3 Bank Combined Math Suggestion : 2018

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3 Bank Combined Math Suggestion 

01. Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?
A. 49
B. 70
C. 140
D. 215
E. 525

Answer: E. 525

Solution:

the sum of 5 consecutive odd integers

and

the sum of 7 consecutive odd integers must be odd.

Eliminate B, and C.
Next ,

the sum of 5 consecutive odd integers can be written as:

(x-4)+(x-2)+x+(x+2)+(x+4)=5x,

so the sum must be odd multiple of 5;
Similarly,

the sum of 7 consecutive odd integers can be written as:

(y-6)+(y-4)+(y-2)+y+(y+2)+(y+4)+(y+6)=7y,

so the sum must be odd multiple of 7;
So, the sum must be odd multiple of 5×7=35.

Only option E meets this condition.

02. A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A. 13
B. 13.5
C. 14
D. 14.5
E. 15

Answer: B. 13.5

Solution:

Let,

Distance from B to C = x.

So, A to B = 2x

Avg miles per gallon,

= total miles covered/total gallon used

= (2x + x)/{(2x/12) + (x/18)}

= 3x/{(2x/9)}

= 13.5

03. What is the smallest possible distance between the point (0, 5) and any point on the line y = -x + 3?
A. 0
B. 1/√2
C. 1
D. √2
E. 2

Answer: D. √2

Solution:

Distance from a point(x,y) from line ax + by + c = 0 is

= |ax+by+c|/(a2+b2)

Given,

y = -x + 3

Or, x+y-3=0

Hence,

Distance,

= |0×1 + 5×1 – 3|/(12 + 12)

= √2

04. A sum of money invested under simple interest, amounts to $1200 in three years and $1500 in five years. What is the rate at which the sum of money was invested?
(A) 10%
(B) 15%
(C) 20%
(D) 25%
(E) 45%

Answer:

Solution:

Firstly,
P + P × r × 3 = 1200
Or, P(1 + 3r) = 1200 ——(1)
Similarly,
P + P × r × 5 = 1500
Or, P(1 + 5r) = 1500 ——-(2)
Dividing (1) by (2):
(1 + 3r)/(1 + 5r) = 1200/1500
Or, (1 + 3r)/(1 + 5r) = 4/5
Or, 5(1 + 3r) = 4(1 + 5r)
Or, 5 + 15r = 4 + 20r
Or, 1 = 5r
Or, r = 1/5 = 20%

05. If x and y are integers, and x/y is not an integer then which of the following must be true?
A. x is odd and y is even.
B. x is odd and y is odd.
C. x is even and y is odd.
D. x<y
E. None of the above

Answer: E. None of the above

3 Bank Combined Math Suggestion

06. Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2?
(A) 36
(B) 81
(C) 91
(D) 112
(E) 116

Answer: (C) 91

Solution:

Total # of outcomes – # of outcomes without 2’s = 63−53=91

07. A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?
A) 1/6
B) 1/5
C) 3/10
D) 1/3
E) 2/5

Answer: E) 2/5

Solution:

The area of the square will be more than 1 if and only if the longer piece of the wire is longer than 4. To produce such a result, the cutting point has to be either on the first meter of the wire or on its last meter. The probability of this is 2/5.

08. When integer x is divided by 4 and integer y is divided by 5, the remainder is 1 and the difference between quotients is 6. What is the sum of x and y?
A. 117
B. 119
C. 126
D. 137
E. 143

Answer: E. 143

Solution:

ATQ,

(x-1)/4-(y-1)/5=6
Or, 5x-4y=121
Or, 5x/11-4y/11=11
if x and y are integers, they must both be multiples of 11 and their sum must be a multiple of 11.
143 is the only multiple of 11.

09. Find the number of factors of 180 that are in the form (4×k + 2), where k is a non-negative integer?
A) 1
B) 2
C) 3
D) 4
E) 6

Answer: E) 6

Solution:

4k + 2 = 2(2k + 1) = 2×odd.

So, we are looking for even factors which are not multiples of 4.
Now,
180 = 22×32×5. Consider the part without 22.

Now, 32×5 has (2 + 1)(1 + 1) = 6 factors: 1, 3, 5, 9, 15, 45.

Any of them paired with 2 will be even factor of 180 which is not a multiple of 4, so 2, 6, 10, 18, 30 and 90.

