mathewhe/aqua_rat huggingface dataset:
Question: For any positive integer n, n>1, thelengthof n is the number of positive primes (not necessary distinct) whose product is n. For ex, the length of 50 is 4, since 50=2x5x5. What is the greatest possible length of a positive integer less than 1000.
(A) 10
(B) 9
(C) 8
(D) 7
(E) 6
Rationale: you are missing something in your post:
For any positive integer n , n > 1, thelengthof n is the number of positive primes (not distinct) whose product is n. For example, the length of 50 is 3 since 50 = (2)(5)(5)
the lenght of 1000 = (2)(5)(2)(5)(2)(5) = 6
but we need n < 1000
using 2 as the base = (2)(2)(2)(2)(2)(2)(2)(2)(2) = 98 the length of 512.
the answer is C
Answer: (C) 8
Question: By selling an shirt for $300, a shop keeper gains 20%. During a clearance sale, the shopkeeper allows a discount of 10% on the marked price. His gain percent during the sale is:
(A) 2
(B) 14
(C) 8
(D) 12
(E) 10
Rationale: C
8%
Marked price = $300
C.P. = 100/120 * 300 = $250
Sale price = 90% of $300 = $270
Required gain % = 20/250 * 100 = 8%.
Answer: (C) 8
Question: A car covers a distance of 720 km in 6 hours. Find its speed?
(A) 104
(B) 7778
(C) 266
(D) 288
(E) 120
Rationale: 720/6 = 120 kmph
Answer:E
Answer: (E) 120
Question: A rectangular grass field is 75 m * 55 m, it has a path of 2.5 m wide all round it on the outside. Find the area of the path and the cost of constructing it at Rs.1 per sq m?
(A) 675
(B) 135
(C) 320
(D) 430
(E) 820
Rationale: Area = (l + b +2d) 2d
= (75 + 55 +2.5 * 2) 2*2.5 => 675
675 * 1 = Rs.675
Answer:A
Answer: (A) 675
Question: By investing in 16 2/3% stock at 64, one earns Rs. 1500. The investment made is:
(A) Rs. 5640
(B) Rs. 5760
(C) Rs. 7500
(D) Rs. 9600
(E) Rs. 8600
Rationale: To earn Rs.50/3,investment =Rs.64
To earn Rs.1500,investment =(64*1500*3)/50=Rs.5760
ANSWER:B
Answer: (B) Rs. 5760
Question: The balance of a trader weighs 8% less than it should. Still the trader marks up his goods to get an overall profit of 30%. What is the mark up on the cost price?
(A) 40%
(B) 19%
(C) 25%
(D) 16.66%
(E) 9%
Rationale: The most natural way to deal with 'weights' questions is by assuming values.
Say the trader's balance shows 100 gms. It is actually 92 gms because it weighs 8% less. Say, the cost price is $92 ($1/gm). Since he gets a profit of 30%, the selling price must be 92+(30/100)*90 = $119
Since the cost price is actually supposed to be $100 (for 100 gms) and the selling price is $119, the mark up is simply 19%
Ans : B.
Answer: (B) 19%
Question: All the page numbers from a book are added, beginning at page 1. However, one page number was mistakenly added twice. The sum obtained was 1000. Which page number was added twice?
(A) 44
(B) 45
(C) 10
(D) 12
(E) 11
Rationale: Explanation :
sum of n natural numbers = n (n+1)/2
Here, n( n + 1)/2 = 1000 - x (where x is the number that is added twice)
=> n(n+1) < 2000 - 2x
we need to find max value n can take . We know for a product of two numbers to be max the number should be closest. We need to find out the least number whose square is higher than 2000. For example, 452 = 2025.
So our number is somewhere near to 45. it can't be more than 45 as our required product is less than 2000.
44 * 45 = 1980. So, we can have n = 44.
=> n (n+1) = 44 * 45 = 1980 = 2000 - 2 * 10.
=> x = 10.
Hence, we have 44 pages and page number 10 was added twice.
