mathewhe/aqua_rat huggingface dataset:
Question: if n>0 and x is n percent of y, then, in terms of n, y is what percent of x?
(A) a)100n
(B) b)1/100n
(C) c)1/n
(D) d)10^4/n
(E) e)10,00/n
Rationale: Given (n*y)/100=x
to find (y/x)*100
Rearranging first equation
y/x*100 = 100*100/n
=10000/n
Answer D
Answer: (D) d)10^4/n
Question: Sandy walks around a circular field at the rate of one round per hour while Molly runs around it at the rate of two rounds per hour. If they start in the same direction from the same point, how many minutes will it take until they meet again?
(A) 30
(B) 45
(C) 60
(D) 75
(E) 90
Rationale: Since Sandy and Molly move in the same direction along the circle, they will first meet each other when there is a difference of one round between the two.
The relative speed of Sandy and Molly is 2 - 1 = 1 round per hour.
The time required to complete one round at this speed 60 minutes.
The answer is C.
Answer: (C) 60
Question: When x is divided by y, the quotient is z with remainder s. Which of the following is equal to s?
(A) yz – x
(B) y/x−z
(C) x – y – z
(D) x – yz
(E) zx – y
Rationale: (x/y)=z(+s)
(18/8)=2(+2)
x-yz
which is 18-(8*2)=2
Answer is D
Answer: (D) x – yz
Question: Cars emerging from a motorway arrive at a junction that splits the road into two separate lanes. The number of cars per hour that continue in either lane is constant. If 70 cars per hour were diverted from the left lane to the right lane, the number of cars entering the right lane per hour would be twice as big as the number of cars entering the left lane per hour. Alternatively, if 70 cars per hour were diverted from the right lane to the left lane, the number of cars entering the left lane per hour would be four times as great as the number of cars entering the right lane per hour. How many cars enter the left lane per hour?
(A) 140
(B) 170
(C) 210
(D) 265
(E) 235
Rationale: as per question 70 cars diverted from left lane to right lane. Thus we get equation R+70=2(L-70)
as per question 70 cars diverted from right lane to left lane .thus we get equation L+70=4(R-70)
by solving two equation i.e R +70 =2L-140 OR. R =2L -210 AND L+70= 4R -280 OR L= 4R -3500
BY SOLVING WE GET L= 4( 2L -210) -350 OR L= 8L -840-350 OR 1190 =7L THUS L =170
B
Answer: (B) 170
Question: The first three terms of a proportion are 4, 12 and 16. The fourth term is?
(A) 22
(B) 28
(C) 27
(D) 36
(E) 48
Rationale: (12*16)/4 = 48
Answer:E
Answer: (E) 48
Question: When 40 per cent of a number is added to another number the second number increases to its 140 per cent. What is the ratio between the first and the second number?
(A) 1 : 1
(B) 4 : 3
(C) 3 : 2
(D) Data inadequate
(E) None of these
Rationale: Let the first and the second numbers be x and y respect then
y + 40% of x = 140% of y
or, y + 0.4x = 1.4y
or, 0.4x = 0.4y
∴ x : y = 0.4 : 0.4 = 1 : 1
Answer A
Answer: (A) 1 : 1
Question: If 3x + 7 = x2 + M = 7x + 5, what is the value of 120% of M?
