mathewhe/aqua_rat huggingface dataset:
Question: Find the minimum value of 5cosA + 12sinA + 12
(A) -1
(B) 1
(C) 2
(D) 3
(E) None of these
Rationale: Solution:
5cosA +12sinA + 12 = 13(5/13 cosA +12/13sinA) + 12
Now, for any values of B we can get sinB =5/13 and we can replace cosB = 12/13.
We see that our assumption is right because we satisfy the condition sin^2B + cos^2B=1 so we get 13(sinBcosA +cosBsinA) + 12 =13(sin(A+B))+12. Therefore we know that minimum value of sinx=-1 and greatest is 1 so the least value of the expression becomes when sin(A+B) =-1 then the value of the whole term becomes 13.(-1) + 12 = -13 +12 = -1
Answer A
Answer: (A) -1
Question: The average (arithmetic mean) of 4 different integers is 70. If the largest integer is 90, what is the least possible value of the smallest integer?
(A) 1
(B) 19
(C) 29
(D) 13
(E) 33
Rationale: Total of Integers=70*4=280
Lowest of the least possible integer is when the middle 2 intergers are at the maximum or equal to the highest possible integer.
But all integers are distinct. So if the largest integer is 90, then the middle 2 will be 88 and 89
Lowest of least possible integer = 280-(90+89+88)=280-267=13
Answer: D
Answer: (D) 13
Question: In a school with 620 students, the average age of the boys is 12 years and that of the girls is 11 years. If the average age of the school is 11 years 9 months, then the number of girls in the school is
(A) 150
(B) 200
(C) 155
(D) 350
(E) None
Rationale: Sol.
Let the number of grils be x.
Then, number of boys = (620 - x).
Then, (11 3/4 × 620)
⇔ 11x + 12(620 - x) ⇔ x = 7440 - 7285 ⇔ 155.
Answer C
Answer: (C) 155
Question: An article costing rs.300 is sold at 10% discount on a mark-up price. What is the selling price after
discount?
(A) 270
(B) 116
(C) 126
(D) 136
(E) 146
Rationale: 300*90/100=270
ANSWER:A
Answer: (A) 270
Question: A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?
(A) 9:1
(B) 9:8
(C) 9:2
(D) 9:7
(E) 9:6
Rationale: Explanation:
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1
Answer: A
Answer: (A) 9:1
Question: The average age of three boys is 15 years. If their ages are in the ratio 3:5:7 the average age of the young boy is ______ years?
(A) 18
(B) 8
(C) 9
(D) 19
(E) 17
Rationale: avg of three boys is 15 years given ratio of their ages are 3:5:7
let x be the age
(3x+5x+7x)/3 =15 implies x=3
then the age of the young boy is 3x3=9
ANSWER:C
Answer: (C) 9
Question: The number of ways in which six men and six women can be seated in a row, so that no two girls sit together is?
(A) 6! * ⁷P₆
(B) 5! * 7
(C) 6! * 7
(D) 4! * 4
(E) None
Rationale: We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six women to be arranged. This can be done in ⁷P₆ ways.
Hence required number of ways = 6! * ⁷P₆
A
Answer: (A) 6! * ⁷P₆
Question: Medals are to be awarded to three teams in a 10-team competition. If one medal is gold, one medal is silver, and one medal is bronze, how many different ways are there to award the three medals to teams in the competition?
(A) 10!/7!
(B) 10!/(3!7!)
(C) 10!/3!
(D) 7!/3!
(E) 7!/94!3!)
Rationale: Choosing 3 teams out of 10 when order of the teams matters - P310=10!7!
Or: choosing which 3 teams out of 10 will get the medals - C310C103 and arranging them - 3!3!, so total - C310∗3!=10!7!
Or:
1-2-3-4-5-6-7-8-9-10 (teams);
G-S-B-N-N-N-N-N-N-N (GSB - medals, N - no medal);
Permutation of 10 letters out of which 7 N's are identical is 10!7!10!7! (so you'll get 10!7!10!7! different ways of assigning the medals to the teams).
Answer: A.
Answer: (A) 10!/7!
Question: In a 1000 m race, A beats B by 40 m and B beats C by 100 m. In the same race, by how many meters does A beat C?
(A) 145 m
(B) 136 m
(C) 268 m
(D) 129 m
(E) 218 m
Rationale: By the time A covers 1000 m, B covers (1000 - 40) = 960 m.