10. If a triangle has sides 8, 10, and z, which of the following cannot be the area of the triangle?
A. 3
B. 6
C. 12
D. 24
E. 46

Answer: E. 46

Solution:

If the triangle is a right triangle with z being the hypotenuse, then the two legs 10 and 8 will be the base and height of the triangle. In that case,

the area of the triangle is ½ x 10 x 8 = 40.

This is the maximum area for a triangle with two sides of 8 and 10;

thus, any area less than 40 is a possible value for the area.

Thus, answer choice E cannot be the area of the triangle.

3 Bank Combined Math Suggestion

11. Which of the following fractions is closest to 1/2?
A) 4/7
B) 5/9
C) 6/11
D) 7/13
E) 9/16

Answer: D) 7/13

12. Train A and Train B began traveling towards each other from opposite ends of a 500-mile long track at 1:00 PM. If Train A traveled at 35 miles per hour and Train B traveled at 25 miles per hour, at what time did the trains meet?
A. 5:40 PM
B. 6:00 PM
C. 7:20 PM
D. 8:00 PM
E. 9:20 PM

Answer: E. 9:20 PM

Solution:

Distance = 500 mile
Speed of train A = 35 mph
Speed of train B = 25 mph
Relative speed of both trains
= 35+ 25 =60 mph
Time = Distance / Speed
= 500/60
=8 hours and 20 mins
Since the trains started at 1 PM , they will meet at 9:20 PM

13. Two cars start at the same point and travel in opposite directions. If one car travels at 45 miles per hour and the other at 60 miles per hour, how much time will pass before they are 210 miles apart?
(A) 0.5 hours
(B) 1 hour
(C) 1.5 hours
(D) 2 hours
(E) 2.5 hours

Answer: (D) 2 hours

Solution:

When cars travel in opposite directions, relative speed,

= 45 + 60 = 105 mph
Speed × time = distance
Time =  distance/speed = 210 miles / 105 mph = 2 hours

14. A runs 25% faster than B and is able to allow B a lead of 7 meters to end a race in dead heat. What is the length of the race?
A. 10 meters
B. 15 meters
C. 25 meters
D. 35 meters
E. 45 meters

Answer: 35 meters

Solution:

Let,

d=length of race
r=B’s rate
because times of A and B are equal,
then d/(5r/4)=(d-7)/r
d=35 meters

15. A’s speed is 20/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?
(A) 1/17
(B) 3/17
(C) 1/10
(D) 3/20
(E) 3/10

Answer: (D) 3/20

Solution:

Picking numbers approach.
From ratio of speed 20/17, when B run 17 miles, A run 20 miles
LCM (17, 20) = 340.

Let,

distance equal to 340 miles
B run that distance by 340/17 = 20 hours
A run this distance by 340/20 = 17 hours
So A should wait 3 hours before start his running.
B will run 3 × 17 = 51 miles by this 3 hours
So A should give head start equal to 51 miles
51/340 = 3/20

3 Bank Combined Math Suggestion

16. Two trains traveling toward each other on parallel tracks at constant rates of 50 miles per hour and 60 miles per hour are 285 miles apart. How far apart will they be 2 hours before their engine meet?
(A) 110
(B) 120
(C) 150
(D) 200
(E) 220

Answer: (E) 220

Solution:

When they travel toward each other,

their relative speed is the sum of the two speeds = (50 + 60) = 110 mph
In 2 hrs, they will together cover a distance of 2×110 = 220 miles.
Hence, 2 hrs before their engines meet, distance between them will be 220 miles.

17. A car completed its first 30 miles of a 60-mile trip at an average speed of 60 miles per hour. At what approximate average speed, in miles per hour, did the car complete the remaining miles to achieve an average speed of 50 miles per hour for the entire 60-mile trip? The car completed its 60-mile trip without a break.
(A) 33
(B) 38
(C) 40
(D) 43
(E) 46

Answer: (D) 43

Solution:

Avg speed = (2×v1×v2)/(v1+v2)
Or, 50 = (2×60×v2)/(60+v2)
Or, 3,000 + 50v2 = 120v2
Or, 3,000 = 70v2
Or, v2 = 3000/70 = 42.8 approximately 43.