Answer : C
Answer: (C) 10
Question: A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
(A) 5%
(B) 10%
(C) 7%
(D) 8%
(E) 9%
Rationale: Explanation :
Let the rate of interest per annum be R%
Simple Interest for Rs. 5000 for 2 years at rate R% per annum +Simple Interest for Rs. 3000 for 4 years at rate R% per annum = Rs.2200
⇒5000×R×2/100+3000×R×4/100=2200
⇒100R + 120R=2200
⇒220R=2200
⇒R=10
i.e, Rate = 10%.
Answer : Option B
Answer: (B) 10%
Question: Three cubes of metal whose edges are 5, 10 and 15 cm respectively, are melted and one new cube is made. Find the edge of the new cube?
(A) 28
(B) 77
(C) 66
(D) 16.5
(E) 99
Rationale: 53 + 103 + 153 = a3 => a
= 16.5
Answer: D
Answer: (D) 16.5
Question: Cubes with each side one inch long are glued together to form a larger cube. The larger cube's face is painted with red color and the entire assembly is taken apart. 26 small cubes are found with no paints on them . How many of unit cubes have at least one face that is painted red?
(A) 64
(B) 72
(C) 86
(D) 98
(E) 99
Rationale: Use the options.
The options which after getting added to 26 shows a cube of a number could be right.
Here
64+26 = 90
72+26 = 98
86+26 = 112
98+26+124
99+26=125 ---(5*5*5)
So we have 99 as the answer!
E
Answer: (E) 99
Question: Fresh grapes contain 85% water by weight and raisins obtained by drying fresh grapes contain 25% water by weight. How many kgs of fresh grapes are needed to get 20kgs of raisins?
(A) 75 kgs
(B) 64kgs
(C) 100 kgs
(D) 65 kgs
(E) 70 kgs
Rationale: The weight of non-water in 20 kg of dried grapes (which is 100-25=75% of whole weight) will be the same as the weight of non-water in x kg of fresh grapes (which is 100-85=15% of whole weight), so 20*0.75 = x*0.15
--> x =100.
Answer: C
Answer: (C) 100 kgs
Question: Find the probability that the sum of n numbers is divisible by n,assuming n is very large no.
(A) 1
(B) 1/2
(C) 3/4
(D) None of these
(E) Can't be determined.
Rationale: sum of n numbers is n(n+1)/2
case 1)when n is odd take 3
Sum is 3*4/3 it is divisible by 3
we can conclude that n=odd sum of n numbers is divisible by n
case2) when n is even take 4
Sum is 4*5/2=>10 10/4 rem is 2 so not divisible
we can conclude that n=even sum of n numbers is not divisible by n
Total of total 2 cases 1 case is possible
So probability=possible/total=1/2
ANSWER:B
Answer: (B) 1/2
Question: Two trains are running in opposite directions with the same speed. If the length of each train is 120 m and they cross each other in 12 sec, then the speed of each train is?
(A) 30
(B) 35
(C) 36
(D) 37
(E) 38
Rationale: Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (120 + 120)/12 => x = 10
Speed of each train = 10 m/sec.
= 10 * 18/5 =- 36 km/hr.
Answer: Option C
Answer: (C) 36
Question: An employee’s annual salary was increased 30%. If her old annual salary equals $30,000 , what was the new salary?
(A) $28,000
(B) $25,000
(C) $20,000
(D) $35,000
(E) $39,000
Rationale: Old annual salary = $30,000
Salary increase = 30%.
Original salary = $30,000 *30/100
= $9000
new salary = $30,000+$9000
=$39,000
Hence E.
Answer: (E) $39,000
Question: Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?
(A) 11
(B) 10
(C) 8
(D) 12
(E) 15
Rationale: Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
Therefore 20x + 25(x - 1) = 110
45x = 135
x = 3.
So, they meet at 10 a.m.
Answer: B
Answer: (B) 10
Question: For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 4 to n, inclusive. If p is the smallest prime factor of h(100) +4, then p is?
(A) between 4 and 20
(B) between 10 and 20
(C) between 20 and 30
(D) between 30 and 40
(E) 4
Rationale: h(100) will have 0 in its unit digit because while doing the calculation of the function we are multiplying by 10 (and 100). So h(100)+4 will have 4 in the units digit so smallest prime factor is 4.
Answer: E
Answer: (E) 4
Question: Johnny makes $2.35 per hour at his work. If he works 5 hours, how much money will he earn?