(A) 8.9
(B) 9.9
(C) 9.98
(D) Cannot be determined
(E) None of these
Rationale: If 3x + 7 = x2 + M = 7x + 5
ie, 3x + 7 = 7x + 5
or, 4x = 2 ∴ x = 1⁄2
and 3x + 7 = x2 + M
or, 1⁄4 + M = 3⁄2 + 7 ⇒ M + 1⁄4 = 8 + 1⁄2
∴ M = 81⁄4
Answer B
Answer: (B) 9.9
Question: Population of a city in 20004 was 1000000. If in 2005 there isan increment of 15% , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45%, then find the population of city atthe end of the year 2007
(A) 354354
(B) 545454
(C) 465785
(D) 456573
(E) 1083875
Rationale: Required population = P (1 + R1/100)(1 - R2/100)(1 + R3/100)
= P (1 + 15/100)(1 - 35/100)(1 + 45/100)
= 1083875
E
Answer: (E) 1083875
Question: Apple costs L rupees per kilogram for first 30kgs and Q rupees per kilogram for each additional kilogram. If the price of 33 kilograms is 168 and for 36kgs of Apples is 186 then the cost of first 20 kgs of Apples is
(A) 3.69
(B) 3.66
(C) 3.6
(D) 100
(E) 3.61
Rationale: Ans: By framing equations we get
30L+3Q=168
30L+6Q=186
Eliminate Q by multiplying the first equation by 2 and subtracting second equation from the first
30 L = 150 => L = 5
Then we get L = 5
Cost of 20 kgs of apples = 20 x 5 =100
Answer:D
Answer: (D) 100
Question: If 9^y = 3^14, what is y?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 7
Rationale: 9^y = 3^2y = 3^14
2y=14
y = 7
The answer is E.
Answer: (E) 7
Question: A train speeds past a pole in 15 sec and a platform 110 m long in 25 sec, its length is?
(A) 50 m
(B) 165 m
(C) 200 m
(D) 300 m
(E) None of these
Rationale: Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 110)/25 = x/15 => x = 165 m.
ANSWER:B
Answer: (B) 165 m
Question: Triangle A’s base is 10% greater than the base of triangle B, and A’s height is 10% less than the height of triangle B. The area of triangle A is what percent less or more than the area of triangle B?
(A) 9 % less
(B) 1 % less
(C) Equal to each other
(D) 1 % more
(E) 9 % more
Rationale: Base of A = 11/10 * Base of B
Height of A = 9/10 * Height of B
Area of A = (1/2) * Base of A * Height of A = 11/10 * 9/10 * Area of B = 99/100 * Area of B
Area of A is 1% less than the area of B.
Answer (B)
Answer: (B) 1 % less
Question: When the integer k is divided by 7, the remainder is 1. Which of the following expressions below when divided by 7, will have a remainder of 5?
I. 4k + 1
II. 6k + 6
III. 8k + 2
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II and III
Rationale: k = 7j + 1
4k + 1 = 28j + 4 + 1 = 28j + 5
6k + 6 = 42j + 6 + 6 = 42j + 7 + 5
8k + 2 = 56j + 8 + 2 = 56j + 7 + 3
The answer is D.
Answer: (D) I and II only
Question: If “&” implies “Add”, “@” implies “multiply”, “#” implies “subtract”, “$” implies “division”,
12 & 3 @ 2 # 32 $ 4 = ?
(A) 12
(B) 10
(C) 13
(D) 14
(E) None of these
Rationale: Explanation:
Given:
12 & 3 @ 2 # 32 $ 4 = ?
By substituting appropriate mathematical symbols, we get,
12 + 3 * 2 – 32/4 = ?
12 + 3 * 2 – 8 = ?
12 + 6 – 8 = ?
18 – 8 = 10
ANSWER IS B
Answer: (B) 10
Question: what is next no. 11 23 47 83 131 ???
(A) 171
(B) 181
(C) 191
(D) 201
(E) 211
Rationale: 11+12*1=23
23+12*2=47
47+12*3=83
83+12*4=131
131+12*5=191
ANSWER:C
Answer: (C) 191
Question: Equal amount of water were poured into two empty jars of different capacities, which made one jar 1/7 full and other jar 1/6 full. If the water in the jar with lesser capacity is then poured into the jar with greater capacity, what fraction of the larger jar will be filled with water?
(A) 1/7
(B) 7/12
(C) 1/2
(D) 2/7
(E) 2/3
Rationale: Same amount of water made bigger jar 1/7 full, then the same amount of water(stored for a while in smaller jar) were added to bigger jar, so bigger jar is 1/7+1/7=2/7 full.
Answer: D.