18. At a speed of 50 miles per hour, a certain car uses 1 gallon of gasoline every 30 miles. If the car starts with a full 12 gallon tank of gasoline and travels for 5 hours at 50 miles per hour, the amount of gasoline used would be what fraction of a full tank?
A. 3/25
B. 11/36
C. 7/12
D. 2/3
E. 25/36

Answer: E. 25/36

Solution:

In 5 hours at 50 miles per hour car will travel 5×50=250 miles

and thus will use 250/30=25/3 gallons of gasoline.

Gasoline used = (25/3)/12=25/36.

19. At the rate of m meters per s seconds, how many meters does a cyclist travel in x minutes?
A. m/sx
B. mx/s
C. 60m/sx
D. 60ms/x
E. 60mx/s

Answer: E. 60mx/s

Solution:

s seconds need to go m meters
1 second needs to go m/s meters
1 minute needs to go 60 ×m/s meters
x minutes need to go 60 mx/s meters

20. If a car had traveled 20 kmh faster than it actually did, the trip would have lasted 30 minutes less. If the car went exactly 60 km, at what speed did it travel?
A. 35 kmh
B. 40 kmh
C. 50 kmh
D. 60 kmh
E. 65 kmh

Answer: B. 40 kmh

Solution:

Let,

Speed = x kmh

ATQ,
60/x – 60/(x + 20) = ½

Or, (60x + 1200 – 60x)/(x2 + 20x) = ½

Or, 1200/(x2 + 20x) = ½

Or, x2 + 20x – 2400 = 0

Or, (x -40)(x+60) = 0

x- 40 = 0 Acceptable

Or, x = 40 kmh

3 Bank Combined Math Suggestion

21. A car goes 12 kilometers on a liter of gas when driven at 80 kilometers per hour. When the car is driven at 50 kilometers per hour, it only goes 75% as far per liter gas how many liters of gas will it take to travel 135 kilometers driving at 50 kilometers per hour?
A. 3
B. 5
C. 9
D. 11
E. 15

Answer: E. 15

Solution:

75% of 12 km/liter=9km/liter.
So the fuel efficiency is 9 km/liter when driving at 50 kmph.
So,

to drive 9 km needs 1 liter
to drive 135 km need 135/9 liter = 15 liter

22. The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 36 ft long. What is the distance traveled by the cart, when the front wheel has done five more revolutions than the rear wheel?
A. 20 ft
B. 25 ft
C. 750 ft
D. 900 ft
E. 1000 ft

Answer: D. 900 ft

Solution:

Circumference of front wheel C1=30 ft,

Circumference of front wheel C2=36 ft
If Rotation1=x

ATQ,
30(x+5)=36x

Or, x=25

So, distance =30(25+5)=30×30=900

23. Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours and Car Y traveled at an average speed that was 50 percent faster than the average speed of Car X, how many hours did it take Car Y to travel the route?
(A) 2/3
(B) 1
(C) 4/3
(D) 8/5
(E) 3

Answer: (C) 4/3

Solution:

The speed of car X is (distance)/(time) = 80/2 = 40 miles per hour.

The speed of car Y = 40 + (50% of 40) = 60 miles per hour

Time needed = (distance)/(speed) = 80/60 = 4/3 hours.

24. A boy is buying two kinds of marbles. One kind costs 25 cents per pound and he purchased 5 pounds of this kind. The other kind of marble costs 32 cents per pound. If the boy spent approximately 29 cents per pound for everything, how many pounds of the more expensive marbles did he buy?
A. 5
B. 6
C. 7
D. 8
E. 9

Answer: C. 7

Solution:

Suppose, the boy bought x pounds of the expensive marbles.
so,
29(5+x) = 25×5 + x×32
Or, 3x = 20
Or, x = 6.67 ≈ 7.

