Ans. (i) … (1) 

 … (2)

Let = p and = q

Putting this in equation (1) and (2), we get

 

 

Multiply both equation by 6, we get

⇒ 3p + 2q = 12 and 2p + 3q = 13

⇒ 3p + 2q – 12 = 0    ………..… (3)

and 2p + 3q – 13 = 0 ……….… (4)

⇒ 

⇒ 

⇒ 

⇒ p = 2 and q = 3

But = p and = q

Putting value of p and q in this we get

x =  and y = 

(ii) … (1)

 … (2)

Let = p and = q

Putting this in (1) and (2), we get

2p + 3q = 2 … (3)

4p − 9q = −1 … (4)

Multiplying (3) by 2 and subtracting it from (4), we get

=> 4p − 9q  – 2 (2p + 3q) = -1 – 2(2)

⇒ 4p − 9q  − 4p − 6q = -1 -4

⇒ −15q = -5

⇒ q = 

Putting value of q in (3), we get

=> 2p + 1 = 2

⇒ 2p = 1

⇒ p = ½

Putting values of p and q in (= p and = q), we get

 and 

⇒ 

⇒ x = 4 and y = 9

(iii) + 3y = 14 … (1)

 − 4y = 23 … (2)

Let = p

we get

4p + 3y = 14 … (3)

3p − 4y = 23 … (4)

Multiplying (3) by 3 and (4) by 4, we get

3 (4p + 3y – 14 =0 ) and, 4 (3p − 4y – 23 = 0)

⇒ 12p + 9y – 42 = 0 … (6) 12p − 16y – 92 = 0 … (7)

Subtracting (7) from (6), we get

9y − (−16y) – 42 − (−92) = 0

⇒ 25y + 50 = 0

⇒ y = −5025−5025= −2

Putting value of y in (4), we get

4p + 3 (−2) = 14

⇒ 4p – 6 = 14

⇒ 4p = 20

⇒ p = 5

Putting value of p in (3), we get

 = 5

⇒ x = 

Therefore, x =  and y = −2

(iv)  … (1)

 … (2)

Let 

Putting this in (1) and (2), we get

5p + q = 2

⇒ 5p + q – 2 = 0 … (3)

And, 6p − 3q = 1

⇒ 6p − 3q – 1 = 0 … (4)

Multiplying (3) by 3 and adding it to (4), we get

3 (5p + q − 2) + 6p − 3q – 1 = 0

⇒ 15p + 3q – 6 + 6p − 3q – 1 = 0

⇒ 21p – 7 = 0

⇒ p = 

Putting this in (3), we get

5 () + q – 2 = 0

⇒ 5 + 3q = 6

⇒ 3q = 6 – 5 = 1

⇒ q = 

Putting values of p and q in (), we get

⇒ 3 = x − 1 and 3 = y – 2

⇒ x = 4 and y = 5

(v) 7x − 2y = 5xy … (1)

8x + 7y = 15xy … (2)

Dividing both the equations by xy, we get

Let = p and = q

Putting these in (3) and (4), we get

7q − 2p = 5 … (5)

8q + 7p = 15 … (6)

From equation (5),

2p = 7q – 5

⇒ p = 

Putting value of p in (6), we get

8q + 7 () = 15

⇒ 16q + 49q – 35 = 30

⇒ 65q = 30 + 35 = 65

⇒ q = 1

Putting value of q in (5), we get

7 (1) − 2p = 5

⇒ 2p = 2

⇒ p = 1

Putting value of p and q in (= p and = q), we get x = 1 and y = 1

(vi) 6x + 3y − 6xy = 0 … (1)

2x + 4y − 5xy = 0 … (2)

Dividing both the equations by xy, we get

Let = p and = q

Putting these in (3) and (4), we get

6q + 3p – 6 = 0 … (5)

2q + 4p – 5 = 0 … (6)

From (5),

3p = 6 − 6q

⇒ p = 2 − 2q

Putting this in (6), we get

2q + 4 (2 − 2q) – 5 = 0

⇒ 2q + 8 − 8q – 5 = 0

⇒ −6q = −3⇒ q = ½

Putting value of q in (p = 2 – 2q), we get

p = 2 – 2 (½) = 2 – 1 = 1

Putting values of p and q in (= p and = q), we get x = 1 and y = 2

(vii)  … (1)