By the time B covers 1000 m, C covers (1000 - 100) = 900 m.
So, the ratio of speeds of A and C =
1000/960 * 1000/900 = 1000/864. So, by the time A covers 1000 m, C covers 864 m.
So in 1000 m race A beats C by 1000 - 864 = 136 m.
Answer:B
Answer: (B) 136 m
Question: In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 times as old as Robert. In which year was Richard 1.75 times as old as Robert?
(A) 1960
(B) 1965
(C) 1970
(D) 1975
(E) 1980
Rationale: In 1950: Ri = 4 Ro -------------- eq1
In 1955: Ri +5 = 3(Ro+5)---------eq2
Thus in 1950, solving eq1 and eq2
Ro= 10 , Ri =40
Now for each year we can calculate:
1960: Ri = 50 , Ro =20
1965: Ri= 55, Ro=25
1980: Ri= 40, Ro = 70
Thus8Ans:E
Answer: (E) 1980
Question: A train covers a distance of 12 km in 10 min. If it takes 10 sec to pass a telegraph post, then the length of the train is?
(A) 178 m
(B) 168 m
(C) 120 m
(D) 168 m
(E) 200 m
Rationale: Speed = (12/10 * 60) km/hr = (72 * 5/18) m/sec
= 20 m/sec.
Length of the train = 20 * 10
= 200 m.
Answer: E
Answer: (E) 200 m
Question: find how many 4 digits no. is less then 1239 such that all 4 digits are distinct.
(A) 68
(B) 69
(C) 70
(D) 71
(E) 72
Rationale: case 1- 1000-1099- there are 56 distinct digits
case 2- 1200-1209- there are 7 distinct digits
case 3- 1230-1239(excluding 1239)- there are 6 distinct digits
hence, 56+7+6=69
ANSWER:B
Answer: (B) 69
Question: A fruit seller sells mangoes at the rate of Rs.14 per kg and thereby loses 15%. At what price per kg, he should have sold them to make a profit of 15%?
(A) Rs.11.81
(B) Rs.12
(C) Rs.18.94
(D) Rs.12.31
(E) None
Rationale: Solution
85 : 14 = 115 : x
x= (14Ã—115/85)
= Rs.18.94
Hence, S.p per Kg	= Rs.18.94
Answer C
Answer: (C) Rs.18.94
Question: The average of six number is 3.95. The average of two of them is 3.4, while the average of the other two is 3.85. what is the average of the remaining three number ?
(A) 4.5
(B) 4.6
(C) 4.7
(D) 4.8
(E) None of these
Rationale: solution:
Sum of the remaining three numbers = (3.95 × 6) - [(3.4 × 2) + (3.85 ×2)]
=23.70 -(6.8 + 7.7) = 23.70 - 14.5 = 9.20.
∴ Required average =9.2/3 = 3.06 Answer E
Answer: (E) None of these
Question: How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits F equal to 28?
(A) 14
(B) 24
(C) 28
(D) 48
(E) 96
Rationale: First, look for all 4 digits without repeat that add up to 28. To avoid repetition, start with the highest numbers first.
Start from the largest number possible 9874.
Then the next largest number possible is 9865.
After this, you'll realize no other solution. Clearly the solution needs to start with a 9 (cuz otherwise 8765 is the largest possible, but only equals 26). With a 9, you also need an 8 (cuz otherwise 9765 is the largest possible, but only equals 27). With 98__ only 74 and 65 work.
So you have two solutions. Each can be rearranged in 4!=24 ways. So F=24+24=48.D
Answer: (D) 48
Question: How many unique positive odd integers less than 80 are equal to the product of a positive multiple of 5 and an odd number?
(A) 4
(B) 8
(C) 11
(D) 12
(E) 15
Rationale: The Question basically asks how many positive odd integers less than 80 are odd multiples of 5
So we have 5,15,25,35,45,55, 65,75
=8
B
Answer: (B) 8
Question: A man can row downstream at 18 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively?
(A) 13,3
(B) 12,6
(C) 15,3
(D) 14,4
(E) None of these
Rationale: Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 18 --- (1)
and x - y = 10 --- (2)
From (1) & (2) 2x = 28 => x = 14, y = 4.
