RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: The average mark of a class of twenty five students is 66. If five students whose marks are 32,28, 30, 26 and 34 are removed , then find the average mark of the remaining students of the class.
A.75
B.74
C.57
D.70
Answer: A.75
Explanation: Exp. Total mark of 25 students = 66*25 = 1650,
Total mark after the removal of 5 students = 1650 –(32 +28+30+26+34 )
= 1650 – 150 = 1500
Average mark = 1500/(25-5) = 1500/ 20 = 75
Answer: A
Question: If 40% of a number is 28, then the number is
A.number is 60
B.number is 70
C.number is 90
D.number is 80
Answer: B.number is 70
Explanation: Solution:
Let "a" be the number we seek. We get
40% × a = 28
2 × 40% × a = 2 × 28 = 56
80% × a = 56
10% × a = 56 : 8 = 7
100% × a = 7 × 10 = 70
a = 70
Answer B
Question: A man purchases 8 pens for Rs.9 and sells 9 pens for Rs.8, how much profit or loss does he make?
A.20.97
B.20.98
C.90.97
D.20.95
Answer: B.20.98
Explanation: 81 ---- 17
100 ----- ? è
20.98%loss
Answer: Option B
ExplanatiB
Question: In the city of San Durango, 90 people own cats, dogs, or rabbits. If 30 people owned cats, 30 owned dogs, 50 owned rabbits, and 14 owned exactly two of the three types of pet, how many people owned all three?
A.1
B.2
C.3
D.4
Answer: C.3
Explanation: Total = C + D + R - (CD + DR + CR) - 2CDR
90 = 30 + 30 + 50 - (14) - 2x
x = 3
So, answer will be C
Question: If the speed of X inch per second is equivalent to the speed of Y foot per hour, what is Y in terms of X? (1 foot =12 inch)
A.120X
B.240X
C.360X
D.300X
Answer: D.300X
Explanation: X inch per second -->
--> 3,600X inch per hour(as there are 3,600 seconds in one hour);
--> 3,600X/12 = 300X foot per hour (as there are 12 inches in one foot).
Answer: D.
Question: The two lines y=x and x=-7 intersect on the coordinate plane. What is the value of the area of the figure formed by the intersecting lines and the x-axis?
A.18.5
B.21.5
C.24.5
D.27.5
Answer: C.24.5
Explanation: The point of intersection is (-7,-7).
The triangle has a base of length 7 and a height of 7.
area = (1/2)*base*height = (1/2)*7*7 = 24.5
The answer is C.
Question: What will come in place of the question mark (?) in the following question?
112.36 + 225.05 + ? = 815.30
A.474.46
B.477.89
C.524.41
D.564.40
Answer: B.477.89
Explanation: Explanation:
815.30 - ( 112.36 + 225.05) = ?
= 477.89
ANSWER: B
Question: An amount of Rs. 65000 is invested in two types of shares. The first yields an interest of 9% p.a and the second, 11% p.a. If the total interest at the end of one year is 9 3/4 %, then the amount invested in each share was?
A.s. 52500; Rs. 47500
B.s. 40625; Rs. 24375
C.s. 72500; Rs. 27500
D.s. 82500; Rs. 17500
Answer: B.s. 40625; Rs. 24375
Explanation: Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (65000 - x). Then,
(x * 9 * 1)/100 + [(65000 - x) * 11 * 1]/100 = (65000 * 39/4 * 1/100)
(9x + 715000 - 11x)/100 = 39000/4 = 12675/2
x = 40625
Sum invested at 9% = Rs. 40625
Sum invested at 11% = Rs. (65000 - 40625) = Rs. 24375.
ANSWER:B
Question: Ajith and Rana walk around a circular course 115 km in circumference, starting together from the same point. If they walk at speed of 4 and 5 kmph respectively, in the same direction, when will they meet ?
