regisss/math_qa huggingface dataset:
Problem: a certain tire company can produce tires at a cost of $ 22500 per batch plus $ 8 per tire . the company can sell tires to the wholesaler at a cost of $ 20 per tire . if a batch of 15000 tires is produced and sold , what is the company ’ s profit per tire ?
Options: a ) $ 9.00 , b ) $ 9.50 , c ) $ 10.50 , d ) $ 13.50 , e ) $ 20.00
Rationale: cp ( 15000 tires ) = $ 22500 + ( $ 8 × 15000 ) = $ 142500 sp ( 15000 tires ) = $ 20 × 15000 = $ 300000 profit = sp - cp = $ 300000 - $ 142500 = $ 157500 profit / tire = $ 157500 / 15000 = $ 10.50 answer c
Answer: c
Problem: at a certain company , two - thirds of the employees had cell phones and two - fifths had pagers . if one - third of the employees had neither cell phones nor pagers then what fraction of the employees had both cell phones and pagers ?
Options: a ) 7 / 5 , b ) 6 / 11 , c ) 2 / 5 , d ) 1 / 3 , e ) 4 / 15
Rationale: let there be 15 employees cell phone = 10 pagers = 6 neither pager nor cell phone = 5 so , 10 = 10 + 6 - x or , x = 6 so , fraction of the employees had both cell phones and pagers = 6 / 15 = > 2 / 5 answer : c
Answer: c
Problem: the original price of a suit is $ 200 . the price increased 30 % , and after this increase , the store published a 30 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 30 % off the increased price , how much did these consumers pay for the suit ?
Options: a ) $ 182 , b ) $ 191 , c ) $ 200 , d ) $ 209 , e ) $ 219
Rationale: "given the foregoing discussion , it may be obvious now the trap - mistake answer is ( c ) . even if you can ’ t remember the correct thing to do , at the very least , learn to spot the trap ! the multiplier for a 30 % increases is 1 + 0.30 = 1.3 , and the multiplier for a 30 % decrease is 1 – 0.30 = 0.70 , so the combined change is 1.3 * 0.7 = 0.91 , 91 % percent of the original , or a 9 % decreases . now , multiply $ 200 * 0.91 = $ 182 . answer = ( a ) ."
Answer: a
Problem: a boat can travel with a speed of 22 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 81 km downstream
Options: a ) 5 hours , b ) 4 hours , c ) 3 hours , d ) 2 hours , e ) none of these
Rationale: "explanation : speed of the boat in still water = 22 km / hr speed of the stream = 5 km / hr speed downstream = ( 22 + 5 ) = 27 km / hr distance travelled downstream = 81 km time taken = distance / speed = 81 / 27 = 3 hours . answer : option c"
Answer: c
Problem: a man can do a piece of work in 5 days , but with the help of his son he can do it in 4 days . in what time can the son do it alone ?
Options: a ) 2 days , b ) 6 days , c ) 12 days , d ) 20 days , e ) none of these
Rationale: "explanation : in this type of question , where we have one person work and together work done . then we can easily get the other person work just by subtracting them . as son ' s one day work = ( 1 / 4 − 1 / 5 ) = ( 5 − 4 ) / 20 = 1 / 20 so son will do whole work in 20 days answer : d"
Answer: d
Problem: the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 149 , what is the difference between the ages of patrick and monica ?
Options: a ) 27 , b ) 64 , c ) 45 , d ) 72 , e ) 18
Rationale: "ages of p and mi = 3 x + 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 47 x = 149 x = 4 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 4 = 64 answer b"
Answer: b
Problem: a dealer offers a cash discount of 15 % and still makes a profit of 36 % when he further allows 25 articles to be sold at the cost price of 20 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ?
Options: a ) 60 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 150 %
Rationale: given cash discount - 15 % profit - 36 % items sold - 25 price sold at = list price of 20 assume list price = $ 10 total invoice = $ 200 - 15 % cash discount = $ 170 let cost price of 25 items be x so total cost = 25 * x given the shopkeeper had a profit of 36 % 25 * x * 136 / 100 = 170 or x = $ 5 which means his products were listed at $ 10 which is a 100 % markup over $ 5 answer c
Answer: c
Problem: if x + y = 2 and x 2 y 3 + y 2 x 3 = 32 , what is the value of xy ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
Rationale: "xy = 4 as x + y = 2 x 2 y 3 + y 2 x 3 = 32 x 2 y 2 ( y + x ) = 32 substituting x + y x 2 y 2 = 16 xy = 4 answer : d"
Answer: d
Problem: a man swims downstream 72 km and upstream 45 km taking 3 hours each time ; what is the speed of the current ?
Options: a ) 1.5 , b ) 3.5 , c ) 4.5 , d ) 2.5 , e ) 5.5
Rationale: "72 - - - 3 ds = 24 ? - - - - 1 45 - - - - 3 us = 15 ? - - - - 1 s = ? s = ( 24 - 15 ) / 2 = 4.5 answer : c"
Answer: c
Problem: there are 4 books on a shelf , of which 3 are paperbacks and 6 are hardbacks . how many possible selections of 4 books from this shelf include at least one paperback ?
Options: a ) a ) 40 , b ) b ) 45 , c ) c ) 50 , d ) d ) 55 , e ) e ) 111
Rationale: "approach 1 at - least 1 paper back = total - no paper back 9 c 4 - 6 c 4 = 111 approach 2 at - least 1 paper back = 1 paper back , 3 hard back or 2 paper back 2 hard back = 3 c 1 * 6 c 3 + 3 c 2 * 6 c 2 + 3 c 3 * 6 c 1 = 111 answer is e"
Answer: e
Problem: if the price of a book is first decreased by 20 % and then increased by 10 % , then the net change in the price will be :
Options: a ) 10 , b ) 12 , c ) 37 , d ) 29 , e ) 22
Rationale: explanation : let the original price be rs . 100 . decreased by 20 % = 80 then increased 10 % on rs 80 = 80 + 8 = 88 net change in price = 100 - 88 = 12 answer : b
Answer: b
Problem: a small company is planning to rent either computer a or computer b to print customer mailing lists . both computer a and computer b must be rented on an hourly basis . the rental fee is based only on the amount of time the computer is turned on . it will cost 40 percent more per hour to rent computer a than to rent computer b . computer b would , however , require 20 hours more than computer a to do the job . if either computer a , or computer b were rented the total cost to rent the computer would be $ 880.00 . what would be the approximate hourly charge to rent computer b ?
