The 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 detailed answers by Cracku. These questions will help you 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>A group of 20 girls has average age of 12 years. Average of first 12 from the same group is 13 years and what is the average age of other 8 girls in the group?<\/p>\n a)\u00a010<\/p>\n b)\u00a011<\/p>\n c)\u00a011.5<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the age of each of the girl in the group be $x_1,x_2,x_3,…..,x_{20}$ years<\/p>\n Average age of 20 girls = 12<\/p>\n => $\\frac{(x_1+x_2+x_3+…..+x_{20})}{20}=12$<\/p>\n =>\u00a0$(x_1+x_2+x_3+…..+x_{20})=12 \\times 20=240$ ————(i)<\/p>\n Average of first 12 girls = 13<\/p>\n => $\\frac{(x_1+x_2+x_2+….+x_{12})}{12}=13$<\/p>\n =>\u00a0$(x_1+x_2+x_3+…..+x_{12})=13 \\times 12=156$ ———–(ii)<\/p>\n Subtracting equation (ii) from (i)<\/p>\n =>\u00a0$(x_1+x_2+x_3+…..+x_{20})$ $-$ $(x_1+x_2+x_3+…..+x_{12}) = (240-156)$<\/p>\n => $(x_{13}+x_{14}+…..+x_{20})=84$<\/p>\n Dividing above equation by 8<\/p>\n => $\\frac{(x_{13}+x_{14}+……+x_{20})}{8}=\\frac{84}{8} = 10.5$<\/p>\n => Ans – (E)<\/p>\n Question 2:\u00a0<\/b>The average of the 5 consecutive even numbers A,B,C,D ,E is 52.what is the product of B & E<\/p>\n a)\u00a02912<\/p>\n b)\u00a02688<\/p>\n c)\u00a03024<\/p>\n d)\u00a02800<\/p>\n e)\u00a0NONE OF THESE<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the five consecutive even numbers\u00a0A,B,C,D ,E = $(x-4) , (x-2) , (x) , (x+2) , (x+4)$ respectively.<\/p>\n Average = $\\frac{A+B+C+D+E}{5} = 52$<\/p>\n => $(x-4) + (x-2) + (x) + (x+2) + (x+4) = 52 \\times 5$<\/p>\n => $5x = 52 \\times 5$<\/p>\n => $x = \\frac{52 \\times 5}{5} = 52$<\/p>\n => $B = 52 – 2 = 50$ and $E = 52 + 4 = 56$<\/p>\n $\\therefore$ Product of B & E = $50 \\times 56 = 2800$<\/p>\n Question 3:\u00a0<\/b>find the average of set scores? 221,231,441,359,665,525<\/p>\n a)\u00a0399<\/p>\n b)\u00a0428<\/p>\n c)\u00a0407<\/p>\n d)\u00a0415<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Set\u00a0:\u00a0221,231,441,359,665,525<\/p>\n Sum =\u00a0221 + 231 + 441 + 359 + 665 + 525 = 2442<\/p>\n => Required average = $\\frac{2442}{6}$<\/p>\n = 407<\/p>\n Question 4:\u00a0<\/b>There are three positive numbers, ${1 \\over 3}$rd of average of all the three numbers is 8 less than the value of the highest number. Average of the lowest and the second lowest number is 8. Which is the highest number?<\/p>\n a)\u00a011<\/p>\n b)\u00a014<\/p>\n c)\u00a010<\/p>\n d)\u00a09<\/p>\n e)\u00a013<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the three positive numbers be $x,y,z$ \u00a0 \u00a0 (where $x < y < z$)<\/p>\n Average of the three numbers = $\\frac{x + y + z}{3}$<\/p>\n Acc. to ques, => $\\frac{1}{3} \\times (\\frac{x + y + z}{3}) = z – 8$<\/p>\n => $x + y + z = 9z – 72$<\/p>\n => $x + y = 8z – 72$<\/p>\n Dividing both sides by 2, we get\u00a0:<\/p>\n => $\\frac{x + y}{2} = 4z – 36$<\/p>\n Also, average of the lowest and the second lowest number is 8, => $\\frac{x + y}{2} = 8$<\/p>\n => $4z – 36 = 8$<\/p>\n => $4z = 8 + 36 = 44$<\/p>\n => $z = \\frac{44}{4} = 11$<\/p>\n Question 5:\u00a0<\/b>The average age of some males and 15 females is 18 years. The sum of the ages of 15 females is 240 years and average age of males is 20 years. Find the number of males.