Give Scientific reson
food on which fungi has grown cannot be eaten?
Answers
(B) 2 ) attached.
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(Q3) 1)
Let, the age of Manish = a
age of Savita = b
Sum of their ages is 31 year:
➡a+b = 31 .. .. (1)
→ a=31-b
3 years ago:
Manish's age = (a - 3) years Savita's age = (b - 3) years
In question: that time
→ Manish = 4 * Savita
(a - 3) = 4(b - 3)
➡a-3 = 4b - 12
➡a-4b = -12 + 3
(31- b) - 4b = -9
{a=31 - b}
→ 31 - 5b = -9
→ 31+ 9 = 5b → 40 = 5b
→ (40/5) = b
⇒b=8
Putting the value of x in equation 1,
→ a=31-8= 23 year
Age of manish = a = 23 years
Age of manish = a =23 yearsAge of savita = b = 8 years
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(Q4) (1)
Let the present age of me be x and my son
be y.
Two years ago,
My age was x-2 and
my son's age was y-2
It is given that two years ago my age was 4 and half times the age of my son. Or in other words, my age was 2 times the age of my son.
So, (x-2) = 4.5(y-2)
⇒ x = (9/2)y - 7
Six years ago,
My age was x-6 and
My son's age was y-6
It is given that my age was twice the square of the age of my son.
So, (x-6) = 2×(y-6)²
Substituting the value of x from (1), we get:
(9/2)y - 7 - 6 = 2(y-6)²
(9/2)y - 13 = 2(y-6)²
→ (9/2)y - 13 = 2y² + 72 - 24y
=> (9/2)y= 2y² - 24y + 85
=> 9y = 4y² - 48y + 170
=>4y² - 57y + 170 = 0
=>4y²-40y - 17y + 170 = 0
=>4y(y - 10) - 17(y - 10) = 0
= (y-10)(4y-17)
By zero product rule, we get y = 10 and y = 17/4 = 4.25 y = 4.25 is discarded as age cannot be fractional.
So, present age of son is y = 10 years.
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(Q5) (2)
Let two numbers be x be the smaller number and y be the greater number. From first condition,
x+y=60
From 2nd condition,
3x+8 =y
3x-y =-8
Adding equation 1 and 2
x+y=60
+3x-y =-8
x=13
substituting x=13 in equation 1
x+y=60
13+y=60
y=60-13
y=47
smaller number is 13 and greater number is 47
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all done bestu bus copy kar do and rest karo tabiyat thik nhi ha na zaida tention mat lo mai hu na apke sath hamesha help karne ke lia ☺✌
question is attached bcoz no space was left so i have given here hope u understand
