regisss/math_qa huggingface dataset:
Problem: in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 30 percent , but profits were 20 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ?
Options: a ) 80 % , b ) 105 % , c ) 140 % , d ) 124.2 % , e ) 138 %
Rationale: "0,14 r = x / 100 * 0.1 r answer c"
Answer: c
Problem: the sum of two numbers is 96 , and one of them is 12 more than the other . what are the two numbers ?
Options: a ) 36 - 48 , b ) 50 - 34 , c ) 60 - 24 , d ) 42 - 54 , e ) 21 - 63
Rationale: "in this problem , we are asked to find two numbers . therefore , we must let x be one of them . let x , then , be the first number . we are told that the other number is 12 more , x + 12 . the problem states that their sum is 96 : word problem = 96 the line over x + 12 is a grouping symbol called a vinculum . it saves us writing parentheses . we have : 2 x = 96 â ˆ ’ 12 = 84 . x = 84 / 2 = 42 . this is the first number . therefore the other number is x + 12 = 42 + 12 = 54 . the sum of 42 + 54 is 96 . d"
Answer: d
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 difference q 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 q will be 8 - 5 = 3 and answer b"
Answer: b
Problem: a motorcyclist started riding at highway marker a , drove 120 miles to highway marker b , and then , without pausing , continued to highway marker c , where she stopped . the average speed of the motorcyclist , over the course of the entire trip , was 20 miles per hour . if the ride from marker a to marker b lasted 3 times as many hours as the rest of the ride , and the distance from marker b to marker c was half of the distance from marker a to marker b , what was the average speed , in miles per hour , of the motorcyclist while driving from marker b to marker c ?
Options: a ) 40 , b ) 135 , c ) 50 , d ) 55 , e ) 60
Rationale: a - b = 120 miles b - c = 60 miles avg speed = 20 miles time taken for a - b 3 t and b - c be t avg speed = ( 120 + 60 ) / total time 20 = 180 / 4 t t = 135 b - c = 135 mph answer b
Answer: b
Problem: the difference between the place values of 9 and 4 in the number 529435 is
Options: a ) 8500 , b ) 8900 , c ) 8600 , d ) 6970 , e ) none
Rationale: sol . = ( place value of 9 ) – ( place value of 4 ) = ( 9000 - 400 ) = 8600 answer c
Answer: c
Problem: at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 450 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ?
Options: a ) 1,108 , b ) 3,700 , c ) 2,108 , d ) 2,124 , e ) 2,256
Rationale: "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 450 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . e . a total of 450 x 8 = 3600 hot dogs . we are left with ~ $ 16 . similarly , let ' s use $ 3 for calculation . we can buy 5 20 - pack hot dogs ( 3 x 5 ) , a total of 20 x 5 = 100 hot dogs . so we have 3700 hot dogs . 2108 looks far - fetched ( since we are not likely to be left with > $ 1.55 ) . hence , ( b ) 3700 ( answer b )"
Answer: b
Problem: from a total of 3 boys and 4 girls , how many 4 - person committees can be selected if the committee must have exactly 2 boys and 2 girls ?
Options: a ) 18 , b ) 24 , c ) 60 , d ) 120 , e ) 240
Rationale: "answer = c = 18 no of 4 person committees that can be formed = 3 c 2 * 4 c 2 = 18 answer a"
Answer: a
Problem: mary is expected to pay back a total sum of $ 819200 to her bank as mortgage . if she is expected to make a payment which is double of the previous months amount , how long will it take her to fully repay the loan if her initial payment was $ 400 .
Options: a ) 13 , b ) 15 , c ) 12 , d ) 10 , e ) 11
Rationale: first payment - - $ 400 total amount to be repaid - - $ 819200 first month payment - - $ 400 second month payment - - $ 400 * 2 = $ 800 third month payment - - $ 800 * 2 = $ 1600 fourth month payment - - $ 1600 * 2 = $ 3200 fifth month payment - - $ 3200 * 2 = $ 6400 sixth month payment - - $ 6400 * 2 = $ 12800 seventh month payment - - $ 25600 * 2 = $ 51200 eight month payment - - $ 51200 * 2 = $ 102400 ninth month payment - - $ 102400 * 2 = $ 204800 tenth month payment - - $ 204800 * 2 = $ 409600 eleventh month payment - - $ 409600 * 2 = $ 819200 answer is e
Answer: e
Problem: irin , ingrid and nell bake chocolate chip cookies in the ratio of 9.18 : 5.17 : 2.05 . if altogether they baked a batch of 148 cookies , what percent of the cookies did nell bake ?
Options: a ) 0.125 % , b ) 1.25 % , c ) 12.5 % , d ) 125 % , e ) 0.152 %
Rationale: "9.18 x + 5.17 x + 2.05 x = 16.4 x = 148 cookies x = 148 / 16.4 = 9 ( approx ) so , tell baked 9 * 2.05 cookies or 19 cookies ( approx ) % share = 19 / 148 = 12.5 approx hence , answer is c ."
Answer: c
Problem: a jar contains a mixture of ab in the ratio 4 : 1 . when 30 l of mixture is replaced with liquid b , ratio becomes 2 : 3 . how many liters of liquid a was present in mixture initially .
Options: a ) 12 , b ) 48 , c ) 16 , d ) 20 , e ) 25
Rationale: "30 litres of mixture that is replaced will contain 24 litres of a and 6 litres of b ( as a : b = 4 : 1 ) let the initial volume of the mixture be 4 k + 1 k = 5 k so by condition , [ 4 k - 24 ] / [ k - 6 + 30 ] = 2 / 3 = > 12 k - 72 = 2 k - 12 + 60 = > 10 k = 120 solve for k which is k = 12 so initial volume of liquid a = 4 k = 48 litres answer : b"
Answer: b
Problem: two trains are moving at 50 kmph and 70 kmph in opposite directions . their lengths are 150 m and 100 m respectively . the time they will take to pass each other completely is ?
Options: a ) 7 1 / 2 sec , b ) 7 1 / 7 sec , c ) 7 8 / 2 sec , d ) 7 1 / 9 sec , e ) 7 2 / 2 sec
Rationale: "70 + 50 = 120 * 5 / 18 = 100 / 3 mps d = 150 + 100 = 250 m t = 250 * 3 / 100 = 15 / 2 = 7 1 / 2 sec answer : d"
Answer: d
Problem: if there are 56 laborers in a crew , and on a certain day , 30 were present . calculate the percentage that showed up for work ? ( round to the nearest tenth ) .
