mathewhe/aqua_rat huggingface dataset:
Question: An alloy of copper and zinc contains copper and zinc in the ratio 3 : 5. Another alloy of copper and zinc contains copper and zinc in the ratio 6 : 2. In what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc?
(A) 1:2
(B) 2:2
(C) 2:5
(D) 2:6
(E) 2:7
Rationale: Let Alloy_1 be x units, and let Alloy_2 be y units.
So, fraction of copper in Alloy_1 = 3x/8, and fraction of zinc in Alloy_1 = 5x/8.
Similarly, fraction of copper in Alloy_2 = 6y/8, and fraction of zinc in Alloy_2 = 2y/8.
Mixing them, we get
copper = 3x/8 +6y/8; zinc = 5x/8 + 2y/8.
So, 3x+6y = 5x+2y
-> 2x = 4y
-> x/y = 2/4 = 1/2
So, they must be mixed in the ratio 1 : 2
Answer : A
Answer: (A) 1:2
Question: By selling 15 pencils for a rupee a man loses 25%. How many for a rupee should he sell in order to gain 25%?
(A) 8
(B) 9
(C) 7
(D) 6
(E) 4
Rationale: 75% --- 15
125% --- ?
75/125 * 15 = 9
Answer: B
Answer: (B) 9
Question: Find the value of 3/11 × [(1 + 1/3) (1 + 1/4) …………. (1 + 1/120)] is
(A) 11
(B) 13
(C) 15
(D) 17
(E) 19
Rationale: 3/11 × [(1 + 1/3) (1 + 1/4) ....... (1 + 1/120)]
3/11 × [4/3 + 5/4 × ........ 121/120]
3/11 × 4/3 × 5/4 × .... 121/120
121/11 = 11.
A)
Answer: (A) 11
Question: A sum of Rs. 2900 amounts to Rs. 3422 in 3 years. Had the rate of interest be 5% higher then what would the sum amount in the same time period?
(A) Rs. 4465
(B) Rs. 3683
(C) Rs. 3596
(D) Rs. 3857
(E) None of these
Rationale: 3422=2900+(2900*R*3/100)
R=6
If R is 5 more.
Then R=11
Sum=2900+(2900*11*3/100)
Sum=3857
ANSWER:D
Answer: (D) Rs. 3857
Question: A school bus travels from Veldhoven to Roosendaal. There are four children in the bus. Each child has four backpacks with him. There are four dogs sitting in each backpack. Every dog has four puppies with her. All these dogs have four legs, with four toes at each leg.
What is the total number of toes in the bus?
(A) 4789
(B) 5170
(C) 4902
(D) 4490
(E) 5538
Rationale: B
5170
The total number of toes is 5170.
The puppies: 4 children × 4 backpacks × 4 dogs × 4 puppies × 4 legs × 4 toes = 4096
Plus the dogs: 4 children × 4 backpacks × 4 dogs × 4 legs × 4 toes = 1024
Plus the children: 4 children × 2 legs × 5 toes = 40
Plus the driver of the school bus: 2 legs × 5 toes = 10
4096+1024+40+10 = 5170
This gives the following total number of toes: 5170
Answer: (B) 5170
Question: An oblong piece of ground measures 19m 2.5dm by 12m 5dm. From the centre of each side a path 2m wide goes across to the centre of the opposite side. Find the cost of paving these paths at the rate of rs.1.32 per sq m?
(A) Rs. 43.6
(B) Rs. 78.65
(C) Rs 54.75
(D) Rs. 344
(E) Rs. 34.4
Rationale: Area of path horizontal=20*192.5=3850 dm^2
Area of path vertical=20*125=2500 dm^2
Since center path is covered twice so total area=(3850+2500)-(20*20)=5950 dm^2
total cost=1.32*59.5 m^2=78.54
ANSWER:B
Answer: (B) Rs. 78.65
Question: x and y are positive integers. When x is divided by 17, the remainder is 3, and when x is divided by 19, the remainder is 1. When y is divided by 14, the remainder is 5, and when y is divided by 21, the remainder is 12. What is the least possible value of y - x?