(A) $30
(B) $54
(C) $28.50
(D) $12
(E) $11.75
Rationale: 2.35*5=11.75. Answer is E.
Answer: (E) $11.75
Question: The number of boys in a class is 2 times the number of girls. Which one of the following numbers cannot represent the total number of children in the class ?
(A) A)47
(B) B)44
(C) C)42
(D) D)40
(E) E)38
Rationale: Explanation:
Let number of girls = x and number of boys = 2x.
Then, 2x + x = 3x = total number of students.
Thus, to find exact value of x, the total number of students must be divisible by 3.
Answer: C
Answer: (C) C)42
Question: On a certain scale of intensity, each increment of 30 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125?
(A) 40
(B) 100
(C) 400
(D) 1000
(E) 10 000
Rationale: Increase of 40 in magnitude corresponds to 10^4 increase in intensity:
If intensity for 125 is x then for 135 it'll be 10*x, for 145 it'll be 10*10*x=10^2*x, for 155 it'll be 10*10*10*x=10^3*x and for 165 it'll be 10*10*10*10*x=10^2*x.
Answer: B.
Answer: (B) 100
Question: The least number which should be added to 720 so that the sum is exactly divisible by 5,6,4 and 3 is :
(A) 3
(B) 13
(C) 48
(D) 33
(E) 43
Rationale: Solution
L.C.M. of 5,6,4 and 3 = 60. on dividing 720 by 60, the remainder is 12.
∴ Number to be added = (60-12) = 48. Answer C
Answer: (C) 48
Question: A certain club has 6 members, including Harry. One of the 6 members is to be chosen at random to be the president, one of the remaining 5 members is to be chosen at random to be the secretary, and one of the remaining 4 members is to be chosen at random to be the treasurer. What is the probability that Harry will either be the secretary or the member chosen to be the treasurer?
(A) 7/3
(B) 5/3
(C) 1
(D) 2/3
(E) 1/3
Rationale: Each member out of 6, including Harry, has equal chances to be selected for any of the positions (the sequence of the selection is given just to confuse us). The probability that Harry will be selected to be the secretary is 1/6 and the probability that Harry will be selected to be the treasurer is also 1/6. So, the probability that Harry will be selected to be either the secretary or the the treasurer is 1/6+1/6=2/6.
Answer: E.
Answer: (E) 1/3
Question: In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
(A) 21/46
(B) 25/117
(C) 1/50
(D) 3/25
(E) 2/25
Rationale: Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S)	= Number ways of selecting 3 students out of 25
= 25C3 `
=	(25 x 24 x 23)/(3 x 2 x 1)
= 2300.
n(E)	= (10C1 x 15C2)
= (10 x	(15 x 14)/(2 x 1))
= 1050.
P(E) =	n(E)	/n(S)	=	1050/2300	=	21/46	.
Answer:A
Answer: (A) 21/46
Question: By selling a house for Rs.45000, it was found that 1/8 of the outlay was gained, what ought the selling to price to have been in order to have lost 5 p.c?
(A) 33377
(B) 38000
(C) 27999
(D) 2776
(E) 198712
Rationale: CP + CP/8 = 45000
CP = 40000
SP = 40000*(95/100) = 38000
Answer:B
Answer: (B) 38000
Question: A number is decreased by 10% and then increased by 10%. The number so obtained is 10 less than the original number. What was the oiginal number ?
(A) 1000
(B) 3899
(C) 1028
(D) 1039
(E) 1002
Rationale: Explanation:
Let the original number be x.
Final number obtained = 110% of (90% of x)
Answer: A) 1000
Answer: (A) 1000
Question: Number of ways in which the letters of word GARDEN can be arranged with vowels in alphabetical order, is
(A) 360
(B) 240
(C) 120
(D) 480
(E) None
Rationale: Solution
Order of vowels of fixed
∴ required number of ways are 6!/2!
Answer A
Answer: (A) 360
Question: If the remainder is 16 when the integer n is divided by 30, what is the remainder when 2n is divided by 15?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
Rationale: n = 30k+16
2n = 2(30k+16) = 4k*15 + 32 = 4k*15 + 2*15 + 2 = 15j + 2
The answer is B.
