ML Aggarwal Solution Class 9 Chapter 6 Problems on Simultaneous Linear Equations Exercise 6.1

 Exercise 6.1

Question 1

The sum of two numbers is 50 and their difference is 16. Find the numbers.

Sol :

Let two numbers are x and y

According to first condition of the problem,

x+y=50..(1)

According to second condition of the problem,

x-y=160..(2)

From equation (1) and (2), we get

Adding ,

$\begin{array}{r}x+y=50 \\x-y=16 \\\hline 2 x=66\end{array}$

⇒$x=\frac{66}{2}=33$

Substituting the value of x in equation (1), we get

⇒33+y=50

⇒y=50-33

⇒y=17

Question 2

The sum of two numbers is 2. If their difference is 20, find the numbers.

Sol :

Let two numbers are x and y 

According to first condition of the problem,

x+y=50...(1)

According to second condition of the problem,

x-y=16..(2)

From equation (1) and (2) , we get

Adding 

$\begin{aligned}x+y&=50 \\x-y&=16 \\\hline 2 x&=66 \\\hline\end{aligned}$

⇒$x=\frac{66}{2}=33$

Substituting the value of x in equation (1), we get

⇒33+y=50

⇒y=50-33

⇒y=17

From equation (1) and equation (2), we get

Adding,

$\begin{aligned}x+y&=2 \\x-y&=20 \\\hline 2 x&=22\end{aligned}$

⇒$x=\frac{22}{2}=11$

Substituting the value of x in equation (1), we get

⇒11+y=2

⇒y=2-11

⇒y=-9

⇒Hence, the numbers are 11 and -9

Question 3

The sum of two numbers is 43. If the larger is doubled and the smaller is tripled, the difference is 36. Find the two numbers.

Sol :

Let the two numbers are x and y 

According to first condition of the problem,

⇒x+y=43..(1)

According to second condition of the problem, 

⇒2x-3y=36...(2)

Multiplying equation (1) by 3, we get

Adding , 

$\begin{array}{l}3 x+3 y=129..(3) \\2 x-3 y=36...(2) \\\hline 5 x=165\end{array}$

$x=\frac{165}{5}$

⇒x=33

Substituting the value of x in equation (1), we get

⇒33+y=43

⇒y=43-33

⇒y=10

Hence the two numbers are 33 and 10

Question 4

The cost of 5 kg of sugar and 7 kg of rice is Rs. 153, and the cost of 7 kg of sugar and 5 kg of rice is Rs. 147. Find the cost of 6 kg of sugar and 10 kg of rice.

Sol :

Let cost of 1 kg sugar Rs x

and let cost of 1 kg rice Rs y 

Then cost of 5 kg of sugar 

=Rs. 5 x cost of 7 kg of Rice = Rs 7 y

Cost of 7 kg of sugar Rs 7 x 

Cost of 5 kg of Rice = Rs. 5 y

According to first condition of the problem

⇒5x+7y=153...(1)

According to second condition of the problem,

⇒7x+5y=147...(2)

Multiplying equation (1)  by 7 and equation (2) by 5

we get

Subtracting,

$\begin{aligned}35 x+49 y&=1071...(3) \\35 x+25 y&=735...(4) \\-\quad-\quad&- \\\hline 24 y&=336\end{aligned}$

⇒$y=\frac{336}{24}$

⇒y=14

Substituting the value of y in equation (1), we get

⇒5x+7×14=153

⇒5x+98=153

⇒5x=153-98

⇒5x=55

⇒$x=\frac{55}{5}$

⇒x=11

⇒Hence, cost of 6kg of sugar=6×11=66

Cost of 10kg of rice=10×14=140

Then cost of 6kg of sugar and 10kg of rice

=66+140=206

Question 5

The class IX students of a certain public school wanted to give a farewell party to the outgoing students of class X. They decided to purchase two kinds of sweets, one costing Rs. 70 per kg and the other costing Rs. 84 per kg. They estimated that 36 kg of sweets were needed. If the total money spent on sweets was Rs. 2800, find how much sweets of each kind they purchased.

Sol :

Let sweets purchased x kg which cost

Rs 70 per kg

and let sweets purchased y kg cost 

Rs 84 per kg

According to first condition of the problem

⇒x+y=36..(1)

According to second condition of the problem,

⇒70x+84y=2800..(2)

Multiplying equation (1) by 70, we get

Subtracting, 

$\begin{array}{r}70 x+70 y=2520...(3) \\70 x+84 y=2800..(2) \\\hline-14 y=-280\end{array}$

⇒$y=\frac{-280}{-14}=20$

Substituting the value of y in equation (1) we get

⇒x+20=36

⇒x=36-20=16

⇒Hence, sweets purchased 16 kg which cost Rs 70 per kg

and sweets purchased 20 kg which cost Rs 84 per kg

Question 6

If from twice the greater of two numbers 16 is subtracted, the result is half the other number. If from half the greater number 1 is subtracted, the result is still half the other number. What are the numbers.

