Average is an important topic in the Quant section of the SNAP Exam. You can also download this Free Average Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Average questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practicing.<\/p>\n Download Average Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>The average age of three friends is 20 years and their ages are in the proportion 2:3:5, the age of the oldest friend is :<\/p>\n a)\u00a012 years<\/p>\n b)\u00a018 years<\/p>\n c)\u00a030 years<\/p>\n d)\u00a040 years<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given average age of three friends is 20 years<\/p>\n Sum of their ages =\u00a0$3\\times\\ 20$ = 60 years<\/p>\n Given, their ages are in ratio 2:3:5. Let their ages be 2x, 3x and 5x<\/p>\n 2x + 3x + 5x = 60<\/p>\n 10x = 60<\/p>\n x = 6<\/p>\n Their ages are 12 years, 18 years and 30 years<\/p>\n Age of oldest friend = 30 years<\/p>\n Answer is option C.<\/p>\n Question 2:\u00a0<\/b>In a batch of 50 students has ratio of Boys to Girls as 4:6. Average marks of Girls are 60 and Average marks of the batch is 62. Average Marks of Boys will be:<\/p>\n a)\u00a061<\/p>\n b)\u00a063<\/p>\n c)\u00a062<\/p>\n d)\u00a065<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number of boys be 4x and number of girls be 6x<\/p>\n Given, 4x+6x = 50<\/p>\n x = 5<\/p>\n Number of boys = 20 and number of girls = 30<\/p>\n Average marks of girls = 60<\/p>\n Total marks scored by girls = 60$\\times\\ $30 = 1800<\/p>\n Average marks score by batch = 62<\/p>\n Total marks scored by batch = 62$\\times\\ $50 = 3100<\/p>\n Marks scored by boys = 3100 – 1800 = 1300<\/p>\n Average marks of boys =\u00a0$\\frac{1300}{20}$ = 65<\/p>\n Answer is option D.<\/p>\n Question 3:\u00a0<\/b>The average temperature of Tuesday, Wednesday and Thursday was $45^\\circ C$. The\u00a0average temperature of Wednesday, Thursday and Friday was $46^\\circ C$. If the temperature\u00a0on Friday was $47^\\circ C$, what was the temperature on Tuesday?<\/p>\n a)\u00a0$44^\\circ C$<\/p>\n b)\u00a0$48^\\circ C$<\/p>\n c)\u00a0$40^\\circ C$<\/p>\n d)\u00a0$46^\\circ C$<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Sum of temperatures on Wednesday, Thursday and Friday= 46*3 = 138<\/p>\n The temperature on friday= 47<\/p>\n So, the sum of temperatures on Wednesday and Thursday= 138-47 = 91<\/p>\n Sum of temperatures on Tuesday, Wednesday and Thursday= 45*3 = 135<\/p>\n The temperature on Tuesday= 135-91= 44<\/p>\n Question 4:\u00a0<\/b>In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is<\/p>\n 4)\u00a0Answer:\u00a010<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let Total matches played be n and in initial n-10 matches his goals be x Question 5:\u00a0<\/b>Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to<\/p>\n a)\u00a026<\/p>\n b)\u00a018<\/p>\n c)\u00a016<\/p>\n d)\u00a020<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.<\/p>\n Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.<\/p>\n Total amount bought = 22kg.<\/p>\n Total amount spent = 100+100+100+50+50 = 400.<\/p>\n Average expense =\u00a0$\\frac{400}{22}=Rs.18.18\\approx\\ 18$<\/p>\n Question 6:\u00a0<\/b>A person goes from X to Y on a cycle at 20 kilometers per hour and returns at 24kilometers per hour. The method of Central Tendency most appropriate to calculate theaverage speed would be____________.<\/p>\n a)\u00a0Geometric Mean<\/p>\n b)\u00a0Harmonic Mean<\/p>\n c)\u00a0Arithmetic Mean<\/p>\n d)\u00a0Weighted Arithmetic Mean<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n If the distance travelled is the same during all the speeds then the average speed for the whole journey is the harmonic mean of the speed.<\/p>\n Question 7:\u00a0<\/b>In a company, the average salary of the boys is Rs. 50000. The average salary of all the employees is Rs. 48000. There are 80 boys in the company and the average salary of the girl is Rs. 40000. What is the number of girls working in the company?