regisss/math_qa huggingface dataset:
Problem: a cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick . the thickness of the bottom is :
Options: a ) 90 cm , b ) 1 dm , c ) 1 m , d ) 1.1 cm , e ) none of these
Rationale: "explanation : let the thickness of the bottom be x cm . then , [ ( 330 - 10 ) × ( 260 - 10 ) × ( 110 - x ) ] = 8000 × 1000 = > 320 × 250 × ( 110 - x ) = 8000 × 1000 = > ( 110 - x ) = 8000 × 1000 / 320 = 100 = > x = 10 cm = 1 dm . answer : b"
Answer: b
Problem: a theater charges $ 12 for seats in the orchestra and $ 8 for seats in the balcony . on a certain night , a total of 380 tickets were sold for a total cost of $ 3,320 . how many more tickets were sold that night for seats in the balcony than for seats in the orchestra ?
Options: a ) 90 , b ) 110 , c ) 120 , d ) 130 , e ) 240
Rationale: "orchestra seats - a balcony seats - b a + b = 380 and 12 a + 8 b = 3320 solving equations simultaneously ( multiply equation 1 with 8 and subtract from second equation ) 4 a = 3320 - 8 * 380 = 3320 - 3040 = 280 i . e . a = 70 and b = 380 - 70 = 310 more seats in balcony than orchestra = b - a = 310 - 70 = 240 answer : option e"
Answer: e
Problem: in an electric circuit , two resistors with resistances x and y are connected in parallel . if r is the combined resistance of these two resistors , then the reciprocal of r is equal to the sum of the reciprocals of x and y . what is r if x is 5 ohms and y is 6 ohms ?
Options: a ) 7 / 30 , b ) 11 / 30 , c ) 11 / 15 , d ) 30 / 11 , e ) 15 / 11
Rationale: 1 / r = 1 / x + 1 / y 1 / r = 1 / 5 + 1 / 6 = 11 / 30 r = 30 / 11 the answer is d .
Answer: d
Problem: in a practice paper at 2 iim . com , questions were given from 5 topics . out of the appearing students , 10 % passed in all topics while 10 % did not pass in any . of the remaining , 20 % passed in one topic only and 25 % in two topics only . if 24 % of the total students passed 4 topics only and 500 students passed in 3 topics only , find the total number of students who appeared in the examination ?
Options: a ) 2500 , b ) 2000 , c ) 1600 , d ) 4545 , e ) 6565
Rationale: detailed solution let the number of appearing students be 100 . pass only in 0 topic – 10 1 – 16 ( 20 % of 80 ) 2 – 20 ( 25 % of 80 ) 3 – 20 % ( 100 – ( 16 + 20 + 24 + 20 ) ) 4 – 24 % 5 – 20 % therefore , 20 % of x = 500 = ) x = 2500 correct answer : a
Answer: a
Problem: a person buys an article at rs . 650 . at what price should he sell the article so as to make a profit of 10 % ?
Options: a ) 600 , b ) 715 , c ) 269 , d ) 261 , e ) 281
Rationale: "cost price = rs . 650 profit = 10 % of 650 = rs . 65 selling price = cost price + profit = 650 + 65 = rs . 715 answer : b"
Answer: b
Problem: wink , inc . follows a certain procedure that requires two tasks to be finished independently in order for a job to be done . on any given day , there is a 7 / 8 probability that task 1 will be completed on time , and a 1 / 5 probability that task 2 will be completed on time . on a certain day , what is the probability that task 1 will be completed on time , but task 2 will not ?
Options: a ) 1 / 20 , b ) 3 / 40 , c ) 13 / 40 , d ) 7 / 10 , e ) 13 / 22
Rationale: "p ( 1 and not 2 ) = 7 / 8 * ( 1 - 1 / 5 ) = 7 / 10 . answer : d ."
Answer: d
Problem: a mixture contains milk and water in the ratio 3 : 2 . on adding 10 litters of water , the ratio of milk to water becomes 2 : 3 . total quantity of milk & water before adding water to it ?
Options: a ) 10 , b ) 30 , c ) 50 , d ) 20 , e ) 30
Rationale: "explanation : milk : water = 3 : 2 after adding 10 liters of water milk : water = 2 : 3 olny water patrs increase when mixture of water milk : wate = 3 : 2 = 2 * ( 3 : 2 ) = 6 : 4 after adding 10 liters of water milk : water = 2 : 3 = 3 * ( 2 : 3 ) = 6 : 9 milk parts always same short cut method : milk : water = 6 : 4 after adding 10 liters of water milk : water = 6 : 9 milk is same but water increse 10 liters then the water ratio is increse 5 parts 5 part - - - - - > 10 liters the quantity of milk in the original mixture is = 6 : 4 = 6 + 4 = 10 10 parts - - - - - > 20 liters ( answer is = 20 ) short cut method - 2 : for only milk problems milk : water 6 : 4 6 : 9 milk ratio same but water ratio 5 parts incress per 10 liters 5 part of ratio - - - - - - - > 10 liters 10 part of ratio - - - - - - > 20 liters answer : option d"
Answer: d
Problem: a drink vendor has 80 liters of maaza , 144 liters of pepsi and 368 liters of sprite . he wants to pack them in cans , so that each can contains the same number of liters of a drink , and does n ' t want to mix any two drinks in a can . what is the least number of cans required ?
Options: a ) 35 , b ) 37 , c ) 42 , d ) 30 , e ) 38
Rationale: "the number of liters in each can = hcf of 80 , 144 and 368 = 16 liters . number of cans of maaza = 80 / 16 = 5 number of cans of pepsi = 144 / 16 = 9 number of cans of sprite = 368 / 16 = 23 the total number of cans required = 5 + 9 + 23 = 37 cans . answer : b"
Answer: b
Problem: ( 112 % of 2348 ) ÷ 4.98 = ?
Options: a ) 505 , b ) 515 , c ) 525 , d ) 538 , e ) 567
Rationale: "explanation : ? = ( 112 x 2348 / 100 ) ÷ 5 = 2630 / 5 = 526 = 525 answer : option c"
Answer: c
Problem: the unit digit in the product 1256 * 6785 * 4587 * 9785 is ?
Options: a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
Rationale: "unit digit in the given product = unit digit in 6 * 5 * 7 * 5 = 0 answer is a"
Answer: a
Problem: a man can row upstream at 60 kmph and downstream at 90 kmph , and then find the speed of the man in still water ?
Options: a ) 75 kmph , b ) 65 kmph , c ) 30 kmph , d ) 73 kmph , e ) 65 kmph
Rationale: us = 60 ds = 90 m = ( 60 + 90 ) / 2 = 75 answer : a
Answer: a
Problem: a train running at the speed of 52 km / hr crosses a pole in 9 sec . what is the length of the train ?