Answer: (D) 2/7
Question: A can do a work in 15 days and B in 20 days. If they work on it together for 5 days, then the fraction of the work that is left is :
(A) 1/3
(B) 7/12
(C) 5/12
(D) 2/3
(E) 1/2
Rationale: Ans is:C
A's 1 day's work =1/15
B's 1 day's work =1/20
(A + B)'s 1 day's work =(1/15+1/20)=7/60
(A + B)'s 5 day's work =(7/60*5)=7/12
Therefore, Remaining work =(1-7/12)=5/12. Answer: C
Answer: (C) 5/12
Question: In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got?
(A) 2800
(B) 2700
(C) 2100
(D) 2500
(E) None of these
Rationale: Explanation :
Total number of votes = 7500
Given that 20% of Percentage votes were invalid
=> Valid votes = 80%
Total valid votes = 7500 × (80/100)
1st candidate got 55% of the total valid votes.
Hence the 2nd candidate should have got 45% of the total valid votes
=> Valid votes that 2nd candidate got = total valid votes × (45/100)
= 7500 × (80/100) × (45/100) = 75 × 45 × 45 = 75 × 4 × 9 = 300 × 9 = 2700
Answer : Option B
Answer: (B) 2700
Question: In the first 10 overs of a cricket game, the run rate was only 3.8. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
(A) 6.25
(B) 6.5
(C) 6.75
(D) 6.1
(E) 8
Rationale: 10 overs - run rate = 3.8
runs scored in first 10 overs = 38
remaining overs 40
total runs to be scored= 282
38 runs already scored 282-38=244
244 runs to be scored in 40 overs
let required runrate be x
40*x=244
x=244/40
x=6.1
the required runrate is 6.1
ANSWER:D
Answer: (D) 6.1
Question: A sum fetched a total simple interest of 4025.25 at the rate of 9 %.p.a. in 5 years. What is the sum?
(A) 5768
(B) 8925
(C) 2345
(D) 8945
(E) 8723
Rationale: Principal = (100 x 4025.25)/(9 x 5)
= 402525/45
= 8945.
ANSWER D
Answer: (D) 8945
Question: A man buys an article for $110. and sells it for $125. Find the gain percent?
(A) 10%
(B) 13.6%
(C) 25%
(D) 20%
(E) 30%
Rationale: C.P. = $110
S.P. = $125
Gain = $15
Gain% = 15/110 *100 = 13.6%
Answer is B
Answer: (B) 13.6%
Question: A car takes 6 hours to cover a distance of 540 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time?
(A) 55 Kmph
(B) 60 Kmph
(C) 65 Kmph
(D) 70 Kmph
(E) 80 Kmph
Rationale: Time = 6
Distence = 540
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 540/9 = 60 Kmph
B)
Answer: (B) 60 Kmph
Question: The operation T is defined by x T y=1/x+1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true.
I. z T (-z)=0
II. z T z/z-1=0
III. 2/z T 2/z=z
(A) a.) I
(B) b.) I and II only
(C) c) I and III only
(D) d) II and III only
(E) e) I, II and III only
Rationale: Statement I is TRUE
Statement II is not TRUE
Statement III is TRUE
Answer : C
Answer: (C) c) I and III only
Question: A factory has three types of machines, each of which works at its own constant rate. If 7 Machine As and 11 Machine Bs can produce 525 widgets per hour, and if 8 Machine As and 22 Machine Cs can produce 600 widgets per hour, how many widgets could one machine A, one Machine B, and one Machine C produce in one 8-hour day?
(A) 400
(B) 475
(C) 550
(D) 625
(E) 600
Rationale: Let Machine A produce A widgets per hour. B produce B widgets per hour and C produce C widgets per hour.
7A+11B=525 ---(1)
8A+22C=600 ---(2)
Dividing (2) by 2
4A+11C=300.....(3)
Adding (1)(3)
11A+11B+11C = 825
A+B+C=75 per hour
So for eight hrs = 75*8 = 600 = Answer = E
Answer: (E) 600
Question: How many odd numbers between 10 and 1,100 are the squares of integers?
(A) 12
(B) 13
(C) 14
(D) 15
(E) 16
Rationale: The square of an odd number is an odd number:
10 < odd < 1,000
10 < odd^2 < 1,000
3.something < odd < 31.something (by taking the square root).
