Practice Test - 2
Class- 8 Subject - Maths Marks : 40 marks
Chapters - Direct and Inverse variations and Linear Equations in one variables.
Section A each carries 3 marks
Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8.
Answer: Solution:
Let one of the numbers be ‘x’
The other number is 184 – x
So, One-third of one part may exceed one-seventh of another part by 8.
x/3 – (184-x)/7 = 8
LCM for 3 and 7 is 21
(7x – 552 + 3x)/21 = 8
By cross-multiplying, we get,
(7x – 552 + 3x) = 8(21)
10x – 552 = 168
10x = 168 + 552
10x = 720
x = 720/10
= 72
∴ One of the numbers is 72, and the other number is 184 – x => 184 – 72 = 112.
A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50. Find the number of notes of each type.
Answer: Solution:
Let the number of 10Rs notes be x
Number of 20Rs notes be 50 – x
Amount due to 10Rs notes = 10 × x = 10x
Amount due to 20Rs notes = 20 × (50 – x) = 1000 – 20x
So the total amount is Rs 800
10x + 1000 – 20x = 800
-10x = 800 – 1000
-10x = -200
x = -200/-10
= 20
∴ The number of 10Rs notes is 20
Number of 20Rs notes are 50 – 20 = 30
Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.
Solution:
Let the age of the son five years ago be x years
The age of man five years ago be 7x years
After five years, the son’s age is x + 5 years
After five years father’s age is 7x + 5 years
So, since five years, the relation in their ages are
7x + 5 + 5 = 3(x + 5 + 5)
7x + 10 = 3x + 15 + 15
7x + 10 = 3x + 30
7x – 3x = 30 – 10
4x = 20
x = 5
∴ Present father’s age is 7x + 5 = 7(5) + 5 = 35 + 5 = 40years
Present son’s age is x + 5 = 5 + 5 = 10years
There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question, one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?
Solution:
Let the number of correct answers be x
The number of questions answered wrong is (180 – x)
Total score when answered right = 4x
Marks deducted when answered wrong = 1(180 – x) = 180 – x
So,
4x – (180 – x) = 450
4x – 180 + x = 450
5x = 450 + 180
5x = 630
x = 630/5
= 126
∴ 126 questions he answered correctly.
A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day he works, and he will be fined Rs 5 for each day he is absent. If he receives Rs 745 in all, how many days he remained absent?
Solution:
Let us consider the number of absent days as x
So, the number of present days is (20 – x)
The wage for one day of work = Rs 60
Fine for absent day = Rs 5
So,
60(20 – x) – 5x = 745
1200 – 60x – 5x = 744
-65x = 744-1200
-65x = -456
x = -456/-65
= 7
∴ For 7 days, the labourer was absent.
Mini inherited Rs 12000.00. She invested part of it at 10% and the rest at 12%. Her annual income from these investments is Rs 1280.00 How much did she invest at each rate?
Answer: Solution:
At a rate of 10%, let the investment be Rs x
At the rate of 12%, the investment will be Rs (12000 – x)
At 10% of rate the annual income will be x × (10/100) = 10x/100
At 12% of rate, the annual income will be (12000 – x) × 12/100 = (144000 – 12x)/100
Total investment = 1280
So, 10x/100 + (144000 – 12x)/100 = 1280
(10x + 144000 – 12x)/100 = 1280
(144000 – 2x)/100 = 1280
By cross-multiplying, we get,
144000 – 2x = 1280(100)
-2x = 128000 – 144000
-2x = -16000
x = -16000/-2
= 8000
∴ At 10% of rate, she invested Rs 8000, and at 12% of the rate she invested Rs (12000 – x) = Rs (12000 – 8000) = Rs 4000
The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 84 cm2 more than that of the given rectangle. Find the length and breadth of the given rectangle.
