mathewhe/aqua_rat huggingface dataset:
Question: if set a ={2,3,5,7} and two numbers are selected at random from A(duplicates allowed), then what is the probability that the sum of the numbers is even?
(A) 1/8
(B) 13/16
(C) 5/12
(D) 5/8
(E) 2/3
Rationale: Sum of the two numbers would be even if
(1) Both digits are even.
(2) Both digit are odd.
(1) only one case is there: 22
(2) Following cases are there: 33, 35, 37 - 3 nos. Similarly 3 nos, with 5 (53, 55, 57) and 3 nos. with 7 (73, 75, 77)
Total 10 nos.
Total nos. that can be formed (repetition is allowed)
{1st no. can be chosen in 4 ways. Similarly 2nd no. can be chosen in 4 ways. Total 4*4 = 16 ways}
Prob : 10/16 = 5/8
D is the answer.
Answer: (D) 5/8
Question: 3 Years ago, the average age of a family of 5 members was 21 years. A baby having been born, the average age of the family is the same today. The present age of the baby is
(A) 1 year.
(B) 1.5 Years.
(C) 2 Years.
(D) 3 Years.
(E) 6 Years.
Rationale: We're told that 3 YEARS AGO, the average age of a family of 5 members was 21 years.
This means that the SUM of their ages was (5)(21) = 105 years.
A baby was born (meaning that there are now 6 family members). We're told that the PRESENT average age of the family members is the SAME.
This means that the sum of the ages is (6)(17) = 102 years.
We're asked for the PRESENT age of the baby.
Since each of the original 5 family members has aged 3 years since the initial average was calculated, the sum of their PRESENT ages is...
105 + (5)(3) = 120.
126-120 =6
E
Answer: (E) 6 Years.
Question: A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price?
(A) 1000
(B) 2876
(C) 1977
(D) 2778
(E) 2711
Rationale: 90%
104%
--------
14% ---- 140
100% ---- ? => Rs.1000
Answer: A
Answer: (A) 1000
Question: C & D started one company. They invest Rs. 100,000/- each. They deposit the money to Bank. Bank give 1% interest per month. After 2 years D left from company and asked C to return the money with interest. C says it will take time to release the FD and he assure D to give the money withinterest @0.5%. D Agreed this. What is the profit of C ?
(A) A)48000
(B) B)12000
(C) C)3600
(D) D)24000
(E) E)30000
Rationale: Total Investment : 10000/-*2=20000/-
Interest rate @ 1%= 20000*1%*12*2= 4800/-
Total : 248000/-
Interest to B : 10000/-*0.5%*12*2 = 1200/-
Paid to B : 10000/- + 1200/- = 11200/-
profit of A : 3600/-
Answer is C
Answer: (C) C)3600
Question: Henry answered 4/5 of the questions he was given for homework in the first hour, 5/17 of the remaining questions in the second hour and the remaining x questions in the third hour where x is an integer. How many questions was Henry given for homework?
(A) x
(B) (85/12)*x
(C) (85/24)*x
(D) (85/36)*x
(E) (85/48)*x
Rationale: Let Q be the number of homework questions.
Q = 4Q/5 + (5/17)(Q/5) + x
Q = 68Q/85 + 5Q/85 + x
12Q/85 = x
Q = (85/12)*x
The answer is B.
Answer: (B) (85/12)*x
Question: All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A?
(A) 1884 liters
(B) 1180 liters
(C) 1984 liters
(D) 1184 liters
(E) 1194 liters
Rationale: B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k - 5k = 3k.
Quantity of water in container C = 8k - 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k - 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.
Answer:D
Answer: (D) 1184 liters
Question: Students of three different classes appeared in common examination. Pass average of 10 students of first class was 70%, pass average of 15 students of second class was 60% and pass average of 25 students of third class was 80% then what will be the pass average of all students of three classes?
(A) 74%
(B) 75%
(C) 69%
(D) 72%
(E) None
Rationale: Solution:
Sum of pass students of first, second and third class,
= (70% of 10) + (60% of 15)+ (80% of 25)
= 7+9+20 = 36
Total students appeared,
= 10+15+25 = 50
Pass average,
= 36*100/50 = 72%.
Answer: Option D
Answer: (D) 72%
Question: In traveling from a dormitory to a certain city, a student went 1/5 of the way by foot, 2/3 of the way by bus, and the remaining 10 kilometers by car. What is the distance, in kilometers, from the dormitory to the city?