<\/p>\n a)\u00a08<\/p>\n b)\u00a07<\/p>\n c)\u00a010<\/p>\n d)\u00a015<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let number of males = $x$<\/p>\n Average age of males = 20 years<\/p>\n => Sum of age of males = $20 \\times x = 20x$ years<\/p>\n Sum of age of females = 240 years<\/p>\n Acc. to ques, => $\\frac{(20 x) + (240)}{15 + x} = 18$<\/p>\n => $20x + 240 = 270 + 18x$<\/p>\n => $20x – 18x = 2x = 270 – 240 = 30$<\/p>\n => $x = \\frac{30}{2} = 15$<\/p>\n 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 6:\u00a0<\/b>Average score of a class of 50 students, in an exam is 34. Average score of the students who have passed is 52 and the average score of students who have failed is 16. How many students have failed in the exam?<\/p>\n a)\u00a025<\/p>\n b)\u00a020<\/p>\n c)\u00a015<\/p>\n d)\u00a018<\/p>\n e)\u00a030<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number of students who failed = $x$<\/p>\n => Number of students who passed = $(50 – x)$<\/p>\n Acc. to ques,<\/p>\n => $\\frac{(52 \\times (50 – x)) + (16 \\times x)}{50} = 34$<\/p>\n => $52 \\times 50 – 52x + 16x = 34 \\times 50$<\/p>\n => $52x – 16x = 50 \\times (52 – 34)$<\/p>\n => $36x = 50 \\times 18$<\/p>\n => $x = \\frac{50 \\times 18}{36} = 25$<\/p>\n Question 7:\u00a0<\/b>In a class, the average weight of 80 boys is 64 kg and that of 75 girls is 70 kg. After a few days, 60% of the girls and 30% of the boys leave. What would be the new average weight of the class (in kg)? Assume that the average weight of the boys and the girls remain constant throughout.<\/p>\n a)\u00a063<\/p>\n b)\u00a066.09<\/p>\n c)\u00a068.5<\/p>\n d)\u00a065.5<\/p>\n e)\u00a057.5<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Initially, number of boys = 80 and number of girls = 75<\/p>\n Average weight of boys = 64 kg and average weight of girls = 70 kg<\/p>\n Now, 60% of the girls and 30% of the boys leave<\/p>\n => Boys left = $\\frac{100 – 30}{100} \\times 80 = 56$<\/p>\n Girls left = $\\frac{100 – 60}{100} \\times 75 = 30$<\/p>\n Since,\u00a0average weight of the boys and the girls remains constant throughout<\/p>\n $\\therefore$ New average weight of the class<\/p>\n = $\\frac{(56 \\times 64) + (30 \\times 70)}{56 + 30} = \\frac{3584 + 2100}{86}$<\/p>\n = $\\frac{5684}{86} = 66.09$ kg<\/p>\n Question 8:\u00a0<\/b>In a class, the average weight of 40 boys is 65 kg and that of 50 girls is 60 kg. After a few days, 40% of the girls and 50% of the boys leave. What would be the new average weight of the class (in kg)? Assume that the average weight of the boys and the girls remains constant throughout.<\/p>\n a)\u00a065<\/p>\n b)\u00a062<\/p>\n c)\u00a068<\/p>\n d)\u00a055<\/p>\n e)\u00a058<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Initially, number of boys = 40 and number of girls = 50<\/p>\n Average weight of boys = 65 kg and average weight of girls = 60 kg<\/p>\n Now,\u00a040% of the girls and 50% of the boys leave<\/p>\n => Boys left = $\\frac{100 – 50}{100} \\times 40 = 20$<\/p>\n Girls left = $\\frac{100 – 40}{100} \\times 50 = 30$<\/p>\n Since,\u00a0average weight of the boys and the girls remains constant throughout<\/p>\n $\\therefore$ New average weight of the class<\/p>\n = $\\frac{(20 \\times 65) + (30 \\times 60)}{20 + 30} = \\frac{1300 + 1800}{50}$<\/p>\n = $\\frac{3100}{50} = 62$ kg<\/p>\n Question 9:\u00a0<\/b>When 9 is subtracted from a two digit number, the number so formed is reverse of the original number. Also, the average of the digits of the original number is 7.5. What is definitely the original number ?