mathewhe/aqua_rat huggingface dataset:
Question: A bowl contains equal numbers of red, orange, green, blue, and yellow candies. Kaz eats all of the green candies and half of the orange ones. Next, he eats half of the remaining pieces of each color. Finally, he eats red and yellow candies in equal proportions until the total number of remaining candies of all colors equals 30% of the original number. What percent of the red candies remain?
(A) 22.5%
(B) 31.5%
(C) 37.5%
(D) 42.5%
(E) 46.5%
Rationale: Let x be the original number of each color so there are a total of 5x candies.
Kaz eats all of the green candies and half of the orange ones.
There are 0 green candies and 0.5x orange candies remaining.
He eats half of the remaining pieces of each color.
There are 0.25x orange candies, and 0.5x each of red, yellow, and blue candies.
He eats red and yellow candies in equal proportions.
orange+blue+red+yellow = 0.75x + red + yellow = 1.5x
red + yellow = 0.75x
red = 0.375x, since red = yellow.
The answer is C.
Answer: (C) 37.5%
Question: Consider two postmen A and B respectively. A is young and can deliver 20 parcels in 3 hours while B is older than A and can deliver only 15 parcels in 4 hours. If the total number of parcels to deliver is 60, how long they will take working together.
(A) 144/22
(B) 144/25
(C) 144/21
(D) 144/221
(E) 144/219
Rationale: Work done by 1st in 1 hour = 20/3 parcels / hour , Work done by 2nd om 1 hour = 15/4 parcels / hour
Total work done by both together per hour = 20/3 + 15/4 = 125/12 parcels/hour
Time to do 60 unit work (ie, parcels)= 60 ÷ 125/12 = 60 × 12/125 = 144/25 hours.
Answer:B
Answer: (B) 144/25
Question: In what time will a train 175 m long cross an electric pole, it its speed be 180 km/hr?
(A) 7.5
(B) 6.5
(C) 5.5
(D) 4.5
(E) 3.5
Rationale: Speed = 180 * 5/18 = 50 m/sec
Time taken = 175/50 = 3.5 sec.
Answer: E
Answer: (E) 3.5
Question: Twenty five percent of Country Y's yearly exports come from fruit exports. One-fifth of all fruit exports from Country Y are orange exports. If country Y generates $10 million from its orange exports, how much money does it generate from its yearly total of exports?
(A) $2125m
(B) $25m
(C) $200m
(D) $300m
(E) $153m
Rationale: 25/100*1/5*(total) = 10
1/20*(total) = 10
(total) = 10*20 = 200
Answer: C
Answer: (C) $200m
Question: On planet korba, a solar blast has melted the ice caps on its equator. 9 years after the ice melts, tiny planetoids called echina start growing on the rocks. Echina grows in the form of circle, and the relationship between the diameter of this circle and the age of echina is given by the formula d = 4*[sqrt(t-9)] for t>9, where d represents the diameter in mm and t the number of years since the solar blast. Jagan recorded the radius of some echina at a particular spot as 7mm. How many years back did the solar blast occur?
(A) 17
(B) 21.25
(C) 12.25
(D) 14.05
(E) 18
Rationale: radius=7mm.
So dia= 14mm.
putting D=14 in equation we can easily calculate
t=21.25
ANSWER:B
Answer: (B) 21.25
Question: Point (m,n) is on the circle represented by m^2+n^2=10, and m, n are integers. How many such points are possible?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
Rationale: m^2+n^2=10 and m, n are integers means that 10 is the sum of two perfect squares.
10 is the sum of only one pair of perfect squares 1 and 9.
So, there can be 8 such points, 4 in each quadrant:
(1, 3);
(1, -3);
(-1, 3);
(-1, -3);
(3, 1);
(3, -1);
(-3, 1);
(-3, -1).
Answer: E.
Answer: (E) 8
Question: In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is
(A) 50 liters
(B) 55 liters
(C) 60 liters
(D) 70 liters
(E) 75 liters
Rationale: Quantity of milk = 60 x 2 litres = 40 litres.
3
Quantity of water in it = (60- 40) litres = 20 litres.
