mathewhe/aqua_rat huggingface dataset:
Question: Richard has to paint a mural with seven horizontal stripes. He only has enough paint for four blue stripes, four white stripes, four black stripes, and four yellow stripes. If his patron wants at most two different colors in the mural, how many different ways can he paint the wall?
(A) 420
(B) 350
(C) 700
(D) 2,520
(E) 5,040
Rationale: Number of ways to choose 2 colors among 4 - 4c2
Number of ways that the chosen colors can be painted for 7 horizontal stripes - 2 (either (4,3) or (3,4))
Now the order,
the number of ways in which a particular color among the 2 chosen colors can be painted for 4 stripes is - 7c4
[Note: We don't need to calculate the ways for the remaining 3 stripes as they are obvious and already counted under this]
Total number of combinations - 4c2 * 2* 7c4 = 420
A
Answer: (A) 420
Question: The parameter of a square is equal to the perimeter of a rectangle of length 16 cm and breadth 14 cm. Find the circumference of a semicircle whose diameter is equal to the side of the square. (Round off your answer to two decimal places)
(A) 28.99
(B) 23.52
(C) 23.51
(D) 23.56
(E) 23.57
Rationale: Let the side of the square be a cm.
Parameter of the rectangle = 2(16 + 14) = 60 cm Parameter of the square = 60 cm
i.e. 4a = 60
A = 15
Diameter of the semicircle = 15 cm
Circimference of the semicircle
= 1/2(∏)(15)
= 1/2(22/7)(15) = 330/14 = 23.57 cm to two decimal places.Answer:E
Answer: (E) 23.57
Question: The C.I. on a certain sum for 2 years at 10% per annum is Rs. 525. The S.I. on the same sum for double the time at half the rate percent per annum is?
(A) Rs.502
(B) Rs.500
(C) Rs.520
(D) Rs.512
(E) Rs.132
Rationale: Let the sum be Rs. P. Then,
[P(1 + 10/100)2 - p] = 525
P[(11/10)2 - 1] = 525
P = (525 * 100) / 21 = 2500
Sum = Rs. 2500
So, S.I. = (2500 * 5 * 4)/100
= Rs.500
Answer: B
Answer: (B) Rs.500
Question: 1000 men have provisions for 18 days. If 400 more men join them, for how many days will the provisions last now?
(A) 12.9
(B) 12.8
(C) 12.6
(D) 12.2
(E) 12.1
Rationale: 1000*18 = 1400*x
x = 12.8
Answer: B
Answer: (B) 12.8
Question: A certain college's enrollment at the beginning of 1992 was 20 percent greater than it was at the beginning of 1991, and its enrollment at the beginning of 1993 was 25 percent greater than it was at the beginning of 1992. The college's enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ?
(A) 47.5%
(B) 44%
(C) 50%
(D) 35%
(E) 38%
Rationale: Suppose enrollment in 1991 was 100
Then enrollment in 1992 will be 120
And enrollment in 1993 will be 120*1.25=150
Increase in 1993 from 1991 = 150-100= 50
Answer : C
Answer: (C) 50%
Question: In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
(A) 21
(B) 21/41
(C) 21/47
(D) 21/45
(E) 21/46
Rationale: Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25
= 25C3 `
= (25 x 24 x 23)
(3 x 2 x 1)
= 2300.
n(E) = (10C1 x 15C2)
= 10 x (15 x 14)
(2 x 1)
= 1050.
P(E) = n(E) = 1050 = 21 .
n(S) 2300 46
Answer: (E) 21/46
Question: A circular ground whose diameter is 35 metres, has a 1.4 metre-broad garden around it. What is the area of the garden in square metres?