Sol :

Let the greater number =x

and let smaller number=y

According to first condition of the problem,

⇒$2 x-16=\frac{y}{2}$

⇒$2 x-\frac{y}{2}=16$

⇒$\frac{4 x-y}{2}=16$

⇒4x-y=32...(1)

According to second condition of the problem

⇒$\frac{x}{2}-1=\frac{y}{2}$

⇒$\frac{x}{2}-\frac{y}{2}=1$

⇒$ \frac{x-y}{2}=1$

⇒x-y=2..(2)

From equation (1) and (2), we get

$\begin{aligned}4 x-y&=32...(1) \\x-y&=2...(2) \\-\quad+&\quad- \\\hline 3 x&=30\end{aligned}$

Subtracting

⇒x=\frac{30}{3}

⇒x=10

Substituting the value of  x in equation (2), we get 

⇒10-y=2

⇒10-2=y

⇒8=y

⇒y=8

Hence , two numbers are 10 and 8

Question 7

There are 38 coins in a collection of 20 paise coins and 25 paise coins. If the total value of the collection is Rs. 8.50, how many of each are there ?

Sol :

Let number of coins of 20 paise =x

and let number of coins of } 25 paise =y

Then, according to first condition of given problem,

⇒x+y=38..(1)

According to second condition of given problem,

⇒20x+25y=850...(2)

[∴Rs 8.50=850 paise]

⇒Multiplying equation (1) by 20, we get

Subtracting, 

$\begin{aligned}20 x+20 y &=760..(3) \\20 x+25 y &=850...(2) \\-\quad-&-\\\hline-5 y=-90\end{aligned}$

⇒$y=\frac{-90}{-5}=18$

Substituting the value of y in equation (1), we get

⇒x+18=38

⇒x=38-18

⇒x=20

Hence , 20 paise 20 coins and 25 paise 18 coins

Question 8

A man has certain notes of denominations Rs. 20 and Rs. 5 which amount to Rs. 380. If the number of notes of each kind is interchanged, they amount to Rs. 60 less as before. Find the number of notes of each denomination.

Sol :

Let number of 20 rupee notes =x

and number of 5 rupee notes =y

According to first condition of given problem,

⇒20x+5y=380..(1)

According to second condition of given problem, 

⇒5x+20y=380-60

⇒5x+20y=320...(2)

Multiplying equation (1) by 4, we get

Subtracting

$\begin{aligned}80 x+20 y&=1520 \\5 x+20 y&=320 \\-\quad-\quad& \quad- \\\hline 75 x \quad=1200\end{aligned}$

⇒$x=\frac{1200}{75}$

⇒x=16

Substituting the value of x in equation (1) , we get

⇒20×16+5y=380

⇒320+5y=380

⇒5y=380-320

⇒5y=60

⇒$y=\frac{60}{5}=12$

Hence , Number of 20 rupee notes=16 and 5 rupee notes=12

Question 9

The ratio of two numbers is $\frac{2}{3}$ .If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.

Sol :

Let two numbers are x and y

Given ratio of two numbers is $=\frac{2}{3}$ reciprocal of the ratio is $=\frac{3}{2}$

According to first condition of given problem,

⇒$\frac{x}{y}=\frac{2}{3}$

⇒3x=2y

⇒3x-2y=0...(1)

According to second condition of given problem,

⇒$\frac{x-2}{y-8}=\frac{3}{2}$

⇒2(x-2)=3

⇒3(y-8)...(1)

⇒2x-4=3y-24

⇒2x-3y=-24+4

⇒2x-3y=-20..(2)

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

Subtracting 

$\begin{aligned}9 x-6 y&=0...(3) \\4 x-6 y&=-40..(4) \\-\quad+\quad &\quad+\\\hline 5x&=40\end{aligned}$

⇒$x=\frac{40}{5}=8$

Substituting the value of x in equation (1) we get

⇒3×8-2y=0

⇒24=2y

⇒2y=24

⇒$y=\frac{24}{2}$

⇒y=12

Hence, the numbers are 8 and 12

Question 10

If 1 is added to the numerator of a fraction, it becomes $\frac{1}{5}$  ; if 1 is taken from the denominator, it becomes $\frac{1}{7}$, find the fraction.