New ratio = 1 : 2
Let quantity of water to be added further be x litres.
Then, milk : water = 40 .
20 + x
Now, 40 = 1
20 + x 2
20 + x = 80
x = 60.
Quantity of water to be added = 60 liters.
C
Answer: (C) 60 liters
Question: In a school 10% of the boys are same in number as 1/4th of the girls. what is the ratio of boys to the girls in the school?
(A) 5:2
(B) 2:3
(C) 1:4
(D) 3:7
(E) 2:5
Rationale: 10% of B = 1/4 G
10B/100 = G/4
B = 5G/2
B/G = 5/2
B:G = 5:2
Answer is A
Answer: (A) 5:2
Question: The length of the rectangular field is double its width. Inside the field there is square shaped pond 8m long. If the area of the pond is 1/8 of the area of the field. What is the length of the field?
(A) 31 m
(B) 32 m
(C) 16 m
(D) 20 m
(E) 22 m
Rationale: A/8 = 8 * 8 => A = 8 * 8 * 8
x * 2x = 8 * 8 * 8
x = 16 => 2x = 32
ANSWER:B
Answer: (B) 32 m
Question: Carol and Jordan draw rectangles of equal area. If Carol's rectangle measures 5 inches by 24 inches and Jordan's rectangle is 8 inches long, how wide is Jordan's rectangle, in inches?
(A) 25
(B) 23
(C) 22
(D) 15
(E) 18
Rationale: Area of Carol's rectangle = 24*5 = 120
Let width of Jordan's rectangle= w
Since , the areas are equal
8w = 120
=>w = 15
Answer D
Answer: (D) 15
Question: At a meeting of the 3 Joint Chiefs of Staff, the Chief of Naval Operations does not want to sit next to the Chief of the National Guard Bureau. How many ways can the 3 Chiefs of Staff be seated around a circular table?
(A) 120
(B) 480
(C) 960
(D) 2520
(E) 5040
Rationale: Bunuel, I'm also a little confused with the number of arrangements of n distinct objects in a circle. Why is it given by (n-1)!. In theveritasanswer they say:answer C (960), should be the number of ways to arrange all 7 without the seating restriction given. Is this incorrect?C
Answer: (C) 960
Question: If in a 20 over cricket match, there were no wide balls, no wides, no extras and no overthrows.
What is the maximum number of runs that a batsman can score in an ideal case scenario?
(A) 663
(B) 720
(C) 625
(D) 640
(E) None
Rationale: Solution:
663
Explanation:
For an ideal case, the batsman will hit a six on each ball. But if he hits six on the last ball of the over, the strike will change in the next over. Thus, the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over. Thus the total runs that he can score in each over:
6 * 5 + 3 = 33
But this will have to go like it only till the 19th over. In the last over, he can hit a six in the last ball as well as that will be the last ball of the match.
Thus runs for the last over will be 6 * 6 = 36.
Hence the maximum runs = 33 * 19 + 36 = 663
Answer A
Answer: (A) 663
Question: Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is
(A) 42
(B) 36
(C) 28
(D) 20
(E) 15
Rationale: Explanation:
Distance covered = 120+120 = 240 m
Time = 12 s
Let the speed of each train = v. Then relative speed = v + v = 2v
2v = distance/time = 240/12 = 20 m/s
Speed of each train = v = 20/2 = 10 m/s
= 10 × 36/10 km/hr = 36 km/hr.
Answer: B
Answer: (B) 36
Question: In a hostel of 12 boys and some girls, an average consumption of rice per month is 12 kg, while the Average consumption for boys is 16 kg per head and for girls 8 kg per head. The number of girls in The hostel?
(A) 12
(B) 13
(C) 14
(D) 15
(E) 16
Rationale: Explanation:
Number of girls — G
(12+G) × 12 = (12 × 16)+(8 × G)
144 + 12G = 192 + 8G
12G − 8G = 192 − 144
4G = 48
G = 12.