Options: a ) $ 9.40 , b ) $ 12.57 , c ) $ 13.20 , d ) $ 17.80 , e ) $ 22.10
Rationale: "pa = price of a pb = price of b ta = time for a to complete the job tb = time for b to complete the job given pa = 1.4 pb ta + 20 = tb pa * ta = pb * tb = 880 1.4 pb * ( tb - 20 ) = pb * tb 1.4 pb tb - pb tb = 1.4 pb * 20 0.4 pbtb = 28 pb tb = 28 / 0.4 = 70 pb = 880 / 70 ~ 12.57 b"
Answer: b
Problem: two men a and b start from place x walking at 4 ½ kmph and 5 ¾ kmph respectively . how many km apart they are at the end of 7 ½ hours if they are walking in the same direction ?
Options: a ) 7 km , b ) 6 km , c ) 9 3 / 8 km , d ) 9 km , e ) 5 km
Rationale: "rs = 5 ¾ - 4 ½ = 1 ¼ t = 7 ½ h . d = 5 / 4 * 15 / 2 = 75 / 8 = 9 3 / 8 km answer : c"
Answer: c
Problem: the regular price per can of a certain brand of soda is $ 0.60 . if the regular price per can is discounted 20 percent when the soda is purchased in 24 - can cases , what is the price of 72 cans of this brand of soda purchased in 24 - can cases ?
Options: a ) $ 16.32 , b ) $ 18.00 , c ) $ 21.60 , d ) $ 34.56 , e ) $ 28.80
Rationale: "the discounted price of one can of soda is ( 0.8 ) ( $ 0.60 ) , or $ 0.48 therefore , the price of 72 cans of soda at the discounted price would be ( 72 ) ( $ 0.48 ) = 34.56 answer : d ."
Answer: d
Problem: the cost price of an article is 36 % of the marked price . calculate the gain percent after allowing a discount of 20 % .
Options: a ) 22.2 % , b ) 36 % , c ) 80 % , d ) 122.2 % , e ) none
Rationale: "sol . let marked price = rs . 100 . then , c . p . = rs . 36 . s . p = rs . 80 . â ˆ ´ gain % = [ 44 / 36 * 100 ] % = 122.2 % . answer d"
Answer: d
Problem: the cost of 2 chairs and 3 tables is rs . 1300 . the cost of 3 chairs and 2 tables is rs . 1200 . the cost of each table is more than that of each chair by ?
Options: a ) 50 , b ) 60 , c ) none of these , d ) 70 , e ) 100
Rationale: "c 100 2 c + 3 t = 1300 - - - ( 1 ) 3 c + 3 t = 1200 - - - ( 2 ) subtracting 2 nd from 1 st , we get - c + t = 100 = > t - c = 100"
Answer: c
Problem: he ratio between the sale price and the cost price of an article is 5 : 1 . what is the ratio between the profit and the cost price of that article ?
Options: a ) 2 : 6 , b ) 2 : 9 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 5
Rationale: "let c . p . = rs . x and s . p . = rs . 5 x . then , gain = rs . 4 x required ratio = 2 x : x = 2 : 1 answer : d"
Answer: d
Problem: the average of first five multiples of 5 is ?
Options: a ) 6 , b ) 8 , c ) 9 , d ) 15 , e ) 7
Rationale: "average = 5 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 75 / 5 = 15 . answer : d"
Answer: d
Problem: mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 19008 , how much amount per month would he have spent on clothes and transport ?
Options: a ) 4038 , b ) 8076 , c ) 9691.2 , d ) 1584 , e ) 1625
Rationale: "∵ amount , he have spent in 1 month on clothes transport = amount spent on saving per month ∵ amount , spent on clothes and transport = 19008 ⁄ 12 = 1584 answer d"
Answer: d
Problem: express 15 mps in kmph ?
Options: a ) 22 , b ) 88 , c ) 90 , d ) 21 , e ) 54
Rationale: "15 * 18 / 5 = 54 kmph answer : e"
Answer: e
Problem: if | 5 x - 30 | = 100 , then find the sum of the values of x ?
Options: a ) 1 , b ) - 2 , c ) 12 , d ) - 3 , e ) 4
Rationale: "| 5 x - 30 | = 100 5 x - 30 = 100 or 5 x - 30 = - 100 5 x = 130 or 5 x = - 70 x = 26 or x = - 14 sum = 26 - 14 = 12 answer is c"
Answer: c
Problem: the class mean score on a test was 60 , and the standard deviation was 5 . if jack ' s score was within 2 standard deviations of the mean , what is the lowest score he could have received ?
Options: a ) 30 , b ) 31 , c ) 45 , d ) 40 , e ) 89
Rationale: "1 sd from the mean is adding and subtrating the amount if standard deviation from the mean one time . 2 sd from the mean is adding and subtracting twice . 1 sd from the mean ranges from 65 to 55 , where 65 is within sd above the mean and 55 within 1 sd below the mean 2 sd = 5 twice = 10 from the the mean , which is 70 to 40 , where 70 is within 2 sd above the mean and 40 is within 2 sd below the mean . answer = d"
Answer: d
Problem: a man buy a book in rs 50 & sale it rs 80 . what is the rate of profit ? ? ?
Options: a ) 10 % , b ) 60 % , c ) 30 % , d ) 25 % , e ) 28 %
Rationale: "cp = 50 sp = 80 profit = 80 - 50 = 30 % = 30 / 50 * 100 = 60 % answer : b"
Answer: b
Problem: find the simple interest on rs . 71,200 at 16 2 / 3 % per annum for 9 months .
Options: a ) s . 8500 , b ) s . 8900 , c ) s . 7500 , d ) s . 7000 , e ) s . 6500
Rationale: "p = rs . 71200 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 71,200 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8900 answer is b ."
Answer: b
Problem: a number is doubled and 9 is added . if resultant is trebled , it becomes 51 . what is that number
Options: a ) 4 , b ) 10 , c ) 12 , d ) 14 , e ) 15
Rationale: "explanation : = > 3 ( 2 x + 9 ) = 51 = > 2 x + 9 = 17 = > x = 4 answer : option a"
Answer: a
Problem: a car was driving at 80 km / h for 40 minutes , and then at 60 km / h for another 40 minutes . what was its average speed ?
Options: a ) 113.33 . , b ) 103.33 , c ) 93.33 , d ) 83.33 , e ) 73.33
Rationale: "driving at 80 km / h for 40 minutes , distance covered = 80 * 2 / 3 = 53.33 km driving at 90 km / h for 40 minutes , distance covered = 90 * 2 / 3 = 60 km average speed = total distance / total time = 113.33 / 1 = 113.33 km / h answer : 113.33 km / h"
Answer: a
Problem: in x game of billiards , x can give y 20 points in 45 and he can give z 30 points in 45 . how many points can y give z in x game of 100 ?
