RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: There are two sections Y and Z of a class, consisting of 288 and 352 students respectively. If the average weight of sections Y is 320 kg and that of sections Z is 280 kg. Find the average weight of the whole class
A.297.5
B.298.0
C.298.5
D.299.0
Answer: B.298.0
Explanation: Solution
Total weight of(288+352)Students	= (288x320+352x280)Kg
= 1,90,720 kg.
Average weight of the whole class	= (190720 / 640)
= 298.00.
Answer B
Question: Of the 100 athletes at a soccer club, 40 play defense and 68 play midfield. If at least 20 of the athletes play neither midfield nor defense, the number of athletes that play both midfield and defense could be any number between
A.10 to 20
B.10 to 40
C.28 to 40
D.30 to 70
Answer: C.28 to 40
Explanation: First of all notice that since only 40 athletes play defense, then the number of athletes that play both midfield and defense cannot possibly be greater than 40.Eliminate D and E.
{Total} = {defense} + {midfield} - {both} + {neither}
100 = 40 + 68 - {both} + {neither}
{both} = {neither} + 8.
Since the least value of {neither} is given to be 20, then the least value of {both} is 20+8=28.Eliminate A and B.
Answer: C.
Question: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 4 students left over. What is the sum of the two smallest possible values of n?
A.30
B.34
C.38
D.42
Answer: C.38
Explanation: n = 4k+1 = 5j+4
Let's start at 1 = 4(0)+1 and keep adding 4 until we find a number in the form 5j+4.
1, 5, 9 = 5(1)+ 4
The next such number is 9 + 4*5 = 29.
9+29 = 38
The answer is C.
Question: Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H
A.153
B.150
C.137
D.120
Answer: D.120
Explanation: let the number be x, now x is halved => x/2..the question asks what is x?
x(number) + x/2 (half of that number) = something
=> 3x/2 = something
=> x = 2 (something) / 3
so the answer choice must be divisible by 3
eliminate C, E right away
check the rest
D satisfies
3x/ 2 = 120
=> x = 86
Verify. 80 + 40= 120
Answer is D
Question: In how many seconds will a train 150 meters long pass an oak tree, if the speed of the train is 54 km/hr?
A.6
B.8
C.10
D.12
Answer: C.10
Explanation: speed = 54 * 5/18 = 15 m/s
time = 150/15 = 10 seconds
The answer is C.
Question: A committee has 5 men and 4 women. What are the number of ways of selecting 2 men and 3 women from the given committee?
A.127
B.40
C.882
D.272
Answer: B.40
Explanation: The number of ways to select two men and three women
= ⁵C₂ * 4C₃
= (5 *4 )/(2 * 1) * (4 * 3 * 2)/(3 * 2)
= 40
Answer: B
Question: What do you get if you add 9 to 311 four times?
A.305
B.310
C.327
D.320
Answer: D.320
Explanation: D
320 , 320 , 320 , 320
Question: Mary's income is 60% more Than Tim's income and Tim's income is 50% less than Juan's income.What % of Juan's income is Mary's income.
A.124%
B.B.120%
C.96%
D.80%
Answer: D.80%
Explanation: Even I got 96%
J = 100
T = 100*0.5 = 50
M = 50*1.6 = 80
If Mary's income is x percent of J
M = J*x/100
x = M*100/J = 80*100/100 = 80
Ans:D
Question: What is the value of x^2yz − xyz^2, if x = 2, y = 1, and z =2?
A. 20
B. 24
C. 0
D. 32
Answer: C. 0
Explanation: 4*1*2-(2*1*4)
=8+8 = 0
Ans: C
Question: Find the missing figures: 0.25% of ? = 0.04
A.16
B.20
C.17
D.12
Answer: A.16
Explanation: Let 0.25% of x = 0.04. Then , 0.25*x/100 = 0.04
X= [(0.04*100)/0.25] = 16.
Answer is A.
Question: The maximum number of girls among them 2923 bags and 3239 eyeliners can be distributed in such a way that each student gets the same number of bags and eyeliners.
