divelab/Sys2Bench huggingface dataset:
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?
Options:
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
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?
Options:
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
Question: A man buys an article for $110. and sells it for $125. Find the gain percent?
Options:
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
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?
Options:
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
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
Options:
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
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?
Options:
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
Question: How many odd numbers between 10 and 1,100 are the squares of integers?
Options:
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
Question: six boys picked up 30 mangoes .In how many ways can they divide them if all mangoes be identical?
Options:
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
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?
Options:
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
Question: If the weight of 12 meters long rod is 14 kg. What is the weight of 6 meters long rod ?
Options:
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
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?
Options:
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
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?
Options:
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
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?
Options:
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
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?
Options:
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
Question: 99,996^2 - 4^2 =
Options:
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
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?
Options:
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
Question: The ratio of the cost price and selling price is 4:5. The profit percent is?
Options:
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
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.
Options:
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
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?
Options:
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
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?
Options:
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
Question: If the sum of two integers is 6, then it must be true that
Options:
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
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?
Options:
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
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?
Options:
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
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.
Options:
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
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)
Options:
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
Question: If the day after tomorrow will be Sunday, when was Thursday?
Options:
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
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.
Options:
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
Question: If â€˜PUNJABâ€™ is coded as â€˜GCEAABâ€™, then â€˜KERALAâ€™ is coded as
Options:
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
Question: The triplicate ratio of 2:4 is?
Options:
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
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?
Options:
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
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?
Options:
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
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
Options:
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
Question: 4, 12, 48, 240, 1440, (...)
Options:
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
Question: The ratio between the radii of two spheres is 1:3. Find the ratio between their volumes?
Options:
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
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?
Options:
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
Question: If 80% of 90 is greater than 70% of a number by 30, what is the number?
Options:
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
Question: In practical applications, battery voltage:
Options:
A)22
B)66
C)76
D)126
E)81
Rationale: Explanation:
No answer description available for this questio
Answer: Option B
Answer: B
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?
Options:
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
Question: A car drove from Town A to Town B without stopping. The car traveled the first 80 miles of its journey at an average speed of 25 miles per hour. What was the car’s average speed, in miles per hour, for the remaining 80 miles if the car’s average speed for the entire trip was 40 miles per hour?
Options:
A)28
B)48
C)50
D)55
E)100
Rationale: KAPLANOFFICIAL SOLUTION:
To solve this problem, you must remember that average speed means total distance divided by total time over an entire journey. This should not be confused with the average of the speeds. The total distance in this problem is 160 miles, but we will need to express the total time in a more complex way.
For the first 80 miles, the car traveled at 25 miles per hour. Therefore, we can say that the time this took was 80/25, because distance divided by rate equals time.
For the last 80 miles, we are trying to solve for the rate, so we can call this speed R. Thus, the time for the final 80 miles is 80/R.
If we put all of this together, knowing that the average speed over the entire journey is 40 miles per hour, we get the equation 160/(80/25 + 80/R) = 40. We can now solve for R and reach our answer.
160/(80/25 + 80/R) = 40
R = 100
100 is choice (E) and that’s our correct answer.
Answer: E
Question: If x and y are positive real numbers such that x + y = 1, which of the following could be the value of 100x + 500y?
I. 90
II. 250
III. 499
Options:
A)II only
B)III only
C)I and II
D)I and III
E)II and III
Rationale: We know that x and y are both between 0 and 1.
The minimum limit of 100x + 500y is 100, when x is close to 1 and y is close to 0.
The maximum limit of 100x + 500y is 500, when x is close to 0 and y is close to 1.
The expression can take on any value between 100 and 500.
The answer is E.
Answer: E
Question: What is the units digit of 2222^(333)*4444^(222)?
Options:
A)0
B)2
C)4
D)6
E)10
Rationale: Each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeatingpatternof the units digits.
Here's another way to organize the information.
We're given [(2222)^333][(4444)^222]
We can 'combine' some of the pieces and rewrite this product as....
([(2222)(3333)]^222) [(2222)^111]
(2222)(3333) = a big number that ends in a 6
Taking a number that ends in a 6 and raising it to a power creates a nice pattern:
6^1 = 6
6^2 = 36
6^3 = 216
Etc.
Thus, we know that ([(2222)(3333)]^222) will be a gigantic number that ends in a 6.
2^111 requires us to figure out thecycleof the units digit...
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
So, every 4powers, the pattern of the units digits repeats (2, 4, 8, 6.....2, 4, 8, 6....).
111 = 27 sets of 4 with a remainder of 3....
This means that 2^111 = a big number that ends in an 8
So we have to multiply a big number that ends in a 6 and a big number that ends in an 8.
(6)(8) = 48, so the final product will be a gigantic number that ends in an 8.
Final Answer:
E
Answer: E
Question: The dimensions of a certain machine are 48*32*52,if the size of the machine is increased proportionally until sum of dimensions equals 156.