Options: a ) 30 , b ) 20 , c ) 25 , d ) 40 , e ) 50
Rationale: "x scores 45 while y score 40 and z scores 30 . the number of points that z scores when y scores 100 = ( 100 * 30 ) / 25 = 60 . in x game of 100 points , y gives ( 100 - 60 ) = 40 points to c . d"
Answer: d
Problem: bill spends two days driving from point a to point b . on the first day , he drove 2 hours longer and at an average speed 5 miles per hour faster than he drove on the second day . if during the two days he drove a total of 680 miles over the course of 18 hours , what was his average speed on the second day , in miles per hour ?
Options: a ) 20 , b ) 25 , c ) 28 , d ) 30 , e ) 35
Rationale: first recognize that we have two pieces of information regarding the time bill spent driving each day . on day 1 , bill drove 2 hours longer than he drove on day 2 . so , let x = # of driving hours on day 2 then x + 2 = # of driving hours on day 1 bill drove a total of 18 hours so , x + ( x + 2 ) = 18 simplify : 2 x + 2 = 18 solve , x = 8 so , bill drove 10 hours on day 1 and he drove 8 hours on day 2 now let ' s solve the question by starting with a word equation . let x = speed driven on day 2 so , x + 5 = speed driven on day 1 ( distance traveled on day 1 ) + ( distance traveled on day 2 ) = 680 distance = ( rate ) ( time ) we get : ( x + 5 ) ( 10 ) + ( x ) ( 8 ) = 680 expand : 10 x + 50 + 8 x = 680 simplify : 18 x + 50 = 680 18 x = 630 x = 35 ( mph ) answer : e
Answer: e
Problem: the expenditure of an woman increase consistently by 30 % per year . if her present expenditure is rs . 1000 then what will her expenditure be after 3 years ?
Options: a ) 2107 , b ) 2197 , c ) 2187 , d ) 2177 , e ) 2167
Rationale: explanation : expenditure = 1000 x 1.3 x 1.3 x 1.3 = 2197 answer : option b
Answer: b
Problem: what percent of 6.3 kg is 25 gms ?
Options: a ) 39 , b ) 66 , c ) 288 , d ) 19 , e ) 17
Rationale: "explanation : required percentage = ( 25 / 63000 * 100 ) % = 39 / 100 % = 0.39 % answer : a ) . 39 %"
Answer: a
Problem: a rectangular floor is covered by a rug except for a strip 4 meters wide along each of the four edge . if the floor is 25 meters by 20 meters , what is the area of the rug in square meters ?
Options: a ) 256
b ) 266
c ) 204
d ) 224
e ) 324
Rationale: a strip of 4 meters is covering the inner rectangular rug for all 4 sides . length of inner rug = 25 - ( 2 * 4 ) breadth of inner rug = 20 - ( 2 * 4 ) area of rug = 17 * 12 = 204 sq . mt â nswer : c
Answer: c
Problem: the average of 45 results is 9 . the average of first 22 of them is 15 and that of last 22 is 20 . find the 23 result ?
Options: a ) 515 , b ) 480 , c ) 750 , d ) 650 , e ) 560
Rationale: "23 th result = sum of 45 results - sum of 44 results 9 * 45 - 15 * 22 + 20 * 22 = 450 - 330 + 440 = 515 answer is a"
Answer: a
Problem: at what rate percent per annum will a sum of money double in 5 years .
Options: a ) 20.0 % , b ) 13.5 % , c ) 11.5 % , d ) 14.5 % , e ) 21.5 %
Rationale: let principal = p , then , s . i . = p and time = 8 years rate = [ ( 100 x p ) / ( p x 5 ) ] % = 20.0 % per annum . answer : a
Answer: a
Problem: the percentage increase in the area of rectangle , if each of its side is increased by 20 % is ?
Options: a ) 30 % , b ) 44 % , c ) 50 % , d ) 55 % , e ) 60 %
Rationale: "let original length = x meters original breadth = y meters original area = xy m ^ 2 new length = 120 x / 100 new breadth = 120 y / 100 = 6 y / 5 new area = 6 x / 5 * 6 y / 5 = 36 xy / 25 m ^ 2 increase percent = 11 xy / 25 * 1 / xy * 100 = 44 % answer is b"
Answer: b
Problem: a cube is painted such that one pair of surfaces is painted brown and the other pair of surfaces is painted orange . the cube is cut in to 64 small cubes of equal size . find how many cubes have both the color brown and orange ?
Options: a ) 22
b ) 16
c ) 11
d ) 12
e ) 10
Rationale: there are 6 surfaces to the cube . in which opposite surfaces are painted brown and other 2 opposite surfaces are painted orange . then 4 edges of the cube have both the colors . since the big cube cut in to 64 smaller cubes , the 4 edges of the cube have both the colors . since the big cube cut into 64 smaller cubes , the 4 edges of the big cube has 4 × 4 = 16 smaller cubes have both brown and orange colors . answer = 16 answer : b
Answer: b
Problem: a bag contains 4 red , 4 blue and 2 green balls . if 2 ballsare picked at random , what is the probability that both are red ?
Options: a ) 2 / 15 , b ) 2 / 21 , c ) 5 / 26 , d ) 3 / 29 , e ) 4 / 27
Rationale: "p ( both are red ) , = 4 c 210 c 2 = 4 c 210 c 2 = 6 / 45 = 2 / 15 a"
Answer: a
Problem: in the xy - coordinate system , if ( m , n ) and ( m + 3 , n + k ) are two points on the line with the equation x = 2 y + 5 , then k =
Options: a ) 1 / 2 , b ) 1 , c ) 2 , d ) 3 / 2 , e ) 4
Rationale: since ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 5 they should satisfy m = 2 n + 5 and m + 3 = 2 * ( n + k ) + 5 . by 1 st equation we have m - 2 n = 5 and by 2 nd equation m - 2 n = 2 k + 2 - - - > 5 = 2 k + 2 - - - > k = 3 / 2 . the answer is , therefore , ( d ) .
Answer: d
Problem: each factor of 210 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 45 ?
Options: a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 4
Rationale: "210 = 2 * 3 * 5 * 7 , so the # of factors 210 has is ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 16 ( see below ) ; 42 = 2 * 3 * 7 , so out of 16 factors only two are multiples of 42 : 42 and 210 , itself ; so , the probability is 2 / 16 = 5 / 42 . answer : b ."
Answer: b
Problem: in a garden , 26 trees are planted at equal distances along a yard 300 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ?