A.80
B.79
C.78
D.81
Answer: B.79
Explanation: Max no of girls=HCF of 2923 and 3239= 79
ANSWER:B
Question: Which of the following functions does not intersect with Q = 3x^2 + 2x + 1
A.Q= 3x^2 + 3x + 1
B.Q = 2x^2 + 3x + 1
C.Q = 3x^2 + 2x + 3
D.Q = 4x^2 + 2x - 3
Answer: C.Q = 3x^2 + 2x + 3
Explanation: Let’s find the actual intersecting point of given function and each choice Q
A. 3x^2 + 2x + 1 = 3x^2 + 3x + 1 --> x=0
B. 3x^2 + 2x + 1 = 2x^2 + 3x + 1 --> x^2 – x =0 --> x=0 or 1
C. 3x^2 + 2x + 1 = 3x^2 + 2x + 3 --> 1=3(X) --> they don’t have any intersect.
D. 3x^2 + 2x + 1 = 4x^2 + 2x - 3 --> x^2 –4=0 --> x= 2 or -2
E. 3x^2 + 2x + 1 = x^2 + 2x + 3 --> 2x^2 –2 =0 --> x= 1 or –1.
So only choice (C) has no intersecting point. The answer is (C).
Question: A bike covers a certain distance at the speed of 64km/h in 8 hrs. If the bike was to cover the same distance in approximately 6 hrs, at what approximate speed should the bike travel?
A.80 km/h
B.85.34 km/h
C.90.14 km/h
D.91.24 km/h
Answer: B.85.34 km/h
Explanation: Ans.(B)
Sol. Total distance = 64 × 8 = 512 km
Now speed = 512/6 = 85.34 km/h
Question: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. Four pentagons are considered different only when the positions of the side lengths are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lengths ?
A. 5
B. 12
C. 24
D. 32
Answer: A. 5
Explanation: CONCEPT: The number of ways to arrange n elements around a circular Manner is (n-1)!.
But the number of ways to arrange n elements in circular manner where the arrangement can be flipped around (e.g. Necklace, Garland, Bracelet) = (n-1)!/2.
The pentagon in the problem above can be flipped over (Clockwise Arrangement of Sides = Counter-clockwise arrangement), the number of distinct ways to arrange the 5 sides = (5-1)!/2 = 5.
Answer: option A
Question: 4 men or 6 women can do a piece of work in 20 days. In how many days will 12 men and 8 women do the same work?
A.12/9 days
B.60/13 days
C.82/8 days
D.22/76 days
Answer: B.60/13 days
Explanation: 4M = 6W ---- 20 days
12M + 8W -----?
18W + 8 W = 26W ---?
6W ---- 20 26 -----?
6 * 20 = 26 * x => x = 60/13 days
Answer:B
Question: A shopkeeper sells two articles at Rs.1000 each, making a profit of 20% on the first article and a loss of 20% on the second article. Find the net profit or loss that he makes?
A.3%
B.8%
C.9%
D.4%
Answer: D.4%
Explanation: SP of first article = 1000
Profit = 20%
CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3
SP of second article = 1000
Loss = 20%
CP = (SP)*[100/(100-L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CP-SP = (6250/3)- 2000 = 250/3
Loss Percent = [(250/3)/(6250/3)]*100
=0.04 * 100
= 4%
Answer: D
Question: A thief is spotted by a policeman from a distance of 200 meters. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8km/hr and that of the policeman 10 km/hr, how far the thief will have run before he is overtaken?
A.350m
B.200m
C.400m
D.800m
Answer: D.800m
Explanation: Relative speed of the policeman = (10-8) km/hr =2 km/hr.
Time taken by police man to cover
(200m/1000) x 1/2 hr = 1/10 hr.
In 1/10 hrs, the thief covers a distance of 8 x 1/10 km = 4/5 km = 800 m
Answer is D.
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 3 ½ hours if they are walking in the same direction?
A.4 3/6
B.4 3/9
C.4 3/8
D.4 3/4
Answer: C.4 3/8
Explanation: RS = 5 3/4 - 4 1/2 = 1 1/4
T = 3 1/2 h.
D = 5/4 * 7/2 = 35/8 = 4 3/8 km
Answer: C
Question: The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?
A.8 : 9
B.21 : 22
C.17 : 18
D.12 : 20
Answer: B.21 : 22
Explanation: Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
(120/100 x 7x)	and	(110/100 x 8x)
42x/5 and 44x/5
The required ratio = (42x/5 : 44x/5)= 21 : 22.
ANSWER B
Question: A, B and C started a business with capitals of Rs. 8000, Rs. 10000 and Rs. 12000 respectively. At the end of the year, the profit share of B is Rs. 1500. The difference between the profit shares of A and C is?
