mathewhe/aqua_rat huggingface dataset:
Question: An athlete runs 200 meters race in 36sec. His speed is?
(A) 20km/hr
(B) 15km/hr
(C) 30km/hr
(D) 25km/hr
(E) 40km/hr
Rationale: speed = 200/36 = 50/9 m/sec
= 50/9 * 18/5 km/hr = 20km/hr
Answer is A
Answer: (A) 20km/hr
Question: When asked what the time is,a person answered that the amount of time left is 1/3 of the time already completed.what is the time.
(A) 2 PM
(B) 9 PM
(C) 3 PM
(D) 8 PM
(E) 6 PM
Rationale: A day has 24 hrs. Assume x hours have passed. Remaining time is (24 - x)
24−x=1/3x⇒x=18
Time is 6 PM
Answer:E
Answer: (E) 6 PM
Question: The speed at which a man can row a boat in still water is 15 km/hr. If he rows downstream, where the speed of current is 3 km/hr, how many seconds will he take to cover 90 meters?
(A) 12
(B) 16
(C) 17
(D) 18
(E) 20
Rationale: The speed of the boat downstream = 15 + 3 = 18 km/hr
18 km/hr * 5/18 = 5 m/s
The time taken to cover 90 meters = 90/5 = 18 seconds.
The answer is D.
Answer: (D) 18
Question: Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
(A) 4:5
(B) 3:5
(C) 6:8
(D) 3:5
(E) 5:4
Rationale: third number be x.
first no = 120% of x = 6x/5
second no = 150% of x = 3x/2
ratio in first two nos = (6x/5:3x/2)=> 4:5
ANSWER A
Answer: (A) 4:5
Question: A is thrice as efficient as B and is, therefore, able to finish a piece of work 10 days earlier than B. In how many days A and B will finish it together?
(A) 14 days
(B) 5 days
(C) 8 days
(D) 4 days
(E) 3 days
Rationale: WC = 3:1
WT = 1:3
x 3x
1/x – 1/3x = 1/8
x = 16/3
3/16 + 1/16 = 1/4 => 4 days
ANSWER:D
Answer: (D) 4 days
Question: A train speeds past a pole in 15 sec and a platform 150 m long in 25 sec, its length is?
(A) 100
(B) 150
(C) 160
(D) 225
(E) 180
Rationale: Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 150)/25 = x/15 => x = 225 m.
Answer: Option D
Answer: (D) 225
Question: A shopkeeper purchased 80 kg of potatoes for Rs. 550 and sold the whole lot at the rate of Rs. 9 per kg. What will be his gain percent?
(A) 22.6 %
(B) 30.9 %
(C) 35.6 %
(D) 39.5%
(E) 23.6 %
Rationale: C.P. of 1 kg = 550/80 = Rs. 6.875S.P. of 1 kg = Rs. 9Gain % = 2.125/6.875 * 100 = 212.5/6.875 = 30.9 %Answer: B
Answer: (B) 30.9 %
Question: We are given the positive numbers a,b,c, which form an arithmetic progresion in the given order. We know that a+b+c=9. The numbers a+1,b+1,c+3 form a geometric progression in the given order. Find c.
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Rationale: Solution:
Since a,b,c form an arithmetic progression, c=b+d=a+2d and a+b+c=a+a+d+a+2d=3a+3d=9, so a+d=b=3 and a+c=6=>c=6−a. Since a+1,b+1,c+3 form a geometric progression, we have (b+1)2=(a+1)(c+3), or 16=(a+1)(6−a+3)=(a+1)(9−a)
16=9a+9−a2−a
a2−8a+7=0, therefore a=7 or a=1. But if a=7, c=6−a=6−7=−1<0 and by definition c is positive, so a=7 is not a solution. Therefore a=1 and c=6−a=6−1=5.
Answer A
Answer: (A) 5
Question: 36 identical chairs must be arranged in rows with the same number of chairs in each row. Each row must contain at least three chairs and there must be at least three rows. A row is parallel to the front of the room. How many different arrangements are possible?
