mathewhe/aqua_rat huggingface dataset:
Question: Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and y are both prime numbers and 0 < x < 40 and 0 < y < 40, which of the following MUST be true?
I. The maximum possible range of the set is greater than 33.
II. The median can never be an even number.
III. If y = 37, the average of the set will be greater than the median.
(A) I only
(B) I and II only
(C) I and III only
(D) III only
(E) I, II, and III
Rationale: Solution
Key Data from the Question Stem
Set S has 6 elements.
The elements of set S are distinct.
x and y are prime numbers. Because 2 is already an element in S, both x and y have to be odd.
0 < x < 40 and 0 < y < 40
I. The maximum possible range of the set is greater than 33
The key word in this entire statement is maximum. We have to determine whether the maximum value possible for the range exceeds 33.
We know x and y are prime numbers. The largest prime number less than 40 is 37.
If either x or y is 37, the largest number in the set will be 37 and the smallest number is 2.
Therefore, the maximum range of the set will be 37 - 2 = 35. It is greater than 33.
Statement I is true. So, eliminate choices that donot contain I. Eliminate choice D
II. The median can never be an even number
There are 6 numbers in the set. Therefore, the median is the arithmetic mean of the 3rd and the 4th term when the numbers are written in ascending or descending order.
The elements are {2, 4, 6, 8, x, y}, where x and y are prime numbers.
If x and y take 3 and 5 as values, the median is 4.5
If x = 3, y = 7 or greater, the median is 5
If x = 5, y = 7 or greater, the median is 5.5
If x = 7, y = 11 or greater, the median is 6.5
If x = 11 or greater and y = 13 or greater, the median is 7.
It is quite clear that the median is either an odd number or is not an interger. So, the median can never be an even integer.
Statement II is true. Eliminate choices that do not contain II. Eliminate choices A and C as well.
III. If y = 37, the average of the set will be greater than the median
If y = 37, the set will be {2, 4, 6, 8, x, 37}, where x is a prime number greater than 2 and less than 37.
The average will be 57+x657+x6 = 9.5 + x6x6
If x = 3, median = 5 and average = 10. Average > median.
If x = 5, median = 5.5 and average = 10.33. Average > median
If x = 7, median = 6.5 and average = 10.66. Average > medain
If x = 11 or greater, the median = 7. Average will be definitely greater than 10. So, Average > Median.
It is true that the average is greater than the median if y = 37.
Statement III is also true.
Statements I, II, and III are true. Choice E
Answer: (E) I, II, and III
Question: A circular mat with diameter 20 inches is placed on a square tabletop, each of whose sides is 30 inches long. Which of the following is closest to the fraction of the tabletop covered by the mat?
(A) 5/12
(B) 0.3489
(C) 1/2
(D) 3/4
(E) 5/6
Rationale: C.
It is a circle inscribed in a square.
Square side=30---> square(table) area=30^2
Circle diameter=20---> Circle area=PiR^2=100Pi (where Pi=~3.14)
covered fraction=100*3.14/30*30=~314/30*30=3.14/9=0.3489
B
Answer: (B) 0.3489
Question: If apples are bought at the rate of 30 for Rs.100. How many apples must be sold for Rs.100 so as to gain 20%?
(A) 28
(B) 25
(C) 20
(D) 22
(E) 23
Rationale: SP for 30 apples to gain 20%= 100*120/100 =120
so 30 apples must be sold for rs 120
rs 120= 30 apples
rs 100= (30/120)*100 =25 apple
ANSWER:B
Answer: (B) 25
Question: How long does a train 110 m long running at the speed of 72 km/hr takes to cross a bridge 132 m length?
(A) 13.1 sec
(B) 12.1 sec
(C) 22.1 sec
(D) 32.1 sec
(E) 13.2 sec
Rationale: Speed = 72 * 5/18 = 20 m/sec
Total distance covered = 110 + 132 = 242 m.
Required time = 242/20 = 12.1 sec.