Options: a ) 53.6 % , b ) 52.6 % , c ) 54.6 % , d ) 55.6 % , e ) 59.6 %
Rationale: 30 / 56 * 100 = 53.57 53.6 % correct answer a
Answer: a
Problem: how many of the positive factors of 24 are not factors of 30
Options: a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 5
Rationale: "factors of 24 - 1 , 2 , 3 , 4 , 6 , 8 , 12,24 factors of 30 - 1 , 2 , 3 , 5 , 6 , 10 , 15 and 30 . comparing both , we have four factors of 24 which are not factors of 30 - 4,8 , 12,24 answer ( c )"
Answer: c
Problem: if f ( x ) = 12 - x ^ 2 / 2 and f ( 2 k ) = 5 k , what is one possible value for k ?
Options: a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 8
Rationale: "first of all , see thisgmat blog postand check the related lesson linked below for some background on function notation . we can plug anything in for x and get a result . you can find f ( 1 ) , for example , by plugging in 1 where x is , and you would get 12 - 1 / 2 = 11.5 . or we could find f ( 2 ) , which would be 12 - 4 / 2 = 10 . so the notation f ( 2 k ) means that we are going to plug a 2 k in for x everywhere in the formula for f ( x ) . that would be : f ( 2 k ) = 12 - ( 2 k ) ^ 2 / 2 = 12 - 2 k ^ 2 . remember that we have to square both the 2 and the k , to get 4 k 2 . now , this expression , the output , we will set equal to 2 k . 12 - 2 k ^ 2 = 2 k - - > k = - 3 or k = 5 . all the answers are positive , so we choose k = 5 . answer = d"
Answer: d
Problem: the average runs scored by a batsman in 20 matches is 40 . in the next 10 matches the batsman scored an average of 13 runs . find his average in all the 30 matches ?
Options: a ) 31 , b ) 67 , c ) 88 , d ) 69 , e ) 12
Rationale: total score of the batsman in 20 matches = 800 . total score of the batsman in the next 10 matches = 130 . total score of the batsman in the 30 matches = 930 . average score of the batsman = 930 / 30 = 31 . answer : a
Answer: a
Problem: p , q and r have rs . 9000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ?
Options: a ) rs . 3000 , b ) rs . 3600 , c ) rs . 2400 , d ) rs . 4000 , e ) none of these
Rationale: "let the amount with r be rs . r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 9000 - r ) = > 3 r = 18000 - 2 r = > 5 r = 18000 = > r = 3600 . answer : b"
Answer: b
Problem: a car gets 60 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 170 kilometers ?
Options: a ) 3.5 gallons , b ) 2.37 gallons , c ) 5.7 gallons , d ) 4.25 gallons , e ) 2.83 gallons
Rationale: "each 60 kilometers , 1 gallon is needed . we need to know how many 60 kilometers are there in 170 kilometers ? 60 ã · 170 = 2.83 ã — 1 gallon = 2.83 gallons correct answer is e ) 2.83 gallons"
Answer: e
Problem: if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 293 pounds . what is jake ’ s present weight , in pounds ?
Options: a ) 182 , b ) 186 , c ) 190 , d ) 194 , e ) 198
Rationale: "j + k = 293 and so k = 293 - j j - 8 = 2 k j - 8 = 2 ( 293 - j ) 3 j = 594 j = 198 the answer is e ."
Answer: e
Problem: the radius of a circle is 4 / π cm then its perimeter is ?
Options: a ) 10
b ) 9
c ) 8
d ) 12
e ) 15
Rationale: perimeter = 2 π r 2 π * 4 / π = 8 cm answer : c
Answer: c
Problem: the effective annual rate of interest corresponding to a nominal rate of 6 % per annum payable half - yearly is ?
Options: a ) 6.99 % , b ) 6.89 % , c ) 6.08 % , d ) 6.09 % , e ) 6.19 %
Rationale: "amount of rs . 100 for 1 year when compounded half - yearly = [ 100 * ( 1 + 3 / 100 ) 2 ] = rs . 106.09 effective rate = ( 106.09 - 100 ) = 6.09 % answer : d"
Answer: d
Problem: find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 5 cm .
Options: a ) 187 cm 2 , b ) 185 cm 2 , c ) 85 cm 2 , d ) 85 cm 2 , e ) 95 cm 2
Rationale: "explanation : area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 5 ) = 95 cm 2 answer : option e"
Answer: e
Problem: from a pack of 52 cards , 1 card is drawn at random . what is the probability that a heart is drawn ?
Options: a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 13 , e ) 3 / 26
Rationale: "the total number of cards is 52 . the number of hearts is 13 . p ( heart ) = 13 / 52 = 1 / 4 the answer is c ."
Answer: c
Problem: if each year the population of the country grows by 50 % , how many years will elapse before the population of the country doubles ?
Options: a ) 3 , b ) 4 , c ) 2 , d ) 6 , e ) 7
Rationale: "till year 2000 , population is 100 . year 2001 : population becomes 150 . . . . . . . . . . . . . 1 year elapsed year 2002 : population > 200 . . . . . . . . . . . . . 2 year elapsed answer : c"
Answer: c
Problem: for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 7 + 3 and * 7 + 7 , inclusive ?
Options: a ) none , b ) one , c ) two , d ) three , e ) four
Rationale: "7 ! = 720 * 7 = 5040 # in question = 5043 , 5044 , 5045 , 5046 , 5047 none of these are prime answer - a"
Answer: a
Problem: two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 12 seconds . find the length of the faster train ?
Options: a ) 270 , b ) 288 , c ) 266 , d ) 120 , e ) 126
Rationale: "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 12 sec = 27 * 10 = 270 m . the length of the faster train = 120 m . answer : d"
Answer: d
Problem: two trains of equal are running on parallel lines in the same direction at 44 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ?
Options: a ) 40 m , b ) 72 m , c ) 80 m , d ) 82 m , e ) 84 m
Rationale: "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 44 - 36 = 8 km / hr . = 8 * 5 / 18 = 20 / 9 m / sec . 2 x / 36 = 20 / 9 = > x = 40 . answer : a"
Answer: a
Problem: a man sold 20 articles for $ 60 and gained 20 % . how many articles should he sell for $ 50 to incur a loss 20 % ?