Answer: (B) 2
Question: Find the one which does not belong to that group ?
(A) 35
(B) 77
(C) 28
(D) 19
(E) 11
Rationale: Explanation:
48, 75, 84 and 57 are divisible by 3 but not 35.
Answer:A
Answer: (A) 35
Question: A student has 8 trousers and 12 shirts. In how many different ways can the student select a trouser and a shirt?
(A) 84
(B) 96
(C) 108
(D) 120
(E) 132
Rationale: The student can select one trouser in 8 ways.
The student can select one shirt in 12 ways.
The number of ways to select one trouser and one shirt is 8 * 12 = 96 ways
The answer is B.
Answer: (B) 96
Question: The diagonals of a rhombus are 15 cm and 20 cm. Find its area?
(A) 158
(B) 129
(C) 150
(D) 123
(E) 117
Rationale: 1/2 * 15 * 20 = 150
Answer:C
Answer: (C) 150
Question: Calculate the mean of each data set {9,10,11,7,13}
(A) 4
(B) 5
(C) 8
(D) 9
(E) 10
Rationale: mean of Data set A = (9+10+11+7+13)/5 = 10
mean of Data set B = (10+10+10+10+10)/5 = 10
mean of Data set C = (1+1+10+19+19)/5 = 10
Option E
Answer: (E) 10
Question: Two highways start from a point P and meet again at the point Q. Between P and Q there are 5 subways, joining the highways at 5 different places. How many different routes are possible for a journey from P to Q?
(A) 12
(B) 16
(C) 24
(D) 32
(E) 64
Rationale: If one starts its journey from point P, one has 2 options ( to go via either highway)... When he reaches 1st subway point he has again, 2 options, to continue on the highway or to go through subway., When he reaches next subway, he will have same two options, of course we are considering that he doesn't return from any point.
Since, there are 5 subways, he has the option of choosing from 2 choices 6 times (one when he just starts his journey). So, total ways he can reach is destination is 2^6 = 64.
ANSWER:E
Answer: (E) 64
Question: If m and n are positive integers and m^2 + n^2 = 89, what is the value of m^3 + n^3?
(A) 72
(B) 224
(C) 320
(D) 512
(E) 637
Rationale: You need to integers which squared are equal 89.
Which could it be ? Let's start with the first integer:
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
Stop. The integers can't be greater than 8 or we will score above 89.
The second integer need to be picked up the same way.
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
The only pair that matches is 8^2 + 5^2 = 89.
So 8^3 + 5^3 = 637.
Answer E.)
Answer: (E) 637
Question: There are 9 people and 7 chairs. In how many different ways people can occupy the chairs?
(A) 405
(B) 504
(C) 404
(D) 505
(E) 506
Rationale: number of ways = 9*8*7 = 504
Answer is B
Answer: (B) 504
Question: Canister C is 1/2 full of water and canister D, which has twice the capacity of canister C, is 1/3 full of water. If the water in canister D is poured in canister C until canister C is completely full of water, canister D will still contain what fraction of its capacity of water?
(A) 0
(B) 1/36
(C) 1/12
(D) 1/6
(E) 1/4
Rationale: Say canister C has a capacity of 6 liters. It's half full, thus there can be poured 3 liters of water.
Canister B is 12 liters and there are 4 liters of water. We can pour 3 liters from D to C and 1 liter will still be left in D, which is 1/12 of its total capacity.
Answer: C.
Answer: (C) 1/12
Question: A cistern 10 meters long and 4 meters wide contains water up to a depth of 1 meter 25 cm. What is the total area of the wet surface?
(A) 45
(B) 55
(C) 65
(D) 75
(E) 85
Rationale: Area of the wet surface = [2(lb + bh + lh) - lb]
= 2(bh + lh) + lb
= [2 (4 x 1.25 + 10 x 1.25) + 10 x 4]
= 75
The answer is D.
Answer: (D) 75
Question: If set A={2,2,2,....,n times} , set B={4,4,4,4....m times} and set C={11,11,11...k time} , then in terms of m,n and k , how many possible subsets from sets A,B, and C can be created ?
