ICSE Class 9 Mathematics · ML Aggarwal Solutions Understanding ICSE Mathematics Class 9 · 52 solved questions
The sum of two numbers is 50 and their difference is 16. Find the numbers.
Let two numbers be x and y.
Given, sum = 50 and difference = 16.
x + y = 50 ........(i)
x - y = 16 ........(ii)
Adding (i) and (ii) we get,
⇒ (x + y) + (x - y) = 50 + 16
⇒ 2x = 66
⇒ x = 33.
Substituting value of x in (i) we get,
⇒ 33 + y = 50
⇒ y = 50 - 33 = 17.
Hence, two numbers are 33 and 17.
The sum of two numbers is 2. If their difference is 20, find the numbers.
Let two numbers be x and y.
Given, sum = 2 and difference = 20.
x + y = 2 ........(i)
x - y = 20 ........(ii)
Adding (i) and (ii) we get,
⇒ (x + y) + (x - y) = 2 + 20
⇒ 2x = 22
⇒ x = 11.
Substituting value of x in (i) we get,
⇒ 11 + y = 2
⇒ y = 2 - 11 = -9.
Hence, two numbers are 11 and -9.
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.
Let the larger number be x and smaller number be y.
Given, sum = 43.
x + y = 43 .......(i)
According to question,
2x - 3y = 36 .......(ii)
Multiplying (i) by 3 we get,
3x + 3y = 129 ......(iii)
Adding (ii) and (iii) we get,
⇒ (2x - 3y) + (3x + 3y) = 36 + 129
⇒ 5x = 165
⇒ x = 33.
Substituting value of x in (i) we get,
⇒ 33 + y = 43
⇒ y = 43 - 33 = 10.
Hence, two numbers are 33 and 10.
The cost of 5 kg of sugar and 7 kg of rice is ₹765 and the cost of 7 kg of sugar and 5 kg of rice is ₹735. Find the cost of 6 kg of sugar and 10 kg of rice.
Let cost of sugar and rice per kg be ₹ x and ₹ y respectively.
So, cost of 5 kg of sugar and 7 kg of rice = 5x + 7y and,
cost of 7 kg of sugar and 5 kg of rice = 7x + 5y.
Given,
Cost of 5 kg of sugar and 7 kg of rice is ₹ 765.
⇒ 5x + 7y = 765 .....................(1)
Cost of 7 kg of sugar and 5 kg of rice is ₹ 735
⇒ 7x + 5y = 735 .....................(2)
Multiplying eq (1) by 7, we get :
⇒ 7(5x + 7y) = 7 × 765
⇒ 35x + 49y = 5355 .....................(3)
Multiplying eq (2) by 5, we get :
⇒ 5(7x + 5y) = 5 × 735
⇒ 35x + 25y = 3675 .....................(4)
Subtracting equation (4) from (3) we get,
⇒ 35x + 49y - (35x + 25y) = 5355 - 3675
⇒ 35x - 35x + 49y - 25y = 5355 - 3675
⇒ 24y = 1680
⇒ y = $\frac{1680}{24}$
⇒ y = 70.
Substituting the value of y in (1) we get,
⇒ 5x + 7(70) = 765
⇒ 5x + 490 = 765
⇒ 5x = 765 - 490
⇒ 5x = 275
⇒ x = $\frac{275}{5}$
⇒ x = 55.
∴ Cost of 6 kg of sugar and 10 kg of rice = 6x + 10y
= 6(55) + 10(70) = 330 + 700 = ₹ 1,030.
Hence, cost of 6kg of sugar and 10 kg of rice = ₹ 1,030.
The class IX students of a certain public School wanted to give a farewell party to the outgoing student of class X. They decided to purchase two kinds of sweets, one costing ₹350 per kg and the other costing ₹ 440 per kg. They estimated that 36 kg of sweets were needed. If the total money spent on sweet was ₹ 14,000, find how much sweets of each kind they purchased.
Let first sweet purchased be x kg and second be y kg.
Then, x + y = 36 or y = 36 - x ......................(1)
Cost of first sweet = ₹ 350 per kg
Cost of second sweet = ₹ 440 per kg
Total cost of sweets will be 350x + 440y.
Total money spent on sweet = ₹ 14,000
⇒ 350x + 440y = 14000
Substituting the value of y from (1) in above equation we get,
⇒ 350x + 440(36 - x) = 14000
⇒ 350x + 15840 - 440x = 14000
⇒ 15840 - 90x = 14000
⇒ 90x = 15840 - 14000
⇒ 90x = 1840
⇒ x = $\frac{1840}{90}$
⇒ x = 20.44 ≈ 20.
Substituting this value of x in eq (1)
⇒ y = 36 - x
⇒ y = 36 - 20
⇒ y = 16.
Hence, the sweet costing ₹350 per kg purchased was 20 kg and the one costing ₹440 per kg was 16 kg.
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?
Let the greater number be x and smaller number be y.
According to question if from twice the greater of two numbers 16 is subtracted, the result is half the other number.
∴ 2x - 16 = $\frac{y}{2}$
⇒ 2(2x - 16) = y
⇒ 4x - y = 32 .......(i)
According to question, if from half the greater number 1 is subtracted, the result is still half the other number.
$$\begin{aligned}\therefore \frac{x}{2} - 1 = \frac{y}{2} \\[1em] \Rightarrow 2\Big(\frac{x}{2} - 1\Big) = y \\[1em] \Rightarrow x - y = 2 .......(ii)\end{aligned}$$
Subtracting eq. (ii) from eq. (i) we get,
⇒ 4x - y - (x - y) = 32 - 2
⇒ 4x -y - x + y = 30
⇒ 3x = 30
⇒ x = $\frac{30}{3}$
⇒ x = 10
Substituting value of x in (ii) we get,
10 - y = 2
⇒ y = 10 - 2
⇒ y = 8
Hence, two numbers are 8 and 10.
There are 38 coins in a collection of 20 paise coins and 25 paise coins. If the total value of the collection is ₹8.50, how many of each are there?
Let x be the no. of 20 paise coins and y be the no. of 25 paise coins.
x + y = 38 .......(i)
Total value of coins = 0.2x + 0.25y.
Given, total value = ₹8.50
⇒ 0.2x + 0.25y = 8.50 .......(ii)
Multiplying (ii) by 4 we get,
⇒ 0.8x + y = 34 .......(iii)
Subtracting (iii) from (i) we get,
⇒ x + y - (0.8x + y) = 38 - 34
⇒ 0.2x = 4
⇒ x = $\frac{4}{0.2}$
⇒ x = $\frac{40}{2}$
⇒ x = 20.
Substituting value of x in (i) we get,
⇒ 20 + y = 38
⇒ y = 38 - 20
⇒ y = 18.
Hence, no. of 25 paise coins = 18 and no. of 20 paise coins = 20.
A man has certain notes of denominations ₹ 200 and ₹ 50 which amount to ₹ 3,800. If the number of notes of each kind is interchanged, they amount to ₹ 600 less as before. Find the number of notes of each denomination.
Let number of ₹ 200 notes be x and number of ₹ 50 notes be y.
So, total amount = 200x + 50y.
Given, total amount = ₹ 3,800.
∴ 200x + 50y = 3800 ....................(1)
On exchanging number of notes, number of ₹ 200 notes will be y and number of ₹ 50 notes will be x.