A.after 20 hours
B.after 115 hours
C.after 115 minutes
D.after 20 minutes
Answer: B.after 115 hours
Explanation: Rana is the faster person. He gains 1 km in 1 hour. So Rana will gain one complete round over
Ajith in 115 hours. i.e. they will meet after 115 hours.
ANSWER:B
Question: How much time will take for an amount of Rs. 360 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
A.5 years
B.6 years
C.7 years
D.12 years
Answer: A.5 years
Explanation: Time = (100 * 81) / (360 * 4.5) = 5 years
ANSWER:A
Question: The average age of a class of 32 students is 16 yrs. if the teacher's age is also included, the average increases by one year. Find the age of the teacher
A.34 Years
B.37 Years
C.49 Years
D.51 Years
Answer: C.49 Years
Explanation: Total age of students is 32X16 = 512 Years
Total age inclusive of teacher = 33X (16+1) = 561
So, Teacher's age is 561-512 = 49 Yrs
There is a shortcut for these type of problems
Teacher's age is 16+(33X1) = 49 Years
C
Question: In a family 7 children dont eat spinach, 6 dont eat carrot, 5 dont eat beans, 4 dont eat spinach & carrots, 3 dont eat carrot & beans, 2 dont eat beans & spinach. One doesnt eat all 3. Find the no. of children.
A.9
B.10
C.11
D.12
Answer: B.10
Explanation: do not spinach only=2
do not carrot only=0
do not beans only=1
do not spinach & carrots only=3
do not beans & carrots only=2
do not spinach & beans only=1
do not all three=1
total 10 children(2+0+1+3+2+1+1)
ANSWER:B
Question: To deliver an order on time, a company has to make 25 parts a day. After making 25 parts per day for 3 days, the company started to produce 5 more parts per day, and by the last day of work 100 more parts than planned were produced. Find how many parts the company made and how many days this took.
A.530
B.675
C.330
D.987
Answer: B.675
Explanation: Let x e the number of days the company worked. Then 25x is the number of parts they planned to make. At the new production rate they made:
3⋅25+(x−3)⋅30=75+30(x−3)
Therefore: 25x=75+30(x−3)−100
25x=75+30x−90−100
190−75=30x−25
115=5x
x=23
So the company worked 23 days and they made 23⋅25+100=675 pieces.
Question: A boy has six trousers and 13 shirts. In how many different ways can he select a trouser and a shirt?
A.238 ways
B.87 ways
C.66 ways
D.78 ways
Answer: D.78 ways
Explanation: Explanation:
The boy can select one trouser in six ways.
The boy can select one shirt in 13 ways.
The number of ways in which he can select one trouser and one shirt is 6 * 13 = 78 ways.
Answer: D
Question: A man 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.45
B.60
C.75
D.90
Answer: D.90
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.
Answer D
Question: From a group of boys and girls, 15 girls leave. There are then left 2 boys for each girl. After this, 45 boys leave. There are then 5 girls for each boy. Find the number of girls in the beginning?
A.40
B.10
C.50
D.30
Answer: A.40
Explanation: Let at present there be x boys.
Then, no of girls at present=5x
Before the boys had left: no of boys=x+45
And no of girls=5x
X+45=2*5x
9x=45
x=5
no of girls in the beginning=25+15=40
Answer is A
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 42 degrees. Find the temperature on Friday?
A.65 degrees
B.73 degrees
C.37 degrees
D.34 degrees
Answer: D.34 degrees
Explanation: M + Tu + W + Th = 4 * 48 = 192
Tu + W + Th + F = 4 * 46 = 184
M = 42
Tu + W + Th = 192 -42 = 150
F = 184 – 150 = 34
Answer:D
Question: If two positive numbers are in the ratio 1/10 : 1/7, then by what percent is the second number more than the first?
A.67%.
B.42.8%.
C.60%.
D.68%.
Answer: B.42.8%.
