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
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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Answer

(C) 10 m


Step 1: Volume of rainfall.
Area of land =1=1 km2=1000×1000=106^2=1000\times1000=10^6 m2^2.
Depth =2=2 cm =0.02=0.02 m.

V=106×0.02=2×104 m3.V=10^6\times0.02=2\times10^4\ m^3.

Step 2: Volume collected.
50%50\% of rainfall =0.5×2×104=104=0.5\times2\times10^4=10^4 m3^3.

Step 3: Base area of pool =100=100 m ×10\times10 m =103=10^3 m2^2.

Step 4: Water level rise.

h=104103=10 m.h=\frac{10^4}{10^3}=10\ m.

(C) 10 m


Step 1: Let legs be aa cm and bb cm.
Perimeter: a+b+26=60a+b+26=60.

a+b=34 cm.a+b=34\ cm.

Step 2: Pythagoras theorem.

a2+b2=262=676.a^2+b^2=26^2=676.

Step 3: Square the sum.

(a+b)2=a2+2ab+b2=342=1156.(a+b)^2=a^2+2ab+b^2=34^2=1156.

Step 4: Substitute.

676+2ab=1156.676+2ab=1156. 2ab=1156676=480,ab=240.2ab=1156-676=480,\quad ab=240.

Step 5: Area.

12ab=2402=120 cm2.\frac{1}{2}ab=\frac{240}{2}=120\ cm^2.

(A) 120 cm²


Step 1: Let rate of 1 man =m=m (work per day).
Rate of 1 child =m2=\frac{m}{2}.

Step 2: Combined rate.
1212 men +8+8 children: 12m+8×m2=12m+4m=16m12m+8\times\frac{m}{2}=12m+4m=16m.

Step 3: Time given.
16m×9=116m\times9=1.

16m=19,m=1144.16m=\frac{1}{9},\quad m=\frac{1}{144}.

Step 4: Rate of 1212 men =12m=12×1144=112=12m=12\times\frac{1}{144}=\frac{1}{12}.

Step 5: Days for 1212 men.

1112=12 days.\frac{1}{\frac{1}{12}}=12\ days.

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.

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

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.

ScanSolve AI Answer
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