(A) 15
(B) 13
(C) 11
(D) 14
(E) 18
Rationale: When x is divided by 17, the remainder is 3: So, the possible values of x are: 3, 20, 37,54 , etc.
When x is divided by 19, the remainder is 1: So, the possible values of x are: 1,20... STOP. Since both lists include 20, the smallest possible value of x is 20.
When y is divided by 14, the remainder is 5: So, the possible values of y are: 5, 19, 33,47 etc.
When y is divided by 21, the remainder is 12: So, the possible values of y are: 12, 33, ...STOP. Since both lists include 33, the smallest possible value of y is 33
Since the smallest possible values of x and y are 20 and 33 respectively, the smallest possible value of y - x is 13. So,B is the correct answer to the original question.
Answer: (B) 13
Question: Insert the missing number
100,91,....,76,70
(A) 82
(B) 83
(C) 84
(D) 81
(E) 80
Rationale: Missing number: 83
Answer: B
Answer: (B) 83
Question: 1/(√2-1)=?
(A) 9+4√5
(B) 9-4√5
(C) √2+1
(D) 9-2√5
(E) 4+9√5
Rationale: This question requires us to rationalize (fix) the denominator.
For more on this technique, seehttps://www.gmatprepnow.com/module/gmat ... video/1044
Given: 1/(√2-1)
Multiply top and bottom by the CONJUGATE of √2-1, which is √2+1
So, 1/(√2-1) = (1)(√2+1)/(√2-1)(√2+1)
= (√2+1)/(1)
= √2+1
=
C
Answer: (C) √2+1
Question: what will be the next number in the following series??
4,18,...,100,180,294
(A) 32
(B) 36
(C) 40
(D) 44
(E) 48
Rationale: 2x2x1=4;
3x3x2=18;
4x4x3=48;
5x5x4=100;
6x6x5=180;
7x7x6=294;
so answer is 48.
ANSWER:E
Answer: (E) 48
Question: Two machines, Y and Z, work at constant rates producing identical items. Machine Y produces 13 items in the same time Machine Z produces 14 items. If machine Y takes 28 minutes to produce a batch of items, how many minutes does it take for machine Z to produce the same number of items?
(A) 26
(B) 9
(C) 9 1/2
(D) 12
(E) 13 1/2
Rationale: rate z/rate y=14/13
time z/time y=13/14
(13/14)*28=26 minutes
A
Answer: (A) 26
Question: A sequence consists of 16 consecutive even integers written in increasing order. The sum of the first 8 of these even integers is 472. What is the sum of the last 8 of the even integers?
(A) 588
(B) 594
(C) 600
(D) 606
(E) 612
Rationale: The sum of the first 8 numbers is: x + (x+2) +...+ (x+14) = 472
The sum of the next 8 numbers is: (x+16) + (x+2+16) +...+ (x+14+16) =
472 + 8(16) = 472 + 128 = 600
The answer is C.
Answer: (C) 600
Question: Last year, a certain public transportation system sold an average (arithmetic mean) of 41,000 tickets per day on weekdays (Monday through Friday) and an average of 15,000 tickets per day on Saturday and Sunday. Which of the following is closest to the total number of tickets sold last year?
(A) 1 million
(B) 1.25 million
(C) 10 million
(D) 12.2 million
(E) 125 million
Rationale: If you type that into the calculator, it results in 12.3 mil
[D], 12.2 mil, is the correct answer... see 2 posts above for quick approx method
Answer: (D) 12.2 million
Question: The sum of all the integers s such that -26 < s < 24 is
(A) 0
(B) -2
(C) -25
(D) -49
(E) -51
Rationale: Easy one -
-25, -24, -23,-22,...... -1,0, 1, 2...., 22, 23
Cancel everyhitng and we're left with - -25 and -24 s= -49.
D is the answer.
Answer: (D) -49
Question: If (x – 2)^2 = 144, which of the following could be the value of 2x?