Given, total amount in this case = 3200 (3800 - 600).
∴ 200y + 50x = 3200 ....................(2)
Multiplying eq. (1) by 4 we get,
⇒ 4(200x + 50y) = 3800 x 4
⇒ 800x + 200y = 15200 ....................(3)
Subtracting eq. (2) from (3) we get,
⇒ 800x + 200y - (200y + 50x) = 15200 - 3200
⇒ 800x - 50x + 200y - 200y = 12000
⇒ 750x = 12000
⇒ x = $\frac{12000}{750}$
⇒ x = 16.
Substituting the value of x in equation 1 we get,
⇒ 200(16) + 50y = 3800
⇒ 3200 + 50y = 3800
⇒ 50y = 3800 - 3200
⇒ 50y = 600
⇒ y = $\frac{600}{50}$
⇒ y = 12.
Hence, number of ₹ 200 notes = 16 and number of ₹ 50 notes = 12.
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.
Let two numbers be x and y.
Given, ratio of two numbers is $\frac{2}{3}$.
$$\begin{aligned}\therefore \frac{x}{y} = \frac{2}{3} \\[1em] \Rightarrow x = \frac{2y}{3} .......(i)\end{aligned}$$
Given, if 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio.
$\therefore \frac{x - 2}{y - 8} = \frac{3}{2}$
Substituting value of x from (i) in above equation,
$$\begin{aligned}\Rightarrow \frac{\frac{2y}{3} - 2}{y - 8} = \frac{3}{2} \\[1em] \Rightarrow \frac{\frac{2y - 6}{3}}{y - 8} = \frac{3}{2} \\[1em] \Rightarrow \frac{2y - 6}{3(y - 8)} = \frac{3}{2} \\[1em] \Rightarrow 2(2y - 6) = 3(y - 8) \times 3 \\[1em] \Rightarrow 4y - 12 = 9y - 72 \\[1em] \Rightarrow 9y - 4y = 72 - 12 \\[1em] \Rightarrow 5y = 60 \\[1em] \Rightarrow y = \frac{60}{5} \\[1em] \Rightarrow y = 12.\end{aligned}$$
Substituting value of y in eq. (i),
$x = \frac{2y}{3} = \frac{2 \times 12}{3}$ = 8.
Hence, the numbers are 8 and 12.
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.
Let numerator be x and denominator be y.
Given, if 1 is added to the numerator of a fraction, it becomes $\frac{1}{5}$.
$\therefore \frac{x + 1}{y} = \frac{1}{5}$
y = 5(x + 1) .......(i)
Given, if 1 is taken from the denominator, it becomes $\frac{1}{7}$.
$\therefore \frac{x}{y - 1} = \frac{1}{7}$
⇒ y - 1 = 7x
⇒ y = 1 + 7x ........(ii)
Equating (i) and (ii) we get,
⇒ 5(x + 1) = 1 + 7x
⇒ 5x + 5 = 1 + 7x
⇒ 7x - 5x = 5 - 1
⇒ 2x = 4
⇒ x = 2.
Substituting value of x in (ii) we get,
y = 1 + 7x = 1 + 7(2) = 1 + 14 = 15.
Original fraction = $\frac{x}{y} = \frac{2}{15}$.
Hence, fraction = $\frac{2}{15}$.
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.
Let numerator be x and denominator be y.
Given, if the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to $\frac{5}{8}$
$\therefore \frac{x + 2}{y + 1} = \frac{5}{8}$
⇒ 8(x + 2) = 5(y + 1)
⇒ 8x + 16 = 5y + 5
⇒ 5y - 8x = 11 .......(i)
Given, if the numerator and denominator are each diminished by 1, the fraction becomes equal to $\frac{1}{2}$
$\therefore \frac{x - 1}{y - 1} = \frac{1}{2}$
⇒ 2(x - 1) = y - 1
⇒ 2x - 2 = y - 1
⇒ y = 2x - 1 ......(ii)
Substituting value of y from (ii) in eq. (i) we get,
⇒ 5(2x - 1) - 8x = 11
⇒ 10x - 5 - 8x = 11
⇒ 2x = 16
⇒ x = 8.
Substituting value of x in eq. (ii) we get,
y = 2x - 1 = 2(8) - 1 = 16 - 1 = 15.
Original fraction = $\frac{x}{y} = \frac{8}{15}.$
Hence, fraction = $\frac{8}{15}$.
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.
Let numerator be x and denominator be y.
Given, fraction becomes $\frac{1}{2}$ when the denominator is increased by 4.
$\therefore \frac{x}{y + 4} = \frac{1}{2}$
⇒ 2x = y + 4
⇒ 2x - y = 4 .......(i)
Given, fraction becomes $\frac{1}{8}$ when the numerator is diminished by 5
$\therefore \frac{x - 5}{y} = \frac{1}{8}$
⇒ 8(x - 5) = y ......(ii)
Substituting value of y from eq. (ii) in (i) we get,
⇒ 2x - [8(x - 5)] = 4
⇒ 2x - 8x + 40 = 4
⇒ -6x = 4 - 40
⇒ 6x = 36
⇒ x = 6.
Substituting value of x in (ii) we get,
⇒ 8(6 - 5) = y
⇒ y = 8(1) = 8.
Original fraction = $\frac{x}{y} = \frac{6}{8}$
Hence, fraction = $\frac{6}{8}$.
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 digits of the original number, find the number.
Let's consider the digits at ten's place as x and let the digit at unit's place be y.
Then according to first condition,
⇒ x + y = 7
⇒ x = 7 - y ......(i)
The number if digits are reversed = 10 × y + x.
Given, number with the order of the digits reversed is 28 greater than twice the unit's digits of the original number
∴ (10 × y + x) - 2y = 28
Substituting value of x from (i) we get,
⇒ 10y + 7 - y - 2y = 28
⇒ 7y = 21
⇒ y = 3.
⇒ x = 7 - y = 7 - 3 = 4.
Original number = 10 × x + y = 10 × 4 + 3 = 43.
Hence, original number = 43.
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.
Let's consider the digits at ten's place as x and let the digit at unit's place be y.
Number = 10 × x + y = 10x + y,
On reversing digits the number is = 10 × y + x = 10y + x.
According to first condition, we have
⇒ 10x + y - [4(x + y)] = 6
⇒ 10x + y - 4x - 4y = 6
⇒ 10x - 4x + y - 4y = 6
⇒ 6x - 3y = 6
⇒ 2x - y = 2 ......(i)
According to second condition, we have
⇒ 10x + y + 9 = 10y + x
⇒ 10x - x + y - 10y = -9
⇒ 9x - 9y = -9
⇒ x - y = -1
⇒ y - x = 1 ......(ii)
Adding eq. (i) and (ii) we get,
⇒ 2x - y + (y - x) = 2 + 1
⇒ x = 3.
Substituting value of x in (ii) we get,
⇒ y - 3 = 1
⇒ y = 4.
Number = 10 × x + y = 10 × 3 + 4 = 30 + 4 = 34.
Hence, number = 34.
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?
Let's consider the digits at ten's place as x and let the digit at unit's place be y.