(A) a) k(n+m+mn)+ k
(B) b) k^2(mn+n/m)
(C) c) (1+n+m+mn)(k+1)
(D) d) kmn(k+m+n)
(E) e) None of the above.
Rationale: a) how many 2's are included?
b) how many 3's are included?
and
c) how many 11's are included?
For the number of 2, we could have zero 2's, or one 2, or two 2's, all the way up to n 2's. That's (n + 1) possibilities for the 2's. Similarly, (m + 1) possibilities for the 3's and (k + 1) possibilities for the 11's. We simply multiply these three numbers.
NOTICE that one set, the set that includes no 2's, no 3's, and no 11's, is included. This is known in mathematics as thenull set, sometimes called the empty set, a set with no members. Technically, this is a subset of every possible set, but that's a technical detail of set theory that goes well beyond what the GMAT would expect students to know. Even the the calculation is not that difficult, some of the technical aspects of this question are not in line with the GMAT's expectations.
The number of subsets is (m + 1)(n + 1)(k + 1). The answer is not given in that form. Instead, the first two factors have been FOILed together:
(mn + m + n + 1)(k + 1).
Mike
C
Answer: (C) c) (1+n+m+mn)(k+1)
Question: Two persons A and B can complete a piece of work in 30 days and 45 days respectively. If they work together, what part of the work will be completed in 2 days?
(A) 2/5
(B) 1/6
(C) 1/4
(D) 1/9
(E) 2/6
Rationale: A's one day's work = 1/30
B's one day's work = 1/45
(A + B)'s one day's work = 1/30 + 1/45 = 1/18
The part of the work completed in 2 days = 2 (1/18) = 1/9.
ANSWER D
Answer: (D) 1/9
Question: A woman named Susila started a business investing Rs. 25,000. Roja joined him after six months with an amount of Rs. 50,000 and Sachin joined them with Rs. 75,000 after another six months. The amount of profit earned should be distributed in what ratio among Susila , Roja and Sachin respectively, 3 years after Aman started the business ?
(A) 7:6:10
(B) 12:15:16
(C) 42:45:56
(D) 3 : 5 : 6
(E) 12:15:18
Rationale: Explanation :
Akash : Roja : Sachin = (25,000 x 36) : (50,000 x 30) : (75,000 x 24)
= 36 : 60 : 72 => 3 : 5 : 6
Answer : D
Answer: (D) 3 : 5 : 6
Question: Tough and Tricky questions: Combinations.
Four students - 2 boys and 2 girls - are to sit side by side for a makeup exam. How many ways could they arrange themselves given that no two boys and no two girls can sit next to one another?
(A) 8
(B) 36
(C) 72
(D) 240
(E) 720
Rationale: We can have 2 arrangements:
BGBG or GBGB
2!*2!*2 = 2*2*2 = 8 ways.
Ans. A) 8
Answer: (A) 8
Question: SCORE INTERVAL----------------NUMBER OF SCORES
50-59-------------------------- 2
60-69--------------------------10
70-79--------------------------16
80-89--------------------------5
90-99--------------------------2
The table above shows the distribution of test scores for a group of management trainees. Which score interval contains the median of the 35 scores?
(A) 16th
(B) 20th
(C) 22th
(D) 24th
(E) 18th(70-79)
Rationale: Total scores= 2+10+16+5+2 = 35, which is odd, therefore the median is the Floor(35/2)+1 = 18th score. And the 18th score is in the 70-79 range, because 50-69 only reference 12 scores.
E
Answer: (E) 18th(70-79)
Question: Ramu bought an old car for Rs. 36400. He spent Rs. 8000 on repairs and sold it for Rs. 68400. What is his profit percent?
(A) 16%
(B) 17%
(C) 18%
(D) 54%
(E) 13%
Rationale: Total CP = Rs. 36400 + Rs. 8000
= Rs. 44400 and SP = Rs. 68400
Profit(%)
= (68400 - 44400)/44400 * 100
= 54%
Answer: D
Answer: (D) 54%
Question: What are the last two digits of 63*35*37*82*71*412
(A)  10
(B)  30
(C)  40
(D)  70
(E)  80
Rationale: We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.
Remainder of (63*35*37*82*71*41)/ 100
Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:
Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.
