Here you can download a free Averages and Ages questions PDF with answers for IBPS PO 2022 by Cracku. These are some tricky questions in the IBPS PO 2022 exam that you need to find the Averages and Ages for the given questions. These questions will help you to do practice and solve the Averages and Ages questions in the IBPS PO exams. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Averages and Ages MCQ<\/strong> PDF for IBPS PO 2022 for free.<\/p>\n Download Averages and Ages Questions for IBPS PO Prelims<\/a><\/p>\n Download IBPS PO Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>Suresh gave 17.39% of his monthly income to his wife and spent 31.57% of the remaining amount for household expenses. Out of the remaining amount he gave 15.38% to the charity. Again from the remaining amount he gave 18.18% to his friend. Finally if Suresh is left with ___ then monthly income of Suresh is ___ . a)\u00a0Only (ii) and (iv)<\/p>\n b)\u00a0Only (i) and (iv)<\/p>\n c)\u00a0All (i), (ii), (iii) and (iv)<\/p>\n d)\u00a0Only (i), (ii) and (iv)<\/p>\n e)\u00a0Only (ii), (iii) and (iv)<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Suresh gave 17.39% of his monthly income to his wife. Suresh spent 31.57% of the remaining amount for household expenses. Out of the remaining amount he gave 15.38% to the charity. Again from the remaining amount he gave 18.18% to his friend. Ratio of the final amount remaining with Suresh and his monthly income respectively = 9p : 23p ii) 8991, 22977 iii) 16083, 41101 iii) 14085, 35995 Question 2:\u00a0<\/b>The average weight of \u2018a-7\u2019 students is \u2018a+7\u2019, \u2018a+5\u2019 students is \u2018a+17\u2019 and \u20182a-50\u2019 students is \u20182a-26\u2019 respectively. What is the value of \u2018a\u2019 if the average weight of all the given students is \u20182a-29\u2019?<\/p>\n a)\u00a02<\/p>\n b)\u00a0Either 2 or 43<\/p>\n c)\u00a041<\/p>\n d)\u00a043<\/p>\n e)\u00a0Either 2 or 41<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average weight of \u2018a-7\u2019 students is \u2018a+7\u2019 The average weight of \u2018a+5\u2019 students is \u2018a+17\u2019 The average weight of \u20182a-50\u2019 students is \u20182a-26\u2019 Total number of students = a – 7 + a + 5 + 2a – 50 = 4a – 52 (a+7)(a-7) + (a+17)(a+5) + (2a-26)(2a-50) = (2a-29)(4a-52) Question 3:\u00a0<\/b>The average weight of 63 students in a class is 55 kg. When 63 students were combined with a teacher and 6 more students whose average weight is 58 kg the average weight increased by 0.5 kg. What is the weight of the teacher?<\/p>\n a)\u00a050 kg<\/p>\n b)\u00a062 kg<\/p>\n c)\u00a066 kg<\/p>\n d)\u00a078 kg<\/p>\n e)\u00a072 kg<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average weight of 63 students in a class is 55 kg. Let the weight of the teacher = w According to the problem, average weight of 70 members = 55 + 0.5 = 55.5 Question 4:\u00a0<\/b>A school distributed sweets to its students in celebration of its 25th anniversary and different students received different number of sweets. The average of the number of sweets received by P and Q is 14.28% more than the average of the number of sweets received by Q and R. Ratio of the number of sweets received by Q and R is 31:25 respectively then which of the following can be the possible number of sweets received by P?<\/p>\n a)\u00a050<\/p>\n b)\u00a062<\/p>\n c)\u00a066<\/p>\n d)\u00a078<\/p>\n e)\u00a072<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Ratio of the number of sweets received by Q and R is 31:25 respectively. The average of the number of sweets received by P and Q is 14.28% more than the average of the number of sweets received by Q and R. Question 5:\u00a0<\/b>If $(x+5) \\text{ of } 11.11\\% = y \\text{ of } 12.5\\% -10$ and $(x-25) = 75\\% \\text{ of } (y-40)$, then find out the value of $40\\% of (y-x)$.