(A) 2
(B) 4
(C) 5
(D) 6
(E) 10
Rationale: Three conditions have to be satisfied.
1. The number of students per row has to be at least 3.
2. Number of row has to be at least 3.
3. Equal number of students has to be seated in a row.
The following arrangements satisfy all 3 conditions.
Arrangement 1: 3 students to a row; 12 rows.
Arrangement 2: 4 students to a row; 9 rows.
Arrangement 3: 6 students to a row; 6 rows.
Arrangement 4: 9 students to a row; 4 rows.
Arrangement 5: 12 students to a row; 3 rows.
You will observe that the number of students in a row is a factor of 36.
So, an alternative and faster approach is to list down factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, and 36.
And then start from 3 and quickly find out if the number of rows is at least 2.
Both the conditions are satisfied for the following factors : 3, 4, 6, 9, and 12. i.e., 5 arrangements.
Option C is answer
Answer: (C) 5
Question: A and B start a business with Rs.16000 and Rs.32000 respectively. Hoe should they share their profits at the end of one year?
(A) 1:2
(B) 3:4
(C) 2:5
(D) 3:7
(E) 2:1
Rationale: They should share the profits in the ratio of their investments.
The ratio of the investments made by A and B =
16000 : 32000 => 1:2
ANSWER:A
Answer: (A) 1:2
Question: How many numbers up to 300 and 700 are divisible by 2, 3 and 7 both together?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
Rationale: (700 – 300)/42 = 9 22/42
=> 9 Numbers
ANSWER:A
Answer: (A) 9
Question: In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
(A) 152
(B) 3^5
(C) 125
(D) 217
(E) 136
Rationale: The first letter can be posted in any of the 3 post boxes. Therefore, we have 3 possibilities.
Similarly, the second, the third, the fourth and the fifth letter can each be posted in any of the 3 post boxes.
Each of the 5 letters has 3 possibilities because we can post any number of letters in all of the boxes.
Therefore, the total number of ways the 5 letters can be posted in 3 boxes is
3 * 3 * 3 * 3 * 3 = 3^5
Ans: B
Answer: (B) 3^5
Question: Compound X contains elements A and B at an approximate ratio, by weight, of 2:10. Approximately how many grams of element B are there in 342 grams of compound X?
(A) 54
(B) 162
(C) 250
(D) 270
(E) 280
Rationale: total number of fractions =2+10=12
element B constitutes =10 out of 12 parts of X
So in 342 gms of X have 342*10/12=280 gms of B
and 342-280=62 gms of A.
cross check:- A/B = 62/280 =2/10 (as given)
Ans E
Answer: (E) 280
Question: Two trains are running in opposite directions with the same speed. If the length of each train is 120 m and they cross each other in 12 sec, then the speed of each train is?
(A) 30
(B) 31
(C) 36
(D) 25
(E) 24
Rationale: Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (120 + 120)/12 => x = 10
Speed of each train = 10 m/sec.
= 10 * 18/5 =- 36 km/hr.
Answer: Option C
Answer: (C) 36
Question: If n and k are integers whose product is 3600, which of the following statements must be true?
(A) n+k>0
(B) n does not equal k
(C) Either n or k is a multiple of 10
(D) If n is even, then k is odd
(E) If n is odd, then k is even
Rationale: If n is odd and k is odd, then the product is odd.
The answer is E.
Answer: (E) If n is odd, then k is even
Question: How many prime numbers exist between 260 and 280?
(A) None
(B) One
(C) Two
(D) Three
(E) Four
Rationale: Straight a way lets exclude all the Even numbers between 260 and 280.
So now the number starts from 261 to 279 (Only ODD)
261 is a divisible of 3 and next odd divisible by 3 will be 261 + 6 = 267 + 6 = 273 + 6 =279.
Also we can eliminate numbers ending with '5'
So in Odd, the excluded numbers are 261,265,267,273,279, which leave us with 263,269,271,277.
