Exercise: 10-B
Q1: Multiple Choice Type
i. If a varies inversely to b and b = 12 when a is 8; the value of b when a = 6 is:
Step 1: a ∝ 1/b ⇒ a × b = constant
Given: a = 8, b = 12 ⇒ 8 × 12 = 96
Now, when a = 6: 6 × b = 96
b = 96 ÷ 6 = 16
Answer: d. 16
ii. 12 men can make a certain number of screws in 30 days. The number of screws, opf the same type will be made by 24 men in:
Step 1: Men and days are inversely proportional (more men ⇒ less days)
12 men × 30 days = 24 men × x days
\[
12 \times 30 = 24 \times x \Rightarrow x = \frac{12 \times 30}{24} = 15
\]Answer: a. 15 days
iii. A school has 8 periods in a days, each of 45 minutes duration. Assuming that the number of school hours to be the same and the school has 9 periods a day, the duration of each period is:
Step 1: Total school time = 8 × 45 = 360 minutes
Now total time is same, but periods = 9
New duration = 360 ÷ 9 = 40 minutes
Answer: c. 40 minutes
iv. From the table, find values of a and b:
x 12 4 10 30 y 5 a 6 b
Step 1: For inverse variation, \( x \times y = \text{constant} \)
Use known pair (x = 12, y = 5):
\[
12 \times 5 = 60
\]
So, \( x \times y = 60 \) for all pairs.
Step 2: Find \( a \) when \( x = 4 \):
\[
4 \times a = 60 \Rightarrow a = \frac{60}{4} = 15
\]Step 3: Find \( b \) when \( x = 30 \):
\[
30 \times b = 60 \Rightarrow b = \frac{60}{30} = 2
\]Answer: d. a = 15 and b = 2
v. A hostel has provisions for 200 students for 30 days. If 100 new students join the hostel, the same provisions will last for:
Step 1: More students ⇒ fewer days (inverse proportion)
200 × 30 = 300 × x ⇒ x = (200 × 30)/300 = 20
Answer: d. 20 days
Q2: Check whether x and y vary inversely or not
i.
x: 4 3 12 1 y: 6 8 2 24
Step 1: In inverse variation, x × y = constant
Calculate each x × y:
4 × 6 = 24
3 × 8 = 24
12 × 2 = 24
1 × 24 = 24
Step 2: Since x × y = 24 in all cases
Answer: Yes, x and y vary inversely.
ii.
x: 30 120 60 24 y: 60 30 30 75
Step 1: Calculate x × y for each pair
30 × 60 = 1800
120 × 30 = 3600
60 × 30 = 1800
24 × 75 = 1800
Step 2: All except one give 1800. One pair (120 × 30 = 3600) does not.
Answer: No, x and y do not vary inversely.
iii.
x: 10 30 60 10 y: 90 30 20 90
Step 1: Calculate x × y for each pair
10 × 90 = 900
30 × 30 = 900
60 × 20 = 1200
10 × 90 = 900
Step 2: One pair (60 × 20 = 1200) differs
Answer: No, x and y do not vary inversely.
Q3: If x and y vary inversely, find the values of l, m and n:
i.
x: 4 8 2 32 y: 4 l m n
Step 1: In inverse variation, x × y = constant
Using x = 4, y = 4 ⇒ 4 × 4 = 16
So for all other pairs, x × y = 16
Step 2: Use x × y = 16 to find missing values:
When x = 8:
8 × l = 16 ⇒ l = 16 ÷ 8 = 2
When x = 2:
2 × m = 16 ⇒ m = 16 ÷ 2 = 8
When x = 32:
32 × n = 16 ⇒ n = 16 ÷ 32 = 0.5
Answer: l = 2, m = 8, n = 0.5
ii.
x: 24 32 m 16 y: l 12 8 n
Step 1: Use x = 32, y = 12 ⇒ x × y = 32 × 12 = 384
So, x × y = 384 for all pairs
When x = 24:
24 × l = 384 ⇒ l = 384 ÷ 24 = 16
When y = 8:
m × 8 = 384 ⇒ m = 384 ÷ 8 = 48
When x = 16:
16 × n = 384 ⇒ n = 384 ÷ 16 = 24
Answer: l = 16, m = 48, n = 24
Q4: 36 men can do a piece of work in 7 days. How many men will do the same work in 42 days?