Answer A
Answer: (A) 12
Question: The speed of a boat in still water is 12 km/hr and the rate of current is 4 km/hr. The distance travelled downstream in 18 minutes is
(A) 1.6 km
(B) 2 km
(C) 3.6 km
(D) 4.8 km
(E) None of these
Rationale: Explanation:
Speed downstreams	=(12 + 4)kmph
= 16 kmph.
Distance travelled	= (16 x 18/60)km
= 4.8km
Option D
Answer: (D) 4.8 km
Question: How many digits are in the number 50^8 × 8^3 × 11^2 x 10^6?
(A) 22
(B) 20
(C) 19
(D) 18
(E) 17
Rationale: Whenever you are asked to find the number of digits, try to bring the number in multiples of 10. This way, we can wasily calculate the umber of 0's through the powers of 10
50^8 × 8^3 × 11^2 x 10^6 = (5^2*2)^8*2^9*11^2 x 10^6 = 5^16*2^17*11^2 x 10 ^6= 2*11^2*10^16 x 10^6= 242*10^22
Hence we would have 16 trailing 0's and the three digits from 242
Total digits = 3 + 22 = 25
Correct Option: E
Answer: (E) 17
Question: If dy > 0 and yz < 0, which of the following must be negative:
(A) dyz
(B) dy(z^2)
(C) d(y^2)z
(D) x(y^2)(z^2)
(E) (x^2)(y^2)(z^2)
Rationale: dy > 0 means that dy is positive.
yz < 0 means that yz is negative.
d(y^2)z = (dy)(yz) = positive*negative = negative. Thus option C is always negative.
Answer: (C) d(y^2)z
Question: If the cost price of 11 pens is equal to the selling price of 10 pens, the gain percent is:
(A) 10%
(B) 26%
(C) 50%
(D) 80%
(E) 34%
Rationale: Solution :
Let cost price of the price be Rs.1
10 pens C.P. = 10
11 pens S.P. = 11
Gain = 1/10*100 = 10 %
Answer : A
Answer: (A) 10%
Question: By walking at 3/4th of his usual speed, a man reaches office 20 minutes later than usual. What is his usual time?
(A) 40min
(B) 50min
(C) 65min
(D) 60min
(E) 70min
Rationale: let T be the usual time.
time spent=4T/3
therefore, 4T/3=T+20
4T=3T+60...therefor T=60min
ANSWER:D
Answer: (D) 60min
Question: Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
(A) 6400
(B) 7300
(C) 6438
(D) 6439
(E) 9473
Rationale: Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).
Then, x x 14 x 2	/100	+ (13900 - x) x 11 x 2	/100	= 3508
28x - 22x = 350800 - (13900 x 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.
Answer:A
Answer: (A) 6400
Question: The average of 6 no.'s is 3.95. The average of 2of them is 3.4, while the average of theother 2 is 3.85. What is the average of the remaining 2 no's?
(A) 3.5
(B) 3.9
(C) 4.2
(D) 4.3
(E) 4.6
Rationale: Sum of the remaining two numbers = (3.95 * 6) - [(3.4 * 2) + (3.85 * 2)]
= 23.70 - (6.8 + 7.7) = 23.70 - 14.5 = 9.20.
Required average = (9.2 / 2) = 4.6.
E
Answer: (E) 4.6
Question: The mean daily profit made by a shopkeeper in a month of 30 days was Rs. 350. If the mean profit for the first fifteen days was Rs. 285, then the mean profit for the last 15 days would be
(A) Rs. 200
(B) Rs. 350
(C) Rs. 275
(D) Rs. 415
(E) None of these
Rationale: Average would be : 350 = (285 + x)/2
On solving, x = 415.
Answer: D
Answer: (D) Rs. 415
Question: Rajeev,Lakshmi and Muthu enter into partnership. Rajeev invests some money at the beginning, Lakshmi invests double the amount after 6 months, and Muthu invests thrice the amount after 8 months. If the annual gain be Rs.4500. Lakshmi's share is?
(A) 1000
(B) 1500
(C) 500
(D) 750
(E) 1250
Rationale: x* 12 : 2x* 6: 3x* 4
1:1:1
1/3 * 4500 = 1500
Answer: B
Answer: (B) 1500
Question: There are 20 girls inhostel whose average age is decreased by 2 months, when one girl 18 yrs is replaced by a new boy.Find the age of new girl.