(A) –26
(B) –22
(C) –11
(D) 12
(E) -20
Rationale: You can also get to the answer by backsolving using answer choices
Starting with x = -10
(x – 2)^2 = 144
(-10– 2)^2 = 144 =====> (-12)^2 = 144 Bingo, Clearly, among the answer choices only -10 will result in 144 in the equation so x must be -10,
Then 2x = 2 * -10 = -20
Answer: E
Answer: (E) -20
Question: 3 workers transfer a tool weighing 120kg in 12 seconds. How many men are required to transfer it in 9 seconds?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Rationale: We have, M1D1H1 = M2D2H2
3*120*12 = M2*120*9
M2 = 3*120*12/120*9 = 4
ANSWER:A
Answer: (A) 4
Question: A batch of cookies was divided amomg 6 tins: 2/3 of all the cookies were placed in either the blue or the green tin, and the rest were placed in the red tin. If 1/4 of all the cookies were placed in the blue tin, what fraction of the cookies that were placed in the other tins were placed in the green tin
(A) 15/2
(B) 9/4
(C) 5/9
(D) 5/7
(E) 9/7
Rationale: This will help reduce the number of variables you have to deal with:
G + B = 2/3
R = 1/6
B = 1/4
We can solve for G which is 5/12
What fraction (let it equal X) of the cookies that were placed in the other tins were placed in the green tin?
So..
X*(G+R)=G
X*(5/12 + 1/6) = 5/12
X = 5/7
ANSWER: D
Answer: (D) 5/7
Question: Each child has 12 crayons and 24 apples. If there are 18 children, how many crayons are there in total?
(A) 220
(B) 65
(C) 216
(D) 219
(E) 230
Rationale: 12*18=216.Answer is C.
Answer: (C) 216
Question: A bag is bought for Rs.400 and sold at a gain of 35% find its selling price
(A) A)Rs.460/-
(B) B)Rs.470/-
(C) C)Rs.480/-
(D) D)Rs.500/-
(E) E)Rs.520/-
Rationale: 100 % ------> 400 (100 * 4 = 400)
135 % ------> 540 (135 * 4 = 540)
Selling price = Rs.540
Answer : E
Answer: (E) E)Rs.520/-
Question: If x and y are different prime numbers, each greater than 8, which of the following must be true? I. x+y is an even integer II. xy is an odd integer III. (x/y) is not an integer
(A) I,II only
(B) I,III only
(C) II,III only
(D) I,II and III
(E) II only
Rationale: Given: x and y are different prime numbers, both greater than 8
Statement I: x+y is an even integer
Since x and y are prime numbers greater than 8, they will be odd
Therefore sum will be even.
Correct
Statement II: xy is an odd integer
Since x and y are prime numbers greater than 8, they will be odd
Therefore multiplication will be odd.
Correct
Statement III: (x/y) is not an integer
Since x and y are prime integers, therefore they will not have any common factor apart from 1
Hence (x/y) will not be an integer
Correct
Hence all three statements I, II and III are correct
Answer : D
Answer: (D) I,II and III
Question: In an examination, 25% of total students failed in Hindi, 35% failed in English and 40% in both. The percentage of these who passed in both the subjects is :
(A) 10%
(B) 20%
(C) 60%
(D) 80%
(E) 50%
Rationale: pass percentage = 100 - (25+35-40) = 100-20 = 80
ANSWER:D
Answer: (D) 80%
Question: Ramesh starting from a fixed point goes 15 km towards North and then after turning to his right he goes 15 km. then he goes 10, 15 and 15 metres after turning to his left each time. How far is he from his starting point?
(A) 8 km
(B) 9 km
(C) 10 km
(D) 11 km
(E) 12 km
Rationale: he will be 10 km. away from his starting point.
ANSWER:C
Answer: (C) 10 km
Question: Copper is 7 times as heavy as water and gold is 13 times as heavy as water. In what ratio should these be mixed to get an alloy 11 times as heavy as water ?
(A) 2 : 1
(B) 1 : 2
(C) 2 : 3
(D) 3 : 2
(E) 3 : 1
Rationale: Explanation :
Solution: C = 7W and G = 13W.
Let 1 gm of copper be mixed with x gm of gold to get (1+x)gm of alloy.
(1gm copper)+(xgm gold) = (x+1)gm of alloy.