(A) 30
(B) 45
(C) 62
(D) 90
(E) 120
Rationale: I believe there is a better way to do it.
Basically one of the options should satisfy the given criteria.
60 did
1/5 *60 =12
2/3*60=40
so total distance
52 + remaining 10 =62
Answer C
Answer: (C) 62
Question: In Set T, the average (arithmetic mean) equals the median Q. Which of the following must be true?
I. Set T consists of evenly spaced numbers.
II. Set T consists of an odd number of terms.
III. Set T has no mode.
IV. None of the above.
(A) I only
(B) I and II
(C) II and III
(D) I, II, and III
(E) IV only
Rationale: This is a very good 'trap question'.
Consider Set A= { 0,0,0,0 } Mean=Median=Mode=0- Eliminates 1,2 and 3 OR
Consider Set B= {1,2,2,3 } Mean=Median=Mode=2- Eliminates 1,2 and 3
If you think that only in 'consecutive integers' the average is equal to the median Q you might fall for 'D'.
Ans: E
Answer: (E) IV only
Question: The average age of four children in a family is 12 years. If the spacing between their ages is 4 years, the age of the youngest child is ?
(A) 6 years
(B) 7 years
(C) 8 years
(D) 9 years
(E) None
Rationale: Answer
Let the age of youngest child be y.
so from question y + (y + 4) + (y + 8) + (y + 12) = 4 x 12
⇒ 4y + 24= 48
∴ y = 6 years
Correct Option: A
Answer: (A) 6 years
Question: Sum of the squares of three numbers is 267 and the sum of their products taken two at a time is 131. Find the sum?
(A) 20
(B) 25
(C) 26
(D) 23
(E) 29
Rationale: (a + b + c)2 = a2 + b2 + c2 + 2(ab +bc + ca) = 267 + 2* 131
a + b + c = √529 = 23
D
Answer: (D) 23
Question: A man can row his boat with the stream at 16 km/h and against the stream in 6 km/h. The man's rate is?
(A) 5 kmph
(B) 6 kmph
(C) 7 kmph
(D) 4 kmph
(E) 9 kmph
Rationale: Explanation:
DS = 16
US = 6
S = ?
S = (16 - 6)/2 =5 kmph
Answer: A
Answer: (A) 5 kmph
Question: In a division sum, the remainder is 0. As student mistook the divisor by 12 instead of 21 and obtained 70 as quotient. What is the correct quotient ?
(A) 0
(B) 12
(C) 13
(D) 40
(E) 25
Rationale: 12*70=840
840%21=40
ANSWER:D
Answer: (D) 40
Question: The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with (m + 2)?
(A) m + 4
(B) n + 6
(C) n + 3
(D) m + 5
(E) n + 4
Rationale: Explanatory Answer
Approach: Work with a numerical example
The fastest way to solve problems of this kind is to take numerical examples.
Let the 5 consecutive integers be 1, 2, 3, 4, and 5.
The average of 5 consecutive integers from 1 to 5 is 3.
Therefore, the value of m is 1 and the value of n is 3.
9 consecutive integers starting from m + 2 will be 3, 4, ... , 11
The average of positive integers from 3 to 11 is 7.
We know n = 3. So, 7 has to be (n + 4)
Choice E is the correct answer.
Answer: (E) n + 4
Question: Two buses each 3125 m long are running in opposite directions on parallel roads. Their speeds are 40 km/hr and 35 km/hr respectively. Find the time taken by the slower bus to pass the driver of the faster one?
(A) 50 sec
(B) 66 sec
(C) 48 sec
(D) 55 sec
(E) 45 sec
Rationale: Relative speed = 40 + 35 = 75 km/hr.
75 * 5/18 = 125/6 m/sec.
Distance covered = 3125+3125 = 6250 m.
Required time = 6250 * 6/125 = 50 sec.
Answer: A
Answer: (A) 50 sec
Question: From the integer n, if n*n*n is odd, then what is true from below options?
(A) n is odd
(B) n is odd and n*n is even
(C) n is even
(D) n*n is even
(E) n*n odd
Rationale: n is odd and n*n is even
Option B
Answer: (B) n is odd and n*n is even
Question: Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is ?
