RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Subtracting 10% from X is the same as multiplying X by what number?
A.12
B.90
C.26
D.27
Answer: B.90
Explanation: X - (10/100) X = X * ?
? = 90%
Answer:B
Question: If the Average (mean ) of 7 positive temperatures is x degrees fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be
A.6x
B.4x
C.5x/3
D.3x/2
Answer: A.6x
Explanation: Let the 5 numbers be 1,2,3,4,5,6,7 (since no restrictions are given). There mean is 4 (x).
Now the sum of greatest three would be 5+6+7= 18
so the answer has to be 6x....that is option A
Question: If p and r are integers, and p^2 = 28r, then r must be divisible by which of the following?
A.2
B.4
C.5
D.7
Answer: D.7
Explanation: Solution: for p to be an int 28 r must be whole square of a number.
28r= 7*2*2*r
to make it whole square we need 7
so r can must be divisible by 7y where y is itself a whole square.
so Answer is D
Question: the efficiency of doing work of A is triple than B.hence,any work finished by A,60 days before than B.both together in how many days will finish the work?
A.20 days
B.45/2 days
C.25 days
D.30 days
Answer: B.45/2 days
Explanation: let any work finished by B in x days.A will take=(x-60) days
work of A in 1 days=3*(work of B in 1 days)
1/(x-60)=3*1/x,x=3(x-60),2x=180,x=90.
so,B take=90 days and A take=(90-60) days=30 days
(A+B) work in 1 days=(1/30+1/90)=4/90=2/45
so, together will finish work in 45/2 days
answer B
Question: If a fair die is rolled twice, what is the probability that a 2 occurs on at least one roll?
A.11/36
B.125/216
C.9/36
D.1/3
Answer: A.11/36
Explanation: Probability of no 2s is 5/6 * 5/6 = 25/36, so the probability of at least one 2 is 1 - (25/36) = 11/36
The correct answer is therefore A.
Question: Suppose you want to buy three loaves of bread that cost $1.75 each
and a jar of peanut butter that costs $3. A jar of jelly is $2.75, but you
donâ€™t need any jelly. You have $24. How much money will you have left
over?
A.$15.75
B.$12.50
C.$13.50
D.$14.50
Answer: A.$15.75
Explanation: The jelly is extra information. 24.00 â€“ 3 x 1.75 â€“ 3.00 =
24.00 â€“ 5.25 â€“ 3.00 = 15.75.
You have $5.75 left.
correct answer A
Question: Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long it will take to fill the tank ?
A.10 mins
B.12 mins
C.15 mins
D.20 mins
Answer: B.12 mins
Explanation: Explanation:
In this type of questions we first get the filling in 1 minute for both pipes then we will add them to get the result, as
Part filled by A in 1 min = 1/20
Part filled by B in 1 min = 1/30
Part filled by (A+B) in 1 min = 1/20 + 1/30
= 1/12
So both pipes can fill the tank in 12 mins.
Option B
Question: N = 1/2 + 5/7
Find N
A.13/14
B.16/18
C.3/4
D.17/14
Answer: D.17/14
Explanation: N = 1/2 + 5/7
Multiply both the denominators 2 * 7 = 14
Multply 14 by both sides
14 * N = 14 * 1/2 + 14 * 5/7
14 * N = 7 + 10
14 * N = 17
N = 17/14
Ans: D
Question: How many digits will be there to the right of the decimal point in the product of 95 and .02554 ?
A.6
B.7
C.4
D.5
Answer: C.4
Explanation: product of 95 and .02554 is 2.4263.
Therefore number of digits to right of decimal point is 4
Answer is C.
Question: If 45% of a class averages 100% on a test, 50% of the class averages 78% on the test, and the remainder of the class averages 65% on the test, what is the overall class average? (Round final answer to the nearest percent).
A.86%
B.87%
C.88%
D.89%
Answer: B.87%
Explanation: This question is a Weighted Average question with a series of dependent variables.
The remaining portion of the class represents 100% - 45% - 50% = 5% of the class
Converting the portions of the class population to decimal weights, we find:
Class Average = 0.45x100 + 0.50x78 + 0.05x63 = 87.15
The class average (rounded) is 87%
Final Answer
B) 87%
Question: 3 ltr of water is added with 9 ltr of a solution containing 57%of alcohol in the water. The % of alcohol in the new mixture is?