Number = 10 × x + y = 10x + y,
According to first condition, we have
$\Rightarrow \frac{10x + y}{x + y} = 8$
⇒ 10x + y = 8(x + y)
⇒ 10x + y = 8x + 8y
⇒ 10x - 8x + y - 8y = 0
⇒ 2x - 7y = 0 .......(i)
According to second condition, we have
x - 3y = 1 ......(ii)
Multiplying eq. (ii) by 2 we get,
2x - 6y = 2 .......(iii)
Subtracting eq. (i) from (iii) we get,
⇒ 2x - 6y - (2x - 7y) = 2 - 0
⇒ 2x - 2x - 6y + 7y = 2
⇒ y = 2.
Substituting value of y in (ii) we get,
⇒ x - 3(2) = 1
⇒ x - 6 = 1
⇒ x = 7.
Number = 10 × x + y = 10 × 7 + 2 = 72.
Hence, number = 72.
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.
Let's consider the digits at ten's place as x and let the digit at unit's place be y.
Number = 10 × x + y = 10x + y,
On reversing digits the number is = 10 × y + x = 10y + x.
According to first condition, we have
$$\begin{aligned}\Rightarrow \frac{10x + y}{10y + x} = 1\frac{3}{4} \\[1em] \Rightarrow \frac{10x + y}{10y + x} = \frac{7}{4} \\[1em] \Rightarrow 4(10x + y) = 7(10y + x) \\[1em] \Rightarrow 40x + 4y = 70y + 7x \\[1em] \Rightarrow 40x - 7x + 4y - 70y = 0 \\[1em] \Rightarrow 33x - 66y = 0 \\[1em] \Rightarrow 11(3x - 6y) = 0 \\[1em] \Rightarrow 3x - 6y = 0 ........(i)\end{aligned}$$
According to second condition, we have
⇒ x + y = 12
⇒ x = 12 - y .......(ii)
Substituting value of x from (ii) in (i) we get,
⇒ 3(12 - y) - 6y = 0
⇒ 36 - 3y - 6y = 0
⇒ 36 - 9y = 0
⇒ 9y = 36
⇒ y = 4.
Substituting value of y in (ii) we get,
⇒ x = 12 - 4 = 8.
Number = 10 × x + y = 10 × 8 + 4 = 84.
Hence, number = 84.
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.
Let's consider the digits at ten's place as x and let the digit at unit's place be y.
Number = 10 × x + y = 10x + y,
On reversing digits the number is = 10 × y + x = 10y + x.
According to first condition we have,
$$\begin{aligned}\Rightarrow \frac{10x + y}{10y + x} = \frac{5}{6} \\[1em] \Rightarrow 6(10x + y) = 5(10y + x) \\[1em] \Rightarrow 60x + 6y = 50y + 5x \\[1em] \Rightarrow 60x - 5x + 6y - 50y = 0 \\[1em] \Rightarrow 55x - 44y = 0 \\[1em] \Rightarrow 11(5x - 4y) = 0 \\[1em] \Rightarrow 5x - 4y = 0 .......(i)\end{aligned}$$
According to second condition we have,
⇒ y - x = 1
⇒ x = y - 1 .......(ii)
Substituting value of x in (i) we get,
⇒ 5(y - 1) - 4y = 0
⇒ 5y - 5 - 4y = 0
⇒ y = 5.
Substituting value of y in (ii) we get,
⇒ x = y - 1 = 5 - 1 = 4.
Number = 10 × x + y = 10 × 4 + 5 = 45.
Hence, number = 45.
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.
Let the digit at ten's place be x and unit's place be y.
Then according to question,
Digit at hundred's place = 4y.
Number = 100 × 4y + 10 × x + y = 400y + 10x + y = 401y + 10x.
Reversing the number becomes = 100 × y + 10 × x + 4y = 100y + 10x + 4y = 104y + 10x.
Given, sum of digits = 14.
∴ 4y + x + y = 14
⇒ x + 5y = 14 ......(i)
According to second condition,
⇒ 401y + 10x = 104y + 10x + 594
⇒ 401y - 104y + 10x - 10x = 594
⇒ 297y = 594
⇒ y = 2.
Substituting value of y in (i) we get,
⇒ x + 5(2) = 14
⇒ x + 10 = 14
⇒ x = 4.
Number = 100 × 4y + 10 × x + y = 100 × 4(2) + 10 × 4 + 2 = 800 + 40 + 2 = 842.
Hence, number = 842.
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.
Let present age of Marina be x years and daughter's age be y years.
Four years ago,
Age of Marina = (x - 4) years
Age of Marina's daughter = (y - 4) years
According to first condition in the problem,
⇒ (x - 4) = 3(y - 4)
⇒ x - 4 = 3y - 12
⇒ 3y - x = 8 .......(i)
After 6 years,
Age of Marina = (x + 6) years
Age of Marina's daughter = (y + 6) years
According to second condition in the problem,
⇒ x + 6 = 2(y + 6)
⇒ x + 6 = 2y + 12
⇒ x - 2y = 6 .......(ii)
Adding eq. (i) and (ii) we get,
⇒ (3y - x) + (x - 2y) = 8 + 6
⇒ 3y - 2y - x + x = 14
⇒ y = 14.
On substituting value of y in (ii) we get,
⇒ x - 2(14) = 6
⇒ x - 28 = 6
⇒ x = 34.
Hence, present age of Marina = 34 years and daughter = 14 years.
On selling a tea set at 5% loss and a lemon set at 15% gain, a shopkeeper gains ₹70. If he sells the tea set at 5% gain and lemon set at 10% gain he gains ₹130. Find the cost price of lemon set.
Let C.P. of tea set = ₹x and C.P. of lemon set = ₹y.
Given, Loss on tea set = 5% and gain on lemon set = 15%, total gain = ₹70.
$$\begin{aligned}\therefore \frac{15}{100}y - \frac{5}{100}x = 70 \\[1em] \Rightarrow \frac{15y - 5x}{100} = 70 \\[1em] \Rightarrow 15y - 5x = 7000 \\[1em] \Rightarrow 3y - x = 1400 ......(i)\end{aligned}$$
Given, Gain on tea set = 5% and gain on lemon set = 10%, total gain = ₹130.
$$\begin{aligned}\therefore \frac{5}{100}x + \frac{10}{100}y = 130 \\[1em] \Rightarrow \frac{5x + 10y}{100} = 130 \\[1em] \Rightarrow 5x + 10y = 13000 \\[1em] \Rightarrow x + 2y = 2600 ......(ii)\end{aligned}$$
Adding eq. (i) and (ii) we get,
⇒ (3y - x) + (x + 2y) = 1400 + 2600
⇒ 5y = 4000
⇒ y = 800.
Hence, C.P. of lemon set = ₹800.
A person invested some money at 12% simple interest and some other amount at 10% simple interest. He received yearly interest of ₹1300. If he had interchanged the amounts, he would have received ₹40 more as yearly interest. How much did he invest at different rates?
Let the amount invested at 12% S.I. be ₹x and at 10% S.I. be ₹y.
According to first condition,
$$\begin{aligned}\Rightarrow \frac{12}{100} \times x + \frac{10}{100} \times y = 1300 \\[1em] \Rightarrow \frac{12x + 10y}{100} = 1300 \\[1em] \Rightarrow 12x + 10y = 130000 \\[1em] \Rightarrow 6x + 5y = 65000 .....(i)\end{aligned}$$
Now, let the amount invested at 12% S.I. be ₹y and at 10% S.I. be ₹x.