A.277
B.266
C.299
D.600
Answer: D.600
Explanation: Ratio of investments of A, B and C is 8000 : 10000 : 12000 = 4 : 5 : 6
And also given that, profit share of B is Rs. 1500
=> 5 parts out of 15 parts is Rs. 1500
Now, required difference is 6 - 4 = 2 parts
Required difference = 2/5 (1500) = Rs. 600
Answer: D
Question: What is the perimeter of a square with area 4p^2/16 ?
A.3p/4
B.3p^2/4
C.2p
D.3p^2
Answer: C.2p
Explanation: Area of square, (side)^2 = (2p/4)^2
Therefore side of the square = 2p/4
Perimeter of square = 4*side = 4* (2p/4) = 2p
Answer is C.
Question: In a store, the profit is 320% of the cost. If the cost increases by 25% but the SP remains constant, approximatelywhat %of the sp is the profit?
A.50%
B.70%
C.80%
D.90%
Answer: B.70%
Explanation: 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
Question: If Rs.7500 are borrowed at C.I at the rate of 4% per annum, then after 2 years the amount to be paid is?
A.2868
B.1718
C.1689
D.8112
Answer: D.8112
Explanation: A = 7500(26/25)2
= 8112
Answer:D
Question: The cost of 20 pens and 12 pencils is Rs.400 and the cost of 6 pens and 4 pencils is Rs.100. Find the cost of each pen?
A.Rs.50
B.Rs.100
C.Rs.150
D.Rs.25
Answer: A.Rs.50
Explanation: 20p + 12q = 400 --- (1)
6p + 4q = 100
8p + 8q = 192 --- (2)
(1) - (2) => 8p = 160
=> p = 50
ANSWER:A
Question: In a shop, 40% socks are white and rest are black. 60% socks are made of cotton and rest are made of wool. 25% white socks are made of cotton and 20 of the black socks are made of wool. How many black socks X are made of cotton?
A.100
B.80
C.60
D.50
Answer: A.100
Explanation: I'll prefer to solve this with plug n play method.
As we know, White Cotton socks are only 25% and Total Cotton socks is 60% - Hence Black Cotton socks X has to be [highlight]HIGH[/highlight]number. Thats why we've to pick high number... say 100 or 80.
100 fits perfectly after relevant calculations. Hence answer A.
Question: For which of the following functions f is f(b) = f(1-b) for all b?
A.f (x) = 1 - x
B.f (b) = 1 - b^2
C.f (b) = b^2 - (1 - b)^2
D.f (b) = b^2*(1 - b)^2
Answer: D.f (b) = b^2*(1 - b)^2
Explanation: I remember seeing this problem in a GMATprep test. I think choice D should be F(b)=(b^2) (1-b)^2. If that is true then substituting (1-b) for will give us the same function back and the answer choice is D, where the function is multiplicative
Question: The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs. 8748, its purchase price was:
A.10000
B.12000
C.14000
D.16000
Answer: B.12000
Explanation: Explanation:
Purchase Price =Rs[8748 / (1- 10/100)3]=Rs.8748 x 10/9 x10/9 x10/9 = Rs.12000
ANSWER IS B
Question: The average of 7 numbers is 20. If each number be multiplied by 5. Find the average of new set of numbers?
A.A)100
B.B)122
C.C)120
D.D)125
Answer: A.A)100
Explanation: Explanation:
Average of new numbers = 20 * 5 = 100
Answer: Option A
Question: A retailer buys 100 pens at the market price of 36 pens from a wholesaler, if he sells these pens giving a discount of 1%, what is the profit % ?
A.110
B.175
C.212
D.218
Answer: B.175
Explanation: let the market price of each pen be $1
then, Cost Price of 100 pens = $ 36 Selling Price of 100 pens =99% of $100=$ 99
profit %=((63*100)/36) %=175%
Answer B
Question: A searchlight on top of the watchtower makes 2 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 10 seconds?
A.5/6
B.3/4
C.2/3
D.1/2
Answer: C.2/3
Explanation: The searchlight completes one revolution every 30 seconds.
The probability that the man's area will be lit up is 10/30=1/3.
The probability that he will stay in the dark is 1-1/3=2/3
The answer is C.
Question: The units digit of (35)^(87) + (3)^(45) is:
A.2
B.8
C.6
D.9
Answer: B.8
Explanation: any power of anything ending in 5 always has a units digit of 5. So the first term has a units digit of 5.
