Exercise: 10-B
Multiple Choice Questions
Q1: If 20 articles cost ₹90, the cost of 9 articles is:
Step 1: Cost of 1 article = ₹90 ÷ 20
\[
= \frac{90}{20} = ₹4.50
\]Step 2: Cost of 9 articles = ₹4.50 × 9 = ₹40.50
Answer: b. ₹40.50
Q2: With ₹1000, one can buy 4 pairs of trousers or 8 shirts. What is the amount required to buy 3 pairs of trousers and 3 shirts?
Step 1: Cost of 1 trouser = ₹1000 ÷ 4
\[
= \frac{1000}{4} = ₹250
\]Step 2: Cost of 1 shirt = ₹1000 ÷ 8
\[
= \frac{1000}{8} = ₹125
\]Step 3: Cost of 3 trousers = ₹250 × 3 = ₹750
Step 4: Cost of 3 shirts = ₹125 × 3 = ₹375
Step 5: Total cost = ₹750 + ₹375 = ₹1125
Answer: d. ₹1125
Q3: If 22.5 metres of a uniform weighs 85.5 kg, what will be the weight of 9 metres of the same rod?
Step 1: Weight per metre = 85.5 ÷ 22.5
\[
= \frac{85.5}{22.5} = 3.8 \text{ kg per metre}
\]Step 2: Weight of 9 metres = 3.8 × 9 = 34.2 kg
Answer: a. 34.2 kg
Q4: On a scale of a map, 0.6 cm represents 6.6 km. If the distance between two points on the map is 80.5 cm, the actual distance between these points is
Step 1: Distance in km per cm = 6.6 ÷ 0.6
\[
= \frac{6.6}{0.6} = 11 \text{ km/cm}
\]Step 2: Distance for 80.5 cm on the map:
\[
= 80.5 \times 11 = 885.5 \text{ km}
\]Answer: d. 885.5 km
Q5: If 15 men can reap a field in 35 days, in how many days will 21 men reap the field?
Step 1: This is an inverse proportion problem (more men ⇒ fewer days)
Step 2: Let required number of days = \( x \)
Use inverse proportion:
\[
15 \times 35 = 21 \times x \\
525 = 21x
\]Step 3: Solve for \( x \):
\[
x = \frac{525}{21} = 25
\]Answer: b. 25 days
Q6: If 20 persons can do a piece of work in 7 days, then the number of persons required to complete the work in 28 days, is
Step 1: Let required number of persons be \( x \)
Step 2: Since more days ⇒ fewer persons needed, it’s inverse proportion:
\[
20 \times 7 = x \times 28 \\
140 = 28x
\]Step 3: Solve for \( x \):
\[
x = \frac{140}{28} = 5
\]Answer: a. 5 persons
Q7: 10 pipes of the same type fill a tank in 24 minutes, If two pipes go out of order, how long will the remaining pipes take to fill the tank?
Step 1: 10 pipes fill the tank in 24 minutes.
Let required time for 8 pipes = \( x \) minutes
Step 2: Fewer pipes ⇒ more time (inverse proportion)
\[
10 \times 24 = 8 \times x \\
240 = 8x \\
x = \frac{240}{8} = 30 \text{ minutes}
\]Answer: b. 30 minutes
Q8: A rope makes 140 cylinder with base radius 14 cm. How many times can it go round a cylinder with base radius 20 cm?
Step 1: Length of rope = number of rounds × circumference
\[
L = 140 \times 2\pi r = 140 \times 2\pi \times 14 \\
L = 140 \times 88 = 12320 \text{ cm} \quad (\text{Since } 2\pi \times 14 = 88)
\]Step 2: Circumference of new cylinder (radius 20 cm):
\[
2\pi r = 2\pi \times 20 = 125.6 \text{ cm (or use } 2\pi \times 20 = 2 \times \frac{22}{7} \times 20 = \frac{880}{7}) \\
\text{So, number of rounds} = \frac{12320}{\frac{880}{7}} = 12320 \times \frac{7}{880} \\
= \frac{12320 \times 7}{880} = \frac{86240}{880} = 98
\]Answer: b. 98 times
Q9: A garrison of 750 men has provisions for 20 weeks. If at the end of 4 weeks they are reinforced by 450 men, how long will the remaining provisions last?
Step 1: Total food available = for 750 men × 20 weeks =
\[
750 \times 20 = 15000
\]Step 2: Food consumed in first 4 weeks by 750 men =
\[
750 \times 4 = 3000
\]Step 3: Remaining food =
\[
15000 – 3000 = 12000
\]Step 4: After 4 weeks, men = 750 + 450 = 1200
Let the remaining food last for \(x\) weeks:
\[
1200 \times x = 12000 \\
\Rightarrow x = \frac{12000}{1200} = 10
\]Answer: a. 10 weeks
Q10: 120 men had provisions for 200 days. After 5 days, 30 men died due to epidemic. The remaining will now last for
Step 1: Total food = 120 men × 200 days = 24000 man-days
Step 2: In 5 days, 120 men consume:
\[
120 \times 5 = 600
\]Step 3: Remaining food = 24000 − 600 = 23400 man-days
Step 4: After 5 days, 30 men died ⇒ remaining men = 90
Let food now last for \( x \) days:
\[
90 \times x = 23400 \\
x = \frac{23400}{90} = 260
\]Answer: d. 260 days
Q11: If \(\frac{4}{5}\)th of a cistern is filled in 1 minute, how much more time will be required to fill the rest of it?
Step 1: \(\frac{4}{5}\)th is filled in 1 minute = 60 seconds
Let’s find time to fill \(\frac{1}{5}\)th
Step 2: Time taken to fill \(\frac{1}{5}\):
\[
\text{If } \frac{4}{5} \text{ is filled in 60 sec, then }
\text{time to fill } \frac{1}{5} = \frac{60 \times 5}{4 \times 5} = 15 \text{ sec}
\]Answer: c. 15 seconds