(A) 1 year
(B) 1 year 2 months
(C) 2 year 5 months
(D) 3 year
(E) None
Rationale: Total decrease = 20 * 2 = 40 month
= 3 years 4 month
E)None
Answer: (E) None
Question: A flagstaff 17.5 m high casts a shadow of length 40.25 m.The height of the building,which casts a shadow of length 28.75 m under similar conditions will be :
(A) 10 m
(B) 12.5 m
(C) 17.5 m
(D) 21.25 m
(E) None of these
Rationale: Solution
14.8
Let the height of the building be x meters.
Less lengthy shadow,Less is the height(Direct Proportion)
∴ 40.25 : 28.75 : : 17.5 : x ⇔ 40.25 × x = 28.75 × 17.5
⇔ x =(28.75x17.50)/ 40.25 ⇔ x = 12.5, Answer B
Answer: (B) 12.5 m
Question: The contents of a bag of candy consist of 11 green pieces of candy and 20 blue pieces of candy. How many blue pieces of candy must be eaten so that 55% of the candies in the bag will be green?
(A) 8
(B) 10
(C) 11
(D) 13
(E) 15
Rationale: This question is rooted in basic arithmetic. Since the question asks how many marbles need to be removed and the answers ARE numbers, we can TEST THE ANSWERS until we find the one thatfits.
To start, we have a bag of candy....
11 green
20 blue
31 total
We're asked to eat enough blue candies until the green candies make up 55% of the bag....
Right now, green marbles make up 11/31 which equals about 35% of the bag, we need to eat blue candy to achieve green candies making up for 55% of the bag. We can test our answers
Let's TEST B....
IF 10 blue candies were eaten, then we'd have...
11 green candies
10 blue candies
21 total
11/21 is about 52%.
Let's TEST E...
IF 11 blue candies were eaten, then we'd have...
11 green candies
9 blue candies
20 total
11/20 IS 55%, so this MUST be the answer!
Final Answer:
C
Answer: (C) 11
Question: John professes to sell his goods at the cost price but he made use of 800 grams instead of a kg, what is the gain percent?
(A) 12 1/3%
(B) 12 1/4%
(C) 12 2/3%
(D) 12 2/1%
(E) 12 1/2%
Rationale: 800 --- 100
100 --- ? => 12 1/2%
Answer:E
Answer: (E) 12 1/2%
Question: If a card is drawn from a well shuffled pack of cards, the probability of drawing a spade or a king is?
(A) 4/16
(B) 4/19
(C) 4/12
(D) 4/13
(E) 4/11
Rationale: P(SᴜK) = P(S) + P(K) - P(S∩K), where S denotes spade and K denotes king.
P(SᴜK) = 13/52 + 4/52 - 1/52 = 4/13
Answer:D
Answer: (D) 4/13
Question: A rope of which a calf is tied is increased from 12 m to 23 m, how much additional grassy ground shall it graze?
(A) 1298
(B) 1228
(C) 1210
(D) 1782
(E) 1762
Rationale: π (232 – 122)
= 1210
Answer:C
Answer: (C) 1210
Question: A cistern is filled by a tap in 3 1/2 hours. Due to leak in the bottom of the cistern, it takes half an hour longer to fill the cistern. If the cistern is full how long will it take the leak to empty it?
(A) 7 hours
(B) 8 hours
(C) 14 hours
(D) 28 hours
(E) 24 hours
Rationale: 2/7 - 1/x = 1/4
x = 28
ANSWER:D
Answer: (D) 28 hours
Question: Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen
(A) 45/78
(B) 15/48
(C) 36/91
(D) 451/75
(E) 55/221
Rationale: WE HAVE N(S)=52C2=(52*51)/(2*1)=1326.
