1.

som added 4 to a number x and then multiplied the sum by 2 .then he subtracted three fourths of a number b from the previous result. write the finial result (number) obtained by him .​

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Solution 4:

(i) x+4=9

x=9-4

x= 5

(II) y-2=8

y=8+2

y=10

(iii) a\times10\ =\ 70a×10 = 70

(iv) \frac{b}{5}=\ 6\

5

b

= 6

b= 30

(v) \frac{3}{4}\times t\ =\ 15\

4

3

×t = 15

t\ =\ 15\times\frac{4}{3}t = 15×

3

4

t = 20

(vi) \frac{b}{5}=\ 6\

5

b

= 6

b\ =\ 6\ \times5\b = 6 ×5

b = 30

(vii) \left(\frac{1}{4}\times x\right)-\ 4\ =\ 4\(

4

1

×x)− 4 = 4

\frac{x}{4}-\ 4\ =\ 4\

4

x

− 4 = 4

\frac{x}{4}=4+4\

4

x

=4+4

x\ =\ 8\times4x = 8×4

x= 32

(viii) 6y-6 = 60

6y = 60+6

6y = 66

y\ =\ \frac{60}{6}y =

6

60

y = 11

(ix) \frac{z}{3}+\ 3\ =\ 30\

3

z

+ 3 = 30

\frac{z}{3\ }=\ 33

3

z

= 33

z\ =\ 33\times3z = 33×3

z = 99


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