<\/p>\n a)\u00a087<\/p>\n b)\u00a092<\/p>\n c)\u00a090<\/p>\n d)\u00a069<\/p>\n e)\u00a096<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the ten’s digit and unit’s digit of the original number be $x$ and $y$ respectively.<\/p>\n => original number = $10x + y$<\/p>\n Average of digits = $\\frac{x + y}{2} = 7.5$<\/p>\n => $x + y = 7.5 \\times 2 = 15$ ————-(i)<\/p>\n When 9 is subtracted from it, => Reverse number = $10y + x$<\/p>\n => $(10x + y) – 9 = 10y + x$<\/p>\n => $9x – 9y = 9$<\/p>\n => $x – y = \\frac{9}{9} = 1$ —————-(ii)<\/p>\n Adding equations (i) & (ii), we get\u00a0:<\/p>\n => $2x = 16$ => $x = \\frac{16}{2} = 8$<\/p>\n Putting it in eqn(i), => $y = 15 – 8 = 7$<\/p>\n $\\therefore$ Original number = $87$<\/p>\n Question 10:\u00a0<\/b>Average weight of three boys P, T and R is 54 1\/3 kgs while the average weight of three boys, T, F and G is 53 kgs. What is the average weight of P, T, R, F and H?<\/p>\n a)\u00a053.8 kgs<\/p>\n b)\u00a052.4 kgs<\/p>\n c)\u00a053.2 kgs<\/p>\n d)\u00a0Can\u2019t be determined<\/p>\n e)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Clearly, there is no information about the weight of H<\/p>\n Thus, we cannot determine their average weights.<\/p>\n Ans – (D)<\/p>\n Question 11:\u00a0<\/b>The average age of a man and his son is 48 years. The ratio of their ages is 11 : 5 respectively. What will be ratio of their ages after 6 years ?<\/p>\n a)\u00a06 : 5<\/p>\n b)\u00a05 : 3<\/p>\n c)\u00a04 : 3<\/p>\n d)\u00a02 : 1<\/p>\n e)\u00a0None of these<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let age of man = $11x$ and age of son = $5x$<\/p>\n Sum of ages of man and son = 48 * 2 = 96 years<\/p>\n => $11x + 5x = 96$<\/p>\n => $x = \\frac{96}{16} = 6$<\/p>\n => Age of man = 11 * 6 = 66 years<\/p>\n Age of son = 5 * 6 = 30 years<\/p>\n => Age of man after 6 years = 66 + 6 = 72 years<\/p>\n Age of son after 6 years = 30 + 6 = 36 years<\/p>\n $\\therefore$ Required ratio = 72\u00a0: 36 =\u00a02\u00a0: 1<\/strong><\/p>\n Question 12:\u00a0<\/b>The average age of a man and his son is 18 years. The ratio of their ages is 5 : 1 respectively. What will be the ratio of their ages after 6 years ?<\/p>\n a)\u00a010 : 3<\/p>\n b)\u00a05 : 2<\/p>\n c)\u00a04 : 3<\/p>\n d)\u00a03 : 1<\/p>\n e)\u00a0None of these<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let age of man = $5x$ and age of son = $x$<\/p>\n Sum of ages of man and son = 18 * 2 = 36 years<\/p>\n => $5x + x = 36$<\/p>\n => $x = \\frac{36}{6} = 6$<\/p>\n => Age of man = 5 * 6 = 30 years<\/p>\n Age of son = 1 * 6 = 6 years<\/p>\n => Age of man after 6 years = 30 + 6 = 36 years<\/p>\n Age of son after 6 years = 6 + 6 = 12 years<\/p>\n $\\therefore$ Required ratio = 36 : 12 =\u00a03 : 1<\/strong><\/p>\n Question 13:\u00a0<\/b>The average age of a man and his son is 54 years. The ratio of their ages is 23 : 13 respectively. What will be ratio of their ages after 6 years?<\/p>\n a)\u00a010 : 7<\/p>\n b)\u00a05 : 3<\/p>\n c)\u00a04 : 3<\/p>\n d)\u00a03 : 2<\/p>\n e)\u00a0None of these<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let age of man = $23x$ and age of son = $13x$<\/p>\n Sum of ages of man and son = 54 * 2 = 108 years<\/p>\n => $23x + 13x = 108$<\/p>\n => $x = \\frac{108}{36} = 3$<\/p>\n => Age of man = 23 * 3 = 69 years<\/p>\n Age of son = 13 * 3 = 39 years<\/p>\n => Age of man after 6 years = 69 + 6 = 75 years<\/p>\n Age of son after 6 years = 39 + 6 = 45 years<\/p>\n $\\therefore$ Required ratio = 75 : 45 =\u00a05\u00a0: 3<\/strong><\/p>\n Question 14:\u00a0<\/b>The average of 5 consecutive even numbers A, B, C. D and E is 52. What is the product of B & E?