So, that odd number could be any odd number from 5 to 31, inclusive: 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, and 31.
15 numbers.
Answer: D.
Answer: (D) 15
Question: six boys picked up 30 mangoes .In how many ways can they divide them if all mangoes be identical?
(A) A)4^30
(B) B)2^30
(C) C)1^30
(D) D)3^30
(E) E)6^30
Rationale: Each Mango can be given to any one of the four people or in other words..1 mango can be divided into 6 ways...so all 30 can be divided in 6^30 ways
ANS:E
Answer: (E) E)6^30
Question: Two trains, each 100 m long, moving in opposite directions, cross other in 8 sec. If one is moving twice as fast the other, then the speed of the faster train is?
(A) 22
(B) 98
(C) 60
(D) 88
(E) 12
Rationale: Let the speed of the slower train be x m/sec.
Then, speed of the train = 2x m/sec.
Relative speed = ( x + 2x) = 3x m/sec.
(100 + 100)/8 = 3x => x = 25/3.
So, speed of the faster train = 50/3 = 50/3 * 18/5 = 60 km/hr.
Answer:C
Answer: (C) 60
Question: If the weight of 12 meters long rod is 14 kg. What is the weight of 6 meters long rod ?
(A) 8 kg.
(B) 10.8 kg.
(C) 7 kg.
(D) 18.0 kg
(E) None
Rationale: Answer
∵Weight of 12 m long rod = 14 kg
∴ Weight of 1 m long rod = 14 / 12 kg
∴ Weight of 6 m long rod = 14 x 6 / 12
= 7 kg
Option: C
Answer: (C) 7 kg.
Question: Cost of 16 mirror and 8 comb is Rs.352 and the cost of 4 mirror and 4 comb is Rs.96. Find the cost of each mirror?
(A) 10
(B) 12
(C) 20
(D) 21
(E) 22
Rationale: Let the cost of each pen and pencil be 'p' and 'q' respectively.
16m + 8c = 352 --- (1)
4m + 4c = 96
8m + 8c = 192 --- (2)
(1) - (2) => 8p = 160
=> m = 20
C
Answer: (C) 20
Question: A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are drawn at random, what is the probability that at least one is green?
(A) 23/35
(B) 23/39
(C) 23/33
(D) 23/31
(E) 23/34
Rationale: Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that at least one green marble can be picked in the random draw of two marbles = Probability that one is green + Probability that both are green
= (⁶C₁ * ⁹C₁)/¹⁵C₂ + ⁶C₂/¹⁵C₂
= (6 * 9 * 2)/(15 * 14) + (6 * 5)/(15 * 14) = 36/70 + 1/7
= 46/70 = 23/35
Answer: A
Answer: (A) 23/35
Question: If a randomly selected non-negative single digit integer is added to {2, 3, 4, 7}. What is the probability that the median of the set will increase but the range still remains the same?
(A) 0.2
(B) 0.3
(C) 0.4
(D) 0.5
(E) 0.6
Rationale: We are selecting from non-negative single digit integers, so from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. These 10 digits represent the total number of outcomes.
Hence, the total number of outcomes is 10.
We need to find the probability that the median of the set will increase but the range still remains the same.
The median of the set is (3 + 4)/2 = 3.5, thus the number selected must be 4 or greater.
For the range to remain the same, the number must be between 2 and 7 inclusive.
To satisfy both conditions, the number selected must be 4, 5, 6, or 7.
The probability is 4/10 = 0.4
The answer is C.
Answer: (C) 0.4
Question: The denominator of a fraction is 2 less than twice the numerator. If the numerator and denominator are both increased by 1, the fraction becomes 3/5. Find the fraction?