Solution:
Let the breadth of the rectangle be x meter
Length of the rectangle be (x + 9) meter
Area of the rectangle length×breadth = x(x +9) m2
When length and breadth increased by 3cm, then,
New length = x + 9 + 3 = x + 12
New breadth = x + 3
So, the area is
(x + 12) (x + 3) = x (x + 9) + 84
x2 + 15x + 36 = x2 + 9x + 84
15x – 9x = 84 – 36
6x = 48
x = 48/6
= 8
∴ The length of the rectangle (x + 9) = (8 + 9) = 17cm, and the breadth of the rectangle is 8cm.
The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to 2/3. What is the original fraction equal to?
Solution:
Let us consider the denominator as x and numerator as (x-6)
By using the formula,
Fraction = numerator/denominator = (x-6)/x
(x – 6 + 3)/x = 2/3
(x – 3)/x = 2/3
By cross-multiplying
3(x-3) = 2x
3x – 9 = 2x
3x – 2x = 9
x = 9
∴ The denominator is x = 9, numerator is (x-6) = (9-6) = 3
And the fraction = numerator/denominator = (x-6)/x = 3/9 = 1/3
Mani has enough money to buy 75 machines worth Rs. 200 each. How many machines can he buy if he gets a discount of Rs. 50 on each machine?
Solution:
Let number of machines be ‘x’ machines
When cost of one machine decreases = 200 – 50 = 150Rs
We know k = xy
200 × 75 = 150 × x
x = (200×75)/150
= 100
∴ Mani can buy 100 machines when he gets a discount of Rs 50 on each machine.
1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted now for 32 more days. How many soldiers left the fort?
Solution:
Total soldiers = 1200
Let us consider the number of soldiers transferred to another fort be ‘x’ soldiers
After four days number of soldiers left the fort = 1200 – x
And for remaining soldiers (1200 – x) the food lasted for 32 more days
We know k = xy
24 × 1200 = 32 × (1200-x)
28800 = 38400 – 32x
32x = 38400 – 28800
32x = 9600
x = 9600/32
= 300
∴ 300 soldiers left for another fort.
A group of 3 friends staying together, consume 54 kg of wheat every month. Some more friends join this group and they find that the same amount of wheat lasts for 18 days. How many new members are there in this group now?
Solution:
Total friends = 3
Let us consider number of friends joined in a group be ‘x’ friends
We know k = xy
3 × 30 = x × 18
x = (3×30)/18
= 5
Joined friends = 5 – 3 = 2
∴ 2 new members joined the group.
A car can finish a certain journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance?
Solution:
Let us consider the speed of the car as ‘x’ km/hr.
We know k = xy
10 × 48 = 8 × x
x = (10×48)/8
= 60
Increased speed = 60 – 48 = 12km/hr.
∴ The speed should be increased by 12 km/hr. to cover the same distance.
In 10 days, the earth picks up 2.6 × 108 pounds of dust from the atmosphere. How much dust will it pick up in 45 days?
Solution:
Let the dust picked up from atmosphere be ‘x’ pounds
10/2.6 × 108 = 45/x
By cross-multiplying
10x = 45 × 2.6 × 108
x = (45 × 2.6 × 108)/10
= 11.7 × 108
∴ The dust picked up from the atmosphere in 45 days is 11.7 × 108 pounds.
A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 12 minutes?
Solution:
Let’s consider distance travelled be ‘x’ meter
We know that one hour = 60minutes
50/60 = x/12
By cross-multiplying
50(12) = 60x
600 = 60x
x = 600/60
= 10
∴ The distance travelled in 12min is 10km
The second class railway fare for 261 km of Journey is Rs 15.00. What would be the fare for a Journey of 139.2 km?
Solution:
Let us consider the fare as Rs ‘x’
261/15 = 139.2/x
By cross-multiplying
261x = 139.2(15)
261x = 2088
x = 2088/261
= 8
∴ The fare for the journey of 139.2 km is Rs 8
No comments:
Post a Comment