Checked the above listed four numbers are divisible by any numbers till 20.
ANSWER:E
Answer: (E) Four
Question: There are k-2 members in a certain band, including Jim and Ellen. Two members are to be selected to attend the Grammy awards ceremony. If there are 15 possible combinations in which Jim and Ellen are not selected, what is the value of k?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
Rationale: There are k-2 members in the band, and k-4 members without Jim and Ellen.
(k-4)C2 = 15
(k-4)(k-5)/2 = 15
(k-4)(k-5) = 30 = 6*5
k = 10
The answer is C.
Answer: (C) 10
Question: Solve the given two equations and find the inequality of X and Y. (a) x^4- 227= 398 and (b) y^2 + 321=346
(A) If x>y
(B) If x>=y
(C) If x<y
(D) If x<=y
(E) x=y or relationship cannot be established
Rationale: Solution:-
X=+5,-5 and Y= +5,-5
Comparing, we get X=Y, X<Y, X>Y.
Therefore, no relationship can be established between X and Y.
E
Answer: (E) x=y or relationship cannot be established
Question: Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?
(A) 177
(B) 132
(C) 176
(D) 278
(E) 267
Rationale: 2 * 22/7 * 14 = 88
88 * 1 1/2
=Rs.132
Answer:B
Answer: (B) 132
Question: Mala and Usha shared their water from a bottle. If mala and usha both drank for the same amount of time, but usha only drank 2/10 of the total bottle, what was the ratio of mala and usha speed?
(A) 1:4
(B) 2:5
(C) 4:1
(D) 3:5
(E) 2:3
Rationale: The answer is suppose to be C. 4:1. It's from the GMATPrep
Answer:option C
Answer: (C) 4:1
Question: A man has $ 10,000 to invest. he invest $4,000 at 5% and $3500 at 4%. In order to have ayearly incom of $500, he must invest the remainder at:
(A) 6%
(B) 5.4%
(C) 6.1%
(D) 4.4%
(E) 6.4%
Rationale: income from $4000 at 5% in one year = $4000of 5%= $200, income from $3500 at 4% in one year = $3500 of 4%= $140,total income= $200+$140= $340. Remaining yearly income amount =$500-$340= $160, remaining investamount= $10000-($4000+$3500)= $2500, I=PNR. 160=2500 x R x 1year, R= 0.064. therefore answer is: 6.4%, correct answer will be (E)
Answer: (E) 6.4%
Question: A train 125 m long passes a man, running at 6 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
(A) 35 km/hr
(B) 50 km/hr
(C) 45 km/hr
(D) 65 km/hr
(E) 51 km/hr
Rationale: Speed of the train relative to man
= (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr
= 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed
= (x - 6) km/hr. x - 6 = 45 ==> x = 51 km/hr.
Answer: E
Answer: (E) 51 km/hr
Question: The velocity of a falling object in a vacuum is directly proportional to the amount of time the object has been falling. If after 5 seconds an object is falling at a speed of 84 miles per hour, how fast will it be falling after 12 seconds?
(A) 204 miles per hour
(B) 212 miles per hour
(C) 216 miles per hour
(D) 1080 miles per hour
(E) 5400 miles per hour
Rationale: Since Velocity is proportional to Time
Velocity =k*Timewhere k is the constant of proportionality
Time= 5 seconds
Velocity = 5k = 84 miles per Hour
i.e. k = 17
i.e. The relation between Velocity and Time becomes
Velocity =17*Time
Time= 12 seconds
Velocity = 17*12 = 204 miles per hour
Answer: Option A
Answer: (A) 204 miles per hour
Question: In a game, a player can win by getting five consecutive tails in six throws of a coin. What is the probability of winning if the player gets a tail on the first two throws?
(A) 1/4
(B) 3/4
(C) 5/8
(D) 3/8
(E) 1/8
Rationale: The only way to win is to get a tail on each of the next three throws.