Options: a ) 25 , b ) 36 , c ) 40 , d ) 50 , e ) 48
Rationale: production cost per article : $ 60 * ( 100 % - 20 % ) / 20 = $ 2.40 required production costs for a loss of 20 % : $ 50 * ( 100 % + 20 % ) = $ 60 number of articles to be sold for $ 60 to incur a 20 % loss : $ 60 / $ 2.40 = 25 thus , solution a is correct .
Answer: a
Problem: each of the 59 members in lourdes school class is required to sign up for a minimum of one and a maximum of 3 academic clubs . the 3 clubs to choose from are the poetry club , the history club , and the writing club . a total of 22 students sign up for the poetry club , 27 students for the history club , and 28 students for the writing club . if 6 students sign up for exactly two clubs , how many students sign up for all 3 clubs ?
Options: a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 9
Rationale: each of the 59 members in lourdes school class is required to sign up for a minimum of one and a maximum of three academic clubs . total = g 1 + g 2 + g 3 - ( # in exactly 2 ) - 2 * ( # in 3 sets ) 59 = 22 + 27 + 28 - ( 6 ) - 2 x so , # in 3 sets = 6 = c
Answer: c
Problem: two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 22 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ?
Options: a ) 565 , b ) 444 , c ) 676 , d ) 456 , e ) 663
Rationale: "1 h - - - - - 6 ? - - - - - - 60 12 h rs = 16 + 22 = 38 t = 12 d = 38 * 12 = 456 answer : d"
Answer: d
Problem: bruce and bhishma are running on a circular track of length 600 m . speed of bruce is 30 m / s and that of bhishma is 20 m / s . they start fro the same point at the same time in the same direction . when will they meet again for the first time ?
Options: a ) 60 sec , b ) 70 sec , c ) 80 sec , d ) 90 sec , e ) 100 sec
Rationale: actually arjun has to make a lead of 600 m , because when arjun will be 600 m ahead of bhishma , they will be together again as a person when completes the total length it starts retracing he same path and thus arjun and bhishma can be together again . since , they make a difference of 10 m in 1 second . so , he will create 600 m difference in 60 second . d
Answer: d
Problem: if a certain computer is capable of printing 4700 monthly credit card bills per hour , while a new model is capable of printing at a rate of 6800 per hour , the old model will take approximately how much longer than the new model to print 10000 bills ?
Options: a ) 19 mins , b ) 29 mins , c ) 39 mins , d ) 38 mins , e ) 37 mins
Rationale: old model ' s 1 hr work = 4700 new model ' s 1 hr work = 6800 old model ' time taken to print 10000 cards for 1 hr - - 4700 cards x - - 10000 x = 10000 / 4700 = 100 / 47 new model ' time taken to print 10000 cards for 1 hr - - 6800 cards y - - 10000 y = 10000 / 6800 = 100 / 68 x = 10000 / 4700 = 100 / 47 = 2 6 / 47 hr difference is 100 / 47 - 100 / 68 = 6800 - 4700 / ( 47 * 68 ) 2100 / 3196 = 0.66 hrs = > 39 minutes old model ll take 39 minutes longer to print 10000 cards than new model ans : 39 mins answer : c
Answer: c
Problem: average between two sets of numbers is closer to the set withmore no ' s ?
Options: a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130
Rationale: "if on a test three people answered 90 % of the questions correctly and two people answered 80 % correctly , then the average for the group is not 85 % but rather 3 × 90 + 2 × 805 = 4305 = 86.3 × 90 + 2 × 805 = 4305 = 86 . here , 90 has a weight of 3 = > it occurs 3 times . whereas 80 has a weight of 2 = > it occurs 2 times . so the average is closer to 90 than to 80 as we have just calculated . b"
Answer: b
Problem: in a graduating class of 232 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number w and the smallest possible number of students that could have taken both geometry and biology ?
Options: a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31
Rationale: "official solution : first of all , notice that since 144 took geometry and 119 took biology , then the number of students who took both geometry and biology can not be greater than 119 . { total } = { geometry } + { biology } - { both } + { neither } ; 232 = 144 + 119 - { both } + { neither } ; { both } = 31 + { neither } . { both } is minimized when { neither } is 0 . in this case { both } = 31 . the greatest possible number w of students that could have taken both geometry and biology , is 119 . thus , the answer is 119 - 31 = 88 . answer : d ."
Answer: d
Problem: rani bought more apples than oranges . she sells apples at ₹ 23 apiece and makes 15 % profit . she sells oranges at ₹ 10 apiece and makes 25 % profit . if she gets ₹ 653 after selling all the apples and oranges , find her profit percentage .
Options: a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 %
Rationale: "given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the value between 11 and 21 , if a = 11 , o = 30 - - > not possible if a = 21 , o = 17 - - > possible cost price = 20 * 21 + 8 * 17 = 420 + 136 = 556 profit = 653 - 556 = 97 profit % = ( 97 / 556 ) * 100 = 17.4 % answer : b"
Answer: b
Problem: a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 70 points . to the nearest tenth of a point , what was the student ’ s average grade for the entire two - year period ?
Options: a ) 79 , b ) 89 , c ) 95 , d ) 86.36 , e ) 97.2
Rationale: "let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 70 * 5 = 350 student ' s average = ( 600 + 350 ) / 11 = 86.36 answer d"
Answer: d
Problem: if m : n is 3 : 9 and n : p is 9 : 12 then m : p is equal to
Options: a ) 1 : 5 , b ) 1 : 4 , c ) 1 : 6 , d ) 1 : 8 , e ) 1 : 7
Rationale: "the two ratios given are having the same number 9 for n in both the ratios . hence - m : n = 3 : 9 n : p = 9 : 12 = > m : p = 3 : 12 = > 1 : 4 answer b"
Answer: b
Problem: the ration of the number of boys and girls in a college is 7 : 8 . if the percentage is increase in the number of boys and girls be 20 % and 10 % respectively . what will be the new ration ?
Options: a ) 5 : 7 , b ) 15 : 21 , c ) 21 : 22 , d ) 13 : 17 , e ) 23 : 42
Rationale: let the number of boys and girls be 7 x and 8 x their increased number is 120 % of 7 x and 110 % of 8 x 7 x * 120 / 100 and 8 x * 110 / 100 42 x / 5 and 44 x / 5 required ratio = 42 x / 5 : 44 x / 5 = 21 : 22 answer is c
Answer: c
Problem: a and b can do a piece of work in 50 days and 50 days respectively . they work together for 10 days and b leaves . in how many days the whole work is completed ?
