Download Averages Questions for CAT<\/a><\/p>\n CAT Online Coaching<\/a><\/p>\n Question 1:\u00a0<\/b>Consider the set S = {2, 3, 4, …., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y ?<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a0(1\/2)*n<\/p>\n d)\u00a0(n+1)\/2n<\/p>\n e)\u00a02008<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The odd numbers in the set are 3, 5, 7, …2n+1<\/p>\n Sum of the odd numbers = 3+5+7+…+(2n+1) = $n^2 + 2n$<\/p>\n Average of odd numbers =\u00a0$n^2 + 2n$\/n = n+2<\/p>\n Sum of even numbers = 2 + 4 + 6 + … + 2n = 2(1+2+3+…+n) = 2*n*(n+1)\/2 = n(n+1)<\/p>\n Average of even numbers = n(n+1)\/n = n+1<\/p>\n So, difference between the averages of even and odd numbers = 1<\/p>\n Question 2:\u00a0<\/b>Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight-member joint family is nearest to a)\u00a023 years<\/p>\n b)\u00a022 years<\/p>\n c)\u00a021 years<\/p>\n d)\u00a025 years<\/p>\n e)\u00a024 years<\/p>\n 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Ten years ago, the total age of the family is 231 years.<\/p>\n Seven years ago, (Just before the death of the first person), the total age of the family would have been 231+8*3 = 231+24 = 255. After the death of one member, the total age is 255-60 =\u00a0195 years.<\/p>\n Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8<\/p>\n Four years ago, (i.e. 6 years after start date)\u00a0one of the member of age 60 dies, Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8<\/p>\n After 4 more years, the current total age of the family is = 8×4 + 159 = 191 years<\/p>\n The average age is 191\/8 = 23.875 years = 24 years (approx)<\/p>\n Alternatively, Question 3:\u00a0<\/b>A college has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the people already solicited. The people already solicited represent 60% of the people the college will ask for donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people to be solicited?<\/p>\n a)\u00a0Rs. 300<\/p>\n b)\u00a0Rs. 250<\/p>\n c)\u00a0Rs. 400<\/p>\n d)\u00a0500<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let there be total 100 people whom the college will ask for donation. Out of these 60 people have already given average donation of 600 Rs. Thus total amount generated by 60 people is 36000. This is 75% of total amount required . so the amount remaining is 12000 which should be generated from remaining 40 people. So average amount needed is 12000\/40 = 300<\/p>\n Question 4:\u00a0<\/b>Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is: a)\u00a0n<\/p>\n b)\u00a0n+1<\/p>\n c)\u00a0kn, where k is a function of n<\/p>\n d)\u00a0n+(2\/7)<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The first five numbers could be n-2, n-1, n, n+1, n+2. The next two number would then be, n+3 and n+4, in which case, the average of all the 7 numbers would be $\\frac{(5n+2n+7)}{7}$ = n+1<\/p>\n Question 5:\u00a0<\/b>The average marks of a student in 10 papers are 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest.<\/p>\n a)\u00a055<\/p>\n b)\u00a060<\/p>\n c)\u00a062<\/p>\n d)\u00a0Cannot be determined<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Total marks = 80 x 10 = 800 Question 6:\u00a0<\/b>Total expenses of a boarding house are partly fixed and partly varying linearly with the number of\u00a0boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600\u00a0when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?<\/p>\n a)\u00a0550<\/p>\n b)\u00a0580<\/p>\n c)\u00a0540<\/p>\n d)\u00a0560<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the fixed income be x and the number of boarders be y.<\/p>\n x + 25y = 17500<\/p>\n x + 50y = 30000<\/p>\n => y = 500 and x = 5000<\/p>\n x + 100y = 5000 + 50000 = 55000<\/p>\n Average expense = $\\frac{55000}{100}$ = Rs.550.