P(three tails)=(1/2)^3=1/8
The answer is E.
Answer: (E) 1/8
Question: A sells a bicycle to B and makes a profit of 50%. B sells the same bicycle to C at a profit of 25%. If the final S.P. of the bicycle was Rs. 225, find out the cost price of the bicycle for A.
(A) 237
(B) 126
(C) 971
(D) 611
(E) 120
Rationale: Explanation:
Let CP be 100
A sells at 50% profit so SP=150
B sells at 25% profit = 150 x (1+25/100) = 187.50
CP --- SP
100--- 187.50
x --- 225
CP = 225 x 100/187.50 = 120
ANSWER: E
Answer: (E) 120
Question: One farmer takes 12 hours to plow his field, but another farmer takes 8 hours to plow the same field with an equal tractor. How long would it take them together to plow the field?
(A) 12.5 hours
(B) 4 hours
(C) 4.8 hours
(D) 5 hours
(E) 5.6 hours
Rationale: First we must find the least common denominator.
1/12 + 1/8 =
2/24+3/24=5/24
They can plow the field 5/24 per hour. Now to find the actual time, Lets use T for time as a variable.
1/T = 5/24
T/1 = 24/5 = 4.8 hours
Answer is C
Answer: (C) 4.8 hours
Question: P and Q can do a work in 30 days. Q and R can do the same work in 24 days and R and P in 20 days. They started the work together, but Q and R left after 8 days. How many days more will P take to finish the work?
(A) 10
(B) 15
(C) 18
(D) 20
(E) 24
Rationale: Let work done by P in 1 day = p,
Work done by Q in 1 day = q,
Work done by R in 1 day = r
p + q = 1/30
q + r = 1/24
r + p = 1/20
Adding all the above, 2p + 2q + 2r = 1/30 + 1/24+ 1/20 = 15/120 = 1/8
=> p + q + r = 1/16
=> Work done by P,Q and R in 1 day = 1/16
Work done by P, Q and R in 8 days = 8 × (1/16) =8/16 = 1/2
Remaining work = 1 = 1/2= 1/2
Work done by P in 1 day = Work done by P,Q and R in 1 day - Work done by Q and R in 1 day
= 1/16 – 1/24 = 1/48
Number of days P needs to work to complete the remaining work = (1/2) / (1/48) = 24
Option E
Answer: (E) 24
Question: Set #1 = {J, K, L, M, E}
Set #2 = {K, L, M, N, O, P}
There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel?
(A) 1/6
(B) 1/8
(C) 1/3
(D) 1/4
(E) 1/9
Rationale: At least questions are best solved by taking the opposite scenario and subtracting it from 1. Probability of choosing no vowel from set 1 is 4/5 and set 2 is 5/6. Multiply these to get 2/3. Therefore, probability of picking at least one vowel = 1-2/3=1/3.
ANSWER : C
Answer: (C) 1/3
Question: Bobby bought 2 shares, and which he sold for $106 each. If he had a profit of 120% on the sale of one of the shares but a loss of 47% on the sale of the other share, then on the sale of both shares Bobby had...
(A) a profit of 10
(B) a profit of 8
(C) a loss of 38
(D) a loss of 10
(E) neither a profit nor a loss
Rationale: Cost of the first stock: 106/2.2=48.18, so profit from it 106-48.18=57.18
Cost of the second stock: 106/0.53=200, so loss from it 106-200=-94
Overall loss 57.18-94=-36.82
Answer: C.
Answer: (C) a loss of 38
Question: When 33 is divided by the positive integer k, the remainder is 3, for how many different values of k is this true?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: This means that 30 must be a multiple of k.
The factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30.
Out of these, k can be 5, 6, 10, 15, and 30.
The answer is E.