(A) 160.16
(B) 6.16
(C) 122.66
(D) Data inadequate
(E) None of these
Rationale: Req. area = π[(17.5 + 1.4)2 – (17.5)2]
= 22⁄7 × (36.4 × 1.4)[since a2 - b2 = (a + b)(a - b)]
= 22 × 36.4 × 0.2 = 160.16 sq m
Answer A
Answer: (A) 160.16
Question: Last year a certain bond yielded 5 percent of its face value in interest. If that interest was approximately 4 percent of the bonds selling price of $8,000, what is the bonds face value?
(A)  $6,000
(B)  $6,750
(C)  $6,400
(D)  $7,500
(E)  $9,375
Rationale: Interest = 0.05*face value = 0.04*8,000 --> face value = 0.04*8,000/0.05 = 6,400.
Answer: C.
Answer: (C)  $6,400
Question: If 100 microns = 1 decimeter, and 10,000 angstroms = 1 decimeter, how many angstroms equal 1 micron?
(A) 100
(B) 10
(C) 0.01
(D) 0.0001
(E) 0.0001
Rationale: Given that 100 microns = 1 decimeter =10,000 angstroms
So, 1 micron = 10,000/100
=100
Answer : A
Answer: (A) 100
Question: The product of all the prime numbers less than 15 is closest to which of the following powers of 10 ?
(A) 10^9
(B) 10^8
(C) 10^5
(D) 10^6
(E) 10^7
Rationale: Product of Prime numbers less than 15 is 2*3*5*7*11*13
2*5=10 .
10*11*3*7*13 = 10*10*3*10*10 (approximately) ~ Product will be >10^4 <10^6 i.e. 10^5
C is the answer
Answer: (C) 10^5
Question: A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If four marbles are picked at random, what is the probability that none is blue?
(A) 33/998
(B) 33/91
(C) 33/99
(D) 33/95
(E) 33/92
Rationale: Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. When four marbles are picked at random, then the probability that none is blue is = ¹²C₄/¹⁵C₄ = (12 * 11 * 10 * 9)/(15 * 14 * 13 * 12) =33/91.Answer: B
Answer: (B) 33/91
Question: If the sum of the first n positive integers is S, what is the sum of the first n negative even integers, in terms of S ?
(A)  S/2
(B)  S
(C)  2S
(D)  2S + 2
(E)  4S
Rationale: Answer is C
Given sum of -1 -2 -3.....-n = S
sum of -2 , -4 , -6 ....-2n = 2[sum(-1 -2 -3....n) ] = 2S
Answer: (C)  2S
Question: A solution contains 8 parts of water for every 7 parts of Lemonade syrup. How many parts of the solution should be removed and replaced with water so that the solution will now contain 10% lemonade syrup?
(A) 1.5
(B) 1.75
(C) 2.14
(D) 2.34
(E) 2.64
Rationale: Let the total solution is 150 L with 80 L water70 L syrup.
To make 10% syrup solution, the result solution must have 135 L syrup and 15 L syrup.
Therefore we are taking 55 L of syrup from initial solution and replacing with water.
using urinary method:
70 L syrup in 150 L solution
55 L syrup in 117.9 L solution
We started by multiplying 10
Now to get to the result we need to divide by 55 => amount of solution to be replaced with water = (117.9/55) = 2.14.
Correct option : C
Answer: (C) 2.14
Question: 12 men take 18 days to complete a job whereas 12 women in 18 days can complete 3â„5 of the same job. How many days will 10 men and 8 women together take to complete the same job?
(A) 6
(B) 13 1â„2
(C) 14 3/5
(D) Data inadequate
(E) None of these
Rationale: 12 M Ã— 18 = 12 W Ã— 18 Ã— 5/3
\ w = 3/5 M
10M + 8W = 10M + 8 Ã— 3â„5M = 14 4/5 M
\14 4/5 men can complete the same work
in 12Ã—18/14 4/5=14 3/5days
Answer C
Answer: (C) 14 3/5
Question: Yearly subscription to professional magazines cost a company $940.00. To Make a 35% cut in the magazine budget, how much less must be spent?