ANSWER:D
Answer: (D) 14,4
Question: Factor: 6x4y3 â€“ 96y3
(A) A) 3y3(x2 + 4)(x + 2)(x -2)
(B) B) 6y3(x2 + 4)(x + 2)(x -2)
(C) C) 3y3(x2 + 4)(x + 2)(x -3)
(D) D) 3y3(x2 + 4)(x + 3)(x -2)
(E) E) 3y3(x2 + 4)(x + 2)(x -4)
Rationale: 6x4y3â€“ 963.
= 6y3(x4 â€“ 16).
= 6y3[(x2)2 - 42].
= 6y3(x2 + 4)(x2 - 4).
= 6y3(x2 + 4)(x2 - 22).
= 6y3(x2 + 4)(x + 2)(x -2).
Answer: (B) 6y3(x2 + 4)(x + 2)(x -2)
Answer: (B) B) 6y3(x2 + 4)(x + 2)(x -2)
Question: Seven children — A, B, C, D, E, F, G and H— are going to sit in seven chairs in a row. The children CF have to sit next to each other, and the others can sit in any order in any remaining chairs. How many possible configurations are there for the children?
(A) 600
(B) 720
(C) 1440
(D) 10080
(E) 4800
Rationale: for such questions , we can treat both as one then total person=7..
these 7 can be arranged in 7! ways.. but within these 7, one consists of two people who can be arranged in 2 ways CF or FC.. so ans =7!*2=10080
ans D
Answer: (D) 10080
Question: A “palindromic integer” is an integer that remains the same when its digits are reversed. So, for example, 43334 and 516615 are both examples of palindromic integers. How many 6-digit palindromic integers are both even and greater than 600,000?
(A) 150
(B) 200
(C) 240
(D) 300
(E) 480
Rationale: The first digit and last digit are the same so the 2 possibilities are 6 or 8.
The second and third digits can be any number from 0 to 9.
The total number of palindromic integers is 2*10*10 = 200
The answer is B.
Answer: (B) 200
Question: The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 19 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?
I. 3
II. 7
III. 12
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
Rationale: B
for 7 we will have mean as 7.16(approx) and median as 7 but for 3 we will have mean and median = 6.5 and for 12 we will have median and mean =8
Answer: (B) II only
Question: A company bought for its 7 offices 3 computers of brand N and 3 computers of brand M. In how many ways could computers be distributed among the offices if each office can have only 1 computer.
(A) 196
(B) 140
(C) 256
(D) 292
(E) 312
Rationale: This problem has to do withcombinations. Here's the general idea: if you have a set of n elements, and you are going to choose r of them (r < n), then the number of combinations of size r one could choose from this total set of n is:
# of combinations = nCr = (n!)/[(r!)((n-r)!)]
where n! is the factorial symbol, which means the product of every integer from n down to 1. BTW, nCr is readn choose r.
In this problem, let's consider first the three computers of brand M. How many ways can three computer be distributed to seven offices?
# of combinations = 7C3 = (7!)/[(3!)(4!)] = (7*6*5*4*3*2*1)/[3*2*1)(4*3*2*1)]
= (7*6*5)/(3*2*1) = (7*6*5)/(6) = 7*5 = 35
There are 35 different ways to distribute three computers to 7 offices. (The massive amount of cancelling that occurred there is very much typical of what happens in the nCr formula.)
One we have distributed those three M computers, we have to distribute 3 N computers to the remaining four offices. How many ways can two computer be distributed to four offices?
# of combinations = 4C3 = 4
For each of the 35 configurations of distributing the M computers, we have 4 ways of distributing the N computers to the remaining offices. Thus, the total number of configurations is 35*4 = 140. Answer choice =B
Answer: (B) 140
Question: A bookstore has a shelf that contains biographies which normally sell for $20 each and mysteries that normally sell for $12 each. During a sale, the biographies and mysteries are discounted at different rates so that a customer saves a total of $19 from the normal price by buying 5 discounted biographies and 3 discounted mysteries. If the sum of the discount rates for the two types of books is 27 percent, what is the discount rate on mysteries?
