ML Aggarwal Solutions Class 9 Math 6th Chapter Problems on Simultaneous Linear Equations Exercise 6
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Sixth Chapter Problems on Simultaneous Linear Equations Exercise 6. APC Solution Class 9 Exercise 6.
(1) The sum of two numbers is 50 and their difference is 16. Find the numbers.
Solution:
Let, the two numbers be x & y
(ATQ), x + y = 50 —- (i)
x – y = 16 —- (ii) [x > y]
Adding equation (i) & (ii) we get,
x + y = 50
x – y = 16
__________
2x = 66
Or, x = 33
Substituting the value of x in equation (ii) we get,
x – y = 16
Or, -y = 16 – 33
Or, -y = -17
Or, y = 17
∴The two numbers are 33 & 17.
(2) The sum of two numbers is 2. If their difference is 20, find the numbers.
Solution:
Let, the greater number be x & smaller number by y.
ATQ, x + y = 2 —- (i), x – y = 20 — (ii)
Adding equation (i) & (ii) we get
x+y = 2
x-y = 20
_________
2x = 22
Or, x = 11
Substituting the value of x equation (i) we get,
11 + y = 2
Or, y = 2 – 11
Or, y = -9
∴The two numbers are 11 & -9.
(5) The class IX students of a class X. They decided to purchase two kinds of sweets, one costing ₹70 per kg and the other costing ₹84 per kg. They estimated that 36kg of sweets were needed. If the total money spent on sweets was ₹2800, find how much sweets of each kind they purchased.
Solution:
Let, sweet of ₹ 70/kg bough be x kg
Sweet of ₹84/kg bought be y kg.
(ATQ) x + y = 36 — (i),
X × 70 + y × 84 = 2800
Or, 7 (10x + 12y) = 2800
Or, 2 (5x + 6y) = 400
Or 5x + by = 200 —- (ii)
Multiplying equation (i) with 5 then subtracting equation (ii) from equation (i) we get,
5x + 5y = 180
5x + by = 200
(-) (-) (-)
______________
y = -20
Or, y = 20kg
Substituting the value of y in equation (i) we get,
x + 20 = 36
Or, x = 16 kg
∴ Amount of ₹70/kg sweets bought = 16kg
Amount of ₹ 84/kg sweets bought = 20 kg
(6) If from twice the greater 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?
Solution:
Let, the greater no. be x, the smaller no. be y.
(ATQ), 2x – 16 = y/2
Or, 4x – 32 = y
or, 4x – y = 32 —- (i)
x/2 – 1 = y/2
Or, x-2/2 = y/2
Or, x – y = 2 —- (ii)
Subtracting equation (ii) from equation (i) we get,
4x – y = 32
x – y = 2
(-) (+) (-)
___________
3x = 30
Or, x = 10
Substituting the value of x from equation (ii) we get,
10 –y = 2
Or, -y = -8
Or, y > 8
∴The greater no. is 10 & the smaller no. is y.
(7) 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?
Solution:
Let, the no. of 25 paise coin be x.
The no. of 20 paise coin be y.
(ATQ) x + y = 38 – — (i)
X × 25 + y × 20 = 8.50 × 100
Or, 25x + 20y = 850
Or, 5 (5x + 4y) = 170
Or, 5x + 4y = 170 — (ii)
Multiplying equation (i) with 4 then subtracting equation (ii) from equation (i) we get,
4x + 4y = 152
5x + 4y = 100
(-) (-) (-)
____________
-x = -18
Or, x = 18
Substituting the value of x from equation (i) we get,
18 + y = 38
Or, y = 20
∴ No. of 25 paise coins = 18, No. of 20 paise coins = 20
(8) A man has certain notes of denominations ₹20 and ₹5 which amount to ₹380. If the number of notes of each kind is interchanged, they amount to ₹60 less as before. Find the number of notes of each denomination.
Solution:
Let, the no. of ₹20 notes be x.
The no. of ₹5 notes be y
(ATQ), 20x + 5y = 380
Or, 5 (4x + y) = 380
Or, 4x + y = 76 — (i)
5x + 20y = 380 – 60
Or, 5 (x + 4y) = 320
Or, x + 4y = 64 – (ii)
Multiplying equation (ii) with 4 then subtracting equation (ii) from equation (i) we get,
4x + y = 76
4x + 16y = 256
(-) (-) (-)
_______________
-15y = -180
Or, y = 12
Substituting the value of y in equation (i) we get,
4x + 12 = 76
Or, 4x = 64
Or, x = 16
∴ No. of ₹20 notes are 16 & no. of ₹5 notes are 12.
