GRE Averages

Understanging GRE averages is pretty easy and problems on GRE averages are pretty easy. Just bu going through our chapter on GRE averages you can solve each and every problem efficiently and effectively.

The average of a number is a measure of the central tendency of a set of numbers. In other words, it is an estimate of where the Centre point of a set of number lies.

The basic formula for the average of n numbers \({ x }_{ 1, }{ \quad x }_{ 2 },\quad { x }_{ 3—— }{ x }_{ n }\) is

\(Average\quad =\quad \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }—-{ x }_{ n } }{ n } \\ \quad \quad \quad \quad \quad \quad \quad =\quad \frac { Sum\quad of\quad n\quad numbers\quad of\quad a\quad set }{ n }\)

 

Concept of weighted average:

When we have two or more groups whose individual averages are known, then to find the combined average of all the elements of all the groups we use weighted average. Thus if we have k groups with average  & having  elements than the weighted average is given by the formula:

\({ A }_{ w }\quad =\quad \frac { { A }_{ 1 }{ n }_{ 1 }+{ A }_{ 2 }{ n }_{ 2 }+{ A }_{ 3 }{ n }_{ 3 }——{ A }_{ k }{ n }_{ k } }{ { n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }———-{ n }_{ k } }\)

Problem 1) If the average of 5 consecutive integers is 15, what is the sum of the least & greatest of the 5 integers?

Sol.  Average of 5 consecutive integers is 15.

Five consecutive integers are 13, 14, 15, 16, and 17.

Sum of the least & greatest of the 5 integers = 30 Ans.

 

Problem 2) The average of five numbers is 30. After one of the number is removed, the average arithmetic mean of the remaining numbers is 32. What number was removed?

Sol.   Average = \(\frac { Sum\quad of\quad Quantities }{ Number\quad of\quad Quantities }\)

 

\( 30\quad =\quad \frac { Sum\quad Of\quad Quantities }{ 5 }\)

        

          Sum of quantities = 150

After one number is removed,

\( \frac { Sum\quad Of\quad Remaining\quad numbers }{ 4 } \quad =\quad 32\)

Sum of remaining number = 128

Removed number was 22.    Ans.

 

Problem 3) Two airplanes leave the same airport at the same time, one traveling west & the other traveling to east. Their average speeds different by 10 miles/hr. After 1.5 hours, they are 520 miles apart. What is the approximate average speed of each plane over the 1.5 hours? Round to the nearest tenth.

Sol.  Let average speed of one airplane be \(x\) miles/hr.

Let average speed of another airplane be \(x+10\) miles/hr.

\(1.5x\quad +\quad 1.5(x+10)\quad =\quad 520\\ 1.5(2x+10)\quad =\quad 520\\ 2x\quad +\quad 10\quad =\quad \frac { 520 }{ 1.5 } \\ x\quad =\quad 168.3\)

Average speed of one airplane is 168.3 miles/hr.

Average speed of another airplane is 178.3miles/hr.

 

Problem 4) If the average of 10 consecutive odd integers is 224, what is the least of these integers?

Sol.   Average of 10 consecutive odd integers is 224.

Therefore, five consecutive odd integers are less than 224 & five consecutive integers are more than 224.

10 consecutive odd integers are:-

215, 217, 219, 221, 223, 225, 227, 229, 231 & 233

Least of these integers = 215                   Ans.

 

Problem 5) A man travels at 60 kmph on the journey from A to B & returns at 100 kmph. Find the average speed for the journey?

Sol.  Let distance be d.

Time taken by man from A to B = \(\frac { d }{ 60 }\)

Time taken by man in returns from B to A = \(\frac { d }{ 100 }\)

Average speed of the journey = \(\frac { Total\quad distance }{ Total\quad time }\) \( =\frac { 2d }{ \frac { d }{ 60 } +\frac { d }{ 100 } } \\ =\frac { 2d }{ d(\frac { 1 }{ 60 } +\frac { 1 }{ 100 } ) } \\ =\frac { 2d }{ d(\frac { 5+3 }{ 300 } ) } \\ =\frac { 2(300) }{ 8 } \\ =\frac { 300 }{ 4 } \\\)