Sol :

Let the fraction $=\frac{x}{y}$

According to given first condition of problem,

⇒$\frac{x+1}{y}=\frac{1}{5}$

⇒5(x+1)=y 

⇒5x+5=y

⇒5x-y=-5...(1)

According to second condition of given problem,

⇒$\frac{x}{y-1}=\frac{1}{7}$

⇒7x=1(y-1)

⇒7x=y-1

⇒7x-y=-1..(2)

From equation (1) and (2) , we get

Subtracting ,

⇒$\begin{aligned}5 x-y&=-5 \\7 x-y&=-1 \\-\quad+&\quad+ \\\hline-2 x&=-4\end{aligned}$

⇒$x=\frac{-4}{-2}=2$

Substituting the value of x in equation (1) , we get

⇒5×2-y=-5

⇒10-y=-5

⇒-y=-5-10

⇒-y=-15

⇒y=15

Hence, fraction $=\frac{x}{y}=\frac{2}{15}$

Question 11

If the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to $\frac{5}{8}$ and if the numerator and denominator are each diminished by 1, the fraction becomes equal to $\frac{1}{2}$ , find the fraction.

Sol :

Let the fraction $=\frac{x}{y}$

⇒According to first condition of given problem,

⇒$\frac{x+2}{y+1}=\frac{5}{8}$

⇒8(x+2)=5(y+1)

⇒8x+16=5y+5

⇒8x-5y=5-16

⇒8x-5y=-11..(1)

According to second condition of given problem,

⇒$\frac{x-1}{y-1}=\frac{1}{2}$

⇒2(x-1)=1(y-1)

⇒2x-2=y-1

⇒2x-y=-1+2

⇒2x-y=1...(2)

Multiplying equation (2), by 5, we get

Subtracting

$\begin{aligned}10 x-5 y&=5 \\8 x-5 y&=-11 \\-\quad+& \quad+ \\\hline 2 x=16\end{aligned}$

⇒$x=\frac{16}{2}$

⇒x=8

Substituting the value of x in equation (2), we get

⇒2×8-y=1

⇒16-y=1

⇒-y=-15

⇒y=15

Hence, fraction  $=\frac{x}{y}=\frac{8}{15}$

Question 12

Find the fraction which becomes $\frac{1}{2}$ when the denominator is increased by 4 and is equal to $\frac{1}{8}$, when the numerator is diminished by 5.

Sol :

Let the fraction $=\frac{x}{y}$

According to first condition of given problem,

⇒$\frac{x}{y+4}=\frac{1}{2}$

⇒2x=1(y+4)

⇒2x=y+4

⇒2x-y=4..(1)

According to second condition of given problem,

⇒$\frac{x-5}{y}=\frac{1}{8}$

⇒8(x-5)=y

⇒8x-40=y

⇒8x-y=40...(2)

From equation (1) and equation (2), we get

Subtracting

$\begin{aligned}2 x-y&=4 \\8 x-y&=40 \\-\quad +&\quad - \\\hline-6 x&=-36\end{aligned}$

⇒$x=\frac{-36}{-6}=6$

Substituting the value of x in equation (1), we get

⇒2×6-y=4

⇒12-y=4

⇒12-4=y

⇒8=y

⇒y=8

Hence, fraction $=\frac{6}{8}$

Question 13

In a two digit number the sum of the digits is 7. If the number with the order of the digits reversed is 28 greater than twice the unit’s digit of the original number, find the number.

Sol :

Let the digits at tens place=x

and let the digits at unit place =y 

According to first condition of problem

⇒x+y=7...(1)

Also, number=10×x+y×1=10x+y

Reversing the number=10×y+x×1=10y+x

According to second condition of problem,

⇒10y+x=2y+28

⇒10y+x-2y=28

⇒x+8y=28..(1)

From equation (1), and (2), we get

Subtracting 

$\begin{aligned}x+y&=7 \\x+8 y&=28 \\-\quad-&\quad - \\\hline-7 y&=-21\end{aligned}$

⇒y=\frac{-21}{-7}=3

Substituting the value of y in equation (1), we get

⇒x+3=7

⇒x=7-3

⇒x=4

Hence, number=10×4+3

=40+3=43

Question 14

A number of two digits exceeds four times the sum of its digits by 6 and it is increased by 9 on reversing the digits. Find the number.

