RD Sharma Solutions Class 10 Pair Of Linear Equations In Two Variables Exercise 3.6

RD Sharma Class 10 Solutions Chapter 3 Ex 3.6 PDF Free Download

Exercise 3.6

LEVEL 1

1. 5 pens and 6 pencils together cost Rs 9 and 3 pens and 2 pencils cost Rs 5. Find the cost of 1 pen and 1 pencil.

Solution:

Let the cost of a pen and pencil be Rs x and Rs y respectively.

Then, according to the question,

5x + 6y = 9 …(i)

3x + 2y = 5 …(ii)

Multiplying equation(i) by 2 and equation (ii) by 6, we get

10x + 12y = 18 …(iii)

18x + 12y = 30 …(iv)

Subtracting equation(iii) from equation(iv), we get

18x-10x+12y-12y=30 – 18

8x=12

rd sharma class 10 maths chapter 2 exercise 3.6-1

Therefore, the cost of one pen = Rs. 1.50

And the cost of one pencil = Rs. 0.25

2. 7 audio cassettes and 3 videocassettes cost Rs. 1110, while 5 audio cassettes and 4 videocassettes cost Rs. 1350. Find the cost of audio cassettes and a video cassette.

Solution:

Let the cost of an audio cassette and that of a video cassette be Rs. X and Rs. y, respectively. Then, according to the question,

7x + 3y = 1110 …(i)

5x + 4y = 1350 …(ii)

Multiplying equation(i) by 4 and equation(ii) by 3,

We get,

28x + 12y = 4440 …(iii)

15x + 4y = 4050 …(iv)

Subtracting equation (iv) from equation(iii),

We get,

28x-13x+12y-12y=4440-4050

13x=390

rd sharma class 10 maths chapter 2 exercise 3.6-2

Therefore, the cost of one audio cassette = Rs. 30

And the cost of one video cassette = Rs. 300

3. Reena has pens and pencils which together are 40 in number. If she has 5 more pencils and 5 fewer pens, then the number of pencils would become 4 times the number of pens. Find the original number of pens and pencils.

Solution:

Let the number of pens and pencils be x and y, respectively.

Then, according to the question,

x+y=40 …(i)

(y+5)=4((x-5)

y+5=4x-20

5+20=4x-y

4x-y=25 …(ii)

Adding equation (i) and (ii),

We get

x+4x=40+25

5x=65

rd sharma class 10 maths chapter 2 exercise 3.6-3

Putting x=13 in equation (i), we get

13+y=40

y=40-13-27

Therefore, the number of pens Reena has is 13

And, the number of pencils Reena has is 27

4. 4 tables and 3 chairs, together, cost Rs. 2250 and 3 tables and 4 chairs cost Rs. 1950. Find the cost of 2 chairs and 1 table.

Solution:

Let the cost of 1 table is Rs. x and cost of 1 chair is Rs. y.

Then, According to the question,

4x+3y=2250 …(i)

3x+4y=1950 …(ii)

Multiplying (i) with 3 and (ii) with 4,

We get,

12x+9y=6750 …(iii)

12x+16y=7800 …(iv)

Now, subtracting equation (iv) from (iii),

We get,

-7y=-1050

y=150

Hence,

rd sharma class 10 maths chapter 2 exercise 3.6-4

Therefore, the total cost of 2 chairs and 1 table is:

2y+x =2×150+450

=300+450

=750

5. 3 bags and 4 pens together cost Rs 257 whereas 4 bags and 3 pens together cost Rs. 324. Find the total cost of 1 bag and 10 pens.

Solution:

Let the cost of a bag and a pen be Rs. x and Rs. y, respectively.

Then,

3x+4y=257 …(i)

4x+3y=324 …(ii)

Multiplying equation (i) by 3 and (ii) by 4,

We get,

9x+12y=770 …(iii)

16x+12y=1296 …(iv)

Subtracting equation (iii) from (iv), we get

16x-9x=1296-771

7x=525

x=525/7=75

Hence, the cost of a bag = Rs. 75

Substituting x = 75 in equation (i),

We get,

3 x 75 + 4y =257

225 + 4y = 257

4y = 257-225

4y=32

y=32/4=8

Hence, th cost of a pen = Rs. 8

i.e, the cost of 10 pens = 8 x 10 = Rs. 80

Therefore, the total cost of 1 bag and 10 pens = 75 + 80 = Rs. 155.

6. 5 books and 7 pens together cost Rs. 79 whereas 7 books and 5 pens together cost Rs. 77. Find the total cost of 1 book and 2 pens.