A.41.5%
B.42.75%
C.43%
D.45%
Answer: B.42.75%
Explanation: We have a 9 litre solution containing 57% of alcohol in the water.
=> Quantity of alcohol in the solution
=9Ã—57/ 100
Now 3 litre of water is added to the solution.
=> Total quantity of the new solution = 9 + 3 = 12
Percentage of alcohol in the new solution
=9Ã—57/100
12
Ã—
100
=9Ã—5710012Ã—100
=9Ã—4.75/100
=42.75%
B
Question: On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $0.48 per glass on the first day, what was the price per glass on the second day?
A. $015
B. $0.20
C. $0.30
D. $0.32
Answer: D. $0.32
Explanation: On the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade;
On the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade;
So, the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3. Naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3.
We are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day.
Say the price of the glass of the orangeade on the second day was $x then 2*0.48=3*x --> x=$0.32.
Answer: D.
Question: Set Q consists of the following five numbers: Q = {5, 7, 13, 21, 34}. Which of the following sets has the same standard deviation as Set Q?
I. {35, 38, 43, 51, 64}
II. {10, 16, 26, 42, 68}
III. {46, 59, 67, 72, 75}
A. I only
B. III
C. IIII
D. IIIII
Answer: B. III
Explanation: (1)	Multiplying all the numbers change Standard Deviation
(2)	Adding and subtracting all the numbers with the same number keeps the standard deviation same.
If you observe Set I and III are added versions of Set Q .
Set I: 5 has been added to the Set Q
Set III: Subtract each element from 80 and you would find a number there in the Set Q.
Set II: elements are multiplied by 2 and standard deviation changes.
Hence the answer is B.
Question: A mixture of 70 litres of milk and water contains 10% water. How many litres of water should be added to the mixture so that the mixture contains 12 1/2% water?
A.2
B.6
C.7
D.9
Answer: A.2
Explanation: Explanation:
Quantity of milk in the mixture = 90/100 (70) = 63 litres.
After adding water, milk would form 87 1/2% of the mixture.
Hence, if quantity of mixture after adding x liters of water, (87 1/2) / 100 x = 63 => x = 72
Hence 72 - 70 = 2 litres of water must be added.
Answer: A
Question: In a sample of 800 high school students in which all students are either freshmen, sophomores, juniors, or seniors, 28 percent are juniors and 75 percent are not sophomores. If there are 160 seniors, how many more freshmen than sophomores are there among the sample of students?
A.16
B.24
C.32
D.40
Answer: A.16
Explanation: 200 are sophomores.
The number of freshmen is 600 - 160 - 0.28(800) = 216
The answer is A.
Question: In a certain sequence, the first term is 6, and each successive term is 1 more than the reciprocal of the term that immediately precedes it. What is the fifth term in this sequence?
A.13/8
B.21/13
C.8/5
D.33/20
Answer: D.33/20
Explanation: let five terms in the sequence be a,b,c,d,e
a=6
b= 1+1/6=7/6
c=1+6/7= 13/7
d= 1+7/13 = 20/13
e= 1+13/20 = 33/20
hence answer should be D.
Question: What is the unit digit in 7^100?
A.0
B.1
C.2
D.3
Answer: B.1
Explanation: Unit digit in 7^100 = unit digit in [(7^4)^25 ]
But unit digit in 7^4=1
unit digit 7^100= 1
Answer is B
Question: How many consonants are present between the first and fifth vowel in the sequence of English alphabet ?
A.12
B.13
C.11
D.16
Answer: D.16
Explanation: Explanation:
Ref : A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
The first and fifth vowels are A and U respectively. There are 19 letters between A and U; out of which 'E' , 'I' and 'O' are vowels. Thus, there are 16 consonants between A and U.
ANSWER D
Question: After decreasing 24% in the price of an article costs Rs.760. Find the actual cost of an article?
A.1218
B.2777
C.1000
D.2688
Answer: C.1000
Explanation: CP* (76/100) = 760
CP= 10 * 100 => CP = 1000
Answer: C
Question: A survey was sent to 60 customers, 7 of whom responded. Then the survey was redesigned and sent to another 72 customers, 9 of whom responded. By approximately what percent did the response rate increase from the original survey to the redesigned survey?