Options: a ) 10 metres , b ) 20 metres , c ) 14 metres , d ) 12 metres , e ) 13 metres
Rationale: "between 26 trees , there are 25 gaps length of each gap = 300 / 25 = 12 i . e . , distance between two consecutive trees = 12 answer : d"
Answer: d
Problem: the product of 4 consecutive even numbers is always divisible by :
Options: a ) 384 , b ) 350 , c ) 400 , d ) 200 , e ) 250
Rationale: the product of 4 consecutive numbers is always divisible by 4 ! . since , we have 4 even numbers , we have an additional 2 available with each number . now , using both the facts , we can say that the product of 4 consecutive even numbers is always divisible by , 2 ^ 4 * 4 ! 16 * 24 = 384 answer a
Answer: a
Problem: workers at a campaign office have 2500 fliers to send out . if they send out 1 / 5 of them in the morning and 1 / 4 of the remaining ones out during the afternoon , how many are left for the next day ?
Options: a ) 300 , b ) 800 , c ) 1500 , d ) 1200 , e ) 1900
Rationale: ( 1 / 5 ) * 2500 = 500 remaining = 2500 - 500 = 2000 ( 1 / 4 ) of remaining = ( 1 / 4 ) * 2000 = 500 remaining now = 2000 - 500 = 1500 answer : option c
Answer: c
Problem: two numbers n and 24 have lcm = 48 and gcf = 8 . find n .
Options: a ) 16 , b ) 56 , c ) 76 , d ) 87 , e ) 24
Rationale: "the product of two integers is equal to the product of their lcm and gcf . hence . 24 * n = 48 * 8 n = 48 * 8 / 24 = 16 correct answer a"
Answer: a
Problem: what is the sum of all the multiples of 8 between 30 and 100 ?
Options: a ) 485 , b ) 72 , c ) 547 , d ) 104 , e ) 624
Rationale: "you first have to know all the multiples of 8 between 30 and 100 . they are 8 , 16 , 24 , 32,40 , 48,56 , 64,72 , 80,88 and 96 . if you add all these numbers together , you get 624 . final answer : e"
Answer: e
Problem: a trailer carries 3 , 4 and 5 crates on a trip . each crate weighs no less than 1250 kg . what is the maximum weight of the crates on a single trip ?
Options: a ) 1250 , b ) 6250 , c ) 600 , d ) 7500 , e ) 375
Rationale: "max no . of crates = 5 . max weight = 1250 kg max . weight carried = 5 * 1250 = 6250 kg = b"
Answer: b
Problem: what is the remainder when 1271 * 1276 * 1277 * 1285 is divided by 12 ?
Options: a ) 0 , b ) 1 , c ) 11 , d ) 9 , e ) 4
Rationale: "1271 * 1275 * 1277 * 1285 = ( 1272 - 1 ) ( 1272 + 4 ) ( 1284 - 7 ) ( 1284 + 1 ) = ( multiple of 12 - 1 ) ( multiple of 12 + 3 ) ( multiple of 12 - 7 ) ( multiple of 12 + 1 ) now , if we expand this expression all terms but the last one will have a multiple of 12 in them , thus will leave no remainder upon division by 12 , the last term will be ( - 1 ) ( 4 ) ( - 7 ) ( 1 ) = 28 . thus the question becomes : what is the remainder upon division of 28 by 12 ? the answer to this question is 4 . answer : e"
Answer: e
Problem: what will be the difference between simple and compound interest @ 10 % per annum on a sum of $ 1000 after 4 years ?
Options: a ) $ 48.12 , b ) $ 52.64 , c ) $ 64.10 , d ) $ 72.85 , e ) $ 60.10
Rationale: "s . i . = 1000 * 10 * 4 / 100 = $ 400 c . i . = 1000 * ( 1 + 10 / 100 ) ^ 4 - 1000 = $ 464.10 difference = 464.10 - 400 = $ 64.10 answer is c"
Answer: c
Problem: if n is a positive integer and n ^ 2 is divisible by 150 , then what is the largest positive integer that must divide n ?
Options: a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50
Rationale: "150 = 2 * 3 * 5 ^ 2 if 150 divides n ^ 2 , then n must be divisible by 2 * 3 * 5 = 30 the answer is c ."
Answer: c
Problem: the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 196 , what is the difference between the ages of patrick and monica ?
Options: a ) 27 , b ) 64 , c ) 45 , d ) 72 , e ) 18
Rationale: "ages of p and mi = 3 x : 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 49 x = 196 x = 4 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 4 = 64 answer b"
Answer: b
Problem: if tim had lunch at $ 50.50 and he gave 20 % tip , how much did he spend ?
Options: a ) $ 30.60 , b ) $ 60.60 , c ) $ 70.60 , d ) $ 40.60 , e ) $ 50.60
Rationale: "the tip is 20 % of what he paid for lunch . hence tip = 20 % of 50.50 = ( 20 / 100 ) * 50.50 = 101 / 100 = $ 10.10 total spent 50.50 + 10.10 = $ 60.60 correct answer b"
Answer: b
Problem: 20 . a certain church bell rings the bell twice at half past the hour and 4 times at the hour plus an additional number of rings equal to what ever time it is . how many rings will the clock make from 6 : 20 in the morning to 09 : 50 in the morning ?
Options: a ) 44 . , b ) 36 . , c ) 42 . , d ) 46 . , e ) 50 .
Rationale: unless i understood the problem correctly , i get 58 @ 6 : 30 - 2 @ 7 - 4 + 7 = 11 @ 7 : 30 - 2 @ 8 - 12 @ 8 : 30 - 2 @ 9 - 13 @ 9 : 30 - 2 totals to a = 44
Answer: a
Problem: if a book is sold at 7 % profit instead of 7 % loss , it would have brought rs 14 more . find out the cost price of the book
Options: a ) 75 , b ) 100 , c ) 60 , d ) 70 , e ) 80
Rationale: "let c . p . of the book be rs . ’ x ’ given , 1.07 x - 0.93 x = 14 = > 0.14 x = 14 = 14 / 0.14 = rs 100 answer : b"
Answer: b
Problem: mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 460 ?
Options: a ) 18 % , b ) 21 % , c ) 25.9 % , d ) 19 % , e ) none of these
Rationale: cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 460 , discount = 621 - 460 = 161 discount % = 161 / 621 * 100 = 25.9 % answer : c
Answer: c
Problem: a shopkeeper has 280 kg of apples . he sells 50 % of these at 20 % profit and remaining 60 % at 30 % profit . find his % profit on total .