(A) 10%
(B) 11%
(C) 12.5%
(D) 13%
(E) 14%
Rationale: Let B be the discount on biographies and M be the discount on mysteries
So., B + M = 0.27 -----(1)
And
(20*5 + 12*3) - (20*5*(1-B) + 12*3*(1-M)) = 19
--> 100(1-(1-B)) + 36(1-(1-M) = 19
100B + 36M = 19 ------(2)
Solving 12., we get M = 0.125 = 12.5%
C
Answer: (C) 12.5%
Question: If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ?
(A) Area is quadrupled
(B) Area is tripled
(C) Area is doubles
(D) Area become half
(E) None of these
Rationale: Explanation:
2πR1=4π
=>R1=22πR2=8π
>R2=4Original Area =4π∗22=16πNew Area =4π∗42=64π
So the area quadruples.
Option A
Answer: (A) Area is quadrupled
Question: Which of the following is equivalent to the pair of inequalities x + 5 > 10 and x - 1 <= 5 ?
(A)  2 < x < 16
(B)  2 <= x < 4
(C)  2 < x <= 8
(D)  4 < x <= 8
(E)  5 < x <= 6
Rationale: Solution:
Let’s isolate x in both inequalities, starting with x + 5 > 10.
x + 5 > 10
x > 5
Next we isolate x in the inequality x – 1  5.
x – 1 <= 5
x <= 6
Bringing these two inequalities together we know:
5 < x <= 6
The answer is E.
Answer: (E)  5 < x <= 6
Question: A man rows his boat 75 km downstream and 45 km upstream, taking 2 1/2 hours each time. Find the speed of the stream?
(A) 5 kmph
(B) 6 kmph
(C) 9 kmph
(D) 8 kmph
(E) 1 kmph
Rationale: Speed downstream = d/t = 75/(2 1/2) = 30 kmph
Speed upstream = d/t = 45/(2 1/2) = 18 kmph
The speed of the stream = (30 - 18)/2 = 6 kmph
Answer:B
Answer: (B) 6 kmph
Question: The length of minute hand of a clock is 5.4 cm. What is the area covered by this in 10 minutes
(A) 15.27
(B) 16.27
(C) 17.27
(D) 18.27
(E) 19.27
Rationale: area of circle is pi*r^2
but in 10 minutes area covered is (10/60)*360=60 degree
so formula is pi*r^2*(angle/360)=3.14*(5.4^2)*(60/360)=15.27 cm^2
ANSWER:A
Answer: (A) 15.27
Question: If the remainder is 12 when the integer n is divided by 22, what is the remainder when 2n is divided by 11?
(A) 0
(B) 2
(C) 3
(D) 6
(E) 7
Rationale: n = 22k+12
2n = 2(22k+12) = 4*11k + 24 = 4*11k + 2*11 + 2 = 11j+2.
The answer is B.
Answer: (B) 2
Question: A particular library has 75 books in a special collection, all of which were in the library at the beginning of the month. These book are occasionally loaned out through an inter-library program. If, by the end of the month, 70 percent of books that were loaned out are returned and there are 66 books in the special collection at that time, how many books of the special collection were loaned out during that month?
(A) 20
(B) 30
(C) 35
(D) 40
(E) 55
Rationale: Total = 75 books.
65% of books that were loaned out are returned --> 100%-70%=30% of books that were loaned out are NOT returned.
Now, there are 66 books, thus 76-68=7 books are NOT returned.
{loaned out}*0.30=7 --> {loaned out}=30.
Answer: B.
Answer: (B) 30
Question: A is twice as good a workman as B and they took 9 days together to do the work B alone can do it in?
(A) 25 days
(B) 88 days
(C) 21 days
(D) 27 days
(E) 13 days
Rationale: WC = 2:1
2x + x = 1/9
x = 1/27 => 27 days
Answer: D
Answer: (D) 27 days
Question: When 2/3 of the garments in the shipment were inspected, 18 of the garments passed inspection and the remaining 2 garments failed. How many of the uninspected garments must pass inspection in order that 70 percent of the garments in the shipment pass?
(A) 10
(B) 3
(C) 8
(D) 7
(E) 5
Rationale: B = 9
2/3x= 20
x=30
for 90% approval we need 21 garments approved.
already approved = 18
we need 3 more.
Answer : B
Answer: (B) 3
Question: David has d books, which is 1/2 times as many as Jeff and 3 as many as Paula. How many books do the three of them have altogether, in terms of d?