(11) If the numerator of certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to 5/8 and if the numerator and denominator are each diminished by 1, the fraction becomes equal to 1/2; find the fraction.
Solution:
Let, the fraction be x/y
(ATQ), x+2/y+1 = 5/8
Or, 8x + 16 = 5y + 5
Or, 8x – 5y = 5 – 16
Or, 8x – 5y = -11 —- (i)
x-1/y-1 = 1/2
Or, 2x – 2 = y – 1
Or, 2x – y = -1 + 2
Or, 2x – y = 1 —- (ii)
Multiplying equation (ii) by 5 then subtracting equation (ii) from equation (i) we get,
8x – 5y = -11
10x – 5y= 5
(-) (+) (-)
_____________
-2x = -16
Or, x = 8
Substituting the value of x in equation (ii) we get,
2 × 8 – y = 1
Or, y = 16-1
Or, y = 15
(12) Find the fraction which becomes 1/2 when the denominator is increased by 4 and is equal to 1/8 when the numerator is diminished by 5.
Solution:
Let, the fraction be x/y
x/y+4 = 1/2
Or, 2x = y + 4
Or, 2x – y = 4 —– (i)
x-5/y = 1/8
Or, 8x – 40 = y
Or, 8x – y = 40 —- (ii)
Substracting equation (ii) from equation (i) we get,
2x – y = 4
8x – y = 40
(-) (-) (-)
___________
-6x = -36
Or, x = 6
Substituting the value of x in equation (i) we get,
2 × 6 – y = 4
Or, – y = 4 – 12
Or, – y = – 8
Or, y = 8
∴ the original fraction is 6/8.
(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.
Solution:
Let, the ten’s digit be x
The unit’s digit be y
∴the original number is 10x + y
(ATQ) x + y = 7 — (i)
10y + x = 28 + 2y
Or, x + 10y – 2y = 28
Or, x + 8y = 28 —- (ii)
Substracting equation (ii) from equation (i) we get,
x + y = 7
x + 8y = 28
(-) (-) (-)
___________
-7y = -21
Or, y = 3
Substituting the value of y from equation (i) we get,
x + 3 = 7
Or, x = 4
∴The original number is 10x + y = 10 × 4 + 3
= 43
(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.
Solution:
Let, the ten’s digit be x
the units digit be y
∴ the original number is 10x + y
(ATQ) 10x + y = 4 (x + y) + 6
Or, 10x + y = 4x + 4y + 6
Or, 6x – 3y = 6
Or, 3 (2x – y) = 6
Or, 2x – y = 2 — (i)
10y + x – (10x + y) = 4
Or, 10y + x – 10x – y = 9
Or, -9x + 9y = 9
Or, 9 (-x + y) = 9
Or, -x + y = 1 — (ii)
Adding equation (i) & (ii) we get,
2x – 4 = 2
– x + y = 1
_____________
x = 3
Substituting the value of x from equation (ii) we get,
-3 + y = 1
Or, y = 4
∴ The original number is 10x + y = 10 × 3 + 4 = 34
(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?
Solution:
Let, the ten’s digit be x, the unit digit be y,
∴ the original number is 10x + y
(ATQ) 10x+y/x+y = 8
Or, 10x + y = 8x + 8y
Or, 10x – 8x + y – 8y = 0
Or, 2x – 7y = 0 —- (i)
x – 3y = 1 —- (ii)
Multiplying equation (ii) with 2 then subtracting equation (ii) from equation (i) we get,
2x – 7y = 0
2x – 6y = 2
(-) (+) (-)
____________
-y = -2
Or, y = 2
Substituting the value of y in equation (ii) we get,
x – 3 × 2 = 1
Or, x = 1 + 6
Or, x = 7
∴ The original number is 10x + y = 10× 7 + 2 = 72
(16) The result of dividing a number of two digits by the number with digits reversed is 1 3/4. If the sum of digits is 12, find the number.
Solution:
Let, the ten’s digit be x, the units digit be y.
