Answer the questions based on the information given below. There is different number of three types of pipes (‘A’, ‘B’ and ‘C’) at four different sites. The given table shows the average number of pies ‘A’ and ‘B’, ratio of number of pipe ‘A’ to that of pipe ‘C’ and difference between number of pipes ‘B’ and ‘C’, at the respective sites.
Sites
Average number of pipes ‘A’ and ‘B’
Ratio of number of pipe ‘A’ to that of pipe ‘C’
Difference between number of pipes ‘B’ and ‘C.
1
110
1 : 2
140
2
225
2 : 1
150
3
230
4 : 5
100
4
180
4 : 5
180
Additional information : a.) Number of pipes of each type (‘A’, ‘B’ or ‘C’) at the respective site is neither more than 350 nor less than 100. b.) Quantity of water pumped at each of the site by each pipe ‘B’ is 20% more than that by each of the pipe ‘A’ but 10 units less than each of the pipe ‘C’. c.) Each unit of water from each pipe can irrigate 5 hectares of land.
Question: 1
If 7500 hectares of land can be irrigated by the amount of water pumped by all the pipe ‘B’ at site 2, then find the difference between amount of water pumped by 50% of number of pipe ‘A’ and 75% of number of pipe ‘C’, at site 2.
Let the number of pipe ‘A’ and pipe ‘C’ be ‘x’ and 2x respectively Number of pipe ‘B’ = 2 × 110 - x = (220 - x) Case 1 : Let the number of pipe ‘B’ be more than pipe ‘C’ Therefore, 220 - x - 2x = 140 Or, 3x = 80 Or, x = 26.66 (not possible) Case 2 : Let the number of pipe ‘C’ be more than pipe ‘B' 2x - (220 - x) = 140 Or, 2x + x = 140 + 220 Or, 3x = 360 Or, x = 120 (possible) Therefore, number of pipe ‘A’ = x = 120 Number of pipe ‘B’ = 220 - x = 100 Number of pipe ‘C’ = 2x = 240 Similarly,
Sites
Number of pipes ‘A’
Number of pipe ‘B’
Number of pipe ‘C’
1
120
100
240
2
200
250
100
3
160
300
200
4
240
120
300
According to the question, At site 2: Let each pipe ‘B’ can pump ‘x’ litres of water Amount of water pumped by all pipe ‘B’ together = 250x Area of field which can be irrigated by water pumped by pipe ‘B’ = 250x × 5 = 1250x hectares Therefore, x = \(\frac{7500}{1250}\) = 6 Amount of water pumped by 75% of number of pipe ‘C’ = 0.75 × 100 × (6 + 10) = 1200 units Amount of water pumped by 50% of number of pipe ‘A’ = 0.5 × 200 × (6/1.2) = 500 units Required difference = 1200 - 500 = 700 units So, the correct option is (C) : 700 units.
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Question: 2
For site 1, electricity bill paid for total amount of water pumped by all pipes ‘C’ together is Rs. 7488 more than electricity bill paid for total amount of water pumped by all pipes ‘B’ together. If each unit of water pumped fetch Rs. 2 as electricity bill, then find the electricity bill to be paid for water pumped by all pipes ‘A’ together is :
Let the number of pipe ‘A’ and pipe ‘C’ be ‘x’ and 2x respectively Number of pipe ‘B’ = 2 × 110 - x = (220 - x) Case 1 : Let the number of pipe ‘B’ be more than pipe ‘C’ Therefore, 220 - x - 2x = 140 Or, 3x = 80 Or, x = 26.66 (not possible) Case 2 : Let the number of pipe ‘C’ be more than pipe ‘B' 2x - (220 - x) = 140 Or, 2x + x = 140 + 220 Or, 3x = 360 Or, x = 120 (possible) Therefore, number of pipe ‘A’ = x = 120 Number of pipe ‘B’ = 220 - x = 100 Number of pipe ‘C’ = 2x = 240 Similarly,
Sites
Number of pipes ‘A’
Number of pipe ‘B’
Number of pipe ‘C’
1
120
100
240
2
200
250
100
3
160
300
200
4
240
120
300
At site 1 : Let each pipe ‘A’ can pump ‘x’ units of water Therefore, each pipe ‘B’ will pump ‘1.2x’ units of water Each pipe ‘C’ will pump = (1.2x + 10) units of water According to the question, 2 × 240 × (1.2x + 10) – 2 × 1.2x × 100 = 7488 Or, 288x + 2400 – 120x = \(\frac{7488}{2}\) Or, 168x = 3744 – 2400 Or, 168x = 1344 Or, x = 8 Therefore, required amount = 2 × 120 × 8 = Rs. 1920 So, the correct option is (A) : Rs. 1920.
