regisss/math_qa huggingface dataset:
Problem: in a restaurant , the profit is 140 % of the cost . if the cost increases by 12 % but the selling price remains constant , approximately what percentage of the selling price is the profit ?
Options: a ) 30 % , b ) 53 % , c ) 90 % , d ) 100 % , e ) none of these
Rationale: "explanation : let c . p . = rs . 100 . then , profit = rs . 140 , s . p . = rs . 240 . new c . p . = 112 % of rs . 100 = rs . 112 new s . p . = rs . 240 . profit = rs . ( 240 - 112 ) = rs . 128 . required percentage = ( 128 / 240 * 100 ) % = 53 % appox answer : b"
Answer: b
Problem: a hemispherical bowl is made of 0.5 cm thick steel . inner radius of the bowl is 4 cm . the volume of steel used in making the bowl is
Options: a ) 55.83 cm 3
b ) 56.83 cm 3
c ) 113.67 cm 3
d ) 58.83 cm 3
e ) none of these
Rationale: inner r = 4 outer r = 4.5 formula = 2 / 3 * 22 / 7 ( r ^ 2 - r ^ 2 ) ( 2 / 3 ) * ( 22 / 7 ) * ( 4.5 * 4.5 * 4.5 - 4 * 4 * 4 ) = 56.83 answer : b
Answer: b
Problem: a large box contains 18 small boxes and each small box contains 28 chocolate bars . how many chocolate bars are in the large box ?
Options: a ) a ) 220 , b ) b ) 490 , c ) c ) 380 , d ) d ) 450 , e ) e ) 504
Rationale: "the only thing you have to do is a simple multiply we have : 18 * 28 = 504 correct answer is : e ) 504"
Answer: e
Problem: if one root of the equation 2 x ^ 2 + 3 x – k = 0 is 5 , what is the value of k ?
Options: a ) 65 , b ) 79 , c ) 68 , d ) 87 , e ) 90
Rationale: "we just enter this root into the equation in order to recieve an equation to find the answer ! 2 * 5 ^ 2 + 3 * 5 - k = 0 k = 50 + 15 = 65 the answer is a"
Answer: a
Problem: from the salary of rahul , 20 % is deducted as house rent , 10 % of the rest he spends on children ’ s education and 10 % of the balance , he spends on clothes . after this expenditure he is left with rs 1377 . his salary is ?
Options: a ) rs 2125 , b ) rs 2040 , c ) rs 2100 , d ) rs 2200 , e ) rs 2,300
Rationale: suppose that his salary = rs 100 house rent = rs 20 , balance = rs 80 expenditure on education = rs ( 10 x 80 ) / 100 = rs 8 balance = rs 72 expenditure on clothes rs ( 10 x 72 ) / 100 = rs 7.2 balance now = rs 64.8 if balance is rs 64.8 , salary = rs 100 if balance is rs 1,377 , salary = rs ( 100 x 1377 ) / 64.8 rs 2,125 answer : a
Answer: a
Problem: the closest approximation of q = ( 69.28 × 0.004 ) / 0.03 is
Options: a ) 0.092 , b ) 0.92 , c ) 9.2 , d ) 92 , e ) 920
Rationale: "q = ( 69.28 × 0.004 ) / 0.03 1 . 0.004 = 4 × 10 ^ ( - 3 ) 2 . 0.03 = 3 × 10 ^ ( - 2 ) 3 . ( a × b ) / c = a × ( b / c ) 4 . 0.004 / 0.03 = 4 × 10 ^ ( - 3 ) / ( 3 × 10 ^ ( - 2 ) ) = 4 × 10 ^ ( - 3 - ( - 2 ) ) / 3 = 4 × 10 ^ ( - 1 ) / 3 = ( 4 / 3 ) × 10 ^ ( - 1 ) = 1.333 × 10 ^ ( - 1 ) therefore , ( 69.28 × 0.004 ) / 0.03 = 69.28 × ( 0.004 / 0.03 ) = 69.28 × 1.33 × 10 ^ ( - 1 ) = 69.28 × 1.33 / 10 = 6.928 * 1.33 now , 7 × 2 = 14 7 × 1 = 7 or better : 6.9 × 1 = 6.9 6.9 × 2 = 13.8 hence , 6.9 < 6.928 × 1.33 < 13.8 9.2 is the only answer that satisfies this condition . c"
Answer: c
Problem: in an office , totally there are 800 employees and 25 % of the total employees are males . 40 % of the males in the office are at - least 50 years old . find the number of males aged below 50 years ?
Options: a ) 120 , b ) 160 , c ) 132 , d ) 130 , e ) none of these
Rationale: "number of male employees = 800 * 25 / 100 = 200 required number of male employees who are less than 50 years old = 200 * ( 100 - 40 ) % = 200 * 60 / 100 = 120 . answer : a"
Answer: a
Problem: if a certain number x is divided by 82 , the reminder is 5 . what is the reminder when x + 13 is divided by 41 ?
Options: a ) 3 , b ) 18 , c ) 6 , d ) 16 , e ) 5
Rationale: "x can be written as 82 k + 5 or x = 5 , 87,169 , etc . x + 13 = 82 k + 5 + 13 = 82 k + 18 or x + 13 = 18,100 , 182 etc . when divided by 41 , we will get the remainder 18 . b"
Answer: b
Problem: find the sum lend at c . i . at 5 p . c per annum will amount to rs . 341 in 2 years ?
Options: a ) 221 , b ) 309 , c ) 400 , d ) 288 , e ) 171
Rationale: "explanation : 341 = p ( 21 / 20 ) 2 p = 309.29 answer : b"
Answer: b
Problem: a sum fetched total simple interest of 4016.25 at the rate of 10 p . c . p . a . in 5 years . what is the sum ?
Options: a ) 8925 , b ) 8032.5 , c ) 4462.5 , d ) 8900 , e ) none of these
Rationale: let the sums be p . now , 50 % of p = 4016.25 or , p = 8032.5 answer b
Answer: b
Problem: in a factory , an average of 50 tv ' s are produced per day for the fist 25 days of the months . a few workers fell ill for the next 5 days reducing the daily avg for the month to 45 sets / day . the average production per day for day last 5 days is ?
Options: a ) 20 , b ) 36 , c ) 48 , d ) 50 , e ) 59
Rationale: "production during these 5 days = total production in a month - production in first 25 days . = 30 x 45 - 25 x 50 = 100 ∴ average for last 5 days = 100 / 5 = 20 a"
Answer: a
Problem: the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 42 ?