Options: a ) 30 days , b ) 35 days , c ) 40 days , d ) 44 days , e ) 45 days
Rationale: "explanation : ( a + b ) ’ s 10 days work = 10 [ 1 / 50 + 1 / 50 ] = 10 [ 1 + 1 / 50 ] = 2 / 5 a complete remaining work in 2 / 5 * 50 = 20 total work = 10 + 20 = 30 days answer : option a"
Answer: a
Problem: of all the homes on gotham street , 1 / 3 are termite - ridden , and 2 / 5 of these are collapsing . what fraction of the homes are termite - ridden , but not collapsing ?
Options: a ) a ) 2 / 15 , b ) b ) 3 / 15 , c ) c ) 4 / 5 , d ) d ) 2 / 5 , e ) e ) 1 / 5
Rationale: let total homes be 15 termite ridden = 1 / 3 ( 15 ) = 5 termite ridden and collapsing = 2 / 5 ( 5 ) = 2 thus homes that are termite ridden , but not collapsing = 5 - 2 = 3 thus required ratio = 3 / 15 = 1 / 5 answer e
Answer: e
Problem: how many 3 - digit numerals begin with a digit that represents a odd prime and end with a digit that represents a prime number ?
Options: a ) 16 , b ) 80 , c ) 160 , d ) 120 , e ) 240
Rationale: "prime digits 2 , 3,5 and 7 . three digit numbers _ _ _ 1 st place can be filled in 3 ways 2 nd place can be filled in 10 ways 3 rd place can be filled in 4 ways total = 3 * 10 * 4 = 120 ans : d"
Answer: d
Problem: a certain mixture of nuts consists of 5 parts almonds to 1 parts walnuts , by weight . what is the number of pounds of almonds in 140 pounds of the mixture ?
Options: a ) 116 , b ) 84 , c ) 40 , d ) 28 , e ) 20
Rationale: "almonds : walnuts = 5 : 1 total mixture has 6 parts in a 140 pound mixture , almonds are 5 / 6 ( total mixture ) = 5 / 6 * 140 = 116 pounds answer ( a )"
Answer: a
Problem: a certain number of workers can do a work in 45 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ?
Options: a ) 25 , b ) 30 , c ) 28 , d ) 24 , e ) 35
Rationale: "number of workers = 10 * ( 45 - 10 ) / 10 = 35 answer is e"
Answer: e
Problem: the smallest number which when diminished by 9 , is divisible 13 , 26 , 39 , 52 and 65 is :
Options: a ) 789 , b ) 769 , c ) 790 , d ) 787 , e ) 788
Rationale: "required number = ( l . c . m . of 13 , 26 , 39 , 52,65 ) + 9 = 780 + 7 = 789 answer : option a"
Answer: a
Problem: sheela deposits rs . 4500 in bank savings account . if this is 28 % of her monthly income . what is her monthly income in ?
Options: a ) 22000 , b ) 27000 , c ) 16071 , d ) 16789 , e ) none of these
Rationale: "explanation : 28 % of income = rs . 4500 100 % of income = 4500 x 100 / 28 = rs . 16071 answer : c"
Answer: c
Problem: the difference between the ages of two persons is 10 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the elder person is ?
Options: a ) 35 yr , b ) 25 yr , c ) 20 yr , d ) 30 yr , e ) 40 yr
Rationale: "let their ages be x years and ( x + 10 ) years then , ( x + 10 ) - 15 = 2 ( x - 15 ) x - 5 = 2 x - 30 x = 25 present age of the elder person = 25 + 10 = 35 years answer is a"
Answer: a
Problem: two tests had the same maximum mark . the pass percentages in the first and the second test were 40 % and 45 % respectively . a candidate scored 216 marks in the second test and failed by 75 marks in that test . find the pass mark in the first test ?
Options: a ) 768 , b ) 243 , c ) 246 , d ) 288 , e ) 224
Rationale: let the maximum mark in each test be m . the candidate failed by 75 marks in the second test . pass mark in the second test = 249 + 75 = 324 45 / 100 m = 324 pass mark in the first test = 40 / 100 m = 40 / 45 * 324 = 288 . answer : d
Answer: d
Problem: simplify : 0.4 * 0.5 + 0.3 * 0.2
Options: a ) 0.52 , b ) 0.26 , c ) 0.48 , d ) 0.64 , e ) 0.46
Rationale: "given exp . = 0.4 * 0.5 + ( 0.3 * 0.2 ) = 0.2 + 0.06 = 0.26 answer is b ."
Answer: b
Problem: elena purchased brand x pens for $ 4.00 apiece and brand y for $ 2.80 apiece . if elena purchased a total of 12 of these pens for $ 40.00 , how many brand x pens did she purchase ?
Options: a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8
Rationale: "4 x + 2.8 y = 42 - - > multiply by 2.5 ( to get the integers ) - - > 10 x + 7 y = 105 - - > only one positive integers solutions x = 8 and y = 5 ( how to solve : 7 y must have the last digit of 5 in order the last digit of the sum to be 5 ) . answer : e ."
Answer: e
Problem: if money is invested at r percent interest , compounded annually , the amount of investment will double in approximately 70 / r years . if pat ' s parents invested $ 10000 in a long term bond that pays 8 percent interest , compounded annually , what will be the approximate total amount of investment 18 years later , when pat is ready for college ?
Options: a ) $ 40000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000
Rationale: since investment doubles in 70 / r years then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 10,000 * 2 ) * 2 = $ 40,000 ( after 9 years investment will become $ 10,000 * 2 = $ 20,000 and in another 9 years it ' ll become $ 20,000 * 2 = $ 40,000 ) . answer : a .
Answer: a
Problem: in a ratio which is equal to 4 : 6 , if the antecedent is 30 , then the consequent is ?
Options: a ) 72 , b ) 85 , c ) 64 , d ) 51 , e ) 45
Rationale: "we have 4 / 6 = 30 / x 4 x = 30 * 6 x = 45 consequent = 45 answer is e"
Answer: e
Problem: a side of beef lost 35 percent of its weight in processing . if the side of beef weighed 560 pounds after processing , how many pounds did it weigh before processing ?