(A) A)2/3
(B) B)3/5
(C) C)4/9
(D) D)5/9
(E) E) None of these
Rationale: Let the numerator and denominator of the fraction be 'n' and 'd' respectively.
d = 2n - 2
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n - 2 ) + 3 => n = -8
d = 2n - 2 => d = -18
Hence the fraction is : 4/9
ANSWER:C
Answer: (C) C)4/9
Question: 99,996^2 - 4^2 =
(A) 10^10 - 2
(B) (10^5 – 2)^2
(C) 10^5(10^5 – 8)
(D) 10^5(10^4 – 2)
(E) 10^5(10^5 – 2)
Rationale: (99,996)^2 - (4)^2 = (100,000 -4)^2 - (4)^2 = (100,000)^2 - 2*4*100,000 + 4^2-4^2=(10^5)^2 - 8*(10^5) = 10^10 - 8*(10^5)=10^5(10^5 - 8) Answer C
Answer: (C) 10^5(10^5 – 8)
Question: Cassandra invested one part of her $10,000 at 7.5% per year and the other part at 8.5% per year. Her income from the two investment was $820. How much did she invest at each rate?
(A) x = $2000 and y = $7000
(B) x = $4000 and y = $7000
(C) x = $3000 and y = $7000
(D) x = $5000 and y = $7000
(E) x = $6000 and y = $7000
Rationale: Let x and y be the amount invested at 7.5% and 8.5% respectively
Income = $820 = 7.5% * x + 8.5% * y
The total amount invested is also known
10,000 = x + y
Solve the system of the equations to find x and y.
x = $3000 and y = $7000
Answer C
Answer: (C) x = $3000 and y = $7000
Question: The ratio of the cost price and selling price is 4:5. The profit percent is?
(A) 20%
(B) 25%
(C) 30%
(D) 50%
(E) 15%
Rationale: Let the C.P. = $4x
Then S.P. = $5x
Then, Gain = 5x-4x = $x
Gain% = x/4x *100 = 25%
Answer is B
Answer: (B) 25%
Question: Reema took a loan of Rs 1100 with simple interest for as many years as the rate of interest. If she paid Rs. 704 as interest at the end of the loan period, what was the rate of interest.
(A) 5%
(B) 6%
(C) 7%
(D) 8%
(E) 9%
Rationale: Explanation:
Let rate = R% then Time = R years.
=>1100∗R∗R /100=704
=>R2=64
=>R=8%
Option D
Answer: (D) 8%
Question: Rs.1170 is divided so that 4 times the first share, thrice the 2nd share and twice the third share amount to the same. What is the value of the third share?
(A) Rs.589
(B) Rs.518
(C) Rs.519
(D) Rs.540
(E) Rs.528
Rationale: A+B+C = 1170
4A = 3B = 2C = x
A:B:C = 1/4:1/3:1/2 = 3:4:6
6/13 * 1170 = Rs.540
Answer: D
Answer: (D) Rs.540
Question: Mrs. Anderson divided $420 evenly among her children. Later, when Mrs. Anderson was away, Daniel, the eldest, took $5 from each of his brothers and sisters. Daniel ended up with $90. If Mrs. Anderson has more than 7 children, how many children does Mrs. Anderson have?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
Rationale: Let no of children be x
Even share as per initial division = 420/x
Daniel took $5 from each of his brother & sisters = 5(x-1)
so, 420/x + 5(x-1) = 90
5x^2 -95x + 420 = 0
x^2 - 19x + 84 = 0
Solving for X, x = 12 or x =7
Since we are given that x > 7, so no of children = 12
Option A.
Answer: (A) 12
Question: If the sum of two integers is 6, then it must be true that
(A) both integers are even
(B) both integers are odd
(C) both integers are positive
(D) if one integer is negative, the other is positive
(E) if one integer is positive, the other is negative
Rationale: We're asked to find a statement that MUST be true. So, if we can find a case where a statement is false, we can ELIMINATE that answer choice.
(A) both integers are even : it could be the case that the numbers are 1 and 5. ELIMINATE A
(B) both integers are odd : it could be the case that the numbers are 2 and 4. ELIMINATE B
(C) both integers are positive : it could be the case that the numbers are -1 and 7. ELIMINATE C
(D) if one integer is negative, the other is positive : Can't think of a counterexample so KEEP D for now
(E) if one integer is positive, the other is negative : it could be the case that the numbers are 1 and 5. ELIMINATE E
By the process of elimination, the correct answer is D
Answer: (D) if one integer is negative, the other is positive
Question: A grocer has a sale of Rs. 5420, Rs. 5660, Rs. 6200, Rs. 6350 and Rs. 6500 for 5 consecutive months. Find the sale he should have in the sixth month, so that he gets an average sale of Rs. 6100?