Options: a ) 130 m , b ) 786 m , c ) 566 m , d ) 546 m , e ) 440 m
Rationale: "speed = 52 * 5 / 18 = 130 / 9 m / sec length of the train = speed * time = 130 / 9 * 9 = 130 m answer : a"
Answer: a
Problem: find the sum lend at c . i . at 5 p . c per annum will amount to rs . 441 in 2 years ?
Options: a ) 387 , b ) 400 , c ) 776 , d ) 268 , e ) 181
Rationale: "441 = p ( 21 / 20 ) 2 p = 400 answer : b"
Answer: b
Problem: solve the quickfire maths brain teaser â ˆ š 36 % = ?
Options: a ) 30 % , b ) 40 % , c ) 60 % , d ) 19 % , e ) 29 %
Rationale: "â ˆ š 36 % = > â ˆ š 36 / â ˆ š 100 = > 6 / 10 = > 60 / 100 = > 60 % c"
Answer: c
Problem: a salesman ’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 900 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth ?
Options: a ) 10000 , b ) 12000 , c ) 12019 , d ) 12197 , e ) 12012
Rationale: "[ 900 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 10000 answer : a"
Answer: a
Problem: jaime earned enough money by selling seashells at 30 cents each to buy several used paperback books at 55 cents each . if he spent all of the money he earned selling seashells to buy the books , what is the least number of seashells he could have sold ?
Options: a ) 5 , b ) 22 , c ) 17 , d ) 25 , e ) 30
Rationale: "let ' s test answer b : 22 seashells . . . with 22 seashells , jamie would have 22 ( 30 ) = 660 cents . this would allow him to buy 12 books for 660 cents total , with no money left over . this is an exact match for what we were told , so this must be the answer . final answer : [ reveal ] spoiler : b"
Answer: b
Problem: a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 180 kilometers ?
Options: a ) 3.5 gallons , b ) 2.7 gallons , c ) 5.7 gallons , d ) 4.5 gallons , e ) 7.5 gallons
Rationale: "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 180 kilometers ? 180 ÷ 40 = 4.5 × 1 gallon = 4.5 gallons correct answer is d ) 4.5 gallons"
Answer: d
Problem: what is the area of a square field whose diagonal of length 18 m ?
Options: a ) 120 sq m , b ) 250 sq m , c ) 200 sq m , d ) 180 sq m , e ) 100 sq m
Rationale: "d 2 / 2 = ( 18 * 18 ) / 2 = 180 answer : d"
Answer: d
Problem: a car covers a distance of 624 km in 3 1 / 5 hours . find its speed ?
Options: a ) 104 kmph , b ) 195 kmph , c ) 109 kmph , d ) 174 kmph , e ) 101 kmph
Rationale: "624 / 3 1 / 5 = 195 kmph answer : b"
Answer: b
Problem: a train which has 400 m long , is running 45 kmph . in what time will it cross a person moving at 9 kmph in same direction ?
Options: a ) 56 sec , b ) 40 sec , c ) 36 sec , d ) 29 sec . , e ) 19 sec .
Rationale: "time taken to cross a moving person = length of train / relative speed time taken = 400 / ( ( 45 - 9 ) ( 5 / 18 ) = 400 / 36 * ( 5 / 18 ) = 400 / 10 = 40 sec answer : b"
Answer: b
Problem: if an integer n is to be selected at random from 1 to 100 , inclusive , what is probability n ( n + 1 ) will be divisible by 5 ?
Options: a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 2 / 5
Rationale: "for n ( n + 1 ) to be a multiple of 5 , either n or n + 1 has to be a multiple of 5 . thus n must be of the form 5 k or 5 k - 1 . the probability is 2 / 5 . the answer is e ."
Answer: e
Problem: each of the integers from 0 to 9 , inclusive , is written on a separate slip of blank paper and the ten slips are dropped into a hat . if 4 of the slips are the drawn , without replacement , what is the probability that all 4 have a odd number written on it ?
Options: a ) 1 / 12 , b ) 1 / 10 , c ) 1 / 8 , d ) 1 / 42 , e ) 5 / 9
Rationale: "key is that there is no replacement , so each successive choice will become more skewed towards picking a neg ( i . e . the pool of positives decreases , while the pool of negatives stay the same ) p ( + on 1 st pick ) = 5 / 10 p ( + on 2 nd pick ) = 4 / 9 p ( + on 3 rd pick ) = 3 / 8 p ( + on 4 rd pick ) = 2 / 7 5 / 10 * 4 / 9 * 3 / 8 * 2 / 7 = 1 / 42 d"
Answer: d
Problem: a superhero flies at an incredible rate of 1000 kilometers per minute . what is her speed in miles per hour ? [ 1 km = 0.6 miles ]
Options: a ) 6000 , b ) 36000 , c ) 60000 , d ) 600 , e ) 3600
Rationale: to calculate the equivalent of miles in a kilometer 0.6 kilometers = 1 mile 1000 kilometers = ( 0.6 ) * 1000 = 600 miles speed in miles per minute = 600 = 600 1 hour = 60 minutes speed in miles per hour = 600 * 60 = 36,000 correct answer - b
Answer: b
Problem: 5100 − ( 102 ÷ 20.4 ) = ?
Options: a ) 5200 , b ) 5150 , c ) 5250 , d ) 6150 , e ) 5095
Rationale: explanation : = 5100 − ( 102 / 204 × 10 ) = 5100 − 5 = 5095 option e
Answer: e
Problem: what least number should be subtracted from 13603 such that the remainder is divisible by 87 ?
Options: a ) 27 , b ) 29 , c ) 28 , d ) 30 , e ) 31
Rationale: "13603 ÷ 87 = 156 , remainder = 31 hence 31 is the least number which can be subtracted from 13603 such that the remainder is divisible by 87 answer is e"
Answer: e
Problem: in a sports club with 30 members , 17 play badminton and 21 play tennis and 2 do not play either . how many members play both badminton and tennis ?
Options: a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
Rationale: "let x play both badminton and tennis so 17 - x play only badminton and 19 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 21 - x ) + x + 2 = 30 40 - 2 x + x = 30 40 - x = 30 x = 10 so 10 members play both badminton and tennis . d"
Answer: d
Problem: in 40 litres of a mixture , the ratio of milk to water is 7 : 1 . in order to make the ratio of milk to water as 3 : 1 , the quantity of water that should be added to the mixture will be ?