Explanation: Given ratio = 1/10 : 1/7 = 7 : 10
Let first number be 7x and the second number be 10x.
The second number is more than first number by 3x.
Required percentage = 3x/7x * 100 = 42.8%.
Answer: B
Question: How many integers between 100 and 180, inclusive, cannot be evenly divided by 3 nor 5?
A.39
B.41
C.43
D.45
Answer: C.43
Explanation: The total numbers between 100 and 180, inclusive, is 81.
3*34=102 and 3*60=180 so the number of multiples of 3 is 27.
5*20=100 and 5*36=180 so the number of multiples of 5 is 17.
However, the multiples of 15 have been counted twice.
15*7=105 and 15*12=180 so the number of multiples of 15 is 6.
Thus the total number is 81-27-17+6=43.
The answer is C.
Question: Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is
A.5/9
B.5/12
C.1/36
D.7/12
Answer: B.5/12
Explanation: Using question number 11 and 12, we get the probability as
1 - (1/36 + 5/9) = 5/12
ANSWER:B
Question: Two persons start running simultaneously around a circular track of length 300 m from the same point at speeds of 15 km/hr and 25 km/hr. When will they meet for the first time any where on the track if they are moving in opposite directions?
A.11
B.10
C.28
D.27
Answer: D.27
Explanation: Time taken to meet for the first time anywhere on the track
= length of the track / relative speed
= 300 / (15 + 25)5/18 = 300* 18 / 40 * 5 = 27 seconds.
Answer: D
Question: A 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform?
A.399
B.267
C.350
D.788
Answer: C.350
Explanation: Speed = [300 / 18] m/sec = 50/3 m/sec.
Let the length of the platform be x meters.
Then, x + 300 / 39 = 50/3
3(x + 300) = 1950 è x = 350m.
Answer:C
Question: How many ltrs of water must beadded to 16ltrs of milk and water containing 10% water to make it 20% water in it?
A.1 liter
B.2 liter
C.3 liter
D.5 liter
Answer: B.2 liter
Explanation: By rule of alligation,
% Concentration
of water in pure water (100) % Concentration
of water in the given mixture (10)
Mean % concentration
(20)
20 - 10 = 10 100 - 20 = 80
=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8
Here quantity of the mixture = 16 litres
=> Quantity of water : 16 = 1 : 8
=> Quantity of water =16×18=2 liter
B
Question: If log12 27 = a, then log6 16 is:
A.(3-a)/4(3+a)
B.(3+a)/4(3-a)
C.4(3+a)/(3-a)
D.4(3-a)/(3+a)
Answer: D.4(3-a)/(3+a)
Explanation: log12 27 = a => log 27/ log 12 = a
=> log 33 / log (3 * 22) =a
=> 3 log 3 / log 3 + 2 log 2 = a => (log 3 + 2 log 2)/ 3 log 3 = 1/a
=> (log 3/ 3 log 3) + (2 log 2/ 3 log 3) = 1/3
=> (2 log 2)/ (3 log 3) = 1/a – 1/3 = (3-a)/ 3a
=> log 2/ log 3= (3-a)/3a => log 3 = (2a/3-a)log2
log16 16 = log 16/ log 6 = log 24/ log (2*3) = 4 log 2/ (log 2 + log 3)
= 4(3-a)/ (3+a)
ANSWER:D
Question: The average age of a class of 12 students is 19 years. The average increased by 2 when the teacher's age also included. What is the age of the teacher?
A.40 years
B.34 years
C.42 years
D.43 years
Answer: B.34 years
Explanation: If age of the teacher was 12, average would not have changed.
Since average increased by 2,
Age of the teacher=12+22×1=34
ANSWER:B
Question: A car is traveling 75 kilometers per hour. How many meters does the car travel in one minute?