Answer:B
Answer: (B) 12.1 sec
Question: Nicky and Cristina are running a 300 meter race. Since Cristina is faster than Nicky, she gives him a 12 second head start. If Cristina runs at a pace of 5 meters per second and Nicky runs at a pace of only 3 meters per second, how many seconds will Nicky have run before Cristina catches up to him?
(A) 40
(B) 50
(C) 30
(D) 60
(E) 45
Rationale: The distance traveled by both of them is the same at the time of overtaking.
3(t+12) = 5t
t=18.
Cristina will catch up Nicky in 18 seconds.
So in 18 seconds Cristina would cover = 18*5 = 90 meter.
Now Time taken my Nicky to cover 90 meter = 90/3 = 30 Seconds.
C
Answer: (C) 30
Question: Ifaequals the sum of the even integers from 2 to 20, inclusive, andbequals the sum of the odd integers from 1 to 19, inclusive, what is the value of (a - b)/2 ?
(A) 1
(B) 10
(C) 19
(D) 5
(E) 2
Rationale: Answer is 10
Yes! there is really a faster way to solve it.
Sum of consecutive odd or even integers = (no. of odd or even ints) * (first int + last int) / 2
Here A = sum of even ints from 2 to 20, inclusive
number of even ints = 10,
first int + last int = 2+20 = 22
A = 10*22 / 2 = 110
B = sum of odd ints from 1 to 19, inclusive
number of odd ints = 10,
first int + last int = 1+19 = 20
A = 10*20 / 2 = 100
(A-B)/2 = 110 - 100 = 10/2=5
Ans :D
Answer: (D) 5
Question: Replace X with the appropriate number in
4 , 5 , 14 , 15 , 24 , 25 , X
(A) 30
(B) 29
(C) 34
(D) 21
(E) 20
Rationale: C
list of consecutive numbers that has an 'f' in the spelling
Answer: (C) 34
Question: What should come in place of the question mark (?) in the following equation?
10000+ 566 = 45+ ?
(A) 73689
(B) 35876
(C) 45943
(D) 10521
(E) 35786
Rationale: 10000+ 566 = 45+ ?
10000+ 566 – 45= 10521
Answer D
Answer: (D) 10521
Question: By selling 15 pencils for a rupee a man loses 20%. How many for a rupee should he sell in order to gain 20%?
(A) 8
(B) 7
(C) 6
(D) 4
(E) 10
Rationale: 80% --- 15
120% --- ?
80/120 * 15
= 10
Answer:E
Answer: (E) 10
Question: A train 500 m long can cross an electric pole in 10 sec and then find the speed of the train?
(A) 140
(B) 150
(C) 160
(D) 170
(E) 180
Rationale: Length = Speed * time
Speed = L/T
S = 500/10
S = 50 M/Sec
Speed= 50*18/5 (To convert M/Sec in to Kmph multiply by 18/5)
Speed = 180 Kmph
Answer:E
Answer: (E) 180
Question: A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.
(A) 1525
(B) 1625
(C) 1725
(D) 1825
(E) 1925
Rationale: let the sum be x=P
then sum(P)+(P*R*T)/100=2502.50
therefore
x+(x*4*13.5)/100=2502.50
x=1625
ANSWER:B
Answer: (B) 1625
Question: Chang Li has pet cockroaches that multiply at the same rate every week.
The first week he had 5, the next week he had 10, the second week he had 20, and the fourth week he had 80. How many did he have on week 3?
(A) 30
(B) 25
(C) 40
(D) 60
(E) 50
Rationale: The answer is C) 40, because the cockroaches are doubling in size every week- half of 80 is 40, and 20 doubled is 40.
Answer: (C) 40
Question: If the positive integer x is a multiple of 5 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following?
I. 8
II. 12
III 18
(A) II only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II and III
Rationale: Ans C.
From question x/4 is an integer.
x/4 = 2*2*q/4 ---> where q is the integer
Also y/6 = p => y/6 = 2*3*p/6
So, xy = 2*2*2*3 *pq
Thus 2*2*2 = 8 and 2*3*2 = 12 satisfy
C
Answer: (C) I and III only
Question: Two trains start from P and Q respectively and travel towards each other at a speed of 50 km/hr and 40 km/hr respectively. By the time they meet, the first train has traveled 100 km more than the second. The distance between P and Q is?