Options: a ) rs . 1000 , b ) rs . 1200 , c ) rs . 1400 , d ) rs . 1600 , e ) rs . 2500
Rationale: explanation : total simple interest received , i = rs . 1600 principal , p = 7000 period , n = 5 years rate of interest , r = ? simple interest , i = p n r 100 i = pnr 100 ⇒ 1600 = 7000 × 5 × r 100 ⇒ r = 1600 × 100 7000 × 5 = 160 35 = 32 7 % ⇒ 1600 = 7000 × 5 × r 100 ⇒ r = 1600 × 1007000 × 5 = 16035 = 327 % by rule of alligation , rate of interest % from part 1 rate of interest % from part 2 6 4 net rate of interest % 32 7 327 32 7 − 4 = 4 7 327 − 4 = 47 6 − 32 7 = 10 7 6 − 327 = 107 = > part 1 : part 2 = 4 7 : 10 7 = 4 : 10 = 2 : 5 = 47 : 107 = 4 : 10 = 2 : 5 given that total amount is rs . 7000 . therefore , the amount lent at 6 % per annum ( part 1 amount ) = 7000 × 2 7 = rs . 2000 d )
Answer: d
Problem: 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ?
Options: a ) 388 , b ) 266 , c ) 350 , d ) 112 , e ) 134
Rationale: speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 39 = 50 / 3 3 ( x + 300 ) = 1950 è x = 350 m . answer : c
Answer: c
Problem: a person want to give his money of $ 12000 to his 5 children a , b , c , d , e in the ratio 2 : 4 : 3 : 1 : 5 . what is the d + e share ?
Options: a ) $ 1800 , b ) $ 2800 , c ) $ 3800 , d ) $ 5800 , e ) $ 4800
Rationale: "d ' s share = 12000 * 1 / 15 = $ 800 d ' s share = 12000 * 5 / 15 = $ 4000 a + d = $ 4800 answer is e"
Answer: e
Problem: an enterprising businessman earns an income of re 1 on the first day of his business . on every subsequent day , he earns an income which is just double of that made on the previous day . on the 20 th day of business , he earns an income of :
Options: a ) rs 219 , b ) rs 220 , c ) rs 20 ² , d ) rs 20 , e ) rs 30
Rationale: "2 nd day he earns = 2 = 2 ( 2 – 1 ) 3 rd day he earns = 2 ( 3 – 1 ) on 20 th day he earns 2 ( 20 - 1 ) = 219 rupees answer : a"
Answer: a
Problem: two cars started from the same point , at 5 am , traveling in opposite directions at 40 and 50 mph respectively . at what time will they be 450 miles apart ?
Options: a ) 8 am , b ) 10 am , c ) 11 am , d ) 12 am , e ) 14 am
Rationale: after t hours the distances d 1 and d 2 , in miles per hour , traveled by the two cars are given by d 1 = 40 t and d 2 = 50 t after t hours the distance d separating the two cars is given by d = d 1 + d 2 = 40 t + 50 t = 90 t distance d will be equal to 450 miles when d = 90 t = 450 miles t = 5 hours . 5 am + 5 hours = 10 am answer : b
Answer: b
Problem: if 8 men or 12 women can do a piece of work in 20 days , in how many days can the same work be done by 6 men and 11 women ?
Options: a ) 10 days , b ) 11 days , c ) 12 days , d ) 15 days , e ) 17 days
Rationale: "8 men = 12 women ( i . e 2 men = 3 women ) 12 women 1 day work = 1 / 20 soln : 6 men ( 9 women ) + 11 women = 20 women = ? 1 women 1 day work = 12 * 20 = 1 / 240 so , 20 women work = 20 / 240 = 1 / 12 ans : 12 days answer : c"
Answer: c
Problem: 3 men can complete a piece of work in 6 days . two days after they started the work , 3 more men joined them . how many days will they take to complete the remaining work ?
Options: a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 7
Rationale: explanation : 3 man 1 day work = 1616 3 man 2 days work = 2626 remaining work = ( 1 – 2626 ) = 2 / 3 parts . 6 man together perform the work in 1 day is = 1616 + 1616 = 2626 parts 2626 parts completed in 1 day 2323 parts will b completed in 2 days answer : a
Answer: a
Problem: 36 people { a 1 , a 2 … a 36 } meet and shake hands in a circular fashion . in other words , there are totally 36 handshakes involving the pairs , { a 1 , a 2 } , { a 2 , a 3 } , … , { a 35 , a 36 } , { a 36 , a 1 } . then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is
Options: a ) 76
b ) 55
c ) 44
d ) 12
e ) 91
Rationale: ans : { a 1 , a 2 } , { a 2 , a 3 } , { a 3 , a 4 } , { a 4 , a 5 } , { a 5 , a 6 } , { a 6 , a 7 } … , { a 35 , a 36 } , { a 36 , a 1 } from the above arrangement , if we separate a 3 , a 6 , a 9 , . . . . . a 36 . total 12 persons the reamining persons must have shaked hand with atleast one person . so answer is 12 . answer : d
Answer: d
Problem: rates for having a manuscript typed at a certain typing service are $ 6 per page for the first time a page is typed and $ 4 per page each time a page is revised . if a certain manuscript has 100 pages , of which 35 were revised only once , 15 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ?
Options: a ) $ 860 , b ) $ 850 , c ) $ 840 , d ) $ 830 , e ) $ 820
Rationale: "50 pages typed 1 x 35 pages typed 2 x ( original + one revision ) 15 pages typed 3 x ( original + two revisions ) 50 ( 6 ) + 35 ( 6 + 4 ) + 15 ( 6 + 4 + 4 ) = 300 + 350 + 210 = 860 answer - a"
Answer: a
Problem: the perimeter of a semi circle is 216 cm then the radius is ?
Options: a ) 26 , b ) 48 , c ) 98 , d ) 37 , e ) 42
Rationale: "36 / 7 r = 216 = > r = 42 answer : e"
Answer: e
Problem: a train 360 metre long runs with a speed of 45 km / hr . what time will it take to pass a platform of 140 metre long ?
Options: a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50
Rationale: speed = 45 km / hr = 45 × 5 / 18 = 25 / 2 m / s distance travelled = ( speed x time ) time = ( distance travelled / speed ) distance travelled = length of the train + length of the platform = 360 + 140 = 500 m = 500 / 25 / 2 = 40 sec answer : c
Answer: c
Problem: the events a and b are independent , the probability that event a occurs is greater than 0 , and the probability that event a occurs is twice the probability that event b occurs . the probability that at least one of events a and b occurs is 18 times the probability that both events a and b occur . what is the probability that event a occurs ?