Options: a ) 191 , b ) 355 , c ) 737 , d ) 861 , e ) 1,560
Rationale: "let weight of side of beef before processing = x ( 65 / 100 ) * x = 560 = > x = ( 560 * 100 ) / 65 = 861 answer d"
Answer: d
Problem: 49 , 98 , 686 , 1372 . . . .
Options: a ) 9052 , b ) 8711 , c ) 9279 , d ) 9604 , e ) 9086
Rationale: "49 x 2 = 98 98 x 7 = 684 684 x 2 = 1372 1372 x 7 = 9604 answer = d"
Answer: d
Problem: a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 20 % , he is only able to save rs . 240 per month . what is his monthly salary ?
Options: a ) 5000 , b ) 2999 , c ) 6000 , d ) 2990 , e ) 2771
Rationale: "income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 * ( 20 / 100 ) = rs . 96 present savings = 100 â € “ 96 = rs . 4 100 - - - - - - 4 ? - - - - - - - - - 240 = > 6000 answer : c"
Answer: c
Problem: a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent q of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ?
Options: a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 %
Rationale: cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is q = . 25 / 1 * 100 = 25 % d is the answer .
Answer: d
Problem: in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 78 per kg ?
Options: a ) 5 : 8 , b ) 4 : 7 , c ) 3 : 7 , d ) 9 : 5 , e ) 9 : 8
Rationale: "( 96 - 78 ) / ( 78 - 68 ) = 18 / 10 = 9 / 5 answer : d"
Answer: d
Problem: arjun started a business with rs . 20000 and is joined afterwards by anoop with rs . 4000 . after how many months did anoop join if the profits at the end of the year are divided equally ?
Options: a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
Rationale: suppose anoop joined after 3 months . then , 20000 * 12 = 40000 * ( 12 – x ) = > = > x = 6 . answer : d
Answer: d
Problem: the length of a rectangle is increased by 20 % and its breadth is decreased by 20 % . what is the effect on its area ?
Options: a ) 1288 , b ) 9600 , c ) 1000 , d ) 10000 , e ) 2887
Rationale: "100 * 100 = 10000 120 * 80 = 9600 answer : b"
Answer: b
Problem: a perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer . how many positive integers n are there such that n is less than 100,000 and at the same time n is a perfect square and a perfect cube ?
Options: a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
Rationale: "if n is a perfect square and a perfect cube , then n = a ^ 6 for some integer a . the numbers are 1 ^ 6 = 1 , 2 ^ 6 = 64 , 3 ^ 6 = 729 , 4 ^ 6 = 4096 , 5 ^ 6 = 15,625 , 6 ^ = 46,656 . the answer is d ."
Answer: d
Problem: what is the sum of the integers from - 90 to 95 , inclusive ?
Options: a ) 450 , b ) 550 , c ) 650 , d ) 750 , e ) 465
Rationale: "in an arithmetic progression , the nth term is given by tn = a + ( n - 1 ) d here tn = 95 , a = - 90 , d = 1 hence , 95 = - 90 + ( n - 1 ) or n = 186 sum of n terms can be calculated by sn = n / 2 ( a + l ) a = first term , l = last term , n = no . of terms sn = 186 * ( - 90 + 95 ) / 2 sn = 186 * 5 / 2 = 465 answer : e"
Answer: e
Problem: a sum fetched a total simple interest of rs . 4016.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ?
Options: a ) rs 4462.50 , b ) rs . 8032.50 , c ) rs . 8900 , d ) rs . 8925 , e ) none
Rationale: "solution principal = rs . ( 100 * 4016.25 / 9 * 5 ) = rs . ( 401625 / 45 ) = rs . 8925 answer d"
Answer: d
Problem: a train 110 m long is running with a speed of 40 km / h . in how many seconds will the train pass a man who is running at 4 km / h in the direction opposite to that in which the train is going ?
Options: a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
Rationale: "the speed of the train relative to the man = 40 + 4 = 44 km / h . 44000 m / h * 1 h / 3600 s = ( 440 / 36 ) m / s ( 110 m ) / ( 440 / 36 m / s ) = ( 110 * 36 ) / 440 = 36 / 4 = 9 seconds the answer is d ."
Answer: d
Problem: vijay sells a cupboard at 16 % below cost price . had he got rs . 1200 more , he would have made a profit of 16 % . what is the cost price of the cupboard ?
Options: a ) 3750 , b ) 3600 , c ) 7400 , d ) 7500 , e ) none of these
Rationale: "explanation : cost price = 1200 / ( 0.16 + 0.16 ) = 1200 / 0.32 = rs . 3750 answer a"
Answer: a
Problem: the dimensions of a field are 16.5 m by 10 m . a pit 10 m long , 5.5 m wide and 4 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field . what will be the rise in the height of field as a result of this operation ?
Options: a ) 2 m , b ) 3 m , c ) 1 m , d ) 2.5 m , e ) 3.5 m
Rationale: "the volume of the earth removed is 10 * 5.5 * 4 = 220 m ^ 3 . the remaining area of the field is 16.5 * 10 - 10 * 5.5 = 110 m ^ 2 . 220 m ^ 3 of the earth evenly spread over the area of 110 m ^ 2 will rise the height by ( height ) = ( volume ) / ( area ) = 220 / 110 = 2 m . answer : a"
Answer: a
Problem: a class of 35 students has an average height of 180 cm . 7 students whose average height is 120 cm , left the class and 7 others whose average height is 140 cm , joined . calculate the new average height of the students of the class ( in cm ) is ?
Options: a ) 204.6 cm , b ) 404.6 cm , c ) 224.6 cm , d ) 205.6 cm , e ) 256.6 cm
Rationale: total height = 180 * 35 = 6300 cm . total height of those who left = 120 * 7 = 840 total height of those who joined = 140 * 7 = 980 new average = [ 6300 - 120 + 980 ] / 35 = 7160 / 35 cm . = 204.6 cm answer : a
Answer: a
Problem: convert the 11 / 36 m / s into kilometers per hour ?
Options: a ) 2.9 kmph , b ) 1.1 kmph , c ) 1.3 kmph , d ) 1.2 kmph , e ) 5.7 kmph
Rationale: "11 / 36 m / s = 11 / 36 * 18 / 5 = 11 / 10 = 1.1 kmph . answer : b"
Answer: b
Problem: a shop produces sarongs . the daily average production is given by 5 n + 20 , where n is the number of workers aside from the owner . in the first k days , 500 units are produced , and then 5 workers are added to the team . after another k days , the cumulative total is 800 . how many workers were part of the latter production run ?
