2 cm of rain has fallen on a sq. km of land. Assuming that 50% of the raindrops could have been collected and…

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2 cm of rain has fallen on a sq. km of land. Assuming that 50% of the raindrops could have been collected and contained in a pool having a 100 m x 10 m base, by what level would the water level in the pool have increased?

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Problem 3 Step 1: Volume of rainfall. Area of land =1 km^2=1000×1000=10⁶ m^2. Depth =2 cm =0.02 m. V=10⁶×0.02=2×10⁴ m³. Step 2: Volume collected. 50% of rainfall =0.5×2×10⁴=10⁴ m^3. Step 3: Base area of pool =100 m ×10 m =10³ m^2. Step 4: Water level rise. h=(10⁴)/(10³)=10 m. (C) 10 m Problem 4 Step 1: Let legs be a cm and b cm. Perimeter: a+b+26=60. a+b=34 cm. Step 2: Pythagoras theorem. a²+b²=26²=676. Step 3: Square the sum. (a+b)²=a²+2ab+b²=34²=1156. Step 4: Substitute. 676+2ab=1156. 2ab=1156−676=480, ab=240. Step 5: Area. 1/2ab=240/2=120 cm². (A) 120 cm² Problem 5 Step 1: Let rate of 1 man =m (work per day). Rate of 1 child =m/2. Step 2: Combined rate. 12 men +8 children: 12m+8×m/2=12m+4m=16m. Step 3: Time given. 16m×9=1. 16m=1/9, m=1/144. Step 4: Rate of 12 men =12m=12×1/144=1/12. Step 5: Days for 12 men. (1)/(1/12)=12 days. 12

2 cm of rain has fallen on a sq. km of land. Assuming that 50% of the raindrops could have been collected and contained in a pool having a 100 m x 10 m base, by what level would the water level in the pool have increased?

Answer

Step-by-step solution

Volume of rainfall.

Volume collected.

Base area of pool $=100$ m $\times10$ m $=10^3$ m $^2$ .

Water level rise.

Let legs be $a$ cm and $b$ cm.

Pythagoras theorem.

Square the sum.

Substitute.

Area.

Let rate of 1 man $=m$ (work per day).

Combined rate.

Time given.

Rate of 12 men $=12m=12\times\frac{1}{144}=\frac{1}{12}$ .

Days for 12 men.

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2 cm of rain has fallen on a sq. km of land. Assuming that 50% of the raindrops could have been collected and contained in a pool having a 100 m x 10 m base, by what level would the water level in the pool have increased?

Answer

Step-by-step solution

Volume of rainfall.

Volume collected.

Base area of pool =100=100=100 m ×10\times10×10 m =103=10^3=103 m2^22.

Water level rise.

Let legs be aaa cm and bbb cm.

Pythagoras theorem.

Square the sum.

Substitute.

Area.

Let rate of 1 man =m=m=m (work per day).

Combined rate.

Time given.

Rate of 12 men =12m=12×1144=112=12m=12\times\frac{1}{144}=\frac{1}{12}=12m=12×1441​=121​.

Days for 12 men.

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More Mathematics Questions

Base area of pool $=100$ m $\times10$ m $=10^3$ m $^2$ .

Let legs be $a$ cm and $b$ cm.

Let rate of 1 man $=m$ (work per day).

Rate of 12 men $=12m=12\times\frac{1}{144}=\frac{1}{12}$ .