Problems on Simultaneous Linear Equations — Chapter Test

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?

Answer

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 = $\dfrac{576}{1000}x$

Cost of cashew kernel = ₹672 per kg

Cost of y gm cashew kernel = $\dfrac{672}{1000}y$

Cost of dry fruit pack = ₹432.

$\therefore \dfrac{576}{1000}x + \dfrac{672}{1000}y = 432\\[1em] \Rightarrow \dfrac{576x + 672y}{1000} = 432\\[1em] \Rightarrow 576x + 672y = 432000 ......................(2)$

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 = $\dfrac{28800}{96}$

⇒ 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?

Answer

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 = $\dfrac{12}{1.5}$

⇒ 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?

Answer

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 = $\dfrac{2080}{26}$

⇒ x = 80.

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

⇒ 40(80) + 7y = 3900

⇒ 3200 + 7y = 3900

⇒ 7y = 700

⇒ y = $\dfrac{700}{7}$

⇒ 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.

Answer

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?

Answer

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.

Answer

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.

Answer

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 $2x + \dfrac{y}{2}, \dfrac{5x}{3} + y + \dfrac{1}{2} \text{ and } \dfrac{2}{3}x + 2y + \dfrac{5}{2}.$ If the triangle is equilateral, find its perimeter.

Answer

Since, triangle is equilateral hence all sides are equal.

$\therefore 2x + \dfrac{y}{2} = \dfrac{5x}{3} + y + \dfrac{1}{2} \\[1em] \Rightarrow 2x - \dfrac{5x}{3} + \dfrac{y}{2} - y = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{6x - 5x}{3} + \dfrac{y - 2y}{2} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{x}{3} - \dfrac{y}{2} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{2x - 3y}{6} = \dfrac{1}{2} \\[1em] \Rightarrow 2x - 3y = 3 .......(i)$

Similarly,

$\Rightarrow \dfrac{5x}{3} + y + \dfrac{1}{2} = \dfrac{2x}{3} + 2y + \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{5x}{3} - \dfrac{2x}{3} + y - 2y = \dfrac{5}{2} - \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{3x}{3} - y = \dfrac{4}{2} \\[1em] \Rightarrow x - y = 2 ........(ii)$

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 = $2x + \dfrac{y}{2} = 2(3) + \dfrac{1}{2} = 6 + 0.5$ = 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?

Answer

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 = $4y - \dfrac{5y}{2} = \dfrac{8y - 5y}{2} = \dfrac{3y}{2}$

Now, according to second condition given in problem,

2x = $\dfrac{3y}{2}$ (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 = $\dfrac{9}{2}$ = 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.