<\/p>\n Question 7:\u00a0<\/b>A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is<\/p>\n a)\u00a09.5<\/p>\n b)\u00a010<\/p>\n c)\u00a04.5<\/p>\n d)\u00a06<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let, the average score of boys in the mid semester exam is A. The average score\u00a0of the entire class is $\\dfrac{(50\\times A + 150)}{50} = A + 3$<\/p>\n wkt, class average increased by 2, class average in final term $= (A+3) + 2 = A + 5$<\/p>\n Given, that score of girls dropped by 3, i.e $(A+5)-3 = A+2$ Total class score in final term $= (A + 5)\\times50 = 50A +\u00a0250$<\/p>\n the total marks scored by the boys is $(50A + 250) – (30A – 60) = 20A + 190$ Hence, the increase in the average marks of the boys is $(A+9.5) – A = 9.5$<\/p>\n Question 8:\u00a0<\/b>In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?<\/p>\n a)\u00a027<\/p>\n b)\u00a025<\/p>\n c)\u00a026<\/p>\n d)\u00a028<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The possible average age of people\u00a0whose ages are below 51 years will be maximum if the average age of the number of people aged\u00a051 years and above is minimum. Hence, we can say that that there are 30 people having same age 51 years.<\/p>\n Let ‘x’ be the maximum average age of people whose ages are below 51.<\/p>\n Then we can say that,<\/p>\n $\\dfrac{51*30+39*x}{30+39} = 38$<\/p>\n $\\Rightarrow$ $1530+39x = 2622$<\/p>\n $\\Rightarrow$ $x = 1092\/39 = 28$<\/p>\n Hence, we can say that option D is the correct answer.<\/p>\n Question 9:\u00a0<\/b>The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?<\/p>\n a)\u00a03.5<\/p>\n b)\u00a05<\/p>\n c)\u00a04.5<\/p>\n d)\u00a04<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n It is given that the average of the 30 integers = 5<\/p>\n Sum of the 30 integers = 30*5=150<\/p>\n There are exactly 20 integers whose value is less than 5.<\/p>\n To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers<\/p>\n So the sum of 10 integers = 10*6=60<\/p>\n The sum of the 20 integers = 150-60= 90<\/p>\n Average of 20 integers =\u00a0$\\ \\frac{\\ 90}{20}$ = 4.5<\/p>\n Question 10:\u00a0<\/b>The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?<\/p>\n a)\u00a03.5<\/p>\n b)\u00a05<\/p>\n c)\u00a04.5<\/p>\n d)\u00a04<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n It is given that the average of the 30 integers = 5<\/p>\n Sum of the 30 integers = 30*5=150<\/p>\n There are exactly 20 integers whose value is less than 5.<\/p>\n To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers<\/p>\n So the sum of 10 integers = 10*6=60<\/p>\n The sum of the 20 integers = 150-60= 90<\/p>\n Average of 20 integers =\u00a0$\\ \\frac{\\ 90}{20}$ = 4.5<\/p>\n Question 11:\u00a0<\/b>Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is<\/p>\n a)\u00a053<\/p>\n b)\u00a051<\/p>\n c)\u00a048<\/p>\n d)\u00a049<\/p>\n
\n[CAT 2007]<\/p>\n
\nThis is because, in 3 years, every person in the family would have aged by 3 years,
\nTotal change in age = 231+24 = 255<\/p>\n
\ntherefore,\u00a0total age of the family is 195+24-60 = 159 years.<\/p>\n
\nSince the number of members is always the same throughout
\nThe 2 older members dropped their age by 60
\nSo, after 10yrs, total age = 231 + 8*10 – 2*60\u00a0= 191
\nAverage age = 191\/8 = 23.875 $\\simeq$ 24<\/p>\n
\n[CAT 2000]<\/p>\n
\nTotal marks except highest and lowest marks = 81 x 8 = 648
\nSo Summation of highest marks and lowest marks will be = 800 – 648 = 152
\nWhen highest marks is 92, lowest marks will be = 152-92 = 60<\/p>\n
\nTherefore, the average score of girls in the mid semester exam be A+5.
\nHence, the total marks scored by the class is $20\\times (A) +\u00a030\\times (A+5)\u00a0= 50\\times A\u00a0+ 150$<\/p>\n
\nTotal score of girls in final term\u00a0$= 30\\times(A+2) = 30A + 60$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/image_TwUupuV.png\")
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\nHence, the average of the boys in the final exam is $\\dfrac{(20G + 190)}{20} = A + 9.5$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/image_i4bX8Hv.png\")
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