Options: a ) 24 % , b ) 25 % , c ) 26 % , d ) 28 % , e ) 35 %
Rationale: "if the total quantity was 100 then 50 x 20 % + 60 x 30 % = 28 this profit will remain same for any total quantity unless the % of products remains the same . hence ' d ' is the answer"
Answer: d
Problem: s ( n ) is a n - digit number formed by attaching the first n perfect squares , in order , into one integer . for example , s ( 1 ) = 1 , s ( 2 ) = 14 , s ( 3 ) = 149 , s ( 4 ) = 14916 , s ( 5 ) = 1491625 , etc . how many digits y are in s ( 99 ) ?
Options: a ) y = 350 , b ) y = 353 , c ) y = 354 , d ) y = 356 , e ) y = 357
Rationale: "focus on the points where the number of digits in squares change : 1 , 2 , 3 - single digit squares . first 2 digit number is 10 . 4 , 5 , . . . 9 - two digit squares . to get 9 , the last number with two digit square , think that first 3 digit number is 100 which is 10 ^ 2 . so 9 ^ 2 must be the last 2 digit square . 10 , 11 , 12 , . . . 31 - three digit squares . to get 31 , think of 1000 - the first 4 digit number . it is not a perfect square but 900 is 30 ^ 2 . 32 ^ 2 = 2 ^ 10 = 1024 , the first 4 digit square . 32 - 99 - four digit squares . to get 99 , think of 10,000 - the first 5 digit number which is 100 ^ 2 . so number of digits in s ( 99 ) = 3 * 1 + 6 * 2 + 22 * 3 + 68 * 4 = 3 + 12 + 66 + 272 = 353 . b"
Answer: b
Problem: a can do a piece of work in 12 days and b can do a piece of work in 18 days how many days will the work be completed if both of them work together ?
Options: a ) 15 , b ) 7.2 , c ) 7.7 , d ) 7.4 , e ) 8
Rationale: "a ' s 1 day work = 1 / 12 , b ' s 1 day work = 1 / 18 ; ( a + b ) 1 day work = 1 / 12 + 1 / 18 = 5 / 36 ; they can finish the work in = 36 / 5 = 7.2 answer : b"
Answer: b
Problem: how many two - digit numbers yield a remainder of 2 when divided by both 7 and 16 ?
Options: a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
Rationale: "easier to start with numbers that are of the form 16 p + 2 - - - > 18,34 , 50,66 , 82,98 . out of these no number is also of the form 7 q + 2 . thus 0 is the answer . a is the correct answer ."
Answer: a
Problem: find the 37.5 % of 976 ?
Options: a ) 368 , b ) 655 , c ) 366 , d ) 437 , e ) 545
Rationale: "37.5 % of 976 = 37.5 / 100 * 976 = 375 / 1000 * 976 = 3 / 8 * 976 = 3 * 122 = 366 . answer : c"
Answer: c
Problem: in track last week , the boys ran 34 laps . the girls ran 20 more laps . each lap is a 1 / 6 th of a mile . how many miles did the girls run ?
Options: a ) 9 miles , b ) 10 miles , c ) 11 miles , d ) 12 miles , e ) 13 miles
Rationale: the girls ran 34 + 20 = 54 laps . 54 x 1 / 6 = 54 / 6 , which reduces to 9 . the girls ran 9 miles correct answer a
Answer: a
Problem: the average weight of a group of boys is 30 kg . after a boy of weight 40 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ?
Options: a ) 14 , b ) 5 , c ) 6 , d ) 7 , e ) 9
Rationale: "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 40 kg joins the group , total weight of boys = 30 x + 40 so 30 x + 40 = 31 ( x + 1 ) = > x = 9 . answer : e"
Answer: e
Problem: in a garment industry , 12 men working 8 hours per day complete a piece of work in 10 days . to complete the same work in 8 days , working 20 hours a day , the number of men required is :
Options: a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9
Rationale: explanation : let the required number of men be x . less days , more men ( indirect proportion ) more working hrs per day , less men ( indirect proportion ) days 8 : 10 working hrs 20 : 8 : : 12 : x = > 8 x 20 x x = 10 x 8 x 12 = > x = 10 x 8 x 12 / ( 8 x 20 ) = > x = 6 answer : c
Answer: c
Problem: find the minimum value of n such that 50 ! is perfectly divisible by 2520 ^ n .
Options: a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
Rationale: 50 ! / 2520 ^ n 2520 - > 2 ^ 3 * 3 ^ 2 * 5 ^ 1 * 7 ^ 1 here 7 is the largest prime factor . . . so in order to find the minimum value of ` ` n ' ' , it is enough to find the minimum power of ` ` 7 ' ' . . . nd for maximum value of ` ` n ' ' , find max power of 7 . . . for max . value of n , find 50 / 7 ^ 1 + 50 / 7 ^ 2 = 7 + 1 = 8 [ quotient only ] max . value of n which is perfectly divisible by 2520 ^ n is ( 8 ) min . value is 1 max value : 8 min value : 1 answer : b
Answer: b
Problem: a can finish a work in 10 days and b can do same work in half the time taken by a . then working together , what part of same work they can finish in a day ?
Options: a ) 1 / 5 , b ) 1 / 6 , c ) 1 / 7 , d ) 3 / 10 , e ) none of these
Rationale: "explanation : please note in this question , we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now , a ' s 1 day work = 1 / 10 b ' s 1 day work = 1 / 5 [ because b take half the time than a ] ( a + b ) ' s one day work = ( 1 / 10 + 1 / 5 ) = 3 / 10 so in one day 3 / 10 work will be done answer : d"
Answer: d
Problem: 81 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ?
Options: a ) a ) 99 , b ) b ) 77 , c ) c ) 66 , d ) d ) 55 , e ) e ) 88
Rationale: "( 81 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 180 180 – 81 = 99 answer : a"
Answer: a
Problem: fog + fog + fog + fog = 1464 . if f , o and g are digits in a 3 - digit number in the preceding equation the f + o + g = ?
Options: a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15
Rationale: given , fog + fog + fog + fog = 1464 4 ( fog ) = 1464 = > fog = 366 face value of f is : 3 face value of o is : 6 face value of g is : 6 so f + o + g = 3 + 6 + 6 = 15 answer : e
Answer: e
Problem: a is an integer greater than 39 but less than 51 , b is an integer greater than 49 but less than 61 , what is the range of a / b ?
Options: a ) 1 / 4 , b ) 1 / 3 , c ) 3 / 4 , d ) 1 , e ) 5 / 4
Rationale: "min value of a / b will be when b is highest and a is lowest - - - > a = 40 and b = 60 so , a / b = 2 / 3 max value of a / b will be when b is lowest and a is highest - - - > a = 50 and b = 50 so , a / b = 1 range is 1 - ( 2 / 3 ) = 1 / 3 . answer should be b ."