Sol :

Let the digit at ten's place =x and let digit at unit place =y

Number =10×x+y×1=10x+y

Reversing the number =10×y+x×1=10y+x

According to first condition of given problem,

⇒10x+y=4(x+y)+6

⇒1x+y=4x+4y+6

⇒3(2x-y)=6

⇒$2x-y=\frac{6}{3}$

⇒2x-y=2...(1)

According to second condition of given problem

⇒10x+y+9=10y+x

⇒10 x+y-10 y-x=-9

⇒9x-9 y=-9

⇒$x-y=\frac{-9}{9}$

⇒x-y=1...(2)

From equation (1) and (2), we get

Subtracting 

$\begin{aligned}2 x-y&=2...(1) \\x-y&=-1..(2) \\-\quad+&\quad+ \\\hline x&=3\end{aligned}$

Substituting the value of x in equation (1), we get

⇒2×3-y=2

⇒6-y=2

⇒-y=2-6

⇒-y=4

⇒y=4

Hence, number=10×3+4=30+4=34

Question 15

When a two digit number is divided by the sum of its digits the quotient is 8. If the ten’s digit is diminished by three times the unit’s digit the remainder is 1. What is the number ?

Sol :

Let the digit at ten's place =x

and let digit at unit place=y

Number=10×x+y×1=10x+y

According to first condition of given problem

⇒$\frac{10 x+y}{x+y}=8$

⇒10x+y=8(x+y)

⇒10x+y=8x-8y

⇒2x-7y=0...(1)

According to second condition of given problem,

⇒x-3y=1...(2)

Multiplying equation (2) by 2, we get

Subtracting,

$\begin{aligned}2x-6 y&=2...(3) \\2 x-7y&=0..(1) \\-\quad+&\quad- \\\hline y=2\end{aligned}$

Substituting the value of y in equation (1), we get

⇒2x-7×2=0

⇒2x=14

⇒$x=\frac{14}{2}$

⇒x=7

Hence, number=10×7+2=70+2

=72

Question 16

The result of dividing a number of two digits by the number with digits reversed is $1\frac{3}{4}$ . If the sum of digits is 12, find the number.

Sol :

Let the digit at ten's place =x

and let digit at unit place =y, number =10 x+y

Reversing the number =10×y+x×1=10y+x

According to first condition of given problem,

⇒$\frac{10x+y}{10 y+x}=1 \frac{3}{4}$

⇒$\frac{10x+y}{10 y+x}=\frac{7}{4}$

⇒4(10x+y)=7(10y+x)

⇒40x+4y=70y+7x

⇒40x+4y-70y-7x=0

⇒33x-66y=0

⇒33(x-2y)=0

⇒x-2y=0...(1)

According to second condition of given problem

⇒x+y=12

From equation (1) and (2), we get

Subtracting 

$\begin{aligned}x-2 y&=0..(1) \\x+y&=12...(2) \\-\quad-& \quad - \\\hline-3 y&=-12\end{aligned}$

⇒$y=\frac{-12}{-3}$

⇒y=4

Substituting the value of y in equation (2), we get

⇒x+4=12

⇒x=12-4

⇒x=8

Hence , the number=10×8+4=80+4=84

Question 17

The result of dividing a number of two digits by the number with the digits reversed is $\frac{5}{6}$ . If the difference of digits is 1, find the number.

Sol :

Let the digit at ten's place =x

and let the digit at unit place =y

⇒Number=10×x+y×1=10x+y

Reversing the number=10×y+1×x=10y+x

According to first condition of given problem,

⇒$\frac{10 x+y}{10 y+x}=\frac{5}{6}$

⇒$\frac{10 x+y}{10 y+x}=\frac{5}{6}$

⇒6(10x+y)=5(10y+x)

⇒60x+6y=50y+5x

⇒60x+6y-50y-5x=0

⇒55x-44y=0

⇒11(5x-4y)=0

⇒5x-4y=0...(1)

According to second condition of given problem,

⇒y-x=1

⇒-x+y=1...(2)

Multiplying equation (2) by 5, we get

Adding 

$\begin{aligned}-5 x+5y&=5...(3) \\5 x-4 y&=0 \\\hline y&=5\end{aligned}$

Substituting the value of y in equation (2), we get

⇒-x+5=1

⇒-x=1-5

⇒-x=-4

⇒x=4

Hence, number=10x+y=10×4+5

=40+5=45

Question 18

A number of three digits has the hundred digit 4 times the unit digit and the sum of three digits is 14. If the three digits are written in the reverse order, the value of the number is decreased by 594. Find the number.