 …(2)

Let 

Putting this in (1) and (2), we get

10p + 2q = 4 … (3)

15p − 5q = −2 … (4)

From equation (3),

2q = 4 − 10p

⇒ q = 2 − 5p … (5)

Putting this in (4), we get

15p – 5 (2 − 5p) = −2

⇒ 15p – 10 + 25p = −2

⇒ 40p = 8⇒ p = 

Putting value of p in (5), we get

q = 2 – 5 () = 2 – 1 = 1

Putting values of p and q in (), we get

⇒ x + y = 5 … (6) and x – y = 1 … (7)

Adding (6) and (7), we get

2x = 6 ⇒ x = 3

Putting x = 3 in (7), we get

3 – y = 1

⇒ y = 3 – 1 = 2

Therefore, x = 3 and y = 2

(viii)  … (1)

 … (2)

Let 

Putting this in (1) and (2), we get

p + q =  and 

⇒ 4p + 4q = 3 … (3) and 4p − 4q = −1 … (4)

Adding (3) and (4), we get

8p = 2 ⇒ p = ¼

Putting value of p in (3), we get

4 (¼) + 4q = 3

⇒ 1 + 4q = 3

⇒ 4q = 3 – 1 = 2

⇒ q = ½

Putting value of p and q in , we get

⇒ 3x + y = 4 … (5) and 3x – y = 2 … (6)

Adding (5) and (6), we get

6x = 6 ⇒ x = 1

Putting x = 1 in (5) , we get

3 (1) + y = 4

⇒ y = 4 – 3 = 1

Therefore, x = 1 and y = 1

Ans. (i) Let speed of rowing in still water = x km/h 

Let speed of current = y km/h

So, speed of rowing downstream = (x + y) km/h

And, speed of rowing upstream = (x − y) km/h

According to given conditions,

⇒ 2x + 2y = 20 and 2x − 2y = 4

⇒ x + y = 10 … (1) and x – y = 2 … (2)

Adding (1) and (2), we get

2x = 12

⇒ x = 6

Putting x = 6 in (1), we get

6 + y = 10

⇒ y = 10 – 6 = 4

Therefore, speed of rowing in still water = 6 km/h

Speed of current = 4 km/h

(ii) Let time taken by 1 woman alone to finish the work = x days

Let time taken by 1 man alone to finish the work = y days

So, 1 woman’s 1-day work = ()th part of the work

And, 1 man’s 1-day work = ()th part of the work

So, 2 women’s 1-day work = ()th part of the work

And, 5 men’s 1-day work = ()th part of the work

Therefore, 2 women and 5 men’s 1-day work = (+)th part of the work… (1)

It is given that 2 women and 5 men complete work in = 4 days

It means that in 1 day, they will be completing th part of the work … (2)

Clearly, we can see that (1) = (2)

⇒  … (3)

Similarly,  … (4)

Let 

Putting this in (3) and (4), we get

2p + 5q =  and 3p + 6q = 

⇒ 8p + 20q = 1 … (5) and 9p + 18q = 1 … (6)

Multiplying (5) by 9 and (6) by 8, we get

72p + 180q = 9 … (7)

72p + 144q = 8 … (8)

Subtracting (8) from (7), we get

36q = 1

⇒ q = 

Putting this in (6), we get

9p + 18 () = 1

⇒ 9p = ½

⇒ p = 

Putting values of p and q in , we get x = 18 and y = 36

Therefore, 1 woman completes work in = 18 days

And, 1 man completes work in = 36 days

(iii) Let speed of train = x km/h and let speed of bus = y km/h

According to given conditions,

Let 

Putting this in the above equations, we get

60p + 240q = 4 … (1)

And 100p + 200q =  … (2)

Multiplying (1) by 5 and (2) by 3, we get

300p + 1200q = 20 … (3)

300p + 600q =  … (4)

Subtracting (4) from (3), we get

600q = 20 − = 7.5

⇒ q = 

Putting value of q in (2), we get

100p + 200 () = 

⇒ 100p + 2.5 = 

⇒ 100p = – 2.5

⇒ p = 

But 

Therefore, x =km/h and y = km/h

Therefore, speed of train = 60 km/h

And, speed of bus = 80 km/h