আপনার কাছে যদি মনে হয় প্রশ্নটি অধ্যায় অনুযায়ী সঠিক নয় তাহলে সঠিক অধ্যায় ও প্রশ্নটি অথবা কোন প্রকার ভুল থাকলে আমাদের কে জানান ইমেইল করে kabirdepart@gmail.com
Let the score in the 5th test be x.
The sum of the first 4 tests = 78 X 4=312
Now it implies ,
312+x/5=80
Therefore, x=88
Keep the average in middle and place two numbers at left and right.
Five Numbers are 10,11,12,13,14
Sum of least and greatest integers =(10+14)=24
(z + x)/2 = 35
z + x = 70 .................... (1)
And,
(x + y)/2 = 25
x + y = 50 ................. (2)
(1) - (2), we get,
z - y = 70 -50 = 20
for first eight hour she is paid= 8d tk
for remaining 4 hour she is paid= 4c tk
so , she is averagely paid = 8d+4c/12= 4(2d+c)/12=(2d+c)/3
If we add or subtract a constant to each term in a set the standard deviation will not change..
If we add (or subtract) a constant to each term in a set the mean and the median will increase (decrease) by the value of that constant.
A= { 2, 3, 5} and
B= {1, 3, 5}
. Any of the 3 numbers in A could be added to nay of the 3 numbers in B, so 9 sums could be formed. However, there could be some duplication. List the sums systematically; first add 1 to each number in A, then 3, and then 5: 3, 4, 6; 5, 6(x), 10; 7, 8, 10(x). there are 7 different sums
If we add or subtract a constant to each term in a set the standard deviation will not change..
If we add (or subtract) a constant to each term in a set the mean and the median will increase (decrease) by the value of that constant.
Total Difference = 32 - (-13) = 45
Average hourly increase = 45/9=5
let,
sum of 6 of them=x=100
The sum of 4 of them=y
so,x+y/10=-10
y=100
so average=-200/4=-50
here,
2x+4x/2 = 12
or,6x/2=12
or,x=4
sum of three number= 24 X 3=72
sum of two numbers are= 21+23=44
so third number is = 72-44=28
at 90 tk price,
total share he bought= 45000/90=500 share
at 60 tk price,
total share he bought= 3000/60=50 share
total investment= 45000+3000 tk= 48000 tk
total share= 500+50=550
average cost=48000/550=87.27 tk
total used= 2800+3200+3600=9600 kwh
total cost= 0.30 X 9600=2880 tk
average= 2880/3=960 tk
The arithmetic mean of 5 observations is $7000.
Therefore, the sum of these 5 observations is 5 * 7000 = $35,000.
The median is $7000.
Let us say their incentives in ascending order are a, b, c, d and e.
So, c = 7000 and a + b + c + d + e = $35,000
The only mode is $12,000. So, the maximum number of observations among the 5 will be in $12,000.
The only possibility is that both d and e got an incentive of $12,000 each.
So, c + d + e = 7000 + 12000 + 12000 = $31,000
Therefore, a + b = 35,000 - 31,000 = 4000.
a and b have to be two different values as the only mode is $12,000.
So, a has to be $1000 and b has to be $3000.
The difference between the highest and the lowest is therefore 12,000 - 1000 = $11,000
The new average weight of the group after including the football coach = 31
As the new average is 1kg more than the old average, old average without including the football coach = 30 kgs.
The total weight of the 30 friends without including the football coach = 30 X 30 = 900.
After including the football coach, the number people in the group increases to 31 and the average weight of the group increases by 1kg.
Therefore,
the total weight of the group after including the weight of the football coach = 31 X 31 = 961 kgs.
Therefore, the weight of the football coach = 961 - 900 = 61 kgs.
here,
yearly sales =150000 tk
so, monthly sales=150000/12=12500 tk.
then,the sales in June, if June sales were half the monthly average=12500/2=6250 tk
৬, ৮, ১০ এর গানিতিক গড়=২৪/৩=৮
এখন, মনে করি ,
সংখ্যাটি "ক"
প্রশ্নমতে,
৭+৯+ক/৩=৮
বা,১৬+ ক= ২৪
বা,ক=২৪-১৬=৮
সুতরাং,ক=৮
Average of 20 numbers = 0.
Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is X then 20th number is (-X).
Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x - 8) + (7x - 8) + (9x - 8) = 56
=> 20x = 80
=> x= 4.
Therefore, Their present ages are 4x = 16 years,7x = 28 years and 9x = 36 years respectively.
Let there be x pupils in the class.
Total increase in marks =( x × ½) = x/2
So, x/2 = (83 - 63)
=> x/2 = 20
=> x= 40.
Required average = ((55 x 50) + (60 x 55) + (45 x 60))/(50+60+45)
=(2750 + 3300 + 2700)/160
= 8750/160
= 54.68
Since the month begins with a Sunday, to there will be five Sundays in the month.
Required average = ((510 x 5) + (240 x 25))/30
= 8550/30
=285
Required average = (50.25 x 16 + 45.15 x 8)/(16+8)
= (804 + 361.20)/24
= (1165.20)/24
= 48.55
Let A, B, C represent their respective weights. Then, we have:
A + B + C = (45 x 3) = 135 .... (i)
A + B = (40 x 2) = 80 .... (ii)
B + C = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31.
B's weight = 31 kg.
Let Ashik's weight by X kg.
According to Ashik, 65 < X < 72
According to Ashik's brother, 60 < X < 70.
According to Ashik's mother, X <= 68
The values satisfying all the above conditions are 66, 67 and 68.
Required average = (66 + 67 + 68)/3
= 201/3 = 67 kg.
Total quantity of petrol
consumed in 3 years = (4000/7.50 + 4000/8 + 4000/8.50) litres
= 4000(2/15 + 1/8 + 2/17) litres
= 76700/51 litres
Total amount spent = Tk. (3 x 4000) = Tk. 12000.
Average cost = Tk. (12000 x 51)/76700
= Tk. 6120/767 = Tk. 7.98
Sum of the present ages of husband, wife and child
= (27 x 3 + 3 x 3) years = 90 years..
Sum of the present ages of wife and child
= (20 x 2 + 5 x 2) years = 50 years..
Husband's present age = (90 - 50) years = 40 years.
Let P, Q and R represent their respective monthly incomes.
Then, we have:
P + Q = (5050 x 2) = 10100 .... (i)
Q + R = (6250 x 2) = 12500 .... (ii)
P + R = (5200 x 2) = 10400 .... (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000
or, P + Q + R = 16500 .... (iv)
Subtracting (ii) from (iv), we get P = 4000.
P's monthly income = Tk. 4000.
Let the average age of the whole team by x years..
=> 11x - (26 + 29) = 9(x -1)
=> 11x - 9x = 46
=> 2x = 46
=> x = 23.
So, average age of the team is 23 years.
Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
Average of 20 numbers = 0.
Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive
and if their sum is a then 20th number is (-a).
Total sale for 5 months = Tk. (6435 + 6927 + 6855 + 7230 + 6562) = Tk. 34009.
So, Required sale = Tk. [(6500 x 6) - 34009]
= Tk. (39000 - 34009)
= Tk. 4991.
Required average = (67 x 2 + 35 x 2 + 6 x 3)/(2+2+3)
= (134 + 70 + 18)/7
= 222/7
= 31(5/7) years.