<\/p>\n a)\u00a039<\/p>\n b)\u00a051<\/p>\n c)\u00a028<\/p>\n d)\u00a013<\/p>\n e)\u00a0None of the above<\/p>\n 5)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(x+5) \\text{ of } 11.11\\% = y \\text{ of } 12.5\\% – 10$ $x = y\\times\\frac{9}{8} – 90 -5$<\/p>\n $x = y\\times\\frac{9}{8} – 95$ Eq.(i)<\/p>\n $(x-25) = 75\\% \\text{ of } (y-40)$ = 26 Question 6:\u00a0<\/b>Ketan spent 41.67% of his monthly salary on house rent. Out of the remaining, he spent 20% on food. Out of the remaining, he spent 10% on education. After that the remaining amount was invested in FD and MF in the ratio of 7:5 respectively. If the amount spent on education by him is Rs. 4480, then find out the difference between the amount spent on house rent and the amount invested on FD from his monthly salary.<\/p>\n a)\u00a0Rs. 15240<\/p>\n b)\u00a0Rs. 18640<\/p>\n c)\u00a0Rs. 16480<\/p>\n d)\u00a0Rs. 12680<\/p>\n e)\u00a0None of the above<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let\u2019s assume the monthly salary of Ketan is 12z. Question 7:\u00a0<\/b>The average weight of a group of \u2018y\u2019 people is \u2018z\u2019 kg. If five people left the group whose average weight is 52 kg, then the average of weight of the group will be increased by 0.25. If five people joined the group whose average weight is 38 kg, then the average of weight of the group will be (z-1) kg. Find out the value of \u2018z\u2019.<\/p>\n a)\u00a054<\/p>\n b)\u00a058<\/p>\n c)\u00a056<\/p>\n d)\u00a052<\/p>\n e)\u00a0None of the above<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average weight of a group of \u2018y\u2019 people is \u2018z\u2019 kg. If five people left the group whose average weight is 52 kg, then the average of weight of the group will be increased by 0.25. Question 8:\u00a0<\/b>The monthly savings of A is 58.33% less than the monthly expenditure of B. The monthly income of A is 25% more than the monthly income of B. If the monthly savings of B is Rs. 760 and the sum of the monthly expenditure of A and B together is Rs. 2490, then find out the difference between the monthly income of A and B.<\/p>\n a)\u00a0Rs. 500<\/p>\n b)\u00a0Rs. 200<\/p>\n c)\u00a0Rs. 300<\/p>\n d)\u00a0Rs. 400<\/p>\n e)\u00a0None of the above<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The monthly savings of A is 58.33% less than the monthly expenditure of B. Question 9:\u00a0<\/b>The sum of the squares of the two odd numbers is 1450 and the product of the numbers is 525, then what is the average of cubes of both the numbers?<\/p>\n a)\u00a025250<\/p>\n b)\u00a023125<\/p>\n c)\u00a020750<\/p>\n d)\u00a027425<\/p>\n e)\u00a0None of the above<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the two odd numbers be p and q Question 10:\u00a0<\/b>The ratio of income of Rahul and Amit is 9:13 respectively. The ratio of savings and expenditure of Amit is 6:7 respectively. If the savings of Amit is 20% more than the savings of Rahul, then the expenditure of Amit is how much percent more or less than the expenditure of Rahul?<\/p>\n a)\u00a075% less<\/p>\n b)\u00a025% more<\/p>\n c)\u00a025% less<\/p>\n d)\u00a075% more<\/p>\n e)\u00a0None of the above<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The ratio of income of Rahul and Amit is 9:13 respectively. Take Free IBPS PO Mock Tests<\/a><\/p>\n Question 11:\u00a0<\/b>There are six numbers and the average of the first four numbers is equal to 35. The average of the last four numbers is 40. The average of the first and sixth number is 45 whereas the average of the second and fifth number is 55. What is the average of the first, third, fourth and sixth numbers?