Options: a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27
Rationale: "21 # 42 = 21 * 24 * 27 * 30 * 33 * 36 * 39 * 42 = 3 ^ 11 * 7 ^ 2 * 2 ^ 7 * 5 ^ 1 * 11 ^ 1 * 13 ^ 1 so total is = 11 + 2 + 7 + 1 + 1 + 1 = 23 answer a"
Answer: a
Problem: ( 1.2 ) ( power 3 ) - ( 0.9 ) ( power 3 ) / ( 1.2 ) ( power 2 ) + 1.08 + ( 0.9 ) ( power 2 ) is :
Options: a ) 0.68 , b ) 0.6 , c ) 0.8 , d ) 0.3 , e ) none of them
Rationale: "given expression = ( 1.2 ) ( power 3 ) - ( 0.9 ) ( power 3 ) / ( 1.2 ) ( power 2 ) + ( 1.2 x 0.9 ) + ( 0.9 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 1.2 - 0.9 ) = 0.3 answer is d"
Answer: d
Problem: a volume of 10996 l water is in a container of sphere . how many hemisphere of volume 4 l each will be required to transfer all the water into the small hemispheres ?
Options: a ) 2812 , b ) 8231 , c ) 2734 , d ) 4222 , e ) 2749
Rationale: "a volume of 4 l can be kept in 1 hemisphere therefore , a volume of 10996 l can be kept in ( 10996 / 4 ) hemispheres ans . 2749 answer : e"
Answer: e
Problem: linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 200 , what were her original savings ?
Options: a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 900
Rationale: "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 200 . so 1 / 4 of her savings is $ 200 . so her original savings are 4 times $ 200 = $ 800 correct answer d"
Answer: d
Problem: a lent rs . 4000 to b for 2 years and rs . 2000 to c for 4 years on simple interest at the same rate of interest and received rs . 2200 in all from both of them as interest . the rate of interest per annum is :
Options: a ) 14 % , b ) 15 % , c ) 12 % , d ) 13.75 % , e ) 14.56 %
Rationale: explanation : let the rate of interest per annum be r % simple interest for rs . 4000 for 2 years at r % + simple interest for rs . 2000 for 4 years at r % = 2200 ( 4000 × r × 2 ) / 100 + ( 2000 × r × 4 ) / 100 = 2200 80 r + 80 r = 2200 160 r = 2200 16 r = 220 4 r = 55 r = 55 / 4 = 13.75 % answer : option d
Answer: d
Problem: what is the 23 th digit to the right of the decimal point in the decimal form of 36 / 66 ?
Options: a ) 3 , b ) 4 , c ) 8 , d ) 9 , e ) 6
Rationale: "we first have to use some long division . this long division allows us to get 36 / 66 in decimal form , which is 0.545454 … where “ 54 ” is repeating . we can see that the 1 st , 3 rd , 5 th digit to the right of the decimal point is a 5 and that the 2 nd , 4 th , 6 th digit to the right of the decimal point is a 4 . in other words , each odd - positioned digit is a 5 , and each even - positioned digit is a 4 . then about the 10 digit to the right of the decimal point and we see that 10 is even , we know that the 10 th digit is a 4 . answer b ."
Answer: b
Problem: a man can row his boat with the stream at 6 km / h and against the stream in 2 km / h . the man ' s rate is ?
Options: a ) 1 kmph , b ) 2 kmph , c ) 5 kmph , d ) 7 kmph , e ) 8 kmph
Rationale: "ds = 6 us = 2 s = ? s = ( 6 - 2 ) / 2 = 2 kmph answer : b"
Answer: b
Problem: a mixture contains alcohol and water in the ratio 4 : 3 . if 5 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture .
Options: a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20
Rationale: "explanation : let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 4 x / ( 3 x + 5 ) = 4 / 5 20 x = 4 ( 3 x + 5 ) 8 x = 20 x = 2.5 quantity of alcohol = ( 4 x 2.5 ) litres = 10 litres . answer : a"
Answer: a
Problem: if the l . c . m of two numbers is 600 and their product is 18000 , find the h . c . f of the numbers .
Options: a ) 50 , b ) 30 , c ) 125 , d ) 25 , e ) none of these
Rationale: "h . c . f = ( product of the numbers ) / ( their l . c . m ) = 18000 / 600 = 30 . answer : b"
Answer: b
Problem: working at a constant rate , p can finish a job in 3 hours . q , also working at a constant rate , can finish the same job in 18 hours . if they work together for 2 hours , how many more minutes will it take p to finish the job , working alone at his constant rate ?
Options: a ) 32 , b ) 36 , c ) 40 , d ) 44 , e ) 48
Rationale: "each hour they complete 1 / 3 + 1 / 18 = 7 / 18 of the job . in 2 hours , they complete 2 ( 7 / 18 ) = 7 / 9 of the job . the time for p to finish is ( 2 / 9 ) / ( 1 / 3 ) = ( 2 / 3 ) hour = 40 minutes the answer is c ."
Answer: c
Problem: at a restaurant , glasses are stored in two different - sized boxes . one box contains 12 glasses , and the other contains 16 glasses . if the average number of glasses per box is 15 , and there are 16 more of the larger boxes , what is the total number of glasses q at the restaurant ? ( assume that all boxes are filled to capacity . )
Options: a ) 96 , b ) 240 , c ) q = 256 , d ) q = 384 , e ) q = 480
Rationale: "most test takers would recognize the system of equations in this prompt and just do algebra to get to the solution ( and that ' s fine ) . the wording of the prompt and the ' spread ' of the answer choices actually provide an interesting ' brute force ' shortcut that you can take advantage of to eliminate the 4 wrong answers . . . . we ' re told that there are 2 types of boxes : those that hold 12 glasses and those that hold 16 glasses . since the average number of boxes is 15 , we know that there must be at least some of each . we ' re also told that that there are 16 more of the larger boxes . this means , at the minimum , we have . . . 1 small box and 17 large boxes = 1 ( 12 ) + 17 ( 16 ) = 12 + 272 = 284 glasses at the minimum since the question asks for the total number of glasses , we can now eliminate answers a , b and c . . . . the difference in the number of boxes must be 16 though , so we could have . . . . 2 small boxes and 18 large boxes 3 small boxes and 19 large boxes etc . with every additional small box + large box that we add , we add 12 + 16 = 28 more glasses . thus , we can justadd 28 suntil we hit the correct answer . . . . 284 + 28 = 312 312 + 28 = 340 340 + 28 = 368 368 + 28 = 396 at this point , we ' ve ' gone past ' answer d , so the correct answer must be answer e . . . . . but here ' s the proof . . . . 396 + 28 = 424 424 + 28 = 452 452 + 28 = 480 final answer : e"
Answer: e
Problem: tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 10 meters and a circumference of 7 meters , and the interior of tank b has a height of 7 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ?