Options: a ) a ) 6 , b ) b ) 10 , c ) c ) 11 , d ) d ) 21 , e ) e ) 23.5
Rationale: "the daily average production is given by 5 n + 20 - given in the first k days , 500 units are produced = ( 5 n + 20 ) k = 500 k = 500 / 5 n + 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5 workers were added = 5 ( n + 5 ) + 20 = 5 n + 45 cumulative is 1250 . . thus for the current period = 800 - 500 = 300 ( 5 n + 45 ) k = 300 k = 300 / 5 n + 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 equate 1 and 2 500 / 5 n + 20 = 300 / 5 n + 45 500 ( 5 n + 45 ) = 300 ( 5 n + 20 ) 2500 n + 22500 = 1500 n + 6000 1000 n = - 16500 n = 16 thus n + 5 = 21 hence d"
Answer: d
Problem: at a certain food stand , the price of each apple is 40 ¢ and the price of each orange is 60 ¢ . mary selects a total of 10 apples and oranges from the food stand , and the average ( arithmetic mean ) price of the 10 pieces of fruit is 56 ¢ . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is 50 ¢ ?
Options: a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
Rationale: "if the average price is 56 , then mary has 8 oranges and 2 apples ( a ratio of 4 : 1 ) . for the average price to be 50 , the ratio should be 1 : 1 . mary must put back 6 oranges . the answer is e ."
Answer: e
Problem: if a lends rs . 3500 to b at 10 % per annum and b lends the same sum to c at 12 % per annum then the gain of b in a period of 3 years is ?
Options: a ) 200 , b ) 210 , c ) 220 , d ) 225 , e ) 230
Rationale: ( 3500 * 2 * 3 ) / 100 = > 210 answer : b
Answer: b
Problem: think of a number , divide it by 5 and add 10 to it . the result is 21 . what is the number thought of ?
Options: a ) 24 , b ) 55 , c ) 297 , d ) 267 , e ) 29
Rationale: explanation : 21 - 10 = 11 11 x 5 = 55 answer : b
Answer: b
Problem: what percent is 5 gm of 1 kg ?
Options: a ) 1.5 % , b ) 0.5 % , c ) 2.5 % , d ) 3.5 % , e ) 4 %
Rationale: "1 kg = 1000 gm 5 / 1000 × 100 = 500 / 1000 = 1 / 2 = 0.5 % b )"
Answer: b
Problem: a sum fetched a total simple interest of 4025.25 at the rate of 9 % . p . a . in 5 years . what is the sum ?
Options: a ) 5768 , b ) 8925 , c ) 2345 , d ) 8945 , e ) 8723
Rationale: "principal = ( 100 x 4025.25 ) / ( 9 x 5 ) = 402525 / 45 = 8945 . answer d"
Answer: d
Problem: steve traveled the first 2 hours of his journey at 40 mph and the last 4 hours of his journey at 80 mph . what is his average speed of travel for the entire journey ?
Options: a ) 53.33 mph , b ) 56.67 mph , c ) 60 mph , d ) 64 mph , e ) 66.67 mph
Rationale: "average speed = total distance / total time = ( 40 * 2 + 80 * 4 ) / ( 2 + 4 ) = 400 / 6 = 66.6 answer : e"
Answer: e
Problem: if each edge of cube increased by 10 % , the percentage increase in
Options: a ) 21 , b ) 22 , c ) 25 , d ) 19 , e ) 15
Rationale: "100 × ( 110 ) / 100 × ( 110 ) / 100 = 121 = > 21 % answer is a ."
Answer: a
Problem: the average of 6 numbers is 3.9 . the average of two of them is 3.4 , while the average of the other two is 3.85 . what is the average of the remaining two numbers ?
Options: a ) 4.7 , b ) 4.45 , c ) 4.2 , d ) 4.1 , e ) 4.9
Rationale: explanation : sum of the remaining two numbers = ( 3.9 * 6 ) - [ ( 3.4 * 2 ) + ( 3.85 * 2 ) ] = 23.40 - ( 6.8 + 7.7 ) = 23.40 - 14.5 = 8.90 . required average = ( 8.9 / 2 ) = 4.45 . answer : b ) 4.45
Answer: b
Problem: a merchant has selected two items to be placed on sale , one of which currently sells for 30 percent less than the other . if he wishes to raise the price of the cheaper item so that the two items are equally priced , by what percentage must he raise the price of the less expensive item ?
Options: a ) 22 % , b ) 30 % , c ) 42 6 / 7 % , d ) 70 % , e ) 130 %
Rationale: "expensive item = $ 100 ; cheap item = $ 70 ; we must increase $ 70 to $ 100 , so by $ 30 , which is approximately 40 % increase : ( 100 - 70 ) / 70 = 3 / 7 = ~ 0.42 . answer : c ."
Answer: c
Problem: a works twice as fast as b . if b can complete a work in 12 days independently , the number of days in which a and b can together finish the work is :
Options: a ) 4 days , b ) 6 days , c ) 8 days , d ) 18 days , e ) none of these
Rationale: solution rates of worktio of raing of a and b = 2 : 1 . so , ratio of times taken = 1 : 2 ∴ a ’ s 1 day ’ s work = 1 / 6 b ’ s 1 day ’ s work = 1 / 12 ( a + b ) ' s 1 day ’ s work = ( 1 / 6 + 1 / 12 ) = 3 / 12 = 1 / 4 so , a and b together can finish the work in 4 days . answer a
Answer: a
Problem: p and q started a business investing rs . 25,000 and rs . 20,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ?
Options: a ) 5 : 4 , b ) 5 : 6 , c ) 4 : 3 , d ) 6 : 5 , e ) 17 : 4
Rationale: "p : q = 25000 : 20000 = 5 : 4 . answer : a"
Answer: a
Problem: the average weight of a , b and c is 45 kg . if the average weight of a and b be 41 kg and that of b and c be 43 kg , then the weight of b is :
Options: a ) 33 kg , b ) 31 kg , c ) 32 kg , d ) 36 kg , e ) 37 kg
Rationale: "a 33 kg let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 . . . . ( i ) a + b = ( 41 x 2 ) = 82 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 168 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 33 . b ' s weight = 33 kg . a"
Answer: a
Problem: the salary of a worker is first increased by 40 % and afterwards reduced by 40 % . what is the net change in the worker ' s salary ?