<\/p>\n a)\u00a035<\/p>\n b)\u00a041<\/p>\n c)\u00a033<\/p>\n d)\u00a031<\/p>\n e)\u00a0None of the above<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the six numbers are a, b, c, d, e and f respectively.<\/p>\n The average of the first four numbers is equal to 35. The average of the last four numbers is 40. The average of the first and sixth number is 45. The average of the second and fifth number is 55. Adding (1) and (2), Average of the first, third, fourth and sixth numbers = $\\frac{a+c+d+f}{4}$ Question 12:\u00a0<\/b>A man spent 20% of his monthly income on house rent. Out of the remaining amount he spent 22% for shopping. Out of the remaining amount one fourth was spent on food and the remaining amount was saved. If the amount spent on shopping was \u20b99504, then what was the amount saved by the man?<\/p>\n a)\u00a0\u20b926732<\/p>\n b)\u00a0\u20b927892<\/p>\n c)\u00a0\u20b924382<\/p>\n d)\u00a0\u20b925272<\/p>\n e)\u00a0None of the above<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the monthly income of man = p Question 13:\u00a0<\/b>Average weight of 45 students in a class is 65 kg. Another six students joined the class and the average weight of the class increased by 3 kg. What is the average weight of the six students who joined later?<\/p>\n a)\u00a080.25<\/p>\n b)\u00a085.50<\/p>\n c)\u00a060.75<\/p>\n d)\u00a070.25<\/p>\n e)\u00a090.50<\/p>\n 13)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, average weight of 45 students = 65 kg Question 14:\u00a0<\/b>In 2010, the number of bears in the zoo was increased by 35% and next year these were decreased by 16.67%. Now the number of bears in the zoo is 405, then find out the 20% of the number of bears in the zoo two years ago.<\/p>\n a)\u00a066<\/p>\n b)\u00a054<\/p>\n c)\u00a060<\/p>\n d)\u00a078<\/p>\n e)\u00a0None of the above<\/p>\n 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let\u2019s assume the number of bears in the zoo two years ago is \u2018y\u2019. Question 15:\u00a0<\/b>The price of icecream has increased by 40%. Lekha has decided to spend only 4% more than what he initially did on buying icecream. What is the percentage decrease in Lekha’s rice consumption?<\/p>\n a)\u00a019%<\/p>\n b)\u00a060%<\/p>\n c)\u00a015%<\/p>\n d)\u00a030%<\/p>\n e)\u00a025%<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the initial price of icecream be Rs. 100 per unit and Lekha’s consumption be 10 units.<\/p>\n \u2234 Initial amount spent = 100 \u00d7 10 = Rs. 1,000<\/p>\n New price of icecream = 140% of 100 = Rs. 140 and new total amount spent = 104% of 1000 = Rs. 1,040 Decrease in consumption =\u00a03 units<\/p>\n % decrease\u00a0\u00a0=$\\ \\frac{\\ 3}{10}\\times\\ 100$ = 30%<\/p>\n Hence answer is option d<\/p>\n Question 16:\u00a0<\/b>A shopkeeper has a sale of Rs. 4475, Rs. 8657, Rs. 4755, Rs. 4230 and Rs. 4542 for 5 consecutive months respectively. How much sale must he have in the sixth month so that he gets an average sale of Rs. 4500?<\/p>\n a)\u00a0320<\/p>\n b)\u00a0348<\/p>\n c)\u00a0341<\/p>\n d)\u00a0478<\/p>\n e)\u00a0321<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total sale for 5 months = Rs. (4475+ 8657+ 4755 + 4230+ 4542) = Rs.26659 Question 17:\u00a0<\/b>Three students has birthday on same day they received\u00a0\u00a0126, 786 and 986 gifts\u00a0respectively from their friends\u00a0\u00a0What percentage of the total gifts\u00a0 did the person with more gifts\u00a0get?