Sol :

Let the digits at ten's place =x

Let the digits at unit place =y

then, digit at hundred place =4y

Number=100×4y+10×x+1×y

=400y+10x+y

=401y+10x

=10x+401y

Reversing the number

=100×y+10×x+1×4y

=100y+10x+4y

=104y+10x

=10x+104y

According to first condition of the given problem,

⇒10x+401y=10x+104y+594

⇒10x+401y-10x-104y=594

⇒401y-104y=594

⇒297y=594

⇒$y=\frac{594}{297}$

⇒y=2

Substituting the value of y in equation (1), we get 

⇒x+5×2=14

⇒x+10=14

⇒x=14-10

⇒x=4

Hence , number=10x+401y

=10×4+401×2

=40+802=842

Question 19

Four years ago Marina was three times old as her daughter. Six years from now the mother will be twice as old as her daughter. Find their present ages.

Sol :

Let the present age of Marina =x years

and let the present age of Marina's daughter=y years

Four years ago age of Marina's=(x-4) years

Four years ago age of Marina's daughter=(y-4) years

According to first condition of given problem, 

⇒x-4=3(y-4)

⇒x-4=3y-12

⇒x-3y=-12+4

⇒x-3y=-8..(1)

Age of Marina's six years from now=(x+6) years

Age of Marina's daughter , six years from now=(y+6) years

According to second condition of given problem,

⇒x+6=2(y+6)

⇒x+6=2y+12

⇒x-2y=12-6

⇒x-2y=6..(2)

From equation (1) and (2), we get

Subtracting

⇒$\begin{aligned}x-3 y&=-8..(1) \\x-2 y&=6..(2) \\-\quad+&\quad- \\\hline-y&=-14\end{aligned}$

⇒y=14

Substituting the value of y in equation (1), we get

⇒x-3×14=-8

⇒x-42=-8

⇒x=-8+42

⇒x=34

Hence , Age of Marina=34 years

Age of Marina's Daughter=14 years

Question 20

On selling a tea set at 5% loss and a lemon set at 15% gain, a shopkeeper gains Rs. 70. If he sells the tea set at 5% gain and lemon set at 10% gain, he gains Rs. 130. Find the cost price of the lemon set.

Sol :

Loss on tea set=5%

and gain on lemon set=15%

Let the C.P. of tea set=Rs x

and C.P. of lemon set=Rs y

According to the conditions,

⇒$\frac{y \times 15}{100}-\frac{x \times 5}{100}=70$

⇒5y-5x=7000

⇒3y-x=1400...(i)

and ⇒$\frac{x \times 5}{100}+\frac{y \times 150}{100}=130$

⇒5x+150y=13000

⇒x+2y=2600...(ii)

Adding (i) and (ii)

⇒5y=4000

⇒$y=\frac{4000}{5}=800$

∴Cost price of lemon set=Rs 800

Question 21

A person invested some money at 12% simple interest and some other amount at 10% simple interest. He received yearly interest of Rs, 1300. If he had interchanged the amounts, he would have received Rs. 40 more as yearly interest. How much did he invest at different rates ?

Sol :

Let amount invested at S.I= Rs x

Rate =12% p.a.

and investment at S.I.=Rs y

Rate=10% p.a.

According to the condition,

⇒$\frac{x \times 12}{100}+\frac{y \times 10}{100}=1300$

⇒12x+10y=130000

⇒6x+5y=65000..(i)

and $\frac{10 x}{100}+\frac{12 y}{100}=1340$

⇒10x+12y=134000

⇒5x+6y=67000..(ii)

Multiplying (i) by 6 and (ii) by 5, we get

⇒$\begin{aligned}36 x+30 y&=390000 \\25 x+30 y&=335000\\-\quad -\quad &\quad-\\ \hline11x&=55000\end{aligned}$

⇒$x=\frac{55000}{11}=5000$

Substituting the value of x in (i) 

⇒6×5000+5y=65000

⇒30000+5y=65000

⇒5y=65000-30000=35000

⇒$y=\frac{35000}{5}=7000$

∴Investment at 12%=Rs 5000

and investment at 10%=Rs 7000

Question 22

A shopkeeper sells a table at 8% profit and a chair at 10% discount, thereby getting Rs. 1008. If he had sold the table at 10% profit and chair at 8% discount, he would have got Rs. 20 more. Find the cost price of the table and the list price of the chair.

Sol :

Profit on table =8%

and discount on chair=10%

Let C.P. of table =Rs x

and C.P. of chair =Rs y

According to the condition,

⇒$\frac{x \times(100+8)}{100}+\frac{y \times(100-10)}{100}=1008$

⇒108x+90y=100800

⇒6x+5y=5600..(i)

Similarly, 

⇒$\frac{x(100+10)}{100}+\frac{y(100-8)}{100}=1028$

⇒110x+92y=102800

⇒55x+46y=51400..(ii)

Multiplying (i) by 55 and (ii) by 6

Subtracts

$\begin{aligned}330x+275 y=308000 \\330 x+276 y=308400 \\-\quad-&\quad- \\\hline-y=-400\end{aligned}$

⇒y=400

Now 6x+5×400=5600

Substituting the value of y in (i)

⇒6x+200=5600

⇒6x=5600-2000

⇒6x=3600

⇒$x=\frac{3600}{6}=600$

∴C.P. of table=Rs 600

and C.P of chair=Rs 400

Question 23

A and B have some money with them. A said to B, “if you give me Rs. 100, my money will become 75% of the money left with you.” B said to A” instead if you give me Rs. 100, your money will become 40% of my money, How much money did A and B have originally ?