Solution:

Let the cost of a book a pen be Rs. x and Rs. y, respectively. Then,

5x+7y=79 …(i)

7x+5y=77 …(ii)

Multiplying equation (i) by 5 and (ii) by 7,

We get,

25+35y=395 …(iii)

49x+35y=539 …(iv)

Subtracting equation (iii) from (iv),

We get,

49x-25x=539-395

24x=144

x=144/24}=6

Hence , the cost of a book = Rs. 6

Substituting, x= 6 in equation (i) ,

We get,

5 x 6+7y=79

30+7y=79

7y=79-30

7y=49

y=49/7=7

Hence, the cost of a pen = Rs. 7

I.e., the cost of 2 pens = 2 x 7 = Rs. 14

Therefore, the total cost of 1 book and 2 pens = 6+14= Rs. 20

7. Jamila sold a table and a chair for Rs 1050, thereby making a profit of 10% on the table and 25% on the chair. If she had taken a profit of 25% on the table and 10% on the chair she would have got Rs 1065. Find the cost price of each.

Solution:

Let the cost price of the table and of the chair be Rs x and Rs y, respectively.

The selling price of the table, when it is sold at a profit of 10% = Rs x + 10x/100 = 110x / 100

The selling price of the chair, when it is sold at a profit of 25% = Rs y + 25y/100 = 125y / 100

Hence, 110x / 100 + 125y / 100 = 1050 …(i)

The selling price of the table, when it is sold at a profit of 25% = Rs(x + 25x/100) = Rs 125x/ 100

The selling price of the chair, when it is sold at a profit of 10% = Rs(y + 10y/100) = Rs 110y / 100

Hence, 125x / 100 + 110y / 100 = 1065 …(ii)

Solving (i) and (ii),

We get, x = 500 and y = 400

Therefore, the cost price of the table is Rs 500 and that of the chair is Rs 400.

8. Susan invested a certain amount of money in two schemes A and B, which offer interest at the rate of 8% and 9% per annum respectively. She received Rs 1860 as annual interest. However, if she had interchanged the amounts of investment in the two schemes, she would have received Rs 20 more as annual interest. How much money did she invest in each scheme?

Solution:

Let Rs x be the amount of money Susan invested in scheme A at the rate of 8% per annum

And Rs y be the amount of money she invested in scheme B at the rate of 9% per annum

According to the question,

8x+9y=1860 …(i)

9x+8y=1860+20 …(ii)

Multiplying (i) by 9 and (ii) by 8, we get

72x+81y=16740 …(iii)

72x+64y=15040 …(iv)

Subtracting (iv) from (iii)

Hence, 17y = 1700

y=1700/17=100

y = 100

Substituting the value of y in equation (i),

We get,

rd sharma class 10 maths chapter 2 exercise 3.6-5

x = 120

Therefore, Susan invested Rs. 120 in the scheme A and Rs. 100 in the scheme B.

9. The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Solution:

Let the cost of each bat = Rs x

And, that of each ball = Rs y

According to the question,

The coach buys 7 bats and 6 balls for Rs 3800.

I.e., 7x + 6y =3800

6y= 3800- 7x

Dividing LHS and RHS by 6 we get

y = (3800 – 7x) /6 … (i)

According to the question, she later buys 3 bats and 5 balls for Rs 1750.

I.e., 3x + 5y = 1750

Substituting the value of y

rd sharma class 10 maths chapter 2 exercise 3.6-6

Therefore, the cost of each bat = Rs 500

And, the cost of each ball is Rs 50

10. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

Let the fixed charge for the first three days be Rs x and the charge thereafter be Rs y.

According to the question,

x + 4y = 27 … (i)

x + 2y = 21 … (ii)

Subtracting equation (ii) from (i),

We get,

2y = 6

y = 3 … (iii)

Substituting y=3 in equation (i),

We get

x + 12 =27

x = 15

Hence, fixed charge = Rs 15

And, the charge per day = Rs 3.

11. The cost of 4 pens and 4 pencil boxes is Rs 100. Three times the cost of a pen is Rs 15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.

Solution:

Let the price of pens and pencil boxes be Rs. x and Rs. y, respectively.

4x + 4y = 100 …(i)

3x- y=15 …(ii)

From equation (i) and (ii), multiplying equation (ii) by 4

We have,

4x+4y=100

(3x-y=15) x 4

So the equation becomes,

4x+4y=100

12x-4y=60

Adding both equations, we have,

16x=160

x= Rs.10

Substituting the value of x in equation(i),

We get,

40+4y=100

y = Rs. 15

LEVEL 2

12. One says, “Give me a hundred, friend! I shall be twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you,”. Tell me the amount of their respective capital.