=>7W+13W*x = (x+1)*11W
=> 7+13x = 11(x+1)
=> 2x = 4 => x = 2.
.'. Ratio of copper and gold = 1 : 2
Answer : B
Answer: (B) 1 : 2
Question: A train running at the speed of 60 km/hr crosses a pole in 12 seconds. What is the length of the train?
(A) 200 metres
(B) 180 metres
(C) 324 metres
(D) 228 metres
(E) 150 metres
Rationale: Speed= 60 x	5/18	m/sec	= 50/3	m/sec.
Length of the train = (Speed x Time).
Length of the train = 50/3	x 9	m = 200 m.
Option A
Answer: (A) 200 metres
Question: The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, what is the weight of B?
(A) 31 kg
(B) 28 1/2 kg
(C) 32 kg
(D) 30 1/2 kg
(E) 32 kg
Rationale: Explanation:
Let the weight of A, B and C are a,b and c respectively.
Average weight of A,B and C = 45
a + b + c = 45 × 3 = 135 --- equation(1)
Average weight of A and B = 40
a + b = 40 × 2 = 80 --- equation(2)
Average weight of B and C = 43
b + c = 43 × 2 = 86 --- equation(3)
equation(2) + equation(3) - equation(1)
=> a + b + b + c - (a + b + c) = 80 + 86 - 135
=> b = 80 + 86 -135 = 166 - 135 = 31
Weight of B = 31 Kg
Answer: Option A
Answer: (A) 31 kg
Question: A milk man sells the milk at the cost price but he mixes the water in it and thus he gains 9.09%. The quantity of water in the mixture of 1 liter is?
(A) 83.33 ml
(B) 12.33 ml
(C) 77.33 ml
(D) 22.33 ml
(E) 18.33 ml
Rationale: Profit (%) = 9.09 % = \inline \frac{1}{11}
Since the ratio of water and milk is 1 : 11,
Therefore the ratio of water is to mixture = 1:12
Thus the quantity of water in mixture of 1 liter =
{1}/{12} = 83.33 ml
Answer: A
Answer: (A) 83.33 ml
Question: Three years ago the average age of a class of six members was 23 years. A boy have been added, the average age of the class is the same today. What is the age of the boy?
(A) A)1
(B) B)2
(C) C)3
(D) D)4
(E) E)5
Rationale: 6 * 26 = 156
7 * 23 = 161
`5
E
Answer: (E) E)5
Question: A train 220 m long is running with a speed of 90 km/hr. In what time will it pass a bus that is running with a speed of 60 km/hr in the direction opposite to that in which the train is going?
(A) 4.37
(B) 5.28
(C) 6.75
(D) 8
(E) 3.1
Rationale: Speed of train relative to bus = 90 + 60 = 150 km/hr.
= 150 * 5/18 = 125/3 m/sec.
Time taken to pass the bus = 220 * 3/125
= 5.28 sec.
Answer: B
Answer: (B) 5.28
Question: 50 litres of diesel is required to travel 600 km using a 800 cc engine. If the volume of diesel required to cover a distance varies directly as the capacity of the engine, then how many litres of diesel is required to travel 800 km using 1200 cc engine?
(A) 80 litres
(B) 90 litres
(C) 100 litres
(D) 170 litres
(E) None of these
Rationale: Explanatory Answer
To cover a distance of 800 kms using a 800 cc engine, the amount of diesel required = 800/600*50 = 66.67 litres.
However, the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity.
i.e., for instance, if the capacity of engine doubles, the diesel requirement will double too.
Therefore, with a 1200 cc engine, quantity of diesel required = 1200/800*66.67 = 100 litres.
Answer C
Answer: (C) 100 litres
Question: There are 5 bags three of which each contains 4 white and 2 black balls, and remaining 2 bags each contains 1 white and 4 black ball; a white ball has been drawn, find the chance that it came from first group.
(A) 2/7
(B) 6/35
(C) 8/25
(D) 5/6
(E) 3/5
Rationale: Please go through my solution and suggest any mistake.