Options: a ) 70 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 %
Rationale: "the radius of tank a is 7 / ( 2 * pi ) . the capacity of tank a is 10 * pi * 49 / ( 4 * pi ^ 2 ) = 245 / ( 2 * pi ) the radius of tank b is 10 / ( 2 * pi ) . the capacity of tank b is 7 * pi * 100 / ( 4 * pi ^ 2 ) = 350 / ( 2 * pi ) tank a / tank b = 245 / 350 = 7 / 10 = 70 % the answer is a ."
Answer: a
Problem: car a runs at the speed of 50 km / hr and reaches its destination in 8 hours . car b runs at the speed of 70 km / h and reaches its destination in 4 hours . what is the ratio of distances covered by car a and car b ?
Options: a ) 3 : 7 , b ) 4 : 9 , c ) 10 : 7 , d ) 5 : 7 , e ) 6 : 11
Rationale: "car a travels 50 × 8 = 400 km car b travels 70 × 4 = 280 km the ratio is 400 : 280 = 40 : 28 = 10 : 7 the answer is c ."
Answer: c
Problem: a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 12700 after a year ?
Options: a ) 3630 , b ) 9232 , c ) 8828 , d ) 2387 , e ) 3810
Rationale: "6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12700 = 3810 answer : e"
Answer: e
Problem: sum of two numbers is 25 and their difference is 13 . find their product .
Options: a ) 104 , b ) 108 , c ) 114 , d ) 124 , e ) 142
Rationale: "explanation : friends , this sort of question is quite important in competitive exams , whenever any question come which have relation between sum , product and difference , this formula do the magic : = > ( x + y ) 2 = ( x − y ) 2 + 4 xy < = > ( 25 ) 2 = ( 13 ) 2 + 4 xy < = > 4 xy = ( 25 ) 2 − ( 13 ) 2 < = > xy = 456 / 4 = 114 option c"
Answer: c
Problem: if 13 = 13 w / ( 1 - w ) , then ( 3 w ) 2 =
Options: a ) 1 / 4 , b ) 1 / 2 , c ) 1 , d ) 2 , e ) 3
Rationale: "13 - 13 w = 13 w 26 w = 13 w = 1 / 2 3 w = 3 / 2 3 w * 2 = 3 / 2 * 2 = 3 answer : e"
Answer: e
Problem: john ' s bank ' s saving amount is decreased 12 % due to loan payment and current balance is rs . 44000 . find the actual balance before deduction ?
Options: a ) 8000 , b ) 50000 , c ) 9000 , d ) 9500 , e ) 10000
Rationale: 12 % decreased 88 % balance = 44000 100 % = 44000 / 88 * 100 = 50000 answer : b
Answer: b
Problem: a 280 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ?
Options: a ) 230 , b ) 288 , c ) 267 , d ) 256 , e ) 220
Rationale: "relative speed = 120 + 80 = 200 km / hr . = 200 * 5 / 18 = 500 / 9 m / sec . let the length of the other train be x m . then , ( x + 280 ) / 9 = 500 / 9 = > x = 220 . answer : e"
Answer: e
Problem: what is the sum of all remainders obtained when the first 300 natural numbers are divided by 9 ?
Options: a ) 397 , b ) 401 , c ) 403 , d ) 405 , e ) 399
Rationale: "a positive integer can give only the following 9 remainders when divided by 9 : 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , and 0 . 1 divided by 9 gives the remainder of 1 ; 2 divided by 9 gives the remainder of 2 ; . . . 8 divided by 9 gives the remainder of 8 ; 9 divided by 9 gives the remainder of 0 . we ' ll have 11 such blocks , since 99 / 9 = 11 . the last will be : 91 divided by 9 gives the remainder of 1 ; 92 divided by 9 gives the remainder of 2 ; . . . 98 divided by 9 gives the remainder of 8 ; 99 divided by 9 gives the remainder of 0 . the last number , 100 , gives the remainder of 1 when divided by 9 , thus the sum of all remainders will be : 11 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 0 ) + 1 = 405 . answer : d ."
Answer: d
Problem: an escalator moves towards the top level at the rate of 12 ft . sec and its length is 150 feet . if a person walks on the moving escalator at the rate of 3 feet per second towards the top level , how much time does he take to cover the entire length .
Options: a ) 10 sec , b ) 20 sec , c ) 15 sec , d ) 16 sec , e ) 14 sec
Rationale: "explanation : time taken to cover the entire length = tot . dist / resultant speed = 150 / ( 12 + 3 ) = 10 sec answer : a"
Answer: a
Problem: the area of rectangular field of length 90 meters and breadth 80 meters is equal to the area of a square plot . what will be the length of the diagonal of the square plot ?
Options: a ) 100 m
b ) 120 m
c ) 95 m
d ) 80 m
e ) 150 m
Rationale: given that the length of the rectangular field l = 90 m and breadth = b = 80 m . then its area lb = 90 x 80 sq . m we know that the area of square = ( 1 / 2 ) x ( diagonal ) 2 from the given question , we have area of square plot = area of rectangular field that is , 90 x 80 = 1 / 2 x diagonal 2 diagonal ^ 2 = 2 x 7200 = 14400 = ( 120 ) 2 diagonal = 120 m . hence b
Answer: b
Problem: in a 100 m sprint race kelly was given a 3 m head start by abel . if abel lost the race to kelly by a distance of 0.50 m , how many meters more would abel need to overtake kelly ?
Options: a ) 20.5 , b ) 18.4 , c ) 17.5 , d ) 19.9 , e ) 21.5
Rationale: distance abel ran before completion of race - - - 100 m - 0.50 m = 99.5 m distance gained on kelly over 99.5 m - - - 3 m - 0.50 m = 2.50 m then abel gains 99.5 m / 2.50 m = 1 m on kelly every 39.8 meters . therefore 39.8 divided by half a meter ( i . e the 0.50 m abel was defeated with ) answer : d . 19.9 m
Answer: d
Problem: the volume of a wall , 5 times as high as it is broad and 8 times as long as it is high , is 12.8 cu . meters . find the breadth of the wall .
Options: a ) 40
b ) 29
c ) 28
d ) 27
e ) 12
Rationale: explanation : let the breadth of the wall be x metres . then , height = 5 x metres and length = 40 x metres . x * 5 x * 40 x = 12.8 = > \ inline \ fn _ jvn x ^ 3 = \ frac { 12.8 } { 200 } = \ frac { 128 } { 2000 } = \ frac { 64 } { 1000 } = > \ inline \ fn _ jvn x = \ frac { 4 } { 10 } m = > \ inline \ fn _ jvn x = \ frac { 4 } { 10 } \ times 100 = 40 cm answer : a ) 40 cm
Answer: a
Problem: a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 % of the distribution lies one standard deviation h of the mean , what percent of the distribution is less than m + h ?