Answer: b
Problem: two sets of 6 consecutive positive integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ?
Options: a ) 14 , b ) 42 , c ) 28 , d ) 12 , e ) it can not be determined from the information given .
Rationale: "a = ( 1,2 , 3,4 , 5,6 ) , sum of this = 21 b = ( 6 , 7,8 , 9,10 , 11,12 ) , sum of this = 63 , the differenct between 63 - 21 = 42 hence , 42 is the answer i . e . b"
Answer: b
Problem: how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 50 m long and 16 1 / 2 m board ?
Options: a ) 99 , b ) 18 , c ) 165 , d ) 10 , e ) 15
Rationale: 50 * 33 / 2 = 5 / 2 * 2 * x = > x = 165 answer : c
Answer: c
Problem: how many digits are required to number a book containing 260 pages ?
Options: a ) 784 , b ) 672 , c ) 492 , d ) 372 , e ) 300
Rationale: "9 pages from 1 to 9 will require 9 digits . 90 pages from 10 to 99 will require 90 * 2 = 180 digits . 260 - ( 90 + 9 ) = 161 pages will require 161 * 3 = 483 digits . the total number of digits is 9 + 180 + 483 = 672 . the answer is b ."
Answer: b
Problem: log 3 n + log 15 n what is 3 digit number n that will be whole number
Options: a ) 3375 , b ) 7292 , c ) 8291 , d ) 3929 , e ) 2727
Rationale: "no of values n can take is 1 15 ^ 3 = 3375 answer : a"
Answer: a
Problem: the sum of squares of 3 numbers is 764 and sum of their products taken two at a time is 500 . what is their sum ?
Options: a ) 48 , b ) 42 , c ) 52 , d ) 32 , e ) 33
Rationale: description : let the numbers be x , y , z . = > x 2 + y 2 + z 2 = 764 = > 2 ( xy + yz + xz ) = 2 * 500 = 1000 = > ( x + y + z ) 2 = ( x 2 + y 2 + z 2 ) + 2 ( xy + yz + xz ) = 1764 = > ( x + y + z ) = sqrt ( 1764 ) = 42 answer b
Answer: b
Problem: a doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds . if the typical dosage is 2 cubic centimeters per 10 pounds of the body weight , by what percent was the prescribed dosage lesser than the typical dosage ?
Options: a ) 8 % , b ) - 37.5 % , c ) 11 % , d ) 12.5 % , e ) 14.8 %
Rationale: typical dosage per 10 pound of the body weight = 2 c . c typical dosage per 120 pound of the body weight = 2 * ( 120 / 10 ) = 2 * 12 = 24 c . c dosage prescribed by doctor for 120 pound patient = 18 c . c % prescribed dosage greater than the typical dosage = ( 18 - 24 / 16 ) * 100 % = ( - 6 / 16 ) * 100 % = - 37.5 % answer b
Answer: b
Problem: the maximum number of students among them 1340 pens and 1280 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ?
Options: a ) 25 , b ) 30 , c ) 20 , d ) 45 , e ) 90
Rationale: "number of pens = 1340 number of pencils = 1280 required number of students = h . c . f . of 1340 and 1280 = 20 answer is c"
Answer: c
Problem: if the remainder is 10 when the integer n is divided by 18 , what is the remainder when 2 n is divided by 9 ?
Options: a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
Rationale: "n = 18 k + 10 2 n = 2 ( 18 k + 10 ) = 4 k * 9 + 20 = 4 k * 9 + 2 * 9 + 2 = 9 j + 2 the answer is c ."
Answer: c
Problem: the sum of two numbers is 15 and their geometric mean is 20 % lower than arithmetic mean . find the numbers .
Options: a ) 11,4 , b ) 12,3 , c ) 9,6 , d ) 3,2 , e ) 2,1
Rationale: "let a and b are two numbers so a + b = 15 . . . . ( 1 ) now gm = am * 80 / 100 or , sqrt ( ab ) = [ ( a + b ) / 2 ] * 80 / 100 squaring both sides . . . and after that putting the value of ( a + b ) . . . we get . . . . . ab = 36 now go through from option then we get 12 * 3 = 36 answer : b"
Answer: b
Problem: the value of ( 34.5 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) is close to
Options: a ) 2 , b ) 1.15 , c ) 2.05 , d ) 2.16 , e ) 2.35
Rationale: "( 34.5 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) = 25.5710895 / 11.845 = 2.16 answer : d"
Answer: d
Problem: in the land of oz only one or two - letter words are used . the local language has 67 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ?
Options: a ) 65 , b ) 66 , c ) 67 , d ) 131 , e ) 134
Rationale: "the answer to the question is indeed e . the problem with above solutions is that they do not consider words like aa , bb , . . . the number of 1 letter words ( x ) that can be made from 67 letters is 67 ; the number of 2 letter words ( xx ) that can be made from 67 letters is 67 * 67 , since each x can take 67 values . total : 67 + 67 * 67 . similarly : the number of 1 letter words ( x ) that can be made from 66 letters is 66 ; the number of 2 letter words ( xx ) that can be made from 66 letters is 66 * 66 , since each x can take 66 values . total : 66 + 66 * 66 . the difference is ( 67 + 67 * 67 ) - ( 66 + 66 * 66 ) = 134 . answer : e ."
Answer: e
Problem: $ 500 is divided amongst a , b and c so that a may get 2 / 3 as much as b and c together , b may get 6 / 9 as much as a and c together , then the share of a is
Options: a ) $ 100 , b ) $ 150 , c ) $ 125 , d ) $ 200 , e ) $ 250
Rationale: "a : ( b + c ) = 2 : 3 a ' s share = 500 * 2 / 5 = $ 200 answer is d"
Answer: d
Problem: in an election between two candidates a and b , the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled . if 20 % of the votes polled were invalid and a total of 5720 votes were polled , then how many valid votes did b get ?
Options: a ) 1859 , b ) 2999 , c ) 2834 , d ) 2777 , e ) 2991
Rationale: "let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 5720 32.5 k = 32.5 k / 100 k * 85720 = 1859 the number of valid votes polled in favour of b is 1859 . answer : a"
Answer: a
Problem: 10 x 1.8 - 2 x 1.5 / 0.3 = ?
Options: a ) 30 , b ) 20 , c ) 15 , d ) 44 , e ) 50
Rationale: given expression = ( 18 - 3.0 ) / 0.3 = 15 / 0.3 = 150 / 3 = 50 answer is e .