Options: a ) 12 / 112 , b ) 6 / 81 , c ) 11 / 61 , d ) 12 / 33 , e ) 3 / 19
Rationale: "let us say probability of a occuring is a . let us say probability of b occuring is b . a = 2 b probability ( either a or b or both ) = 8 times probability ( a and b ) a * ( 1 - b ) + b * ( 1 - a ) + ab = 18 * ab substituting a = 2 b in the second equation : 2 b * ( 1 - b ) + b * ( 1 - 2 b ) + 2 b * b = 18 * 2 b * b 3 b - 2 b ^ 2 = 36 b ^ 2 3 b = 38 b ^ 2 b = 3 / 38 so , a = 2 b = 6 / 38 = 3 / 19 the correct answer is e ."
Answer: e
Problem: a student got twice as many sums wrong as he got right . if he attempted 24 sums in all , how many did he solve correctly ?
Options: a ) 12 , b ) 16 , c ) 8 , d ) 24 , e ) 26
Rationale: explanation : suppose the boy got x sums right and 2 x sums wrong . then , x + 2 x = 24 3 x = 24 x = 8 . answer : c
Answer: c
Problem: a 300 m long train crosses a platform in 42 sec while it crosses a signal pole in 18 sec . what is the length of the platform ?
Options: a ) 287 , b ) 350 , c ) 828 , d ) 277 , e ) 400
Rationale: "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 42 = 50 / 3 3 x + 900 = 2200 = > x = 400 m . answer : e"
Answer: e
Problem: a fashion designer sold a pair of jeans to a retail store for 40 percent more than it cost to manufacture the pair of jeans . a customer bought the pair of jeans for 10 percent more than the retailer paid for them . the price the customer paid was what percent greater than the cost of manufacturing the jeans ?
Options: a ) 54 % , b ) 70 % , c ) 75 % , d ) 85 % , e ) 95 %
Rationale: "find the product of the two increases : ( 1.4 ) * ( 1.1 ) , which is 1.54 and a 54 % increase . a"
Answer: a
Problem: the time it took car p to travel 900 miles was 2 hours less than the time it took car r to travel the same distance . if car p ’ s average speed was 10 miles per hour greater than that of car r , what was car r ’ s average speed , in miles per hour ?
Options: a ) 40 , b ) 54 , c ) 62 , d ) 70 , e ) 80
Rationale: "let speed of car r be = x then speed of car p = x + 10 a / q , ( 900 / x ) - ( 900 / ( x + 10 ) ) = 2 solving for x = 62 miles \ hr . c"
Answer: c
Problem: how many seconds will a 400 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ?
Options: a ) 20 , b ) 30 , c ) 40 , d ) 24 , e ) 60
Rationale: "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 400 * 3 / 50 = 24 sec . answer : option d"
Answer: d
Problem: a certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales in 1997 were up 7 percent from 1996 . if total revenues from car sales and truck sales in 1997 were up 1 percent from 1996 , what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996 ?
Options: a ) 1 : 2 , b ) 4 : 5 , c ) 1 : 1 , d ) 3 : 2 , e ) 5 : 3
Rationale: "let c = revenue from car sales in 1996 let t = revenue from truck sales in 1996 equating the total revenue in 1997 with the individual revenue from selling cars and trucks in 1997 , c ( 0.89 ) + t ( 1.07 ) = ( c + t ) ( 1.01 ) dividing throughout by t , ( c / t ) ( 0.89 ) + 1.07 = ( c / t + 1 ) ( 1.01 ) = > ( c / t ) ( 0.12 ) = 0.06 = > ( c / t ) = 1 / 2 option ( a ) ."
Answer: a
Problem: a sum was put at simple interest at a certain rate for 10 years . had it been put at 5 % higher rate , it would have fetched rs . 150 more . what was the sum ?
Options: a ) s . 120 , b ) s . 130 , c ) s . 140 , d ) s . 150 , e ) s . 300
Rationale: at 5 % more rate , the increase in s . i for 10 years = rs . 150 ( given ) so , at 5 % more rate , the increase in si for 1 year = 150 / 10 = rs . 15 / - i . e . rs . 15 is 5 % of the invested sum so , 1 % of the invested sum = 15 / 5 therefore , the invested sum = 15 × 100 / 5 = rs . 300 answer : e
Answer: e
Problem: 12 men can complete a piece of work in 4 days , while 15 women can complete the same work in 4 days . 6 men start working on the job and after working for 2 days , all of them stopped working . how many women should be put on the job to complete the remaining work , if it so to be completed in 3 days ?
Options: a ) 15 women , b ) 85 women , c ) 75 women , d ) 65 women , e ) 35 women
Rationale: "1 man ' s 1 day work = 1 / 48 ; 1 woman ' s 1 day work = 1 / 60 . 6 men ' s 2 day ' s work = 6 / 48 * 2 = 1 / 4 . remaining work = ( 1 - 1 / 4 ) = 3 / 4 now , 1 / 60 work is done in 1 day by 1 woman . so , 3 / 4 work will be done in 3 days by ( 60 * 3 / 4 * 1 / 3 ) = 15 women . answer : a"
Answer: a
Problem: in a stream running at 2 kmph , a motor boat goes 10 km upstream and back again to the starting point in 55 minutes . find the speed of motor boat in still water ?
Options: a ) 11 , b ) 15 , c ) 22 , d ) 18 , e ) 25
Rationale: "let the speed of motor boat instill water be x kmph then , speed in downstream = ( x + 2 ) km and . speed in upstream = ( x - 2 ) kmph time taken to row 10 km & back = ( 10 / x + 2,10 / x - 2 ) 10 / x + 2 + 10 / x - 2 = 55 / 60 11 x 2 - 240 x - 44 = 0 ( x - 22 ) ( 11 x + 2 ) = 0 x = 22 or x = - 2 / 11 then x = 22 kmph answer is c ."
Answer: c
Problem: a train 100 m long can cross an electric pole in 20 sec and then find the speed of the train ?
Options: a ) 19 kmph , b ) 18 kmph , c ) 72 kmph , d ) 17 kmph , e ) 91 kmph
Rationale: "length = speed * time speed = l / t s = 100 / 20 s = 5 m / sec speed = 5 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 18 kmph answer : b"
Answer: b
Problem: if the sales tax be reduced from 4 ( 1 / 5 ) % to 3 ( 1 / 4 ) % , then what difference does it make to a person who purchases a bag with marked price of rs . 5400 ?