Average speed of journey = 75 kmph

 

Problem 6) A school has only 3 classes that contain 10, 20 & 30 student respectively. The pass percentages of these classes are 20%, 30% & 40% respectively. Find the pass percentage of the entire school.

a) \(5\frac { 3 }{ 16 } \\\)

b) \(3\frac { 3 }{ 16 } \\\)

c) \(16\frac { 5 }{ 3 } \\\)

d) \(3\frac { 16 }{ 5 } \\\)

Sol.  Total students = 60

Total students who passed the examination are

\(=\frac { 20 }{ 100 } (10)\quad +\quad \frac { 30 }{ 100 } (20)\quad +\quad \frac { 40 }{ 100 } (30)\\ =\quad 2\quad +\quad 6\quad +\quad 12\\ =\quad 20\)

Pass percentage of the entire school = \(\frac { 20 }{ 60 } \times 100\quad =\quad 33.33%\)

 

Problem 7) Find the average of four numbers

\(2\frac { 3 }{ 4 } ,\quad 5\frac { 1 }{ 3 } ,\quad 5\frac { 1 }{ 6 } ,\quad 8\frac { 1 }{ 2 }\) ?

Sol.  Four numbers are \(\frac { 11 }{ 4 } ,\quad \frac { 16 }{ 3 } ,\quad \frac { 25 }{ 6 } ,\quad \frac { 17 }{ 2 }.\)

Average \(=\quad (\frac { \frac { 11(3)\quad +\quad 16(4)\quad +\quad 25(2)\quad +\quad 17(6) }{ 12 } }{ 4 } )\\ =\quad (\frac { 33\quad +\quad 64\quad +\quad 50\quad +\quad 102 }{ 48 } )\\ =\quad \frac { 249 }{ 48 } \\ =\quad \frac { 83 }{ 16 } \\ =\quad 5\frac { 3 }{ 16 }\)

 

Problem 8)  Find the average increase rate if increase in the population in the first year is 30% & that in the second year is 40%?

Sol.   Let population be p.

Population after the end of first year = \(p(\frac { 130 }{ 100 } )\)

Population after the end of second year = \(p(\frac { 130 }{ 100 } )(\frac { 140 }{ 100 } )\)

Let average increase rate be x.

\(p(1\quad +\quad \frac { x }{ 100 } )\quad =\quad p(\frac { 130 }{ 100 } )(\frac { 140 }{ 100\\ } )\\ p(100\quad +\quad x)\quad =\quad p(13)(14)\\ x\quad =\quad 13\times 14\quad -\quad 100\quad =\quad 82%\)

 

Problem 9) A man covers half of his journey by train at 60 km/hr. half of the remainder by bus at 30 km/hr. & the rate by cycle at 10 km/hr. Find his average speed during the entire journey.

a) 36 kmph

b) 30 kmph

c) 14 kmph

d) 18 kmph

Sol.   Let total distance be d.

Distance covered by train = \(\frac { d }{ 2 }\)

Distance covered by bus = \(\frac { d }{ 2 }\)

Distance covered by cycle = \(\frac { d }{ 2 }\)

Average speed during the entire journey =

\(=\quad (\frac { d }{ \frac { d }{ 2(60) } +\frac { d }{ 4(30) } +\frac { d }{ 4(10) } } )\\ =\quad \frac { 1 }{ \frac { 1 }{ 4(30) } +\frac { 1 }{ 30(4) } +\frac { 1 }{ 4(10) } } \\ =\quad \frac { 4 }{ \frac { 1 }{ 30 } +\frac { 1 }{ 30 } +\frac { 1 }{ 10 } } \\ =\quad \frac { 4 }{ \frac { 2 }{ 30 } +\frac { 3 }{ 30 } } \\ =\quad \frac { 4(30) }{ 5 } \\ =\quad 4(6)\\ =\quad 24Kmph\)

 

Problem 10) The average of 71 results is 48. If the average of the 59 results is 46 & that of the last 11 is 52. Find the 60th result.

Sol.