Required run rate = (282 - (3.2 x 10))/40
= 250/40 = 6.25
Average, (7+13+p+q) /4 = 17
Or, 20+p+q = 68
or, p + q =48
Again, average = (p+11)+(q-9)/2
= (p+q+2)/2
= (48+2)/2
= 25
Average = (30×83.5)+(5×94) / (30+5)
=85 or, 85%
Let no of people in class X,Y,Z be x,y,z respectively,
total x = 83x
total y = 76y
total z = 85z
total(x+y) = 79(x+y)
total(y+z) = 81(y+z)
implies
83x + 76y = 79(x+y) --------1
76y + 85z = 81(y+z) --------2
1 is 3y=4x
2 is 5y=4z
average of 3 classes = total( x + y + z )/( x + y + z)
= (83x+76y+85z)/( x + y + z)
= 978/12
= 81.5
(5a+4m)=36*9=324
(7a+8m)=48*15=720
Thurs + Fri + Sat = 24 * 3 = 72
Sat = 27
So, Thurs + Fri = 72 - 27 = 45
Again,
Wed + Thurs + Fri = 25 * 3 = 75
or, Wed = 75 - (Thurs + Fri)
or Wed = 75 - 45
= 30
Let the average weight of the 59 students be A.
So the total weight of the 59 of them will be 59 * A.
The questions states that when the weight of this student who left is added, the total weight of the class = 59A + 45
When this student is also included, the average weight decreases by 0.2 kgs.
Thus,
(59A + 45) / 60 = A - 0.2
=> 59A + 45 = 60A - 12
=> 45 + 12 = 60A - 59A
=> A = 57
The average of 5 quantities is 6.
Therefore, the sum of the 5 quantities is 5 * 6 = 30.
The average of three of these 5 quantities is 8.
Therefore, the sum of these three quantities = 3 * 8 = 24
The sum of the remaining two quantities = 30 - 24 = 6.
Average of these two quantities = 6 / 2 = 3
To solve an averages problem, you should multiply for the products of the average.
In this case, a 3.0 GPA after 3 semesters gives us 9. (3 times 3.)
We also know that after the fourth semester, the cumulative GPA is supposed to be a 3.1.
Hence, this multiplier’s product is 12.4. (3.1 times 4.)
Subtracting 9 from 12.4 gives you GPA 3.4.
Total marks of 40 students = 45 * 40
= 1800
Total marks of 30 boys = 50 * 30
= 1500
Total marks of 10 girls = 1800 - 1500
= 300
Average marks of 10 girls = 300/10
&
For April,
Cost of electric bill = (2800 * 0.30) tk
= 840 tk
For May,
Cost of electric bill = (3200 * 0.30) tk
= 960 tk
For June,
Cost of electric bill = (3600 * 0.30) tk
= 1080 tk
Total Cost of 3 Months = (840 + 960 + 1080) tk
&n
(X + Y) / 2 = 40
or, (X + Y) = 80
Again,
(Y + Z) / 2 = 35
or, (Y + Z) = 70
Now,
(X + Y) - (Y + Z) = 80 - 70
or, X - Z = 10
Each of the 10 number is being multiplied by 8. Thus, multipying the average by 8 will seek out the answer.
Average 7 * 8 = 56
Total of 7 numbers = 12x7 = 84
Total of 6 numbers = 11x6 = 66
Discarded number = 18
(M + 2M + 3 + 3M - 5 + 5M + 1) / 4 = 63
or, (11M - 1) = 63 * 4
or, 11M = 252 + 1
or, 11M = 253
or, M = 253 / 11
or, M = 23
Total mark of first 15 students = 15 * 10 = 150
Total mark of first 10 students = 15 * 10 = 150
Total mark of 25 students = 150 + 150 = 300
Average = 300 / 25 = 12
x+x+x+x+x+1+2+3+4+5 = 265
or, 5x + 15 = 265
or, 5x = 265 - 15 = 250
or, x = 50
The largest 5 member of the set is,
50+6 = 56
50+7 =57
50+8 = 58
50+9= 59
50+10= 60
Then,
Average = (56+57+58+59+60) / 5
= 290/5
= 58
30*39 = 1170
29.8*40 = 1192
Weight of new student = 1192 - 1170
= 22
Total of 5 number is: 25*5 = 125
Total of 4 number is: 31*4 = 124
Thus, the removed number is= 125-124 = 1
Current average = M
Total Mark is 100
So the new average is (100+M)/2 = 50+M/2
here,
a+b/2=45
or,a+b=90
again,
b+c/2=35
or,b+c=70
so,
a+b-b-c=90-70=20
sum of six numbers = 6 x 3.95 =23.7
sum of two of them = 2 x 3.4 = 6.8
sum of other two = 2 x 3.85 =7.7
So Sum of four numbers = 14.5
So the remaining two numbers' sum = 9.2
Avg of the two = 4.6
(a + b + c) / 3 = 40
or, a + b + c = 120
or, 3 * (a + b + c) = 120 * 3
or, 3a + 3b + 3c = 360
or, 3a + 3b + 3c + 30 = 360 + 30
or, (3a + 10) + (3b + 10) + (3c + 10) = 390
or, [(3a + 10) + (3b + 10) + (3c + 10)] / 3 = 390/3
= 130
Total of 5 number is = 6.9 * 5 = 34.5
Total of 4 number is = 4.4 * 4 = 17.6
Thus, the deleted number is = (34.5 - 17.6)
= 16.9
3 years ago, their total age was = 81 - (3*3) = 72
Average of 3 person's age = 72/3
= 24
Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
M=10x+y
N=10y+x
M+N=11x+11y=11(x+y)
In other words the answer is a multiple of 11.