Sol :

Let A has money=x

and B has money=y

According to the condition.

⇒$x+100=(1-100) \times \frac{75}{100}$

⇒$x+100=(y-100) \times \frac{3}{4}$

⇒4x+400=3y-300

⇒4x-3y=-300-400...(i)

Again, $x-100=\frac{40}{100}(y+100)$

⇒$x-100=\frac{2}{5}(y+100)$

⇒5x-500=2y+200

⇒5x-2y=200+500=700..(ii)

Multiplying (i) by 2 and (ii) by 3,

Subtracting 

$\begin{aligned}8 x-6 y&=-1400 \\15 x-6 y&=2100 \\-\quad~+&\quad- \\\hline-7 x&=-3500\end{aligned}$

⇒$x=\frac{-3500}{-7}=500$

Substituting the value of x in (i)

⇒4×500-3y=-700

⇒2000-3y=-700

⇒3y=-700-2000=-2700

⇒$y=\frac{-2700}{-3}=900$

Hence A has money=Rs 500

and B has money=Rs 900

Question 24

The students of a class are made to stand in (complete) rows. If one student is extra in a row, there would be 2 rows less, and if one student is less in a row, there would be 3 rows more. Find the number of students in the class.

Sol :

Let the number of students in one row =x

and let the number of rows=y

Then total number of students=xy

According to first condition of given problem

⇒(x+1)(y-2)=xy

⇒x(y-2)+1(y-2)=xy

⇒xy-2x+y-2=xy

⇒-2x+y-2=0

⇒-2x+y=2..(1)

According to second condition of given problem,

⇒(x-1)(y+3)=xy

⇒x(y+3)-1(y+3)=xy

⇒xy+3x-y-3=0

⇒3x-y-3=0

⇒3x-y=3..(2)

From equation (1) and equation (2), we get

Adding ,

$\begin{aligned}-2 x+y&=2 \\3 x-y&=3 \\\hline x&=5\end{aligned}$

Substituting the value of x in (1), we get 

⇒-2×5+y=2

⇒-10+y=2

⇒y=2+10

⇒y=12

Hence, Number of students=xy=5×12=60

Question 25

A jeweller has bars of 18-carat gold and 12- carat gold. How much of each must be melted together to obtain a bar of 16-carat gold weighing 120 grams ? (Pure gold is 24 carat)

Sol :

Let x gm of 18 carat gold and Let y gm of 12 carat gold 

Then according to first condition of given problem, x+y=120

Pure gold is 24 carat

Then purity of 18 carat gold $=\frac{18}{24} \times 100 \%$

$=\frac{3}{4} \times 100=75 \%$

Purity of 12 carat gold$=\frac{12}{24} \times 100 \%$

$=\frac{1}{2} \times 100 \%=50 \%$

Purity of 16 carat gold $=\frac{16}{24} \times 100 \%$

$=\frac{2}{3} \times 100 \%=\frac{200}{3} \%$

According to second condition of given problem,

⇒$75 x+50 y=\frac{200}{3} \times 120$

⇒75x+50y=200×40

⇒75x+50y=8000...(2)

Multiplying equation (1) by 50 , we get

Subtracting

⇒$\begin{aligned}50 x+50 y&=6000 \\75 x+50 y&=8000 \\-\quad-\quad&\quad- \\\hline-25 x&=-2000\end{aligned}$

⇒$x=\frac{-2000}{-25}$

⇒x=80

Substituting the value of x in equation (1), we get

⇒80+y=120

⇒y=120-80

⇒y=40

Hence , 80 gm of 18 carat gold and 40 gm of 12 carat gold

Question 26

A and B together can do a piece of work in 15 days. If A’s one day work is 1 $\frac{1}{2}$ times the one day’s work of B, find in how many days can each do the work.

