RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: A shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 20% while buying and by 10% while selling. What is his percentage profit?
A.10.22%
B.20.22%
C.21.22%
D.20%
Answer: D.20%
Explanation: The owner buys 100kg but actually gets 120kg;
The owner sells 100kg but actually gives 90kg;
Profit: (120-90)/90*100=20%
Answer: D.
Question: If the average marks of three batches of 60, 60 and 45 students respectively is 50, 55 and 60, then the average marks of all the students is
A.53.33
B.54.54
C.55
D.57
Answer: B.54.54
Explanation: Sol.
Required average = [60x50 + 60x55 + 45x60 / 60 + 60 + 45]
= [3000 + 3300 + 2700 / 165]
= 9000 / 165 = 54.54.
Answer B
Question: How many 3-digit numerals begin with a digit that represents a prime number?
A.16
B.80
C.400
D.180
Answer: C.400
Explanation: prime digits 2,3,5 and 7.
three digit numbers _ _ _
1st place can be filled in 4 ways
2nd place can be filled in 10 ways
3rd place can be filled in 10 ways
Total=4*10*10=400
Ans : C
Question: A man sells two articles for Rs.3600 each and he gains 30% on the first and loses 30% on the next. Find his total gain or loss?
A.9%
B.8%
C.2%
D.5%
Answer: A.9%
Explanation: (30*30)/100
= 9%loss
Answer: A
Question: If 133,414 cans of paint need to be placed on crates and each crate can contain at most 9 cans of paint. How many cans of paint will be on the last unfilled crate?
A.1 bottle
B.3 bottles
C.5 bottles
D.7 bottles
Answer: D.7 bottles
Explanation: The number of cans of paint that can be placed on crates of 9 cans is the first number that is divisible by 9 that occurs before 133,414.
In order to divide the sum in 9 parts, the amount must be divisible by 9
Divisibility rule of 9: The sum of the digits must be divisible by 9
Sum of digits of 133,414 = 16 and 9 is divisible by 9.
Hence, we need to remove 7 to this number for it to be divisible by 9
Correct Option: D
Question: A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is:
A.100 m
B.125 m
C.130 m
D.150 m
Answer: D.150 m
Explanation: Explanation:
Let the length of the train be x meters and its speed be y m/sec.
They, x / y = 15 => y = x/15
x + 100 / 25 = x / 15
x = 150 m.
ANSWER IS D
Question: In a college, the ratio of the number of boys to girls is 8 : 5. If there are 225 girls, the total number of
students in the college is
A.420
B.520
C.585
D.620
Answer: C.585
Explanation: Explanation:
Let the boy are 8x and Girls are 5x
=> 5x = 225
=> x = 45
Total students = 8x+5x = 13x = 13(45) = 585
Answer: C
Question: The annual interest rate earned by an investment increased by 10 percent from last year to this year. If the annual interest rate earned by the investment this year was 9 percent, what was the annual interest rate last year?
A.1%
B.1.1%
C.8.18%
D.10%
Answer: C.8.18%
Explanation: 9=1.1*x
X=8.18%
Answer C)
Question: The perimeter of a rhombus is 68 cm and one of its diagonals is 16 cm. Find its area?
A.290
B.240
C.2870
D.289
Answer: B.240
Explanation: 4a = 68 => a = 17
172 – 82 = 152
1/2 * 16 * 30 = 240
Answer: B
Question: In a single throw with two dice, find the probability that their sum is a multiple either of 3 or 4.
A.22
B.29
C.28
D.10
Answer: B.29
Explanation: Answer:B
Question: A and B walk around a circular track. They start at 9 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 11.30 a.m.?
A.8
B.17
C.16
D.15
Answer: B.17
Explanation: Sol.
Relative speed = (2 + 3) = 5 rounds per hour.
So, they cross each other 5 times in an hour and 2 times in half an hour.
Hence, they cross each other 17 times before 11.30 a.m.
Answer B
Question: A pipe can fill a cistern in 14 minutes whereas the cistern when fill can be emptied by a leak in 28 minutes. When both pipes are opened, find when the cistern will be full?
A.17
B.16
C.28
D.13
Answer: C.28
Explanation: 1/14 - 1/28 = 1/28
28 minutes
Answer: C
Question: There are two sections C and D of a class, consisting of 18 and 22 students respectively. If the average weight of sections C is 20 kg and that of sections D is 18 kg. Find the average weight of the whole class
A.18.7
B.18.8
C.18.9
D.19.9
Answer: C.18.9
Explanation: Solution
Total weight of(18+22)Students	= (18x20+22x18)Kg
= 756 kg.