(A) 767 km
(B) 975 km
(C) 678 km
(D) 900 km
(E) 546 km
Rationale: At the time of meeting, let the distance traveled by the second train be x km. Then, distance covered by the first train is (x + 100) km.
x/40 = (x + 100)/50
50x = 40x + 4000 => x = 400
So, distance between P and Q = (x + x + 100)km =900 km.
Answer: D
Answer: (D) 900 km
Question: All of the students of the Music High School are in the band, the orchestra, or both. 90 percent of the students are in only one group. There are 84 students in the band. If 60 percent of the students are in the band only, how many students are in the orchestra only?
(A) 24
(B) 27
(C) 30
(D) 33
(E) 36
Rationale: 60% of students are in the band only, so 30% of students are in the orchestra only.
Since 10% of students are in both, then 70% of students are in the band which is 84 students.
The number of students in the orchestra is (30/70)*84 = 36.
The answer is E.
Answer: (E) 36
Question: Danny bought 2Q steaks for W dollars. Jerome buys R steaks for a 50% discount, how much will the steaks cost him in cents?
(A) 50RW/Q.
(B) 50QR/W.
(C) 25RQ/W.
(D) 25RW/Q.
(E) RW/(4Q).
Rationale: Danny bought 2Q steaks for W dollars, so 1 steak = W/2Q
Jerome buys R steaks for a 50% discount: r* (W/4Q)
in cents the answer will be : r* (100 W/4Q)= 25 RW/Q
Answer: (D) 25RW/Q.
Question: The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is?
(A) 23
(B) 21
(C) 52
(D) 56
(E) 12
Rationale: Let the total number of workers be x. Then,
8000x
= (12000 * 7) + 6000(x - 7)
= 2000x = 42000
= x =21.
Answer:B
Answer: (B) 21
Question: When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?
(A) 1296
(B) 1369
(C) 1681
(D) 1764
(E) 2500
Rationale: Let’s call the two perfect squares x^2 and y^2, respectively. Then the given information translates as x^2+148=y^2. Subtracting x^2 gives 148=y^2−x^2, a difference of squares. This, in turn, factors as (y+x)(y−x)=148.
The next step is tricky. It begins with factoring 148, which breaks down as 2∗2∗37. Since we’re dealing with perfect squares, x and y are positive integers, and (y+x) and (y−x) must be paired integer factors of 148. The options are 148∗1,74∗2, and 37∗4. But our number properties establish that (y+x) and (y−x) must be either both odd or both even, so only 74∗2 is an actual possibility. And because for any positive integers (y+x)>(y−x), we can conclude that y+x=74 and y–x=2. Solving by elimination, 2y=76, y=38, and x=36.
Finally, we just need to square 36 . But rather than multiplying it out, note that 36^2 ends in 6 – as does only one answer, choice A. This answer must be the one we want.
Answer: (A) 1296
Question: If population of certain city increases at the rate of 5%. If population in 1981 was 138915, then population in 1978 was?
(A) 120000
(B) 120029
(C) 187899
(D) 178968
(E) 789698
Rationale: X * (105/100) * (105/100) * (105/100) = 138915
X = 138915/1.157625
X = 120000
Answer:A
Answer: (A) 120000
Question: If G = 775 × 778 × 781, what is the remainder when G is divided by 14?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Rationale: G 775/14 leaves a remainder 5
778/14 leaves a remainder 8
781/14 leaves a remainder 11
5*8*11 =440
So the remainder will be the remainder of 440/14 which is 6
Ans A
Answer: (A) 6
Question: How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?
(A) 1245
(B) 4175
(C) 5412
(D) 4712
(E) 6400
Rationale: Number of bricks = Volume of the wall/Volume of 1 brick = (800 x 600 x 22.5) /(25 x 11.25 x 6)=6400.