2nd part
3^45 has a units digit of 3
Therefore, the second term has a units digit of 3.
Of course 5 + 3 = 8, so something with a units digit of 5 plus something with a units digit of 3
Answer = B
Question: The area of a square field 3136 sq m, if the length of cost of drawing barbed wire 3 m around the field at the rate of Rs.1.50 per meter. Two gates of 1 m width each are to be left for entrance. What is the total cost?
A.299
B.277
C.999
D.255
Answer: C.999
Explanation: a2 = 3136 => a = 56
56 * 4 * 3 = 672 – 6 = 666 * 1.5
= 999
Answer: C
Question: A theater charges $12 for seats in the orchestra and $8 for seats in the balcony. On a certain night, a total of 370 tickets were sold for a total cost of $3,320. How many more tickets were sold that night for seats in the balcony than for seats in the orchestra?
A. 90
B. 190
C. 120
D. 130
Answer: B. 190
Explanation: Orchestra Seats - a
Balcony Seats - b
a+b = 370
and 12a + 8b = 3320
Solving equations simultaneously (Multiply Equation 1 with 8 and subtract from second equation)
4a = 3320 - 8*370 = 3320 - 2960 = 360
i.e. a = 90
and b = 370-90 = 280
More seats in balcony than orchestra = b-a = 280 - 90 = 190
Answer: option B
Question: A boat running downstream covers a distance of 16 km in 2 hours while for covering the same distance upstream,it takes 4 hour.What is the speed of the boat in still water ?
A.4 km/hr
B.6 km/hr
C.8 km/hr
D.Data inadequate
Answer: B.6 km/hr
Explanation: Solution
Rate downstream =(16/2) kmph = 8 kmph,Rate upstream = (16/4) kmph = 4 kmph.
∴ Speed in still water =1/2(8+4) kmph = 6 kmph. Answer B
Question: An article is bought for Rs.600 and sold for Rs.500, find the loss percent?
A.16 6/3%
B.16 2/8%
C.16 3/3%
D.16 2/3%
Answer: D.16 2/3%
Explanation: 600 ---- 100
100 ---- ? => 16 2/3%
Answer: D
Question: A cube of side one meter length is cut into small cubes of side 10 cm each. How many such small cubes can be obtained?
A.1017
B.1029
C.1000
D.1028
Answer: C.1000
Explanation: Along one edge, the number of small cubes that can be cut
= 100/10 = 10
Along each edge 10 cubes can be cut. (Along length, breadth and height). Total number of small cubes that can be cut = 10 * 10 * 10 = 1000
Answer: C
Question: Beginning in Town A, Biker Bob rides his bike 12 miles west, 6 miles north, 4 miles east, and then 12 miles north, to Town B. How far apart are Town A and Town B? (Ignore the curvature of the earth.)
A.18.97 miles
B.20 miles
C.22 miles
D.23 miles
Answer: A.18.97 miles
Explanation: Using Pythagoras we have one side i,e total distance traveled in north direction = 12+6=18m
other being the base ie distance traveled west- distance traveled eat=12- 6= 6 m
now this third side or the distance between town A and Town B=18^2+ 6^2=sq root 360=18.97 mile
Answer : A
Question: The height of a triangle is decreased by 15% and its base increased by 50%. Its area will
A.decrease by 12.5%
B.increase by 12.5%
C.decrease by 25%
D.Increase by 27.5%
Answer: D.Increase by 27.5%
Explanation: Explanation :
Let the initial height be 100 and the base be 200 units.
âˆ´ Area = 1/2 x 100 x 200 = 10000 sq units
Now, height = 85 units and base = 300 units
âˆ´ New area = 1/2 x 85 x 300 = 12750
âˆ´ Area is increased by (12750 - 10000)/10000 x 100
= 27.5%
Answer : Option D
Question: A cubical block of metal weighs 6 pounds. How much will another cube of the same metal weigh if its sides are twice as long?
A.48
B.99
C.266
D.28
Answer: A.48
Explanation: Explanation:
If you double the sides of a cube, the ratio of the surface areas of the old and new cubes will be 1: 4. The ratio of the volumes of the old and new cubes will be 1: 8.
Weight is proportional to volume. So, If the first weighs 6 pounds, the second weighs 6x8 pounds =48.