<\/p>\n a)\u00a052%<\/p>\n b)\u00a055%<\/p>\n c)\u00a075%<\/p>\n d)\u00a090%<\/p>\n e)\u00a080%<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total number of gifts = (126+786+986) = 1898<\/p>\n Required percentage =\u00a0\u00a0$\\ \\frac{\\ 986}{1898}\\times\\ 100$<\/p>\n => 51.95% which is approximately 52%<\/p>\n hence answer is option a<\/p>\n Question 18:\u00a0<\/b>The average marks of 40 students is 67.875. If two more students whose marks are (y-3) and (y+3) added then the average will be 67.5. Find out the value of \u2018y\u2019.<\/p>\n a)\u00a057<\/p>\n b)\u00a060<\/p>\n c)\u00a063<\/p>\n d)\u00a054<\/p>\n e)\u00a0None of the above<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average marks of 40 students is 67.875. Question 19:\u00a0<\/b>The average age of a group of 18 people is 54 years. Two more people joined the group in which one\u2019s age is 12 years more than the other\u2019s age. If after joining the two people in the group, the average age will be 52 years, then find out the age of the youngest person among those people who joined the group later.<\/p>\n a)\u00a044 years<\/p>\n b)\u00a040 years<\/p>\n c)\u00a028 years<\/p>\n d)\u00a032 years<\/p>\n e)\u00a0Cannot be determined<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average age of a group of 18 people is 54 years. Question 20:\u00a0<\/b>If 35% of a number P is 82 less than the 45% of a number Q and the sum of numbers P & Q is 1320, then find out the sum of the 55% of P and 25% of Q.<\/p>\n a)\u00a0546<\/p>\n b)\u00a0568<\/p>\n c)\u00a0584<\/p>\n d)\u00a0522<\/p>\n e)\u00a0None of the above<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n 35% of P = 45% of Q – 82
\nWhich of the following values can fill in the blanks appropriately?
\ni) 10035, 25645
\nii) 8991, 22977
\niii) 16083, 41101
\niv) 14085, 35995<\/p>\n
\nLet the monthly income of Suresh = 23p
\nAmount given to his wife = $\\frac{4}{23}\\times$23p = 4p
\nAmount remaining = 23p – 4p = 19p<\/p>\n
\nAmount spent for household expenses = $\\frac{6}{19}\\times$19p = 6p
\nAmount remaining = 19p – 6p = 13p<\/p>\n
\nAmount given charity = $\\frac{2}{13}\\times$13p = 2p
\nAmount remaining = 13p – 3p = 11p<\/p>\n
\nAmount given to friend = $\\frac{2}{11}\\times$11p = 2p
\nFinal amount remaining = 11p – 2p = 9p<\/p>\n
\n= 9 : 23
\n
\n<\/strong>i) 10035, 25645
\n<\/strong>Ratio of the final amount remaining with Suresh and his monthly income respectively = 10035 : 25645
\n= 9 : 23
\nGiven values satisfy the condition of the question.<\/p>\n
\n<\/strong>Ratio of the final amount remaining with Suresh and his monthly income respectively = 8991 : 22977
\n= 9 : 23
\nGiven values satisfy the condition of the question.<\/p>\n
\n<\/strong>Ratio of the final amount remaining with Suresh and his monthly income respectively = 16083 : 41101
\n= 9 : 23
\nGiven values satisfy the condition of the question.<\/p>\n
\n<\/strong>Ratio of the final amount remaining with Suresh and his monthly income respectively = 14085 : 35995
\n= 9 : 23
\nGiven values satisfy the condition of the question.
\nHence, the correct answer is Option C<\/p>\n
\nSum of the weights of \u2018a-7\u2019 students = (a+7)(a-7)<\/p>\n
\nSum of the weights of \u2018a+5\u2019 students = (a+17)(a+5)<\/p>\n
\nSum of the weights of \u20182a-50\u2019 students = (2a-26)(2a-50)<\/p>\n
\nOverall average weight = 2a – 29
\nTotal weight of all students = (2a-29)(4a-52)<\/p>\n
\na$^2$ – 49 + a$^2$ + 22a + 85 + 4a$^2$ – 152a + 1300 = 8a$^2$ – 220a + 1508
\n6a$^2$ – 130a + 1336 = 8a$^2$ – 220a + 1508
\n2a$^2$ – 90a + 172 = 0
\na$^2$ – 45a + 86 = 0
\na$^2$ – 2a – 43a + 86 = 0
\na(a – 2) – 43(a – 2) = 0
\n(a – 2)(a – 43) = 0
\na = 2, 43
\nIf a = 2 then number of students(2a – 50) is negative which is not possible.
