A solution contains 25% sugar by weight is made sweeter by doubling the amount of sugar. The percent of sugar, by weight, in the new solution is-
A
33.33%
B
32%
C
40%
D
23.27%
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Question: A solution contains 25% sugar by weight is made sweeter by doubling the amount of sugar. The percent of sugar, by weight, in the new solution is-
Solution:
let, solution is 100 unit
amount of sugar = 100 × 25%
= 100 × (25/100)
= 25 unit
by doubling, amount of sweet = 50 unit
∴ solution now = 100 + 25 = 125 unit
∴ percent of sugar = {(50 × 100)/125}% = 40%
0
Updated: 1 month ago
P(x) = x3 + Zx2 - 6x - 9; Z এর মান কত হলে P(3) = 0 হবে?
Created: 1 month ago
A
3
B
0
C
1
D
- 1
প্রশ্ন: P(x) = x3 + Zx2 - 6x - 9; Z এর মান কত হলে P(3) = 0 হবে?
সমাধান:
P(3) = 0
P(x) = x3 + Zx2 - 6x - 9
∴ P(3) = (3)3 + Z(3)2 - 6 × 3 - 9 = 0
⇒ 27 + 9Z - 18 - 9 = 0
⇒ 27 + 9Z - 27 = 0
⇒ 9Z = 0
∴ Z = 0
0
Updated: 1 month ago
Created: 1 month ago
A
1
B
2
C
√2
D
4
সমাধান:
দেওয়া আছে,
a2 + 1 - √5a = 0
⇒ a2 + 1 = √5a
⇒ (a2 + 1)/a = √5a/a
⇒ (a2/a) - (1/a) = √5
∴ a + (1/a) = √5
আমরা জানি,
{a - (1/a)}2 = {a + (1/a)}2 - 4 · a · (1/a)
= (√5)2 - 4
= 5 - 4
= 1
0
Updated: 1 month ago
(35x -
6 · b4x - 7)/3(x + 2) = b(4x - 7) এবং b > 0 হলে, x এর মান কত?
Created: 1 month ago
A
- 5/2
B
2
C
7/4
D
0
প্রশ্ন: (35x - 6 · b4x - 7)/3(x + 2) = b(4x - 7) এবং b > 0 হলে, x এর মান কত?
সমাধান:
(35x - 6 · b4x - 7)/3(x + 2) = b(4x - 7)
⇒ (35x - 6/3(x + 2) = b(4x - 7)/b(4x - 7)
⇒ 3(5x - 6) = 3(x + 2)
⇒ 5x - 6 = x + 2
⇒ 5x - x = 2 + 6
⇒ 4x = 8
∴ x = 2
0
Updated: 5 days ago