A.1250m/min
B.1360m/min
C.2560m/min
D.1670m/min
Answer: A.1250m/min
Explanation: Convert hour into minutes ( 1 hour = 60 minutes) and kilometers into meters (1 km = 1000 m) and simplify
75 kilometers per hour = 75 km/hr
= (75 × 1000 meters) / (60 minutes) = 1,250 meters / minute
correct answer is A)1250m/min
Question: For how many prime pairs (p, q) does there exist an integer n such that
(p2 + 1)*(q2 + 1) = n2 + 1 ?
A.2
B.4
C.6
D.8
Answer: A.2
Explanation: First, we show that one of p and q is 2. If both p and q are odd, then n2 + 1 = (p2 + 1)(q2 + 1)
is even so that n is also odd. Hence, there is an integer t such that n = 2t + 1 so that n2 + 1 =
4t2+4t+2. But this means n2+1 is not divisible by 4 whereas (p2+1)(q2+1) clearly is. Thus, we
must have one of p or q is 2. We now have n2 + 1 = 5(x2 + 1) where either x = p or x = q. Note
that x 3. We deduce that 5x2 = n2 − 4 = (n − 2)(n + 2). Since x is prime and n + 2 > n − 2,
we obtain that n−2 2 {1, 5, x} and n+2 = 5x2/(n−2). Since also (n+2)−(n−2) = 4, we get
that one of 5x2 − 1, x2 − 5 and 4x equals 4. As x is prime, we obtain x = 3. Thus, (p, q) = (2, 3)
or (3, 2). Each pair gives a solution to the equation n2+1 = (p2+1)(q2+1), implying the answer.
correct answer A
Question: Each shelf of a bookcase contained 11 books. If the librarian took out 30 books and rearranged the remaining books so that all shelves but the last one contained 7 books and that last shelf contained 5 books, how many shelves does the bookcase have?
A.5
B.6
C.7
D.8
Answer: C.7
Explanation: Let x be the number of shelves.
11x - 30 = 7(x-1) + 5
4x = 28
x = 7
The answer is C.
Question: There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
A. 5
B. 21
C. 33
D. 60
Answer: B. 21
Explanation: Given the constraints of this question, we have to consider 6 possible outcomes...
1 green ball: 6 options
GRRRRR
RGRRRR
RRGRRR
RRRGRR
RRRRGR
RRRRRG
2 green balls: 5 options
GGRRRR
RGGRRR
RRGGRR
RRRGGR
RRRRGG
3 green balls: 4 options
GGGRRR
RGGGRR
RRGGGR
RRRGGG
4 green balls: 3 options
GGGGRR
RGGGGR
RRGGGG
5 green balls: 2 options
GGGGGR
RGGGGG
6 green balls: 1 option
GGGGGG
6+5+4+3+2+1 = 21 options.
B
Question: A and B rent a pasture for 10 months. A put in 80 cows for 7 months. How many can B put in for the remaining 3 months, if he pays half as much again as A?
A.120
B.180
C.200
D.280
Answer: D.280
Explanation: 80* 7: x* 3 = 1:1 1/2
560: 3x = 2: 3
x = 280
ANSWER:D
Question: If 40% of a number is equal to 2/3 of another number, what is the ratio of first number to the second number?
A.5/3
B.3/5
C.3/7
D.5/7
Answer: A.5/3
Explanation: Let 40% of A = 2/3 B
Then, 40*A/100 = 2 *B /3
A/B = 5/3
Ans- A
Question: If n is the square of a positive integer, which of the following must be equal to the square of the next positive integer?
.
A.√n + 1
B.n + 1
C.n^2 + 1
D.n + 2√n + 1
Answer: D.n + 2√n + 1
Explanation: Let a= +ve integer such that
n= a^2
(a+1)= next +ve integer
(a+1)^2 = a^2 + 1 + 2a
and a=n^(1/2)
Therefore ,
(a+1)^2=n+1+2*n^(1/2)
Answer D
Question: A bag contains 5 red, 6 blue and 4 green balls. If 2 ballsare picked at random, what is the probability that both are red?