According to second condition,
$$\begin{aligned}\Rightarrow \frac{12}{100}y + \frac{10}{100}x = 1340 \\[1em] \Rightarrow \frac{12y + 10x}{100} = 1340 \\[1em] \Rightarrow 12y + 10x = 134000 \\[1em] \Rightarrow 6y + 5x = 67000 ......(ii)\end{aligned}$$
Multiplying (i) by 6 and (ii) by 5, we have
⇒ 36x + 30y = 390000 .......(iii)
⇒ 30y + 25x = 335000 .......(iv)
Subtracting eq. (iv) from (iii) we get,
⇒ (36x + 30y) - (30y + 25x) = 390000 - 335000
⇒ 36x - 25x + 30y - 30y = 55000
⇒ 11x = 55000
⇒ x = 5000.
On substituting the value of x in (i) we get,
⇒ 6(5000) + 5y = 65000
⇒ 30000 + 5y = 65000
⇒ 5y = 65000 - 30000
⇒ 5y = 35000
⇒ y = 7000.
Hence, the investment at 12% is ₹5000 and the investment at 10% is ₹7000.
A shopkeeper sells a table at 8% profit and a chair at 10% discount, thereby getting ₹1008. If he had sold the table at 10% profit and chair at 8% discount, he would have got ₹20 more. Find the cost price of the table and the list price of the chair.
Let the cost price of table be ₹x and list price of chair be ₹y.
According to first condition,
$$\begin{aligned}\Rightarrow \frac{108}{100}x + \frac{90}{100}y = 1008 \\[1em] \Rightarrow \frac{108x + 90y}{100} = 1008 \\[1em] \Rightarrow 108x + 90y = 100800 \\[1em] \Rightarrow 18(6x + 5y) = 100800 \\[1em] \Rightarrow 6x + 5y = \frac{100800}{18} \\[1em] \Rightarrow 6x + 5y = 5600 ......(i)\end{aligned}$$
According to second condition,
$$\begin{aligned}\Rightarrow \frac{110}{100}x + \frac{92}{100}y = 1028 \\[1em] \Rightarrow \frac{110x + 92y}{100} = 1028 \\[1em] \Rightarrow 110x + 92y = 102800 \\[1em] \Rightarrow 2(55x + 46y) = 102800 \\[1em] \Rightarrow 55x + 46y = 51400 ......(ii)\end{aligned}$$
Multiplying (i) by 55 and (ii) by 6, we have
⇒ 330x + 275y = 308000 ........(iii)
⇒ 330x + 276y = 308400 ........(iv)
Subtracting eq. (iii) from (iv) we get,
⇒ (330x + 276y) - (330x + 275y) = 308400 - 308000
⇒ 330x - 330x + 276y - 275y = 400
⇒ y = 400.
On substituting value of y in (i) we get,
⇒ 6x + 5(400) = 5600
⇒ 6x + 2000 = 5600
⇒ 6x = 3600
⇒ x = 600.
Hence, cost of table = ₹600 and cost of chair = ₹400.
A and B have some money with them. A said to B, 'if you give me ₹100, my money will become 75% of the money left with you'. "B said to A" instead if you give me ₹100, your money will become 40% of my money. How much money did A and B have originally?
Let A have ₹x and B have ₹y.
According to first condition,
$\Rightarrow x + 100 = \frac{75}{100}(y - 100)$
⇒ 100(x + 100) = 75(y - 100)
⇒ 100x + 10000 = 75y - 7500
⇒ 75y - 100x = 10000 + 7500
⇒ 75y - 100x = 17500
⇒ 25(3y - 4x) = 25 × 700
⇒ 3y - 4x = 700 .......(i)
According to second condition,
$\Rightarrow \frac{40}{100}(y + 100) = x - 100$
⇒ 40(y + 100) = 100(x - 100)
⇒ 40y + 4000 = 100x - 10000
⇒ 100x - 40y = 14000
⇒ 20(5x - 2y) = 20 × 700
⇒ 5x - 2y = 700 .......(ii)
Multiplying (i) by 2 and (ii) by 3 we get,
6y - 8x = 1400 ......(iii)
15x - 6y = 2100 .....(iv)
Adding (iii) and (iv) we get,
⇒ (6y - 8x) + (15x - 6y) = 1400 + 2100
⇒ 6y - 6y - 8x + 15x = 3500
⇒ 7x = 3500
⇒ x = 500.
Substituting value of x in (i) we get,
⇒ 3y - 4(500) = 700
⇒ 3y - 2000 = 700
⇒ 3y = 2700
⇒ y = 900.
Hence, A has ₹500 while B has ₹900.
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.
Let number of students in a row be x and no. of rows be y.
Total no. of students = xy.
According to first condition we have,
⇒ (x + 1)(y - 2) = xy
⇒ xy - 2x + y - 2 = xy
⇒ xy - xy + y - 2x = 2
⇒ y - 2x = 2 .........(i)
According to second condition,
⇒ (x - 1)(y + 3) = xy
⇒ xy + 3x - y - 3 = xy
⇒ 3x - y = 3 + xy - xy
⇒ 3x - y = 3 ........(ii)
Adding (i) and (ii) we get,
⇒ (y - 2x) + (3x - y) = 2 + 3
⇒ y - y - 2x + 3x = 5
⇒ x = 5.
Substituting value of x in (i) we get,
⇒ y - 2(5) = 2
⇒ y - 10 = 2
⇒ y = 12.
xy = 5 × 12 = 60.
Hence, there are 60 students in the class.
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).
Let us assume the quantity of 18 carat gold as x gm and 12 carat gold as y gm.
Given, total weight of new bar = 120 gm.
x + y = 120 ......(i)
Pure gold is 24 carat.
So, the purity of 18 carat gold = $\frac{18}{24} \times 100% = \frac{3}{4} \times 100% = 75%.$
Purity of 12 carat gold = $\frac{12}{24} \times 100% = 50%.$
Purity of 16 carat gold = $\frac{16}{24} \times 100% = \frac{200}{3}%$
According to condition,
$$\begin{aligned}75% \text{ of } x + 55% \text{ of } y = \frac{200}{3}% \text{ of } 120 \\[1em] \Rightarrow \frac{75}{100}x + \frac{50}{100}y = \frac{200}{3 \times 100} \times 120 \\[1em] \frac{3}{4}x + \frac{1}{2}y = 80 \\[1em] \frac{3x + 2y}{4} = 80 \\[1em] 3x + 2y = 320 \space ......(ii)\end{aligned}$$
Multiplying (i) by 2 we get,
⇒ 2x + 2y = 240 .......(iii)
Subtracting eq (iii) from eq (ii)
⇒ (3x + 2y) - (2x + 2y) = 320 - 240
⇒ 3x - 2x + 2y - 2y = 80
⇒ x = 80.
Substituting value of x in (i) we get,
⇒ 80 + y = 120
⇒ y = 40.
Hence, jeweller requires 80 gm of 18 carat gold and 40 gm of 12 carat gold to obtain a bar of 16 carat gold weighing 120 gm.
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.
Let A's one day work be x and B's one day work be y.
According to first condition given in the problem,
x = $\frac{3}{2}$y
2x = 3y
2x - 3y = 0 .......(i)
Also given, A and B together can do a piece of work in 15 days.