Answer: e
Problem: there are 1000 students in a school and among them 20 % of them attends chess class . 10 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ?
Options: a ) 1 , b ) 10 , c ) 100 , d ) 50 , e ) 20
Rationale: "20 % of 1000 gives 200 . so 200 attends chess and 10 % of 200 gives 20 . so 20 enrolled for swimming answer : e"
Answer: e
Problem: an object thrown directly upward is at a height of h feet after t seconds , where h = - 15 ( t - 2 ) ^ 2 + 150 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ?
Options: a ) 72 , b ) 90 , c ) 124 , d ) 132 , e ) 140
Rationale: "we see that h will be a maximum h = 150 when t - 2 = 0 , that is when t = 2 . at t = 4 , h = - 15 ( 4 - 2 ) ^ 2 + 150 = - 15 ( 4 ) + 150 = 90 the answer is b ."
Answer: b
Problem: a trader sells 66 meters of cloth for rs . 660 at the profit of rs . 5 per metre of cloth . what is the cost price of one metre of cloth ?
Options: a ) 26 , b ) 88 , c ) 90 , d ) 42 , e ) 5
Rationale: "sp of 1 m of cloth = 660 / 66 = rs . 10 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 10 - rs . 5 = rs . 5 answer : e"
Answer: e
Problem: what is the sum of the first 15 terms of an a . p whose 11 th and 7 th terms are 5.25 and 3.25 respectively
Options: a ) 56.25 , b ) 60 , c ) 52.5 , d ) 5.25 , e ) none of these
Rationale: a + 10 d = 5.25 , a + 6 d = 3.25 , 4 d = 2 , d = ½ a + 5 = 5.25 , a = 0.25 = ¼ , s 15 = 15 / 2 ( 2 * ¼ + 14 * ½ ) = 15 / 2 ( 1 / 2 + 14 / 2 ) = 15 / 2 * 15 / 2 = 225 / 4 = 56.25 answer : a
Answer: a
Problem: the average marks of a class of 30 students is 40 and that of another class of 50 students is 70 . find the average marks of all the students ?
Options: a ) 52.5 , b ) 52.9 , c ) 52.1 , d ) 58.75 , e ) 42.5
Rationale: "sum of the marks for the class of 30 students = 30 * 40 = 1200 sum of the marks for the class of 50 students = 50 * 70 = 3500 sum of the marks for the class of 80 students = 1200 + 3500 = 4700 average marks of all the students = 4700 / 80 = 58.75 answer : d"
Answer: d
Problem: the ratio of the length and the width of a rectangle is 4 : 3 and the area of the rectangle is 6348 sq cm . what is the ratio of the width and the area of the rectangle ?
Options: a ) 1 : 96 , b ) 1 : 99 , c ) 1 : 94 , d ) 1 : 92 , e ) 1 : 91
Rationale: "let the length and the width be 4 x and 3 x respectively . area = ( 4 x ) ( 3 x ) = 6348 12 x ^ 2 = 6348 x ^ 2 = 529 x = 23 the ratio of the width and the area is 3 x : 12 x ^ 2 = 1 : 4 x = 1 : 92 the answer is d ."
Answer: d
Problem: rates for having a manuscript typed at a certain typing service are $ 5 per page for the first time a page is typed and $ 3 per page each time a page is revised . if a certain manuscript has 100 pages , of which 30 were revised only once , 20 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ?
Options: a ) $ 430 , b ) $ 620 , c ) $ 650 , d ) $ 680 , e ) $ 710
Rationale: "for 100 - 30 - 20 = 50 pages only cost is 5 $ per page for the first time page is typed - 50 * 5 = 250 $ ; for 30 pages the cost is : first time 5 $ + 3 $ of the first revision - 30 * ( 5 + 3 ) = 240 $ ; for 20 pages the cost is : first time 5 $ + 3 $ of the first revision + 3 $ of the second revision - 20 ( 5 + 3 + 3 ) = 220 $ ; total : 250 + 240 + 220 = 710 $ . answer : e ."
Answer: e
Problem: three unbiased coins are tossed . what is the probability of getting at least 2 heads ?
Options: a ) 1 / 4 , b ) 1 / 2 , c ) 1 / 3 , d ) 1 / 8 , e ) 1 / 6
Rationale: "here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting at least two heads = { thh , hth , hht , hhh } p ( e ) = n ( e ) / n ( s ) = 4 / 8 = 1 / 2 answer b ."
Answer: b
Problem: the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 2 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ?
Options: a ) 23 years , b ) 22 years , c ) 25 years , d ) 26 years , e ) 27 years
Rationale: "explanation let the average age of the whole team by x years . 11 x â € “ ( 26 + 28 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 45 2 x = 45 x = 22 . so , average age of the team is 22 years . answer b"
Answer: b
Problem: if josh , doug , and brad have a total of $ 68 between them , and josh has two times as much money as brad but only 3 - fourths as much as doug , how much money does brad have ?
Options: a ) $ 10 , b ) $ 24 , c ) $ 26 , d ) $ 12 , e ) $ 36
Rationale: josh + doug + brad = 68 ; josh = 2 brad , josh = 3 / 4 doug josh + 1 / 2 josh + 4 / 3 josh = 68 ( substituted the given values ) josh = 24 . 24 = 3 / 4 doug = > doug = 32 josh + doug + brad = 68 24 + 32 + brad = 68 brad = 12 answer is d .
Answer: d
Problem: mahesh can do a piece of work in 45 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ?
Options: a ) 54 , b ) 25 , c ) 37 , d ) 41 , e ) 30
Rationale: work done by mahesh in 45 days = 20 * 1 / 45 = 4 / 9 remaining work = 1 - 4 / 9 = 5 / 9 5 / 9 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 9 / 5 = 54 days answer is a
Answer: a
Problem: the speed of a car is 100 km in the first hour and 30 km in the second hour . what is the average speed of the car ?
Options: a ) 50 kmph , b ) 65 kmph , c ) 75 kmph , d ) 85 kmph , e ) 65 kmph
Rationale: "explanation : s = ( 100 + 30 ) / 2 = 65 kmph e )"
Answer: e
Problem: the hour hand of a watch rotates 30 degrees every hour . how many complete rotations does the hour hand make in 6 days ?
Options: a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16
Rationale: explanation : there are 360 degrees in a complete circle , so 360 / 30 = 12 hours to make one full circle . in 6 days there are 24 hours x 6 = 144 hours total . the total number of rotations will be 144 / 12 = 12 . answer : ( c )
Answer: c
Problem: find the remainder when 6 ^ 23 is divided by 5 ?