A.2%
B.7%
C.14%
D.28%
Answer: B.7%
Explanation: Rate of first survey = 7/60
Rate of second survey = 9/72
% Response rate increase (9/72-7/60)/(7/60) = 7%
Answer is B = 7%
Question: If the product of 3 consecutive integers is 210 then the sum of the two smaller integers is:
A.17
B.10
C.11
D.12
Answer: C.11
Explanation: let the numbers be x-1 , x , x+1
(x-1)(x)(x+1)=210
210=2*105
=2*3*35
=2*3*5*7
this can be written as 6*5*7
=>x-1=5
x=6
x+1=7
sum of two smallest numbers=6+5=11
ANSWER:C
Question: How much more would Rs.20000 fetch, after two years, if it is put at 20% p.a. compound interest payable half yearly than if is put at 20% p.a. compound interest payable yearly?
A.482
B.887
C.299
D.277
Answer: A.482
Explanation: 20000(11/10)4 - 20000(6/5)2 = 482
Answer: A
Question: 8 : 18 : : 24 : ?
A.38
B.32
C.44
D.43
Answer: A.38
Explanation: 3^2-1=8
4^2+2=18
5^2-1=24
there for
6^2+2=38
ANSWER:A
Question: Rajesh can finish 1/4 of his home work in one hour. Seema can finish 3/7 of
her homework in 30 minutes and Ramya can finish 3/4 of her homework in
two and a half hours. If all of them start their home work at 12.00 PM and can
go to play as soon as they all finish their homework. When can they start to
play, if they take a break at 3.30 PM for 30 minutes?
A.4:30pm
B.5:30pm
C.6:30pm
D.5:10pm
Answer: A.4:30pm
Explanation: ajesh can finish his work in 4 hr,
ramya can finish her work in 1 hr
seema can finish in 3.3 hr
so 4:30pm they can go to play
ANSWER:A
Question: Nine animals of a circus has to be placed inside in nine cages one in each cage. If 4 of the cage are too small for 6 of the animal then find the number of ways of caging the animal.
A.A.6048
B.B.10825
C.C.14400
D.D.784200
Answer: C.C.14400
Explanation: ways for cages that are too small = 5*4*3*2=120
since we have 5 cages and 5 animal left therefore ways for 5 cages =5!
total ways =120*5!=14400
Answer C
Question: Find the average of first 20 natural numbers.
A.20.5
B.18
C.19.5
D.10.5
Answer: D.10.5
Explanation: Exp. Sum of first n natural numbers = n( n+1)/2
So, sum of first 20 natural numbers = 20*21/2 = 210
Required average =210/20 =10.5
Answer:D
Question: Akash leaves Chennai at 6am & reaches Bangalore at 10am. Prakash leaves Hyderabad at 8am and reaches Chennai at 11:30am. At what time do they cross each other?
A.6 : 32am
B.7 : 56am
C.8 : 56am
D.3 : 56am
Answer: C.8 : 56am
Explanation: Time taken by Akash = 4 h
Time taken by Prakash = 3.5 h
For your convenience take the product of times taken by both as a distance.
Then the distance = 14km
Since, Akash covers half of the distance in 2 hours(i.e at 8 am)
Now, the rest half (i.e 7 km) will be coverd by both prakash and akash
Time taken by them = 7/7.5 = 56 min
Thus , they will cross each other at 8 : 56am.
C
Question: A certain auto manufacturer sold 5% fewer vehicles in 2007 than in 2006. If the manufacturer sold 2.2 million vehicles in 2006, how many vehicles, to the nearest 10,000, did the manufacturer sell in 2007?
A.17,00,000
B.18,00,000
C.19,00,000
D.21,00,000
Answer: D.21,00,000
Explanation: Let's assume in 2006 the manufacturer sold X.
In 2007 he sold X-5/100(X)
If sales in 2006,X= 2.2 million find X-5/100(X)
(2.2 x 10^6)-5/100(2.2 x 10^6)
2200000-1,10,000=20,90,000-----> 21,00,000(approx)
Answer-D
Question: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability T that both selected dogs are NOT littermates?