∴ x + y = $\frac{1}{15}$
⇒ 15(x + y) = 1
⇒ 15x + 15y = 1 .......(ii)
Multiplying (i) by 5 we get,
10x - 15y = 0 ......(iii)
Adding (ii) and (iii) we get,
⇒ 15x + 15y + 10x - 15y = 1 + 0
⇒ 25x = 1
⇒ x = $\frac{1}{25}$.
Substituting x in (i) we get,
$$\begin{aligned}\Rightarrow 2 \times \frac{1}{25} - 3y = 0 \\[1em] \Rightarrow 3y = \frac{2}{25} \\[1em] \Rightarrow y = \frac{2}{75}.\end{aligned}$$
Since, A's one day work is x and B's one day work is y, so A can do complete work in $\frac{1}{x}$ and B can do work in $\frac{1}{y}$ days.
$\frac{1}{x} = \frac{1}{\frac{1}{25}} = 25$ days.
$\frac{1}{y} = \frac{1}{\frac{2}{75}} = \frac{75}{2} = 37\frac{1}{2}$ days.
Hence, A will do the work in 25 days and B will do the work in $37\frac{1}{2}$ days.
2 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?
Let's assume that 1 man takes x days to do the work and y days for a women.
So, the amount of work done by 1 man in 1 day = $\frac{1}{x}$ and,
the amount of work done by 1 woman in 1 day = $\frac{1}{y}$.
∴ The amount of work done by 2 men in 1 day = $\frac{2}{x}$ and,
The amount of work done by 5 woman in 1 day = $\frac{5}{y}$.
According to the given conditions,
$\frac{2}{x} + \frac{5}{y} = \frac{1}{4}$ .......(i)
$\frac{1}{x} + \frac{1}{y} = \frac{1}{12}$ ......(ii)
Multiplying (ii) by 5 we get,
$\frac{5}{x} + \frac{5}{y} = \frac{5}{12}$ .......(iii)
Subtracting (i) from (iii) we get,
$$\begin{aligned}\Rightarrow \frac{5}{x} + \frac{5}{y} - \Big(\frac{2}{x} + \frac{5}{y}\Big) = \frac{5}{12} - \frac{1}{4} \\[1em] \Rightarrow \frac{5}{x} - \frac{2}{x} = \frac{5 - 3}{12} \\[1em] \Rightarrow \frac{3}{x} = \frac{2}{12} \\[1em] \Rightarrow \frac{3}{x} = \frac{1}{6} \\[1em] \Rightarrow x = 18.\end{aligned}$$
Hence, one man can do the work in 18 days.
A train covered a certain distance at a uniform speed. If the train had been 30 km/h faster, it would have taken 2 hours less than scheduled time. If the train were slower by 15km/h , it would have taken 2 hours more than the scheduled time. Find the length of the journey.
Let the actual speed of the train be x km/h and scheduled time be y hours.
Distance = Speed × Distance = xy.
According to first condition,
⇒ (x + 30)(y - 2) = xy
⇒ xy - 2x + 30y - 60 = xy
⇒ -2x + 30y = 60 ......(i)
According to second condition,
⇒ (x - 15)(y + 2) = xy
⇒ xy - 15y + 2x - 30 = xy
⇒ 2x - 15y = 30 .......(ii)
Adding (i) and (ii) we get,
(-2x + 30y) + (2x - 15y) = 60 + 30
15y = 90
y = 6.
Substituting value of y in (i) we get,
⇒ -2x + 30(6) = 60
⇒ -2x + 180 = 60
⇒ -2x = -120
⇒ x = 60.
Distance = xy = 60 × 6 = 360 km.
Hence, distance of journey = 360 km.
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.
Let speed of boat in still water be x km/h and speed of stream be y km/h.
Speed of boat downstream = (x + y) km/h
Speed of boat upstream = (x - y) km/h
Given, boat takes 2 hours to go 40 km down the stream
$\therefore \frac{40}{x + y} = 2$
⇒ 2(x + y) = 40
⇒ x + y = 20 .......(i)
Given, boat returns in 4 hours
$\therefore \frac{40}{x - y} = 4$
⇒ 4(x - y) = 40
⇒ x - y = 10 .......(ii)
Adding (i) and (ii) we get,
⇒ (x + y) + (x - y) = 20 + 10
⇒ 2x = 30
⇒ x = 15.
Substituting value of x in (i) we get,
⇒ 15 + y = 20
⇒ y = 5.
Hence, speed of boat in still water = 15 km/h and speed of stream = 5 km/h.
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.
Let speed of boat in still water be x km/h and speed of current be y km/h.
Speed of boat in the current = (x + y) km/h
Speed of boat against current = (x - y) km/h
Given, boat takes 4 hours to go 44 km with the current
$\therefore \frac{44}{x + y} = 4$
⇒ 4(x + y) = 44
⇒ x + y = 11 .......(i)
Given, boat takes 4 hours 48 minutes longer
$$\begin{aligned}\therefore \frac{44}{x - y} = 4 + 4 + \frac{48}{60} \\[1em] \Rightarrow \frac{44}{x - y} = 4 + 4 + \frac{4}{5} \\[1em] \Rightarrow \frac{44}{x - y} = \frac{20 + 20 + 4}{5} \\[1em] \Rightarrow \frac{44}{x - y} = \frac{44}{5} \\[1em] \Rightarrow x - y = \frac{44 \times 5}{44} \\[1em] \Rightarrow x - y = 5 .......(ii)\end{aligned}$$
Adding (i) and (ii) we get,
⇒ (x + y) + (x - y) = 11 + 5
⇒ 2x = 16
⇒ x = 8.
Substituting value of x in (i) we get,
⇒ 8 + y = 11
⇒ y = 3.
Hence, speed of boat in still water = 8 km/h and speed of current = 3 km/h.
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.
Let speed of aeroplane be x km/h and speed of wind be y km/h.
Speed of aeroplane in direction of wind = (x + y) km/h
Speed of aeroplane against wind = (x - y) km/h
Given, aeroplane takes 3.5 hours to go 1680 km against the wind
$\therefore \frac{1680}{x - y} = 3.5$
⇒ 3.5(x - y) = 1680
⇒ x - y = 480 .......(i)
Given, aeroplane takes 3 hours to go 1680 km with the wind
$\therefore \frac{1680}{x + y} = 3$
⇒ 1680 = 3(x + y)
⇒ x + y = 560 .......(ii)
Adding (i) and (ii) we get,
⇒ (x - y) + (x + y) = 480 + 560
⇒ 2x = 1040
⇒ x = 520.
Substituting value of x in (i) we get,
⇒ 520 - y = 480
⇒ y = 40.
Hence, speed of aeroplane = 520 km/h and speed of wind = 40 km/h.
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 ₹2600 as hostel charges; whereas when Divya takes food for 26 days, she pays ₹3020 as hostel charges. Find the fixed charges and the cost of food per day.
Let fixed charge be ₹x and per day food charge be ₹y.
Given, Bhawana pays ₹2600 when she takes food for 20 days.
x + 20y = 2600 .......(i)
Given, Divya pays ₹3020 when she takes food for 26 days.
x + 26y = 3020 .......(ii)
Subtracting (i) from (ii) we get,
(x + 26y) - (x + 20y) = 3020 - 2600
6y = 420
y = 70.