Options: a ) 8 % decrease , b ) 8 % increase , c ) 16 % decrease , d ) 16 % increase , e ) no change
Rationale: "let x be the original salary . the final salary is 0.6 ( 1.4 x ) = 0.84 x the answer is c ."
Answer: c
Problem: a tank with a volume of 20 cubic feet has one inlet pipe and 2 outlet pipes . the inlet pipe fills water into the tank at the rate of 5 cubic inches / min and the 2 outlet pipes empty it out at the rates of 9 cubic inches / min and 8 cubic inches / min respectively . if all 3 pipes are opened when the tank is full , how many minutes does it take to empty the tank ? ( 1 foot = 12 inches )
Options: a ) 2110 , b ) 2220 , c ) 2440 , d ) 2660 , e ) 2880
Rationale: "the tank is emptied at this rate : 9 + 8 - 5 = 12 cubic inches / min the tank has a volume of 20 * 12 * 12 * 12 = 34560 cubic inches . the time it takes to empty the tank is 34560 / 12 = 2880 minutes . the answer is e ."
Answer: e
Problem: if q = ( x – 5 ) ^ 2 + ( x + 1 ) ^ 2 – 6 , then y is least when x =
Options: a ) - 2 , b ) - 1 , c ) 0 , d ) 2 , e ) none of the above
Rationale: let us transform the formula : q = ( x – 5 ) ² + ( x + 1 ) ² – 6 = x ² – 10 x + 25 + x ² + 2 x + 1 – 6 = 2 x ² – 8 x + 20 = 2 × ( x ² – 4 x + 10 ) = 2 × ( ( x ² – 4 x + 4 ) + 6 ) = 2 × ( ( x – 2 ) ² + 6 ) any square is greater or equal 0 . therefore the formula possess the least value when ( x – 2 ) ² = 0 . x – 2 = 0 x = 2 the correct answer is choice ( d ) .
Answer: d
Problem: what least number should be added to 1024 , so that the sum is completely divisible by 25 ?
Options: a ) 1 , b ) 3 , c ) 2 , d ) 0 , e ) 5
Rationale: "1024 ã · 25 = 40 with remainder = 24 24 + 1 = 25 . hence 1 should be added to 1024 so that the sum will be divisible by 25 answer : option a"
Answer: a
Problem: a student was asked to find the arithmetic mean of the numbers 3 , 117 , 915 , 138 , 1917 , 2114 and x . he found the mean to be 12 . what should be the remainder in the place of x ?
Options: a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9
Rationale: we have ( 3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x ) / 12 = 12 137 + x = 144 x = 7 answer is c
Answer: c
Problem: in an election between the two candidates , the candidates who gets 70 % of votes polled is winned by 200 vote ’ s majority . what is the total number of votes polled ?
Options: a ) 750 , b ) 500 , c ) 800 , d ) 850 , e ) none of these
Rationale: "explanation : note : majority ( 40 % ) = difference in votes polled to win ( 70 % ) & defeated candidates ( 30 % ) 40 % = 70 % - 30 % 40 % - - - - - > 200 ( 40 * 5 = 200 ) 100 % - - - - - > 500 ( 100 * 5 = 500 ) answer : option b"
Answer: b
Problem: if w is 40 percent less than u , u is 40 percent less than y , and z is 46 percent less than y , then z is greater than w by what percent of w ?
Options: a ) 4 % , b ) 18 % , c ) 36 % , d ) 40 % , e ) 50 %
Rationale: given w = 0.6 u , u = 0.6 y , z = 0.54 y , substituting , w = 2 / 3 z - - - - > z = 1.5 w and thus z is 50 % greater than w . e is the correct answer .
Answer: e
Problem: the salary of a person was reduced by 15 % . by what percent should his reduced salary be raised so as to bring it at par with his original salary ?
Options: a ) 10 % , b ) 17.6 % , c ) 13.7 % , d ) 15.1 % , e ) 12.3 %
Rationale: "let the original salary be $ 100 new salary = $ 85 increase on 85 = 15 increase on 100 = 15 / 85 * 100 = 17.6 % answer is b"
Answer: b
Problem: which of the numbers 1 , 2 , 3 , 4 , or 5 is nearest in value to the sum 2007 / 2999 + 8001 / 5998 + 2001 / 3999 = ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
Rationale: the sum is a little more than ( 2 / 3 ) + ( 8 / 6 ) + ( 2 / 4 ) = 2.5 . correct answer c
Answer: c
Problem: if jake loses 12 pounds , he will weigh twice as much as his sister . together they now weigh 156 pounds . what is jake ’ s present weight , in pounds ?
Options: a ) 108 , b ) 135 , c ) 139 , d ) 147 , e ) 188
Rationale: "lets say j is the weight of jack and s is the wt of his sister . if he loses 8 pounds , he s twice as heavy as his sister . j - 12 = 2 * s also , together they weight 156 pounds j + s = 156 solvong the 2 equation , we get j = 108 pounds ! a"
Answer: a
Problem: during a sale of 20 % on everything in a store , a kid is successful in convincing the store manager to give him 40 candies for the discounted price of 14 candies . the store still makes a profit of 12 % on this sale . what is the mark up percentage on each candy ?
Options: a ) 100 % , b ) 80 % , c ) 75 % , d ) 66 + 2 / 3 % , e ) 55 %
Rationale: hi akhil , i can share the way i solved it . . let ' s say marked price = m so , there ' s a discount of 20 % on m so , new s . p . = 80 % of m now , the child convinces the owner to sell 20 candies for the price of 14 candies . let ' s say each candy after discount is 1 $ . so , s . p . of 20 candies = 20 $ . the child bought it for 14 $ so , he got a discount of 6 / 20 * 100 = 30 % so , the latest s . p . = 70 % of 80 % of m = 0.7 * 0.8 m now , we are given that the shopkeeper still makes a profit of 12 % . so we have , 0.7 * 0.8 * m = 1.12 c . p so , we get , m = 2 c . p . i . e . marked price was kept 100 % above c . p . e
Answer: e
Problem: the perimeter of an equilateral triangle is 60 . if one of the sides of the equilateral triangle is the side of an isosceles triangle of perimeter 65 , then how long is the base of isosceles triangle ?
