RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Which symbol would be exactly at the middle of the fifth symbol from the right and ninth symbol from the left end of the following sequence of symbols ?
& * ∧ % $ # @ + - ! ∼ μ Ω Δ ± ∈ ≈ ∫ → √ ∞ ∧ @ ↓ ≥ ∈ $ + @
A.+
B._
C.≈
D."
Answer: C.≈
Explanation: Explanation:
& * ∧ % $ # @ + - ! ∼ μ Ω Δ ± ∈ ≈ ∫ → √ ∞ ∧ @ ↓ ≥ ∈ $ + @
The fifth symbol from the right end of the given sequence is “≥”, while the ninth symbol from the left end of the sequence is “-”.
The symbol at the middle of “≥” and “-” is “ ≈ ”.
ANSWER:C
Question: In a 1000 m race Usha beats Shiny by 50 m. In the same race, by what time margin Shiny beat Mercy.
who runs at 4 m/s ?
A.100 sec.
B.50 sec
C.25 sec
D.Data not sufficient
Answer: B.50 sec
Explanation: Explanation:
Speed of Shiny = 50/ 10 = 5m/s
Time taken by shiny to complete the race is B = 1000/5 = 200 sec.
Time taken by Baley to complete the race is D = 1000/4 = 250 sec.
Hence, D-B = 50 sec
Answer : B
Question: The perimeter of a rhombus is 68 cm and one of its diagonals is 16 cm. Find its area?
A.299
B.240
C.277
D.266
Answer: B.240
Explanation: 4a = 68 => a = 17
172 – 82 = 152
1/2 * 16 * 30 = 240
Answer: B
Question: One fourth of one third of two fifth of a number is 20. What will be40% of that number
A.A)140
B.B)150
C.C)180
D.D)240
Answer: D.D)240
Explanation: Explanation:
(1/4) * (1/3) * (2/5) * x = 20
then x = 20 * 30 = 600
40% of 600 = 240
Answer: Option D
Question: Find the area of a parallelogram with base 24 cm and height 16 cm.
A.200
B.384
C.345
D.244
Answer: B.384
Explanation: Area of a parallelogram = base * height = 24 * 16 = 384 cm2
Answer: Option B
Question: 18 litres of mixture contains 20% alcohol and the rest water. If 3litres of water be mixed with it, the percentage of alcohol in the new mixture would be?
A.16.67%
B.17.14%
C.18.3%
D.19.75%
Answer: B.17.14%
Explanation: Alcohol in the 18litres of mix. = 20% of 18litres = (20*18/100) = 3.6litres
water in it = 18-3.6 = 14.4litres
new quantity of mix. = 18+3 = 21 litres
Quantity of alcohol in it = 3.6litres
Percentage of alcohol in new mix. = 3.6*100/21 = 17.14%
Answer is B
Question: a, b, and c are integers and a<b<c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4)*b. The median of set Q is (6/8)*c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?
A. 3/8
B. 1/2
C. 11/16
D. 5/8
Answer: D. 5/8
Explanation: The answer isC: 11/16.
The key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean. For example, the mean and median of a set consisting of x, x+1, x+2, ... ,y will always be (x+y)/2.
For set S, consisting of numbers (a, a+1,...,b), the median is given to be 3/4*b:
(a+b)/2 = (3/4)*b
a = b/2
For set Q, consisting of numbers (b, b+1,...,c), the median is given to be 6/8*c:
(b+c)/2 = (6/8)*c
b = (1/2)*c
For set R, consisting of numbers (a, a+1,...c), the median needs to be found:
a = b/2 = (1/2*c)/2 = (1/4)*c
Median = (a + c)/2 = (1/4*c + c)/2 = (5/4)*c/2 = (5/8)*c (ANSWER D)
Question: A jogger running at 9 kmph along side a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
A.3.6 sec
B.18 sec
C.36 sec
D.72 sec
Answer: C.36 sec
Explanation: Speed of train relative to jogger
= (45 – 9) km/h = 36 km/h
= (36 × 5⁄18) m/sec = 10 m/sec
Distance to be covered = (240 + 120) m = 360 m.
∴ Time taken = (360⁄10) sec = 36 sec.
Answer C
Question: A monkey ascends a greased pole 20 meters high. He ascends 2 meters in the first minute and then slips down 1 meter in the alternate minute. If this pattern continues until he climbs the pole , in how many minutes would he reach at the top of the pole ?
