Muscle fatigue is caused by the accumulation of:

Solution
Pyruvic acid
Who was the first woman Chief Justice of a High Court of a state in lndia?

Solution
Leila Seth
Insert the missing number:

Solution
The sum of squares of the outside numbers is equal to central number.
(3)^{2}+ (2)^{2} + (2)^{2}+ (4)^{2}
= 9 + 4 + 4 + 16 = 33
(3)^{2}+ (2)^{2} + (5)^{2}+ (4)^{2}
= 9 + 4 + 25 + 16 = 54
similarly
(6)^{2}+ (5)^{2} + (4)^{2}+ (3)^{2}
36 + 25 + 16 + 9 = 86
If 81 × 32 × 56 = 1159
36 × 23 × 11 = 259
then 42 × 18 × 63 = ?

Solution
If ‘+’ stands for division. ‘x’ stands for addition, ‘‘ stands for multiplication and ‘+’ stands for subtraction, then which of the following equations is correct?

Solution
Option (1)
36 × 6 + 3 +5  3
or, 36 × 2 + 5  3
or, 72 + 5  3 = 74
The perimeter of a rhombus is 100 cm. If one of its diagonals is 14 cm, then the area of the rhombus is

Solution
side of rhombus
\(=\sqrt{\left ( \frac{d_{1}}{2} \right )^{1}+\left ( \frac{d_{2}}{2} \right )^{2}}\)
\(=\sqrt{d_{1}^{2}+d_{2}^{2}}\)
Perimeter of rhombus
\(4a=2\sqrt{d_{1}^{2}+d_{2}^{2}}\)
Where d_{1} and d_{2} are diagonals
\(2\sqrt{d_{1}^{2}+d_{2}^{2}}=100\)
\(\Rightarrow \sqrt{d_{1}^{2}+d_{2}^{2}}=50\)
\(\Rightarrow d_{1}^{2}+d_{2}^{2}= 2500\)
\(\Rightarrow (14)^{2}+d_{2}^{2}= 2500\)
\(\Rightarrow d_{2}=\sqrt{2304}= 48\)
∴ Area of the rhombus
\(=\frac{1}{2}\times 14\times 48= 336 sq.cm.\)
2 men and 1 woman together can complete a piece of work in 14 days, while 4 women and 2 men together can do it in 8 days. If a man gets Rs. 600 per day, how much should a woman get per day?

Solution
According to the question,
(2 × 14) men + 14 women
= 2 × 8 men + 4 × 8 women
⇒ (2816) men
=(3214) women
⇒ 12 men = 18 women
⇒ 2 men = 3 women
∴ 1 women = ^{2}⁄_{3} men
∴ Amount received by 1 woman
per day = ^{2}⁄_{3} × 600 = Rs. 400
A takes twice as much time as B and thrice as much as C to completen a piece of work. They together complete the work in 1 day. In what time, will A alone complete the work.

Solution
Let time taken by C to complete
the work = x days
∴ Time taken by A to complete
the work = 3x days
and time taken by B to complete
the work = \(\frac{3x}{2}\) days
According to the question
\(\frac{1}{3x}+\frac{1}{\frac{3x}{2}}+\frac{1}{x}=1\)\(\Rightarrow \frac{1}{3x}+\frac{2}{3x}+\frac{1}{x}=1\)
\(=\frac{1+2+3}{3x}=1\)
\(\Rightarrow \frac{6}{3x}=1\Rightarrow \frac{2}{x}=1\Rightarrow x=2\)
∴ Time taken by A
=3×=3×2=6days
The product of two numbers is 120 and the sum of their squares is 289. The sum of the two numbers is

Solution
Let the numbers be a and b.
According to the question,
ab = 120 ... (i)
and a^{2} + b^{2} = 289 ... (ii)
∴ (a + b)^{2} = a^{2} + b^{2} + 2ab
= 289 + 2 × 120
= 289 + 240 = 529
∴ a + b = \(\sqrt{529}\) = 23
The value of\(\frac{137\times 137+137\times 133+133\times 133}{137\times 137\times 137133 \times 133\times 133}\) is

Solution
Let 137 = a and 133 = b
∴ Expression
\(\frac{a\times a+a\times b+b\times b}{a\times a\times ab\times b\times b}\)\(=\frac{a^{2}+ab+b^{2}}{a^{3}b^{3}}\)
\(\frac{a^{2}+ab+b^{2}}{(a  b)(a^{2} + ab + b^{2}) }=\frac{1}{a  b}\)
\(=\frac{1}{137 133}=\frac{1}{4}\)