(A) 5/16
(B) 5/12
(C) 5/19
(D) 5/17
(E) 5/15
Rationale: Using question number 11 and 12, we get the probability as
1 - (1/36 + 5/9) = 5/12
Answer: B
Answer: (B) 5/12
Question: Find the HCF of 2.1, 1.05 and 0.63
(A) 0.44
(B) 0.64
(C) 0.21
(D) 0.24
(E) None of above
Rationale: Explanation:
To solve this question quickly, first remove decimal by multiplying each term with 100,
Then terms become 210, 105, 63
Then HCF of above terms is 21,
So Answer is 0.21
Answer: Option C
Answer: (C) 0.21
Question: There are 4 red shoes & 6 green shoes . If two of red shoes are drawn what is the probability of getting red shoes
(A) 1/13
(B) 1/14
(C) 1/2
(D) 2/15
(E) 2/16
Rationale: taking 4 red shoe the probablity is 4C2
from 10 shoes probablity of taking 2 red shoe is 4C2/10C2=2/15
ANSWER:D
Answer: (D) 2/15
Question: The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 41 kg, then the weight of B is:
(A) 27 kg
(B) 20 kg
(C) 26 kg
(D) 31 kg
(E) 35 kg
Rationale: let d sum of a, b, c is 3*45=135
and sum of a and b s 2*40=80
sum of b and c is 2*41=82
hence 80+82-135=27
ans=27
ANSWER:A
Answer: (A) 27 kg
Question: In the junior basketball league there are 18 teams, 2/3 of them are bad and ½ are rich. What can't be the number of teams that are rich and bad
(A) 4.
(B) 6.
(C) 7.
(D) 10.
(E) 8.
Rationale: total teams =18
bad teams=(2/3)*18=12
rich teams=9
so maximum value that the both rich and bad can take will be 9. so E= 10 can not be that value.
ans D.
Answer: (D) 10.
Question: In a public show 75 % of the seats were filled. If there were 700 seats in the hall, how many seats were vacant?
(A) 100
(B) 110
(C) 120
(D) 175
(E) 150
Rationale: 75 % of 600
= 75/100 × 700
= 450
Therefore, the number of vacant seats = 700 - 450 = 175.
Answer:D
Answer: (D) 175
Question: The average of 9 observations was 9, that of the 1st of 5 being 10 and that of the last 5 being 8. What was the 5th observation?
(A) 9
(B) 11
(C) 13
(D) 15
(E) 16
Rationale: 1 to 9 = 9 * 9 = 81
1 to 5 = 5 * 10 = 50
5 to 9 = 5 * 8 = 40
5th = 50 + 40 = 90 – 81 = 9
A)
Answer: (A) 9
Question: Replace the ? by the correct Mathematics symbol to make the expression true
16 ? 10 ? 4 ? 5 = 45
(A) 16 ÷ 10 x 4 - 5 = 45
(B) 16 + 10 ÷ 4 - 5 = 45
(C) 16 x 10 + 4 - 5 = 45
(D) 16 x 10 ÷ 4 + 5 = 45
(E) 16 x 10 ÷ 4 - 5 = 45
Rationale: Answer : D
Answer: (D) 16 x 10 ÷ 4 + 5 = 45
Question: A 270 m long train running at the speed of 120 km/hr crosses another train running in opposite direction at the speed of 80 km/hr in 9 sec. What is the length of the other train?
(A) 230
(B) 228
(C) 268
(D) 279
(E) 270
Rationale: Relative speed = 120 + 80 = 200 km/hr.
= 200 * 5/18 = 500/9 m/sec.
Let the length of the other train be x m.
Then, (x + 270)/9 = 500/9 => x
= 230.
Answer: A
Answer: (A) 230
Question: There are 5 identical pens and 8 identical books. In how many different ways can a person select at least one object from this set?
(A) 27
(B) 35
(C) 41
(D) 47
(E) 53
Rationale: We can select from 0 to 5 pens and from 0 to 8 books.
Thus the total possible ways to select objects is 6*9 = 54
We must subtract 1 for the case with 0 pens and 0 books.
The answer is E.
Answer: (E) 53
Question: How many paying stones, each measuring 2 1/2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1/2 m board?
(A) 99
(B) 18
(C) 26
(D) 17
(E) 12
Rationale: 30 * 33/2 = 5/2 * 2 * x => x
= 99
Answer: A
Answer: (A) 99
Question: A 11 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 4 if the original wire was cut at an arbitrary point?