A.30th minute
B.31st minute
C.37th minute
D.32nd minute
Answer: C.37th minute
Explanation: The money is climbing 1meter in 2min. This pattern will go on till he reaches 10meters. I mean this will continue for first 18*2=36mins. he would have reached 18meters. After that he will climb 2meters and he will reach the pole. So total time taken = 36 + 1 = 37mins.
So, asnwer will be C
Question: The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?
A.1m
B.5m
C.8m
D.2m
Answer: D.2m
Explanation: Let the sides of the rectangle be l and b respectively.
From the given data,
(√l2 + b2) = (1 + 108 1/3 %)lb
=> l2 + b2 = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l2 + b2)/lb = 25/12
12(l2 + b2) = 25lb
Adding 24lb on both sides
12l2 + 12b2 + 24lb = 49lb
12(l2 + b2 + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)2 = 49lb
=> 12(14)2 = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l - b = 8 - 6 =2m.
Answer: D
Question: There are 9 baskets numbered from 1 to 9 and filled with apples. 10 children are asked to pick apples one by one from each basket such that the number of apples picked by each child from each basket is equal to the number marked on the basket. If there were 1000 apples in total and the baskets were filled in such a way that none of the basket became empty in the process, how many apples were left in the end?
A.440
B.550
C.660
D.770
Answer: B.550
Explanation: Each child takes a total of 1+2+3+4+5+6+7+8+9 = 9*10/2 = 45 apples.
The total number of apples taken by 10 children is 10*45 = 450
The number of apples left in the end is 1000 - 450 = 550 apples.
The answer is B.
Question: A certain plant was 11 feet long on the day it was plantedthe length grows by a constant amount everyday(inc. the first day it was planted).
if this plant grew by 60% b/w the 4th day and 10 th day after it was planted ,how many feet per day does the plant grow
A.a) 2
B.b) 4.5
C.c) 1.83
D.d) 5.5
Answer: C.c) 1.83
Explanation: If the plant is 11 feet tall when it is planted, it will be 11 + k feet tall at the end of the day (where k is the constant length it grows by every day).
Similarly, at the end of the 4th day the plant must be 11 + 4k feet tall and at the end of the 10th day it must be 11 + 10k feet tall.
Therefore 6k = 0.6 (11 + 4k)
=> k = 1.83
Therefore the plant grows by 1.83 feet every day.
Option (C)
Question: The equation y = ax*b* + c produces the
following (x, y) coordinate pairs: (0, 2),
(1, 7), and (2, 42). What is the value of abc?
A.10
B.20
C.30
D.40
Answer: C.30
Explanation: To begin, plug x = 0 and y = 2 into the equation y = axb + c. Note that the first term
drops out:
2 = a(0)b + c
2 = c
Now you can substitute 2 for c in the original equation, giving you y = axb + 2. Next, plug in
x = 1 and y = 7. Notice that everything drops out except for the coefficient of the first term:
7 = a(1)b + 2
5 = a
You can now substitute 5 for a in the equation, giving you y = 5xb + 2. Now plug in x = 2
and y = 42:
42 = 5(2)b + 2
40 = 5(2)b
8 = 2b
23 = 2b
3 = b
Finally, you can see that abc = (5)(3)(2) = 30.
correct answer C)30
Question: Sheela deposits Rs. 3800 in bank savings account. If this is 22% of her monthly income. What is her monthly income in?
A.22000
B.20000
C.25123
D.17272
Answer: D.17272
Explanation: Explanation :
22% of income = Rs. 3800
100% of income = 3800 X 100/22 = Rs. 17272
Answer : D
Question: Three distinct single digit no A,B and c are in G.M ,if abs(x) for real x is the absolute value of x(x , if x is positive or zero,and -x if x is negative), then the number of different possible values of abs (A+B -C) is;
A.3
B.6
C.5
D.4
Answer: B.6
Explanation: 1,2,4
2,4,8
4,2,1
8,4,2
1,3,9
9,3,1
total=6
ANSWER:B
Question: The average age of three boys is 120 years and their ages are in proportion 2:6:12. What is the age in years of the youngest boy?