Average weight of the whole class	= (756 /40 )
= 18.90.
Answer C
Question: A certain library assesses fines for overdue books as follows. On the first day that a book is overdue, the total fine is $0.05. For each additional day that the book is overdue, the total fine is either increased by $0.30 or doubled, whichever results in the lesser amount. What is the total for a book on the fifth day it is overdue?
A.$0.60
B.$0.70
C.$0.80
D.$0.90
Answer: B.$0.70
Explanation: 1st day fine - 0.05
2nd day fine - 0.05*2 = 0.1 (as doubling gives lower value)
3rd day fine - 0.1*2 = 0.2 (as doubling gives lower value)
4th day fine - 0.2 *2 = 0.4 (as doubling gives lower value we add 0.3 this time)
5th day fine - 0.4 +0.3 = 0.7 (as doubling gives higher value we add 0.3 this time)
Answer: B.
Question: A can do a job in 20 days and B can do it in 30 days. A and B working together will finish thrice the amount of work in ------- days?
A.22 days
B.22 days
C.36 days
D.72 days
Answer: C.36 days
Explanation: 1/20 + 1/30 =5/60 = 1/12
12/1 = 12 *3= 36 days
Answer: C
Question: Points M(-15,2) and N(5,-5) lie on the xy - plane. What is the midpoint of line segment MN?
A.(5,0)
B.(0,5)
C.(0, (136^(1/2))/2)
D.(-5,-3/2)
Answer: D.(-5,-3/2)
Explanation: The midpoint of a segment is exactly halfway between its endpoints and it's location can be found using the Midpoint Theorem, which states:
• The x-coordinate of the midpoint is the average of the x-coordinates of the two endpoints.
• Likewise, the y-coordinate is the average of the y-coordinates of the endpoints.
So, x=(-15+5)/2=-5 and y=(2-5)/2=-3/2 --> coordinates of the midpoint are (-5, -3/2).
Answer: D.
Question: Siddharth wants to borrow Rs.9000 at rate of interest 6% p.a. at S.I and lend the same amount at C.I at same rate of interest for two years. What would be his income in the above transaction?
A.s.21.60
B.s.21.29
C.s.32.40
D.s.21.26
Answer: C.s.32.40
Explanation: Amount of money Siddharth borrowed at S.I at 6% p.a. for two years = Rs.9,000
He lend the same amount for C.I at 6% p.a. for two years.
=> Siddharth's income = C.I - S.I
= p[1 + r/ 100]n - p - pnr/100
= p{ [1 + r/ 100]2 - 1 - nr/100
= 9,000{ [1 + 6/100]2 - 1 - 12/100}
= 9,000 {(1.06)2- 1 - 0.12} = 9,000(1.1236 - 1 - 0.12)
= 9,000 (0.0036) = 9 * 3.6 = Rs.32.40
Answer: C
Question: The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
A.4
B.5
C.6
D.8
Answer: A.4
Explanation: Sol.
Let the ten’s digit be x and unit’s digit be y,
Then, (10x + y) - (10y + x) = 36 ⇔ 9 (x - y) = 36
⇔ x - y = 4
Answer A
Question: The numbers 1, 2, 3, .....n are written in the natural order. Numbers in odd places are struck off to form a new sequence. This process is continued till only one number is left.
Q 1: If n = 1997, the number left is?
Q 2: If the number left is 512, the maximum possible value of n is
A.1023
B.3667
C.267
D.288
Answer: A.1023
Explanation: The given sequence is 1, 2, 3, ....n
After step 1, we have 2, 4, 6, 8, ......
After step 2, we have 4, 8, 12, 16, ....
After step 3, we have 8, 16, 24, 32, ......
After the last step (m th step) we have the highest power of 2m ≤ n.
1. If n = 1997, after step 10, we are left with only 1024 (as 210 ≤ 1997)
2. If the last number left is 512, the maximum possible value of n is 1023.
Answer:A
Question: Find the fourth proportion to 2,3,6
A.18
B.12
C.9
D.4
Answer: C.9
Explanation: Explanation:
2:3 :: 6:x
=> 2/3 = 6/x
=> x = 18/2
=> x = 9
Answer:C
Question: What was the day of the week on 27th April, 2006?
A.Monday
B.Sunday
C.Saturday
D.Thursday
Answer: D.Thursday
Explanation: 28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days
Jan. Feb. March April
(31 + 28 + 31 + 27 ) = 117 days
117 days = (16 weeks + 5 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 5) = 11 0 odd day.