Options: a ) rs . 100.2 , b ) rs . 50.2 , c ) rs . 20.5 , d ) rs . 100 , e ) none of these
Rationale: "explanation : required difference = ( 4 ( 1 / 5 ) of rs . 5400 ) - ( 3 ( 1 / 4 ) of rs . 5400 ) = ( 21 / 5 – 13 / 4 ) % of rs . 5400 = ( 19 / 20 ) x ( 1 / 100 ) x 5400 = rs . 51.30 answer e"
Answer: e
Problem: there are 2 sections a and b in a class , consisting of 30 and 20 students respectively . if the average weight of section a is 40 kg and that of section b is 35 kg , find the average of the whole class ?
Options: a ) 35 kg , b ) 37 kg , c ) 42 kg , d ) 38 kg , e ) 29.78 kg
Rationale: "total weight of 36 + 44 students = 30 * 40 + 20 * 35 = 1900 average weight of the class is = 1900 / 50 = 38 kg answer is d"
Answer: d
Problem: a person can swim in still water at 6 km / h . if the speed of water 2 km / h , how many hours will the man take to swim back against the current for 14 km ?
Options: a ) 3.5 , b ) 4.5 , c ) 5 , d ) 9 , e ) 6
Rationale: "m = 6 s = 2 us = 6 - 2 = 4 d = 42 t = 14 / 4 = 3.5 answer : a"
Answer: a
Problem: two numbers are respectively 12 % and 60 % more than a third number . the percentage that is first of the second is ?
Options: a ) 50 % , b ) 60 % , c ) 70 % , d ) 80 % , e ) 90 %
Rationale: "i ii iii 112 160 100 160 - - - - - - - - - - 112 100 - - - - - - - - - - - ? = > 70 % answer : c"
Answer: c
Problem: buses a and b start from a common bus stop x . bus a begins to travel in a straight line away from bus b at a constant rate of 40 miles per hour . one hour later , bus b begins to travel in a straight line in the exact opposite direction at a constant rate of 80 miles per hour . if both buses travel indefinitely , what is the positive difference , in minutes , between the amount of time it takes bus b to cover the exact distance that bus a has covered and the amount of time it takes bus b to cover twice the distance that bus a has covered ?
Options: a ) 36 , b ) 72 , c ) 132 , d ) 144 , e ) 180
Rationale: "1 st part : - in 1 hr , bus a covers 30 miles . relative speed of bus abus b is ( 80 - 30 ) = 50 mph . so time required for bus b to cover the exact distance as a is 50 * t = 30 t = 3 / 5 = 36 min 2 nd part 80 * t = 2 d - b has to cover twice the distance 30 * ( t + 1 ) = d - a traveled 1 hr more and has to travel only only d so d / 30 - 2 d / 80 = 1 d = 120 t = 3 hrs = 180 min question asks for + ve difference between part 1 and part 2 in minutes = 180 - 36 = 72 min b"
Answer: b
Problem: a train 125 m long passes a man , running at 8 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is :
Options: a ) 53 , b ) 50 , c ) 99 , d ) 288 , e ) 12
Rationale: speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 8 ) km / hr . x - 8 = 45 = = > x = 53 km / hr . answer : a
Answer: a
Problem: when tossed , a certain coin has an equal probability of landing on either side . if the coin is tossed 5 times , what is the probability that it will land on the same side each time ?
Options: a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 12 , d ) 1 / 16 , e ) 1 / 32
Rationale: "on the first toss , the coin will land on one side or the other . on the second toss , the probability of landing on the same side is 1 / 2 . on the third toss , the probability of landing on the same side is 1 / 2 . on the fourth toss , the probability of landing on the same side is 1 / 2 . on the fifth toss , the probability of landing on the same side is 1 / 2 . p ( same side all five times ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 . the answer is d ."
Answer: d
Problem: the closest approximation of d ( 69.28 × 0.004 ) / 0.03 is
Options: a ) 0.092 , b ) 0.92 , c ) 9.2 , d ) 92 , e ) 920
Rationale: "d ( 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 a 1000 m race , a beats b by 200 meters or 25 seconds . find the speed of b ?
Options: a ) 5 m / s , b ) 8 m / s , c ) 9 m / s , d ) 4 m / s , e ) 2 m / s
Rationale: b 8 m / s since a beats b by 200 m or 25 seconds , i t implies that b covers 200 m in 25 seconds . hence speed of b = 200 / 25 = 8 m / s .
Answer: b
Problem: robert ' s salary was decreased by 50 % and subsequently increased by 50 % . how much percentage does he lose ?
Options: a ) 10 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 50 %
Rationale: "let original salary be $ 100 new final salary = 150 % of ( 50 % of $ 100 ) = ( 150 / 100 ) * ( 50 / 100 ) * 100 = $ 75 decrease = 25 % correct option is c"
Answer: c
Problem: the average age of 10 members of a committee is the same as it was 4 years ago , because an old member has been replaced by a young member . find how much younger is the new member ?
Options: a ) 35 , b ) 15 , c ) 40 , d ) 52 , e ) 14
Rationale: c let the sum of nine member ( total ) = x and the age of old one = z so its average 4 yrs before = ( x + z ) / 10 . after 4 yrs let z be replaced by y . so now avg = ( x + 4 * 10 + y ) / 10 now ( x + z ) / 10 = ( x + 40 + y ) / 10 so after solving it found z = y + 40 . so old person is 40 yrs older than young one .
Answer: c
Problem: tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 157 per kg , the price of the third variety per kg will be
Options: a ) rs . 179.50 , b ) rs . 1700 , c ) rs . 175.50 , d ) rs . 180 , e ) none
Rationale: solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 157 / 22.50 = 1 = â € º x - 157 = 22.50 = â € º x = 179.50 . hence , price of the third variety = rs . 179.50 per kg . answer a
Answer: a
Problem: if 75 percent of 600 is 50 percent of x , then x = ?
Options: a ) 800 , b ) 900 , c ) 1000 , d ) 1100 , e ) 1200
Rationale: "0.75 * 600 = 0.5 * x x = 7.5 / 5 * 600 = 900"
Answer: b
Problem: there were 35 students in a hostel . due to the admission of 7 new students , the expenses of mess were increased by 42 per day while the average expenditure per head diminished by 1 . what was the original expenditure of the mess ?
Options: a ) 400 , b ) 400 , c ) 445 , d ) 465 , e ) none of the above
Rationale: let the original average expenditure be ` x . then , 42 ( x – 1 ) – 35 x = 42 ⇔ 7 x = 84 ⇒ x = 12 . ∴ original expenditure = ( 35 × 12 ) = 420 . answer b
Answer: b
Problem: a person can row at 10 kmph and still water . he takes 5 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 2 kmph ?