A.8
B.9
C.7
D.12
Answer: D.12
Explanation: 2x + 6x + 12x = 120
x =6
2x = 12
Answer:D
Question: Village X has a population of 76000, which is decreasing at the rate of 1200 per year. Village Y has a population of 42000, which is increasing at the rate of 800 per year .in how many years will the population of the two villages be equal?
A.15
B.17
C.11
D.18
Answer: B.17
Explanation: Let the population of two villages be equal after p years
Then, 76000-1200p=42000+800p
2000p=34000
p=17
Answer is B.
Question: Tough and Tricky questions: Combinations.
Six students - 3 Girls and 3 boys - are to sit side by side for a makeup exam. How many ways could they arrange themselves given that no two boys and no two girls can sit next to one another?
A.12
B.72
C.34
D.240
Answer: B.72
Explanation: Six students - 3 boys and 3 girls - are to sit side by side for a makeup exam. How many ways could they arrange themselves given that no two boys and no two girls can sit next to one another?
A) 12
B) 36
C) 72
D) 240
E) 720
The boys (B) and girls(G) can either be arranged as BGBGBG or GBGBGB.
Within the arrangement BGBGBG or GBGBGB, the boys can be arranged 3! ways and girls can be arranged 3! ways.
Therefore total number of ways boys and girls can be arranged = 2*3!*3!= 72 ways
Answer:B
Question: If in a $150 mln portfolio of stock, twice as much is invested in stock A as is invested in all the other stocks, how much is invested in all the other stocks?
A.50mln
B.60mln
C.70mln
D.80mln
Answer: A.50mln
Explanation: Let x denote the amount invested in all the other stocks. Then the amount invested in stock A is 2x. As we have the $150 mln portfolio, x+2x=3x=150 mln from where x=50 mln.
Answer: A
Question: On a sum of money, the S.I. for 2 years is Rs. 660, while the C.I. is Rs. 696.30, the rate of interest being the same in both the cases. The rate of interest is?
A.12%
B.32%
C.72%
D.11%
Answer: D.11%
Explanation: Difference in C.I. and S.I for 2 years = (696.30 - 660) = Rs. 36.30.
S.I for one year = Rs. 330.
S.I. on Rs. 330 for 1 year = Rs. 36.30
Rate = (100 * 36.30)/(330 * 1) = 11%
Answer:D
Question: All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A?
A.1289
B.1198
C.281
D.1184
Answer: D.1184
Explanation: Explanation:
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k - 5k = 3k.
Quantity of water in container C = 8k - 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k - 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.
Answer: Option D
Question: A and B can do a piece of work in 8 days. B and C can do it in 12 days and A and C in 16 days. Working together they will complete the work in how many days?
A.3.33 days
B.8.33 days
C.3.99 days
D.7.38 days
Answer: D.7.38 days
Explanation: A + B = 1/8
B + C = 1/12
C + A = 1/16
-------------------------------
2(A + B +C) = 1/8 + 1/12 + 1/16 = 13/48
A + B +C = 13/96
96/13 = 7.38 days
Answer:D
Question: A gives B as many as rupees B has and C as many as rupees C has.Similarly B gives A as many as rupees A has and C as many as rupees C has, C similarly gives B as many as rupees B has and A as many as rupees A has..Now finally each of them has 32 rupees.how much rupees does A has initially?
A.52
B.56
C.60
D.62
Answer: A.52
Explanation: initially
A
B
C
after A gives
A has ->A-B-C
B has ->2B
C has ->2C
after B gives
A has ->2A-2B-2C
B has ->-A+3B-C
C has ->4C
AFTER C gives
A has ->4A-4B-4C=32---------1
B has ->-2A+6B-2C=32----------2
C has ->-A-B-7C=32-----------3
ON solving 1,2,3 A=52
ANSWER:A
Question: What least no. must be subtracted from 50248 so that remaining no.is divisible by 20?
A.3
B.2
C.4
D.8
Answer: D.8
Explanation: Explanation:
On dividing 50248 by 20 we get the remainder 8, so 8 should be subtracted
Option D
Question: On the shelf, there are four books of poetry, four novels, and four reference works. Suppose we picked two books at random. What is the probability that we pick one novel and one reference work?
A.2/15
B.4/21
C.6/25
D.8/33
Answer: D.8/33
Explanation: We could pick one novel and then one reference work, or, we could choose one reference work and then one novel.
