This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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?
Mathematics

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Answer
(C) 10 m
Step 1: Volume of rainfall.
Area of land km m.
Depth cm m.
Step 2: Volume collected.
of rainfall m.
Step 3: Base area of pool m m m.
Step 4: Water level rise.
(C) 10 m
Step 1: Let legs be cm and cm.
Perimeter: .
Step 2: Pythagoras theorem.
Step 3: Square the sum.
Step 4: Substitute.
Step 5: Area.
(A) 120 cm²
Step 1: Let rate of 1 man (work per day).
Rate of 1 child .
Step 2: Combined rate.
men children: .
Step 3: Time given.
.
Step 4: Rate of men .
Step 5: Days for men.
12
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Quick Answer
Problem 3 Step 1: Volume of rainfall. Area of land =1 km^2=1000×1000=10^6 m^2.
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Problem 3 Step 1: Volume of rainfall. Area of land =1 km^2=1000×1000=10^6 m^2. Depth =2 cm =0.02 m. V=10^6×0.02=2×10^4\ m^3. Step 2: Volume collected. 50\% of rainfall =0.5×2×10^4=10^4 m^3. Step 3: Base area of pool =100 m ×10 m =10^3 m^2. Step 4: Water level rise. h=(10^4)/(10^3)=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. a^2+b^2=26^2=676. Step 3: Square the sum. (a+b)^2=a^2+2ab+b^2=34^2=1156. Step 4: Substitute. 676+2ab=1156. 2ab=1156-676=480, ab=240. Step 5: Area. (1)/(2)ab=(240)/(2)=120\ cm^2. (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