Options: a ) 32 km , b ) 25 km , c ) 28 km , d ) 24 km , e ) 20 km
Rationale: let the distance between a and b be x km . total time = x / ( 10 + 2 ) + x / ( 10 - 2 ) = 5 = > x / 12 + x / 8 = 5 = > ( 2 x + 3 x ) / 24 = 5 = > x = 24 km . answer : d
Answer: d
Problem: a tour group of 25 people paid a total of $ 945 for entrance to a museum . if this price included a 5 % sales tax , and all the tickets cost the same amount , what was the face value of each ticket price without the sales tax ? choices
Options: a ) $ 22 , b ) $ 23.94 , c ) $ 36 , d ) $ 25.20 , e ) $ 30
Rationale: soln : - 945 / 25 = x + 0.05 x 945 / 25 = 1.05 x x = 36 answer : c
Answer: c
Problem: carrie likes to buy t - shirts at the local clothing store . they cost $ 8.55 each . one day , she bought 9 t - shirts . how much money did she spend ?
Options: a ) $ 150 , b ) $ 248.75 , c ) $ 200 , d ) $ 171.6 , e ) $ 73.1025
Rationale: $ 8.55 * 9 = $ 73.1025 . answer is e .
Answer: e
Problem: what is the average of the first 10 multiples of 11 ?
Options: a ) 55.5 , b ) 61.5 , c ) 53.5 , d ) 78.5 , e ) 60.5
Rationale: "( 11 + 22 + . . . + 110 ) / 10 = 11 ( 1 + 2 + . . . + 10 ) / 10 = 11 ( 10 ) ( 11 ) / ( 10 * 2 ) = 121 / 2 = 60.5 option e"
Answer: e
Problem: the difference between the squares of two consecutive numbers is 37 . find the numbers .
Options: a ) 38 and 19 .
b ) 28 and 19 .
c ) 18 and 19 .
d ) 18 and 29 .
e ) none
Rationale: solution : if the difference between the squares of two consecutive numbers is x , then the numbers are ( x - 1 ) / 2 and ( x + 1 ) / 2 ( 37 - 1 ) / 2 and ( 37 + 1 ) / 2 36 / 2 and 38 / 2 therefore the required answer = 18 and 19 . answer c
Answer: c
Problem: a miniature roulette wheel is divided into 10 equal sectors , each bearing a distinct integer from 1 to 10 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun four times , approximately what is the probability that the product of the four winning sectors ’ integers will be even ?
Options: a ) 50 % , b ) 67 % , c ) 88 % , d ) 94 % , e ) 98 %
Rationale: "the only way to have an odd product is if all 4 integers are odd . p ( odd product ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 p ( even product ) = 1 - 1 / 16 = 15 / 16 which is about 94 % the answer is d ."
Answer: d
Problem: the diagonals of a rhombus are 25 cm and 30 cm . find its area ?
Options: a ) 250 , b ) 300 , c ) 375 , d ) 550 , e ) 750
Rationale: "area = 0.5 x 25 x 30 = 375 cm â ² answer : c"
Answer: c
Problem: what is 120 % of 13 / 24 of 840 ?
Options: a ) 436 , b ) 478 , c ) 512 , d ) 546 , e ) 620
Rationale: "120 % * 13 / 24 * 360 = 1.2 * 13 * 35 = 546 the answer is d ."
Answer: d
Problem: the avg weight of a , b & c is 84 kg . if d joins the group , the avg weight of the group becomes 80 kg . if another man e who weights is 5 kg more than d replaces a , then the avgof b , c , d & e becomes 79 kg . what is the weight of a ?
Options: a ) 45 , b ) 65 , c ) 77 , d ) 89 , e ) 90
Rationale: "a + b + c = 3 * 84 = 252 a + b + c + d = 4 * 80 = 320 - - - - ( i ) so , d = 68 & e = 68 + 5 = 73 b + c + d + e = 79 * 4 = 316 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 320 – 316 = 4 a = e + 4 = 73 + 4 = 77 c"
Answer: c
Problem: at golds gym class can be divided into 8 teams with an equal number of players on each team or into 12 teams with an equal number of players on each team . what is the lowest possible number of students in the class ?
Options: a ) 20 , b ) 24 , c ) 36 , d ) 48 , e ) 96
Rationale: we are given that golds gym class can be divided into 8 teams or 12 teams , with an equal number of players on each team . translating this into two mathematical expressions we can say , where g is the total number of students in the gym class , that : g / 8 = integer and g / 12 = integer this means that g is a multiple of both 8 and 12 . we are asked to determine the lowest number of students in the class , or the lowest value for variable “ g ” . because we know that g is a multiple of 8 and of 12 , we need to find the least common multiple of 8 and 12 . although there are technical ways for determining the least common multiple , the easiest method is to analyze the multiples of 8 and 12 until we find one in common . starting with 8 , we have : 8 , 16 , 24 , 32 for 12 , we have : 12 , 24 for the multiples of 12 , we stopped at 24 , because we see that 24 is also a multiple of 8 . thus , 24 is the least common multiple of 8 and 12 , and therefore we know that the lowest possible number of students in the gym class is 24 . answer b .
Answer: b
Problem: how many internal diagonals does a hexagon ( 6 sided polygon ) have ?
Options: a ) 5 , b ) 6 , c ) 8 , d ) 9 , e ) 10
Rationale: number of diagonals in any polygon can be found using this formula : n ( n - 3 ) / 2 here n = 6 no . of diagonals = 6 ( 6 - 3 ) / 2 = 9 ans d
Answer: d
Problem: a man sells an article at a profit of 25 % . if he had bought it at 20 % less and sold it for rs . 14.70 less , he would have gained 30 % . find the cost of the article .
Options: a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70
Rationale: "let c . p = 100 gain = 25 % s . p = 125 supposed c . p = 80 gain = 30 % s . p = ( 130 * 80 ) / 100 = 104 diff = ( 125 - 104 ) = 21 diff 21 when c . p = 100 then diff 14.70 when c . p = ( 100 * 14.70 ) / 21 = 70 answer : e"
Answer: e
Problem: find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1568 ?
Options: a ) 1000 , b ) 2777 , c ) 2889 , d ) 1400 , e ) 2771
Rationale: "1568 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1400 . answer : d"
Answer: d
Problem: the average ( arithmetic mean ) of the even integers from 0 to 200 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 0 to 100 inclusive ?
Options: a ) 10 , b ) 25 , c ) 50 , d ) 100 , e ) 200
Rationale: "the sum of even numbers from 0 to n is 2 + 4 + . . . + n = 2 ( 1 + 2 + . . . + n / 2 ) = 2 ( n / 2 ) ( n / 2 + 1 ) / 2 = ( n / 2 ) ( n / 2 + 1 ) the average is ( n / 2 ) ( n / 2 + 1 ) / ( n / 2 + 1 ) = n / 2 the average of the even numbers from 0 to 200 is 200 / 2 = 100 the average of the even numbers from 0 to 100 is 100 / 2 = 50 the answer is c ."