Ans: E
Answer: (E) 6400
Question: ABCDEFGHI is a regular polygon with nine sides. What is the measure in degrees of angle ACB?
(A) 18
(B) 20
(C) 25
(D) 27
(E) 30
Rationale: the formula of sum of angles in a polygon of sides n=(n-2)*180
so sum of angles=7*180..
therefore each angle ,angle ABC=7*180/9=140..
when we join A and C to make a triangle ABC, it becomes an isosceles triangle with AB and BC equal..
so angle ACB=angleCAB=(180-140)/2=20..
ans B
Answer: (B) 20
Question: An uneducated retailer marks all his goods at 40% above the cost price and thinking that he will still make 25% profit, offers a discount of 25% on the marked price. What is his actual profit on the sales?
(A) 12.50 %
(B) 13.50 %
(C) 5 %
(D) 14.50 %
(E) None
Rationale: Sol.
Let C.P. = Rs. 100. Then, marked price = Rs. 140.
S.P. = 75% of Rs. 140 = Rs. 105.
∴ Gain% = 5%.
Answer C
Answer: (C) 5 %
Question: 21 ball numbered 1 to 21. A ballis drawn and then another ball is drawn without replacement.
(A) 1/15
(B) 3/53
(C) 9/42
(D) 6/15
(E) 7/12
Rationale: The probability that first toy shows the even number,
=1021=1021
Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.
Hence, probability that second toy shows the even number,
=920=920
Required probability,
=(1021)×(920)=(1021)×(920)
=9/42
C
Answer: (C) 9/42
Question: The maximum numbers of students among them 451 pens and 410 toys can be distributed in such a way that each student gets the same number of pens and same number of toys is
(A) 41
(B) 910
(C) 1001
(D) 1911
(E) None
Rationale: olution
Required number of students.
= H.C.F of 451 and 410.
â€¹=â€º41.
Answer A
Answer: (A) 41
Question: If you cut a 18ft piece of wood into two pieces making one piece 4ft longer than the other. What size is the smaller piece?
(A) 4ft
(B) 6ft
(C) 8ft
(D) 7ft
(E) 12ft
Rationale: Total length is 18ft, one piece is 4ft longer (x+4), leaving the other piece to figure out (x).
(x)+ (x+4)=18
x+x+4-4=18-4
2x=14
2x/2=14/2
x=7
The piece is D) 7ft.
Answer: (D) 7ft
Question: When is |x-14| = 14-x?
(A) x=14
(B) x=0
(C) x>14
(D) x<=14
(E) x< 0
Rationale: When is |x-14| = 14-x?
Choice A: X=14, it is true but X cannot be always 14
Choice B: X=0, it is also true, but X cannot be always 0
Choice C: X>14, it is false, for e.g. X=6, then one side of equation is 2 and the other side is -2
Choice D: X<=14, this choice encapsulate Choice A, Choice B and for all other conditions and is true for above said equation. Hence the answer choice is D.
Answer: (D) x<=14
Question: The average age of a husband, wife and their child 5 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is?
(A) 10.1 years
(B) 54 years
(C) 11.4 years
(D) 12.6 years
(E) 14 years
Rationale: Sum of the present ages of husband, wife and child
= (23 * 2 + 5 * 2) = 57 years.
Required average
= 57/5
= 11.4 years.
Answer:C
Answer: (C) 11.4 years
Question: What least no. must be subtracted from 427398 so that remaining no.is divisible by 10?
(A) 8
(B) 9
(C) 7
(D) 6
(E) 3
Rationale: Explanation:
On dividing 427398 by 10 we get the remainder 8, so 8 should be subtracted
Option A
Answer: (A) 8
Question: If the selling price of 50 articles is equal to the cost price of 30 articles, then the loss or gain percent is:
(A) 40%
(B) 23%
(C) 20%
(D) 60%
(E) 56%
Rationale: C.P. of each article be Re. 1.
Then, C.P. of 50 articles = Rs. 50;
S.P. of 50 articles = Rs. 30.