\nSo a = 43
\nHence, the correct answer is Option D<\/p>\n
\nSum of the weight of 63 students = 55 x 63 = 3465 kg<\/p>\n
\nAverage weight of 6 new students = 58 kg
\nSum of the weight of 6 new students = 58 x 6 = 348 kg<\/p>\n
\nSum of the weight of 70 members = 55.5 x 70 = 3885 kg
\nSum of the weight of 63 students + Weight of the teacher + Sum of the weight of 6 new students = 3885
\n3465 + w + 348 = 3885
\n3813 + w = 3885
\nw = 72 kg
\nWeight of the teacher = w = 72 kg
\nHence, the correct answer is Option E<\/p>\n
\nLet the number of sweets received by Q and R are 31a and 25a respectively.
\nQ = 31a
\nR = 25a<\/p>\n
\n$\\frac{P+Q}{2}$ = $\\frac{Q+R}{2}$ + $\\frac{1}{7}\\times\\frac{Q+R}{2}$
\n$\\frac{P+Q}{2}$ = $\\frac{8}{7}\\times\\frac{Q+R}{2}$
\n7P + 7Q = 8Q + 8R
\n7P – Q – 8R = 0
\n7P – 31a – 8(25a) = 0
\n7P – 31a – 200a = 0
\n7P = 231a
\nP = 33a
\nNumber of sweets received by P is multiple of 33.
\nThe only multiple of 33 from the options is 66.
\nHence, the correct answer is Option C<\/p>\n
\n$(x+5)\\times\\frac{1}{9} = y\\times\\frac{1}{8} – 10$
\n$(x+5) = y\\times\\frac{9}{8} – 90$<\/p>\n
\n$(x-25) = \\frac{3}{4} \\times (y-40)$
\nPut the value of \u2018x\u2019 in the above equation from Eq.(i).
\n$(y\\times \\frac{9}{8}-95-25)=\\frac{3}{4}\\times (y-40)$
\nAfter solving the above equation, y = 240.
\nPut the value of \u2018y\u2019 in Eq.(i).
\n$x = 240\\times\\frac{9}{8} – 95$
\nx = 270-95
\n= 175
\nvalue of $40\\% of (y-x)$ = $40\\% of (240-175)$
\n= 40% of 65<\/p>\n
\nHence, option e is the correct answer.<\/p>\n
\nKetan spent 41.67% of his monthly salary on house rent.
\nAmount spent on house rent = 12z of (5\/12) = 5z
\nOut of the remaining, he spent 20% on food.
\nAmount spent on food = (12z-5z) of 20%
\n= 7z of 20%
\n= 1.4z
\nOut of the remaining, he spent 10% on education.
\nAmount spent on education = (7z-1.4z) of 10%
\n= 5.6z of 10%
\n= 0.56z
\nIf the amount spent on education by him is Rs. 4480.
\n0.56z = 4480
\nz = 8000
\n<\/b>After that the remaining amount was invested in FD and MF in the ratio of 7:5 respectively.
\nAmount invested on FD = 5.04z of (7\/12)
\n= 2.94z
\nDifference between the amount spent on house rent and the amount invested on FD from his monthly salary = 5z – 2.94z
\n= 2.06z
\nPut the value of \u2018z\u2019 in the equation.
\n= $2.06\\times8000$
\n= Rs. 16480
\nHence, option c is the correct answer.<\/p>\n
\n$yz-5\\times52 = (y-5)(z+0.25)$
\n$yz-260 = (y-5)(z+0.25)$
\nAfter solving the above equation, we obtain an equation which is given below.
\ny = (20z-1035) Eq.(i)
\nIf five people joined the group whose average weight is 38 kg, then the average of weight of the group will be (z-1) kg.
\n$yz+5\\times38 = (y+5)(z-1)$
\n$yz+190 = (y+5)(z-1)$
\nAfter solving the above equation, we obtain an equation which is given below.
\ny = (5z-195) Eq.(ii)
\nEq.(i) = Eq.(ii)
\n(20z-1035) = (5z-195)
\n15z = 1035-195 = 840
\nz = 56
\nHence, option c is the correct answer.<\/p>\n
\nLet\u2019s assume the monthly expenditure of B is 12y.