25. How many liters of water must be added to a 20 liter solution of water and acid, which is 24% acid, to make a solution that is 8% acid?
A. 20
B. 40
C. 48
D. 60
E. 170

Answer: B. 40

Solution:

Let,

x be amount of water to be added to the original 20 liters
ATQ,
0.24*20 + x = 0.08×(20+x)
Or, 4.8 + x = 1.6 + 0.08x
Or, 0.08x = 3.2
Or, x=40

3 Bank Combined Math Suggestion

26. Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporate from 8 kilograms of solutions Y and 2 kilograms of solution Y are added to the remaining 6 kilograms of liquid, what percent of this new liquid solution is liquid X?
A. 30%
B. 33 1/3%
C. 37 1/2%
D. 40%
E. 50%

Answer: C. 37 1/2%

Solution:

Liquid x in 8 kg = 8×30%=12/5
Liquid x in next 2 kg = 2×30% = 3/5
Total x = 12/5 + 3/5 = 15/5 = 3
so, (3/8)×100= 37.5%

27. There are 100 apples in a bag of which 98% are green and the rest red. How many green apples do you need to remove so that only 96% of the apples are green?
A. 40
B. 8
C. 15
D. 25
E. 50

Answer: E. 50

Solution:

Let,

a=total apples remaining after green apples removed
0.04a=2 red apples
a=50 total apples
50-2=48 green apples remaining
98-48=50 green apples removed

28. A solution of 60 ounces of sugar and water is 20% sugar. How much water must be added to make a solution that is 5% sugar?
A. 20 ounces
B. 80 ounces
C. 100 ounces
D. 120 ounces
E. 180 ounces

Answer: E. 180 ounces

Solution:

20% sugar in 60 ounce solution;

Sugar = 60×20/100 = 12 ounce
Water = 60-12 = 48 ounce

Now; let’s add x ounce of water in the solution;
Total solution = 60+x ounce
Sugar = 12 ounce
Or, 12 = (60+x) × 5/100
Or, 60+x = 240
Or, x = 240 – 60 = 180 ounce

29. If xy+z=x(y+z), which of the following must be true?
A. x=0 and y=0
B. x=1 and y=1
C. y=1 and z=0
D. x=1 or y=0
E. x=1 or z=0

Answer: E. x=1 or z=0

Solution:

Xy + z = x(y+z)
Or, xy+z = xy + xz
Or, z = xz
Or, z(x−1)=0

Or, z(x−1)=0
Either z=0

OR

x=1

30. In a school, 80% of the students play either football or cricket, and the rest don’t play any sport. If 35% of the students, who play any sport, play only cricket, then what percent of students, in the school, play football?
A. 42%
B. 48%
C. 52%
D. 52.5%
E. 65%

Answer: C. 52%

Solution:

So say there are 100 students, 80 out of these play games.
35% of 80 play cricket, so remaining 65% of 80 play football =80×65/100=52
So 52 out of 100 play football,

in % it is 100×52/100=52%

3 Bank Combined Math Suggestion

31. The purchase price of an article is $48. In order to include 15% of cost for overhead and to provide $12 of net profit, the markup should be
A. 15%
B. 25%
C. 35%
D. 40%
E. 45%

Answer: D. 40%

Solution:

Cost price of article = 48$
% of overhead cost = 15
Net profit = 12 $
Net profit as % of cost price = (12/48)×100 = 25%
Total markup should be = 25 + 15 = 40%

32. When 1/10 percent of 5,000 is subtracted from 1/10 of 5,000, the difference is
(A) 0
(B) 50
(C) 450
(D) 495
(E) 500

Answer: (D) 495

Solution:

(5000/10) – 5000/(100×10)

= 500 – 5

= 495

33. If $1000 is placed into account X, yielding 10% interest compounded annually and $1000 is placed into account Y using 10% simple annual interest, how much more will be in account X than in account Y at the end of 5 years?
A. $0
B. $100
C. $110.51
D. $133.31
E. $146.41

Answer: C. $110.51

Solution:

S.I = P×T×R / 100 = 1000×5× 0.1 = 500
C.I = P×( 1 + R/100 )nt – P = 1000 × ( 1 + 0.1 )5 – 1000 = 1610.51 – 1000 = 610.51
C.I – S.I = 610.51 – 500 = 110.51
37. 70% of 66 is 10% more than what number
A. 50.82
B. 46.2
C. 43
D. 42
E. 41.58

Answer: D. 42

Solution:

Let the number is X

ATQ,

X + 10% of x = 70% of 66

Or, x + 10x/100 = (70×66)/100

Or, 110x/100 = (70×66)/100

Or, 110x = 70×66

Or, x = 42

34. If y is 80% greater than x, than x is what % less than y?
(A) 20
(B) 25
(C) 33 1/3
(D) 44 4/9
(E) 80

Answer: (D) 44 4/9

Solution:

y=(180/100)×x

x=(100/180)×y
Percentage =100×(100/180)=500/9=55.55
Percentage less = 100 – 55.55 = 44.45