So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.
Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.
We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.
So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.
We need the
Remainder of (63*7*37*41*71*41*5*2)/10*5*2
Remainder of (63*7*37*41*71*41)/10
Now using concept 2, let’s write the numbers in form of multiples of 10
Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10
Remainder of 3*7*7*1*1*1/10
Remainder of 147/10 = 7
Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.
When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 80.
Answer (E)
Answer: (E)  80
Question: A train 360 m long runs with a speed of 45 km/hr. What time will it take to pass a platform of 240 m long?
(A) 38 sec
(B) 35 sec
(C) 44 sec
(D) 48 sec
(E) none of these
Rationale: Speed = 45 km/hr = 45 Ã— (5/18) m/s
= 150/12 = 50/4 = 25/2 m/s
Total distance = length of the train + length of the platform
= 360 + 240 = 600 meter
Time taken to cross the platform = 600/(25/2) = 600 Ã— 2/25 = 48 seconds
ANSWER:D
Answer: (D) 48 sec
Question: P and Q invested in a business. The profit earned was divided in the ratio 2 : 3. If P invested Rs 30000, the amount invested by Q is
(A) 30000
(B) 50000
(C) 45000
(D) 20000
(E) 60000
Rationale: Let the amount invested by Q = q
30000:q=2:3
⇒30000×3=2q
⇒q=(30000×3)/2=45000
Answer is C.
Answer: (C) 45000
Question: If n = 2×3×5×7×11×23, then which of the following statements must be true?
I. n2 is divisible by 600
II. n + 19 is divisible by 19
III. is even n+4 / 2 = even
IV. n is divisible by 35
(A) I only
(B) b- II only
(C) none and IV only
(D) I and III only
(E) e None
Rationale: 2×3×5×7×11×23 =53130
I. not divisible by 600
II 53130+19=39149 not divisible by 19
III. 53130+4/2= odd= 26567
iv. divisible by 35
C
Answer: (C) none and IV only
Question: Working together, Jose and Jane can complete an assigned task in 15 days. However, if Jose worked alone and complete half the work and then Jane takes over the task and completes the second half of the task, the task will be completed in 45 days. How long will Jose take to complete the task if he worked alone? Assume that Jane is more efficient than Jose
(A) 26 days
(B) 31 days
(C) 50 days
(D) 65 days
(E) 36 days
Rationale: Assume:
Jose does 1 job in x days, so Jose does 1/x job in a day
Jane does 1 job in y days, so Jane does 1/y job in a day
Together, they does (x+y)/xy job in a day. This is equals to 1/20. So
(x+y)/xy = 1/15
15(x+y) = xy
Next, we're told 1 job takes 45 days to complete if Jose and Jane each does half the work. So since Jose does 1 job in x days, he wil need x/2 days to do half the job. Jane similarly will need y/2 days to do the other half.
x/2 + y/2 = 45
x+y = 90
So
xy = 1350
The answer choices are:
25 days
30 days
60 days
65 days
36 days
From the answer choices, So I'll go for 50 days for Jose and 27 days for Jane.
Answer : C
Answer: (C) 50 days
Question: The average of 65 numbers is 40. If two numbers, 83 and 60 are discarded, then the average of the remaining numbers is nearly
(A) 28.32
(B) 39
(C) 38.78
(D) 29
(E) 30
Rationale: Explanation :
Total sum of 63 numbers = (65 40) (83 +60) = 2600 143 = 2457
Average = 2457/63 = 39
Answer : B
Answer: (B) 39
Question: If the price of the commodity is increased by 50% by what fraction must its consumption be reduced so as to keep the same expenditure on its consumption?
(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) None
Rationale: Solution: Let the initial price of the commodity be 100.
After 50% increase in price, It will become,
100 ------50% increase----> 150.
Now, we have to reduce the consumption to keep expenditure 100.
Increase in price= 150 - 100 = 50
We have to reduce the consumption,
= (50/150)*100 = 1/3 or 33.33%
Other Method:
Here, we use, Final product constant graphic.
100 ==50% up== 150===33.33% down===>100.
Consumption Reduce = 33.33% = 1/3.
Answer: Option B
Answer: (B) 1/3