Substituting value of y in (i) we get,
x + 20(70) = 2600
x + 1400 = 2600
x = 1200.
Hence, fixed charge = ₹1200 and the cost of food per day = ₹70.
Sum of digits of a two digit number is 8. If the number obtained by reversing the digits is 18 more than the original number, then the original number is
Let the digit at ten's place be x and digit at one's place be y.
Given, sum of digits = 8.
x + y = 8 .......(i)
Number = 10 × x + y = 10x + y
On reversing digits number = 10 × y + x = 10y + x.
Given, number obtained by reversing the digits is 18 more than the original number.
⇒ 10y + x = 10x + y + 18
⇒ 10y - y + x - 10x = 18
⇒ 9y - 9x = 18
⇒ y - x = 2 .........(ii)
Adding (i) and (ii) we get,
⇒ x + y + (y - x) = 8 + 2
⇒ 2y = 10
⇒ y = 5.
Substituting value of y in (i) we get,
⇒ x + 5 = 8
⇒ x = 3.
Number = 10x + y = 10(3) + 5 = 35.
Hence, Option 1 is the correct option.
The sum of two natural numbers is 25 and their difference is 7. The numbers are
Let the two numbers be x and y.
Given, sum = 25.
x + y = 25 ........(i)
Given, difference = 7.
x - y = 7 .......(ii)
Adding (i) and (ii) we get,
⇒ (x + y) + (x - y) = 25 + 7
⇒ 2x = 32
⇒ x = 16.
Substituting value of x in (i) we get,
⇒ 16 + y = 25
⇒ y = 9.
Hence, Option 2 is the correct option.
The sum of two natural numbers is 240 and their ratio is 3 : 5. Then the greater number is
Let the two numbers be x and y.
Given, sum = 240.
∴ x + y = 240
⇒ x = 240 - y ........(i)
$\therefore \frac{x}{y} = \frac{3}{5}$ ......(ii)
Substituting value of x from (i) in (ii) we get,
$$\begin{aligned}\Rightarrow \frac{240 - y}{y} = \frac{3}{5} \\[1em] \Rightarrow 5(240 - y) = 3y \\[1em] \Rightarrow 1200 - 5y = 3y \\[1em] \Rightarrow 8y = 1200 \\[1em] \Rightarrow y = 150.\end{aligned}$$
Substituting value of y in (i) we get,
⇒ x = 240 - 150 = 90.
Hence, Option 3 is the correct option.
The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the number get reversed. The number is
Let the digit at ten's place be x and digit at one's place be y.
Given, sum of digits = 9.
⇒ x + y = 9 .......(i)
Number = 10 × x + y = 10x + y
On reversing digits, number = 10 × y + x = 10y + x.
⇒ 10x + y + 27 = 10y + x
⇒ 10x - x + y - 10y + 27 = 0
⇒ 9x - 9y + 27 = 0
⇒ 9(x - y + 3) = 0
⇒ x - y + 3 = 0
⇒ y - x = 3 .......(ii)
Adding (i) and (ii) we get,
⇒ x + y + (y - x) = 9 + 3
⇒ 2y = 12
⇒ y = 6.
Substituting value of y in (i) we get,
⇒ x + 6 = 9
⇒ x = 3.
Number = 10 × x + y = 10 × 3 + 6 = 36.
Hence, Option 4 is the correct option.
The sum of the digits of a two digit number is 12. If the number is decreased by 18, its digits get reversed. The number is
Let the digit at ten's place be x and digit at one's place be y.
Given, sum of digits = 12.
x + y = 12 .......(i)
Number = 10 × x + y = 10x + y
On reversing digits, number = 10 × y + x = 10y + x.
⇒ 10x + y - 18 = 10y + x
⇒ 10x - x + y - 10y - 18 = 0
⇒ 9x - 9y = 18
⇒ 9(x - y) = 18
⇒ x - y = 2 .......(ii)
Adding (i) and (ii) we get,
⇒ x + y + (x - y) = 12 + 2
⇒ 2x = 14
⇒ x = 7.
Substituting value of x in (i) we get,
⇒ 7 + y = 12
⇒ y = 5.
Number = 10 × x + y = 10 × 7 + 5 = 75.
Hence, Option 4 is the correct option.
Aruna has only ₹1 and ₹2 coins with her. If the total number of coins that she has is 50 and the amount of the money with her is ₹75, then the number of ₹1 and ₹2 coins are respectively
Let ₹1 coins be x and ₹2 coins be y.
According to first condition,
x + y = 50 .......(i)
According to second condition,
1.x + 2.y = 75
x + 2y = 75 ......(ii)
Subtracting (i) from (ii) we get,
⇒ x + 2y - (x + y) = 75 - 50
⇒ 2y - y = 25
⇒ y = 25.
Substituting value of y in (i) we get,
⇒ x + 25 = 50
⇒ x = 25.
Hence, Option 4 is the correct option.
The age of a woman is four times the age of her daughter. Five years hence, the age of the woman will be three times the age of her daughter. The present age of the daughter is
Let age of daughter be x years so, the age of woman is 4x years.
After 5 years,
Age of daughter = (x + 5) years
Age of woman = (4x + 5) years.
According to question,
⇒ 4x + 5 = 3(x + 5)
⇒ 4x + 5 = 3x + 15
⇒ 4x - 3x = 15 - 5
⇒ x = 10.
Hence, Option 4 is the correct option.
Father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present age in years of the son and the father are, respectively,
Let age of son be x years so, the age of father is 6x years.
After 4 years,
Age of son = (x + 4) years
Age of father = (6x + 4) years.
According to question,
⇒ 6x + 4 = 4(x + 4)
⇒ 6x + 4 = 4x + 16
⇒ 2x = 16 - 4
⇒ 2x = 12
⇒ x = 6.
Age of father = 6x = 36.
Hence, Option 3 is the correct option.
Consider the following two statements:
Statement 1: A husband is 2 years older than his wife, and sum of their ages is 52 years. Then the wife is 25 years old.
Statement 2: A father is twice as old as his daughter, and difference of their ages is 26 years. Then the father is 50 years old.
Which of the following is valid?
Let x years be the husband's age and y years be the wife's age.
Given,
Husband is 2 years older than his wife.
⇒ x = y + 2 ..................(1)
Sum of husband's age and wife's age = 52 years
⇒ x + y = 52 ..................(2)
Substituting the value of x from equation (1) in equation (2), we get
⇒ (y + 2) + y = 52
⇒ 2y + 2 = 52
⇒ 2y = 52 - 2
⇒ 2y = 50
⇒ y = $\frac{50}{2}$
⇒ y = 25 years.
∴ Statement 1 is true.
Let the age of daughter be a years.
Given,
Father's age is twice that of daughter's age.
Father's age = 2a years
Difference of father's age and daughter's age = 26 years
⇒ 2a - a = 26
⇒ a = 26
Father's age = 2a = 2 x 26 = 52 years.
∴ Statement 2 is true.
∴ Both the statements are true.
Hence, option 1 is the correct option.
Assertion (A): Difference between ages of two brothers is 5 years, while sum of their ages is 25 years. Then the younger is 10 years old.
Reason (R): The difference between age of two brothers remains constant, even when they grow older.
Let x years be the age of the elder brother.
