Practice Test
Class- 8 Subject - Maths Marks : 45 marks
Chapters - Time and Work, Percentage, Profit and Loss,Algebraic Expressions, Direct and Inverse variations and Linear Equations in one variable.
Section A each carries 3 marks
A tricycle is sold at a gain of 16%. Had it been sold for Rs 100 more, the gain would have been 20%. Find the C.P. of the tricycle.
Solution: the cost price of tricycle be = Rs x
Selling price of the tricycle be = Rs x
Gain% = 16% of 100 = 16/100
Selling price of tricycle = x + x×16/100
= (100x+16x)/100
= 116x/100
= 29x/25
if selling price is Rs 100 more
New Selling price = 29x/25 + 100
Gain% = 20%
Gain % = (gain/cost price) × 100 [by using Gain = SP – CP]
20 = [((29x/25)+100) – x] / x × 100
20x/100 = (29x + 2500 – 25x)/25
x/5 = (29x + 2500 – 25x)/25
5x = 4x + 2500
x = 2500
∴ Cost price of tricycle is Rs 2500
Ramesh bought two boxes for Rs 1300. He sold one box at a profit of 20% and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.
Solution :
C.P of two boxes = Rs 1300
let C.P of one box be Rs x
C.P of other box = Rs 1300 – x
S.P of first box = x + x × ${20}/{100}$ = x + ${x}/{5}$ = Rs 6x/5
S.P of second box = (1300 – x) – (1300 – x) × 12/100
= Rs (28600 – 22x)/25
6x/5 = (28600 – 22x)/25
150x = 28600 × 5 – 110x
150x + 110x = 28600 × 5
260x = 28600 × 5
x = (28600 × 5)/260 = 550
∴ C.P of first box = Rs. 550
C.P of second box = Rs 1300 – Rs 550 = Rs 750
3.Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food intake.
Solution:
Amount of calorie daily needed = 2600 calorie
Amount of protein needed = 12% of 2600
= (12/100) × 2600
= 312 calorie
Amount of fats needed = 25% of 2600
= (25/100) × 2600
= 650 calorie
Amount of carbohydrate needed = 63% of 2600
= (63/100) × 2600
= 1638 calorie
4. A motorist travelled 122 kilometers before his first stop. If he had 10%of his journey to complete at this point, how long was the total ride?
Solution:
details are,
Motorist total distance travelled before first stop = 122 km
Journey completed at first stop = 10 %
total ride to be travelled be ‘x’ km
(x/100) × 10 = 122
x/100 = 122/10
x = (122 × 100)/10
= 1220 km
∴ Motorist total ride is 1220 km
5.A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work?
Solution:
A can do a piece of work in 20 hours
Work done by A in 1 hour = 1/20
B can do same work in = 24 hours
Work done by B in 1 hour = 1/24
A, B and C working together can do the same work in = 8 hours
Work done by A, B, C together in 1 hour = 1/8
Work done by C in 1 hour = (work done by A,B and C in 1 hour) – ( work done by A And B in 1 hr.)
= 1/8 – (1/20 + 1/24)
= 1/8 – 11/120
= (15-11)/120 (by taking LCM for 8 and 120 which is 120)
= 4/120
= 1/30
∴ Time taken by C alone to complete the work = 1/(1/30) = 30hours.
6.A and B can polish the floors of a building in 10 days. A alone can do ¼th of it in 12 days. In how many days can B alone polish the floor?