Options: a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 %
Rationale: "16 % ________________________________________________ m + h 34 % ________________________________________________ m 34 % ________________________________________________ m - h 16 % since 68 % lies one standard deviation from mean m , = > 50 % of 68 % lies on either side as it is symmetric about m . thus 16 % lie below m - h and 16 % lie above m + h now below m + h = 16 + 34 + 34 = 84 % hence d"
Answer: d
Problem: what is the 4 digit number whose second digit is thrice the first digit and 3 rd digit is sum of 1 st and 2 nd and last digit is thrice the second digit .
Options: a ) 2674 , b ) 1349 , c ) 3343 , d ) 3678 , e ) 3679
Rationale: let four digits is a , b , c & d . . . so by given b is 3 a c is 4 a d is 9 a so ans is 1349 answer : b
Answer: b
Problem: how many of the positive factors of 20 are not factors of 24
Options: a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 5
Rationale: "factors of 20 - 1 , 2 , 4,5 , 10 , 20 factors of 24 - 1 , 2 , 3 , 4 , 6 , 8 , 12 and 24 comparing both , we have 3 factors of 20 which are not factors of 24 - 5 , 10,20 answer : b"
Answer: b
Problem: rs . 525 among a , b and c such that b may get 2 / 3 of a and c together get . find the share of c ?
Options: a ) 127 , b ) 150 , c ) 177 , d ) 167 , e ) 122
Rationale: a + b + c = 590 5 a = 6 b = 8 c = x a : b : c = 1 / 5 : 1 / 6 : 1 / 8 = 24 : 20 : 15 15 / 59 * 590 = rs . 150 answer : b
Answer: b
Problem: kamal obtained 81 , 68 , 82 , 69 and 90 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ?
Options: a ) 65 , b ) 69 , c ) 72 , d ) 75 , e ) none
Rationale: "sol . average = 81 + 68 + 82 + 69 + 90 / 5 ) = ( 390 / 5 ) = 78 . answer e"
Answer: e
Problem: a farmer spent $ 33 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 20 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have spent on the chicken feed and goat feed combined ?
Options: a ) $ 36.30 , b ) $ 38.50 , c ) $ 39.20 , d ) $ 39.50 , e ) $ 40.60
Rationale: "a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 33 = $ 13.2 on chicken feed , thus he spent the remaining 33 - 13.2 = $ 19.8 on goat feed . now , since he bought chicken feed at a 20 % discount then the original price of it was x * 0.8 = $ 13.2 - - > x = $ 16.5 . therefore if the farmer had paid full price for both the chicken feed and the goat feed , then he would he have spent 16.5 + 19.8 = $ 36.3 . answer : a"
Answer: a
Problem: the difference in compound interest earned on a deposit ( compounded annually ) in year 1 and year 2 is $ 40 . had the interest rate been 3 times its present value , the difference c would have been how much ?
Options: a ) 40 / 3 , b ) 40 , c ) 120 , d ) 360 , e ) 420
Rationale: case 1 : deposit = $ x ; rate of increase = r . interest yearned in 1 year = xr . deposit in 1 year = x + xr . interest yearned in 2 year = ( x + xr ) r . the difference c = ( x + xr ) r - xr = xr ^ 2 = 40 . case 2 : deposit = $ x ; rate of increase = 3 r . interest yearned in 1 year = x ( 3 r ) . deposit in 1 year = x + 3 xr . interest yearned in 2 year = ( x + 3 xr ) 3 r . the difference = ( x + 3 xr ) 3 r - 3 xr = 9 xr ^ 2 . since from case 1 we know that xr ^ 2 = 40 , then 9 xr ^ 2 = 9 * 40 = 360 . answer : d .
Answer: d
Problem: the sum of the squares of two consecutive positive integers exceeds their product by 91 . find the integers ?
Options: a ) 9 , 10 , b ) 10 , 11 , c ) 11 , 12 , d ) 12 , 13 , e ) none of these
Rationale: let the two consecutive positive integers be x and x + 1 x 2 + ( x + 1 ) 2 - x ( x + 1 ) = 91 x 2 + x - 90 = 0 ( x + 10 ) ( x - 9 ) = 0 = > x = - 10 or 9 . as x is positive x = 9 hence the two consecutive positive integers are 9 and 10 . answer : a
Answer: a
Problem: in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 48 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ?
Options: a ) 36 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 38 , e ) 47
Rationale: working backward from the day it ' s covered : day 48 : fully covered day 47 : half covered so 47 days answer : e
Answer: e
Problem: find compound interest on rs . 5000 at 12 % per annum for 1 year , compounded half - yearly .
Options: a ) rs . 600 , b ) rs . 618 , c ) rs . 525 , d ) rs . 620 , e ) none
Rationale: "principal = rs . 5000 , rate = 6 % per half year time = 1 year = 2 half - years amount = rs . { 5000 x ( 1 + 6 / 100 ) 2 ] = rs . [ 5000 x 53 / 50 x 53 / 50 ] = rs . 5618 c . i . = rs . ( 5618 – 5000 ) = rs . 618 answer b"
Answer: b
Problem: if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 35 men in 3 days ?
Options: a ) 16 mtr . , b ) 52 mtr , c ) 70 mtr . , d ) 78 mtr . , e ) 98 mtr .
Rationale: "20 men is 6 days can build 112 metres 35 men in 3 days can build = 112 * ( 35 / 20 ) x ( 3 / 6 ) = 98 meters answer : e ."
Answer: e
Problem: how many of the positive factors of 85 , 105 and how many common factors are there in numbers ?
Options: a ) 5 , b ) 4 , c ) 2 , d ) 3 , e ) 1
Rationale: "factors of 85 - 1 , 5 , 17 , 85 factors of 105 - 1 , 3 , 5 , 7 , 15 , 21 , 35 , 105 comparing both , we have two common factors of 85 and 105 - 1,5 answer ( c )"
Answer: c
Problem: sakshi can do a piece of work in 20 days . tanya is 25 % more efficient than sakshi . the number of days taken by tanya to do the same piece of work is ?
Options: a ) 4 days , b ) 6 days , c ) 10 days , d ) 16 days , e ) 22 days
Rationale: ratio of times taken by sakshi and tanya = 125 : 100 = 5 : 4 . suppose tanya takes x days to do the work . 5 : 4 : : 20 : x x = 4 x 20 / 5 x = 16 days . hence , tanya takes 16 days to complete the work . answer : d
Answer: d
Problem: if a / b = 1 / 3 , b / c = 2 , c / d = 1 / 2 , d / e = 3 and e / f = 1 / 6 , then what is the value of abc / def ?
Options: a ) 27 / 4 , b ) 27 / 8 , c ) 3 / 4 , d ) 3 / 8 , e ) 1 / 4
Rationale: "say a = 2 . then : a / b = 1 / 3 - - > b = 6 ; b / c = 2 - - > c = 3 ; c / d = 1 / 2 - - > d = 6 ; d / e = 3 - - > e = 2 ; e / f = 1 / 6 - - > f = 12 . abc / def = ( 2 * 6 * 3 ) / ( 6 * 2 * 12 ) = 1 / 4 . answer : e ."