Options:
A)56*40*50
B)56*40*40
C)56*40*30
D)56*40*20
E)56*40*60
Rationale: Given dimensions of a machine are 48*32*52
if the size is raised proportionally untill the sum of dimensions
sum of given dimensions is 48+32+52=132
difference of actual and increased=156-132=24
24/3=8
this means each dimension has to raise for 8 units
thus the new dimensions are 56*40*60
ANSWER:E
Answer: E
Question: A multiple choice test consists of 4 questions, and each question has 5 answer choices. In how many R ways can the test be completed if every question is unanswered?
Options:
A)24
B)120
C)625
D)720
E)1024
Rationale: 5 choices for each of the 4 questions, thus total R of 5*5*5*5 = 5^4 = 625 ways to answer all of them.
Answer: C.
Answer: C
Question: x and y are positive integers. If xy + x is odd, then which of the following must be even?
.
Options:
A)x
B)y
C)x + y
D)xy − x
E)x^2 − y
Rationale: Two possibilities:
(1) even + odd = odd
(2) odd + even= odd
In this case, x can not be even as it will make xy also even .
Therefore, (1) possibility present => y must be even.
Answer: B
Answer: B
Question: A sum amount to Rs.1344 in two years at simple interest. What will be the compound interest on the same sum at the same rate of interest for the same period?
Options:
A)10.75%
B)10.28%
C)10.25%
D)16.25%
E)20.25%
Rationale: 100
5 ---- I
5
0.25 --- II
-----------
10.25%.Answer:C
Answer: C
Question: The prices of tea and coffee per kg were the same in june. In july the price of coffee shot up by 20% and that of tea dropped by 20%. if in july , a mixture containing equal quantities of tea and coffee costs 60/kg. how much did a kg of coffee cost in june ?
Options:
A)40
B)45
C)50
D)55
E)60
Rationale: Let the price of tea and coffee be x per kg in June.
Price of tea in July = 1.2x
Price of coffee in July = 0.8x .
In July the price of 1/2 kg (500gm) of tea and 1/2 kg (500gm) of coffee (equal quantities) = 50
1.2x(1/2) + 0.8x(1/2) = 60
=> x =60
E
Answer: E
Question: X does a work in 10 days. Y does the same work in 15 days. In how many days they together will do the same work?
Options:
A)10
B)15
C)3
D)6
E)12
Rationale: X's 1day's work= 1/10
Y's 1 day's work=1/15
(X+Y)'s 1day's work= (1/10+1/15)=1/6
both together will finish the work in 6 days.
correct option is D
Answer: D
Question: Bob finishes the first half of an exam in one-thirds the time it takes him to finish the second half. If the whole exam takes him an hour, how many minutes does he spend on the first half of the exam?
Options:
A)20
B)24
C)27
D)30
E)40
Rationale: Many times, it is easiest to think of problems like this conceptually (as opposed to formulaically). Conceptually, if the first half of the exam takes 1/3's the time of the second half (3/3's), we can see that the hour the entire exam took to finish can be broken down into fifth's. (Another way to look at this problem is via ratio's - first half is 1:5 and the second half is 3:5). With each fifth of an hour being 15 minutes (60/4), the first half of the exam would have taken 30 minutes. Correct answer is D.
Answer: D
Question: In a store, the profit is 320% of the cost. If the cost increases by 25% but the SP remains constant, approximately what %of the sp is the profit?
Options:
A)50%
B)70%
C)80%
D)100%
E)120%
Rationale: Let C.P. = Rs. 100. Then, profit = Rs. 320, S.P. = Rs. 420
New C.P. = 125% of Rs. 100 = Rs. 125.
New S.P. = Rs. 420
Profit = 420 - 125 = Rs. 295
Required percentage = 295/420 * 100 = 1475/21 = 70%
B
Answer: B
Question: The ratio of ages of Aman, Bren, and Charlie are in the ratio 5:8:7 respectively. If 8 years ago, the sum of their ages was 76, what will be the age of Charlie 9 years from now?
Options:
A)44
B)52
C)57
D)65
E)80
Rationale: Let the present ages of Aman, Bren, and Charlie be 5x, 8x and 7x respectively.
5x-8 + 8x-8 + 7x-8 = 76
x = 5
Present age of Charlie = 7*5 = 35
Charlie's age 9 years hence = 35+9 = 44
Answer = A
Answer: A
Question: A shopkeeper sold an article for Rs 2591.36. Approximately what was his profit percent if the cost price of the article was Rs 2400
Options:
A)4%
B)5%
C)6%
D)8%
E)10%
Rationale: Explanation:
Gain % = (191.36*100/2400) = 7.97 % = 8% approx
Option D
Answer: D
Question: A team has to design a flag. The team has two yellow strips of cloth and five green strips of cloth that they must use as is to make the flag. How many different flags can the team design with the materials at hand?
Options:
A)21
B)121
C)240
D)720
E)1040
Rationale: The # of permutations of total 2+5=7 strips where 2 are identical yellow and 5 are identical green is 7!/(2!*5!)=21 (note that they must use all the materials at hand, also I guess that the strips must be either only vertical or only horizontal).
Answer: A.
Answer: A