(A) Rs. 5870
(B) Rs. 5991
(C) Rs. 6470
(D) Rs. 6850
(E) None of these
Rationale: Explanation:
Total sale for 5 months = Rs. (5420 + 5660 + 6200 + 6350 + 6500) = Rs. 30,130
Therefore, required sale = Rs. [(6100 * 6) – 30,130] = Rs. (36600 – 30,130) = Rs. 6470
ANSWER C
Answer: (C) Rs. 6470
Question: The probability that a visitor at the mall buys a pack of candy is 40%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?
(A) .288
(B) .147
(C) .189
(D) .063
(E) .027
Rationale: One case is: candy - candy - no candy
The probability is 4/10*4/10*6/10 = 96/1000
There are 3 such cases, so we should multiply this probability by 3.
P(exactly 2 buy candy) = 96/1000 * 3 = 288/1000 = 0.288
The answer is A.
Answer: (A) .288
Question: A bus and a truck are available to cross a jungle. The speed of the truck is thrice that of the bus. The capacity of the truck is 50 persons and that of bus is 30 persons. The average occupancy of the bus is twice that of the truck. The tickets for the bus and the truck cost Re 1 and Re 1.50 respectively. What is the ratio of the average rupee collection of the truck to that of the bus in a day? Assume there is no wastage time between trips and the occupancy of the bus/truck is defined as the ratio of the actual number of persons boarding it and its capacity.
(A) 15:2
(B) 15:4
(C) 15:8
(D) 15:7
(E) None of these
Rationale: Explanation :
Average Rupee collection = Speed × capacity × Occupancy × Ticket rate.
The ratio of average Rupee collection of truck to that of bus is :-
=>(3×50×1×1.5):(1×30×2×1)=15:4.
Answer : B
Answer: (B) 15:4
Question: The markup on a shirt is 50 percent of the cost. The markup is what percent of the selling price?
(markup = selling price - cost)
(A)  18%
(B)  20%
(C)  33 1/3%
(D)  45%
(E)  56 2/3%
Rationale: MP = 0.5 CP
SP = CP + 0.5 CP = 1.5 CP
hence MP = 0.5/1.5 SP = 1/3 SP. Hence MP is 33.33 % of SP
Answer C
Answer: (C)  33 1/3%
Question: If the day after tomorrow will be Sunday, when was Thursday?
(A) Today
(B) Tomorow
(C) Yesterday
(D) The day before yesterday
(E) none
Rationale: Explanation:
Day after tomorrow will be Sunday
=>Tomorrow will be Saturday
=> Today is a Friday
=> Yesterday was a Thursday
Answer: Option C
Answer: (C) Yesterday
Question: Median H of price of the car is 52,000. which could be the median of the other two.
35000, 44000, x, y, 57000.
(A) 38000
(B) 47000
(C) 48000
(D) 51,000
(E) 56,000
Rationale: Median of the car is 52,000 therefore 52,000 is the middle number. Assuming Y >= X, X will have to be the median value i.e. 52,000. Therefore the remaining value would have to be between 52,000 and 57,000 inclusive. The only answer choice that fits H is 56,000.E
Answer: (E) 56,000
Question: If â€˜PUNJABâ€™ is coded as â€˜GCEAABâ€™, then â€˜KERALAâ€™ is coded as
(A) BEICCA
(B) BIECAA
(C) BEIACA
(D) BIEACA
(E) BEICAA
Rationale: M=13,1+3=4=D
A=1=A
SIMILARLY,
K=11=1+1=2=B
E=5=E
R=18=9=I
A=1=A
L=12=3=C
A=I=A
BEIACA
ANSWER:C
Answer: (C) BEIACA
Question: The triplicate ratio of 2:4 is?
