
R gives incorrect explanation for inverse variation ❌
Answer: c. A is true, but R is false.

x. Assertion (A): The cost of 16 bulbs is ₹144. The number of bulbs that can be bought for ₹270 is 30.
Reason (R): In inverse variation, the ratio of one kind of like terms is equal to the inverse ratio of the second kind of like terms.

Step 1: Cost per bulb = ₹\( \frac{144}{16} = ₹9 \)
No. of bulbs for ₹270 = \( \frac{270}{9} = 30 \) ✅ A is correct
Step 2: In inverse variation, \( p \times q = k \), a constant. As mentioned \(p1 : p2 = q2 : q1 \Rightarrow \frac{p1}{p2} = \frac{q2}{q1}\)
By cross multiply, \(p1 \times q1 = p2 \times q2\)
So Reason is True.
Answer: b. Both A and R are correct, and R is not the correct explanation for A.

Q2: If x varies directly as y and x = 150 when y = 50. Find:

i. x, when y = 12.5

Step 1: Since x ∝ y, we write: \[ \frac{x}{y} = \text{constant} = \frac{150}{50} = 3 \]Step 2: When y = 12.5: \[ x = 3 \times 12.5 = 37.5 \]Answer: x = 37.5

ii. y, when x = 75

Step 1: Again, \( \frac{x}{y} = 3 \Rightarrow \frac{75}{y} = 3 \)
Step 2: Solving for y: \[ y = \frac{75}{3} = 25 \]Answer: y = 25

Q3: If x varies inversely as y and y = 300 when x = 60. Find:

i. x, when y = 90

Step 1: In inverse variation, \( x \times y = \text{constant} \)
Given: \( 60 \times 300 = 18000 \)
Step 2: Let x be the new value when y = 90: \[ x \times 90 = 18000 \Rightarrow x = \frac{18000}{90} = 200 \]Answer: x = 200

ii. y, when x = 300

Step 1: Using same constant from earlier: \[ x \times y = 18000,\quad 300 \times y = 18000 \\ \Rightarrow y = \frac{18000}{300} = 60 \]Answer: y = 60

Q4: Total length of 153 iron bars is 680 m. What will be the total length of 135 similar bars?

Step 1: Since all bars are similar, this is a direct variation problem.
Number of bars ∝ Total length
Step 2: Let the required length be \( x \) meters. \[ \frac{153}{680} = \frac{135}{x} \]Step 3: Cross-multiply: \[ 153x = 135 \times 680 \\ \Rightarrow x = \frac{135 \times 680}{153} \]Step 4: Simplify: \[ x = \frac{91800}{153} = 600 \]Answer: The total length of 135 bars is 600 meters.

Q5: 12 men can repair a road in 25 days. How long will 30 men take to do so?

Step 1: This is a case of inverse variation:
More men ⇒ Less time
Step 2: Men × Days = Constant \[ 12 \times 25 = 30 \times x \\ \Rightarrow 300 = 30x \\ \Rightarrow x = \frac{300}{30} = 10 \]Answer: 30 men will take 10 days to repair the road.

Q6: The price of oranges is ₹90 per dozen. Manoj can buy 12 dozen oranges with the money he has if the price of oranges is increased by ₹30, how many oranges can Manoj buy?

Step 1: Total money Manoj has = ₹90 × 12 = ₹1080
Step 2: New price per dozen = ₹90 + ₹30 = ₹120
Step 3: Number of dozens he can now buy = ₹1080 ÷ ₹120 = 9 dozens
Step 4: Total oranges = 9 dozens × 12 = 108 oranges
Answer: Manoj can now buy 108 oranges.

Q7: A and B can do a work in 8 days, B and C in 12 days, and A and C in 16 days. In what time can they do it, all working together?

Step 1: Let the total work be 1 unit.
We calculate one-day work from given pairs:
\[ \text{A + B’s 1 day work} = \frac{1}{8} \\ \text{B + C’s 1 day work} = \frac{1}{12} \\ \text{A + C’s 1 day work} = \frac{1}{16} \]Step 2: Add all three equations: \[ (A + B) + (B + C) + (A + C) = \frac{1}{8} + \frac{1}{12} + \frac{1}{16} \]Left-hand side becomes: \[ 2(A + B + C) \]Now solve the RHS: \[ \frac{1}{8} + \frac{1}{12} + \frac{1}{16} = \frac{6 + 4 + 3}{48} = \frac{13}{48} \]So, \[ 2(A + B + C) = \frac{13}{48} \Rightarrow A + B + C = \frac{13}{96} \]Step 3: One-day work of all three together = \(\frac{13}{96}\)
So, total time to finish the work: \[ \text{Time} = \frac{1}{\frac{13}{96}} = \frac{96}{13} = 7\frac{5}{13} \text{ days} \]Answer: \(\frac{96}{13}\) days or \(7\frac{5}{13}\) days.

Q8: A and B complete a piece of work in 24 days. B and C do the same work in 36 days, and A, B and C together finish it in 18 days. In how many days will:

i. A alone,
ii. C alone,
iii. A and C together complete the work?