LET A=EVENT OF GETTING BOTH BLACK CARDS
B=EVENT OF GETTING BOTH QUEENS
A∩B=EVENT OF GETTING QUEEN OF BLACK CARDS
N(A)=26C2=(26*25)/(2*1)=325,
N(B)=4C2=(4*3)/(2*1)=6 AND
N(A∩B)=2C2=1
P(A)=N(A)/N(S)=325/1326;
P(B)=N(B)/N(S)=6/1326 AND
P(A∩B)=N(A∩B)/N(S)=1/1326
P(A∩B)=P(A)+P(B)-P(A∩B)=(325+6-1/1326)=330/1326=55/221
Ans: E
Answer: (E) 55/221
Question: Eighteen years ago, a father was three times as old as his son. Now the father is only twice as old his son. Then the sum of the present ages of the son and the father is:
(A) 54
(B) 72
(C) 105
(D) 108
(E) 51
Rationale: Answer: Option D
Let the present ages of the father and son be 2x and x years respectively.
Then, (2x - 18) = 3(x - 18) => x = 36
Required sum = (2x + x) = 108 years.
Answer: (D) 108
Question: The speed of a boat in still water in 42 km/hr and the rate of current is 4 km/hr. The distance travelled downstream in 44 minutes is:
(A) 86.6 km
(B) 46.6 km
(C) 33.7 km
(D) 35.6 km
(E) 26.6 km
Rationale: Speed downstream = (42 + 4) = 46 kmph
Time = 44 minutes =44/60 hour = 11/15 hour
Distance travelled = Time × speed =11/15×46 = 33.7 km
ANSWER:C
Answer: (C) 33.7 km
Question: If the compound interest (compounded Yearly) on a certain sum for 2 years at 3% is Rs.101.50 then what will be the corresponding simple interest?
(A) Rs.98.25
(B) Rs.101.00
(C) Rs.90.00
(D) Rs.95.50
(E) Rs.75.50
Rationale: simple interest=50+50
compound interest=50+(50+50*0.3)=101.50
ANSWER:B
Answer: (B) Rs.101.00
Question: It takes 60 identical printing machines 4 hours to print 200,000 cards. How long would it take 50 of these machines to print the same total?
(A) 10 hr
(B) 5 hr
(C) 6 hr
(D) 4 hr 30 min
(E) 5 hr 12 min
Rationale: 50% of 60 = 40% or 0.30
4 hr x 60 min = 240 min
240 min x 0.30 = 72 min or 1 hr and 12 min
1 hr and 12 min + 4 hr = 5 hr and 12 min
Answer is E
Answer: (E) 5 hr 12 min
Question: For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is?
(A) Between 2 and 20
(B) Between 10 and 20
(C) Between 20 and 30
(D) Between 30 and 40
(E) Greater than 40
Rationale: h(n) = the product of all even integers from 2 to n inclusive
=> h(100) + 1 = (2 * 4 * 6 * 8 * ....... * 100) + 1
= 2^50 * (1*2*3*4*......*50) + 1
As all the numbers from 2 to 50 are factors of 2^50 * (1*2*3.....*50), none of these can be factors of 2^50 *
(1*2*3*4*......*50) + 1 (i.e. of h(100)+1)
Therefore the smallest prime factor of h(100) + 1 is greater than 50.
This corresponds to option (E)
Answer: (E) Greater than 40
Question: A train running at the speed of 70 km/hr crosses a pole in 9 sec. What is the length of the train?
(A) 298 m
(B) 175 m
(C) 208 m
(D) 988 m
(E) 299 m
Rationale: Speed = 70 * 5/18 = 175/9 m/sec
Length of the train = speed * time = 175/9 * 9
= 175 m
Answer:B
Answer: (B) 175 m
Question: The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in six colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?
(A) 24
(B) 32
(C) 48
(D) 60
(E) 192
Rationale: I have no idea how you got 8C8, 7C6 and 4C1.
I could have understood 8C1*6C1*4C1 (I have explained why this doesn't work in the post above)
The best way to solve it is by first selecting 3 colors out of the given 4 in 4C3 ways. (say, you got black, red and green)
Now for each color, you have 2 choices - model A or B
So you select a model in 2 ways.