Loss % = 20/50 * 100 = 40%
ANSWER A
Answer: (A) 40%
Question: If 4 spiders make 3 webs in 9 days, then how many days are needed for 1 spider to make 1 web?
(A) 9
(B) 12
(C) 27
(D) 24
(E) 16
Rationale: Explanation:
Let, 1 spider make 1 web in x days.
More spiders, Less days (Indirect proportion)
More webs, more days (Direct proportion)
Hence we can write as
(spiders)4:1
(webs) 1:9}::x:3
â‡’4Ã—1Ã—9=1Ã—3 Ã— x
â‡’x=12
Answer: Option B
Answer: (B) 12
Question: A man on tour travels first 160 km at 72 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:
(A) 31.11 km/hr
(B) 35.55 km/hr
(C) 71.11 km/hr
(D) 36 km/hr
(E) 75.8 km/hr
Rationale: Total time taken = 160/72 + 160/80 = 3040/720 = 38/9 hours
--> Average speed = 320 x 9/38 = 75.8 km/hr.
Answer : E.
Answer: (E) 75.8 km/hr
Question: Find out the C.I on Rs.9000 at 4% p.a. compound half-yearly for 1 1/2 years
(A) 550.87
(B) 506.07
(C) 506.04
(D) 506.03
(E) 306.01
Rationale: A = 9000(51/50)3
= 9550.87
9000
-----------
550.87
Answer:A
Answer: (A) 550.87
Question: A customer bought a product at the shop. However, the shopkeeper increased the price of the product by 10% so that the customer could not buy the required amount of the product. The customer managed to buy only 80% of the required amount. What is the difference in the amount of money that the customer paid for the second purchase compared to the first purchase?
(A) 10%
(B) 12%
(C) 14%
(D) 16%
(E) 18%
Rationale: Let x be the amount of money paid for the first purchase.
The second time, the customer paid 0.8(1.1x)=0.88x.
The difference is 12%.
The answer is B.
Answer: (B) 12%
Question: The population of a town increased from 75,000 to 2,25,000 in a decade. The average percent increase of population per year is :
(A) 20%
(B) 5%
(C) 6%
(D) 8.75%
(E) None of these
Rationale: Solution
Increase in 10 year= (225000 - 75000) = 150000.
Increase% =( 150000/75000x 100 )% = 200%
âˆ´ Required average = ( 200/10 )% = 20%
Answer A
Answer: (A) 20%
Question: In how many ways can 4 men & 3 women can be made from a gang of 6 men & 7 women ?
(A) 10 hr
(B) 50
(C) 20
(D) 250
(E) 260
Rationale: Explanation:
6C4 * 7C3 = 50
B
Answer: (B) 50
Question: The side of a square is increased by 10% then how much % does its area increases?
(A) 40.00 %
(B) 10.00 %
(C) 25.00 %
(D) 21.00 %
(E) 15.00 %
Rationale: a = 100 a2 = 10000
a = 110 a2 = 12100
----------------
10000 --------- 2100
100 -------? => 21.00 %
Answer:D
Answer: (D) 21.00 %
Question: What is the sum of the different positive prime factors of 550?
(A) 10
(B) 11
(C) 15
(D) 16
(E) 18
Rationale: I think answer is E: 27
550 = 55 * 10 = (5 * 11) * (2 * 5)
Sum of theDIFFERENTpositive prime factors = 2 + 5 + 11 = 18
Answer: (E) 18
Question: Two dice are tossed. The probability that the total score is a prime number is:
(A) 5/12
(B) 5/10
(C) 6/12
(D) 9/12
(E) 4/12
Rationale: Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.
Then E	= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),
(5, 2), (5, 6), (6, 1), (6, 5) }
n(E) = 15.
P(E) = n(E)/n(S) = 15/36	= 5/12.
Answer:A
Answer: (A) 5/12
Question: The average (arithmetic mean) of y numbers is x. If 10 is added to the set of numbers, then the average will be x - 5. What is the value of y in terms of x ?