Answer: e
Problem: a and b invests rs . 4000 and rs . 4000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ?
Options: a ) 3 : 5 , b ) 3 : 8 , c ) 3 : 2 , d ) 9 : 9 , e ) 3 : 1
Rationale: "( 4 * 6 + 8 * 6 ) : ( 4 * 12 ) 72 : 48 = > 3 : 2 answer : c"
Answer: c
Problem: find the value of x ? ( 2 / 7 ) ^ - 8 * ( 7 / 2 ) ^ - 2 = ( 2 / 7 ) ^ x
Options: a ) - 5 , b ) - 6 , c ) 4 , d ) 0 , e ) - 4
Rationale: ( 2 / 7 ) ^ - 8 * ( 7 / 2 ) ^ - 2 = ( 2 / 7 ) ^ x now , ( 7 / 2 ) ^ 8 * ( 2 / 7 ) ^ 2 = ( 2 / 7 ) ^ x on solving , ( 7 / 2 ) ^ 6 = ( 2 / 7 ) ^ x ( 2 / 7 ) ^ - 6 = ( 2 / 7 ) ^ x thus , x = - 6 . answer : b
Answer: b
Problem: in a maths test , students were asked to find 5 / 16 of a certain number . one of the students by mistake found 5 / 6 th of that number and his answer was 200 more than the correct answer . find the number .
Options: a ) 384 , b ) 76 , c ) 26 , d ) 28 , e ) 11
Rationale: "explanation : let the number be x . 5 * x / 6 = 5 * x / 16 + 200 25 * x / 48 = 200 x = 384 answer : a"
Answer: a
Problem: how many seconds does sandy take to cover a distance of 500 meters , if sandy runs at a speed of 18 km / hr ?
Options: a ) 80 , b ) 100 , c ) 120 , d ) 140 , e ) 160
Rationale: "18 km / hr = 18000 m / 3600 s = 5 m / s time = 500 / 5 = 100 seconds the answer is b ."
Answer: b
Problem: 8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 10 men , 4 women and 18 children together complete the piece of work ?
Options: a ) 5 days , b ) 15 days , c ) 28 days , d ) 6 days , e ) 7 days
Rationale: "explanation : 1 man ’ s 1 day ’ s work = 1 / 8 × 12 = 1 / 96 10 men ’ s 1 day ’ s work = 1 × 10 / 96 = 5 / 48 1 woman ’ s 1 day ’ s work = 1 / 192 4 women ’ s 1 day ’ s work = 1 / 192 × 4 = 1 / 48 1 child ’ s 1 day ’ s work = 1 / 240 18 children ’ s 1 day ’ s work = 3 / 40 therefore , ( 10 men + 4 women + 18 children ) ’ s 1 day ’ s work = 5 / 48 + 1 / 48 + 3 / 40 = 1 / 5 the required no . of days = 5 days answer : option a"
Answer: a
Problem: the circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 35 ft long . what is the distance traveled by the cart , when the front wheel has done five more revolutions than the rear wheel ?
Options: a ) 20 ft , b ) 25 ft , c ) 750 ft , d ) 900 ft , e ) 1050 ft
Rationale: "point to note : both the wheels would have traveled the same distance . now consider , no . of revolutions made by back wheel as x , which implies that the number of revolutions made by the front wheel is ( x + 5 ) . equating the distance traveled by front wheel to back wheel : ( x + 5 ) * 30 = x * 35 . ( formula for calculating the distance traveled by each wheel is : # of revolutions * circumference . ) solving this eqn . gives x = 30 sub x = 30 either in ( x + 5 ) * 30 or in x * 35 to get the distance , which is 1050 . so the correct choice is e ."
Answer: e
Problem: a frog can climb up a well at 3 ft per min but due to slipperiness of the well , frog slips down 2 ft before it starts climbing the next minute . if the depth of the well is 57 ft , how much time will the frog take to reach the top ?
Options: a ) 29 , b ) 27 , c ) 55 , d ) 17 , e ) 10
Rationale: explanation : as per given , in 1 min , frog climbs up 3 ft and slips down by 2 ft . so the frog climbs only 1 ft in 1 min so after 54 mins , it would have climbed 54 ft . at the end of 55 mins it climbs up 3 ft to make it 57 ft and come out of the well . once it had reached the destination , it will not slip . so the frog will take only 55 minutes to climb up the well . answer : c
Answer: c
Problem: if 50 % of x is 20 less than 5 % of 500 , then x is ?
Options: a ) 50 , b ) 100 , c ) 75 , d ) 25 , e ) 10
Rationale: 50 % of x = x / 2 ; 5 % of 500 = 5 / 100 * 500 = 25 given that , x / 2 = 25 - 20 = > x / 2 = 5 = > x = 10 . answer : e
Answer: e
Problem: a rectangular room 14 m long , 12 m broad is surrounded by a varandah , 3 m wide . find the area of the varandah ?
Options: a ) 170
b ) 180
c ) 190
d ) 210
e ) 150
Rationale: area of varandah = ( l + b + 2 p ) 2 p = ( 14 + 12 + 6 ) 6 = 180 m ( power ) 2 answer is b .
Answer: b
Problem: calculate the largest 4 digit number which is exactly divisible by 98 ?
Options: a ) 9800 , b ) 9898 , c ) 9702 , d ) 9604 , e ) 9996
Rationale: largest 4 digit number is 9999 after doing 9999 ÷ 98 we get remainder 3 hence largest 4 digit number exactly divisible by 98 = 9999 - 3 = 9996 e
Answer: e
Problem: 20 6 years ago , a father was 3 times as old as his son . now the father is only twice as old as his son . then the sum of the present ages of the son and the father is :
Options: a ) 120 , b ) 156 , c ) 108 , d ) 105 , e ) 86
Rationale: let the present ages of the father and son be 2 x and x years respectively . then , ( 2 x - 26 ) = 3 ( x - 26 ) x = 52 . required sum = ( 2 x + x ) = 3 x = 156 years . answer is b
Answer: b
Problem: the compound and the simple interests on a certain sum at the same rate of interest for two years are rs . 11730 and rs . 10200 respectively . find the sum ?
Options: a ) rs . 17029 , b ) rs . 17028 , c ) rs . 17000 , d ) rs . 17008 , e ) rs . 17067
Rationale: "the simple interest for the first year is 10200 / 2 is rs . 5100 and compound interest for first year also is rs . 5100 . the compound interest for second year on rs . 5100 for one year so rate of the interest = ( 100 * 1530 ) / ( 5100 * 1 ) = 30 % p . a . so p = ( 100 * 10200 ) / ( 30 * 2 ) = rs . 17000 answer : c"
Answer: c
Problem: x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 6 will be equal to :
Options: a ) 3 , b ) 6 , c ) 1 / 4 , d ) 1 / 3 , e ) 9
Rationale: "explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 6 ^ 2 = 1 / 4 answer : c"
Answer: c
Problem: each child has 3 crayons and 12 apples . if there are 6 children , how many crayons are there in total ?