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Question: 3
If the amount of water pumped by each pipe ‘A’ at site 1 is 4 units less than that at site 3 and sum of total quantity of water pumped by all the pipe ‘A’ at site 1 and 3 together is 2320 units, then find the difference between amount of water pumped by all the pipe ‘C’ at given two sites.
Let the number of pipe ‘A’ and pipe ‘C’ be ‘x’ and 2x respectively Number of pipe ‘B’ = 2 × 110 - x = (220 - x) Case 1 : Let the number of pipe ‘B’ be more than pipe ‘C’ Therefore, 220 - x - 2x = 140 Or, 3x = 80 Or, x = 26.66 (not possible) Case 2 : Let the number of pipe ‘C’ be more than pipe ‘B' 2x - (220 - x) = 140 Or, 2x + x = 140 + 220 Or, 3x = 360 Or, x = 120 (possible) Therefore, number of pipe ‘A’ = x = 120 Number of pipe ‘B’ = 220 - x = 100 Number of pipe ‘C’ = 2x = 240 Similarly,
Sites
Number of pipes ‘A’
Number of pipe ‘B’
Number of pipe ‘C’
1
120
100
240
2
200
250
100
3
160
300
200
4
240
120
300
Let amount of water pumped by each pipe ‘A’ at site 1 be ‘a’ units Therefore, amount of water pumped by each pipe ‘A’ at site 3 = (a + 4) units According to the question, 120a + 160(a + 4) = 2320 Or, 120a + 160a + 640 = 2320 Or, 280a = 1680 Or, a = 6 Therefore, amount of water pumped by each pipe ‘C’ at site 1 = 1.2 × 6 + 10 = 17.2 units Amount of water pumped by each pipe ‘C’ at site 3 = 1.2 × 10 + 10 = 22 units Require difference = 22 × 200 - 17.2 × 240 = 4400 - 4128 = 272 units So, the correct option is (D) : 272 units.
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Question: 4
At site 4, if the sum of amount of water pumped by each pipe ‘A’, each pipe ‘B’ and each pipe ‘C’ is 61 units, then find the average area of land which can be irrigated by water pumped by all the three pipes at site 4.
Let the number of pipe ‘A’ and pipe ‘C’ be ‘x’ and 2x respectively Number of pipe ‘B’ = 2 × 110 - x = (220 - x) Case 1 : Let the number of pipe ‘B’ be more than pipe ‘C’ Therefore, 220 - x - 2x = 140 Or, 3x = 80 Or, x = 26.66 (not possible) Case 2 : Let the number of pipe ‘C’ be more than pipe ‘B' 2x - (220 - x) = 140 Or, 2x + x = 140 + 220 Or, 3x = 360 Or, x = 120 (possible) Therefore, number of pipe ‘A’ = x = 120 Number of pipe ‘B’ = 220 - x = 100 Number of pipe ‘C’ = 2x = 240 Similarly,
Sites
Number of pipes ‘A’
Number of pipe ‘B’
Number of pipe ‘C’
1
120
100
240
2
200
250
100
3
160
300
200
4
240
120
300
At site 4: Let amount of water pumped by each pipe ‘A’ be ‘p’ litres Therefore, amount of water pumped by each pipe ‘B’ = 1.2p litres Amount of water pumped by each pipe ‘C’ = (1.2p + 10) litres According to the question, 1.2p + 1.2p + 10 + p = 61 Or, 3.4p = 51 Or, p = 15 Therefore, area of land which can be irrigated by water pumped by pipe ‘A’ = 15 × 240 × 5 = 18000 hectares Area of land which can be irrigated by water pumped by pipe ‘B’ = 1.2 ×15 × 120 × 5 = 10800 hectares Area of land which can be irrigated by water pumped by pipe ‘C’ = (1.2 × 15 + 10) × 300 × 5 = 42000 hectares Required average = {(18000 + 10800 + 42000)/3} = 23600 hectares So, the correct option is (D) : 23600 hectares.
Let the number of pipe ‘A’ and pipe ‘C’ be ‘x’ and 2x respectively Number of pipe ‘B’ = 2 × 110 - x = (220 - x) Case 1 : Let the number of pipe ‘B’ be more than pipe ‘C’ Therefore, 220 - x - 2x = 140 Or, 3x = 80 Or, x = 26.66 (not possible) Case 2 : Let the number of pipe ‘C’ be more than pipe ‘B' 2x - (220 - x) = 140 Or, 2x + x = 140 + 220 Or, 3x = 360 Or, x = 120 (possible) Therefore, number of pipe ‘A’ = x = 120 Number of pipe ‘B’ = 220 - x = 100 Number of pipe ‘C’ = 2x = 240 Similarly,
Sites
Number of pipes ‘A’
Number of pipe ‘B’
Number of pipe ‘C’
1
120
100
240
2
200
250
100
3
160
300
200
4
240
120
300
Required average= \(\frac{660}{3}\) = 220 So, the correct option is (D) : 220.