Step 1: Let the total work = 1 unit.
We’ll calculate one-day work for each group:
A + B’s one-day work = \(\frac{1}{24}\)
B + C’s one-day work = \(\frac{1}{36}\)
A + B + C’s one-day work = \(\frac{1}{18}\)
Step 2: C’s one day work = One day work of (A + B + C) – One day work of (A + B): \[ (A + B + C) – (A + B) = \frac{1}{18} – \frac{1}{24} \\ \Rightarrow C = \frac{4 – 3}{72} = \frac{1}{72} \]Step 3: One day work of B = One day work of (B + C) – One day work of C: \[ B + C = \frac{1}{36},\ \text{and}\ C = \frac{1}{72} \\ \Rightarrow B = \frac{1}{36} – \frac{1}{72} = \frac{2 – 1}{72} = \frac{1}{72} \]Step 4: One day work of A = One day work of (A + B) – One day work of B: \[ A + B = \frac{1}{24},\ B = \frac{1}{72} \\ \Rightarrow A = \frac{1}{24} – \frac{1}{72} = \frac{3 – 1}{72} = \frac{2}{72} = \frac{1}{36} \]Step 5: Now we have:
A’s 1-day work = \(\frac{1}{36}\) ⇒ A alone takes 36 days
C’s 1-day work = \(\frac{1}{72}\) ⇒ C alone takes 72 days
A + C = \(\frac{1}{36} + \frac{1}{72} = \frac{2 + 1}{72} = \frac{3}{72} = \frac{1}{24}\) ⇒ A and C together take 24 days
Answer:i. A alone = 36 days
ii. C alone = 72 days
iii. A and C together = 24 days

Q9: A and B can do a piece of work in 40 days, B and C in 30 days, and C and A in 24 days.

i. How long will it take them to do the work, working together?

Step 1: Calculate one-day work:
A + B = \(\frac{1}{40}\)
B + C = \(\frac{1}{30}\)
C + A = \(\frac{1}{24}\)
Step 2: Add all three:
(A + B) + (B + C) + (C + A) = \(\frac{1}{40} + \frac{1}{30} + \frac{1}{24}\)
Left side becomes: 2(A + B + C)
Take LCM of 40, 30, and 24 = 120: \[ \frac{1}{40} = \frac{3}{120},\\ \frac{1}{30} = \frac{4}{120},\\ \frac{1}{24} = \frac{5}{120} \\ \Rightarrow 2(A + B + C) = \frac{3 + 4 + 5}{120} = \frac{12}{120} = \frac{1}{10} \\ \Rightarrow A + B + C = \frac{1}{20} \]Step 3: So, all 3 together can complete the work in: \[ \frac{1}{A + B + C} = \frac{1}{1/20} = 20\ \text{days} \]Answer 20 days

ii. In what time can each finish it working alone?

Step 4: Find C’s one-day work:
C = (A + B + C) – (A + B) = \(\frac{1}{20} – \frac{1}{40} = \frac{2 – 1}{40} = \frac{1}{40}\)
So C alone takes: 40 days
Answer C = 40 days

Step 5: Find B’s one-day work:
B = (B + C) – C = \(\frac{1}{30} – \frac{1}{40}\) \[ = \frac{4 – 3}{120} = \frac{1}{120} \\ \Rightarrow \text{B alone takes } 120\ \text{days} \] Answer B = 120 days

Step 6: Find A’s one-day work:
A = (A + B + C) – (B + C) = \(\frac{1}{20} – \frac{1}{30} = \frac{3 – 2}{60} = \frac{1}{60}\)
So A alone takes: 60 days
Answer A = 60 days

Q10: A can do a piece of work in 10 days, B in 12 days and C in 15 days. All begin together but A leaves the work after 2 days and B leaves 3 days before the work is finished. How long did the work last?

Step 1: Work done in 1 day:
A = \(\frac{1}{10}\), B = \(\frac{1}{12}\), C = \(\frac{1}{15}\)
Step 2: Let the total time taken to complete the work be x days.
Then, A worked for 2 days.
B worked for (x − 3) days (since B left 3 days early).
C worked for x days (C worked the full duration).
Step 3: Total work = sum of work done by A, B, and C. \[ \text{Total work} = A’s work + B’s work + C’s work \\ = 2 \times \frac{1}{10} + (x – 3) \times \frac{1}{12} + x \times \frac{1}{15} \]Step 4: Add all and set equal to 1: \[ \frac{2}{10} + \frac{x – 3}{12} + \frac{x}{15} = 1 \\ \Rightarrow \frac{1}{5} + \frac{x – 3}{12} + \frac{x}{15} = 1 \]Step 5: Take LCM of 5, 12, and 15 = 60: \[ \frac{1}{5} = \frac{12}{60},\\ \frac{x – 3}{12} = \frac{5(x – 3)}{60},\\ \frac{x}{15} = \frac{4x}{60} \\ \Rightarrow \frac{12 + 5(x – 3) + 4x}{60} = 1 \\ \Rightarrow 12 + 5x – 15 + 4x = 60 \\ \Rightarrow 9x – 3 = 60 \\ \Rightarrow 9x = 63 \\ \Rightarrow x = 7 \\ \]Answer:The work was completed in 7 days.

Q11: Two pipes P and Q would fill an empty cistern in 24 minutes and 32 minutes respectively. Both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern may be just filled in 16 minutes.

Step 1: One minute work of each pipe:
Pipe P fills = \(\frac{1}{24}\) of the cistern in 1 minute
Pipe Q fills = \(\frac{1}{32}\) of the cistern in 1 minute
Step 2: Let P be turned off after x minutes. Then:
P works for x minutes
Q works for full 16 minutes
Total work done = \[ x \times \frac{1}{24} + 16 \times \frac{1}{32} = 1 \]Step 3: Simplify: \[ \\ \Rightarrow \frac{x}{24} + \frac{16}{32} = 1 \\ \Rightarrow \frac{x}{24} + \frac{1}{2} = 1 \\ \Rightarrow \frac{x}{24} = 1 – \frac{1}{2} = \frac{1}{2} \\ \Rightarrow x = \frac{1}{2} \times 24 = 12 \]Answer:Pipe P must be turned off after 12 minutes.