A.2/15
B.2/21
C.5/26
D.3/29
Answer: B.2/21
Explanation: P(Both are red),
=5C215C2=5C215C2
=10/105=2/21
B
Question: For a certain set of numbers, if x is in the set, then x - 2 is also in the set. If the number 1 is in the set , which of the following must also be in the set ?
a) -4
b) -1
c) -5
A.a only,
B.b only,
C.c only,
D.a and b only
Answer: D.a and b only
Explanation: Notice that the question asks which of the following MUST be in the set, not COULD be in the set.
Since 1 is in the set, then so must be 1-2=-1. Similarly, since -1 is in the set, then so must be -1-3=-4. Could 4 and -1 be in the set? Certainly, but we don't know that for sure.
Answer: D.
Regarding your question: we don't know which is the source integer in the set, if it's 1, then 4 won't be in the set but if the source integer is say 7, then 4 will be in the set. So, 4 may or may not be in the set.
Question: A rectangular field 30 m long and 20 m broad. How much deep it should be dug so that from the earth taken out, a platform can be formed which is 8 m long, 5.5 m broad and 1.5 m high where as the earth taken out is increase by 10/5?
A.37 cm
B.10 cm
C.17 cm
D.88 cm
Answer: B.10 cm
Explanation: 30 * 20 * x = (8 * 5.5 * 1.5)/2
\Answer:B
Question: If the price of a certain computer increased 30 percent from d dollars to 351 dollars, then 2d =
A.540
B.570
C.619
D.649
Answer: A.540
Explanation: Before Price increase Price = d
After 30% Price increase Price = d+(30/100)*D = 1.3d = 351 (Given)
i.e. d = 351/1.3 = $270
i.e. 2d = 2*270 = 540
Answer: option A
Question: If a town of 35,000 people is growing at a rate of approx. 1% per year, the population of the town in 5 years will be closest to?
A.a.36000
B.b.37000
C.c.38000
D.d.39000
Answer: A.a.36000
Explanation: 1% is quite small and hence the answer is A)
Question: We have a rectangular metallic piece of paper that covers exactly the area of a cube. The length of the piece of paper is 120 inches and the width is 108 inches. What is the volume of the cube in cubic feet is 1 feet is 12 inches?
A.a 125
B.b 120
C.c 70
D.d 40
Answer: A.a 125
Explanation: l = 120 /12 = 10 ft
w = 108 / 12 = 9 ft
Area of paper = 90
Area of cube = 10 * side^2
side of cube = 5
V of cube = 125
Question: A shipment of 1500 heads of cabbage, each of which was approximately the same size was purchased for $600.The day the shipment arrived 2/3 of the heads were sold, each at 25% above the cost per head.The following day the rest were sold at a price per head equal to 10% less than the price each head sold for the day before.what was the gross profit Q on this shipment?
A.a) $100
B.b) $115
C.c) $125
D.d) $130
Answer: C.c) $125
Explanation: I recommend one should use fractions to solve this one rather than converting it into decimals
Sol:
1500 heads -> $600
1 head -> $600/1500
1 head -> $(2/5)
25% more of (2/5) -> 125/100 * 2/5 = $(1/2)
He sold 2/3*1500 = 1000 heads for $(1/2) per head
Total revenue by selling 1000 heads = 1000 * 1/2 = $500
Heads left: 500
Cost per head: 90% of the previous price: 90/100 * 1/2 = $(9/20)
Total revenue by selling 500 heads = 9/20 * 500 = 225
Total revenue after selling 1500 cabbage heads - 225+500 = $725
Money spent on the purchase: $600
Profit Q= 725-600 = $125
Ans:C
Question: Two unbiased coins are tossed. What is the probability of getting two heads?
A.1/2
B.1/4
C.2/3
D.1/5
Answer: B.1/4
Explanation: S = {HH, HT, TH, TT}
E = event of getting 2 heads
E = {TT}
P(E) = 1/4
Answer is B
Question: If r is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of r ?