Answer: A) 48
Question: Which of the following Equation is NOT equivalent to 25x^2 = y^2 - 25?Which of the following Equation is NOT equivalent to 25x^2 = y^2 - 4?
A.25x^2 + 4 = y^2
B.75x^2 = 3y^2 - 12
C.5x = y - 5
D.25x^2 = (y + 2)(y - 2)
Answer: C.5x = y - 5
Explanation: 1) Take the original equation and try to manipulate it so that it becomes 4 of the 5 answers
2) Take each answer choice and see if you can manipulate it to make it into the original equation.
With Answer D, we have...
5X = Y - 5
If we square both sides, we end up with....
(5X)^2 = (Y-5)^2
(5X)(5X) = (Y-5)(Y-5)
25X^2 = Y^2 - 10Y + 25
This is NOT a match for the original equation, so it's the answer that is NOT equivalent. C
Question: In how many different ways can pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents) be combined for a total of $1.10 (110 cents), if at least one of each type of coin must be included?
A.36
B.51
C.70
D.87
Answer: D.87
Explanation: I. Case I = Four Quarters = this is impossible, because if we already have a dollar in quarters, then we either have to have one dime or nickels & pennies. We can't have all four coins represented if we have four quarters.
II. Case II = Three Quarters (75 cents)
This allows for two or one dime. Three dimes would bring us up to $1.05, and we wouldn't have room for both pennies & nickels.
Subcase 1 = 3Q, 2 D (95 cents)
Here, we could have have
1) two nickles
2) one nickle
(Notice that once the number of Q & D & N are determined, the number of pennies is fixed and doesn't have to be considered.
Subcase 2 = 3Q, 1 D (85 cents)
Here we could have
3) 4 nickles
4) 3 nickles
5) 2 nickles
6) 1 nickle
Case II allows for a total of six cases.
III. Case III = Two Quarters (50 cents)
This allows for 1-5 dimes.
Subcase 1 = 2Q, 5 D = we could have 1 nickle (1 case)
Subcase 2 = 2Q, 4 D = we could have 1-3 nickles (3 cases)
Subcase 3 = 2Q, 3 D = we could have 1-5 nickles (5 cases)
Subcase 4 = 2Q, 2 D = we could have 1-7 nickles (7 cases)
Subcase 5 = 2Q, 1 D = we could have 1-9 nickles (9 cases)
Case III allows for a total of 25 cases
IV. Case IV = One Quarter (25 cents)
This allows for 1-7 dimes
Subcase 1 = 1 Q, 7D = we could have 1-2 nickles (2 cases)
Subcase 2 = 1 Q, 6D = we could have 1-4 nickles (4 cases)
Subcase 3 = 1 Q, 5D = we could have 1-6 nickles (6 cases)
Subcase 4 = 1 Q, 4D = we could have 1-8 nickles (8 cases)
Subcase 5 = 1 Q, 3D = we could have 1-10 nickles (10 cases)
Subcase 6 = 1 Q, 2D = we could have 1-12 nickles (12 cases)
Subcase 7 = 1 Q, 1D = we could have 1-14 nickles (14 cases)
Case IV allows for a total of 56 cases.
There's no other case, because we have to have at least one quarter. The total over the cases equals
Total = 6 + 25 + 56 = 87
ANSWER:D
Question: What comes next?
120, 60, 40, 30, 24, ?
A.10
B.20
C.25
D.28
Answer: B.20
Explanation: Factors of 120 are arranging in descending order.
So, next would be 20
Final series is like : 120, 60, 40, 30, 24, 20
ANSWER:B
Question: In a friendship gang Andrew has 8 gang, in how many ways can he invite one or more of the gang to his house?
A.230
B.255
C.259
D.265
Answer: B.255
Explanation: Andrew can select one or more than one of his 8 gang.
=> Required number of ways = 28 – 1= 255.
B
Question: 2598 Successive discounts of 20% and 15% are equal to a single discount of
A.30%
B.32%
C.34%
D.35%
Answer: B.32%
Explanation: Successive discounts of 20% and 15% are equal to a single discount of 32 %.
As final value after discount is 0.8*0.85 = 0.68 of original value.
so discount is 0.32 of original value. means 32%.
ANSWER:B
Question: Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both BG, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations R are there for the children?
A.240
B.480
C.720
D.1440
Answer: A.240
Explanation: MAGOOSHOFFICIAL SOLUTION:
First, we will consider the restricted elements — children ABG have to be in three seats in a row. How many “three in a row” seats are there in a row of seven seats?