A.1/6
B.2/9
C.5/6
D.7/9
Answer: C.5/6
Explanation: We have three pairs of dogs for the 6 with exactly one littermate, and one triplet, with each having exactly two littermates.
So, in fact there are two types of dogs: those with one littermate - say A, and the others with two littermates - B.
Work with probabilities:
Choosing two dogs, we can have either one dog of type B or none (we cannot have two dogs both of type B).
The probability of choosing one dog of type B and one of type A is 3/9 * 6/8 * 2 = 1/2 (the factor of 2 for the two possibilities BA and AB).
The probability of choosing two dogs of type A which are not littermates is 6/9 * 4/8 = 1/3 (choose one A, then another A which isn't the previous one's littermate).
The required probability is 1/2 + 1/3 = 5/6.
Find the probability for the complementary event: choose AA or BB.
Probability of choosing two dogs of type A who are littermates is 6/9 * 1/8 = 1/12.
Probability of choosing two dogs of type B (who necessarily are littermates) is 3/9 * 2/8 = 1/12.
Again, we obtain 1 - (1/12 + 1/12) T= 5/6.
Answer: C
Question: The tax on a commodity is diminished by 20% and its consumption increased by 15%. The effect on revenue is?
A.2%
B.8%
C.5%
D.6%
Answer: B.8%
Explanation: 100 * 100 = 10000
80 * 115 = 9200
-----------
10000-----------800
100-----------? => 8% decrease
Answer: B
Question: What is the lowest positive integer that is divisible by each of the odd integers between 11 and 17, inclusive?
A.3*5*13*17*11
B.5×17×19×23
C.7×15×17×19
D.7×15×19×21
Answer: A.3*5*13*17*11
Explanation: We should find the LCM of 11,13,15 = 3*5, 17,-> LCM = 3*5*13*17*11 .
Answer: A.
Question: In digging a pond 20 m * 10 m * 5 m the volumes of the soil extracted will be?
A.100 cu.m
B.1000 cu.m
C.10000 cu.m
D.2000 cu.m
Answer: B.1000 cu.m
Explanation: 20 * 10 * 5 = 1000
ANSWER:B
Question: By selling 12 pencils for a rupee a man loses 20%. How many for a rupee should he sell in order to gain 20%?
A.8
B.9
C.6
D.5
Answer: A.8
Explanation: 80% --- 12
120% --- ?
80/120 * 12 = 8
Answer:A
Question: A bag of potatoes weighs 60 lbs divided by half of its weight. How much does the bag of potatoes weight?
A.20 lb
B.11 lb
C.10 lb
D.15 lb
Answer: B.11 lb
Explanation: Sol.
66÷6 = 11.
Answer : B
Question: The bankerâ€™s discount on Rs. 1600 at 10% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. The time is :
A.2 months
B.4 months
C.6 months
D.7 months
Answer: C.6 months
Explanation: Sol.
S.I. on Rs. 1600 = R.D. on Rs. 1680.
âˆ´ Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is S.I. on Rs. 1600 at 10%.
âˆ´ Time = [100 * 80 / 1600 * 10] year = 1/2 year = 6 months.
Answer C
Question: A and B are two numbers. Let
@(A,B)= Average of A and B,
/(A,B)= Product of A and B, and
*(A,B)= The result of dividing A by B.
The sum of A and B is given by
A./(@(A,B),2)
B.@(*(A,B),2)
C.@(/(A,B),2)
D.None of these
Answer: A./(@(A,B),2)
Explanation: Explanation :
Assume some values for A and B and substitute in the options to get the answer.
Here, Let A = 1 and B = 2.
So, /(@(A,B),2) = /(@(1,2),2)
= /(1.5,2) = 3
= Sum of A and B.
Answer : A
Question: Two men A and B start from place X walking at 4 ½ kmph and 5 ¾ kmph respectively. How many km apart they are at the end of 7 ½ hours if they are walking in the same direction?
A.7 km
B.6 km
C.9 3/8 km
D.9 km
Answer: C.9 3/8 km
Explanation: RS = 5 ¾ - 4 ½ = 1 ¼
T = 7 ½ h.