(A) x/6 - 6
(B) x/6 - 5
(C) x/7 - 5
(D) x/5 - 3
(E) x/5 - 6
Rationale: (A1 + A2 + .. + Ay)/y = x
(A1 + A2 + .. + Ay + 10)/(y+1) = x - 5
=> (xy + 10)/(y+1) = x - 5
=> xy +10 = yx -5y + x - 5
=> 15 = x - 5y
=> 5y = x - 15
=> y = x/5 - 3
Answer - D
Answer: (D) x/5 - 3
Question: The length of a rectangle is two - fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 3025 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units?
(A) 140
(B) 150
(C) 160
(D) 170
(E) 220
Rationale: Given that the area of the square = 3025 sq.units
=> Side of square = √3025 = 55 units
The radius of the circle = side of the square = 55 units
Length of the rectangle = 2/5 * 55 = 22 units
Given that breadth = 10 units
Area of the rectangle = lb = 22 * 10 = 220 sq.units
Answer: Option E
Answer: (E) 220
Question: The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected new mean is?
(A) 36.3
(B) 36.7
(C) 36.5
(D) 36.2
(E) 36.1
Rationale: Correct sum = (36 * 50 + 48 - 23)
= 1825.
Correct mean = 1825/50
= 36.5
Answer:C
Answer: (C) 36.5
Question: A division of a company consists of six men and five women. If two of these employees are randomly selected as representatives of the division, what is the probability that both representatives will be female?
(A) 1/6
(B) 2/5
(C) 2/11
(D) 5/12
(E) 7/15
Rationale: Total number of ways to select 2 members = 11C2 = 55
Total number of ways to select 2 females = 5C2 = 10
P(two females) = 10/55 = 2/11
The answer is C.
Answer: (C) 2/11
Question: HOW MANY KGS OF TEA WORTH RS.20/KG MUST BE BLENDED WITH 30 KGS OF TEA WORTH RS.30/KG SO THAT BY SELLING THE BLENDED VARIETY AT RS.30/KG THERE SHOULD BE A GAIN OF 10%
(A) 3
(B) 4
(C) 6
(D) 5
(E) 2
Rationale: (x+30)*30=(110/100)(20x+30*30)
Solving we get 3.
Answer is A.
Answer: (A) 3
Question: If the reciprocals of two consecutive positive integers are added together, what is the sum in terms of the greater integer d?
(A) (3d-1)/(d^2- d)
(B) (2d-1)/(d^4- d)
(C) d^2- d
(D) 2d-1
(E) (2d-1)/(d^2- d)
Rationale: Let two consecutive positive integers be d and d-1 (Greater integer is d)
So,
(1/d) + [1/(d-1)]
= (2d-1)/d(d-1)
=(2d-1)/(d^2- d)
Answer : E
Answer: (E) (2d-1)/(d^2- d)
Question: Find the natural number nearest to 9217 which completely divides 88 without giving any remainder.
(A) 9240
(B) 9064
(C) 9184
(D) 9152
(E) 9169
Rationale: Explanation:
These types of questions are to be solved by trial and error. Make sure minimum amount of time is wasted and your calculations are quick. Devote maximum 1 minute to such questions.
Here in this case clearly 9240 is the closest number which divides 88.
ANSWER: A
Answer: (A) 9240
Question: If 15 students in a class average 70% on an exam and 10 students average 90% on the same exam, what is the average in percent for all 25 students?
(A) 78%
(B) 79%
(C) 80%
(D) 81%
(E) 82%
Rationale: (15*70 + 10*90) / 25 = 78%
The answer is A.
Answer: (A) 78%
Question: If X,Y is 24%,30% smaller than Z then how much percentage is Y greater than X?
(A) 7 17/19 %
(B) 13%
(C) 13 1/2%
(D) 14%
(E) 15%
Rationale: X = 76%
Y = 70 %
Z = 100 %
X-Y/X × 100 = 6/76 × 100 = 7 17/19 % greater than X
A)
Answer: (A) 7 17/19 %
Question: The difference between a number and its two-fifth is 510. What is 40% of that number?
