We have
\(\frac{-4}{5} \times \frac{3}{7} \times \frac{15}{16} \times\left(\frac{-14}{9}\right)\) 
\(=\left(-\frac{4 \times 3}{5 \times 7}\right) \times\left(\frac{15 \times(-14)}{16 \times 9}\right)\) 
\(=\frac{-12}{35} \times\left(\frac{-35}{24}\right)=\frac{-12 \times(-35)}{35 \times 24}=\frac{1}{2}\) 
We can also do it as.
\(\frac{-4}{5} \times \frac{3}{7} \times \frac{15}{16} \times\left(\frac{-14}{9}\right)\) 
\(=\left(\frac{-4}{5} \times \frac{15}{16}\right) \times\left[\frac{3}{7} \times\left(\frac{-14}{9}\right)\right]\) (Using commutativity and associativity)
\(=\frac{-3}{4} \times\left(\frac{-2}{3}\right)=\frac{1}{2}\) 

Length: 60 m; Breadth: 40 m

A: 20 years, B: 10 years

85 (or 58)

47 (or 74)

Rs 50 \(\rightarrow \) 6, Rs 20 \(\rightarrow \) 10, Rs 10 \(\rightarrow \) 19

Let the number of five-rupee coins that Bansi has be x. Then the number of two-rupee coins he has is 3 times x or 3x.
The amount Bansi has:
(i) from 5 rupee coins, RS. 5 x x = RS. 5x
(ii) from 2 rupee coins, RS. 2 x 3x = RS. 6x
Hence the total money he has = RS. 11x
But this is given to be  RS.77; therefore,
11x = 77
or \(x=\frac{77}{11}=7\)
Thus, number of five-rupee coins = x = 7
and number of two-rupee coins = 3x = 21

If x is a multiple of 11, the next multiple is x + 11. The next to this is x + 11 + 11 or x + 22. So we can take three consecutive multiples of 11 as x, x + 11 and x + 22.
It is given that the sum of these consecutive multiples of 11 is 363. This will give the following equation:
x + (x + 11) + (x + 22) = 363
or x + x + 11 + x + 22 = 363
or 3x + 33 = 363
or 3x = 363 – 33
or 3x = 330
\(x=\frac{330}{3}\)
= 110
Hence, the three consecutive multiples are 110, 121, 132

18, 19 and 20

We have \(\frac { 1 }{ x } -\frac { 3 }{ x } \frac { 5 }{ 2x } -3\Rightarrow \frac { 1-3 }{ x } -\frac { 5 }{ 2x } =-3\)
\(\Rightarrow\) \(\frac { -2 }{ x } -\frac { 5 }{ 2x } =-3\quad \Rightarrow \frac { -4-5 }{ 2x } =-3\)
\(\Rightarrow\) \(\frac { -9 }{ 2x } =-3\)
\(\Rightarrow\) \(x=\frac { -9 }{ 2 } x\left( \frac { -1 }{ 3 } \right) \Rightarrow x=\frac { 3 }{ 2 } \)

Let x be the requires number 
\(\therefore\) \(3x\left( \frac { -13 }{ 4 } \right) -x=\frac { 5 }{ 8 } \)
\(\Rightarrow\)  \(\frac { -39 }{ 4 } -x=\frac { 5 }{ 8 } \Rightarrow \frac { -39 }{ 4 } -\frac { 5 }{ 8 } =x\)
\(\Rightarrow\) \(\frac { -78-5 }{ 8 } =5x\Rightarrow x=-\frac { 83 }{ 8 } \)

y = 19

There are infinitely many rational numbers greater than -2.
Five of them are \({-3\over2},{-1},{-1\over2},0,{1\over2}\)

we have,\({-5\over8}\times{-3 \over7}=​​{-5\over8}\times{-3 \over7}={(-5)\times(-3)\over 8\times7}={15\over56}\)
\(\therefore\)The multiplicative inverse of \({-5\over8}\times{-3 \over7}\) is \(56\over15\)

60 yr

We have, x =\(11\over15\)
LHS = -(-x) = -\(({11\over15})=​​​​{11\over15}=X=RHS\)
S0, - (-x) = x is verified for x =\(11\over15\)

 \(\frac { 2x }{ 3 } +1=\frac { 7x }{ 15 } +3\) 
\(\Rightarrow\) \(\frac { 2x }{ 3 } +1=\frac { 7x }{ 15 } - 3\) [transposing \(\frac { 7x }{ 15 } \) to LHS and 1 to RHS]
\(\Rightarrow\)\(\frac { 10x-7x }{ 15 } =2\) [LCM of 3 and 15 = 15]
\(\Rightarrow\) \(\frac { 3x }{ 15 } =2\Rightarrow \frac { 3x }{ 15 } x5=2x5\)
\(\Rightarrow\) \(\frac { 15x }{ 15 } \)=2 x 5
x = 2 x 5 \(\Rightarrow\) x = 10
which is the required solution.

We have,  8x + 4 = 3 (x -1) + 7
\(\Rightarrow\) 8x + 4 = 3x - 3 + 7
\(\Rightarrow\) 8x + 4 =3x + 4
\(\Rightarrow\) 8x -3x = 4- 4  [transposing3x to LHS and 4 to RHS]
\(\Rightarrow\) 5x = 0 \(\Rightarrow\) x = \(\frac { 0 }{ 5 } \) [dividing both sides by 5]
x = 0, which is the required solution

We have 1.6 =\(\frac { y }{ 1.5 } \) \(\Rightarrow\) 1.6 x 1. 5 = \(\frac { y }{ 1.5 } \) x 1.5 [multiplying both sides by 1.5]
1.6 x 1.5 = y \(\Rightarrow\) 2.40 = y
y = 2.4, Which is the required solution

We have, 6x = 12 \(\frac { 6x }{ 6 } =\frac { 12 }{ 6 } \) [dividing both sides by 6]
\(\Rightarrow \ x=\frac { 12 }{ 6 } =2\) which is the required solution