(A) 654
(B) 655
(C) 611
(D) 657
(E) 658
Rationale: total cost 940
940*35/100=329
so the cut in amount is 329
the less amount to be spend is 940-329=611
ANSWER:C
Answer: (C) 611
Question: The entire contents of a full sack of flour can be used to make 15 cupcakes and 8 pizzas. The same full sack of flour can be used to make 9 cupcakes and 14 pizzas. If a full sack of flour is used to make only pizzas, how many pizzas can be made?
(A) 18 1/4
(B) 18 1/2
(C) 18 3/4
(D) 19
(E) 23
Rationale: 15x+8y=9x+14y
x=y. Answer E
Answer: (E) 23
Question: How many two-digit numbers are there whose remainder when divided by 13 is 1, and whose remainder when divided by 6 is 5?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: 2 digit numbers whose remainder when divided by 10 is 1 are
11 , 21 , 31 , 41 , 51 , 61 , 71 , 81 , 91
Out of above , numbers whose remainder when divided by 6 is 5 are
11 , 41 and 71
Answer D
Answer: (D) 6
Question: A boat can travel with a speed of 16 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 60 km downstream.
(A) 2 hours.
(B) 3 hours.
(C) 4 hours.
(D) 5 hours.
(E) 6 hours.
Rationale: Speed downstream = (16 + 4) km/hr = 20 km/hr.
Time taken to travel 60 km downstream = 60/20 = 3 hours.
Answer :B
Answer: (B) 3 hours.
Question: The average age of 8 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men. The average age of the two new men is
(A) 22
(B) 30
(C) 99
(D) 38
(E) 27
Rationale: total age increased = (8 * 2) years = 16 years.
Sum of ages of two new men = (21 + 23 + 16) years = 60 years
Average age of two new men = (60/2) years = 30 years.
Answer: B
Answer: (B) 30
Question: Look at this series: XXIV, XX, __, XII, VIII, ... What number should fill the blank?
(A) XV
(B) XVI
(C) XII
(D) VII
(E) VI
Rationale: B
XVI
This is a simple subtraction series; each number is 4 less than the previous number.
Answer: (B) XVI
Question: Rs.160 contained in a box consists of one rupee, 50 paisa and 25 paisa coins in the ratio 4:5:6. What is the number of 25 paisa coins?
(A) 197
(B) 219
(C) 120
(D) 165
(E) 287
Rationale: 4x 5x 6x
100 50 25
400x + 350x + 150x = 16000
x = 20
6x =120
Answer: C
Answer: (C) 120
Question: A cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and there by decreases his average by 0.4. The number age of the family now is?
(A) 65
(B) 67
(C) 87
(D) 85
(E) 65
Rationale: Let the number of wickets taken till the last match be x. Then,
(12.4x + 26)/(x + 5) = 12
= 12.4x + 26 = 12x + 60
= 0.4x = 34
= x = 340/4 =85
Answer: D
Answer: (D) 85
Question: How many seconds will a 500 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?
(A) 11 sec
(B) 30 sec
(C) 77 sec
(D) 14 sec
(E) 12
Rationale: Speed of train relative to man = 63 - 3 = 60 km/hr.
= 60 * 5/18 = 50/3 m/sec.
Time taken to pass the man = 500 * 3/50
= 30 sec.
Answer: B
Answer: (B) 30 sec
Question: A woman has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
(A) 80
(B) 90
(C) 100
(D) 110
(E) None
Rationale: Explanation:
Let number of notes of each denomination be x.
Then x + 5x + 10x = 480
16x = 480
x = 30.
Hence, total number of notes = 3x = 90.
B)
Answer: (B) 90
Question: At a reception, one-third of the guests departed at a certain time. Later two-fifths of the guests departed. Even later two-thirds of the remaining guests departed. If six people were left, how many were originally present at the party?