No of ways of selecting the 3 cars = 6C3 * 2 * 2 * 2 = 192
Well, my logic was
The number of ways of selecting r objects from n different objects is nCr.
We have 8 options and can select either of the 8 as the first - 8C8
One car is gone and one becomes a forbidden choice so we have 7 options and need to select either of the 6 as the first - 7C6
Two cars are gone and two becomes a forbidden choice so we have 6 options and need to select either of the 4 as the first - 6C4
E
Answer: (E) 192
Question: A garrison of 1000 men has provisions for 60 days. At the end of 15 days, a reinforcement arrives, and it is now found that the provisions will last only for 20 days more. What is the reinforcement?
(A) 1000
(B) 2250
(C) 60
(D) 45
(E) 1250
Rationale: 1000 ---- 60
1000 ---- 45
x ----- 20
x*20 = 1000*45
x = 2250
1000
-------
1250
Answer: E
Answer: (E) 1250
Question: Two trains of equal lengths take 10 sec and 8 sec respectively to cross a telegraph post. If the length of each train be 120 m, in what time will they cross other travelling in opposite direction?
(A) 22
(B) 12
(C) 9
(D) 99
(E) 21
Rationale: Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/8 = 15 m/sec.
Relative speed = 12 + 15 = 27 m/sec.
Required time = (120 + 120)/27 = 9 sec.
Answer: C
Answer: (C) 9
Question: The average weight of 7 person's increases by 6.2 kg when a new person comes in place of one of them weighing 76 kg. What might be the weight of the new person?
(A) 160 kg
(B) 175 kg
(C) 180.7 kg
(D) 119.4 kg
(E) 190.8 kg
Rationale: Total weight increased = (7 x 6.2) kg = 43.4 kg.
Weight of new person = (76 + 43.4) kg = 119.4 kg
Option D
Answer: (D) 119.4 kg
Question: Sides of a rectangular park are in the ratio 3: 2 and its area is 3750 sq m, the cost of fencing it at 50 ps per meter is?
(A) 288
(B) 277
(C) 298
(D) 125
(E) 272
Rationale: 3x * 2x = 3750 => x = 25
2(75 + 50) = 250 m
250 * 1/2 = Rs.125.Answer:C
Answer: (C) 298
Question: A boatman can row 96 km downstream in 8 hr. If the speed of the current is 4 km/hr, then find in what time he will be able to cover 8 km upstream ?
(A) 1.5 hrs
(B) 1 hrs
(C) 2.5 hrs
(D) 2 hrs
(E) 3 hrs
Rationale: Explanation:
Speed in downstream = 96/8 = 12 kmph
Speed of current = 4 km/hr
Speed of the boatman in still water = 12 â€“ 4 = 8 kmph
Speed in upstream = 8 â€“ 4 = 4 kmph
Time taken to cover 8 km upstream = 8/4 = 2 hours.
ANSWER IS D
Answer: (D) 2 hrs
Question: How many integer values T are there for x such that 1 < 3x + 5 < 17?
(A) Two
(B) Three
(C) Four
(D) Five
(E) Six
Rationale: 1 < 3x + 5 < 17
=> -4 < 3x < 12
=> -4/3 < x < 4
x can take integer values T -1,0 , 1 , 2 , 3
Answer D
Answer: (D) Five
Question: In a kilometer race, A beats B by 80 meters or 10 seconds. What time does A take to complete the race?
(A) 199 sec
(B) 190 sec
(C) 726 sec
(D) 127 sec
(E) 115 sec
Rationale: Time taken by B run 1000 meters
= (1000 * 10)/80 = 125 sec.
Time taken by A = 125 - 10
= 115 sec.
Answer: E
Answer: (E) 115 sec
Question: What is the maximum number of pieces of birthday cake of size 5” by 5” that can be cut from a cake 15” by 15”?
(A) 5
(B) 10
(C) 16
(D) 20
(E) 9
Rationale: The prompt is essentially asking for the maximum number of 5 x 5 squares that can be cut from a larger 15 by 15 square. Since each 'row' and each 'column' of the larger square can be sub-divided into 3 'pieces' each, we have (3)(3) = 9 total smaller squares (at maximum).