Answer: c
Problem: by selling 22 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ?
Options: a ) 83 , b ) 63 , c ) 53 , d ) 33 , e ) 13
Rationale: "80 % - - - 22 120 % - - - ? 80 / 120 * 22 = 33 answer : d"
Answer: d
Problem: solve the equation for x : 19 ( x + y ) + 17 = 19 ( - x + y ) - 21
Options: a ) 1 , b ) 2 , c ) - 1 , d ) - 3 , e ) - 4
Rationale: "c - 1 19 x + 19 y + 17 = - 19 x + 19 y - 21 38 x = - 38 = > x = - 1"
Answer: c
Problem: how many seconds will a 700 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ?
Options: a ) 23 sec , b ) 30 sec , c ) 27 sec , d ) 36 sec , e ) 42 sec
Rationale: "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 700 * 3 / 50 = 42 sec . answer : e"
Answer: e
Problem: two concentric circles form a ring . the inner and outer circumferences of ring are ( 352 / 7 ) m and ( 518 / 7 ) m respectively . find the width of the ring .
Options: a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) none of them
Rationale: "let the inner and outer radii be r and r meters . then 2 ( 22 / 7 ) r = ( 352 / 7 ) = r = ( ( 352 / 7 ) x ( 7 / 22 ) x ( 1 / 2 ) ) = 8 m . 2 ( 22 / 7 ) r = ( 528 / 7 ) = r = ( ( 528 / 7 ) x ( 7 / 22 ) x ( 1 / 2 ) ) = 12 m . therefore , width of the ring = ( r - r ) = ( 12 - 8 ) m = 4 m . answer is b ."
Answer: b
Problem: a person decided to build a house in 100 days . he employed 100 men in the beginning and 100 more after 25 days and completed the construction in stipulated time . if he had not employed the additional men , how many days behind schedule would it have been finished ?
Options: a ) 7 , b ) 5 , c ) 2 , d ) 3 , e ) 1
Rationale: "200 men do the rest of the work in 100 - 25 = 75 days 100 men can do the rest of the work in 75 * 200 / 100 = 75 days required number of days = 75 - 80 = 5 days answer is b"
Answer: b
Problem: the initial price of an article is rs . 4000 which increases 40 % increse in its price in the first year , a 20 % decrease in the second year and a 20 % increase in the next year . what is the final price of the article ?
Options: a ) rs . 2928 , b ) rs . 5928 , c ) rs . 3928 , d ) rs . r 4928 , e ) rs . 6928
Rationale: "the initial price of the article is rs . 4000 . in the 1 st year , price of the article = 4000 + 1600 = rs . 5600 . in the 2 nd year , price = 5600 - 20 % of 5600 = 5600 - 1120 = rs . 4480 . in the 3 rd year , price = 4480 + 10 % of 4480 = 4480 + 448 = rs . 4928 required price = = rs . 4928 . answer : d"
Answer: d
Problem: if 3 men or 4 women can reap a field in 10 days how long will 5 men and 6 women take to reap it ?
Options: a ) 19 / 60 , b ) 9 / 60 , c ) 2 / 3 , d ) 5 / 18 , e ) 6 / 24
Rationale: explanation : 3 men reap 1 / 10 field in 1 day 1 man reap 1 / ( 3 x 10 ) 4 women reap 1 / 10 field in 1 day 1 woman reap 1 / ( 10 x 4 ) 5 men and 6 women reap ( 5 / ( 3 x 10 ) + 6 / ( 4 x 10 ) ) = 19 / 60 in 1 day 5 men and 6 women will reap the field in 19 / 60 days answer : option a
Answer: a
Problem: qames can eat 25 marshmallows is 20 minutes . dylan can eat 25 in one hour . in how much time will the two eat 150 marshmallows ?
Options: a ) 40 minutes . , b ) 1 hour and 30 minutes . , c ) 1 hour . , d ) 1 hour and 40 minutes . , e ) 2 hours and 15 minutes .
Rationale: rate = output / time qames rate = 25 / 20 = 5 / 4 dylan rate = 25 / 60 = 5 / 12 combined rate = 5 / 4 + 5 / 12 = 20 / 12 combinedrate * combinedtime = combinedoutput 20 / 12 * t = 150 t = 90 mins = > 1 hr 30 min
Answer: c
Problem: ajay and balu together can do a piece of work in 12 days . balu and charan together can do the same work in 16 days . after ajay has been working at it for 5 days and balu for 7 days , charan finishes it in 13 days . in how many days will charan alone be able to do the work ?
Options: a ) 16 days , b ) 24 days , c ) 36 days , d ) 48 days , e ) 58 days
Rationale: let the total work be 48 units . ajay and balu ’ s one day work = 48 / 12 = 4 units . balu and charan ’ s one day work = 48 / 16 = 3 units . ajay ’ s one day work = 4 - balu ’ s one day work . balu ’ s one day work = 3 - charan ’ s one day work . therefore , ajay ’ s one day work = 1 + charan ’ s one day work . also given , 5 x ajay ’ s one day work + 7 x balu ’ s one day work + 13 x charan ’ s one day work = 48 equating all the equations formed , we get charan ’ s one day work is 2 units . hence , charan can finish the whole work in 48 / 2 = 24 days . answer : b
Answer: b
Problem: city a to city b , john drove for 1 hour at 40 mph and for 3 hours at 60 mph . what was the average speed for the whole trip ?
Options: a ) 57 , b ) 57.5 , c ) 58.2 , d ) 59 , e ) 55
Rationale: the total distance is 1 × 40 + 3 × 60 = 220 and the total time is 4 hours . hence , average speed = ( total distance / total time ) = 220 / 4 = 55 answer : e
Answer: e
Problem: if each digit in the set a = { 1 , 2 , 3 , 4 , 5 , 6 } is used exactly once , in how many ways can the digits be arranged ?
Options: a ) 720 , b ) 830 , c ) 940 , d ) 1050 , e ) 1160
Rationale: 6 ! = 720 the answer is a .
Answer: a
Problem: find the average of all prime numbers between 12 and 24 .
Options: a ) 30 , b ) 33.6 , c ) 18 , d ) 20 , e ) none
Rationale: "sol . there are five prime numbers between 12 and 24 . they are 13 , 17 , 19 , 23 â ˆ ´ required average = [ 13 + 17 + 19 + 23 / 4 ] = 72 / 4 = 18 answer c"
Answer: c
Problem: a ’ s speed is 20 / 19 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ?