Solution:
A and B can polish a building in = 10 days
Work done by A and B in one day = 1/10
A alone can do 1/4th of work in = 12 days
A’s 1 day work = 1/(4×12) = 1/48
B’s 1 day work = (A+B)’s 1 day work – A’s 1 day work
= 1/10 – 1/48
= (48-10)/480 (by taking LCM for 10 and 48 which is 480)
= 38/480 = 19/240
∴ B alone can polish the floor in = 1/(19/240) = 240/19 =12 12/19 days
7. A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely?
Solution:
Inlet A can fill the cistern in = 12 hours
Inlet A can fill the cistern in 1 hour = 1/12
Inlet B can fill the cistern in = 15 hours
Inlet B can fill the cistern in 1 hour = 1/15
Outlet pipe can empty the cistern in = 10 hours
Outlet pipe can empty the cistern in 1 hour = 1/10
(1/12 + 1/15) – 1/10
= (9/60) – 1/10
= (9-6)/60
= 3/60
= 1/20 part
∴ When all 3 pipes are opened together to empty the cistern, time taken to fill the cistern completly = 1/(1/20) = 20 hours
8.The amount of extension in an elastic string varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm?
Solution:
Extension (cm) 2.9 17.4
Weight (gm) 150 x
2.9/150 = 17.4/x
2.9x = 150(17.4)
2.9x = 2610
x = 2610/2.9
= 900
∴ 900gm of weight is produced for an extension of 17.4cm
9.In a hostel of 50 girls, there are food provisions for 40 days. If 30 more girls join the hostel, how long will these provisions last?
Solution:
Total number of girls are 50 + 30 = 80 girls
Girls 50 80
Time (days) 40 x
We know k = xy
50 × 40 = 80 × x
x = (50×40)/80 = 25
∴ The food provisions lasts for 25 days if 30 more girls are joined.
10. Three spraying machines working together can finish painting a house in 60 minutes. How long will it take for 5 machines of the same capacity to do the same job?
Solution:
Time (min) 60 x
No. of spraying machines 3 5
60 × 3 = x × 5
x = (60×3)/5
= 36
∴ 36 minutes is required to do the same job by 5 spraying machines of the same capacity.
11. Find the following products and verify the results for x = -1, y = -2: (3x – 5y) (x + y)
Solution:
(3x – 5y) × (x + y)
(3x – 5y) × (x + y)
x (3x – 5y) + y (3x – 5y)
3x2 – 5xy + 3xy – 5y2
3x2 – 2xy – 5y2
3x2 – 2xy – 5y2
3 (-1)2 – 2 (-1) (-2) – 5 (-2)2
3 – 4 – 20 = – 21
∴ the expression is verified.
12. Simplify: x² (x + 2y) (x – 3y)
Solution:
x2 (x + 2y) (x – 3y)
x2 (x2 – 3xy + 2xy – 3y2)
x2 (x2 – xy – 6y2)
x4 – x3y – 6x2y2
13. The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 1/2. Find the rational number.
Solution:
Le the denominator be x and the numerator be (x – 3)
= (x – 3)/x
(x – 3 + 2)/(x + 5) = 1/2
(x – 1)/(x + 5) = 1/2
2(x – 1) = x + 5
2x – 2 = x + 5
2x – x = 2 + 5
x = 7
∴ Denominator is x = 7, numerator is (x – 3) = 7 – 3 = 4
And the fraction = numerator/denominator = 4/7
14. I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?
Solution:
Let the number of five rupee notes be x
The number of ten rupee notes be (x + 10)
Amount due to five rupee notes = 5 × x = 5x
Amount due to ten rupee notes = 10 (x + 10) = 10x + 100
The total amount = Rs 1000
5x + 10x +100 = 1000
15x = 900
x = 900/15
= 60
∴ the number of five rupee notes is x = 60
The number of ten rupee notes is x + 10 = 60+10 = 70
15. I am currently 5 times as old as my son. In 6 years time, I will be three times as old as he will be then. What are our ages now?
Solution:
Let the present son’s age be x years
Present father’s age be 5x years
Son’s age after 6 years = (x + 6) years
Fathers’ age after 6 years = (5x + 6) years
5x + 6 = 3(x + 6)
5x + 6 = 3x + 18
5x – 3x = 18 – 6
2x = 12
x = 12/2
= 6
∴ present son’s age is x = 6years
Present father’s age is 5x = 5(6) = 30years
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