(A) 5/8
(B) 1/5
(C) 6/11
(D) 3/11
(E) 2/5
Rationale: A square with an area of 4 has a perimeter of 8.
For the area to be >4, the longer piece must be >8.
The wire must be cut within 3 meters from either end.
The probability of this is 6/11.
The answer is C.
Answer: (C) 6/11
Question: If (x + 2)^2 = 144, which of the following could be the value of 2x?
(A) –26
(B) –22
(C) –11
(D) 12
(E) 14
Rationale: You can also get to the answer by backsolving using answer choices
Starting with choice C = -14
(x + 2)^2 = 144
(-14 + 2)^2 = 144 =====> (-12)^2 = 144 Bingo, Clearly, among the answer choices only -14 will result in 144 in the equation so x must be -14,
Then 2x = 2 * -18 = -22
Answer: E
Answer: (E) 14
Question: The size of the obtuse angle of a rhombus is twice the size of its acute angle. The side length of the rhombus is equal to 10 feet. Find its area.
(A) 90.56 ft.sq
(B) 87.5 ft.sq
(C) 86.6 ft.sq
(D) 75.85 ft.sq
(E) 92.52 ft.sq
Rationale: A rhombus has 2 congruent opposite acute angles and two congruent opposite obtuse angles. One of the properties of a rhombus is that any two internal consecutive angles are supplementary. Let x be the acute angle. The obtuse angle is twice: 2x. Which gives the following equation.
x + 2 x = 180 degrees.
Solve the above equation for x.
3x = 180 degrees.
x = 60 degrees.
We use the formula for the area of a triangle that uses the side lengths and any one of the angles then multiply the area by 2.
area of rhombus = 2 (1 / 2) (10 feet) 2 sin (60 degrees)
= 86.6 feet square (rounded to 1 decimal place)
Answer is C
Answer: (C) 86.6 ft.sq
Question: Anand starts from a point P towards point Q, where PQ = 100 km. After 1 hour, Ram starts from P and catches up with Anand after 2 more hours. After meeting they continue to travel towards Q. On reaching Q, Ram reverses his direction and meets Anand 6 hours after the first meeting. Find Anand's speed.
(A) (45/7) kmph
(B) (60/7) kmph
(C) (40/7) kmph
(D) (30/7) kmph
(E) (200/21) kmph
Rationale: Let the first meeting point be at a distance of x km from P, the remaining distance until Q would be (100-x) km
Anand traveled this x kms in 3 hours, making his speed x/3 kmph
Ram traveled the same x kms in 2 hours, making his speed x/2 kmph
So, in 6 hours:
Anand will cover=6x/3=2x km
Ram will cover=6x/2=3x km
And between their first meeting point and second, they both together covered a distance of 2(90-x) km.
2x+3x=2(100-x)
5x=200-2x
7x=200
x=200/7 km
Anand's speed=x/3=200/(3*7)=200/21 kmph
Ans:E
Answer: (E) (200/21) kmph
Question: Of the b students at a certain college, x are studying French and y are studying German. If w are studying both French and German, which of the following expresses the number of students at the college not studying either French or German ?
(A)  b + w- x- y
(B)  b - w- x- y
(C)  b- w - x + y
(D)  w + x + y- z
(E)  w - x - y - z
Rationale: {Total} = {French} + {German} - {Both} + {Neither}
b = x + y - w + {Neither}
{Neither} = b + w - x - y.
Answer: A.
Answer: (A)  b + w- x- y
Question: The average amount with a group of seven numbers is Rs. 20. If the newly joined member has Rs. 50 with him, what was the average amount with the group before his joining the group?
(A) 28
(B) 27
(C) 15
(D) 19
(E) 12
Rationale: Total members in the group = 7
Average amount = Rs. 20
Total amount with them = 7 * 20 = Rs. 140
One number has Rs. 50. So, the amount with remaining 6 people = 140 - 50 = Rs. 90
The average amount with them = 90/6 = Rs. 15.
Answer: C
Answer: (C) 15
Question: If (55^55+55) is divided by 56, then the remainder is:?
(A) 54
(B) 55
(C) 53
(D) 56
(E) 57
Rationale: Solution: (x^n+1) is divisible by (x+1), when n is odd.
.'. (55^55+1) is divisible by (55+1)=56. when (55^55+1)+54 is divided by 56,
the remainder is 54.