Options: a ) 22 , b ) 65 , c ) 18 , d ) 36 , e ) 10
Rationale: 3 * 6 = 18 . answer is c .
Answer: c
Problem: a man can row 6 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ?
Options: a ) 5.79 , b ) 5.77 , c ) 5.76 , d ) 5.72 , e ) 5.71
Rationale: "m = 6 s = 1.2 ds = 7.2 us = 4.8 x / 7.2 + x / 4.8 = 1 x = 2.88 d = 2.88 * 2 = 5.76 . answer : c"
Answer: c
Problem: if the l . c . m of two numbers is 660 and their product is 18480 , find the h . c . f of the numbers .
Options: a ) 50 , b ) 30 , c ) 125 , d ) 25 , e ) none of these
Rationale: h . c . f = ( product of the numbers ) / ( their l . c . m ) = 18480 / 660 = 28 . answer : e
Answer: e
Problem: a train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds . its length is :
Options: a ) 100 m , b ) 125 m , c ) 130 m , d ) 150 m , e ) 160 m
Rationale: "explanation : let the length of the train be x meters and its speed be y m / sec . they , x / y = 15 = > y = x / 15 x + 100 / 25 = x / 15 x = 150 m . answer is d"
Answer: d
Problem: show timer statistics to apply for the position of photographer at a local magazine , veronica needs to include 3 or 4 of her pictures in an envelope accompanying her application . if she has pre - selected 5 photos representative of her work , how many choices does she have to provide the photos for the magazine ?
Options: a ) 5 , b ) 10 , c ) 12 , d ) 15 , e ) 50
Rationale: veronica can choose 3 photos in 5 c 3 = 10 ways she can choose 4 photos in 5 c 4 = 5 ways total number of ways = 10 + 5 = 15 answer : d
Answer: d
Problem: a train 300 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ?
Options: a ) 298 m , b ) 268 m , c ) 500 m , d ) 267 m , e ) 1000 m
Rationale: "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 300 + x ) / 60 = 65 / 3 x = 1000 m . answer : e"
Answer: e
Problem: a consignment of 20 picture tubes contains 5 defectives . two tubes are selected one after the other at random . the probability that both are defective assuming that the first tube is not replaced before drawing the second , would be :
Options: a ) 1 / 16 , b ) 1 / 19 , c ) 1 / 4 , d ) 1 / 3 , e ) none of these
Rationale: probability of drawing a defective tube in first draw is 5 / 20 probability of drawing a defective tube in second draw ( without replacing first tube ) is 4 / 19 . therefore probability of getting both defective is = ( 5 / 20 ) * ( 4 / 19 ) = 1 / 19 answer : b
Answer: b
Problem: two goods trains each 125 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ?
Options: a ) 40 , b ) 12 , c ) 48 , d ) 51 , e ) 44
Rationale: "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 125 + 125 = 250 m . required time = 250 * 6 / 125 = 12 sec . answer : option b"
Answer: b
Problem: a sum of $ 500 amounts to 600 in 2 years at simple interest . if the interest rate is increased by 10 % it would amount to how much ?
Options: a ) $ 500 , b ) $ 700 , c ) $ 800 , d ) $ 600 , e ) $ 300
Rationale: s . i = 600 - 500 = 100 p = $ 500 t = 2 years r = 100 * 100 / 500 * 2 = 10 % new rate = 10 + 10 = 20 % new s . i . = 500 * 20 * 2 / 100 = $ 200 new amount = 500 + 200 = $ 700 answer is b
Answer: b
Problem: if p = 775 × 778 × 781 , what is the remainder when p is divided by 14 ?
Options: a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
Rationale: "p 775 / 14 leaves a remainder 5 778 / 14 leaves a remainder 8 781 / 14 leaves a remainder 11 5 * 8 * 11 = 440 so the remainder will be the remainder of 440 / 14 which is 6 ans a"
Answer: a
Problem: the present ratio of students to teachers at a certain school is 70 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ?
Options: a ) 2 , b ) 8 , c ) 10 , d ) 12 , e ) 15
Rationale: "we are given that the ratio of students to teacher is 70 to 1 . we can rewrite this using variable multipliers . students : teachers = 70 x : x we are next given that student enrollment increases by 50 and the number of teachers increases by 5 . with this change the new ratio becomes 25 to 1 . we can put all this into an equation : students / teachers  25 / 1 = ( 70 x + 50 ) / ( x + 5 ) if we cross multiply we have : 25 ( x + 5 ) = 70 x + 50 25 x + 125 = 70 x + 50 1.667 = x x ~ 2 since x is the present number of teachers , currently there are 2 teachers . answer a ."
Answer: a
Problem: find out the square of a number which when doubled exceeds its one fifth by 9 ?
Options: a ) 16 , b ) 25 , c ) 19 , d ) 26 , e ) 17
Rationale: a let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p / 5 ) + 9 = > 10 p = p + 45 = > p = 5 p ^ 2 = 5 ^ 2 = 25 . answer : b
Answer: b
Problem: a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 18 , what is the total amount ?
Options: a ) 115 , b ) 116 , c ) 117 , d ) 78 , e ) 119
Rationale: "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 18 39 - - - ? = > 78 answer : d"
Answer: d
Problem: the l . c . m . of 2 numbers is 30 . the numbers are in the ratio 2 : 3 . find their sum ?
Options: a ) 36 , b ) 25 , c ) 48 , d ) 32 , e ) 56
Rationale: "let the numbers be 2 x and 3 x l . c . m . = 6 x 6 x = 30 x = 5 the numbers are = 10 and 15 required sum = 10 + 15 = 25 answer is b"
Answer: b
Problem: if 15 horses eat 15 bags of gram in 15 days , in how many days will one horse eat one bag of grain ?
Options: a ) 15 days , b ) 1 / 15 days , c ) 1 day , d ) 30 days , e ) 225 days
Rationale: horse * days = bags 15 * 15 = 15 and 1 * days = 1 ( 15 * 15 ) / ( 1 * days ) = 15 / 1 days = 15 answer : a
Answer: a
Problem: in a certain game , each player scores either 2 points or 5 points . if n players score 2 points and m players score 5 points , and the total number of points scored is 50 , what is the least possible positive t difference between n and m ?