(A) 28
(B) 37
(C) 20
(D) 45 people
(E) 27
Rationale: Explanation:
Let Original members be x
First One third guest departed i.e x/3
Remaining guests = x – (x/3) = 2x/3
Now from the remaining (2x/3) two-fifths departed = 2/5(2x/3) = 4x/15
i.e. Now remaining guests will be (2x/3 – 4x/15) = 2x/5
Now from remaining (2x/5) two-thirds departed = 2/3(2x/5) = 4x/15
Now remaining guests = (2x/5 – 4x/15) = 2x/15
Given 2x/15 = 6 ⇒ x = 45
Answer: D
Answer: (D) 45 people
Question: Three students appear at an examination of Mathematics. The probability of their success are 1/3, 3/4, 1/5 respectively. Find the probability of success of at least two.
(A) 11/30
(B) 12/50
(C) 13/30
(D) 3/50
(E) 4/50
Rationale: The probability of success of at least two students will involve the following possibilities.
The first two students are successful, the last two students are successful, the first and third students are successful and all the three students are successful.
Therefore, the required probability = 1/3 x 3/4 x 4/5 + 3/4 x 1/5 x 2/3 + 1/3 x 1/5 x 1/4 + 1/3 x 3/4 x 1/5 = 11/30
ANSWER:A
Answer: (A) 11/30
Question: Find the value of 4 x [(3.6 x 0.48 x 2.50) / (0.12 x 0.09 x 0.5)]
(A) 800
(B) 3200
(C) 900
(D) 1600
(E) None
Rationale: Answer
4 x [(3.6 x 0.48 x 2.50) / (0.12 x 0.09 x 0.5)]
= 4 x [(36 x 48 x 250) / (12 x 9 x 5)]
= 4 x 4 x 4 x 50 = 3200
Correct Option: B
Answer: (B) 3200
Question: A drink vendor has 20 liters of Maaza, 144 liters of Pepsi and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required?
(A) 135
(B) 137
(C) 142
(D) 130
(E) 133
Rationale: The number of liters in each can = HCF of 20, 144 and 368 = 4 liters.
Number of cans of Maaza = 20/4 = 5
Number of cans of Pepsi = 144/4 = 36
Number of cans of Sprite = 368/4 = 92
The total number of cans required = 5 + 36 + 92 = 133 cans.
ANSWER:E
Answer: (E) 133
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) 29:32
(B) 21:34
(C) 21:22
(D) 31:20
(E) 21:39
Rationale: 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).
i.e., (120/100 * 7x) and (110/100 * 8x)
i.e., 42x/5 and 44x/5
Required ratio = 42x/5 : 44x/5 = 21:22
Answer: C
Answer: (C) 21:22
Question: nine persons went to a hotel for taking their meals 8 of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine.What was the total money spent by them?
(A) 132
(B) 451
(C) 175
(D) 145
(E) 117
Rationale: Sol: Let the average expenditure of all nine be Rs.x
Then 12*8+(x+8)=9x or 8x=104 or x=13.
Total money spent = 9x=Rs.(9*13)=Rs.117.
Ans: E
Answer: (E) 117
Question: (a-b)/(c/2)
In the expression above, a, b, and c are different numbers and each is one of the numbers 1, 2, or 4. What is the least possible value of the expression?
(A) 1/4`
(B) 2/4
(C) 4/2
(D) 1/2
(E) 3/4
Rationale: (a-b)/(c/2)
= 2(a-b)/c
The expression will have the least value when numerator 2(a-b) is the smallest.
= 2(2-1)/4
=2/4
Answer B
Answer: (B) 2/4
Question: Sum of the squares of three numbers is 138 and the sum of their products taken two at a time is 131. Find the sum?
(A) 20
(B) 25
(C) 26
(D) 28
(E) 29
Rationale: (a + b + c)2 = a2 + b2 + c2 + 2(ab +bc + ca) = 138 + 2* 131
a + b + c = √400 = 20
A
Answer: (A) 20
Question: A certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $130,000 and variables costs averaging $10 per item. If the manufacturer’s selling price per item is $15, how many items the manufacturer produce and sell to earn an annual profit of $150,000?