Answer A
Answer: (A) 54
Question: If 3^z is a factor of 30!, which of the following could be the value of z?
(A) 10
(B) 12
(C) 13
(D) 14
(E) 16
Rationale: 3^z is a factor of 30!, what values can z take.
Powers of 3 in 30! = [30/3] + [30/9] +[30/27] , where [x] is the integral part of x
Powers of 3 in 30! = 10 + 3 + 1 = 14
Hence the maximum value of 3^z in 30! = 3^14
Correct Option: D
Answer: (D) 14
Question: 161?85?65?89=100
then use + or - in place of ? and take + as m,- as n then find value of m-n
(A) 3
(B) -1
(C) 1
(D) -3
(E) 2
Rationale: 161-85-65+89=100
so m=1, n=2 => (m-n)=-1
ANSWER:B
Answer: (B) -1
Question: A train 100 m long crosses a platform 125 m long in 15 sec; find the speed of the train?
(A) 36
(B) 77
(C) 54
(D) 33
(E) 99
Rationale: D = 100 + 125 = 225
T = 15
S = 225/15 * 18/5 = 54 kmph
Answer: C
Answer: (C) 54
Question: The speed of a car is 20 km in the first hour and 30 km in the second hour. What is the average speed of the car?
(A) 25 kmph
(B) 85 kmph
(C) 34 kmph
(D) 23 kmph
(E) 45 kmph
Rationale: S = (20 + 30)/2
= 25 kmph
Answer:A
Answer: (A) 25 kmph
Question: What will be the LCM of 8, 24, 36 and 54
(A) 54
(B) 108
(C) 216
(D) 432
(E) 476
Rationale: Explanation:
LCM of 8-24-36-54 will be
2*2*2*3*3*3 = 216
Option C
Answer: (C) 216
Question: The ratio between the number of sheep and the number of horses at the Stewart farm is 5 to 7, If each horse is fed 230 ounces of horse food per day and the farm needs a total 12,880 ounces of horse food per day, what is the number of sheep in the farm?
(A) 18
(B) 28
(C) 40
(D) 56
(E) 60
Rationale: Let the number of sheeps and horses be 4x and 7x.
Now total number of horses = total consumption of horse food/consumption per horse = 12880/230 = 56, which is equal to 7x. => x = 8
Sheeps = 5x = 5*8 = 40. Hence C.
Answer: (C) 40
Question: In a Question paper there are 3 multiple choice questions. Each question has 4 choices with only one choice as the correct answer. What is the total number of ways in which a candidate will not get all the 3 answers correct?
(A) 19
(B) 85
(C) 63
(D) 54
(E) 64
Rationale: A candidate can answer the test in 4^3 number of ways (each question has 4 choices and we have total of 3 questions). Now, out of these cases there will be only one case when the candidate answered all the four questions correct. Therefore the total number of ways in which a candidate will NOT get all the four answers correct is 4^3-1=63.
Answer: D.
Answer: (C) 63
Question: An inspector rejects 0.08% of the meters as defective, How many meters he examine to reject 2 meteres
(A) 1200
(B) 2400
(C) 1400
(D) 2500
(E) None of these
Rationale: Explanation:
It means that 0.08% of x = 2
=>(8/ 100×100×x)=2=>x=2×100×100/ 8=>x=2500
Option D
Answer: (D) 2500
Question: 7^6n -1^ 6n When n is an integer > 0 is divisible by
(A) 119
(B) 127
(C) 134
(D) 145
(E) 154
Rationale: 127
B
Answer: (B) 127
Question: Henry professes to sell his goods at the cost price but he made use of 750 grams instead of a kg, what is the gain percent?
(A) 12 1/3%
(B) 13 1/3%
(C) 14 1/3%
(D) 15 1/3%
(E) 16 1/3%
Rationale: 750 --- 100
100 --- ? => 13 1/3%
Answer:B
Answer: (B) 13 1/3%
Question: After doing 3/5 of the Biology homework on Monday night, Sanjay did 1/3 of the remaining homework on Tuesday night. What fraction of the original homework would Sanjay have to do on Wednesday night to complete the Biology assignment ?