Given day is Thursday.
Answer: Option D
Question: 4 men and 3 women are lined up in a row. What is the number of cases where they stand with each other in turn? (The number of cases in which men (or women) do not stand next to each other)
A.12
B.144
C.18
D.21
Answer: B.144
Explanation: The list should be WMWMW. Hence, from women 3! And men 4!, we get (3!)(4!)=144. Therefore, the correct answer is B.
Question: A and B invests Rs.10000 each, A investing for 8 months and B investing for all the 12 months in the year. If the total profit at the end of the year is Rs.25000, find their shares?
A.15088
B.15086
C.15017
D.15000
Answer: D.15000
Explanation: The ratio of their profits A:B = 8:12 = 2:3
Share of A in the total profit = 2/5 * 25000 = Rs.10000 Share of A in the total profit = 3/5 * 25000 = Rs.15000.Answer: D
Question: The Italian size of a suit jacket is N centimeters, where N is the linear distance between the armpits when the jacket is laid on a flat surface. The American size of a suit jacket is P inches, where P is twice the linear distance between the armpits when the jacket is laid on a flat surface. If, for a particular jacket, N = P + 15, which of the following is closest to the value of N? (One inch is approximately 2.54 centimeters.)
A.36
B.47
C.58
D.68
Answer: D.68
Explanation: Since one inch is approximately 2.54 centimeters, then N centimeters is N/2.54 inches.
Next, since the American size (P inches) is twice the Italian size (N/2.54 inches), then P=2*N/2.54.
As given that N = P + 15, then N=2*N/2.54+15 --> N=~68.
Answer: D.
Question: A can do a piece of work in 4 days of 8 hours each and B alone can do it in 8 days of 8 hours each.how long will they take it to do working together
10 2/3 hours a day?
A.4 days
B.3 days
C.2 days
D.5 days
Answer: C.2 days
Explanation: A's work per hour=1/32
B's work per hour=1/64
A & B's work per hour together=(1/32)+(1/64)=3/64
so A & B together complete the work in 64/3 hours... if they work 10 2/3=32/3 hours a day,
it will take (64/3)/(32/3) days=2 days...
ANSWER: C
Question: The average temperature for Monday, Tuesday, Wednesday and Thursday was 48 degrees and for Tuesday, Wednesday, Thursday and Friday was 46 degrees. If the temperature on Monday was 40 degrees. Find the temperature on Friday?
A.65 degrees
B.32 degrees
C.37 degrees
D.34 degrees
Answer: B.32 degrees
Explanation: M + Tu + W + Th = 4 * 48 = 192
Tu + W + Th + F = 4 * 46 = 184
M = 40
Tu + W + Th = 192 -40 = 152
F = 184 – 152 = 32
Answer:B
Question: A rectangular plot has a concrete path running inner side of the plot is used as a lawn, which has an area of 432 sq.m. If the width of the path is 4 m and the length of the plot is greater than its breadth by 2m, what is the area of the rectangular plot?
A.980 m^2
B.975 m^2
C.984 m^2
D.960 m^2
Answer: D.960 m^2
Explanation: Let width of plot be 'b', length = b+2.
Then b(b+2) - (b-8)(b-6) = 432
Thus width = 30 m, length = 32 m and area = 960 sq.m
ANSWER:D
Question: Look at this series: 8, 43, 11, 41, __, 39, 17, ... What number should fill in the blank?
A.8
B.14
C.43
D.44
Answer: B.14
Explanation: Explanation: This is a simple alternating addition and subtraction series. The first series begins with 8 and adds 3; the second begins with 43 and subtracts 2.
Answer: Option B
Question: If the area of a square with sides of length 7 centimeters is equal to the area of a rectangle with a width of 4 centimeters, what is the length of the rectangle, in centimeters?
A.4
B.8
C.12
D.13
Answer: D.13
Explanation: Let length of rectangle = L
7^2 = L*4
=> L = 49/4 = 13
Answer D
Question: When positive integer x is divided by positive integer y, the remainder is 7. If x/y = 86.1, what is the value of y?
A.72
B.71
C.70
D.30
Answer: C.70
Explanation: When positive integer x is divided by positive integer y, the remainder is 7 --> x=qy+7;
x/y=86.1 --> x=86y+0.1y (so q above equals to 86);
0.1y=7 --> y=70.
Answer: C.