(A) 2,858
(B) 18,667
(C) 21,429
(D) 35,000
(E) 56,000
Rationale: Let the items manufactured or sold bex
130000+10x = 15x-150000
5x = 280000
x =56000
Ans:E
Answer: (E) 56,000
Question: The HCF of two numbers is 40 and the other two factors of their LCM are 11 and 12. What is the largest number.
(A) 462
(B) 450
(C) 488
(D) 480
(E) 555
Rationale: Explanation:
HCF of the two numbers = 40
HCF will be always a factor of LCM
40 is factor of LCM
other two factors are 11 & 12
Then the numbers are (40 *11) and (40 x 12)
= 440 and 480
Answer: Option D
Answer: (D) 480
Question: The speed at which a man can row a boat in still water is 10 kmph. If he rows downstream, where the speed of current is 2 kmph, what time will he take to cover 60 metres?
(A) 18 seconds
(B) 34 seconds
(C) 14 seconds
(D) 12 seconds
(E) 15 seconds
Rationale: Speed of the boat downstream = 10 + 2 = 12 kmph
= 12 * 5/18 = 10/3 m/s
Hence time taken to cover 60 m = 60*3/10
= 18 seconds.
Answer: A
Answer: (A) 18 seconds
Question: Maxwell leaves his home and walks toward Brad's house at the same time that Brad leaves his home and runs toward Maxwell's house. If the distance between their homes is 80 kilometers, Maxwell's walking speed is 4 km/h, and Brad's running speed is 6 km/h, what is the distance traveled by Brad?
(A)  16
(B)  48
(C)  20
(D)  24
(E)  30
Rationale: Time taken = Total Distance / Relative Speed
Total Distance = 80 kms
Relative Speed (Opposite side) (As they are moving towards each other speed would be added ) = 6+ 4 = 10 kms/hr
Time taken = 80 / 10 = 8 hrs
Distance traveled by Brad = Brad's speed * time taken = 6 * 8 = 48 kms... Answer - B
Answer: (B)  48
Question: Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
Rationale: Make sure to follow posting guidelines (link in my signatures). Title of the thread must mention the first few words of the question itself.
As for the question, the most straightforward way is to list out the factors (this is fine for this question as 180 is a relatively small number). Number of factors of 180 (=2^2*3^2*5) = 18. So listing the factors gives us:
1, 180
2, 90
3, 60
4, 45
5, 36
6, 30
9, 20
10, 18
12, 15
As the question asks us about the factors of the form 4k+2 ---> eliminate all ODD factors. You are now left with
2,4,6,10,12,18,20,30,36,60,90,180
For the form 4k+2, the first few values are 2, 6, 10, 14, 18, 22, 26, 30 ... already giving you 5 factors (=2,6,10,18,30). You can stop here and mark 6 (6th one = 90) as the correct answer as 5 is not given.
For checking any particular factor whether they conform to the form 4k+2---> do the following
4k+2 = 36 ---> 4k = 34 --> this will NOT give you an integer value for k ---> reject. Proceed with others until you find all the values asked.
E is thus the correct answer.
Answer: (E) 6
Question: What is the greatest number which divides 29, 32, 38 and leaves same remainder in each case?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Rationale: Explanation:
HCF of differences between given numbers is the required answer
32 - 29 = 3
38 - 32 = 6
38 - 29 = 9
HCF of 3, 6, 9 is 3
Answer: Option B
Answer: (B) 3
Question: How many positive integers will divide evenly into 420?
(A) 8
(B) 12
(C) 16
(D) 18
(E) 20
Rationale: 520 = 10*13*4 = 8*5*13 = 2^3*5*13
We need to positive integers will divide evenly into 520 or simply the factors of 520
If a number is of the form 2^a*3^b*5^c,
Total factors = (a+1)(b+1)(c+1)
Therefore, number of factors of 520 = (3+1)*(1+1)*(1+1) = 4*2*2 = 12
Option CB
Answer: (B) 12
Question: A student chose a number, multiplied it by 3, then subtracted 138 from the result and got 102. What was the number he chose?