Sum of Quantity= \(71\quad \times \quad 48\) \(71\times 48\quad =\quad 59\times 46\quad +\quad 11\times 52\quad +\quad { (60 }^{ th }\quad result)\\ { (60 }^{ th }\quad result)\quad =\quad 71\times 48\quad -\quad 59\times 46\quad -\quad 11\times 52\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 3408-2714-572\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 122\) Ans

 

Problem 11) The average age of a group of 14 persons is 27 years & 9 months. Two persons, each person years old, left the group. What will be the average age of the remaining persons in the group?

Sol.   9 months = 0.75 year

Average age = 27.75 year

\(Average\quad =\quad \frac { Sum\quad of\quad Quantities }{ No.\quad of\quad Quantities } \\ 27.75\quad =\quad \frac { Sum\quad of\quad Quantities }{ 14 } \\ Sum\quad of\quad Quantities\quad =\quad 27.75\times 14\\\)

Average age of remaining persons = \((\frac { 27.75\times 14\quad -\quad 42(2) }{ 12 } )\\\)

= 25.375 years

 

Problem 12)  Find the average of f(x), g(x), h(x), d(x) at x=10.

\( f(x)\quad =\quad { x }^{ 2 }+2\\ g(x)\quad =\quad 5{ x }^{ 2 }-3\\ h(x)\quad =\quad log({ x }^{ 2 })\\ d(x)\quad =\quad (\frac { 4 }{ 5 } )\quad { x }^{ 2 }\\\)

a) 170

b) 170.25

c) 70.25

d) 70

Sol.   Average of f(x), g(x), h(x) and d(x)  at x = 10

\(=(\frac { { x }^{ 2 }+2+5{ x }^{ 2 }-3+\log { { x }^{ 2 } } +(\frac { 4 }{ 5 } ){ x }^{ 2 } }{ 4 } )\\ =(\frac { { { x }^{ 2 } }(1+\frac { 4 }{ 5 } +5)-1+\log { { x }^{ 2 } } }{ 4 } )\\ =(\frac { 100(1+\frac { 4 }{ 5 } +5)-1+2 }{ 4 } )\\ =(\frac { 100(\frac { 25+4+5 }{ 5 } )+1 }{ 4 } )\\ =100(\frac { 34 }{ 20 } )\quad +\quad \frac { 1 }{ 4 } \\ =5(34)\quad +\quad 0.25\\ =170.25\).

Therefore, Average = 170.25

 

Problem 13)  There are five boxes in a cargo hold the weight of the 1st box is 200kg & the weight of the 2nd box is 20% higher than the weight of the 3rd box; whose weight is 25% higher than the 1st box’s weight. The 4th box at 350kg is 30% lighter than the 5th box. Find the difference in the average weight of the four heaviest boxes & the four lightest boxes.

Sol.   1st box = 200kg

2nd box = \(Third(\frac { 120 }{ 100 } )=200(\frac { 125 }{ 100 } )(\frac { 120 }{ 100 } )=300kg\)

3rd box = \(200(\frac { 125 }{ 100 } )=250Kg\)

4th box = 350kg

5th box = 500kg

\(350=(\frac { 70 }{ 100 } )(Fifth\quad box)\)

Difference in the average weight of the four heaviest boxes & the four lightest boxes.

\(=\frac { 1 }{ 4 } (500+350+250+300)-\frac { 1 }{ 4 } (200+250+300+350)\\ =\frac { 1 }{ 4 } (300)\\ =75Kg\)

 

Problem 14)  With an average speed of 40kmph, a train reaches its destination in time. If it goes with an average speed of 35kmph, it is late by 15 minutes. The length of the total journey is

a) 40km

b) 70km

c) 30km

d) 80km

Sol.   Let distance be d & time taken be t.

\((\frac { d }{ 40 } )\quad =\quad t\\ (\frac { d }{ 35 } )\quad =\quad t\quad +\quad \frac { 15 }{ 60 } \\ d(\frac { 1 }{ 35 } -\frac { 1 }{ 40 } )\quad =\quad \frac { 1 }{ 4 } \\ \frac { d }{ 5 } (\frac { 1 }{ 56 } )\quad =\quad \frac { 1 }{ 4 } \\ d\quad =\quad \frac { 56\times 5 }{ 4 } \\ d\quad =\quad 70Km\)

 

 

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