\nmonthly savings of A = (5\/12) of 12y = 5y
\nThe monthly income of A is 25% more than the monthly income of B.
\nLet\u2019s assume the monthly income of B is 4z.
\nmonthly income of A = 5z
\nmonthly expenditure of A = (5z-5y)
\nmonthly savings of B = (4z-12y)
\nThe monthly savings of B is Rs. 760.
\n(4z-12y) = 760
\nz-3y = 190
\nz = (190+3y) Eq.(i)
\nThe sum of the monthly expenditure of A and B together is Rs. 2490.
\n(5z-5y)+12y = 2490
\n(5z+7y) = 2490
\nPut Eq.(i) in the above equation.
\n$5\\times(190+3y)+7y = 2490$
\n950+15y+7y = 2490
\n15y+7y = 2490-950 = 1540
\n22y = 1540
\ny = 70
\nPut the value of \u2018y\u2019 in Eq.(i).
\nz = (190+210) = 400
\nDifference between the monthly income of A and B = (5z-4z) = 400
\nHence, option d is the correct answer.<\/p>\n
\nThe product of the two odd numbers is 525$\\Rightarrow$ pq = 525\u2026\u2026\u2026\u2026(1)
\nThe sum of the squares of the two odd numbers is 1450
\n$\\Rightarrow$ p$^2$ + q$^2$ = 1450
\n$\\Rightarrow$ (p + q)$^2$ – 2pq = 1450
\n$\\Rightarrow$ (p + q)$^2$ – 2(525) = 1450
\n$\\Rightarrow$ (p + q)$^2$ – 1050 = 1450
\n$\\Rightarrow$ (p + q)$^2$ = 2500
\n$\\Rightarrow$ p + q = 50\u2026\u2026..(2)
\nAverage of cubes of the two odd numbers = $\\frac{1}{2}\\times$(p$^3$ + q$^3$)
\n= $\\frac{1}{2}\\times$[(p+q)$^3$-3pq(p+q)]
\n= $\\frac{1}{2}\\times$[(50)$^3$-3(525)(50)]
\n= $\\frac{1}{2}\\times$[125000-78750]
\n= $\\frac{1}{2}\\times$46250
\n= 23125
\nHence, the correct answer is Option B<\/p>\n
\nLet the income of Rahul and Amit are 9p and 13p respectively.
\nThe ratio of savings and expenditure of Amit is 6:7.
\nLet the savings and expenditure of Amit are 6q and 7q respectively.
\nIncome of Amit = 13p
\n$\\Rightarrow$ 6q + 7q = 13p
\n$\\Rightarrow$ 13q = 13p
\n$\\Rightarrow$ q = p\u2026\u2026..(1)
\nThe savings of Amit is 20% more than the savings of Rahul.
\nLet the savings of Rahul = t
\nSavings of Amit = $\\frac{120}{100}$t
\n$\\Rightarrow$ 6q = $\\frac{6}{5}$t
\n$\\Rightarrow$ t = 5q
\n$\\Rightarrow$ t = 5p
\nSavings of Rahul = t = 5p
\nExpenditure of Rahul = Income of Rahul – Savings of Rahul
\n= 9p – 5p
\n= 4p
\nExpenditure of Amit = 7q = 7p
\nRequired percentage = $\\frac{7p-4p}{4p}\\times$100
\n= $\\frac{3p}{4p}\\times$100
\n= 75% more
\nHence, the correct answer is Option D<\/p>\n
\n$\\Rightarrow$ $\\frac{a+b+c+d}{4}$ = 35
\n$\\Rightarrow$ a + b + c + d = 140\u2026\u2026\u2026\u2026(1)<\/p>\n
\n$\\Rightarrow$ $\\frac{c+d+e+f}{4}$ = 40
\n$\\Rightarrow$ c + d + e + f = 160\u2026\u2026\u2026\u2026(2)<\/p>\n
\n$\\Rightarrow$ $\\frac{a+f}{2}$ = 45
\n$\\Rightarrow$ a + f = 90\u2026\u2026\u2026\u2026(3)<\/p>\n
\n$\\Rightarrow$ $\\frac{b+e}{2}$ = 55
\n$\\Rightarrow$ b + e = 110\u2026\u2026\u2026\u2026(4)<\/p>\n
\na + b + c + d + c + d + e + f = 140 + 160
\n(a + f) + (b + e) + 2(c + d) = 300
\n90 + 110 + 2(c + d) = 300
\n2(c + d) = 100
\nc + d = 50\u2026\u2026\u2026\u2026..(5)<\/p>\n
\n= $\\frac{a+f+c+d}{4}$
\n= $\\frac{90+50}{4}$
\n= $\\frac{140}{4}$
\n= 35
\nHence, the correct answer is Option A<\/p>\n
\nMan spent 20% of his monthly income on house rent.