(A) 1/15
(B) 2/15
(C) 4/15
(D) 2/5
(E) None of these
Rationale: Sol.(c) Remaining homework on Monday night
= 1 - (3/5) = 2/5
Work done on Tuesday night
= 1/3 of 2/5 = 2/15
Remaining homework to complete the biology assignment
= 2/5 - 2/15 = (6 - 2)/15 = 4/15
Answer C
Answer: (C) 4/15
Question: In one hour,a boat goes 16 km along the stream and 6 km against the stream.The speed of the boat in still water (in km/hr) is :
(A) 3
(B) 11
(C) 8
(D) 9
(E) 10
Rationale: Solution
Speed in still water =1/2(16+6)kmph.= 11 kmph. Answer B
Answer: (B) 11
Question: The average score in an examination of 10 students of a class is 55. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 40 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper is
(A) 97
(B) 99
(C) 98
(D) 96
(E) 94
Rationale: The average score in an examination of 10 students of a class is 55 --> the total score = 10*55 = 550;
The 5 smallest scores have an average of 50 --> the total score of the 5 smallest scores = 250.
From above, the total score of the 5 largest scores = 550 - 250 = 300.
Say the 5 largest scores are a, b, c, d, and e (where a<b<c<d<e, since each of the top five scorers had distinct scores). We want to maximize e. To maximize e, we need to minimize a, b, c, and d. The least value of a, is 50 (The least score of the top 5 , a, should be equal to the highest of the bottom 5 and to minimize the highest of the bottom 5, all scores from the bottom 5 should be equal). In this case the least values of b, c, and d are 56, 57, and 58 respectively:
a + b + c + d + e = 50 + 51 +52 + 53 + e = 300;
e = 94.
Answer: E.
Answer: (E) 94
Question: A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y houses at a rate of x houses per week. Realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. The total time C it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week?
(A)  0.8(80 – y)
(B)  0.8 + 0.0025y
(C)  80/y – 1.25
(D)  80/1.25y
(E)  80 – 0.25y
Rationale: This may be a case of bad numberpicking but it worked for me.
80 houses. Let's say that y=40 houses are painted at the rate x=10 houses per week <=> 1/10 week per house. 40*1/10 = 4 houses per week will be painted at this rate.
80-y = 80-40 = 40 houses are to be painted at the faster rate. X*1,25=12,5 houses per week <=> 1/12,5 weeks per house * 40 houses = 40/12,5 = 80/25 = 320/100 = 3,2 weeks.
Which means finishing all houses at normal rate C=> 2*4 = 8 weeks. Faster rate = 4+3,2 = 7,2 weeks.
7,2/8 = 9/10 = 0,9. Insert y=40 in equations and it is clear that only (B) gives us 0,9.
Answer: (B)  0.8 + 0.0025y
Question: X can finish a work in 21 days. Y can finish the same work in 15 days. Yworked for 10 days and left the job. how many days does X alone need to finish the remaining work?
(A) 3
(B) 5
(C) 7
(D) 8
(E) 9
Rationale: Work done by X in 1 day = 1/21
Work done by Y in 1 day = 1/15
Work done by Y in 10 days = 10/15 = 2/3
Remaining work = 1 – 2/3 = 1/3
Number of days in which X can finish the remaining work = (1/3) / (1/21) = 7
C
Answer: (C) 7
Question: There were 36,000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold 9 times as many paperback copies as hardback copies were sold. If a total of 436,000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45,000
(B) 360,000
(C) 364,500
(D) 392,000
(E) 396,900
Rationale: Say x was the # of hardback copies sold from the time the first paperback copy was sold.
Then the total # of paperback copies sold was 9x;
Hence the total # of copies sold was (hardback)+(paperback)=(36+x)+(9x)=436 --> x=40.
So, the total # of paperback copies sold was 9x=9*40=360 or 360,000.
Answer: B.
Answer: (B) 360,000
Question: If Rs. 561 be divided among A, B, C in such a way that A gets 2/3 of what B gets and B gets 1/4 of what C gets, then their shares are respectively?
(A) s.300
(B) s.360
(C) s.396
(D) s.368
(E) s.323
Rationale: (A = 2/3 B and B = 1/4 C) = A/B = 2/3 and B/C = 1/4
A:B = 2:3 and B:C = 1:4 = 3:12
A:B:C = 2:3:12
A;s share = 561 * 2/17 = Rs. 66
B's share = 561 * 3/17 = Rs. 99
C's share = 561 * 12/17 =Rs.396.
Answer:C
Answer: (C) s.396