E
Answer: (E) 9
Question: The average of marks obtained by 120 boys was 36. If the average of marks of passed boys was 39 and that of failed boys was 15, the number of boys who passed the examination is?
(A) 105
(B) 110
(C) 120
(D) 130
(E) 140
Rationale: Let the number of boys who passed = X.
Then, 39 x X + 15 x (120 - X) = 120 x 36
24X = 4320 - 1800
=> X = 2520/24
X = 105.
Hence, the number of boys passed = 105.
ANSWER:A
Answer: (A) 105
Question: A bike covers a certain distance at the speed of 66km/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) 85 km/h
(B) 85.34 km/h
(C) 87.67 km/h
(D) 90.25 km/h
(E) 88 km/h
Rationale: Ans.(E)
Sol. Total distance = 66 × 8 = 528 km
Now speed = 528/6 = 88 km/h
Answer: (E) 88 km/h
Question: The distance light travels in one year is approximately 5,860,000,000,000 miles. The distance light travels in 100 years is
(A) A) 587 × 108 miles
(B) B) 587 × 1010 miles
(C) C) 587 × 10-10 miles
(D) D) 586 × 1012 miles
(E) E) 587 × 10-12 miles
Rationale: The distance of the light travels in 100 years is:
5,860,000,000,000 × 100 miles.
= 586,000,000,000,000 miles.
= 586 × 1012 miles.
The answer is (D)
Answer: (D) D) 586 × 1012 miles
Question: The difference between the simple interest received from two different sources on Rs.1900 for 3 years is Rs.13.50. The difference between their rates of interest is
(A) 0.13%
(B) 0.23%
(C) 0.33%
(D) 0.43%
(E) 0.53%
Rationale: (1900xR1x3)/100-(100xR2x3)/100 =13.50
5700(R1-R2)=1350
R1-R2 = 1350/5700=0.23%
answer : B
Answer: (B) 0.23%
Question: In August,a cricket team that played 120 matches won 24% of the games it played.After a continuous winning streak,this team raised its average to 52%.How many matches did the team win to attain this average?
(A) 40
(B) 52
(C) 68
(D) 70
(E) 98
Rationale: let the no of matches played more=x
so,
(120+x)*52/100=28.8+x
by solving we get x=70
ANSWER:D
Answer: (D) 70
Question: Three candidates contested an election and received 1136, 7636 and 12000 votes respectively. What percentage of the total votes did the winning candidate get?
(A) 46%
(B) 58%
(C) 74%
(D) 63%
(E) 68%
Rationale: B
12000/(1136 + 7636 + 12000) = 58%.
Answer: (B) 58%
Question: How many numbers from 29 to 79 are exactly divisible by 11?
(A) 4
(B) 6
(C) 8
(D) 5
(E) 7
Rationale: Numbers=(79/11)-(29/11)
=7-2
=5 numbers
Option D is answer
Answer: (D) 5
Question: John bought 2 shares and sold them for $97 each. If he had a profit of 20% on the sale of one of the shares but a loss of 20% on the sale of the other share, then on the sale of both shares John had
(A)  a profit of $10
(B)  a profit of $8
(C)  a loss of $6
(D)  a loss of $10
(E)  neither a profit nor a loss
Rationale: $ Profit of the share sold at 20% profit =
120-97
20-?
$17 profit on the sale of 1 share.
$ Loss of the share sold at 20% loss=
80-97
20-?
$24 loss on the sale of 1 share.
17-23=Loss of $6
Answer: C
Answer: (C)  a loss of $6
Question: If 4 (A's capital) = 6 (B's capital) = 10 (C's capital), then out of a profit of Rs. 31, C will receive
(A) 6
(B) 5
(C) 8
(D) 10
(E) 12
Rationale: Explanation:
Let 4A = 6B = 1OC = k. Then, A = k/4, B = k/6, and C =k/10 .
A : B :C = k/4 : k/6 : k/10 = 15 : 10 : 6.
Hence, C's share (31 * 6/31) = Rs, 6.
Answer: A) Rs.6
Answer: (A) 6