(A)  5/6*d
(B)  10/3*d
(C)  7/3*d
(D)  7/2*d
(E)  9/2*d
Rationale: Although we could plug in a real value for d, the problem can be just as easily solved by setting up equations. However, let’s start by defining some variables. Since we are given that David has d books, we can use variable d to represent how many books David has.
number of books David has = d
number of books Jeff has = j
number of books Paula has = p
We are given that David has 3 times as many books as Jeff. We can now express this in an equation.
d = 3p
d/3 = p
We are also given that David has ½ as many books as Paula. We can also express this in an equation.
d = (1/2)j
2d = j
Notice that we immediately solved forj in terms of d and p in terms of d. Getting j and p in terms of d is useful when setting up our final expression. We need to determine, in terms of d, the sum of the number of books for David, Jeff, and Paula. Thus, we have:
d + d/3 + 2d
Getting a common denominator of 3, we have:
3d/3 + d/3 + 6d/3 = 10d/3 = 10/3*d
The answer is B
Answer: (B)  10/3*d
Question: 2 : 3 : : 23 : ?
(A) 25
(B) 28
(C) 22
(D) 29
(E) 30
Rationale: D
29
3 is the next prime number after 2. Similarly, 29 is the next prime number after 23.
Answer: (D) 29
Question: If two-third of a bucket is filled in 100 seconds then the time taken to fill the bucket completely will be .
(A) 90 seconds
(B) 150 seconds
(C) 60 seconds
(D) 100 seconds
(E) 120 seconds
Rationale: 2/3 filled in 100 seconds
1/3 filled in 50 secs
then 2/3+1/3=100 + 50 seconds = 150 seconds
ANSWER: B
Answer: (B) 150 seconds
Question: Find the product of the place value and face value of 3 in 5769354
(A) 900
(B) 9000
(C) 90
(D) 9
(E) 0.9
Rationale: Explanation:
Place value = Local value
Face value = Absolute value
The place value of 3 in 5769354 is 3 x 100 = 300
The face value of 3 in 5769354 is nothing but 3.
=> 300 x 3 = 900
Answer: Option A
Answer: (A) 900
Question: Find the length of the wire required to go 15 times round a square field containing 69696 m2.
(A) 15840
(B) 3388
(C) 2667
(D) 8766
(E) 66711
Rationale: a2 = 69696 => a = 264
4a = 1056
1056 * 15 = 15840
Answer: A
Answer: (A) 15840
Question: If 25 percent of p is equal to 10 percent of q, and pq#0, then p is what percent of q ?
(A) 2.5%
(B) 15%
(C) 20%
(D) 35%
(E) 40%
Rationale: p/4 = q/10
p = 0.4q (40%)
Answer (E).
Answer: (E) 40%
Question: If x and y are positive odd integers, then which of the following must also be an odd integer? I. x^(y+2) II. x(y+4) III. (y+1)^(x-1) + 5
(A) I only
(B) II only
(C) III only
(D) I and III
(E) All options are applicable
Rationale: let x= 5, y= 3
I. 5^5 = > Odd
II. 5*7 = > Odd
III. 4^4 +1 = > Odd
Answer : E
Answer: (E) All options are applicable
Question: Anthony and Michael sit on the six-member board of directors for company T. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
(A) 20%
(B) 30%
(C) 40%
(D) 50%
(E) 60%
Rationale: Soln: Let the commitees be I and II
Assuming that Anthony and Michael T go into commitee I, there is just one more place left to be filled in that commitee and it can be taken by any of the 4 remaining people. Thus 4 ways.
Since the Anthony and Michael can also go into commitee II, we get 4 ways for that commitee also.
So in total = 8 ways
Now total number of ways of choosing 3 from 6 people is
= 6C3
= 6 * 5 * 4/3!
= 20 ways
Therefore Probability is = (8 /20) * 100 = 40%
Answer: (C) 40%
Question: The cash difference between the selling prices of an article at a profit of 10% and 6% is Rs. 3. The ratio of the two selling prices is?
(A) 52:56
(B) 55:53
(C) 52:50
(D) 22:56
(E) 52:51
Rationale: Let C.P. of the article be Rs. x.
Then, required ratio = 110% of x / 106% of x
= 110/106 = 55/53 = 55:53
Answer: B
Answer: (B) 55:53
Question: If n = 2gh, where g and h are distinct prime numbers greater than 2, how many different positive even divisors does n have, including n ?