Now the question becomes " which of the following is NOT a multiple of 11?"
answer----->181----->choice A
(a + b)/2 = 40
or, (a + b) = 80 .............................................. (1)
and,
b + c = 70 .................................. (2)
(1) - (2), we get,
a + b - b - c = 80 - 70
or, a - c = 10
Let the average age of the whole team by x years.
11x - (26 + 29) = 9(x -1)
11x - 9x = 46
2x = 46
x = 23.
So, average age of the team is 23 years
Total of 5 positive integer = 60*5 = 300
Total of 3 of these integer = 67*3 = 201
Then, The other two number is = (300 - 201) = 99
Thus, if one of the number has to the greatest, another has to be the least, and the least positive integer is 1.
So, the greatest possible value = (99 - 1) = 98
(11 + 17 + x + y)/4 = 19
or, (11 + 17 + x + y) = 19*4 = 76
or, x + y = 76 - 11 - 17
or, x + y = 48
Now,
[(x+5) + (y-7)]/2 = (x+5+y-7)/2 = [(x+y) -2]/2 = (48 -2)/2 = 46/2 = 23
(105 + 117 + 125 + x)/4 = 115
347 + x = 460
x = 460 - 347 = 113
x= age of the youngest child
x+x+3+x+6+x+9+x+12=50
x=4
All1
লসাগু=১৮০
x+y/2+(y+z)/2+(z+x)/2=2+3+4
x+y+z=9
avg=(x+y+z)/3=3
(8+11+x)/3 = 12
or, 8+11+x = 36
or, x = 36 - 8 - 11
or, x = 17
8.5 * 6 = 51
7.2 * 5 = 36
The omitted number is (51 - 36) = 15
(84 X 5) - (80 X 4)
420-320 = 100 (Ans)
65+(8X2.5) = 65+20= 85 (ans.)
The average of 6 numbers is 9 --> the sum of 6 numbers = (mean)*(number of terms) = 9*6 = 54.
The average of 5 numbers is 7.2 --> the sum of 5 numbers = (mean)*(number of terms) = 7.2*5 = 36.
The discarded number = 54 - 36 = 18.
P+Q= 72
R=42
Then,
P+Q+R= 72+42 = 114
Average of P,Q,R = 114/3 = 38
ধরি,
জামালের আয় = ক টাকা
কামালের আয় = ক - ১০ টাকা
সালামের আয় = ২*ক টাকা
এখন,
ক + ক - ১০ + ২*ক = ৯০
বা, ৪*ক = ৯০+ ১০
বা, ৪*ক = ১০০
বা, ক = ১০০/৪ = ২৫ টাকা
অতএব,
জামালের আয় = ২৫ টাকা
কামালের আয় = ২৫ - ১০ = ১৫ টাকা
তাদের গড় আয় = (২৫ + ১৫) / ২ = ৪০/২ = ২০ টাকা
ধরে নেই, সংখ্যাটি ক।
তাহলে,
৫২*৪ + ক + ৩৮*৫ = ৪৬২
বা, ক = ৪৬২ - ২০৮ - ১৯০
বা, ক = ৬৪
Let, the numbers be x, x+1, x+2, x+3, x+4, x+5, x+6
Sum of 7 consecutive number be= 20X7= 140
A.T.Q.
x, x+1, x+2, x+3, x+4, x+5, x+6= 140
7x+21 = 140
7x = 119
.:. x= 17 .:. The largest number= x+6= 17+6 = 23 (Ans.)