The age of the younger brother = (x - 5) years
Sum of their ages = 25 years
⇒ x + (x - 5) = 25
⇒ x + x - 5 = 25
⇒ 2x - 5 = 25
⇒ 2x = 25 + 5
⇒ 2x = 30
⇒ x = $\frac{30}{2}$
⇒ x = 15 years.
The age of the younger brother = 15 - 5 = 10 years.
∴ Assertion (A) is true.
The difference between age of two brothers remains constant, even when they grow older.
Both individuals age by the same amount over any given period, effectively cancelling out the increase in their individual ages when calculating the difference.
∴ Reason (R) is true.
∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is not the correct reason (or explanation) for Assertion (A).
Hence, option 4 is the correct option.
Assertion (A): Perimeter of a garden is 24 cm, while the difference between its length and width is 2 units. Then its area is 35 sq. cm.
Reason (R): If length of a rectangle is doubled, while width remains the same, then the perimeter also gets doubled.
Let l cm be the length of a garden and b cm be the width of the garden.
Given,
Difference between length and width is 2 units.
⇒ l - b = 2
⇒ l = 2 + b ..........(1)
Given,
Perimeter = 24 cm
⇒ 2(l + b) = 24 cm
⇒ 2[2 + b + b] = 24 [From equation (1)]
⇒ 2(2b + 2) = 24
⇒ 4b + 4 = 24
⇒ 4b = 24 - 4
⇒ 4b = 20
⇒ b = $\frac{20}{4}$
⇒ b = 5 cm.
The length of a garden (l) = 2 + b = 2 + 5 = 7 cm.
Area of garden = l x b
= 7 x 5 sq. cm.
= 35 sq. cm.
∴ Assertion (A) is true.
Length of a rectangle is doubled, while width remains the same.
⇒ New length = 2l and New width = b
New perimeter = 2(New length + New width)
= 2(2l + b)
So, the perimeter does not gets double.
∴ Reason (R) is false.
∴ Assertion (A) is true, Reason (R) is false.
Hence, option 1 is the correct option.
A 700 g dry fruit pack costs ₹ 432. It contains some almonds and the rest cashew kernel. If almonds cost ₹ 576 per kg and cashew kernel cost ₹ 672 per kg, what are the quantities of the two dry fruits separately?
Let the quantity of almonds be x gm and that of cashew kernel be y gm.
∴ x + y = 700
⇒ x = 700 - y ...................(1)
Cost of almonds = ₹576 per kg
Cost of x gm almonds = $\frac{576}{1000}x$
Cost of cashew kernel = ₹672 per kg
Cost of y gm cashew kernel = $\frac{672}{1000}y$
Cost of dry fruit pack = ₹432.
$$\begin{aligned}\therefore \frac{576}{1000}x + \frac{672}{1000}y = 432\\[1em] \Rightarrow \frac{576x + 672y}{1000} = 432\\[1em] \Rightarrow 576x + 672y = 432000 ......................(2)\end{aligned}$$
Substituting the value of x from equation (1) in (2) we get,
⇒ 576(700 - y) + 672y = 432000
⇒ 403200 - 576y + 672y = 432000
⇒ 403200 + 96y = 432000
⇒ 96y = 432000 - 403200
⇒ 96y = 28800
⇒ y = $\frac{28800}{96}$
⇒ y = 300.
Substituting the value of y in equation (1), we get :
⇒ x = 700 - y = 700 - 300 = 400.
Hence, almonds = 400 gm and cashew kernel = 300 gm in dry fruits pack.
Drawing pencils cost ₹ 4 each and coloured pencils cost ₹ 5.50 each. If altogether two dozen pencils cost ₹108, how many coloured pencils are there?
Let number of drawing pencils be x and number of coloured pencils be y.
Total pencils = 24 (two dozens).
⇒ x + y = 24
⇒ x = 24 - y ......................(1)
Cost of drawing pencils = ₹ 4 each
Total cost of drawing pencils = ₹4x
Cost of coloured pencils = ₹ 5.50 each
Total cost of coloured pencils = ₹5.50y
Total cost = ₹ 108
⇒ 4x + 5.50y = 108
Substituting the value of x from equation (1), we get
⇒ 4(24 - y) + 5.50y = 108
⇒ 96 - 4y + 5.50y = 108
⇒ 96 + 1.5y = 108
⇒ 1.5y = 108 - 96
⇒ 1.5y = 12
⇒ y = $\frac{12}{1.5}$
⇒ y = 8.
Hence, number of coloured pencils = 8.
Shikha works in a factory. In one week she earned ₹ 3,900 for working 47 hours, of which 7 hours were overtime. The next week she earned ₹ 4,160 for working 50 hours, of which 8 hours were overtime. What is Shikha's hourly earning rate?
Let's assume earning's of Shikha be ₹ x per hour for regular hour and ₹ y per hour for overtime.
Given,
In one week she earned ₹ 3,900 for working 47 hours, of which 7 hours were overtime.
Shikha works for 40 hours regular and 7 hours overtime.
40x + 7y = 3900 ....................(1)
Given,
The next week she earned ₹ 4,160 for working 50 hours, of which 8 hours were overtime.
Thus, she works for 42 hours regular and 8 hours overtime.
42x + 8y = 4160 ....................(2)
Multiplying equation (1) by 8 , we get :
⇒ 8(40x + 7y) = 3900 x 8
⇒ 320x + 56y = 31200 ....................(3)
Multiplying equation (2) by 7, we get :
⇒ 7(42x + 8y) = 4160 x 7
⇒ 294x + 56y = 29120 ....................(4)
Subtracting equation (4) from (3) we get,
⇒ 320x + 56y - (294x + 56y) = 31200 - 29120
⇒ 320x - 294x + 56y - 56y = 31200 - 29120
⇒ 26x = 2080
⇒ x = $\frac{2080}{26}$
⇒ x = 80.
Substituting the value of x in (1) we get,
⇒ 40(80) + 7y = 3900
⇒ 3200 + 7y = 3900
⇒ 7y = 700
⇒ y = $\frac{700}{7}$
⇒ y = 100.
Hence, Shikha's earning is ₹ 80 per regular hour and ₹ 100 per hour for over time.
The sum of digits of a two digit number is 7. If the digits are reversed, the new number increased by 3 equals 4 times the original number. Find the number.
Let digit at ten's place be x and unit's digit be y.
Number = 10 × x + y = 10x + y.
Reverse number = 10 × y + x = 10y + x.
Given, sum of digits is 7
∴ x + y = 7
⇒ x = 7 - y ......(i)
According to question,
⇒ 4(10x + y) = 10y + x + 3
⇒ 40x + 4y = 10y + x + 3
⇒ 40x - x + 4y - 10y = 3
⇒ 39x - 6y = 3
⇒ 3(13x - 2y) = 3
⇒ 13x - 2y = 1 .......(ii)
Substituting value of x from (i) in (ii) we get,
⇒ 13(7 - y) - 2y = 1
⇒ 91 - 13y - 2y = 1
⇒ 91 - 15y = 1
⇒ 15y = 91 - 1
⇒ 15y = 90
⇒ y = 6.
Substituting value of y in (i) we get,
x = 7 - y = 7 - 6 = 1.
∴ Number = 10x + y = 10(1) + 6 = 16.
Hence, number = 16.