(A)  Two
(B)  Three
(C)  Four
(D)  Six
(E)  Eight
Rationale: Answer - 'C' - Four.
For a number 2gh with primes gh,there will be four even divisors - 2,2*g,2*h,2*g*h
Answer: (C)  Four
Question: A sum of Rs.63000 is divided into three parts such that the simple interests accrued on them for three, six and nine years respectively may be equal. Find the amount deposited for 3 years.
(A) 6000
(B) 2000
(C) 6500
(D) 4500
(E) 3000
Rationale: Let the amounts be x, y, z in ascending order of value. As the interest rate and interest accrued are same for 3 years 6 years and 9 years i.e. 3x = 6y = 9z = k.
L.C.M. of 3,6,9 = 18 So x:y:z: = 6000 : 3000 :2000
The amount deposited for 3 years = 6000
ANSWER:A
Answer: (A) 6000
Question: A train travelled from station P to Q in 8 hours and came back from station Q to P is 6 hours. What would be the ratio of the speed of the train while traveling from station P to Q to that from station Q to P?
(A) 3:6
(B) 3:1
(C) 3:8
(D) 3:4
(E) 3:2
Rationale: Since S # 1/t
S1 : S2 = 1/t1 : 1/t2
= 1/8 : 1/6
= 3 : 4
Answer: D
Answer: (D) 3:4
Question: Evaluate
30!/28!
(A) 970
(B) 870
(C) 770
(D) 670
(E) None of these
Rationale: Explanation:
=30!/28!
=30∗29∗28!/28!=30∗29=870
Option B
Answer: (B) 870
Question: The cost of 10 kg of mangos is equal to the cost of 24 kg of rice. The cost of 6 kg of flour equals the cost of 2 kg of rice. The cost of each kg of flour is $20.50. Find the total cost of 4 kg of mangos, 3 kg of rice and 5 kg of flour?
(A) 347.4
(B) 987.4
(C) 877.4
(D) 637.4
(E) 667.4
Rationale: C
$877.40
Let the costs of each kg of mangos and each kg of rice be $a and $r respectively.
10a = 24r and 6 * 20.50 = 2r
a = 12/5 r and r = 61.5
a = 147.6
Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5
= 590.4 + 184.5 + 102.5 = $877.40
Answer: (C) 877.4
Question: Krishan and Nandan jointly started a business. Krishan invested four times as Nandan did and invested his money for double time as compared to Nandan. Nandan earned Rs. 4000. If the gain is proportional to the money invested and the time for which the money is invested then the total gain was?
(A) Rs.20000
(B) Rs.24000
(C) Rs.28000
(D) Rs.36000
(E) Rs.35000
Rationale: 4:1
2:1
------
8:1
1 ----- 4000
9 ----- ? => Rs.36,000
Answer: D
Answer: (D) Rs.36000
Question: This week, Florry bought the same number of music albums she bought last week, but for a third of the total price. This week, the average (arithmetic mean) price per album
(A) is twice last week's average
(B) is three times last week's average
(C) is a half of last week's average
(D) is a third of last week's average
(E) is the same as last week's average
Rationale: let the number of music albums brought by Florry last week be '3'
let the total price of the music albums brought by Florry last week be $30 (i.e., $10 per album)
The average price per album purchased last week= $10
Florry brought same number of albums this week but for a total price of $10 ( i.e., one third the total price last week)
Therefore this weeks average price per album= $10/3 one-third of last week's average
Hence D
Answer: (D) is a third of last week's average
Question: On Sunday morning, Pugsley and Wednesday are trading pet spiders. If Pugsley were to give Wednesday two of his spiders, Wednesday would then have eight times as many spiders as Pugsley does. But, if Wednesday were to give Pugsley ten of her spiders, Pugsley would now have fourteen fewer spiders than Wednesday had before they traded. How many pet spiders does Pugsley have before the trading game commences?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
Rationale: If Pugsley were to give Wednesday four of his spiders, Wednesday would then have five times as many spiders as Pugsley does:
(w + 2) = 8(p - 2)
If Wednesday were to give Pugsley four of her spiders, Pugsley would now have four fewer spiders than Wednesday had before they traded:
p + 14 = w - 10
Solving gives p = 6 and w = 30.
Answer: B.
Answer: (B) 6