Answer: (E) 5
Question: If 1st October is Sunday, then 1st November will be
(A) Saturday
(B) Thursday
(C) Wednesday
(D) Tuesday
(E) None of these
Rationale: Explanation :
Given that 1st October is Sunday
Number of days in October = 31
31 days = 3 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)
Hence 1st November = (Sunday + 3 odd days) = Wednesday
Answer : Option C
Answer: (C) Wednesday
Question: In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
(A) 1
(B) 61
(C) 63
(D) 64
(E) 120
Rationale: We need to select 5 men from 7 men and 2 women from 3 women.
Number of ways to do this
= 7C5 × 3C2
= 7C2 × 3C1 [∵ nCr = nC(n-r)]
= 21 * 3 =63
C
Answer: (C) 63
Question: Which one the following fractions has the highest value?
(A) 23/77
(B) 23/78
(C) 23/84
(D) 23/87
(E) 46/150
Rationale: Among options (A) , (B) ,(C) & (D)
23/77 > 23/78 > 23/84 > 23/87
SO, (A) is highest..
Now compare (A) with (E)
(A) 23/77 ~ 29.xx%
(E) 46/150 ~ 31%
So, (E) > (A)
Hence correct answer will be (E)
Answer: (E) 46/150
Question: What is the probability of getting exactly three heads on five flips of a fair coin?
(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2
Rationale: 5 Flips of a fair coin to get = HHHTT = no. of ways this can be achieved = 5!/3!x2! = 10
Probability to get any of the above 10 arrangements (HHHTT) = (1/2)^5 = 1/32
Total probability = 1/32 x 10 = 5/16
Answer:D
Answer: (D) 5/16
Question: The average of first seven multiples of 8 is:
(A) 9
(B) 16
(C) 15
(D) 32
(E) 10
Rationale: Explanation :
(8(1+2+3+4+5+6+7)/7
= 8x28/7
= 32
Answer : D
Answer: (D) 32
Question: A rectangular field has area equal to 150 sq m and perimeter 50 m. Its length and breadth must be?
(A) 11
(B) 99
(C) 89
(D) 10
(E) 12
Rationale: lb = 150
2(l + b) = 50 => l + b = 25
l – b = 5
l = 15 b = 10
Answer:D
Answer: (D) 10
Question: A and B can complete a job in 30 and 20 days respectively. They start working together and B leaves 5 days before the work is finished. In how many days is the total work finished?
(A) 15
(B) 6
(C) 5
(D) 4
(E) 3
Rationale: A and B can do the job in 30 and 20 days individually, so together they will take 30*20/(30 + 20) = 600/50 = 12 days.
The work that A can do in 5 days is 1/6th. So, A left when 5/6th of the work was done or A left after (5/6)*12 = 10 days
Total time taken = 10 + 5 = 15 days.
ANSWER:A
Answer: (A) 15
Question: A man can row 6 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and black. What is the total distance traveled by the man?
(A) 5.98
(B) 8.98
(C) 5.76
(D) 5.87
(E) 5.82
Rationale: M = 6
S = 1.2
DS = 7.2
US = 4.8
x/7.2 + x/4.8 = 1
x = 2.88
D = 2.88 * 2 = 5.76
Answer:C
Answer: (C) 5.76
Question: The length of rectangle is thrice its breadth and its perimeter is 80 m, find the area of the rectangle?
(A) 432
(B) 300
(C) 252
(D) 992
(E) 212
Rationale: 2(3x + x) = 80
l = 30 b = 10
lb = 30 * 10 = 300
Answer:B
Answer: (B) 300
Question: The average (arithmetic mean) of six numbers is 9. If 3 is subtracted from each of four of the numbers, what is the new average?
(A) 1.5
(B) 2
(C) 7
(D) 4
(E) 4.5
Rationale: Sum of 6 numbers = 6*9 =54
If 3 is subtracted from each of four of the numbers , we subtract 3*4=12 from the total sum
Sum of 6 number after subtracting 3 from each of four of the numbers = 54 - 12 = 42
New average = 42/6 = 7
Answer C
Answer: (C) 7
Question: The speed of a boat in upstream is 50 kmph and the speed of the boat downstream is 80 kmph. Find the speed of the boat in still water and the speed of the stream?