Three years hence a man's age will be three times his son's age, and 7 years ago he was seven times as old as his son. How old are they now?
Let present age of son be x and man's age be y.
According to first condition,
⇒ (y + 3) = 3(x + 3)
⇒ y + 3 = 3x + 9
⇒ y - 3x = 9 - 3
⇒ y - 3x = 6 ......(i)
According to second condition,
⇒ (y - 7) = 7(x - 7)
⇒ y - 7 = 7x - 49
⇒ y - 7x = -49 + 7
⇒ y - 7x = -42
⇒ 7x - y = 42 ......(ii)
Adding (i) and (ii) we get,
⇒ y - 3x + 7x - y = 6 + 42
⇒ 4x = 48
⇒ x = 12.
Substituting value of x in (i) we get,
⇒ y - 3(12) = 6
⇒ y - 36 = 6
⇒ y = 42.
Hence, son's age = 12 years and man's age = 42 years.
Rectangles are drawn on line segments of fixed lengths. When the breadths are 6m and 5m respectively the sum of the areas of the rectangles is 83 m2. But if the breadths are 5m and 4m respectively the sum of the areas is 68 m2. Find the sum of the areas of squares drawn on the line segments.
Let length of first fixed line segment be x and second line segment be y.
In first case, when the breadths are 6m and 5m the sum of the areas = 83 m2.
6x + 5y = 83 ......(i)
In second case, when the breadths are 5m and 4m the sum of the areas = 68 m2.
5x + 4y = 68 ......(ii)
Multiplying (i) by 4 and (ii) by 5 we get,
24x + 20y = 332 ......(iii)
25x + 20y = 340 ......(iv)
Subtracting (iii) from (iv) we get,
25x + 20y - (24x + 20y) = 340 - 332
⇒ x = 8.
On substituting value of x in (i) we get,
⇒ 6(8) + 5y = 83
⇒ 48 + 5y = 83
⇒ 5y = 83 - 48
⇒ 5y = 35
⇒ y = 7.
Sum of areas of squares on these two line segments = x2 + y2 = 82 + 72 = 64 + 49 = 113 m2.
Hence, sum of areas of squares on these two line segments = 113 m2.
If the length and breadth of a room are increased by 1 metre each, the area is increased by 21 square meters. If the length is decreased by 1 meter and the breadth is increased by 2 meters, the area is increased by 14 square meters. Find the perimeter of the room.
Let length be x and breadth be y meters.
So, area = xy m2.
According to first condition,
⇒ (x + 1)(y + 1) = xy + 21
⇒ xy + x + y + 1 = xy + 21
⇒ xy - xy + x + y = 21 - 1
⇒ x + y = 20 .......(i)
According to second condition,
⇒ (x - 1)(y + 2) = xy + 14
⇒ xy + 2x - y - 2 = xy + 14
⇒ 2x - y = xy - xy + 14 + 2
⇒ 2x - y = 16 .......(ii)
Adding (i) and (ii) we get,
⇒ x + y + (2x - y) = 20 + 16
⇒ 3x = 36
⇒ x = 12.
Substituting value of x in (i) we get,
⇒ 12 + y = 20
⇒ y = 8.
Perimeter = 2(length + breadth) = 2(x + y) = 2(12 + 8) = 2(20) = 40 meters.
Hence, perimeter of room = 40 meters.
The lengths (in meters) of the sides of a triangle are $2x + \frac{y}{2}, \frac{5x}{3} + y + \frac{1}{2} \text{ and } \frac{2}{3}x + 2y + \frac{5}{2}.$ If the triangle is equilateral, find its perimeter.
Since, triangle is equilateral hence all sides are equal.
$$\begin{aligned}\therefore 2x + \frac{y}{2} = \frac{5x}{3} + y + \frac{1}{2} \\[1em] \Rightarrow 2x - \frac{5x}{3} + \frac{y}{2} - y = \frac{1}{2} \\[1em] \Rightarrow \frac{6x - 5x}{3} + \frac{y - 2y}{2} = \frac{1}{2} \\[1em] \Rightarrow \frac{x}{3} - \frac{y}{2} = \frac{1}{2} \\[1em] \Rightarrow \frac{2x - 3y}{6} = \frac{1}{2} \\[1em] \Rightarrow 2x - 3y = 3 .......(i)\end{aligned}$$
Similarly,
$$\begin{aligned}\Rightarrow \frac{5x}{3} + y + \frac{1}{2} = \frac{2x}{3} + 2y + \frac{5}{2} \\[1em] \Rightarrow \frac{5x}{3} - \frac{2x}{3} + y - 2y = \frac{5}{2} - \frac{1}{2} \\[1em] \Rightarrow \frac{3x}{3} - y = \frac{4}{2} \\[1em] \Rightarrow x - y = 2 ........(ii)\end{aligned}$$
Multiplying (ii) by 2 we get,
2x - 2y = 4 .......(iii)
Subtracting (i) from (iii) we get,
⇒ 2x - 2y - (2x - 3y) = 4 - 3
⇒ -2y + 3y = 1
⇒ y = 1.
Substituting value of y in (i) we get,
⇒ 2x - 3(1) = 3
⇒ 2x = 6
⇒ x = 3.
Side 1 = $2x + \frac{y}{2} = 2(3) + \frac{1}{2} = 6 + 0.5$ = 6.5 m
∵ Triangle is equilateral, its other two sides will also be 6.5 m each.
∴ Perimeter = 3 x 6.5 = 19.5 m.
Hence, perimeter of triangle = 19.5 meters.
On Diwali eve, two candles, one of which is 3 cm longer than the other, are lighted. The longer one is lighted at 5.30 p.m. and the shorter at 7 p.m. At 9.30 p.m. they both are of same length. The longer one burns out at 11.30 p.m. and the shorter one at 11 p.m. How long was each candle originally?
Let's assume that the longer candle shorten at rate of x cm/hr when burning and the smaller candle shorter at rate of y cm/hr.
Given, the longer candle burns out completely in 6 hours and the smaller candle in 4 hours,
So, their lengths are 6x cm and 4y cm respectively.
According to first condition,
⇒ 6x = 4y + 3
⇒ 6x - 4y = 3 .....(i)
At 9:30 p.m. length of longer candle = (6x - 4x) = 2x cm.
At 9:30 p.m. the length of smaller candle = $4y - \frac{5y}{2} = \frac{8y - 5y}{2} = \frac{3y}{2}$
Now, according to second condition given in problem,
2x = $\frac{3y}{2}$ (As both candles have same length at 9:30 p.m.)
⇒ 4x = 3y
4x - 3y = 0 ......(ii)
Multiplying (i) by 3 and (ii) by 4 we get,
18x - 12y = 9 .......(iii)
16x - 12y = 0 .......(iv)
Subtracting (iv) from (iii) we get,
⇒ 18x - 12y - (16x - 12y) = 9 - 0
⇒ 2x = 9
⇒ x = $\frac{9}{2}$ = 4.5 cm/hr.
On substituting value of x in (ii) we get,
4x - 3y = 0
4(4.5) - 3y = 0
18 - 3y = 0
3y = 18
y = 6 cm/hr.
Length of longer candle = 6x = 6(4.5) = 27 cm.
Length of smaller candle = 4y = 4(6) = 24 cm.
Hence, length of longer candle = 27 cm and length of smaller candle = 24 cm.
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