Step 1: This is a case of inverse variation (more days ⇒ fewer men).
So, men × days = constant
Step 2: Use the formula:
\[
\text{Men}_1 \times \text{Days}_1 = \text{Men}_2 \times \text{Days}_2
\]Let required number of men = x
36 × 7 = x × 42
Step 3: Solve the equation:
\[
252 = 42x \Rightarrow x = \frac{252}{42} = 6
\]Answer: 6 men are required to complete the same work in 42 days.
Q5: 12 pipes, all of the same size, fill a tank in 42 minutes. How long will it take to fill the same tank, if 21 pipes of the same size are used?
Step 1: This is an inverse variation case (more pipes ⇒ less time).
So, Pipes × Time = constant
Step 2: Use the formula:
\[
\text{Pipes}_1 \times \text{Time}_1 = \text{Pipes}_2 \times \text{Time}_2
\]Let the required time = x minutes
12 × 42 = 21 × x
Step 3: Solve for x:
\[
504 = 21x \Rightarrow x = \frac{504}{21} = 24
\]Answer: It will take 24 minutes to fill the tank using 21 pipes.
Q6: In a fort, 150 men had provisions for 45 days. After 10 days, 25 men left the fort. How long would the food last at the same rate?
Step 1: First, calculate how much food is left after 10 days.
Originally, food was enough for 150 men for 45 days.
10 days of food is consumed ⇒ Remaining = 45 – 10 = 35 days (for 150 men)
Step 2: Now 25 men leave, so remaining men = 150 – 25 = 125
Step 3: Use inverse variation:
Men × Days = Constant
So, 150 men × 35 days = 125 men × x days
Step 4: Solve for x:
\[
150 \times 35 = 125 \times x \Rightarrow 5250 = 125x \\
x = \frac{5250}{125} = 42
\]Answer: The food will last for 42 more days.
Q7: 72 men do a piece of work in 25 days. In how many days will 30 men do the same work?
Step 1: This is an inverse variation problem (fewer men ⇒ more days).
So, Men × Days = Constant
Step 2: Use the formula:
\[
\text{Men}_1 \times \text{Days}_1 = \text{Men}_2 \times \text{Days}_2
\]Given:
72 × 25 = 30 × x
Step 3: Solve for x:
\[
1800 = 30x \Rightarrow x = \frac{1800}{30} = 60
\]Answer: 30 men will complete the work in 60 days.
Q8: If 56 workers can build a wall in 180 hours, how many workers will be required to do the same work in 70 hours?
Step 1: This is a case of inverse variation (less time ⇒ more workers).
So, Workers × Time = Constant
Step 2: Use the formula:
\[
\text{Workers}_1 \times \text{Time}_1 = \text{Workers}_2 \times \text{Time}_2
\]Given:
56 × 180 = x × 70
Step 3: Solve for x:
\[
10080 = 70x \Rightarrow x = \frac{10080}{70} = 144
\]Answer: 144 workers are required to complete the work in 70 hours.
Q9: A car takes 6 hours to reach a destination by traveling at the speed of 50 km per hour. How long will it take when the car travels at the speed of 75 km per hour?
Step 1: This is an inverse variation case (more speed ⇒ less time).
So, Speed × Time = Constant
Step 2: Use the formula:
\[
\text{Speed}_1 \times \text{Time}_1 = \text{Speed}_2 \times \text{Time}_2
\]Given:
50 × 6 = 75 × x
Step 3: Solve for x:
\[
300 = 75x \Rightarrow x = \frac{300}{75} = 4
\]Answer: The car will take 4 hours to reach the destination at 75 km/h.