(A) 10 kmph.
(B) 17 kmph.
(C) 15 kmph.
(D) 11 kmph.
(E) 12 kmph.
Rationale: Speed of the boat in still water = (50+80)/2 = 65 kmph. Speed of the stream = (80-50)/2
= 15 kmph.
Answer:C
Answer: (C) 15 kmph.
Question: A and B run a 1 km race. If A gives B a start of 50m, A wins by 14 seconds and if A gives B a start of 22 seconds, B wins by 20 meters. Find the time taken by A to run 1 km.
(A) 222
(B) 100
(C) 287
(D) 2976
(E) 111
Rationale: To solve these type of questions, always keep in your mind that, the ratio of the speeds of two contestents never change.
A gives B a start of 50 m means, A runs 1000 m and B runs only 950. By the time A reaches the target, B has to take 22 seconds to reach the target.
ab=1000950−14b=9801000−22bab=1000950−14b=9801000−22b
50,000 -1100b = 46550 -686b
Solving we get b = 25/3
Now Assume A's speed = x
1000950−14(25/3)=x25/31000950−14(25/3)=x25/3
x = 10
So x takes 1000/10 = 100 seconds.
Answer:B
Answer: (B) 100
Question: How many multiples of 4 are there between 16 and 112, inclusive?
(A) 21
(B) 22
(C) 23
(D) 24
(E) 25
Rationale: The multiples of 4 are from 4*4 up to 4*28.
28-4+1 = 25.
The answer is E.
Answer: (E) 25
Question: The value of 1 + 1/(4 x 3) + 1/(4 x 32) +1/(4 x 33) up to four places of decimals is ?
(A) 1.1202
(B) 1.1203
(C) 1.1204
(D) None of these
(E) Cannot be determined
Rationale: Answer
Given expression = (108 + 9 + 3 +1) / 108
= 121 / 108
= 1.1203
Correct Option: B
Answer: (B) 1.1203
Question: The sum of two positive integers multiplied by the bigger number is 204, and their difference multiplied by the smaller number is 35. the numbers are:
(A) 7,2
(B) 8,9
(C) 5,12
(D) 2,7
(E) 12,5
Rationale: x(x+y)=204
x^2+xy=204....(1)
y(x-y)=35
xy-y^2=35....(2)
x^2+Y^2=169
Thus, x=12, y=5
correct answer E
Answer: (E) 12,5
Question: A number when divided by 342 gives a remainder 47. When the same number ift divided by 19, what would be the remainder ?
(A) 9
(B) 7
(C) 6
(D) 5
(E) 3
Rationale: On dividing the given number by 342, let k be the quotient and 47 as remainder.
Then, number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.
The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.
Answer is A.
Answer: (A) 9
Question: (540 - ?) × 1/7 + 12 = 25
(A) A)227
(B) B)449
(C) C)32
(D) D)600
(E) E)3
Rationale: Explanation:
=> (540 - ?) × 1/7 + 12 = 25
=> (540 - ?) × 1/7 = 25 - 12 = 13
=> 540 - ? = 13 × 7 = 91
=> ? = 540 - 91 = 449
Answer: Option B
Answer: (B) B)449
Question: A housewife wishes to purchase three articles A, B and C from a sum of Rs. 200. The unit prices of the articles A, B and C are Rs. 20, Rs. 35 and Rs. 25 respectively. If she spends the entire amount by purchasing 5 numbers of articles of type C, what is the ratio of the number of articles purchased of type A to that of, type B?
(A) 1 : 2
(B) 2 : 1
(C) 1 : 1
(D) None of these
(E) Cannot be determined
Rationale: Explanation:
After spending Rs. 125 (25 * 5) for articles of type C, the housewife is left with Rs. 75 (200 – 125). Since this amount has to be spent in totality, she must have purchased 2 articles of type A (equivalent to Rs. 40) and 1 article of type B (equivalent to Rs. 35). Thus, the required ratio is 2 : 1.
