$\frac{11}{15}$
Give total rod? =25
| Color | Total | Twisted | Clear |
| Blue | 35 | 25 | 10 |
| Red | 40 | 30 | 10 |
Now. we W1″ find the probability.
A. probability of a blue rod from the box =$\frac{Tota.blue.rods}{Total.rods}$ =$\frac{35}{75}=\frac{7}{15}$
B. Probability of a clear rod from the box = $\frac{Tota.clear.rods}{Total.rods}$ = $\frac{10+10}{75}$ =$\frac{4}{15}$
C. Probability of a blue twisted rod = $\frac{Tota.biue.twisted.rods}{Total.rods}$ = $\frac{25}{75}$ =$\frac{1}{3}$
D.Probability of a clear red rod = $\frac{Tota.clear.red.rods}{Total.rods}$ = $\frac{10}{75}$ = $\frac{2}{15}$
E. Probability of a twisted rod = $\frac{Tota.twisted.reds}{Total.rods}$ = $\frac{25+30}{75}$ = $\frac{11}{15}$
60 seconeds
Let, A took t’ seconds
B took ( 170 – x) seconds.
According to question. 2(170 – x) – 10 + (I70 -x) = 170
340-2x – 10+ 170-x =170
=> 330 = 3x
3x = 330
x= $\frac{330}{3}$ = 110
x = 110
A took 10 seconds and B took (I70 – 110) = 60 seconds.
30 years old
Let. Fahim’s age be x year
Shakib is 2x years old.
4 years ago Fihim was (x – 4) years old and 4 years ago shakib was (2x-4) years old
According to question
3(x-4)-6=2x-1
=> 3x- 12-6 = 2x – 4
=> 3x-18 = 2x-4
=> 3x-2x = 18-4 =14
so, x =14
so, After 2 years. Fahim will be = 14+2 = 16 years old
And Shakib will be = (2×14)+2 = 28+2 = 30 years old
5 tk.
Total amount of charity = 20 x 4 = 80 tk.
Here. maximum 3 students can give = 25 Tk
Total = 3 x 25 = 75 Tk.
Minimum amount contributed by one student = 80 – 75 = 5 tk.
$\frac{193}{15}$ is not an integer.
consecutive integers.
So. we let, the numbers are (x – 7), (x- 6), (x- 5), (x – 4), (x- 3), (x – 2), (x – 1), (x+ I), (x+ 2), (X + 3), (x + 4), (x+ 5), (x+ 6), (x + 7)
According to question
x+(x+1) +(x+2)+(x+3) +(x+4) +(x+ 5) + (x+6)+ (x+7)+(x-7)+(x-6)+(x-5) +(x-4) +(x-3) + (x- 2)+(x- 1) = 88
=> 15x + 28 – 28 = 88
=>15x = 88
x = $\frac{88}{15}$
The largest number is (x + 7) = $\frac{88}{15}$ +7 = $\frac{88+105}{15}$ = $\frac{193}{15}$
But. here $\frac{193}{15}$ is not an integer.
So the data given in the question may not be consistent.
Question 6:
3 < x < 9. ($x^{2}$-12x + 27) < 0 will be true.
$x^{2}-12x+27\lt 0$
=> $x^{2}-9x-3x+27<0
=>x(x-9)-3(x-3)$\lt $0
=> (x-9)(x-3)$\lt $0
=> Here if we take x <3
Then ($x^{2}$-12x + 27) < 0 will not be true and if we take x < 9.
Then (x-12x + 27) < 0. will also not be true.
so, x must lie between 3 < x < 9
Besause for each value between 3 < x < 9. ($x^{2}$-12x + 27) < 0 will be true.
Question 7:
$3\lt X\lt 9$
$x^{2}-13x+40\ge 0$
=> $x^{2}-8x-5x+40\ge 0$
=> $x(x-8)-5(x-8)\ge 0$
=> (x-8)(x-5)$\ge 0$
Here the product of (x-8) and (x-5) is grater than zero or equal to zero. so the value of these
When both positive.(x-8)$\ge$0 =>x$\ge$8:(x-5)$\ge$0 => x$\ge$5
When both positive, -(x-8)$\ge$0 => x$\le$8:
-(x-5)$\ge$0=> x$\le$5:
$3\lt X\lt 9$
1:1
The number of seat mom = x
so, The number of 20 sent room = 40 – x
According to question
10x + 20(40 – x) = 600
=> 10x + 800 – 2x = 600
=> -10x =600- 800 = – 200
so, x = $\frac{-200}{-10}$ = 20
Number of 10 seat room = 20 and number of 20 sent room = 40 – 20 =20.
The ratio of the number of 1O seat room to the number of 20 seat room = 20 : 20 = 1:1
$\frac{16}{85}$
Given. total football players 85
Do not score a goal = 85 – 42 = 43 players
Do not receive a yellow card = 85 – 54 = 31 players
Do not score a goal or don‘t receive yellow card or do not do either = 43 + 31 – 5 = 69
Number of players who scored a goal and received a yellow end = 85 – 69 = 16
Required fraction = $\frac{16}{85}$.