(A) 80
(B) 120
(C) 130
(D) 140
(E) 150
Rationale: Solution:
Let xx be the number he chose, then
3⋅x−138=102
3x=240
x=80
Answer A
Answer: (A) 80
Question: Ajay spends 25 per cent of his salary on house rent, 5 percent on food, 10 percent travel, 10 percent on clothes and the remaining amount of Rs.20000 is saved. What is Ajayâ€™s income?
(A) Rs.40000
(B) Rs.80500
(C) Rs.60700
(D) Rs.70500
(E) None of these
Rationale: Explanation:
Savings = Income â€“ expenditure
100 â€“ (25+5+10+10) = 50%
50% of income = Rs.20000
:. Income = 20000 x 100/50 = Rs. 40000.
Answer: Option A
Answer: (A) Rs.40000
Question: if you invested $25000 and received $35251 after 10 years. what was the interest rate?
(A) 0.15%
(B) 0.25%
(C) 0.04%
(D) 0.50%
(E) 0.40%
Rationale: P=25000
I= 35251-25000=10251
t=10 years
10251=(25000)(r)(10)
10251=250000r
10251/250000
=.04
ANSWER:0.40%
Answer: (E) 0.40%
Question: The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 14 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?
I. 3
II. 7
III. 12
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
Rationale: E
for 7 we will have mean as 7.16(approx) and median as 7 but for 3 we will have mean and median = 6.5 and for 12 we will have median and mean =8
Answer: (E) I, II, and III
Question: What is the number of integers from 1 to 1100 (inclusive) that are divisible by neither 11 nor by 35?
(A) 884
(B) 890
(C) 892
(D) 910
(E) 945
Rationale: Normally, I would use the method used by Bunuel. It's the most accurate. But if you are looking for a speedy solution, you can use another method which will sometimes give you an estimate. Looking at the options (most of them are spread out), I wont mind trying it. (Mind you, the method is accurate here since the numbers start from 1.)
In 1000 consecutive numbers, number of multiples of 11 = 1000/11 = 90 (Ignore decimals)
In 1000 consecutive numbers, number of multiples of 35 = 1000/35 = 28
Number of multiples of 11*35 i.e. 385 = 1000/385 = 2
Number of integers from 1 to 1000 that are divisible by neither 11 nor by 35 = 1000 - (90 + 28 - 2) {Using the concept of sets here) = 945
Think: Why did I say the method is approximate in some cases?
Think what happens if the given range is 11 to 1010 both inclusive (again 1000 numbers)
What is the number of multiples in this case?
E
Answer: (E) 945
Question: The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. The ratio of their present ages is:
(A) 5 : 2
(B) 7 : 3
(C) 9 : 2
(D) 13 : 4
(E) 13 : 6
Rationale: Let the present ages of father and son be x and y years respectively.
Then (x-10) = 3 (y-10) or
3y-x = 20 ------ (1)
and (x+10) = 2 (y+10) or
x-2y = 10 ----- (2)
(1) + (2) -> y = 30
Substituting y = 30 in equation (1) we get x = 70
Ratio of their ages = 70 : 30 or 7:3
ANSWER:B
Answer: (B) 7 : 3
Question: A vendor sold two things at same cost of 15 RS with one item at 25%profit and other at 20%loss,by this transaction he made profit or loss by how much
(A) 0.65 loss
(B) 0.75 loss
(C) 0.80 loss
(D) 0.55 loss
(E) 0.40 loss
Rationale: CP when profit of 25% = 12
CP when loss of 20% = 18.75
Total CP = 30.75
Total SP = 30.00
Total loss = 0.75
ANSWER:B
Answer: (B) 0.75 loss
Question: If 3 pounds of dried apricots that cost x dollars per pound are mixed with 4 pounds of prunes that cost y dollars per pound, what is the cost, in dollars, per pound of the mixture?