X X X _ _ _ _
_ X X X _ _ _
_ _ X X X _ _
_ _ _ X X X _
_ _ _ _ X X X
There are five different “three in a row” locations for these three children. Now, for any given triplet of seats, we know A has to be in the middle, so the children could be seated B-A-G or G-A-B — just those two orders. This means the total number of configurations for these three children is 5*2 = 10.
Now, consider the non-restricted elements, the other four. Once ABG are seated, the remaining four children can be seated in any order among the remaining four seats — that’s a permutation of the 4 items —- 4P4 = 4! = 24. For any single configuration of ABG, there are 24 ways that the other children could be seated in the remaining seats.
Finally, we’ll combine with the Fundamental Counting Principle. We have 10 ways for the first three, and 24 ways for the remaining four. That’s a total number of configurations R of 24*10 = 240.
Answer = A
Question: Length of a rectangular plot is 20 mtr more than its breadth. If the cost of fencin gthe plot at 26.50 per meter is Rs. 5300, what is the length of the plot in mtr?
A.46 m
B.60 m
C.58 m
D.78 m
Answer: B.60 m
Explanation: Let breadth = x metres.
Then, length = (x + 20) metres.
Perimeter = 5300 m = 200 m.
26.50
2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m
B
Question: When running a mile during a recent track meet, Nuria was initially credited with a final time of 4 minutes, 44 seconds. Shortly after her run, officials realized that the timing mechanism malfunctioned. The stopwatch did not begin timing her until 11/25 of a minute after she began to run. If the time was otherwise correct, how long did it actually take Nuria to run the mile?
A.5 minutes, 17.6 seconds
B.5 minutes, 21.8 seconds
C.5 minutes, 43.56 seconds
D.5 minutes, 10.4 seconds
Answer: D.5 minutes, 10.4 seconds
Explanation: One approach:
The watch starts to work after Nuria began his running. It means the time should be greater than credited 4 minutes, 44 seconds. The only number is 5 minutes, 10.4 seconds.
Another approach:
11/25 close to 30 second when added to the 4 minutes, 44 seconds, it means it passes 5 minute.
Answer: D
Question: What number has a 3:4 ratio to the number 400?
A.130
B.100
C.200
D.300
Answer: D.300
Explanation: 3:4 = x: 200
4x = 3*400
x=300
ANSWER:D
Question: My brother is 4 years elder to me. My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born. If my sister was 5 years of age of when my brother was born, then, what was the age of my father and mother respectively when my brother was born?
A.32 yrs, 23 yrs
B.32 yrs, 29 yrs
C.35 yrs, 29 yrs
D.35 yrs, 22 yrs
Answer: D.35 yrs, 22 yrs
Explanation: Sol.
Clearly, my brother was born 5 years before I was born and 4 years after my sister was born.
So, father's age when brother was born = (28 + 4) = 32 years.
mother's age when was born = (26 - 4) years = 22 years.
Answer D
Question: AMOUNT OF BACTERIA PRESENT
Time Amount
1:00 P.M. 12.0 grams
4:00 P.M. x grams
7:00 P.M. 16.4 grams
Data for a certain biology experiment are given in the table above. If the amount of bacteria present increased by the same fraction during each of the two 3-hour periods shown, how many grams of bacteria were present at 4:00 P.M.?
A.12.0
B.14.0
C.14.2
D.12.3
Answer: B.14.0
Explanation: The question says that bacteria increased by same fraction, not by same amount in 2 intervals of 3 hours.
Let X represent the amount of bacteria present at 4:00 PM. Since the fractional increase must remain constant from 1 to 4pm as it is from 4pm to 7pm:
Fractional increase from 1 PM to 4 PM = X / 12.0
Fractional increase from 4 PM to 7 PM = 16.4 / X
X \ 12 = 16.4 \ X
X^2 = 16.4 * 12
X = 14
answer : B
Question: Two trains travelling in the same direction at 50.8 and 14.8 kmph completely pass off another in 45 sec.. If the length of the first train is 112 m, what is the length of the second train?
A.338 m
B.287 m
C.450 m
D.383 m
Answer: A.338 m
Explanation: RS = 50.8 â€“ 14.8= 36* 5/18 =10 mps
T = 45 sec
D = 10* 45= 450 m
112
--------
338 m
Answer: A