<\/p>\n a)\u00a02916<\/p>\n b)\u00a02988<\/p>\n c)\u00a03000<\/p>\n d)\u00a02800<\/p>\n e)\u00a0None of these<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the 5 consecutive even numbers be = $(x) , (x+2) , (x+4) , (x+6) , (x+8)$<\/p>\n Sum of these 5 consecutive numbers = 52 * 5 = 260<\/p>\n => $(x) + (x+2) + (x+4) + (x+6) + (x+8) = 260$<\/p>\n => $5x = 260 – 20 = 240$<\/p>\n => $x = 48$<\/p>\n => B = $x + 2$ = 48 + 2 = 50<\/p>\n E = $x + 8$ = 48 + 8 = 56<\/p>\n $\\therefore$ B $\\times$ E = 50 * 56 = 2800<\/p>\n Question 15:\u00a0<\/b>Find the average of the following set of scores : a)\u00a0399<\/p>\n b)\u00a0428<\/p>\n c)\u00a0407<\/p>\n d)\u00a0415<\/p>\n e)\u00a0None of these<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Sum of scores = 221 + 231 + 441 + 359 + 665 + 525<\/p>\n = 2442<\/p>\n => Average score = 2442\/6 = 407<\/p>\n Question 16:\u00a0<\/b>The ratio of roses and lillies in a garden is 3 : 2 respectively. The average number of roses and lillies is 180. What is the number of lillies in the garden?<\/p>\n a)\u00a0144<\/p>\n b)\u00a0182<\/p>\n c)\u00a0216<\/p>\n d)\u00a0360<\/p>\n e)\u00a0None of these<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number of roses and lillies in the garden be $3x$ and $2x$ resp<\/p>\n Acc to ques,<\/p>\n => $\\frac{3x + 2x}{2} = 180$<\/p>\n => $5x = 360$<\/p>\n => $x = 360\/5 = 72$<\/p>\n $\\therefore$ Number of lillies = 2 * 72 = 144<\/p>\n Question 17:\u00a0<\/b>The average of five positive numbers is 308. The average of first two numbers is 482.5 and the average of last two numbers is 258.5. What is the third number?<\/p>\n a)\u00a0224<\/p>\n b)\u00a058<\/p>\n c)\u00a0121<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the numbers be $A_1 , A_2 , A_3 , A_4 , A_5$<\/p>\n => $A_1 + A_2 + A_3 + A_4 + A_5 = 308 * 5 = 1540$<\/p>\n Also, $A_1 + A_2 = 482.5 * 2 = 965$<\/p>\n and $A_4 + A_5 = 258.5 * 2 = 517$<\/p>\n => $A_3 = 1540 – 965 – 517 = 58$<\/p>\n Question 18:\u00a0<\/b>What will be the average of the following set of scores (Rounded off to the nearest integer) ? a)\u00a045<\/p>\n b)\u00a059<\/p>\n c)\u00a062<\/p>\n d)\u00a048<\/p>\n e)\u00a053<\/p>\n 18)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Sum of the scores = 46 + 54 + 62 + 68 + 56 + 29 + 58 = 373<\/p>\n => Required Average = $\\frac{373}{7}$<\/p>\n = $53.28 \\approx 53$<\/p>\n Question 19:\u00a0<\/b>What will be the average of the followings set of scores? a)\u00a062<\/p>\n b)\u00a066<\/p>\n c)\u00a075<\/p>\n d)\u00a052<\/p>\n e)\u00a059<\/p>\n 19)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Sum of the scores<\/p>\n = 59 + 84 + 44 + 98 + 30 + 40 + 58 = 413<\/p>\n => Average = $\\frac{413}{7}$ = 59<\/p>\n Question 20:\u00a0<\/b>The average of five positive numbers is 213. The average of the first two numbers is 233.5 and the average of last two numbers is 271. What is the third number?<\/p>\n a)\u00a064<\/p>\n b)\u00a056<\/p>\n c)\u00a0106<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the numbers be $A_1 , A_2 , A_3 , A_4 , A_5$<\/p>\n => $A_1 + A_2 + A_3 + A_4 + A_5 = 213 * 5 = 1065$<\/p>\n Also, $A_1 + A_2 = 233.5 * 2 = 467$<\/p>\n and $A_4 + A_5 = 271 * 2 = 542$<\/p>\n => $A_3 = 1065 – 467 – 542 = 56$<\/p>\n
\n221, 231, 441, 359, 665, 525<\/p>\n
\n46, 54, 62, 68, 56, 29, 58<\/p>\n
\n59, 84, 44, 98, 30, 40, 58<\/p>\n