Chances of selecting a white ball from group 1: 2/3---------------------Chances of selecting a white ball from group 2: 1/5
Thus combined probability of section of white ball from group 1:
3/5 x 2/3 = 2/5
Thus combined probability of section of white ball from group 2:
2/5 x 1/5 = 2/25
Out of these chances, chance of occurrence of first case : (2/5) / (2/5 + 2/25) = 5/6
D
Answer: (D) 5/6
Question: In an office in Singapore there are 60% female employees. 50 % of all the male employees are computer literate. If there are total 62% employees computer literate out of total 1400 employees, then the no. of female employees who are computer literate ?
(A) 588
(B) 674
(C) 672
(D) 960
(E) None
Rationale: Solution: Total employees,
= 1400
Female employees, 60% of 1400.
= (60 *1400) /100 = 840.
Then male employees,
= 560
50 % of male are computer literate,
= 280 male computer literate.
62 % of total employees are computer literate,
= (62 *1400) /100 = 868 computer literate.
Thus, Female computer literate = 868 - 280
= 588.
Answer: Option A
Answer: (A) 588
Question: 9 - 3 ÷ 1/3 + 5 = ?
(A) 5
(B) 3
(C) 6
(D) 9
(E) 12
Rationale: 9 - 3 1/3 + 5
=9-3÷1/3+5
=9-(3X3) +5
=9-9+5
=5
CORRECT ANSWER : A
Answer: (A) 5
Question: a product of ages of ankit and nikita is 240. if twice the age of nikita is more than ankit's by 4 years. what is the nikita's age?
(A) 12yrs
(B) 13yrs
(C) 14yrs
(D) 15yrs
(E) 11yrs
Rationale: if ankit age=x
nikita age=240/x
then (2*(240/x))-x=4 so,x=20
ankit age=20yrs
nikita age=(240/20)=12yrs
ANSWER:A
Answer: (A) 12yrs
Question: The average weight of 22 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.
(A) 48.89
(B) 42.25
(C) 50
(D) 51.25
(E) 52.25
Rationale: Explanation:
Average weight of 22 boys = 50.25
Total weight of 22 boys = 50.25 × 22
Average weight of remaining 8 boys = 45.15
Total weight of remaining 8 boys = 45.15 × 8
Total weight of all boys in the class = (50.25 × 16)+ (45.15 × 8)
Total boys = 22 + 8 = 30
Average weight of all the boys = ((50.25×22)+(45.15×8))/30
=48.89
Answer: Option A
Answer: (A) 48.89
Question: An seller earns an income of Re 3 on the first day of his business. On every subsequent day, he earns an income which is just thrice of that made on the previous day. On the 15th day of business, he earns an income of:
(A) 19
(B) 22
(C) 25
(D) 20
(E) 36
Rationale: 2nd day he earns = 3(2 – 3)
3rd day he earns = 3(3 – 3)
On 15th day he earns 3(15 -3) = 36 rupees
ANSWER:E
Answer: (E) 36
Question: It rained as much as on Wednesday as on all the other days of the week combined. If the average rainfall for the whole week was 3 cms, How much did it rain on Wednesday?
(A) 10.9
(B) 10.5
(C) 10.2
(D) 10.1
(E) 10.6
Rationale: Explanation:
Let the rainfall on wednesday = 6x.
∴ Rainfall on the remaining days = 6x
Given,
(6x + 6x )/7 = 3
⇒12x = 21
⇒6x = 10.5
Answer: B
Answer: (B) 10.5
Question: Find the value of x from the below equation? : 3x^2+5x+2 = 0
(A) -1
(B) 1
(C) 2
(D) 3
(E) 4
Rationale: a = 3, b = 5, c = 2
x1,2 = (-5 ± √(52 - 4×3×2)) / (2×3) = (-5 ± √(25-24)) / 6 = (-5 ± 1) / 6
x1 = (-5 + 1)/6 = -4/6 = -2/3
x2 = (-5 - 1)/6 = -6/6 = -1
A
Answer: (A) -1
Question: Of the science books in a certain supply room, 50 are on botany, 55 are on zoology, 90 are on physics, 50 are on geology, and 110 are on chemistry. If science books are removed randomly from the supply room, how many must be removed to ensure that 80 of the books removed are on the same area science?