Options: a ) 10
b ) 15
c ) 20
d ) 25
e ) 30
Rationale: the base of the isosceles triangle is 65 - 20 - 20 = 25 units the answer is d .
Answer: d
Problem: the average age of a class of 19 students is 20 years . the average increased by 1 when the teacher ' s age also included . what is the age of the teacher ?
Options: a ) 39 , b ) 41 , c ) 40 , d ) 42 , e ) 43
Rationale: total age of all students = 19 ã — 20 total age of all students + age of the teacher = 20 ã — 21 age of the teacher = 20 ã — 21 â ˆ ’ 20 ã — 19 = 20 ( 21 â ˆ ’ 19 ) = 20 ã — 2 = 40 answer is c .
Answer: c
Problem: the sum of all the integers d such that - 26 < d < 24 is
Options: a ) 0 , b ) - 2 , c ) - 25 , d ) - 49 , e ) - 51
Rationale: "easy one - - 25 , - 24 , - 23 , - 22 , . . . . . . - 1,0 , 1 , 2 . . . . , 22 , 23 cancel everyhitng and we ' re left with - - 25 and - 24 d = - 49 . d is the answer ."
Answer: d
Problem: there are 9 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ?
Options: a ) 15 , b ) 16 , c ) 28 , d ) 56 , e ) 72
Rationale: "every team plays with 8 teams . . . so total no of matches = 9 x 8 = 72 . now , each match is played twice = > 72 x 2 but 2 teams play a match = > 72 x 2 / 2 = 72 . answer : e"
Answer: e
Problem: | x + 1 | – | 4 - x | = | 8 + x | how many solutions will this equation have ?
Options: a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
Rationale: "you have | x + 3 | - | 4 - x | = | 8 + x | first , look at the three values independently of their absolute value sign , in other words : | x + 3 | - | 4 - x | = | 8 + x | ( x + 3 ) - ( 4 - x ) = ( 8 + x ) now , you ' re looking at x < - 8 , so x is a number less than - 8 . let ' s pretend x = - 10 here to make things a bit easier to understand . when x = - 10 i . ) ( x + 3 ) ( - 10 + 3 ) ( - 7 ) ii . ) ( 4 - x ) ( 4 - [ - 10 ] ) ( double negative , so it becomes positive ) ( 4 + 10 ) ( 14 ) iii . ) ( 8 + x ) ( 8 + - 10 ) ( - 2 ) in other words , when x < - 8 , ( x + 3 ) and ( 8 + x ) are negative . to solve problems like this , we need to check for the sign change . here is how i do it step by step . i . ) | x + 3 | - | 4 - x | = | 8 + x | ii . ) ignore absolute value signs ( for now ) and find the values of x which make ( x + 3 ) , ( 4 - x ) and ( 8 + x ) = to zero as follows : ( x + 3 ) x = - 3 ( - 3 + 3 ) = 0 ( 4 - x ) x = 4 ( 4 - 4 ) = 0 ( 8 + x ) x = - 8 ( 8 + - 8 ) = 3 d"
Answer: d
Problem: an electric pump can fill a tank in 6 hours . because of a leak in the tank , it took 12 hours to fill the tank . if the tank is full , how much time will the leak take to empty it ?
Options: a ) 10 hours , b ) 12 hours , c ) 8 hours , d ) 5 hours , e ) 15 hours
Rationale: "work done by the leak in 1 hour = 1 / 6 - 1 / 12 = 1 / 12 the leak will empty the tank in 12 hours answer is b"
Answer: b
Problem: m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 4 ) both lie on the line defined by the equation x = ( y / 2 ) - ( 2 / 5 ) , what is the value of p ?
Options: a ) 4 / 5 , b ) 1 , c ) 2 , d ) 5 , e ) 8
Rationale: "( m , n ) m = x n = y m = n / 2 - 2 / 5 10 m = 5 n - 4 ( multiplied by 10 to simplify equation ) ( m + p , n + 4 ) m + p = x n + 4 = y m + p = ( n + 4 ) / 2 - 2 / 5 multiplied by 10 again 10 m + 10 p = 5 n + 20 - 4 use both equations to solve for p and you get 10 p = 20 p = 2 answer : c"
Answer: c
Problem: lilly has 10 fish and rosy has 14 fish . in total , how many fish do they have in all ?
Options: a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25
Rationale: "10 + 14 = 24 the answer is d ."
Answer: d
Problem: 3 years ago , the ratio of ages of p and q was 4 : 3 . 6 years from now , the ratio of their ages will be 7 : 6 . what is p ' s age at present ?
Options: a ) 9 , b ) 12 , c ) 15 , d ) 18 , e ) 21
Rationale: 3 ( p - 3 ) / 4 = ( q - 3 ) . then q = 3 ( p - 3 ) / 4 + 3 6 ( p + 6 ) / 7 = q + 6 6 ( p + 6 ) / 7 = 3 ( p - 3 ) / 4 + 9 24 p + 144 = 21 p + 189 3 p = 45 p = 15 the answer is c .
Answer: c
Problem: a student scored an average of 55 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ?
Options: a ) 86 , b ) 165 , c ) 76 , d ) 155 , e ) 26
Rationale: "given m + p + c = 55 * 3 = 165 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 165 = 155 answer : d"
Answer: d
Problem: a school has received 30 % of the amount it needs for a new building by receiving a donation of $ 500 each from people already solicited . people already solicited represent 60 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ?
Options: a ) $ 200 , b ) $ 250 , c ) $ 100 , d ) $ 350 , e ) $ 875
Rationale: "let us suppose there are 100 people . 60 % of them donated $ 30000 ( 500 * 60 ) $ 30000 is 30 % of total amount . so total amount = 30000 * 100 / 60 remaining amount is 70 % of total amount . 70 % of total amount = 30000 * ( 100 / 60 ) * ( 70 / 100 ) = 35000 this amount has to be divided by 40 ( remaining people are 40 ) so per head amount is 35000 / 40 = $ 875 ; answer : e"
Answer: e
Problem: if x / y = 4 and y is not ' 0 ' what % of x is 2 x - y
Options: a ) 150 % , b ) 175 percent , c ) 200 % , d ) 250 % , e ) 350 %
Rationale: let z be the percentt x * z / 100 = 2 x - y zx = 7 y * 100 z = 700 y / 4 y z = 350 / 2 z = 175 % answer : b
Answer: b