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Algebra: Linear Equation Test-3
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Question 1 |
There were 35 students in a hostel. If the number of students increases by 7, the expenses of the mess increase by Rs.42 per day while the average expenditure per head diminishes by 1. Find the original expenditure of the mess.
Rs.480 | |
Rs.520 | |
Rs.420 | |
Rs.460 |
Question 1 Explanation:
Suppose the original average expenditure be Rs. x .
Then acc. to question
42(x-1)-35x=42
Then x=12,
Original expenditure would be 12 x 35=420
Then acc. to question
42(x-1)-35x=42
Then x=12,
Original expenditure would be 12 x 35=420
Question 2 |
In a family, each daughter has the same number of brothers as she has sisters and each son has twice as many sisters as he has brothers. How many sons are there in the family?
4 | |
3 | |
2 | |
5 |
Question 2 Explanation:
Suppose there are d girls & b boys
Then from the given statement
(d - 1)=b and
2(b-1) =d
Then b=3 & d=4
Therefore the number of boys is 3 and the number of girls = 4
Hence option b is the right answer
Then from the given statement
(d - 1)=b and
2(b-1) =d
Then b=3 & d=4
Therefore the number of boys is 3 and the number of girls = 4
Hence option b is the right answer
Question 3 |
If the ages of P and R are added to twice the age of 38. Q, the total becomes 59. If the ages of Q and R are added to thrice the age of P, the total becomes 68 and if the age of P is added to thrice the age of Q and thrice the age of R, the total becomes 108. What is the age of P?
17 yr | |
19 yr | |
15 yr | |
12 yr |
Question 3 Explanation:
According to question
P+R+2Q=59,
Q+R+3P=68…………………………………….ii
P+3Q+3R=108 ………………………………….iii
Multiply equation (ii) by 3
And subtract equation (iii) from it,
Then P=12 year.
P+R+2Q=59,
Q+R+3P=68…………………………………….ii
P+3Q+3R=108 ………………………………….iii
Multiply equation (ii) by 3
And subtract equation (iii) from it,
Then P=12 year.
Question 4 |
A father's age is two times the sum of the ages of his two children, but 10 yr hence his age will be equal to the sum of their ages. Then, the father's age is
30 yr | |
20 yr | |
25 yr | |
15 yr |
Question 4 Explanation:
$ \displaystyle \begin{array}{l}Therefore\,\,\,\,\left( {{s}_{1}}+{{s}_{2}} \right)=\frac{f}{2}\\and\,\,\,\,\left( {{s}_{1}}+{{s}_{2}} \right)+10+10=f+10\\\Rightarrow \,\,\frac{f}{2}+20=f+10\\\Rightarrow f=20\,yr\end{array}$
Question 5 |
A body of 7300 troops is formed of 4 battalions so that 1/2 of the first, 2/3 of the second, 3/4 of the third and 4/5 of the fourth are all composed of the same number of men. Find the same number.
2400 | |
1200 | |
1300 | |
1350 |
Question 5 Explanation:
Let 7300 troops divided in x,y,x & p.
Then x + y + z + p=7300,
According to question x/2=2y/3=3z/4=4p/z.
Put these values in equation then
x=2400,
x/2=1200
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