∴ The original number I 10x + y, reversed number 10y + x,
(ATQ)
10x+y/10y+x = 1 3/4
Or, 10x+y/10y+x = 7/4
Or, 40x+4y = 70y + 7x
Or, 35x – 66y = 0
Or, 33 (x – 2y) = 0
Or, x – 2y = 0 —- (ii)
x + y = 12 —- (ii)
Multiplying equation (ii) with 2 the adding both equation we get,
x – 2y = 0
2x + 2y = 24
_________________
3x = 24
Or, x = 8
Substituting the value of x from equation (ii) we get,
8 + y = 12
Or, y = 4
∴The original number is 10x + y = 10 ×8 + 4 = 84
(17) The result of dividing a number of two digits by the number with the digits reversed is 5/6. If the difference of digits is 1, find the number.
Solution:
Let, the ten’s digit be x, the unit’s digit be y.
∴ Original number is 10x + y, reversed number is 10y+ x.
(ATQ)
(18) A number of three digits has the hundred digit 4 times the unit digit and the sum of three digit 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 unit’s digit be x,
∴ The founder’s digit is 4x
the ten’s digit be y
(ATQ) 4x + x + y = 14
Or, 5x + y = 14 — (i)
∴The original number is 100 × 4x + 10y + x
= 410x + 10y + x
= 401x + 10y
Reversed digit is 100x + 10y + 4x
= 104x + 10y
401x + 10y – (104x + 10y) = 594
Or, 297x = 594
Or, x = 594/297
Or, x = 2
Substituting the values of x in equation (i) we get,
5 × 2 + y = 14
Or, y = 14 – 10
Or, y = 4
∴The original number is 401×2 + 10 × 4
= 802 + 40
= 842
(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.
Solution:
Let, Marina’s age be x
Marina’s daughter’s age be y
(AB) x – 4 = 3 (y – 4)
Or, x – 4 = 3y – 12
Or, x – 3y = -12 + 4
Or, x – 3y = -8 — (i)
x + 6 = 2 (y + 6)
Or, x + 6 = 2y + 12
Or, x – 2y = 6 —- (ii)
Subtracting equation (ii) from equation (i) we get,
x – 3y = -8
x – 2y = 6
(-) (+) (-)
____________
-y = -14
Or, y = 14
Substituting the value of y in equation (ii) we get,
x – 2 × 14 = 6
Or, x = 6 + 28
Or, x = 34
∴ Present age of marine is 34 years.
Present age of marine daughter is 14 years
(20) 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 the lemon set.
Solution:
If, the c.p. of set x.
The cp of lemon set be y.
(ATQ) s.p of tea set = x – 5/100 × x
= 19x/20
S.P of lemon set = y + 15/100 × y
= 23y/20
∴ Total S.P 19x/20 + 23y/20 = 19+23y/20
Total C.P = x+y
(ATQ) 19x+23y/20 – (x+y) = 70
Or, 19x+23y-20x-20y/20 = 70
Or, -x + 3y = 1400 —- (i)
Now, S.P of tea set = x + 5/100 × x
= 21x/20
S.P of lemon set = y + 10/100 × y
= 11y/10
Total S.P = 11y/10 + 21x/20 = 21x+22y/20
Total C.P = x+y
(ATQ), 21x+22y/20 – (x+y) = 130
Or, 21x+22y – 20x-20y/20 = 130
Or, x + 2y = 2600 —- (ii)
Adding equation (i) & (ii) we get,
-x + 3y = 1400
x + 2y = 2600
____________
5y = 4000
Or, y = 800
∴The cost price of lemon set is = ₹800
(21) 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?
Solution:
Let, the 1st amount be x, 2nd amount be y.