<\/p>\n a)\u00a012<\/p>\n b)\u00a015<\/p>\n c)\u00a020<\/p>\n d)\u00a024<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let girls be x Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 8:\u00a0<\/b>In a company there are 252 engineers, in which the ratio of the number of electronics engineers and computer engineers is 2 : 1. The ratio of the number of electronics engineers and computer engineers becomes 1:1 after recruitment of some more computer engineers. The average age of all the engineers is now 22 years and the average age of the computer engineers is 2 years less than the average age of electronics engineers. a)\u00a0(A) and (E)only<\/p>\n b)\u00a0(B) and (C) only<\/p>\n c)\u00a0(D) and (E) only<\/p>\n d)\u00a0(E) only<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of electronics engineers initially =\u00a0$\\frac{2}{3}\\times\\ 252\\ =\\ 168$ Question 9:\u00a0<\/b>Sumit stays in Noida in a joint family comprising of his parents, uncle, aunt, an elder sister and a a)\u00a031<\/p>\n b)\u00a035<\/p>\n c)\u00a038<\/p>\n d)\u00a041<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let Sumit’s age at the time of graduating MBA [in 2013] be $X$ years. Thus, his wife’s age that year will be $X-3$ years.<\/p>\n The average age of the\u00a0family in 2013 = 41 years.<\/p>\n Thus, the total age of the family in 2013 = 7 x 41 = 287<\/p>\n Sumit got married in 2015. Thus, his wife’s age in 2015 will added to the family’s total.<\/p>\n The total age of family in 2015 = 287 + 14 + $X$ – 3 + 2 = $X$ + 300<\/p>\n In 2017, Sumit had a daughter.<\/p>\n The total age of the family in 2017 = $X$ + 300 + 8(2) = $X$ + 316<\/p>\n The total age of the family in 2019 = $X$ + 316 + 9(2) = $X$ + 334<\/p>\n We are given that the average age of the family in 2019 is the same as that in 2013.<\/p>\n Thus, $X$ + 334 = 41 x 9 = 369<\/p>\n $X$ = 369 – 334 = 35 years<\/p>\n Thus, the age of Sumit at the time of graduating MBA is 35 years.<\/p>\n Hence, the answer is option B.<\/p>\n Question 10:\u00a0<\/b>A batsman played n + 2 innings and got out on all occasions. His\u00a0average score in these n + 2 innings was 29 runs and he scored 38 and\u00a015 runs in the last two innings. The batsman scored less than 38 runs\u00a0in each of the first n innings. In these n innings, his average score was\u00a030 runs and lowest score was x runs. The smallest possible value of x\u00a0is<\/p>\n a)\u00a04<\/p>\n b)\u00a03<\/p>\n c)\u00a02<\/p>\n d)\u00a01<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given, $\\frac{\\text{sum of scores\u00a0in n matches+38+15}}{n+2}=29$<\/p>\n Given,\u00a0$\\frac{\\text{sum of scores\u00a0in n matches}}{n}=30$<\/p>\n => 30n + 53 = 29(n+2) => n=5<\/p>\n Sum of the scores in 5 matches = 29*7 – 38-15 = 150<\/p>\n Since the batsmen scored less than 38, in each of\u00a0the first 5 innings. The value of x will be minimum when remaining four values are highest<\/p>\n => 37+37+37+37 + x = 150<\/p>\n => x = 2<\/p>\n Question 11:\u00a0<\/b>Five years ago, the average age of A,B,C and D was 45 years. By including X in the present lot their present average changes to 49 years. What is the present age of X?<\/p>\n a)\u00a040 years<\/p>\n b)\u00a045 years<\/p>\n c)\u00a049 years<\/p>\n d)\u00a048 years<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Sum of ages of A, B,\u00a0C and D five years ago = $45\\times4=180$<\/p>\n Sum of their present ages = $180+(5\\times4)=200$<\/p>\n Sum of present ages of all five (including X) = $49\\times5=245$<\/p>\n => Present age of X = $245-200=45$ years<\/p>\n => Ans – (B)<\/p>\n Question 12:\u00a0<\/b>The average of 5 consecutive numbers is n. If the next two numbers are also included the average will<\/p>\n a)\u00a0remain the same<\/p>\n b)\u00a0increase by 1<\/p>\n c)\u00a0increase by 1.4<\/p>\n d)\u00a0increase by 2<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Average of 5 consecutive number =\u00a0$\\frac{\\left(n\\right)+\\left(n+1\\right)+\\left(n+2\\right)+\\left(n+3\\right)+\\left(n+4\\right)}{5}$ = n+2<\/p>\n If 2 more consecutive number are added then its average =\u00a0$\\frac{\\left(n\\right)+\\left(n+1\\right)+\\left(n+2\\right)+\\left(n+3\\right)+\\left(n+4\\right)\\ +\\left(n+5\\right)+\\left(n+6\\right)}{7}$ = n+3<\/p>\n Hence average is increased by 1<\/p>\n Question 13:\u00a0<\/b>The average of nine numbers is M and the average of three of these is P. If the average of remaining numbers is N, then<\/p>\n a)\u00a0M = N + P<\/p>\n b)\u00a02M = N + P<\/p>\n c)\u00a03M = 2N + P<\/p>\n d)\u00a03M = 2P + N<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n This can be solved easily by using allegations<\/p>\n $p=\\frac{\\left(p1q1+p2q2\\right)}{q1+q2}$<\/p>\n Here p = M , p1=P , p2=N<\/p>\n q1=3 , q2=6<\/p>\n $\\ \\therefore\\ M=\\frac{\\left(3P+6N\\right)}{3+6}$<\/p>\n On solving 3M = P+2N<\/p>\n Question 14:\u00a0<\/b>In a retail outlet, the average revenue was Rs.10,000 per day over a 30 day period. During this period, the average revenue on weekends (total 8 days) was Rs. 20,000 per day. What was the average daily revenue on weekdays?