ANSWER B
Answer: (B) 2 : 1
Question: What is the value of (12 + 22 + 32 + 42 + ----- + 102)
(A) 2237
(B) 269
(C) 268
(D) 385
(E) 291
Rationale: Explanation:
(12 + 22 + ….. + n2) = (1/6) n(n + 1) (2n + 1)
Here, n = 10
Therefore,
(12 + 22 + ….. + 102) = (1/6) 10 (10 + 1) ( 2 x 10 + 1)
= (1/6) x 10 x 11 x 21
= 385
ANSWER: D
Answer: (D) 385
Question: A money lender lent a total of $1900 to two of his customers. He charged at the rate of 5%p.a. to one of them and 6% p.a. to the other. If he earned an average of 5.67% on $1900, how much did he lend to each of the two customers?
(A) 700; 1100
(B) 1273; 627
(C) 1000; 800
(D) 1200; 800
(E) None of the above
Rationale: The method given above is the one given in the book.
However the method I used was
(1*x*5)/100 + (1900-x)*6/100 = 1900*5.67/100
Simplifying we get x=627
B
Answer: (B) 1273; 627
Question: There are three numbers. 5/7th of the first number is equal to 48% of the second number. The second number is 1/9th of the third number. If the third number is 1125, then find 25% of the first number?
(A) 168
(B) 84
(C) 42
(D) 21
(E) None of these
Rationale: Let the first number and the second number be F and S respectively.
5/2 F = 48/100 S ----> (1)
S = 1/9 * 1125 = 125
(1) => 5/7 F = 48/100 * 125
=> F = 84
25% of F = 1/4 * 84 = 21
ANSWER:D
Answer: (D) 21
Question: A reduction of 25% in the price of oil enables a house wife to obtain 5kgs more for Rs.1000, what is the reduced price for kg?
(A) s.50
(B) s.46
(C) s.49
(D) s.41
(E) s.42
Rationale: 1000*(25/100) = 250 ---- 5
? ---- 1
=> Rs.50
Answer: A
Answer: (A) s.50
Question: A customer using a certain telephone calling plan pays a fee of $25 per month, and then receives a discount of 40% on the regular charge for all calls made to country W. If calls to country W are regularly charged at $1.60 per minute for the first 3 minutes, and $0.80 per minute for each minute thereafter, what is the maximum the customer could have saved over regular prices if he was charged for 1 hour of calls made to country W in a certain month?
(A) $8.75
(B) $12
(C) $13.40
(D) $17.40
(E) $24.40
Rationale: Discount = 40% of spending
We want to maximise discount so we will need to maximise spending. We do that by assuming that 60 calls were made of 1 min each because the first minute is the most expensive.
Max discount = (40/100)* 60 * 1.6 = (40/100)*96
This would be slightly less than 40. About 38 to 39.
Saving = (Slightly less than 40) - 25 = Slightly less than 15 charged in W country
Answer (C)
Answer: (C) $13.40
Question: For any positive integer n, the sum of the first n positive integers equals [n(n+1)]/2.
What is the sum of all the even integers between 99 and 161?
(A) 1090
(B) 2060
(C) 3080
(D) 4030
(E) 5040
Rationale: 100 + 102 +...+ 160 = 31*100 + (2+4...+60) = 31*100 + 2(1+2+...+30) =
31*100 + 2(30)(31)/2 = 130*31 = 4030
The answer is D.
Answer: (D) 4030
Question: How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?
(A) 321
(B) 65
(C) 120
(D) 100
(E) 80
Rationale: 1 million distinct 3 digit initials are needed.
Let the number of required alphabets in the language be ‘n’.
Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.
Note distinct initials is different from initials where the digits are different.
For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
This n3 different initials = 1 million
i.e. n3 = 106 (1 million = 106)
=> n3 = (102)3 => n = 102 = 100
Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.
Ans: D
Answer: (D) 100