Sol :

Let A's one day work be =x

and B's one day work be =y

Then according to first condition of given problem,

⇒$x=\frac{3}{2}y$

⇒$x-\frac{3}{2} y=0$

⇒$\frac{2 x-3 y}{2}=0$

⇒2x-3y=0...(1)

Also given that, In 15 days, A and B together can do a piece of work,

Then in 1 days A and B together can do $\frac{1}{15}$ piece of work,

Then, according to second condition of given problem,

⇒$x+y=\frac{1}{15}$

⇒15(x+y)=1

⇒15x+15y=1...(2)

Multiplying equation (1) by } 5,we get

Adding 

$\begin{aligned}10 x-15 y&=0 \\15 x+15 y&=1 \\\hline25 x&=1\end{aligned}$

Substituting the value of x in equation (1), we get

⇒$2 \times \frac{1}{25}-3 y=0$

⇒$\frac{2}{25}=3 y$

⇒$y=\frac{2}{25 \times 3}=\frac{2}{75}$

Hence, Man A do the work in days $=\frac{1}{x}=\frac{1}{\left(\frac{1}{25}\right)}$

=25 days

Man B do the work in days, $=\frac{1}{y}=\frac{1}{\left(\frac{2}{75}\right)}$

$=\frac{75}{2}=37 \frac{1}{2}$ days

Question 27

Men and 5 women can do a piece of work in 4 days, while one man and one woman can finish it in 12 days. How long would it take for 1 man to do the work ?

Sol :

Let 1 man to do the work in x days and let 1 woman to do the work in y days,

Then In 1 days 1 man do the work$=\frac{1}{x}$

In 1 days 1 woman do the work$=\frac{1}{y}$

In 1 days 2 woman do the work$=\frac{2}{x}$

1 days 5 women do the work $=\frac{5}{y}$

According to first condition of given problem,

⇒$\frac{2}{x}+\frac{5}{y}=\frac{1}{4}$..(1)

According to second condition of given problem,

⇒$\frac{1}{x}+\frac{1}{y}=\frac{1}{12}$..(2)

Multiplying equation (2) by 5, we get

Subtracting 

$\begin{array}{l}\frac{5}{x}+\frac{5}{y}=\frac{5}{12} \\\frac{2}{x}+\frac{5}{y}=\frac{1}{4}\\-\quad-\quad~- \\\hline\frac{3}{x}=\frac{5}{12}-\frac{1}{4}\end{array}$

⇒$\frac{3}{x}=\frac{5-3}{12}$

⇒$\frac{3}{x}=\frac{2}{12}$

⇒3×12=12×x 

⇒2×x=3×12

⇒$x=\frac{3 \times 12}{2}=\frac{3 \times 6}{1}$

=18 days

Hence, 1 man do the work in 18 days

[If we find the value of y then substitute the value of x in equation (1) or (2)]

Question 28

A train covered a certain distance at a uniform speed. If the train had been 30 km/hr faster, it would have taken 2 hours less than the scheduled time. If the train were slower by 15 km/hr, it would have taken 2 hours more than the scheduled time. Find the length of the journey.

Sol :

Let the due speed of the train be x km/hr and scheduled time be y hours.

Then, (Distance) Length of the journey=Speed×Time

=xy

According to first condition of given problem,

⇒(x+30)(y-2)=xy

⇒x(y-2)+30(y-2)=xy

⇒xy-2x+30y-60=xy

⇒-2(x-15 y)=60

⇒2(x-15 y)=60

⇒$x-15y=\frac{60}{-2}$

⇒x-15 y=-30..(1)

According to second condition of given problem, 

⇒(x-15)(y+2)=xy

⇒x(y+2)-15(y+2)=xy

⇒xy+2 x-15 y-30=xy

⇒2x-15y-30=0

⇒2x-15y=30

From equation (1) and equation (2), we get

Subtracting

$\begin{aligned}x-15 y&=-30 \\2 x-15 y&=30 \\-\quad+~&\quad- \\\hline-x&=-60\end{aligned}$

⇒x=60

Substituting the value of x in equation (1), we get 

⇒60-15y=-30

⇒-15y=-30-60

⇒-15y=-90

⇒$y=\frac{-90}{-15}$

⇒y=6

Hence, length of the journey=60×6km=360km

Question 29

A boat takes 2 hours to go 40 km down the stream and it returns in 4 hours. Find the speed of the boat in still water and the speed of the stream.