∴ S.I for 1year on x amount = x × 12 × 1/100
= 3x/25
S.I for 1 year on y amount = y×10×1/100
= y/10
(ATQ) 3x/25 + y/10 = 1300
Or, 6x+5y/50 = 1300
Or, 6x + 5y = 65000 — (i)
Observing the interest rates we get,
S.I for 1 year on x amount = x×10×1/100
= x/10
S.I for 1 year on y amount = y×12×1/100
= 3y/25
(ATQ) x/10 + 3y/25 = 1300+40
Or, 5x+6y/50 = 1340
Or, 5x+6y = 67000 — (ii)
Adding equation (i) & (ii) we get,
6x + 5y = 65000
5x + 6y = 67000
_________________
11x + 11y = 132000
Or, 11 (x + y) = 132000
Or, x + y = 12000 —- (iii)
Substracting equation (ii) from equation (i) we get,
6x + 5y = 65000
5x + 4y = 67000
(-) (-) (-)
_______________
x – y = -2000 —– (ii)
Adding equation (iii) & (iv) we get,
x + y = 12000
x – y = -2000
____________
2x = 10000
Or, x = 5000
∴ at 12% rate ₹5000 was invested
at 10% rate ₹7000 was invested
Substituting the value of x in equation (iii) we get,
5000 + y = 12000
Or, y = 7000
(22) 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.
Solution:
Let, the cost price of the table be x.
The list price of the chair be y
∴S.P of the table = x + 8/100 × x
= 27x/25
∴ S.P of the chair = y – 10/100 × y
= 9y/10
(ATQ) 27x/25 + 9y/10 = 1008
Or, 54x+45y/50 = 1008
Or, 54x + 45y = 50400
Or, 9 (6x + 5y) = 50400
Or, 6x + 5y = 5600 —- (i)
Now, at different rates we get,
S.P of the table = x + 10/100 × x
= 11x/10
S.P of the chair = y – 8/100 × 100y
= 23y/25
(ATQ) 11x/10 + 23y/25 = 1008+20
Or, 55x+46y/50 = 1028
Or, 55x + 46y = 51400 —- (ii)
Multiplying equation (i) with equation (i) with 55 and (ii) with 6 the substracting equation (i) from equation (ii) we get,
330x + 276y = 308400
330x + 275y = 308000
(-) (-) (-)
_____________________
y = 400
Substituting the value of y in equation (ii) we get,
6x + 5 × 400 = 5600
Or, 6x = 5600 – 2000
Or, x = 3600/6
Or, x = 600
∴The cost price of tables is ₹600
The list price of chair is ₹400
(23) 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 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?
Solution:
Let, amount of A is money be x
Amount of B is money be y
(ATQ) x + 100 = 75/100 × (y – 100)
Or, 4x + 400 = 3y – 300
Or, 4x – 3y = -700 — (i)
x – 100 = 40/100 × (y + 100)
Or, 5x – 500 = 2y + 200
Or, 5x – 2y = 700 — (ii)
Multiplying equation (i) with 2 & equation (ii) with 3 then Substracting equation (i) from equation (ii) we get,
15x – 6y = 2100
8x – 6y = -1400
(-) (+) (+)
________________
7x = 3500
Or, x = 500
∴Substituting the value of x from equation (i) we get
4 × 500 – 3y = -700
Or, -3y = -700 – 2000
Or, -3y = -2700
Or, y = 2700/3
Or, y = 900
∴ A has ₹500
3 has ₹900
(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.
Solution:
Let, the no. of rows be x.
The no. students in each row be y
∴Total no. of students = xy
(ATQ) (y+1) (x-2) = xy
Or, xy – 2y + x – 2 = xy
Or, x – 2y = 2 – (i)
Again, (y – 1) (x + 3) = xy
Or, xy + 3y – x – 3 = xy
Or, -x +3y = 3 —- (ii)
Adding equation (i) & (ii) we get,
x – 2y = 2
-x + 3y = 3
___________
y = 5
Substituting the value of y in equation (i) we get,
x – 2 × 5 = 2
Or, x = 2 + 10
Or, x = 12
∴ Total no. of students = xy = 5×12 = 60
(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 16-carat gold weighing 120 grams? (Pure gold is 24-carat).
Solution:
Let, weight of 18 carat gold be x (24 carat pure gold)
Weight of 12 carat gold be y
(ATQ) x + y = 120 — (i)
18x/24 + 12y/24 = 120×16/24
Or, 18x+12y/24 = 120×16/24
Or, 6 (3x + 24) = 120 × 16
Or, 3x + 2y = 325 —- (ii)
Multiplying equation (ii) with 2 them substracting equation (i) from (ii) we gets
3x + 2y = 320
2x + 2y = 240
(-) (-) (-)
______________
x = 80
∴ Weight of 18 carat gold = 80g
Weight of 12 carat gold = 40g
Substituting the value of x in equation (i) we get,
80 + y = 120
Or, y = 40
(26) A and B together can do a piece of work in 15 days. If A’s one day work is 1 1/2 times the one day’s work of B find in how many days can each do the work.