Options: a ) 1 / 19 , b ) 3 / 19 , c ) 1 / 10 , d ) 1 / 20 , e ) 3 / 10
Rationale: "let d be the full distance . let x be the fraction of the distance that b runs . let v be the speed at which b runs . the time should be the same for both runners . time = d / ( 20 v / 19 ) = xd / v ( 19 / 20 ) * d / v = x * d / v x = 19 / 20 b should have a head start of 1 / 20 of the full distance . the answer is d ."
Answer: d
Problem: sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 504 per week . how much does she earn in dollars per hour ?
Options: a ) 2 , b ) 8 , c ) 9 , d ) 14 , e ) 24
Rationale: explanation : total hours worked = 8 x 3 + 6 x 2 = 36 total earned = 504 . hourly wage = 504 / 36 = 14 answer : d
Answer: d
Problem: if 1 = 6 , 2 = 12 , 3 = 18 , 4 = 24 , 5 = 30 , then 6 = ?
Options: a ) 5 , b ) 3 , c ) 1 , d ) 7 , e ) 9
Rationale: solution : 1 as stated 1 = 6 = > 6 = 1 answer c
Answer: c
Problem: 3 partners a , b , c start a business . twice a ’ s capital is equal to thrice b ’ s capital and b ’ s capital is 4 times c ’ s capital . out of a total profit of rs . 16500 at the end of the year , b ’ share is :
Options: a ) 2000 , b ) 4000 , c ) 5000 , d ) 6000 , e ) none
Rationale: sol . let c = x . then , b = 4 x and 2 a = 3 * 4 x = 12 x or a = 6 x . ∴ a : b : c = 6 x : 4 x : x = 6 : 4 : 1 . so , b ’ s capital = rs . [ 16500 * 4 / 11 ] = rs . 6000 . answer d
Answer: d
Problem: a certain college ' s enrollment at the beginning of 1992 was 20 percent greater than it was at the beginning of 1991 , and its enrollment at the beginning of 1993 was 25 percent greater than it was at the beginning of 1992 . the college ' s enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ?
Options: a ) 47.5 % , b ) 44 % , c ) 50 % , d ) 35 % , e ) 38 %
Rationale: suppose enrollment in 1991 was 100 then enrollment in 1992 will be 120 and enrollment in 1993 will be 120 * 1.25 = 150 increase in 1993 from 1991 = 150 - 100 = 50 answer : c
Answer: c
Problem: a is three times as fast as b . if b alone can do a piece of work in 12 days , in what time can a and b together complete the work ?
Options: a ) 5 , b ) 8 , c ) 3 , d ) 2 , e ) 12
Rationale: "a can do the work in 12 / 3 i . e . , 4 days . a and b ' s one day ' s work = 1 / 4 + 1 / 12 = ( 3 + 1 ) / 12 = 1 / 3 so a and b together can do the work in 3 days . answer : c"
Answer: c
Problem: by selling 10 pencils for a rupee a man loses 15 % . how many for a rupee should he sell in order to gain 15 % ?
Options: a ) 8.39 , b ) 6.39 , c ) 7.39 , d ) 9.39 , e ) 2.39
Rationale: 85 % - - - 10 115 % - - - ? 85 / 115 * 10 = 7.39 answer : c
Answer: c
Problem: a trader bought a car at 20 % discount on its original price . he sold it at a 100 % increase on the price he bought it . what percent of profit did he make on the original price ?
Options: a ) 18 % , b ) 13 % , c ) 60 % , d ) 40 % , e ) 15 %
Rationale: "original price = 100 cp = 80 s = 80 * ( 200 / 100 ) = 160 100 - 160 = 60 % answer : c"
Answer: c
Problem: a can complete a certain job in 7 days . b is 40 % more efficient than a . in how many days can b complete the same job ?
Options: a ) 5 , b ) 6.25 , c ) 7 , d ) 7.5 , e ) 4.8
Rationale: "let , total work unit = 70 units a can finish in 7 days = 70 unit work i . e . a can finish in 1 days = 10 unit work i . e . b can finish in 1 days = 10 + ( 40 / 100 ) * 10 = 14 unit work days in which b will complete the work alone = 70 / 14 = 5 days answer : option a"
Answer: a
Problem: the figure above shows the dimensions of a semicircular cross section of a one - way tunnel . the single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel . if vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane , what should be the limit t on the height of vehicles that are allowed to use the tunnel ?
Options: a ) 5 ½ ft
b ) 7 ½ ft
c ) 8 ½ ft
d ) 9 ½ ft
e ) 10 ft
Rationale: let ' s label the midpoint of the circle o . since the base of the semi - circle is 20 , we know that the diameter is 20 and , accordingly , the radius is 10 . we also know that the traffic lane is 12 feet long and there ' s an equal amount of space on either side , so the traffic lane extends 6 feet on either side of o . let ' s call the leftmost point on the base of the traffic lane a . so , the distance oa is 6 . now draw a line straight up from a to the top of the tunnel . let ' s label the point at which the line intersects the circle b . the answer to the question will , therefore , be the height ab - . 5 feet ( we need to leave . 5 feet of clearance ) . here ' s the key to solving the question : if we draw a line from o to b , that line is a radius of the circle and , therefore , has length 10 . we now have right triangle oab ( the right angle is at point a ) , with leg oa = 6 and hypotenuse ob = 10 . we can now solve for leg ab = 8 ( either by applying the pythagorean theorum or by applying the 3 / 4 / 5 special right triangle ratio ) . finally : ab = 8 , so the correct answer t is 8 - . 5 = 7.5 . . . choose ( b ) ! from a strategic guessing point of view , as soon as we realize that the height of the tunnel is 10 in the middle , we should quickly eliminate ( d ) and ( e ) as too big ; worse case you have a 1 / 3 shot at picking up the points . b
Answer: b
Problem: if a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster , he would have covered 70 more miles than he actually did . how many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day ?
Options: a ) 100 , b ) 120 , c ) 140 , d ) 150 miles , e ) 160
Rationale: let v and t be initial parameter vt = d from the stem ( t + 1 ) ( v + 5 ) = 70 + d = > v + 5 t = 65 - - - - - - - 1 ) we need the additional distance traveled second time at v + 10 and t + 2 ( t + 2 ) ( v + 10 ) = vt + 2 v + 10 t + 20 but vt = d hence d + 2 ( v + 5 t ) + 20 but v + 5 t = 65 hence the additional distance traveled is 2 ( v + 5 t ) + 20 = 2 * 65 + 20 = 130 + 20 = 150 d
Answer: d