A. 10
B. 12
C. 14
D. 16
Answer: C. 14
Explanation: Answer is C.
numbers in format of 3^k in the series from 1 to 30 inclusive r are: 3*1, 3*2, 3*3, 3*4, 3*5, 3*3*2, 3*7, 3*8, 3*3*3, 3*10. Total number of 3 = 13. So k=14.
Question: What is the total cost of 2 sandwiches at $2.45 each and 4 sodas at $0.87 each?
A.$8.38
B.$6.85
C.$8.46
D.$10.08
Answer: A.$8.38
Explanation: Answer = A
2 * 2.45 + 4 * 0.87
= 2 (2.50 - 0.05) + 4 (1.00 - 0.13)
= 5 + 4 - 0.1 - 0.52
= 9 - 0.62
= 8.38
Question: A,B,C can complete a piece of work in 3,9and 15 days respectively. Working together, they will complete the same work in how many days?
A.45/23
B.55/23
C.61/57
D.12/13
Answer: A.45/23
Explanation: (A+B+C)'s 1day work = (1/3)+(1/9)+(1/115) = 23/45
A,B,C together will complete the work in 45/23 days
Answer is A
Question: A person got Rs.48 more when he invested a certain sum at compound interest instead of simple interest for two years at 8% p.a. Find the sum?
A.Rs.7500
B.Rs.7503
C.Rs.7227
D.Rs.7528
Answer: A.Rs.7500
Explanation: Explanation:
P = (d * 1002) / R2
=> (48 * 100 * 100) / 8 * 8 = Rs.7500
Answer: A
Question: What is the remainder when the number T=14^2 * 15^8 is divided by 5?
A.0
B.1
C.2
D.4
Answer: A.0
Explanation: 14^2 has units digit 6
15^8 has units digit 5
Thus T=14^2*15^8 has units digit 0 and will be divisible by 5. The remainder will be zero
Answer: (A)
Question: How many 2 digit number contain number 3 ?
A.18
B.28
C.38
D.68
Answer: A.18
Explanation: total 2 digit no. = 9*10* = 90
not containing 3 = 8*9 = 72
total 2 digit number contain 3 = 90 - 72 = 18
ANSWER:A
Question: Half a number plus 6 is 11.What is the number?
A.8
B.9
C.10
D.11
Answer: C.10
Explanation: Let x be the number. Always replace "is" with an equal sign
(1/2)x + 6 = 11
(1/2)x = 11 - 6
(1/2)x = 5
x = 10
correct answer is C
Question: If A and B are Two +ve integers and A ≠0 and B ≠0 which can be the following CAN equal to Zero?
A.A+B
B.A-B
C.A/B
D.A*B
Answer: B.A-B
Explanation: (A) The option A should be the answer.
If A=B,A-B=0
The other Options can't be answer. Because A and B must be an +ve numbers and not be zero.
Question: What is the present worth of Rs. 264 due in 2 years at 5% simple interest per annum?
A.128
B.277
C.120
D.240
Answer: D.240
Explanation: Let the present worth be Rs. x. Then,
S.I. = (264 - x)
(x * 5 * 2) / 100 = (264 - x)
10x = 26400 - 100x
110x = 26400 => x = 240.
Answer: D
Question: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has five iron coins and fiveteen copper coins, how many different sums from 1¢ to 160¢ can he make with a combination of his coins?
A.80
B.85
C.81
D.160
Answer: C.81
Explanation: The total sum is 5*2 + 15*5 = 85¢. If you can make each sum from 1 to 85 (1¢, 2¢, 3¢, ..., 85¢), then the answer would be 85 (maximum possible).
Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 84¢ and 82¢ (since total sum is 85¢ we cannot remove 1¢ or 3¢ to get 84¢ or 82¢).