Solution:
Let, A’s one day work be x.
B’s one day work be y
∴ (ATQ) x + y = 1/15
Or, 15x + 15y = 1 —- (i)
x = 1 1/2 y
Or, x = 3/2 y = 0 — (ii)
Multiplying equation (ii) with 5 then adding them we get,
15x + 15y = 1
10x + -15y = 0
_________________
25a = 1
Or, x = 1/25
Substituting the value of x in equation (ii) we get,
2 × 1/25 – 3y = 0
Or, y = 2/25×3
Or, y = 2/75
∴ A can complete the work in 1/x days
= 25 days
B can complete the work in 1/y days
= 75/2 days
(27) 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?
Solution:
(28) 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 the scheduled time. If the train were slower by 15 km/h, it would have taken 2 hours more than the scheduled time. Find the length of the journey.
Solution:
Let, the speed of train be x km/hr
The time taken be y hr.
∴ Length of journey = xy km
(ATQ) (x + 30) (y – 2) = xy
Or, xy – 2x + 30y – 60 = xy
Or, -2x + 30y = 60
Or, -x + 15y = 30 — (i)
Also, (x – 15) (y + 2) = xy
Or, xy + 2x – 15y = 30 = xy
Or, 2x – 15y = 30 — (ii)
Adding equation (i) & (ii) we get,
-x + 15y = 30
2x – 15y = 30
__________________
x = 60
Substituing the value of x from equation (i) we get,
-60 + 15y = 30
Or, y = 90/15
Or, y = 6
∴ Length of journey = xy = 60×6 = 360 km.
(29) A boat takes 2 hours to go 40km down the stream and it returns in 4 hours. Find the speed of the boat in still water and the speed of the stream.
Solution:
Let, the speed of boat in still water be x:
The speed of steam be y.
(ATQ) (x + y) 2 = 40
Or, x + y = 20 – (i)
(x – y) 4 = 40
Or, x – y = 10 — (ii)
Adding (i) & (ii) we get,
x + y = 20
x + y = 10
__________
20 = 30
Or, x = 15
Substituting the value of x in equation (i) we get, 15+y = 20
Or, y = 5
∴ Speed of boat in still water is 15km/hr
Speed of steam = 5 km/hr
(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.
Solution:
Let, the speed of boat in still water be x.
Speed of stream be y.
(ATQ) (x + y) 4 = 44
Or, x + y = 11 —- (i)
(x – y) (4 + 40/60 + 4) = 44
Or, (x – y) (24/5 + 4) = 44
Or, (x – y) 44/5 = 44
Or, x – y = 5 —- (ii)
Adding equation (i) & (ii) we get,
x + y = 11
x – y = 5
__________
2x = 16
Or, x = 8
Substituting the value of x in equation (i) we get,
8 + 4 = 11
Or, y = 3
∴ Speed of boat in still water is 8km/hr.
Speed of stream is 3 km/hr.
(31) An aeroplane files 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.
Solution:
Let, the speed of plane be x.
Speed of wind be y.
(ATQ) (x – y) 3.5 = 1680
Or, (x – y) 35/10 = 1680
Or, x – y = 480 —- (i)
(x + y) 3 = 1680
Or, x + y = 560 — (ii)
Adding equation (i) & (ii) we get,
x – y = 480
x + y = 560
________________
2x = 1040
Or, x = 520
Substituting the value of x in equation (i) we get,
520 + y = 560
Or, x = 40
∴ Speed of plane in air 520 km/hr.
Speed of wind = 40 km/hr
(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 food per day.
Solution:
Let, the timed charges be x.
Cost of food per day be y.
(ATQ) x + 20y = 2600 —- (i) x + 26y = 3020 – (ii)
Subtracting equation (ii) from equation (i) we get,
x + 20y = 2600
x + 26y = 3020
(-) (-) (-)
______________
-6y = -420
Or, y = 70
Substituting the value of y in equation (i) we get,
x + 20 × 70 = 2600
Or, x = 2600 – 1400
Or, x = 1200
∴ The fixed charge is RS 1200
The cost of food per day is RS 70.