Solution:

Le one friend have Rs x and the other friend have Rs y with them.

According to the question,

x + 100 = 2(y − 100)

x + 100 = 2y − 200

x − 2y = −300 …(i)

And, 6(x − 10) = (y + 10)

6x − 60 = y + 10

6xy = 70 …(ii)

Multiplying equation (ii) by 2,

We get,

12x − 2y = 140 …(iii)

Subtracting equation (i) from (iii),

We get,

11x = 140 + 300

11x = 440

x = 40

Substituting x=40 in equation (i),

We get,

40 − 2y = −300

40 + 300 = 2y

2y = 340

y = 170

Therefore, one of the friends had Rs 40 and the other had Rs 170 with them respectively.

13.A and B have a certain number of mangoes. A says to B “If you give 30 of your mangoes I will have twice as many left with you.” B replies, ” If you give me 10, I will have thrice as many as left with you. How many mangoes does each have?

Solution:

Let A and B have x mangoes and y mangoes respectively.

(x+30) = 2(y-30)…….. (i)

x=2y-90

3(x-10) = (y+10)…….. (ii)

3x = y+40

Now, substituting x=2y-90 from equation (i), we get,

3(2y-90) = y+40

6y-y=310

y=62

And, x=2y-90

=(2×62)-90

=124-90

=34

Therefore, A has 34 mangoes and B has 62 mangoes.

14. Vijay had some bananas and he divided them into 2 lots A and B. He sold the first lot Rs 2 for 3 bananas and the 2nd lot Rs 1 per banana and got a total of Rs 400. If he had sold the first lot Rs 1 per banana and the 2nd lot Rs 4 for 5 bananas, his total collection would have been Rs 460. Find the total no. of bananas he had?

Solution:

Let the number of bananas in lot A be a and in lot B be b.

First Condition I:

In lot A, cost of 3 bananas = Rs. 2

i.e., the cost of a bananas in lot A =Rs. a x (⅔)= Rs. 2a/3

In lot B, cost of 1 banana =Rs. 1

i.e., the cost of a bananas in lot B = Rs. b

Total cost of bananas in lot A and lot B = Rs. 400

⅔ a+b=400

2a+3b = 1200 x 3

2a+3b = 1200 …(i)

Second Condition II:

In lot A, cost of 1 banana = Rs. 1

i.e., the cost of x bananas in lot A = Rs. a

In lot B, cost of 5 bananas = Rs. 4

i.e., the cost of b bananas in lot A = Rs. b x ⅘ =Rs. ⅘ b

Total cost of bananas in lot A and lot B = Rs. 460

a+ ⅘ b=460

5a + 4b = 460 x 5

5a + 4b = 2300 …(ii)

To equate equation (i) and (ii), We have to multiply equation (i) by 5 and (ii) by 2,

We get

10a +15 b = 6000 …(iii)

10a + 8b = 4600 …(iv)

Subtracting equation (iii) from (iv)

We get,

-7b = -1400

b= 1400/7=200

Substituting the value of b in (i),

We get,

2a + (3 x 200) = 1200

2a = 1200 – 600

a=600/2=300

Hence, the number of bananas in lot A = 300

And the number of bananas in lot B = 200

Therefore, the total number of bananas in lot A and B = 300 + 200 = 500

15. On selling a T.V. at 5% gain and a fridge at 10% gain shopkeeper gain Rs. 2000 but if he sells the T.V. at 10% gain in the fridge at 5% loss. He gains Rs. 1500 on the transaction. Find the actual price of T.V. and fridge.

Solution:

Let price of T.V. and that of the fridge be x and y respectively.

According to the question,

First Condition:

5% gain on chair= 5/100x= 0.05x

10% gain on fridge= 10/100y =0.1y

Hence,

5x+10y=200000 …(i)

0.05x + 0.1 y = 2000

Second Condition:

10% gain on T.V. = 10/100x =0.1x

5%loss on fridge= -5/100y =-0.05y

Hence,

5y= 10x-150000

y = (10x-150000)/5

y=2x-30000 …(ii)

0.1x – 0.05y = 1500

Solving equations (i) and (ii),

We get,

y= 2x-30000

5x+10y=200000

5x + 10(2x-30000)=200000

5x + 20x -300000= 200000

25x = 500000

x = 20000

Substituting x=20000 in Equation (ii),

We get,

y = 2x-30000

=2(20000)-30000

=40000-30000

=10000

Therefore, the price of the T.V. = Rs. 20,000

And, the price of the fridge= Rs. 10,000

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