So, out of 85 sums 4 are for sure not possible, so the answer must be 85 - 4 = 81 sums or less. Only A fits.
Answer: C.
Question: A ladder 30 feet long is leaning against a wall that is perpendicular to level ground. The bottom of the ladder is 10 feet from the base of the wall. If the top of the ladder slips down 5 feet, how many feet will the bottom of the ladder slip?
A.14
B.15
C.18.92
D.19
Answer: C.18.92
Explanation: 30^2-10^2=800
it means that the height is equal to 28.28.
since the top of the ladder slips down 5 feet, then the height of the wall =28.28-5=23.28
the bottom =sqrt(30^2-23.28^2)=sqrt(900-541.95)=18.92
ans is C
Question: Rs. 6000 is lent out in two parts. One part is lent at 7% p.a simple interest and the other is lent at 10% p.a simple interest. The total interest at the end of one year was Rs. 450. Find the ratio of the amounts lent at the lower rate and higher rate of interest?
A.5:1
B.5:6
C.5:2
D.5:8
Answer: A.5:1
Explanation: Let the amount lent at 7% be Rs. x
Amount lent at 10% is Rs. (6000 - x)
Total interest for one year on the two sums lent
= 7/100 x + 10/100 (6000 - x) = 600 - 3x/100
=> 600 - 3/100 x = 450 => x = 5000
Amount lent at 10% = 1000
Required ratio = 5000 : 1000 = 5:1
Answer:A
Question: The current of a stream at 1 kmph. A motor boat goes 35 km upstream and back to the starting point in 12 hours. The speed of the motor boat in still water is?
A.6
B.9
C.5
D.1
Answer: A.6
Explanation: S = 1
M = x
DS = x + 1
US = x - 1
35/(x + 1) + 35/(x - 1) = 12
x = 6
Answer: A
Question: In the seaside summer camp there are 50 children. 90% of the children are boys and the rest are girls. The camp administrator decided to make the number of girls only 5% of the total number of children in the camp. How many more boys must she bring to make that happen?
A.50.
B.45.
C.40.
D.30.
Answer: A.50.
Explanation: Given there are 50 students In the seaside summer camp , 90% of 50 = 45 boys and remaining 5 girls.
Now here 90% are Boys and 10% are Girls.
Now question is asking about how many boys do we need to add, to make the girls percentage to 5 or 5%..
If we add 50 to existing 45 then the count will be 95 and the girls number will be 5 as it.
Now boys are 95% and girls are 5%. ( out of 100 students = 95 boys + 5 girls ).
IMO option A is correct.
Question: If in a game of 80, P can give 16 points to Q and R can give 20 points to P, then in a game of 150, how many points can R give to Q?
A.60
B.28
C.26
D.17
Answer: A.60
Explanation: Explanation:
When P scores 80, Q scores 64.
When R scores 80, P scores 60
Hence, when R scores 150, Q scores (60 * 64 * 150) / (80 * 80) = 90
Therefore, in a game of 150, R can give 60 points to Q.
ANSWER: A
Question: The remainder when 1!+2!+3!...+50! divided by 8! will be
A.21
B.5193
C.77
D.9275
Answer: B.5193
Explanation: The remainder when the terms greater than 8! are divided by 8! becomes 0 so we need to consider the terms upto 7!.
So remainder will be whatever is obtained by dividing 1!+2!+3!+4!+5!+6!+7! with 8!.
So remainder is obtained by dividing (1+2+6+24+120+720+5040)= 5193 with 8! ( 40320)
So remainder is 5193.
Answer:B
Question: In Township K each property is taxed at 10 percent of its assessed value. If the assessed value of a property in Township K is increased from $20,000 to $24,000, by how much will the property tax increase?
A. $32
B. $50
C. $320
D. $400
Answer: D. $400
Explanation: Increase in house value = $24,000 - $20,000 =$4000
So, tax increase = 10% of$4000= $400
Answer: D