Question: Alex can do a piece of in 120 days and peter can do same work in 180 days,both are working together for 30 days after that peter left the job. then in how many days Alex alone complete the remaining work?
A.17.5 DAYS
B.25 DAYS
C.15 DAYS
D.12 DAYS
Answer: A.17.5 DAYS
Explanation: First take LCM of (120,180) = 3600; Total work =3600; Alex's one day work = 3600/120 =30; peter's one day work =3600/180 = 20; One day work of both Alex+peter= 30+20 = 50; Total work completed in 30 days = 30*50=1500; remaining work is = 3600 - 1500 = 2100; remaining work completed by Alex in = 2100/120= 17.5 days. Answer =A
Question: In an IT company, there are a total of 70 employees including 50 programmers. The number of male employees is 80, including 35 male programmers. How many employees must be selected to guaranty that we have 3 programmers of the same sex?
A.10
B.25
C.55
D.35
Answer: B.25
Explanation: You could pick 70 non-programmers, 2 male programmers, and 2 female programmers, and still not have 3 programmers of the same sex. But if you pick one more person, you must either pick a male or a female programmer, so the answer is 25.
B
Question: What is the sum of all possible solutions to |x - 4|^2 + |x - 4| = 20 ?
A.4
B.5
C.6
D.7
Answer: D.7
Explanation: Denote |x - 4| as y: y^2 + y = 20 --> y = -5 or y = 4. Discard the first solution since y = |x - 4|, so it's an absolute value and thus cannot be negative.
y = |x - 4| = 4 --> x = 8 or x = -1. The sum = 7.
Answer: D.
Question: The perimeter of an isosceles right triangle is 12 + 12 sq rt 2. What is the length of the hypotenuse of the triangle?
A.5
B.6
C.7
D.8.485
Answer: D.8.485
Explanation: side of triangle is a then perimeter = a+ a +a.sqrt2 ( right angle and pythagorus) =2a +a .sqrt 2 = 12 + 12 Sqrt2 or,
a. ( 2+ sqrt2) = 12(1 + sqrt2), a= 12*(1+sqrt2)/2+sqrt2 =12*2.414/3.414 = then hypotenuse = 8.485
D
Question: A person spends 40% of his salary on food, 25% on house rent, 15% on entertainment and 5% on conveyance. If his savings at the end of the month is Rs. 3000, then his salary per month in rupees is:
A.20000
B.6000
C.8000
D.10000
Answer: A.20000
Explanation: Total expenditure = 40 + 25 + 15 + 5 = 85%
Saving = (100 - 85) = 15%
15/100 × Salary = 3000, Salary = 20000 Rs.
Answer:A
Question: If (x + 1) times (4x + 1) is an odd integer, then x must be:
A.an odd integer
B.an even integer
C.a prime number
D.a composite number
Answer: B.an even integer
Explanation: Solution:
(x + 1) times (4x + 1) means (x + 1)(4x + 1)
(x + 1)(4x + 1) = 4x^2+5x+1
For 4x^2+5x+1 to be odd, 5x+1 must be odd since 4x^2 is always even.
So 5x must be even, hence x must be even.
Therefore x is an even integer.
Answer: B
Question: Q is not equal to zero and k = (Q x n - s)/2 find n?
A.(2 x k + s)/Q
B.(2 x s x k)/Q
C.(2 x k - s)/Q
D.(2 x k + s x Q)/Q
Answer: A.(2 x k + s)/Q
Explanation: k = (Q x n - s)/2
=> Q x n - s = 2 x k
=> Q x n = 2 x k + s
=> n = (2 x k + s)/Q
ANSWER:A
Question: The average age of a committee of 9 members is 42 years. A member aged 63 years retired and his place was taken by another member aged 54 years . The average age of
present committee is;
A.39 years
B.38 years
C.36 years
D.41 years
Answer: D.41 years
Explanation: Total age of the committee = 42*9 = 378,
Total age when a member is retired
and a new one was joined = 378-63 +54, =369
Average age of present committee = 369/9 = 41.
ANSWER:D
Question: A certain university will select 1 of 6 candidates eligible to fill a position in the mathematics department and 2 of 9 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?
A.216
B.212
C.214
D.210
Answer: A.216
Explanation: 1C6*2C9 = 6*36 = 216
the answer is (A)
Question: How many ways the letters of the word ‘ARMNUR’ can be arranged?
A.230
B.235
C.360
D.380
Answer: C.360
Explanation: Total number of letters = 6 and R is repeated twice.