Sol :

Let the speed of the boat in still water =x km/hr

and the speed of the stream =y km/hr

Speed of boat in the direction of the stream =(x+y) km/hr

Speed of boat in the opposite direction of the stream =(x-y) km/hr

Stream=(x-y) km/hr

Distance=Speed×Time

According to first condition of given problem,

⇒40=(x+y)×2

⇒2(x+y)=40

⇒$x+y=\frac{40}{2}$

⇒x+y=20..(1)

According to second condition of given problem, 

Distance =speed×time

⇒40=(x-y)×2

⇒2(x+y)=40

⇒$x+y=\frac{40}{2}$

⇒x+y=20...(1)

According to second condition of given problem,

Distance =speed×time

⇒40=(x-y)×4

⇒4(x-y)=40

⇒$x-y=\frac{40}{4}$

⇒x-y=10...(2)

From equation (1) and (2), we get

Adding 

$\begin{array}{l}x+y=20 \\x-y=10 \\\hline 2 x=30\end{array}$

⇒$x=\frac{30}{2}$

⇒x=15

Substituting the value of x in equation (1), we get

⇒15+y=20

⇒y=20-15

⇒y=5

Hence, speed of boat in still water=15 km/hr

Speed of the stream=5 km/hr

Question 30

A boat sails a distance of 44 km in 4 hours with the current. It takes 4 hours 48 minutes longer to cover the same distance against the current. Find the speed of the boat in still water and the speed of the current.

Sol :

Let the speed of the boat= x km/hr

and the speed of the current=y km/hr

Speed of the boat in the direction of current=(x+y) km/hr

Speed of boat in the opposite direction of current=(x+y) km/hr

Distance=Speed×Time

According to first condition of given problem,

⇒44=(x+y)×4

⇒4(x+y)=44

⇒$x+y=\frac{44}{4}$

⇒x+y=11...(1)

According to second condition of given problem,

⇒$44=(x-y) \times\left(4 \frac{48}{60}+4\right)$ $\left(\because 4 \text{hours} ~48\text{min}=4 \frac{48}{60} \mathrm{hr}\right)$

⇒$44=(x-y) \times\left(4 \frac{4}{5}+4\right)$

⇒$44=(x-y) \times\left(\frac{24}{5}+4\right)$

⇒$(x-y) \times \frac{44}{5}=44$

⇒$x-y=\frac{44 \times 5}{44}$

⇒$x-y=\frac{11 \times 5}{11}$

⇒x-y=5..(2)

From equation (1) and equation (2), we get

Adding,

$\begin{array}{l}x+y=11 ...(1)\\x-y=5..(2) \\\hline 2 x=16\end{array}$

⇒$x=\frac{16}{2}$

⇒x=8

Substituting the value of x in equation (1), we get

⇒8+y=11

⇒y=11-8

⇒y=3

Hence, Speed of the boat in still water=8 km/hr 

Speed of the current=3 km/hr

Question 31

An aeroplane flies 1680 km with a head wind in 3.5 hours. On the return trip with same wind blowing, the plane takes 3 hours. Find the plane’s air speed and the wind speed.

Sol :

Let the speed of plane=x km/hr

and let the speed of wind= y km/hr

then  speed of aeroplane in the direction of wind=(x+y) km/hr

speed of aeroplane in the opposite direction of wind =(x-y) km/hr

According to first condition of given problem,

Distance=Speed×Time

⇒1680=(x-y)×3.5

⇒3.5(x-y)=1680

⇒$x-y=\frac{1680}{3.5}$

⇒$x-y=\frac{1680 \times 10}{35}$

⇒$x-y=\frac{1680 \times 2}{7}$

⇒x-y=240×2

⇒x-y=480..(1)

According to second condition of given problem,

Distance=Speed×Time

⇒1680=(x+y)×3

⇒3(x+y)=1680

⇒$x+y=\frac{1680}{3}$

⇒x+y=560..(2)

From equation (1) and (2), we get

Adding ,

$\begin{array}{l}x-y=480 \\x+y=560 \\\hline 2 x=1040\end{array}$

⇒$x=\frac{1040}{2}=520$

Substituting the value of x in equation (2), we get,

⇒520+y=560

⇒y=560-520=40

Hence, speed of aeroplane=520 km/hr

and speed of wind =40 km/hr

Question 32

A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When Bhawana takes food for 20 days, she has to pay Rs. 2600 as hostel charges; whereas when Divya takes food for 26 days, she pays Rs. 3020 as hostel charges. Find the fixed charges and the cost of food per day.

Sol :

Let fixed charges =Rs x and charges per day =Rs y 

According to the condition.

Subtracting

$\begin{aligned}x+20 y&=2600..(i) \\x+26 y&=3020..(ii) \\-\quad-~&\quad-\\\hline-6y&=-420\end{aligned}$

⇒$y=\frac{-420}{-6}=70$

Substituting the value of y in (i)

⇒x+20×70=2600

⇒x+1400=2600

⇒x=2600-1400=1200

∴Fixed charges=Rs 1200

and daily charges=Rs 70