<\/p>\n a)\u00a06364<\/p>\n b)\u00a05250<\/p>\n c)\u00a06570<\/p>\n d)\u00a08060<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total revenue of the 30-day period = Rs.10000 x 30 days = Rs.300000<\/p>\n Total revenue of the weekends (8 days) = Rs.20000 x 8 days = Rs. 160000<\/p>\n Total revenue of weekdays = (300000-160000) = Rs. 140000<\/p>\n Avg. daily revenue on weekdays = 140000\/22 = Rs.6363.63\u00a0$\\approx\\ $ Rs.6364<\/p>\n Question 15:\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 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Assume the average of 21 students other than Ramesh = a<\/p>\n Sum of the scores of 21 students other than Ramesh = 21a<\/p>\n Hence the average of 22 students = a+1<\/p>\n Sum of the scores of all 22 students = 22(a+1)<\/p>\n The score of Ramesh = Sum of scores of all 22 students – Sum of the scores of\u00a021 students other than Ramesh = 22(a+1)-21a=a+22\u00a0 = 82.5\u00a0 \u00a0(Given)<\/p>\n => a = 60.5<\/p>\n Hence,\u00a0sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353<\/p>\n Now the sum of the scores of students other than Gautam = 21*62 = 1302<\/p>\n Hence the score of Gautam = 1353-1302=51<\/p>\n Question 16:\u00a0<\/b>A cricket team has 11 players and each of them has played 20 matches till date. Virat, Rohit,\u00a0Mahendra, Rahul and Shikhar have scored runs at an average of 60, 55, 50, 45 and 40\u00a0respectively. Rest of the players have scored at an average of 25 each. In the next 10 matches,\u00a0Virat and Rohit each scored 900 runs whereas Mahendra scored twice that of Rahul. After\u00a030 matches, if Virat\u2019s new average score is twice that of Rahul, what is the approximate\u00a0average score of Mahendra ?<\/p>\n a)\u00a049<\/p>\n b)\u00a041<\/p>\n c)\u00a043<\/p>\n d)\u00a045<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average scores of\u00a0Virat, Rohit, Mahendra, Rahul and Shikhar in the first 20 matches are\u00a060, 55, 50, 45,40 respectively.<\/p>\n In the next 10 matches, Virat and Rohit each scored 900 runs whereas Mahendra scored twice that of Rahul.<\/p>\n Virat’s score in 30 matches = 60*20+900=2100<\/p>\n Let us consider Rahul scored x runs in last 10 matches, then Mahendra scores 2x runs.<\/p>\n Rahul’s score in 30 matches =\u00a045*20+x=900+x<\/p>\n Mahendra’s score in 30 matches = 50*20+2x = 1000+2x<\/p>\n It is given that<\/p>\n After 30 matches, Virat\u2019s new average score is twice that of Rahul.<\/p>\n 2100=2(900+x)<\/p>\n x=150<\/p>\n Mahendra’s score in 30 matches =1300<\/p>\n Average score of Mahendra = 43<\/p>\n C is the correct answer.<\/p>\n
\nso we get\u00a0$\\frac{\\left(x+1\\right)}{n}=0.15$
\nwe get x+1 =0.15n \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 (1)
\nFrom condition (2) we get :
\n$\\frac{\\left(x+2\\right)}{n}=0.2$
\nwe get x+2 = 0.2n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (2)
\nSubtracting (1) and (2)
\nwe get 1 =0.05n
\nn =20
\nSo initially he played n-10 =10 matches<\/p>\n
\nTotal employees will be 80+x
\nnow total value of money will be same :
\ni.e Average of 1employee (Number of employees ) will be equal to (Average of boys)* Number of boys +(Average of girls)*Number of girls
\nwe get 48000(80+x) = 50,000(80)+40,000(x)
\n48(80+x) = 50(80) +40x
\nwe get 8x =160
\nx=20<\/p>\n
\n(A) The average age (in years) of electronics engineers is 21
\n(B) The average age (in years) of computer engineers is 21
\n(C) The average age (in years) of electronics engineers is 23
\n(D) The average age (in years) of electronics engineers is 25
\n(E) The average age (in years) of computer engineers is 23
\nChoose the correct answer from the options given below:<\/p>\n
\nSo number of computer engineers initially = 252-168 =84
\nNow number of computer engineers recruited to make ratio 1:1 = 168-84 =84
\nLet average age of computer engineers be x
\nso total age\u00a0 will be 168x
\nNow average age of electronics engineers will be : x+2
\nSo total age will be : (x+2) 168
\nNow as per given
\nTotal age = 336(22)
\nEquating we get :
\n336(22) = 336(x+1)
\nwe get x=21
\nso B and C are correct<\/p>\n
\ncousin two years younger than him. Sumit completed his MBA in 2013, from a reputed B-School.
\nThat year, the average age of Sumit’s family was 41. Sumit got married in 2015 and two years after he
\nbecame father. If the average age of Sumit’s family in 2019 remains same as in 2013, and Sumit is
\nolder than his wife by 3 years, at what age did Sumit graduate MBA ?<\/p>\n