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## Algebra: Linear Equation Test-4

Congratulations - you have completed Algebra: Linear Equation Test-4. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
 Question 1
1 yr ago, a mother was 4 times older to her son. After 6 yr. her age becomes more than double her son's age by 5 yr. The present ratio of their age will be
 A 13 : 12 B 3 : 1 C 11 : 3 D 25 : 7
Question 1 Explanation:
Let present age of the mother is m & age of her son is s year.
According to question:
(m -1)=4(s-1)
m = 4s-3……………………………………a
& (m+6)=2(s+6)+5
m = 2s + 11………………………………..b
From equation a and b
4s – 3 = 2s +11
s = 7
and m = 25 years
 Question 2
Students of a class are made to stand in rows. If 4 students are extra in each row, there would be 2rows less. If 4 students are less in each row, there would be 4 more rows. The number of students in the class is
 A 90 B 94 C 92 D 96
Question 2 Explanation:
Suppose there are x rows & y students in each row
Then according to question
(x-2)(y+4)=xy &
(x+4)(y-4)=xy ,
Solving these equations
We get x=8 & y=12 then total students=96
 Question 3
$\displaystyle If\,\,\frac{x}{\left( 2x+y+z \right)}=\frac{y}{\left( x+2y+z \right)}=\frac{z}{\left( x+y+2z \right)}=a,\,\,\,then\,\,\,find\,\,\,a,\,\,if\,\,\,x+y+z\ne 0$
 A $\displaystyle \frac{1}{3}$ B $\displaystyle \frac{1}{4}$ C $\displaystyle \frac{1}{8}$ D $l \displaystyle \frac{1}{2}$
Question 3 Explanation:
Put x=y=z=1 then a=1/4
 Question 4
One-fourth of a herd of cows is in the forest. Twice the square root of the herd has gone to mountains and the remaining 15 are on the banks of a river. The total number of cows is
 A 6 B 100 C 63 D 36
Question 4 Explanation:
Let total no of cows is x,
then according to question
x-(x/4+2 √x)=15
then 3x-8√x+60=0
Now if we look at the options then we see that there are two answer
options which are perfect square and for the value of √x we need
a perfect square so if the value of x is 100 then the equation will become
3 x 100 – 8 x 10 + 60 = 300 – 80 + 60
which is not equal to zero then the value of x is 36.
 Question 5
$\displaystyle The\,\,\,\,value\,\,\,of\,\,\,\left( \frac{1}{{{x}^{2}}} \right)+\left( \frac{1}{{{y}^{2}}} \right),\,\,\,where\,\,x=2+\sqrt{3}\,\,\,\,and\,\,\,y=2-\sqrt{3},\,\,\,is$
 A 14 B 12 C 10 D 16
Question 5 Explanation:
$\displaystyle \begin{array}{l}\frac{1}{{{x}^{2}}}+\frac{1}{{{x}^{2}}}=\frac{1}{{{\left( 2x+\sqrt{3} \right)}^{2}}}+\frac{1}{{{\left( 2-\sqrt{3} \right)}^{2}}}\\=\frac{1}{\left( 7+4\sqrt{3} \right)}+\frac{1}{\left( 7-4\sqrt{3} \right)}\\=\frac{14}{49-16\times 3}=14\end{array}$
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