- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.
Algebra Level 1 Test 3
Congratulations - you have completed Algebra Level 1 Test 3.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
Your answers are highlighted below.
Question 1 |
A boy was asked of his age by his friend. The boy said, "The number you get when you subtract 25 times my age from twice the square of my age will be thrice your age." If the friend's age is 14, then the age of the boy is
28 yr | |
21 yr | |
14 yr | |
25 yr |
Question 1 Explanation:
Let the age of boy be a yr.
2a2 – 25a = 3 x 14
2a2 – 25a - 42 = 0
2a(a - 14) +3(a – 14)= 0
2a+3 = 0
a – 14 = 0
a = 14
so the right option for the question is option c
2a2 – 25a = 3 x 14
2a2 – 25a - 42 = 0
2a(a - 14) +3(a – 14)= 0
2a+3 = 0
a – 14 = 0
a = 14
so the right option for the question is option c
Question 2 |
The product of the present ages of Sarita and Gauri is 320. Eight years from now, Sarita's age will be three times the age of Gauri. What was the age of Sarita when Gauri was born?
40 yr | |
32 yr | |
48 yr | |
36 yr |
Question 2 Explanation:
Let the present ages of Sarita and Gauri be a and b.
Then according to the question
ab = 320
(a + 8) = 3(b + 8)
a – 3b = 16
a – 3(320/a) = 16
a2 – 16a - 960 = 0
by solving the equation
i.e. a = 40 and b = 8
So Sarita was 32 at the time of Gauri born.
Then according to the question
ab = 320
(a + 8) = 3(b + 8)
a – 3b = 16
a – 3(320/a) = 16
a2 – 16a - 960 = 0
by solving the equation
i.e. a = 40 and b = 8
So Sarita was 32 at the time of Gauri born.
Question 3 |
In a class, the number of girls is one less than the number of the boys. If the product of the number of boys and that of girls is 272, then the number of girls in the class is
15 | |
14 | |
16 | |
17 |
Question 3 Explanation:
Let the number of girls and boys be a and b.
Then according to the question
a – 1 = b
ab = 272
Then,
a (a - 1) = 272
a2 - a - 272 = 0
by solving this
we get a = 16
Then according to the question
a – 1 = b
ab = 272
Then,
a (a - 1) = 272
a2 - a - 272 = 0
by solving this
we get a = 16
Question 4 |
Children were fallen-in for a drill. If each row contained 4 children less, 10 more rows would have been made. But if 5 more children were accommodated in each row the number of rows would have reduced by S. The number of children in the school is
200 | |
150 | |
300 | |
100 |
Question 4 Explanation:
Let the number of rows and columns be a and b.
(b - 4)(a + 10)= ab
10b – 4a = 40
And (b + 5)(a - 5) = ab
5a – 5b = 25
a – b = 5
a = 15
b = 10
Therefore the number of students = 15 x 10 = 150
(b - 4)(a + 10)= ab
10b – 4a = 40
And (b + 5)(a - 5) = ab
5a – 5b = 25
a – b = 5
a = 15
b = 10
Therefore the number of students = 15 x 10 = 150
Question 5 |
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Sanchit paid Rs. 45 for a book kept for 7 days, while Karan paid Rs. 25 for the book he kept for 5 days. The fixed charge and the charge for each extra day is
Rs. 5 and Rs. 10 | |
Rs. 10 and Rs. 10 | |
Rs. 15 and Rs. 5 | |
Rs. 5 and Rs. 15 |
Question 5 Explanation:
Let a be the fix charges for first three days
And b be the charges of extra days
a+ 4b=45………………………………………………1
a +2b=25……………………………………………….2
so from 1 and 2
a = Rs. 5
b = Rs. 10
And b be the charges of extra days
a+ 4b=45………………………………………………1
a +2b=25……………………………………………….2
so from 1 and 2
a = Rs. 5
b = Rs. 10
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 5 questions to complete.
List |