(A) 1:64
(B) 1:8
(C) 1:34
(D) 3:64
(E) 5:64
Rationale: 2^3: 4^3 = 8:64=1:8
ANSWER:B
Answer: (B) 1:8
Question: A certain set of numbers has an average (arithmetic mean) of 10 and a standard deviation of 2.5. If X, a number in the set, is within 2 standard deviations from the average, then which of the following CANNOT be the value of X?
(A) 2
(B) 5.5
(C) 6
(D) 10
(E) 14
Rationale: average (arithmetic mean) of 10 and a standard deviation of 2.5
Since, X, a number in the set, is within 2 standard deviations
i.e. X, a number in the set, between 10+2*2.5 and 10-2*2.5
i.e. X, a number in the set, between 15 and 5
Option A (2) is not within 2 SD therefore
Answer: Option A
Answer: (A) 2
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 3600 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units?
(A) 240
(B) 250
(C) 260
(D) 270
(E) 280
Rationale: Given that the area of the square = 3600 sq.units
=> Side of square = √3600 = 60 units
The radius of the circle = side of the square = 60 units
Length of the rectangle = 2/5 * 60 = 24 units
Given that breadth = 10 units
Area of the rectangle = lb = 24 * 10 = 240 sq.units
Answer: Option A
Answer: (A) 240
Question: The sum of two numbers is 33. Five times one number is equal to 6 times the other. The bigger of the two numbers is
(A) 10
(B) 18
(C) 15
(D) 22
(E) 24
Rationale: Sol.
Let the numbers be x and (33 - x).
Then, 5x = 6(33 - x)
⇔ 11x = 198⇔ x = 18.
So, the numbers are 18 and 15.
Answer B
Answer: (B) 18
Question: 4, 12, 48, 240, 1440, (...)
(A) 10080
(B) 12000
(C) 24000
(D) 14400
(E) 40000
Rationale: Go on multiplying the given numbers by 3, 4, 5, 6, 7
Answer is A
Answer: (A) 10080
Question: The ratio between the radii of two spheres is 1:3. Find the ratio between their volumes?
(A) 1:27
(B) 27:1
(C) 3:1
(D) 1:3
(E) 1:9
Rationale: r1 : r2 = 1:3
r13 : r23 = 1:27
ANSWER A
Answer: (A) 1:27
Question: Set A = {2,3,4,5}
Set B = {4,5,6,7,8}
Two integers will be randomly selected from the sets above, one integer from Set A and one integer from Set B. What is the probability that the sum of the two integers will equal 9?
(A) 0.20
(B) 0.25
(C) 0.30
(D) 0.33
(E) 0.39
Rationale: Explanation :
Total numbers in n(S)=4×5=20.
=>n(E)=4. [(2,7), (3,6), (4,5), (5,4)].
The required probability :-
=>P(E)=n(E)/n(S).
=>4/20.
=>0.20.
Answer : A
Answer: (A) 0.20
Question: If 80% of 90 is greater than 70% of a number by 30, what is the number?
(A) 70
(B) 60
(C) 10
(D) 75
(E) 85
Rationale: Explanation:
80/100 * 90 - 70/100 * x = 30
72 - 70/100 * x = 30
72 - 30 = 70/100 *x
42 = 70/100 * x
42 * 100 / 70 = x
60 = x
Answer: Option B
Answer: (B) 60
Question: In practical applications, battery voltage:
(A) 22
(B) 66
(C) 76
(D) 126
(E) 81
Rationale: Explanation:
No answer description available for this questio
Answer: Option B
Answer: (B) 66
Question: A tin of oil was 4/5 full. when 6 bottles of oil were taken out and four bottles of oil were poured into it, it was ¾ full. how many bottles of oil can the tin contain?
(A) 20
(B) 10
(C) 40
(D) 50
(E) 20
Rationale: Suppose x bottles can fill the tin completely
Then 4/5x-3/4x=6-4
X/20=2
X=40
Therefore required no of bottles =40
Answer is C.
Answer: (C) 40