Required arrangements = (6! / 2!) = 360
C
Question: The population of a city increases at the rate of 4% p.a. but there is an additional annual increase of 1% in the population due to some job seekers. The percentage increase in the population after 2 years is?
A.10.85%
B.10.85%
C.10.25%
D.10.20%
Answer: C.10.25%
Explanation: 100
5 ---- I
5
0.25 --- II
-----------
10.25%
Answer:C
Question: Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards.
What is the probability that first two cards are queens and the third card drawn is an ace?
A.2
B.2/5525
C.2/5527
D.3
Answer: B.2/5525
Explanation: Let Q denote the event that the card drawn is queen and A be the event that
the card drawn is an ace. Clearly, we have to find P (QQA)
Now P(Q) = 4/52
Also, P (Q|Q) is the probability of second queen with the condition that one queen has
already been drawn. Now there are three queen in (52 - 1) = 51 cards.
Therefore P(Q|Q) = 3/51
P(A|QQ) is the probability of third drawn card to be an ace, with the condition
that two queens have already been drawn. Now there are four aces in left 50 cards.
Therefore P(A|QQ) = 4/50
By multiplication law of probability, we have
P(QQA) = P(Q) P(Q|Q) P(A|QQ)
= 4/52 × 3/51 × 4/50
= 2/5525.
Answer: Option 'B'
Question: A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?
A.111.12
B.111.18
C.131.12
D.211.12
Answer: A.111.12
Explanation: A runs 1000 m while B runs 900 m and C runs 800 m.
The number of meters that C runs when B runs 1000 m,
= (1000 * 800)/900 = 8000/9 = 888.88 m.
B can give C = 1000 - 888.88 = 111.12 m.
Answer:A
Question: A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
A.80 feet
B.85 feet
C.83 feet
D.88 feet
Answer: D.88 feet
Explanation: Given that area of the field = 680 sq. feet
=> lb = 680 sq. feet
Length(l) = 20 feet
=> 20 × b = 680
=> b = 680/20 = 34 feet
Required length of the fencing = l + 2b = 20 + (2 × 34) = 88 feet
Answer is D.
Question: The average of first 8 prime numbers is?
A.10.11
B.9.625
C.12.11
D.13.11
Answer: B.9.625
Explanation: Sum of first 8 prime nos. = 2+3+5+7+11+13+17+19=77
Average = 77/8= 9.625
Answer : B
Question: The price of 10 bat is equal to that of 4 balls. The price of 15 balls and 2 bats together is Rs. 4000. The total price of 12 chairs and 3 balls is:
A.Rs. 3200
B.Rs. 3500
C.Rs. 3700
D.Rs. 3900
Answer: D.Rs. 3900
Explanation: Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.
Then, 10x = 4y or y = 5 x.
2
15x + 2y = 4000
15x + 2 x 5 x = 4000
2
20x = 4000
x = 200.
So, y = 5 x 200 = 500.
2
Hence, the cost of 12 chairs and 3 tables = 12x + 3y
= Rs. (2400 + 1500)
= Rs. 3900.
D)
Question: The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 867 sq m, then what is the breadth of the rectangular plot?
A.65 m
B.17 m
C.16 m
D.76 m
Answer: B.17 m
Explanation: Let the breadth of the plot be b m.
Length of the plot = 3 b m
(3b)(b) = 867
3b2 = 867
b2 = 289 = 172 (b > 0)
b = 17 m.
Answer:B
Question: A money lender lends Rs.2000 for six months at 20% p.a. rate. It the interest is reckoned quarterly than find the amount given after time limit.
A.22075
B.2622
C.2677
D.2666
Answer: A.22075
Explanation: Explanation:
A = 2000(21/20)2 = 2205
Answer: A
Question: You can collect as many rubies and emeralds as you can. Each ruby is worth Rs. 4 crore and each emerald is worth Rs. 5 crore. Each ruby weighs 0.3 kg. And each emerald weighs 0.4 kg. Your bag can carry at the most 12 kg. What should you collect to get the maximum wealth?
A.20 rubies and 15 emeralds
B.40 rubies
C.28 rubies and 9 emeralds
D.None of these
Answer: B.40 rubies
Explanation: Explanation :
The limits our wealth is the capacity of bag which is 12 Kg. So we should make each unit of weight to worth the max.
Rubee: 0.3 Kg worth 4 Crore,
so 1 Kg of Rubee is worth 4/(0.3) = 40/3 = 13.33 Crore.
Emerald: 0.4 Kg worth 5 crore
so 1 Kg of Emerald is worth 5/(0.4) = 50/4 = 12.5 Crore.