(A) 81
(B) 159
(C) 166
(D) 285
(E) 314
Rationale: The worst case scenario is to remove all the books on botany, zoology, and geology.
Also we can remove 79 books each from physics and chemistry.
The total is 50 + 55 + 50 + 79 + 79 = 313.
The next book removed must either be physics or chemistry, ensuring 80 books in that subject.
The answer is E.
Answer: (E) 314
Question: The average weight of 8 girlsincreases by 2.5 kg when a new gitl comes in place of one of them weighing 65 kg. What might be the weight of the new girl?
(A) 65 kg
(B) 75 kg
(C) 85 kg
(D) 90 kg
(E) 92 kg
Rationale: Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg
C
Answer: (C) 85 kg
Question: If money is invested at r percent interest, compounded annually, the amount of investment will double in approximately 70/r years. If Pat's parents invested $ 7000 in a long term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of investment 18 years later, when Pat is ready for college?
(A) $28000
(B) $15000
(C) $12000
(D) $10000
(E) $9000
Rationale: Since investment doubles in 70/r years then for r=8 it'll double in 70/8=~9 years (we are not asked about the exact amount so such an approximation will do). Thus in 18 years investment will double twice and become ($7,000*2)*2=$28,000 (after 9 years investment will become $7,000*2=$14,000 and in another 9 years it'll become $14,000*2=$28,000).
Answer: A.
Answer: (A) $28000
Question: From a Group of 9 People, Including George and Nina, 4 people are to be selected at random to work on a certain project. What is the probability that 4 people selected will include George but not Nina
(A) 5/56
(B) 9/56
(C) 15/56
(D) 35/120
(E) 25/56
Rationale: Number of ways of selecting 4 people out of 9 people = 9C4
In the three members George will always be there in the team. At this step we have vacancy for 3 more members and 8 members are available. Nina cannot be there in the team. So 3 members have to be selected and the number of available members = 8 - Nina = 7
Number of ways to form a 4 member team that includes George and excludes Nina = 7C3
Probability = 7C3/9C4 = 35/120
Answer: D
Answer: (D) 35/120
Question: A train crosses a platform in 25 sec, same train crosses another platform of length 400 m in 30 sec. If the length of the train is 200m , find the length of the platform?
(A) 280m
(B) 180m
(C) 350m
(D) 300m
(E) 267m
Rationale: Length of the platform be ‘X’
(X + 200)/25 = (200 + 400)/30
6X + 1200 = 3000
X = 300m
Answer: D
Answer: (D) 300m
Question: In a class, the average age of 30 boys is 13 years and the average of 20 girls is 12 years. what is the average age of the whole class?
(A) 12.9
(B) 12.1
(C) 12.3`
(D) 12.6
(E) 12.5
Rationale: Total age of 50 students
(30X 13+20 X 12) = 630
Average = 630/50 = 12.6 Years
Answer: D
Answer: (D) 12.6
Question: A number x is 6 times another number y. The percentage that y is less than x is
(A) 12.5%
(B) 83.3%
(C) 80%
(D) 11%
(E) 1%
Rationale: Say y=1 and x=6.
Then y=1 is less than x=6 by (6-1)/6*100=5/6*100=83.3%.
Answer: B.
Answer: (B) 83.3%
Question: A football player scores 4 goals in his fifth match thus increasing his average goals score by 0.2. The total number of goals in his 5 matches would be
(A) 4
(B) 6
(C) 8
(D) 10
(E) 16
Rationale: While this question can be solved with a rather straight-forward algebra approach (as the other posters have noted), it can also be solved by TESTing THE ANSWERS. One of those numbers MUST be the total number of goals...
From a tactical standpoint, it's best to TEST either Answer B or Answer D, so if the answer is not correct, then you would have a gauge for whether you should gohigherorlowerwith your next TEST.
Here, I'll start with Answer E = 16 goals
If....
Total goals = 16 goals
5th game = 4 goals
1st 4 games = 12 goals
Avg. for 1st 4 games = 12/4 = 3 goal/game
Avg. for all 5 games = 6/5 = 3.2 goals/game
This is an EXACT MATCH for what we're told in the prompt, so Answer E MUST be the answer.