P(one novel and one reference work) = 4/12*4/11 + 4/12*4/11 = 8/33
The answer is D.
Question: A, B and C can do a piece of work in 7 days, 14 days and 28 days respectively. How long will they taken, if all the three work together?
A.4 days
B.3 days
C.6 days
D.8 days
Answer: A.4 days
Explanation: A
4 days
1/7 + 1/14 + 1/28 = 7/28 = 1/4 => 4 days
Question: Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 10 minutes earlier than usual. How much time did the man spend walking?
A.45 minutes
B.50 minutes
C.55 minutes
D.52minutes
Answer: C.55 minutes
Explanation: As they arrived 10 minutes earlier than usual, they saved 10 minutes on round trip from home to station (home-station-home) --> 5 minutes in each direction (home-station) --> wife meets husband 5 minutes earlier the usual meeting time --> husband arrived an hour earlier the usual meeting time, so he must have spent waking the rest of the time before their meeting, which is hour-5minutes=55 minutes.
Answer: C
Question: In a race of 1000 m, A can beat by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m?
A.127.7 m
B.127.5 m
C.197.5 m
D.177.5 m
Answer: B.127.5 m
Explanation: When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.
When B runs 900 m, distance that C runs = (900 * 700)/800 = 6300/8 = 787.5 m.
In a race of 1000 m, A beats C by (1000 - 787.5) = 212.5 m to C.
In a race of 600 m, the number of meters by which A beats C = (600 * 212.5)/1000 = 127.5 m.
Answer: B
Question: It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. What is the ratio of the speed of the train to that of the car?
A.3:4
B.3:0
C.3:8
D.3:2
Answer: A.3:4
Explanation: Let speed of the train = x kmph and speed of the car = y kmph
Time needed for traveling 600 km if 120 km by train and the rest by car = 8 hr
⇒120x+(600−120)y=8⇒120x+480y=8.............(Equation 1)⇒15x+60y=1.............(Equation 1)
Time needed for traveling 600 km if 200 km by train and the rest by car = 8 hr 20 min
⇒200x+(600−200)y=82060=813=253⇒200x+400y=253⇒8x+16y=13⇒24x+48y=1.............(Equation 2)Solving Equation1 and Equation2Here Equation1 = Equation2 = 1⇒15x+60y=24x+48y⇒12y=9x⇒4y=3x⇒xy=34⇒x:y=3:4
Answer : A
Question: In a group of cows and hens, the number of legs are 14 more than twice the number of heads. The number of cows is
A.5
B.7
C.10
D.12
Answer: B.7
Explanation: Explanation:
Let the number of cows be x and the number of hens be y.
Then, 4x + 2y = 2 (x + y) + 14 4x + 2y = 2x + 2y + 14 2x = 14 x = 7.
Answer: B
Question: Two trains, each 150 m long, moving in opposite directions, cross other in 8 sec. If one is moving twice as fast the other, then the speed of the faster train is?
A.90
B.98
C.60
D.88
Answer: A.90
Explanation: Let the speed of the slower train be x m/sec.
Then, speed of the train = 2x m/sec.
Relative speed = ( x + 2x) = 3x m/sec.
(150 + 150)/8 = 3x => x = 25/2.
So, speed of the faster train = 50/2 = 25 * 18/5 = 90 km/hr.
Answer:A
Question: The sum of three consecutive odd numbers is always divisible by:
I. 2 II. 3 III. 5 IV. 6
A.Only I
B.Only II
C.Only I and II
D.Only II and IV
Answer: B.Only II
Explanation: Let the three consecutive odd numbers be (2x+1), (2x+3) and (2x+5). Their sum = (6x=9)= 3 (2x+3), which is always divisible by 3.
Answer: B
Question: 70, 75, 80, 85, 90, 105, 105, 130, 130, 130
The list shown consists of the times, in seconds, that it took each of 10 schoolchildren to run a distance of 400 on of meters. If the standard devastation of the 10 running times is 27.4 seconds, rounded to the nearest tenth of a second, how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times?