D = 5/4 * 15/2 = 75/8
= 9 3/8 km
Answer:C
Question: 21 ball numbered 1 to 21. A ballis drawn and then another ball is drawn without replacement.
A.2/23
B.3/25
C.9/42
D.1/21
Answer: C.9/42
Explanation: 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
Question: If a jogger jogs 16 km/hr instead of 12 km/hr, he would have jogged 10 km more. The actual distance he jogged is
A.50 km
B.40 km
C.30 km
D.20 km
Answer: C.30 km
Explanation: Explanation :
The actual distance jogged be d.
=> d/12 = (d+10)/16
=> 16d = 12d+120
=> 4d = 120
=> d = 30
Answer : C
Question: Printer P can print one lakh books in 8 hours. Printer Q can print the same number of books in 10 hours while Printer R can print the same in 12 hours. All the Printers started printing at 9 A.M. Printer P is stopped at 11 A.M. and the remaining 2printers complete work. Approximately at what time will the printing of 1 lac books be completed?
A.1 pm
B.2 pm
C.3 pm
D.4 pm
Answer: A.1 pm
Explanation: Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the Printer were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm
A
Question: Of the three- digit integers greater than 700, how many have two digits T that are equal to each other and the remaining digit different from the other two?
A.90
B.82
C.80
D.45
Answer: C.80
Explanation: three-digit integers greater than 700: 701 to 999, inclusive.
possible values for hundreds-digit--> 7,8,9
possible values for tens-digit and ones-digit --> 0, 1,2,3,4,5,6,7,8,9
when hundreds-digit and tens-digit are the same: (3x1x10)-3=27 ---> we minus three to exclude 777, 888 and 999
when hundreds-digit and ones-digit are the same: (3x10x1)-3=27 ---> we minus three to exclude 777, 888 and 999
when tens-digit and hundreds-digit are the same:[(3x10x1)-3]-1=26 ---> we minus three to exclude 777, 888 and 999; we minus one to exclude 700
T=27+27+26 = 80
Answer: C
Question: Find the number of divisors of 1728.?
A.26
B.27
C.28
D.29
Answer: C.28
Explanation: 1728= 2^6 * 3^3
Hence the Number of factors = (6+1) x (3+1) = 7 x 4 = 28.
We know that if a number represented in standard form (a^m *b^n) , then the number of factors Is given by (m+1)(n+1).
Answer is 28
ANSWER:C
Question: Company C sells a line of 25 products with an average retail price of $1,200. If none of these products sells for less than $400, and exactly 10 of the products sell for less than $1,000, what is the greatest possible selling price of the most expensive product?
A.12000
B.13000
C.14000
D.15000
Answer: A.12000
Explanation: The average price of 25 products is $1,200 means that the total price of 25 products is 25*1,200=$30,000.
Next, since exactly 10 of the products sell for less than $1,000, then let's make these 10 items to be at $400 each (min possible).
Now, the remaining 14 items cannot be priced less than $1,000, thus the minimum possible price of each of these 14 items is $1,000.
Thus the minimum possible value of 24 products is 10*400+14*1,000=$18,000.
Therefore, the greatest possible selling price of the most expensive product is $30,000-$18,000=$12,000.
Answer: A.
Question: The edge of three cubes of metal is 3 dm, 4 dm and 5 dm. They are melted and formed into a single cube. Find the edge of the new cube?
A.7
B.5
C.8
D.6
Answer: D.6
Explanation: Explanation:
33 + 43 + 53 = a3 => a = 6
Answer: Option D
Question: A teacher prepares a test. She gives 8 objective type questions out of which 4 have to be answered. Find the total ways in which they can be answered if the first 5 questions have 3 choices and the last 3 have 4 choices.
A.6255
B.5280
C.5192
D.5100
Answer: B.5280
Explanation: Two cases.
First Case: 2 Questions - 3 Choices, 2 Questions - 4 Choices
5C2 * 3C2 ways to select the questions
3C1 * 3C1 * 4C1 * 4C1 ways to answer the questions
= 10 * 3 * 3 * 3 * 4 * 4
= 4320
Second Case: 1 Question - 3 Choices, 3 Questions - 4 Choices
5C1 * 3C3 ways to select the questions
3C1 * 4C1 * 4C1 * 4C1 ways to answer the questions
= 5 * 1 * 3 * 4 * 4 * 4
= 960
Total = 4320 + 960
=5280
B
Question: What is the units digit of the product of the first 15 odd numbers?