35. Due to a 25% increase in the price of diesel, a person got 10 liters less quantity for $50 than he was getting before the increase. What was the initial price per liter of diesel?
(A) $1.00
(B) $1.50
(C) $2.25
(D) $2.50
(E) $3.00

Answer: (A) $1.00

Solution:

Let,

liter of diesel = x

With 25% increase, price is 62.5
hence,

50x = 62.5(x-10)
x = 50

so price per litre is 50/50 = 1

3 Bank Combined Math Suggestion

36. If the sides of a square are doubled in length, the area of the original square is now how many times as large as the area of the resultant square?
A) 25%
B) 50%
C) 100%
D) 200%
E) 400%

Answer: A) 25%

Solution:

Say, side of original square = 1
Area=12=1
New square side is doubled = 2
Area =22=4
Area of original square is 1/4 times area of new square
Percentage =(1/4)×100=25%

37. Thirty percent of forty percent of fifty is sixty percent of what percent of two-hundred ?
A. 0.5%
B. 1%
C. 5%
D. 50%
E. 500%

Answer: C. 5%

Solution:

(30/100) × (40/100) × 50 = (60/100)× X ×200

Or, 120x = 6

Or, x = 1/20

Or, x = 5%

38. An investment is made at 12.5% annual simple interest. The number of years it will take for the cumulative value of the interest to equal the original investment is equal to which of the following?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Answer: (E) 8

Solution:

Simple Interest
I = pnr
Given,
I = P
So,

I = I × n × r

Or, nr = 1

Or, (12.5/100)×n = 1

Or, n = 100/12.5

Or, n = 8

39. Two workers A and B are engaged to do a work. A working alone takes 8 hours more to complete the job than if both worked together. If B worked alone, he would need 4.5 hours more to complete the job than they both working together. What time would they take to do the work together ?
A. 4 hours
B. 5 hours
C. 6 hours
D. 7 hours
E. 8 hours

Answer: C. 6 hours

Solution:

Let

A works at the rate of ‘a’ units/hr and

B works at the rate of ‘b’ units/hr.

A and B complete the job in x hrs.
A working alone takes 8 hours more to complete the job than if both worked together.
So, a(x+8) = (a+b)x
Or, ax + 8a = ax + bx
Or, 8a = bx

Or, a = bx/8 —(1)
If B worked alone, he would need 4.5 hours more to complete the job than they both working together
b(x+4.5) = (a+b)x
Or, bx +4.5b = ax + bx
Or, 4.5b = ax or a = 4.5b/x —(2)
From (1) and (2), we have
bx/8=4.5bx

Or, x2=4.5×8

Or, x=√36

Or, x=6

40. A contractor undertakes to do a job within 100 days and hires 10 people to do it. After 20 days, he realizes that one fourth of the work is done so he fires 2 people. In how many more days will the work get over?
(A) 60
(B) 70
(C) 75
(D) 80
(E) 100

Answer: (C) 75

Solution:

Let the work done by one person per day be x
Then, work done by 10 persons per day is 10x.
work done by 10 persons in 20 days is 20×10x

which is equal to 1/4th of total work.
Therefore, x = 1/800.
Work done by 8 persons in n days(n×8x) is 3/4th of total work.
Therefore, n = 3/(8×4) = 2400/32 = 75.

3 Bank Combined Math Suggestion

41. A pipe takes 3 hours to fill (2/7) of a tank. What is the time remaining (in hours) till the tank is completely filled?
A. 3
B. 3.5
C. 10.5
D. 7.5
E. 4.5

Answer: D. 7.5

Solution:

Time taken to fill 2/7 of tank 3 hours
Time taken to fill 1 tank =(7/2)∗3=21/2 hours
The time remaining (in hours) till the tank is completely filled =(21/2)−3 =15/2=7.5 Hours

42. A, B and C, each working alone can complete a job in 6, 8 and 12 days respectively. If all three of them work together to complete a job and earn $2340, what will be C’s share of the earnings?
A. $520
B. $630
C. $1080
D. $1100
E. $1170