A.one
B.two
C.three
D.four
Answer: B.two
Explanation: The most time consuming part in this question is to define the mean. Under exam pressure and time pressure it is very easy to make mistake.
it is easier to group numbers: 130*3=390; 105*2=210; 75+85=160; 70+80=150; 90;
Next stage combine results, again using more convenient ways to calculate: 390+210=600; 160+150=310; 90. 600+310+90=1000. Since there are 10 numbers the mean is 100. Questions asks to find the quantity of numbers one SD BELOW the mean, which is 100-27,4=72,6. There are only two numbers below 72,6. The answer is B
Question: The average of first 10 odd numbers is?
A.11
B.10
C.19
D.16
Answer: B.10
Explanation: Sum of 10 odd no. = 100
Average = 100/10 = 10
Answer: B
Question: What percent is 2 minutes 24 seconds of an hour?
A.6%
B.2%
C.4%
D.5%
Answer: C.4%
Explanation: %=144/60×60 = 4%
Answer: C
Question: What approximate value should come in place of the question mark (?) in the following equation?
1.542 × 2408.69 + 2134.632 = ?
A.4600
B.4800
C.5800
D.6400
Answer: C.5800
Explanation: Explanation 3700 + 2100 = 5800
Answer C
Question: Solve for x and check: 15x = 165
A.98
B.165
C.180
D.253
Answer: B.165
Explanation: Solution : Dividing each side by 15, we obtain
(15x/15)=(165/15)
Therefore : x=11
Check :
15x=165
(15*11)=165
165=165
ANSWER : B
Question: If in a certain sequence of consecutive multiples of 50, the median is 825, and the greatest term is 1000, how many terms that are smaller than 825 are there in the sequence?
A.6
B.4
C.8
D.12
Answer: B.4
Explanation: Since the median is 825 we know there must be a even number of integers because 50 is not a multiple of 825.
So the list around 825 must go. 700 750 800 (825) 850 900 950 1000
Since we know there are 4 numbers greater than 825 then there must be 4 numbers less then 825.
answer:B
Question: A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 4320 interest only for the period. The principal amount borrowed by him was:
A.s. 12,000
B.s. 10,000
C.s. 15,000
D.s. 18,000
Answer: A.s. 12,000
Explanation: Principal = Rs. (100 x 4320)/(12X3) = Rs. 12000. Answer: Option A
Question: Rs.8000 become Rs.9261 in a certain interval of time at the rate of 5% per annum of C.I. Find the time?
A.8
B.7
C.6
D.3
Answer: D.3
Explanation: 9261 = 8000(21/20)N
(21/20)3 = (21/20)N => N = 3
Answer:D
Question: If 6^k + 6^k = (6^9)^(6^9) - 6^k, then k = ?
A. 11/3
B. 11/2
C. 242
D. 6^11 - 1
Answer: D. 6^11 - 1
Explanation: 6^k + 6^k = (6^9)^6^9 - 6^k
6*(6^k) = 6 ^ (36 * 6^9 ) = 6^(6^2 *6^9) = 6^(6^11)
6^k+1 = 6^(6^11)
So k+1 = 6^11
So k = 6^11 -1
Answer is D
Question: In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
A.100
B.150
C.50
D.250
Answer: B.150
Explanation: Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.
Then, sum of their values = Rs. ((25x/100) + (10 x 2x/100)+(5 x 3x)/100= Rs. 60x/100
Therefore, 60x/100 = 30
x = 30 x 100/60 = 50.
Hence, the number of 5 p coins = (3 x 50) = 150.
Answer is B.
Question: A 300 m long train crosses a platform in 30 sec while it crosses a signal pole in 18 sec. What is the length of the platform?
A.286 m
B.350 m
C.200 m
D.278 m
Answer: C.200 m
Explanation: Speed = 300/18 = 50/3 m/sec.
Let the length of the platform be x meters.
Then, (x + 300)/30 = 50/3
x = 200 m.
Answer:C
Question: If 4 men and 1 boy working together, can do two times as much work per hour as a man and a boy together. Find the ratio of the work done by a man and that of a boy for a given time?
A.3:2
B.2:1
C.3:3
D.1:2
Answer: D.1:2
Explanation: 4M + 1B =2(1M + 1B)
4M + 1B = 2M + 2B
2M = 1B
The required ratio of work done by a man and a boy = 1:2
D
Question: What annual installment will discharge a debt of Rs.1092 due in 3 years at 12% simple interest?
A.442
B.441
C.440
D.320
Answer: A.442
Explanation: Let the installment be Rs. X
If he keeps the total amount for 3 years, he have to pay the interest
of Rs. (1092*12*3)/100.