Answer: (E) 16
Question: Rs.1200 divided among P, Q and R. P gets half of the total amount received by Q and R. Q gets one-third of the total amount received by P and R. Find the amount received by R ?
(A) 377
(B) 500
(C) 287
(D) 179
(E) 112
Rationale: Let the amounts to be received by P, Q and R be p, q and r.
p + q + r = 1200
p = 1/2 (q + r) => 2p = q + r
Adding 'p' both sides, 3p = p + q + r = 1200
=> p = Rs.400
q = 1/3 (p + r) => 3q = p + r
Adding 'q' both sides, 4q = p + q + r = 1200
=> q = Rs.300
r = 1200 - (p + q) => r = Rs.500.
Answer: B
Answer: (B) 500
Question: Rahul can do a work in 3 days while Rajesh can do the same work in 2days. Both of them finish the work together and get $170. What is the share of Rahul?
(A) $68
(B) $40
(C) $60
(D) $100
(E) $90
Rationale: Rahul's wages: Rajesh's wages = 1/3 : 1/2 = 2:3
Rahul's share = 170*2/5 = $68
Answer is A
Answer: (A) $68
Question: The function F(n) is defined as the product of all the consecutive positive integers between 2 and n^2, inclusive, whereas the function G(n) is defined as the product of the squares of all the consecutive positive integers between 1 and n, inclusive. The exponent on 2 in the prime factorization of F(3)/G(3) is
(A) 	1
(B) 	2
(C) 	3
(D) 	4
(E) 	5
Rationale: F(3)/G(3)
=product(1 to 3 ^2) / 1.2^2.3^2
=1.2.3.4.5.6.7.8.9/1.4.9
=1.2.3.(2^2).5.(2.3).7.(2^3).9/1.(2^2).9
=1.(2^7).3.5.7.9/1.(2^2).9
Loof for 2^7/2^2=2^5 ----Exponent 3
Answer: C
Answer: (C) 	3
Question: S is a set of 85 consecutive multiples of 5. If the smallest number in S is 90, then the greatest number in S is
(A) 158
(B) 597
(C) 599
(D) 402
(E) 510
Rationale: last term =first term +(total no. of terms-1)consecutive difference
S is a set of 85 consecutive multiples of 5. If the smallest number in S is 90, then the greatest number in S is
first term = 90 ; total terms =85 ; difference = 5
90+(84)5 = 510
Ans E
Answer: (E) 510
Question: Positive integer y is 50 percent of 50 percent of positive integer x, and y percent of x equals 9. What is the value of x?
(A) 20
(B) 60
(C) 100
(D) 120
(E) 150
Rationale: y = 0.5*0.5*x = x/4
y%*x = 9
(y/100)*x = 9
(x/400)*x = 9
x^2 = 9*400
x = 60
The answer is B.
Answer: (B) 60
Question: One day a car rental agency rented 3/4 of its cars, including 2/3 of its cars with CD players. If 3/5 of its cars have CD players, what percent of the cars that were not rented had CD players?
(A) 1/2
(B) 2/3
(C) 3/4
(D) 4/5
(E) 5/6
Rationale: The cars with CD players which were not rented is (1/3)(3/5) = 1/5 of all the cars.
The cars which were not rented is 1/4 of all the cars.
The percent of non-rented cars which had CD players is (1/5) / (1/4) = 4/5
The answer is D.
Answer: (D) 4/5
Question: To produce an annual income of Rs. 1200 from a 12 % stock at 90,the amount of stock needed is :
(A) Rs. 10,000
(B) Rs. 10,800
(C) Rs. 14,400
(D) Rs. 16,000
(E) Rs. 46,000
Rationale: Solution
For an income of Rs. 12, stock needed = Rs. 100.
For an income of Rs. 1200, stock needed = Rs. 10/12x12000=10,000
Answer A
Answer: (A) Rs. 10,000
Question: .007/?= .01
(A) .7
(B) .07
(C) .007
(D) .0007
(E) None of them
Rationale: Let .007/x = .01;
Then x = .007/.01
= .7/1
= .7
Answer is A
Answer: (A) .7