\nRemaining amount after spending on house rent = 80% of p
\n= $\\frac{80}{100}$p
\nAmount spent on shopping was \u20b99504.
\n22% of $\\frac{80}{100}$p = 9504
\n$\\frac{22}{100}\\times\\frac{80}{100}$p = 9504
\np = \u20b954000
\nAmount saved by the man = 54000 of (100-20)% of (100-22)% of (3\/4)
\n= 54000 of 80% of 78% of (3\/4)
\n= 54000$\\times\\frac{80}{100}\\times\\frac{78}{100}\\times\\frac{3}{4}$
\n= \u20b925272
\nHence, the correct answer is Option D<\/p>\n
\n$\\Rightarrow$ Sum of the weights of 45 students = 65 x 45 = 2925 kg
\nAverage weight of 51 students = 65 + 3 = 68 kg
\n$\\Rightarrow$ Sum of the weights of 51 students = 68 x 51 = 3468 kg
\nSum of weights of the 6 students who joined later = 3468 – 2925 = 543 kg
\n$\\therefore$ Average weight of the 6 students who joined later = $\\frac{543}{6}$ = 90.5 kg
\nHence, the correct answer is Option E<\/p>\n
\ny of (100+35)% of (100-16.67)% = 405
\n$y\\times1.35\\times\\frac{5}{6} = 405$
\n6.75y = 2430
\ny = 360
\n20% of the number of bears in the zoo two years ago = 20% of 360
\n= 72
\nHence, option e is the correct answer.<\/p>\n
\ntherefore , the new consumption will be\u00a0$\\ \\frac{\\ 1040}{150}$ = 6.93 (let us take approximate value as 7)<\/p>\n
\nRequired sale = Rs. [ (4500 x 6) – 26659 ]
\n= Rs. (27000 – 26659)
\n= Rs. 341<\/p>\n
\nTotal marks of 40 students = $67.875\\times40$
\n= 2715
\nIf two more students whose marks are (y-3) and (y+3) added then the average will be 67.5.
\n2715 + (y-3) + (y+3) = $67.5\\times42$
\n2715 + 2y = 2835
\n2y = 2835-2715
\n2y = 120
\ny = 60
\nHence, option b is the correct answer.<\/p>\n
\nTotal age of the group initially = $18\\times54$ = 972
\nIf after joining the two people in the group, the average age will be 52 years.
\nTotal age of the group after joining the two people = $20\\times52$ = 1040
\nSum of the age of those two people = 1040-972 = 68
\nTwo more people joined the group in which one\u2019s age is 12 years more than the other\u2019s age.
\nz + z+12 = 68
\n2z = 68-12 = 56
\nYoungest person among those people who joined the group later = z = 28
\nHence, option c is the correct answer.<\/p>\n
\n0.35P = 0.45Q – 82 Eq.(i)
\nSum of numbers P & Q is 1320.
\nP+Q = 1320
\nQ = 1320-P Eq.(ii)
\nPut Eq.(ii) in Eq.(i).
\n0.35P = 0.45(1320-P) – 82
\n0.35P = 594-0.45P – 82
\n0.35P+0.45P = 594 – 82
\n0.8P = 512
\nP = 640
\nPut the value of P in Eq.(ii).
\nQ = 1320-640 = 680
\nSum of the 55% of P and 25% of Q = 55% of 640 + 25% of 680
\n= 352+170
\n= 522
\nHence, option d is the correct answer.<\/p>\n