(A) 342
(B) 340
(C) 344
(D) 346
(E) 348
Rationale: Let the number be x. Then,
x - 2/5 x = 510
x = (510 * 5)/3 = 850
40% of 850 = 340.
Answer:B
Answer: (B) 340
Question: If a boat goes 7 km upstream in 42 minutes and the speed of the stream is 4 kmph, then the speed of
the boat in still water is
(A) 12 kmph
(B) 13 kmph
(C) 14 kmph
(D) 15 kmph
(E) None of these
Rationale: Explanation:
Rate upstream = (7/42)*60 kmh = 10 kmph.
Speed of stream = 4 kmph.
Let speed in still water is x km/hr
Then, speed upstream = (x ï¿½4) km/hr.
x-4 = 10 or x = 14 kmph
Answer: C
Answer: (C) 14 kmph
Question: If A:B = 1/3: 1/2 B:C = 1/3:1/2 then A:B:C?
(A) 4:6:9
(B) 6:6:9
(C) 7:6:9
(D) 6:8:9
(E) 8:4:9
Rationale: A:B = 1/3:1/2 = 2:3
B:C = 1/3:1/2 = 2:3
--------------------
A:B:C = 4:6:9
Answer: A
Answer: (A) 4:6:9
Question: There 3 kinds of books in the library physics, chemistry and biology. Ratio of physics to chemistry is 3 to 2; ratio of chemistry to biology is 4 to 3, and the total of the books is more than 3000. Which one of following can be the total R of the book?
(A) 3003
(B) 3027
(C) 3024
(D) 3021
(E) 3018
Rationale: First, you have to find the common ratio for all 3 books. You have:
P:C:B
3:2 --> multiply by 2 (gives you row 3)
4:6
6:4:3
Hence:
P:C:B:T(Total)R
6:4:3:13 ----> This means, the Total Number must be a multiple of 13.
Answer A is correct since 299 is divisible by 13, hence is 2990 and so is 3003 (2990+13).
Answer: (A) 3003
Question: If x, y, and z are consecutive odd integers, with x < y < z, then which of the following must be true?
I. x + y is odd
II. (x+z)/y is an integer
III. xz is even
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
Rationale: consecutive odds are:
2x+1, 2x+3, 2x+5
I. 2x+1+2x+3=4x+4=2(2x+2), so divisible by 2 and even - NOT TRUE
II. 2x+1+2x+5=4x+6/2x+3=2(2x+3)/2x+3=2, so integer - TRUE
III. (2x+1)*(2x+5)=4x^2+10x+2x+5=4x^2+12x+5=(2x(x+6)+5)/2 is not integer, so NOT TRUE
ANSWER: B
Answer: (B) II only
Question: If A = c% of y and B = y% of c, then which of the following is true?
(A) 1) A is smaller than B.
(B) 2) A is greater than B
(C) 3) Relationship between A and B cannot be determined.
(D) 4) If x is smaller than y, then A is greater than B.
(E) 5) None of the above
Rationale: Basically A = B = cy/100
None of the options mentions this; hence E.
Answer: (E) 5) None of the above
Question: A table is bought for Rs.1020/- and sold at Rs.750/- find the loss percentage
(A) 26% Loss
(B) 10% Gain
(C) 20% Loss
(D) 20% Gain
(E) 30% Gain
Rationale: Formula = (Selling price ~ Cost price)/Cost price * 100
= (750 ~ 1020)/1020 = 26% Loss
A
Answer: (A) 26% Loss
Question: Riya and Priya set on a journey. Riya moves eastward at a speed of 25kmph and Priya moves westward at a speed of 40 kmph.How far will be priya from Riya after 15 minutes
(A) 18kms
(B) 16kms
(C) 50kms
(D) 30kms
(E) 40kms
Rationale: total eastward distance= 25kmph*1/4hr=6.25 km
total westward distance= 40kmph*1/4hr=10 km
total distn betn them= 6.25+10=16.25km
ans 16km
ANSWER:B
Answer: (B) 16kms