But he paid Rs. X after 1 year.
So he need not to pay interest for Rs. X for remaining 2 years. i.e he
need not to pay Rs. (X*12*2)/100.
And he paid another Rs. X after 2 years.
So he need not to pay interest for for remaining 1 years. i.e. Rs.
(X*12*1)/100.
So The total interest he has to pay is Rs. ((1092*12*3)/100) - [
((X*12*2)/100) + ((X*12*1)/100) ].
So, The total amount he has to pay is Rs. 1092 + Rs.((1092*12*3)/100) -
[ ((X*12*2)/100) + ((X*12*1)/100) ].
The total amount he paid = 3X. ( because he paid Rs.X for three times)
1092 + ((1092*12*3)/100) - [ ((X*12*2)/100) + ((X*12*1)/100) ] = 3X
BY solving, we will get X = 442.
ANSWER:A
Question: Divide Rs. 2379 into 3 parts so that their amounts after 2, 3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple interest. The first part is?
A.3378
B.627
C.276
D.828
Answer: D.828
Explanation: Let the parts be x, y and [2379 - (x + y)]
x + (x * 2 * 5/100) = y + (y * 3 * 5/100) = z + (z * 4 * 5/100)
11x/10 = 23y/20 = 6z/5 = k
x = 10k/11, y = 20k/23, z = 5k/6
But x + y + z = 2379
10k/11 + 20k/23 + 5k/6 = 2379
k = (2379 * 11 * 23 * 6)/3965 = (3 * 11 * 23 * 6)/5
x = [10/11 * (3 * 11 * 23 * 6)/5] = 828
Hence, the first part is Rs. 828.
Answer: D
Question: How many numbers are there between 200 and 800 which are divisible by both 5 and 7?
A.35
B.16
C.17
D.can’t be determined
Answer: C.17
Explanation: Answer C
Solution
In the given range, the last number which is divisible by both 5 and 7. i.e., 35 is 210 and the highest number is 770. So the total number of numbers between 200 and 800 which are divisible by both 5 and 7 is
(770-210/35)+ 1 = 17
Hence option C
Question: What is the greatest value of n such that 16^n is a factor of 18! ?
A.1
B.2
C.3
D.4
Answer: A.1
Explanation: Another way to solve this question :
We know that 18 = 2 *3^2.
Hence, we need to find total number of 3s in 18! ( As out of 3 and 2 , we will have count of 3 least)
No. of 3s = 18/3 =6/3 =2.
Add the numbers in Bold, you will get total number of 3s = 8.
Hence, total number of 2^1 = 2/2 =1
Hence, answer is 1. A
Question: The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is?
A.51.4
B.55
C.56.4
D.52.2
Answer: A.51.4
Explanation: Let the required mean score be x
20*80 + 25*31 + 55*x = 52*100
1600+775+55x = 5200
x = 51.4
Answer is A
Question: The expression x#y denotes the product of the consecutive multiples of 3 between x and y, inclusive. What is the sum of the exponents in the prime factorization of 21#33?
A.12
B.13
C.14
D.15
Answer: C.14
Explanation: First, let's translate the expression 21#33, using the definition given:
21#33 = 21×24×27×30×33
We need the prime factorization of this product.
Let's factor out 3 from each multiple.
21#33 = 3^6(7×8×9×10×11)
Now let's replace each consecutive integer with its prime factorization:
21#33 = 3^5(7×2^3×3^2×(2×5)×11)
Let's group the prime bases:
21#33 = 2^4×3^7×5×7×11
The sum of the exponents is 4 + 7 + 1 + 1 + 1 = 14
The answer is C.
Question: What is the equation of a circle of radius 8 units centered at (3, 2)?
A.x^2 + y^2 + 6x – 4y = 23
B.x^2 + y^2 - 6x + 4y = 23
C.x^2 + y^2 + 6x + 4y = 23
D.x^2 + y^2 - 6x – 4y = 51
Answer: D.x^2 + y^2 - 6x – 4y = 51
Explanation: the equation of a circle with center at (a,b) and radius R is (X-a)^2+(Y-b)^2=R^2
=> answer = (X-3)^2+(Y-2)^2=8^2
solving,,we get the equation in answer choice D