Options: a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9
Rationale: "we have equation 2 n + 5 m = 50 we have factor 2 in first number and we have factor 5 in second number . lcm ( 2 , 5 ) = 10 so we can try some numbers and we should start from 5 because it will be less list than for 2 2 * 5 = 10 and n should be equal 20 4 * 5 = 20 and n should be equal 15 6 * 5 = 30 and n should be equal 10 8 * 5 = 40 and n should be equal 5 10 * 5 = 50 and n should be equal 0 third variant give us the mininal difference n - m = 10 - 6 = 4 and there is some mistake in my way of thinking because we do n ' t have such answer ) if we change the task and will seek for difference between m and n than minimal result t will be 8 - 5 = 3 and answer b"
Answer: b
Problem: if the average of 201 , 202 , 204 , 205 , 206 , 209 , 209 , 210 , 212 and x is 207 , what is the value of x ?
Options: a ) 207 , b ) 209 , c ) 211 , d ) 212 , e ) 213
Rationale: sum of the deviations of the numbers in the set from the mean is always zero 201 , 202 , 204 , 205 , 206 , 209 , 209 , 210 , 212 mean is 207 so the list is - 6 - 5 - 3 - 2 - 1 + 2 + 2 + 3 + 5 . . . this shud total to zero but this is - 5 , hence we need a number that is 5 more than the mean to get a + 5 and make it zero hence the answer is 207 + 5 = 212 d
Answer: d
Problem: in a group of cows and chickens , the number of legs was 20 more than twice the number of heads . the number of cows was :
Options: a ) 5 , b ) 7 , c ) 10 , d ) 12 , e ) 14
Rationale: "let the number of cows be x and their legs be 4 x . let the number of chicken be y and their legs be 2 x . total number of legs = 4 x + 2 y . total number of heads = x + y . the number of legs was 20 more than twice the number of heads . therefore , 2 × ( x + y ) + 20 = 4 x + 2 y . or , 2 x + 2 y + 20 = 4 x + 2 y . or , 2 x + 20 = 4 x [ subtracting 2 y from both sides ] . or , 20 = 4 x – 2 x [ subtracting 2 x from both sides ] . or , 20 = 2 x . or , x = 10 [ dividing by 2 on both sides ] . therefore , the number of cows = 10 . correct answer : c ) 10"
Answer: c
Problem: the average marks in mathematics scored by the pupils of a school at the public examination were 39 . if 4 of these pupils who actually scored 25 , 12 , 15 and 19 marks at the examination had not been sent up , the average marks for the school would have been 44 . find the number of pupils sent up for examination from the school ?
Options: a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24
Rationale: 39 x = 25 + 12 + 15 + 19 + ( x – 4 ) 44 39 x = 71 + 44 x - 176 5 x = 105 x = 21 answer : b
Answer: b
Problem: amanda sees a sale for 30 % off all items , she sees a dress on sale that originally cost $ 50 . how much will it cost amanda to buy the dress after the sale amount of 30 % has been take off ?
Options: a ) $ 40 , b ) $ 35 , c ) $ 50 , d ) $ 65 , e ) $ 15
Rationale: final number = original number - 30 % ( original number ) = 50 - 30 % ( 50 ) = 50 - 15 = $ 35 . answer b
Answer: b
Problem: a cricket bat is sold for $ 900 , making a profit of $ 150 . the profit percentage would be
Options: a ) 24 % , b ) 20 % , c ) 30 % , d ) 36 % , e ) 40 %
Rationale: 150 / ( 900 - 150 ) = 150 / 750 = 0.2 = 20 % . answer : b
Answer: b
Problem: of the 3,600 employees of company x , 1 / 3 are clerical . if the clerical staff were to be reduced by 1 / 4 , what percent of the total number of the remaining employees would then be clerical ?
Options: a ) 25 % , b ) 28.1 % , c ) 20 % , d ) 12.5 % , e ) 11.1 %
Rationale: "welcome , just post the question and the choices let ' s see , the way i did it was 1 / 3 are clerical out of 3600 so 1200 are clerical 1200 reduced by 1 / 4 is 1200 * 1 / 4 so it reduced 300 people , so there is 900 clerical people left but since 300 people left , it also reduced from the total of 3600 so there are 3300 people total since 900 clerical left / 3200 people total answer : b"
Answer: b
Problem: raja spends 60 % of his monthly income on household items , 10 % of his monthly income on buying cloths , 10 % of his monthly income on medicines and saves the remaining amount which is rs . 5000 . find his monthly income .
Options: a ) rs . 40000 , b ) rs . 36000 , c ) rs . 50000 , d ) rs . 25000 , e ) none of these
Rationale: "savings 20 % - 5000 expenditure 80 % - 20000 total - 25000 answer : d"
Answer: d
Problem: 22.085 ã — 0.001 = ?
Options: a ) 0.22085 , b ) 2.2085 , c ) 0.022085 , d ) 0.0022085 , e ) none of these
Rationale: "22.085 ã — 0.001 = ? or , ? = 0.022085 answer c"
Answer: c
Problem: 0.99999 + 0.11111 = ?
Options: a ) 1 , b ) 1.0001 , c ) 1.0021 , d ) 1.111 , e ) 1.1111
Rationale: "0.99999 + 0.00001 = 1 0.1111 + 1 = 1.1111 correct answer is e ."
Answer: e
Problem: a certain company has records stored with a record storage firm in 15 - inch by 12 - inch by 10 - inch boxes . the boxes occupy 1.08 million cubic inches of space . if the company pays $ 0.8 per box per month for the record storage , what is the total amount that the company pays each month for record storage ?
Options: a ) a . 480 , b ) b . 300 , c ) c . 600 , d ) d . 410 , e ) e . 240
Rationale: "volume per box : 15 x 12 x 10 = 1,800 total volume : 1 , 080,000 number of boxes : total volume / volume per box = 1 , 080,000 / 1,800 = 600 price per month : number of boxes * price per box = 600 * 0.8 = 480 answer : a"
Answer: a
Problem: the sector of a circle has radius of 35 cm and central angle 135 o . find its perimeter ?
Options: a ) 91.5 cm , b ) 92.2 cm , c ) 152.5 cm , d ) 29.2 cm , e ) 98.2 cm
Rationale: "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 35 ) + 2 ( 35 ) = 82.5 + 70 = 152.5 cm answer : c"
Answer: c
Problem: for every even positive integer m , f ( m ) represents the product of all even integers from 2 to m , inclusive . for example , f ( 12 ) = 2 x 4 x 6 x 8 x 10 x 12 . what is the greatest prime factor of f ( 26 ) ?
Options: a ) 23 , b ) 19 , c ) 17 , d ) 13 , e ) 11
Rationale: "f ( 26 ) = 2 * 4 * 6 * 8 * 10 * 12 * 14 * 16 * 18 * 20 * 22 * 24 * 26 the greatest prime factor in this list is 13 . the answer is d ."
Answer: d
Problem: what is the remainder when 6 ^ 381 is divided by 5 ?