Answer: A. $520

Solution:

Let work done by A in a day = 1/6
work done by B in a day = 1/8
work done by C in a day = 1/12
total work done by them in a day = 1/6 + 1/8 + 1/12
= (4+3+2)/24
=9/24
Fraction of work done by C in a day= 2/9
fraction of total work by each of them will be same as the work done by each individual in a day.
Total earning on completion of job = 2340 $
C’s share of earning = (2/9)×2340
=520$

43. Mark can paint twice as fast as Joe. If Mark starts to paint at one end of a mile-long wall and Joe starts to paint at the other end of the wall, how far away from Mark’s end of the wall will they meet?
A. 3/4 mile
B. 2/3 mile
C. 1/2 mile
D. 1/3 mile
E. 1/4 mile

Answer: B. 2/3 mile

Solution:

Since Mark can paint twice as fast as Joe, he will paint twice as much of the wall (length-wise) as Joe will. Let,

x = the length of wall Joe paints, then 2x = the length of wall Mark paints.

Therefore:
2x + x = 1
Or, 3x = 1
Or, x = 1/3
Thus, Mark paints 2/3 of a mile of the wall, i.e., he is 2/3 of a mile from his end of the wall when he meets Joe.

44. A can complete the job in 2 hours and B can complete the same job in 3 hours. A works for 1 hour and then B joins and both complete the job. What fraction of the job did B complete
A. 1/5
B. 3/10
C. 1/2
D. 5/6
E. 8/9

Answer: A. 1/5

Solution:

A’s rate = 1/2 W/h
B’s rate = 1/3 W/h
In 1 hour A finished 1/2 W.
Only half of the work is left to be completed.
Together A and B can finish 1 work in;
1/2+1/3=1/t
t = 6/5 hours
So, they would have finished (1/2)W in half the time;
(6/5)×(1/2) = 6/10 =(3/5)hour
B does 1/3 W in 1 hour
So, in 3/5 hour (3/5)×(1/3) = 1/5 Work
49. How many of the factors of 72 are divisible by 2?
A. 4
B. 5
C. 6
D. 8
E. 9

Answer: E. 9

Solution:

72=23×32 then factors of 72 is (3+1)(2+1)=12. Out of which only 3 are odd 1, 3, and 9, so rest or 12-3=9 are even.

45. The average weight of 15 items is 8 kg. The least average weight of 5 items of these 15 is 5 kg. If no item weighs less than 5 kg, what is the maximum number of items that can have the same weight?
A. 3
B. 5
C. 6
D. 10
E. 14

Answer: E. 14

Solution:

total weight = 15×8 =120 kg
least average weight of 5 items = 5×5 = 25 kg
weight remaining = 95 kgs
items remaining = 10
let 9 items have weight 5 kg and the remaining = 95-45=50 kg
no of items having same weight = 9 + 5 =14

46. If x, a, and b are positive integers such that when x is divided by a, the remainder is b and when x is divided by b, the remainder is a-2, then which of the following must be true?
A. a is even
B. x+b is divisible by a
C. x-1 is divisible by a
D. b=a-1
E. a+2=b+1

Answer: D. b=a-1

Solution:

When x is divided by a, the remainder is b

So, x=aq+b

When x is divided by b, the remainder is a−2

So, x=bp+(a−2)

x=bp+(a−2)

So we have that: a−2<b<a, as a and b are integers,

then it must be true that b=a−1

b=a−1

(there is only one integer between a−2 and a, which is a−1

and we are told that this integer is b,

hence b=a−1

47. If the average of a, b, c, 14 and 15 is 12. What is the average value of a, b, c and 29
A. 12
B. 13
C. 14
D. 15
E. 16

Answer: D. 15

Solution:

(a+b+c+14+ 15)/5 = 12
Or, a+b+c=31
(a+b+c+29)/4 = x
4x=31+29
x=15

আপনার যে কোন প্রয়োজনে, চাকরির সার্কুলার ও আপডেট, এক্সাম ডেট, আপনার কোন কিছু জানার থাকলে নির্দ্বিধায় যোগদান করুন একঝাঁক প্রাণোচ্ছল তরুণের ফেসবুক গ্রুপ: BCS Limelight এ।

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Updated: May 16, 2019 — 2:56 pm

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