(A) (3x + 4y)/5
(B) (3x + 4y)/(x + y)
(C) (3x + 4y)/(xy)
(D) 5(3x + 4y)
(E) 3x + 4y
Rationale: Total Cost = Weight(in pounds) * Price/pound;
To find Total cost/pound, Divide by total pounds.
Cost of Dried apricots = 3x;
Cost of prunes = 4y;
Cost per pound = (3x + 4y)/5;
Ans is (A).
Answer: (A) (3x + 4y)/5
Question: What will be the cost of building a fence around a rectangular plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 50?
(A) A)Rs.3944
(B) B)Rs.3948
(C) C)Rs.3519.5
(D) D)Rs.3949
(E) E)Rs.3923
Rationale: Let the side of the rectangle plot be a ft.
a*b = 289 =>1.7 b*b = 289
b = 13.03
a = 22.17
Length of the fence = Perimeter of the plot = 2(a+b) = 70.39 ft.
Cost of building the fence = 70.39 * 50 = Rs.3519.5
Answer: C
Answer: (C) C)Rs.3519.5
Question: Find the sum The difference between the compound and S.I. on a certain sum of money for 2 years at 10% per annum is Rs.20 of money?
(A) 1500
(B) 1992
(C) 9921
(D) 2798
(E) 2000
Rationale: P = 20(100/10)2
=> P = 2000
Answer: E
Answer: (E) 2000
Question: What is the last number in a sequence of 5 consecutive integers, given that their sum is 335?
(A) 69
(B) 70
(C) 68
(D) 66
(E) 67
Rationale: The median is: 335/5 = 67. Middle position.
Middle position is also: 5/2 = 2.5, rounded up: 3rd position
3rd position is: 5-3 = 2 positions away from the last position
67 + 2 = 69
Answer: A
Answer: (A) 69
Question: If two numbers are in the ratio 5:2. If 10 is added to both of the numbers then the ratio becomes 5:4 then find the smallest number?
(A) 2
(B) 3
(C) 4
(D) 6
(E) 8
Rationale: 5:2
5x + 10 : 2x + 10 = 5 : 4
5[2x + 10] = 4[5x + 10]
10x + 50 = 20x + 40
20x - 10x = 50 - 40
x = 1
Then the second number is = 2X
2x = 2
Option A
Answer: (A) 2
Question: Amy invested Rs. 30000 @ 8% per annum for two years. If the interest is compounded quarterly, then the amount received by Amy at the end of the year will be?
(A) 35,147.89
(B) 35,149.87
(C) 35,148.97
(D) 35,147.98
(E) 35,149.78
Rationale: P = Rs. 30000; R = 8% p.a. = 2% per quarter; T = 2 years = 8 quarters
Amount = [30000 * (1 + 2/100)8]
= (30000 * (1.02)8) = Rs. 35,149.78
Answer: E
Answer: (E) 35,149.78
Question: Albert invested Rs. 8000 in a scheme for 2 years at compound interest rate 5% p.a. How much amount will Albert get on maturity of the fixed deposit?
(A) 8820
(B) 8828
(C) 8628
(D) 8891
(E) 8812
Rationale: Amount = [8000 * (1 + 5/100)2]
= 8000 * 21/20 * 21/20
= Rs.8820
Answer: C
Answer: (C) 8628
Question: The difference between simple and compound interest on Rs. 1200 for one year at 10% per annum reckoned half-yearly is?
(A) 2
(B) 3
(C) 4
(D) 6
(E) 7
Rationale: Explanation:
S.I. = (1200 * 10 * 1)/100 = Rs. 120
C.I. = [1200 * (1 + 5/100)2 - 1200] = Rs. 123 Difference = (123 - 120) = Rs. 3.
Answer:B
Answer: (B) 3