Options: a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
Rationale: i also agree that the remainder is ' 1 ' ( using the last digit of the powers of 7 ) . could we have the official answer please ? b
Answer: b
Problem: two trains of equal length are running on parallel lines in the same directions at 46 km / hr . and 36 km / hr . the faster trains pass the slower train in 108 seconds . the length of each train is :
Options: a ) 82 m , b ) 50 m , c ) 72 m , d ) 150 m , e ) none of these
Rationale: "explanation : the relative speed of train is 46 - 36 = 10 km / hr = ( 10 x 5 ) / 18 = 25 / 9 m / s 10 × 518 = 259 m / s in 108 secs the total distance traveled is 108 x 25 / 9 = 300 m . therefore the length of each train is = 300 / 2 = 150 m . answer d"
Answer: d
Problem: a team won 40 percent of its first 30 games in a particular season , and 80 percent of its remaining games . if the team won a total of 50 percent of its games that season , what was the total number of games that the team played ?
Options: a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80
Rationale: 50 % is 10 % - points above 40 % and 30 % - points below 80 % . thus the ratio of ` ` the first 30 games ' ' to ` ` remaining games ' ' is 3 : 1 . so the team played a total of 30 + 10 = 40 games . the answer is a .
Answer: a
Problem: the parameter of a square is equal to the perimeter of a rectangle of length 22 cm and breadth 16 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places )
Options: a ) 77.14 cm , b ) 47.14 cm , c ) 84.92 cm , d ) 29.85 cm , e ) 23.57 cm
Rationale: "let the side of the square be a cm . parameter of the rectangle = 2 ( 22 + 16 ) = 76 cm parameter of the square = 76 cm i . e . 4 a = 76 a = 19 diameter of the semicircle = 19 cm circimference of the semicircle = 1 / 2 ( ∏ ) ( 19 ) = 1 / 2 ( 22 / 7 ) ( 19 ) = 418 / 14 = 29.85 cm to two decimal places answer : d"
Answer: d
Problem: the ratio of the arithmetic mean of two numbers to one of the numbers is 5 : 9 . what is the ratio of the smaller number to the larger number ?
Options: a ) 1 : 5 , b ) 1 : 6 , c ) 1 : 7 , d ) 1 : 8 , e ) 1 : 9
Rationale: "for two numbers , the arithmetic mean is the middle of the two numbers . the ratio of the mean to the larger number is 5 : 9 , thus the smaller number must have a ratio of 1 . the ratio of the smaller number to the larger number is 1 : 9 . the answer is e ."
Answer: e
Problem: if | 5 x - 10 | = 100 , then find the sum of the values of x ?
Options: a ) 1 , b ) - 2 , c ) 3 , d ) - 3 , e ) 4
Rationale: "| 5 x - 10 | = 100 5 x - 10 = 100 or 5 x - 10 = - 100 5 x = 110 or 5 x = - 90 x = 22 or x = - 18 sum = 21 - 18 = 3 answer is c"
Answer: c
Problem: a is twice as good a workman as b and together they finish a piece of work in 18 days . in how many days will a alone finish the work ?
Options: a ) 31 days , b ) 25 days , c ) 27 days , d ) 29 days , e ) 19 days
Rationale: "if a takes x days to do a work then b takes 2 x days to do the same work . - - > 1 / x + 1 / 2 x = 1 / 18 - - > 3 / 2 x = 1 / 18 - - > x = 27 days . hence , a alone can finish the work in 27 days . answer : c ."
Answer: c
Problem: it takes avery 2.5 hours to build a brick wall while tom can do it in 5 hours . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ?
Options: a ) 15 minutes . , b ) 30 minutes . , c ) 1 hour and 30 minutes . , d ) 1 hour and 40 minutes , e ) 2 hours
Rationale: "avery ' s efficiency is 100 / 2.5 = 40 % tom ' s = 100 / 5 = 20 % they worked together for 1 hour and finished 60 % of the job remaining = 40 % tom will complete 20 % in 60 minutes , 40 % in 120 minutes time taken by tom to finish the remaining on his own = 120 minutes answer : e"
Answer: e
Problem: a , b and c invests rs . 4000 , rs . 2000 and rs . 6000 in a business . after one year a removed his money ; b and c continued the business for two more year . if the net profit after 3 years be rs . 3300 , then a ' s share in the profit is ?
Options: a ) 6000 , b ) 4000 , c ) 2000 , d ) 600 , e ) 300
Rationale: 4 * 12 : 2 * 36 : 4 * 36 2 : 3 : 6 2 / 11 * 3300 = 600 answer : d
Answer: d
Problem: the avg weight of a , b & c is 80 kg . if d joins the group , the avg weight of the group becomes 82 kg . if another man e who weights is 3 kg more than d replaces a , then the avgof b , c , d & e becomes 81 kg . what is the weight of a ?
Options: a ) 56 , b ) 65 , c ) 75 , d ) 89 , e ) 95
Rationale: a + b + c = 3 * 80 = 240 a + b + c + d = 4 * 82 = 328 - - - - ( i ) so , d = 88 & e = 88 + 3 = 91 b + c + d + e = 81 * 4 = 324 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 328 – 324 = 4 a = e + 4 = 91 + 4 = 95 answer : e
Answer: e
Problem: the base of a triangle is 15 cm and height is 12 cm . the height of another triangle of double the area having the base 20 cm is :
Options: a ) 22 cm
b ) 20 cm
c ) 18 cm
d ) 10 cm
e ) none of these
Rationale: explanation : area of triangle , a 1 = 1 / 2 ∗ base ∗ height = 1 / 2 ∗ 15 ∗ 12 = 90 cm 2 area of second triangle = 2 ∗ a 1 = 180 cm 2 1 / 2 ∗ 20 ∗ height = 180 = > height = 18 cm option c
Answer: c
Problem: rectangular tile each of size 20 cm by 30 cm must be laid horizontally on a rectangular floor of size 100 cm by 150 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is :
Options: a ) 40 , b ) 15 , c ) 25 , d ) 32 , e ) 45
Rationale: "area of tile = 20 * 30 = 600 area of floor = 100 * 150 = 15000 no of tiles = 15000 / 600 = 25 so , the no of tile = 25 answer : c"
Answer: c
Problem: 5000 was divided into two parts such a way that when first part was invested at 3 % and the second at 5 % , the whole annual interest from both the investments is rs . 144 , how much was put at 3 % ?
Options: a ) 10600 , b ) 11600 , c ) 12600 , d ) 10680 , e ) 20600
Rationale: "( x * 3 * 1 ) / 100 + [ ( 5000 - x ) * 5 * 1 ] / 100 = 144 3 x / 100 + ( 25000 – 5 x ) / 100 = 144 = > x = 10600 answer : a"
Answer: a