Options: a ) 3 , b ) 2 , c ) 1 , d ) 0 , e ) 4
Rationale: "6 ^ 23 = 6 ^ ( 4 * 5 + 3 ) = 6 ^ ( 20 ) * 6 ^ 3 last 3 digits = 216 216 / 5 remainder = 1 answer : c"
Answer: c
Problem: a vessel of capacity 75 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ?
Options: a ) 23.89 , b ) 72.9 , c ) 58.08 , d ) 78.3 , e ) 79.3
Rationale: "explanation : let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] n * t for the given problem , t = 75 , y = 9 and n = 2 . hence , quantity of milk finally in the vessel = [ ( 75 - 9 ) / 75 ] ^ 2 ( 75 ) = 58.08 litres . answer : option c"
Answer: c
Problem: riya and priya set on a journey . riya moves eastward at a speed of 25 kmph and priya moves westward at a speed of 40 kmph . how far will be priya from riya after 15 minutes
Options: a ) 18 kms , b ) 16 kms , c ) 50 kms , d ) 30 kms , e ) 40 kms
Rationale: "total eastward distance = 25 kmph * 1 / 4 hr = 6.25 km total westward distance = 40 kmph * 1 / 4 hr = 10 km total distn betn them = 6.25 + 10 = 16.25 km ans 16 km answer : b"
Answer: b
Problem: the product of two successive numbers is 9506 . which is the smaller of the two numbers ?
Options: a ) 78 , b ) 68 , c ) 88 , d ) 97 , e ) 37
Rationale: "d 97 from the given alternatives , 97 × 98 = 9506 ∴ smaller number = 97"
Answer: d
Problem: if 2 ^ 4 , 3 ^ 3 , and 10 ^ 3 are factors of the product of 1,452 and w , where w is a positive integer , what is the smallest possible value of w ?
Options: a ) 198 , b ) 288 , c ) 363 , d ) 360 , e ) 484
Rationale: i will go with d ( pending elements to match is 2 ^ 2 * 3 ^ 2 * 10 ^ 1 = 360
Answer: d
Problem: the l . c . m . of 2 numbers is 100 . the numbers are in the ratio 4 : 1 . find their product ?
Options: a ) 1200 , b ) 2000 , c ) 2500 , d ) 3200 , e ) 1250
Rationale: "let the numbers be 4 x and x l . c . m . = 4 x 4 x = 100 x = 25 the numbers are = 100 and 25 required product = 100 * 25 = 2500 answer is c"
Answer: c
Problem: a , b , c subscribe rs . 50,000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 30,000 , b receives :
Options: a ) s . 10,000 , b ) s . 10,200 , c ) s . 10,400 , d ) s . 10,700 , e ) s . 10,800
Rationale: "let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . b ' s share = rs . ( 30000 x 17 / 50 ) = rs . 10,200 . b"
Answer: b
Problem: the difference between a positive proper fraction and its reciprocal is 9 / 20 . the fraction is :
Options: a ) 4 / 5 , b ) 5 / 4 , c ) 3 / 5 , d ) 2 / 5 , e ) 3 / 4
Rationale: "let the required fraction be x . then 1 / x - x = 9 / 20 ( 1 - x ^ 2 ) / x = 9 / 20 = = > 20 - 20 x ^ 2 = 9 x 20 x ^ 2 + 9 x - 20 = 0 20 x ^ 2 + 25 x - 16 x - 20 = 0 5 x ( 4 x + 5 ) - 4 ( 4 x + 5 ) = 0 ( 4 x + 5 ) ( 5 x - 4 ) = 0 x = 4 / 5 ( neglecting - ve value ) answer a ) 4 / 5"
Answer: a
Problem: a man is 20 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is :
Options: a ) 14 years , b ) 18 years , c ) 20 years , d ) 22 years , e ) 16 years
Rationale: "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 20 ) + 2 = 2 ( x + 2 ) x + 22 = 2 x + 4 x = 18 . answer : b"
Answer: b
Problem: a mixture contains milk and water in the ratio 5 : 4 . on adding 30 litters of water , the ratio of milk to water becomes 5 : 7 . total quantity of milk & water before adding water to it ?
Options: a ) 90 , b ) 120 , c ) 80 , d ) 70 , e ) 60
Rationale: "explanation : milk : water = 5 : 4 5 x : 4 x + 30 = 5 : 7 7 [ 5 x ] = 5 [ 4 x + 30 ] 35 x = 20 x + 150 35 x - 20 x = 150 15 x = 150 x = 10 the quantity of milk in the original mixture is = 5 : 4 = 5 + 4 = 9 9 x = 90 short cut method : milk : water = 5 : 4 after adding 30 liters of water milk : water = 5 : 7 milk is same but water increse 30 liters then the water ratio is increse 3 parts 3 part - - - - - > 30 liters the quantity of milk in the original mixture is = 5 : 4 = 5 + 4 = 9 9 parts - - - - - > 9 liters ( answer is = 90 ) short cut method - 2 : for only milk problems milk : water 5 : 4 5 : 7 milk ratio same but water ratio 3 parts incress per 30 liters 3 part of ratio - - - - - - - > 30 liters 1 part of ratio - - - - - - - > 10 liters 9 part of ratio - - - - - - > 90 liters answer : option a"
Answer: a
Problem: in a class , 30 students pass in english and 20 students in math , while some students among these pass in both . how many more students do only english as compared to those doing only maths ?
Options: a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) indeterminate
Rationale: let us consider tht x student are tohse who passed in both english and maths . . . so first we remove x student from both of them therefore , english = 30 - x maths = 20 - x now , number of students more in english = ( 30 - x ) - ( 20 - x ) = 30 - x - 20 + x = 10 answer : a
Answer: a
Problem: a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 550 bolts ?
Options: a ) 36 , b ) 72 , c ) 120 , d ) 132 , e ) 180
Rationale: "old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 550 bolts 550 * 6 / 25 = 132 mins ans d"
Answer: d
Problem: in a village of 2,700 people , 900 people are over 70 years old and 1200 people are female . it is known that 40 percent of the females are younger than 70 years old . if no one in the village is 70 years old , what is the probability that a person chosen at random is either a female or younger than 70 years old ?
Options: a ) 7 / 8 , b ) 14 / 15 , c ) 29 / 30 , d ) 44 / 45 , e ) 131 / 135
Rationale: "the number of people younger than 70 years old is 2700 - 900 = 1800 the number of females older than 70 years old is 0.6 * 1200 = 720 the number of people who are either female or younger than 70 is 1800 + 720 = 2520 . p ( a person is younger than 70 or male ) = 2520 / 2700 = 14 / 15 the answer is b ."
Answer: b