A.0
B.5
C.1
D.2
Answer: B.5
Explanation: 1*3*5*7 ................... will end up in 5 in the units place
Answer : B
Question: If s,u, and v are positive integers and s=2u+2v, which of the following must be true?
i. s=u
ii. u is not equal to v
iii. s > v
A.None
B.I only
C.II only
D.III only
Answer: A.None
Explanation: Notice two things: 1. we are asked to find out which of the following MUST be true, not COULD be true and 2. s, u, and v are positive integers.
Given: s=2u+2v --> s/2=u+v. Now, since s, u, and v are positive integers then s is more than either u or v, so I is never true and III is always true. As for II: it's may be necessarily true,
Answer: A
Question: If n divided by 9 has a remainder of 8, what is the remainder when 2 times n is divided by 9?
A.1
B.7
C.3
D.5
Answer: B.7
Explanation: As per question => N=9P+8 for some integer p
hence 2N => 18Q + 16
But again, 16 can be divided by 9 to get remainder 7 for some integer Q
hence B
Question: On a certain scale of intensity, each increment of 20 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125?
A.40
B.100
C.400
D.1000
Answer: B.100
Explanation: Increase of 40 in magnitude corresponds to 10^4 increase in intensity:
If intensity for 125 is x then for 135 it'll be 10*x, for 145 it'll be 10*10*x=10^2*x, for 155 it'll be 10*10*10*x=10^3*x and for 165 it'll be 10*10*10*10*x=400*x.
Answer: B.
Question: The present value of a bill due at the end of 2 years is Rs.1250. If the bill were due at the end of 2 years and 11 months, its present worth would be Rs.1200. Find the rate of interest and the sum.
A.rs.1175
B.rs.1375
C.rs.1475
D.rs.1575
Answer: B.rs.1375
Explanation: Explanation:
PW=100xAmount/100+(Rx T)
Answer: B
Question: Average of 13 results is 65. If the average of first six results is 61 and average of last six results is 59 Then find the seventh result?
A.120
B.123
C.125
D.128
Answer: C.125
Explanation: 125
Option 'C'
Question: Last year’s receipts from the sale of greeting cards during the week before Mother’s Day totaled $195 million, which represented 10 percent of total greeting card sales for the year. Total greeting card sales for the year totaled how many million dollars?
A.17,010
B.2,100
C.1,890
D.1,950
Answer: D.1,950
Explanation: 10% ---- 195 millions
for 100% => ( 195 * 100% )/10% = 1950.
Option D.
Question: A man can row downstream at 20 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively?
A.13, 5
B.15, 2
C.15, 5
D.14, 5
Answer: C.15, 5
Explanation: Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 20--- (1)
and x - y = 10 --- (2)
From (1) & (2) 2x = 30 => x = 15, y = 5.
Answer:C
Question: A 24 month project had a total budget of $60,000. After eleven months, the project had spent $26,300. At this point, how much was the project under budget?
A.$800
B.$1000
C.$1200
D.$1400
Answer: C.$1200
Explanation: Each month, the project should spend $60,000 / 24 = $2500.
In 11 months, the project should spend 11*$2500=$27,500.
The project is under budget by $27,500 - $26,300 = $1200.
The answer is C.
Question: Two numbers are in the ratio 3:5. If 9 be subtracted from each, they are in the ratio of 2:5. The first number is:
A.A)33
B.B)98
C.C)34
D.D)35
Answer: A.A)33
Explanation: (3x-9):(5x-9) = 2:5
x = 11 => 3x = 33
Answer:A
Question: A person spends 40% of his salary on food, 25% on house rent, 15% on entertainment and 15% on conveyance. If his savings at the end of the month is Rs. 1200, then his salary per month in rupees is:
A.4000
B.24000
C.8000
D.10000
Answer: B.24000
Explanation: Total expenditure = 40 + 25 + 15 + 15 = 95%
Saving = (100 - 95) = 5%
5/100 × Salary = 1200, Salary = 24000 Rs.
Answer:B