As value per kilogram is more for Rubee we should pick max rubees for max wealth.
Here we can accomadate the whole 12Kg with 40 rubees which is worth 160 Crore.
Answer : B
Question: Evaluate : 980 x 436 + 980 x 764
A.1176000
B.968000
C.978000
D.117000
Answer: A.1176000
Explanation: 980 x 436 + 980 x 764 = 986 x (436 + 664) = 986 x 1200 = 117600.
Answer is A.
Question: A train overtakes 2 girls who are walking inthe opposite direction in which the train is going at the rate of 3 km/hour & 6km/hour and passes them completely in 36sec & 30 sec respectively. Find the length of the train is?
A.120 m
B.130 m
C.150 m
D.200 m
Answer: C.150 m
Explanation: Let the length of the train e x meter, and let the speed of train be y km/h, then
\inline x=\left ( y+3 \right )\frac{5}{18}\times 36 ........(1)
and \inline x=\left ( y+6\right )\frac{5}{18}\times 30 ........(2)
From eq (1) and (2), we get
\inline (y+3)\times 36 = ( y+6)\times 30
y = 12 km/h
\inline \therefore \inline x=(y+3)\times \frac{5}{18}\times 36
x= 150 m
C
Question: Find the principle on a certain sum of money at 7% per annum for 2 2/5 years if the amount being Rs.1120?
A.939.6
B.958.9
C.927.78
D.926.82
Answer: B.958.9
Explanation: 1120 = P [1 + (7*12/5)/100]
P= 958.90
Answer: B
Question: The ratio of the present age of Viju to that of Aju is 7:2. Four years from now, the ratio of the ages of Viju to Aju will be 5:2. What was Viju age 6 years ago ?
A.10 years
B.15 years
C.20 years
D.25 years
Answer: B.15 years
Explanation: Let the present age of Viju be 7x years and that of Aju be 2x years.
Then, 4 years from now
7x + 4 / 2x + 4 = 5 / 2
or 4x = 12
or x = 3
Viju present age = 7 * 3 = 21 years
Viju age 6 years ago = 21 - 6 = 15 years
answer :B
Question: A straight line in the xy- plane has y-intercept of 12 and a slope of 4/7. Given the x-coordinate of the point is 49, find it's y-coordinate.
A.27
B.16
C.29.7
D.28
Answer: D.28
Explanation: Eq of line = Y = MX +C
M=4/7
C=12
X=49
Substitute the givens:
Y=(4/7 * 49) + 12,
Y=16 + 12 = 28
Correct option is D
Question: The average age of a husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
A.20
B.17
C.21
D.19
Answer: D.19
Explanation: Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 years.
ANSWER D
Question: The cost price of an article is 64% of the marked price. Calculate the gain percent after allowing a discount of 15%?
A.32.8%
B.37.6%
C.38.5%
D.17.5%
Answer: A.32.8%
Explanation: Let marked price = Rs. 100.
Then, C.P. = RS. 64, S.P. = Rs. 85
Gain % = 21/64 * 100 = 32.8%.
Answer: A
Question: A goods train runs at the speed of 72 km/hr and crosses a 280 m long platform in 26 sec. What is the length of the goods train?
A.240
B.277
C.276
D.270
Answer: A.240
Explanation: Speed = 72 * 5/18 = 20 m/sec.
Time = 26 sec.
Let the length of the train be x meters.
Then, (x + 280)/26 = 20
x = 240 m.
Answer: A
Question: Find the number of square tiles to cover the floor of a room measuring 4.5 m * 9.5 m leaving 0.25 m space around the room. A side of square tile is given to be 25 cms?
A.422
B.576
C.429
D.428
Answer: B.576
Explanation: Floor area to be covered by tiles = 4 * 9 = 36
tiles area = 0.25 * 0.25 =0.0625
No. of tiles = 36/0.0625 = 576
Answer : B
Question: The railway track length, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
A.300 m
B.245 m
C.350 m
D.250 m
Answer: B.245 m
Explanation: speed = (45*5/18) => 25/2 m/sec
time = 30 sec
length of bridge => x
=>130+x/30 = 25/2
2(130+x)=750
x=245 m
ANSWER B
Question: An athlete runs 200 metres race in 20 seconds. What is his speed?
A.23
B.38
C.37
D.36
Answer: D.36
Explanation: Speed=Distance/Time = 200/20 = 10 m/s = 10 * 18/5 = 36 km/hr
Answer :D