3a+5b=10 ..................... (1)
5a + 3b=30 ....................... (2)
We get from (1) + (2),
8a + 8b = 40
or, a + b = 5 (Dividing both sides by 8)
Now, The average would be,
(a + b)/2 = (5)/2 = 2.5
(17+1)./2=9
So,9th number in the series is 42
1st number in the series=42-8x2=26
3rd number in the series=26+2x2=30
Let,numbers are,x,x+2,x+4,x+6,x+8,x+10 and x+12
(7x+42)/7=33
x=27
avg(1, 3, 5, 7 ) = 16/4 = 4
avg(15,17,19,21) = 72/4 = 18
(3 + 13 + 23 + 33 + 43 + 53 + 63 +73 + 83 + 93 )/10
Sum= n/2(2a +(n-1)d)
= 10/2 (6 + 9X10) = 480
Avg = 48
5+6+7+w = 8×4
=> 18+w = 32
=> w = 32-18 = 14 (Ans.)
*সঠিক উত্তর নেই
Total number of share = 4500/90 + 30000/60 = 50+500 = 550
.:. Average price per share = (4500+30000)/550 = 62.73 (Ans)
Let, Karim is x years old
So, Rahim = (x-10) years old
According to question,
(x+5) = 2(x-10+5)
=> x+5 = 2x-10
=> x=15
So, After 3 years Rahim will be = x-10+3 = 8 years (Ans.)
21 X 44 = 924
924 + 66 = 990
990/22 = 45
Let their initial investments be x, 3x and 5x respectively.
Then
X : Y : Z = (x * 4 + 2x * 8) : (3x * 4 + 3x/2 * 8 ): (5x * 4 + 5x/2 * 8 ) = 20x : 24x : 40x= 5 : 6 : 10
First,
you need to solve for x:
(x + x + 2 + x + 4 + x + 6 + x + 8) / 5 = 11
x + x + 2 + x + 4 + x + 6 + x + 8 = 55
5x + 20 = 55
5x = 35
x = 7
Now, substitute for x in each of the last three terms:
x + 4 + x + 6 + x + 8 = (7 + 4) + (7 + 6) + (7 + 8) = (11) + (13) + (15) = 39 / 3 = 13
The mean is 13.
Sum of the ages of (15+5) = 20 employees => 20×(24+2) = 520 years
Sum of the ages of 15 employees => 15×24 = 360
Sum of the ages of new 5 employees = (520-360) = 160
Average = 160/5 = 32.
Average of 12 innings = x
According to the question:
(12x+96)/13 = x+5
x = 96-65 = 31.
Average in 13 innings = 31+5 = 36
w+ x + x + y + y +w = -4 + 25 + 15
Or, 2w+ 2x + 2y = 36
Or, 2 (w + x + y) = 36
Or, w + x + y =18
Or (w + x + y)/3 = 18/3 = 6
So, average of w, x, y = 6
The sum of six numbers =(14 The sum of 4 numbers = .'. Average remaining two numbers
Sum of 6 numbers = (6×25) = 150.
Sum of 3 additional numbers = (3 × 22) = 66.
Sum of (6 + 3) =9 numbers = (150+66)=216
.:. average of 9 numbers = 216/9 = 24
Let, Additional match = x
Now,
30% of 60+ x = 50% of (60+x)
Or, 18+x = 30+0.5x
Or, x = 24
Ans: 117th
Ans: 16th
Ans: 19th
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