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Algebra: Basics Test-2

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Question 1
If $ \displaystyle \frac{2x+y}{x+4y}=3$,then find the value of $ \displaystyle \frac{p+q}{p+2q}$
A
$ \displaystyle \frac{15}{9}$
B
$ \displaystyle \frac{12}{9}$
C
$ \displaystyle \frac{10}{9}$
D
$ \displaystyle \frac{8}{9}$
Question 1 Explanation: 
$ \displaystyle \begin{array}{l}\frac{2p+q}{p+4q}=3\,\,\\\Rightarrow 2p+q=3p+12q\\\Rightarrow 3p-2p=q-12q\\\Rightarrow p=-11q\\Then,\,\,\frac{p+q}{p+2q}=\frac{-11q+q}{-11q+2q}\\=\frac{-10q}{-9q}=\frac{10}{9}\end{array}$
Question 2
If a =4, 965 b =2.343 and c = 2.622, then the value of a3−b3−c3-3xyz is equal to
A
0
B
−3
C
1
D
1.93
Question 2 Explanation: 
We are given by the values of x ,y and z i.e.
a =4, 965 b=2.343 and c = 2.622
If a + b + c = 0
Then a3+ b3+ c = 3xyz
$ \displaystyle \begin{array}{l}\therefore \,When\,\,a-b-c=0,\\{{a}^{3}}-{{b}^{3}}-{{c}^{3}}=3abc\\i.e.,\,{{a}^{3}}-{{b}^{3}}-{{c}^{3}}-3abc=0\end{array}$
$ \displaystyle \begin{array}{l}Here,\\a=4.965,b=2.343,c=2.6222\\\therefore a-b-c=4.965-2.343-2.622=0\\Hence,\,{{a}^{3}}-{{b}^{3}}-{{c}^{3}}-3abc=0\end{array}$
Question 3
p x q= p+ q+ pq, then 3×4 −2×3 is equal to:
A
6
B
4
C
8
D
9
Question 3 Explanation: 
We have
$ \displaystyle \begin{array}{l}p\times q=p+q+pq\\\therefore 3\times 4-2\times 3\\=\left( 3+4+3\times 4 \right)-\left( 2+3+2\times 3 \right)\\=\left( 7+12 \right)-\left( 5+6 \right)=19-11=8\end{array}$
Question 4
If p= 1.21, q= 2.12 and r = −3.33 then, the value of p3+ q3 +r3 -3pqr is
A
1
B
2
C
3
D
0
Question 4 Explanation: 
We have p= 1.21, q= 2.12 and r = −3.33
$ \displaystyle \begin{array}{l}p+q+r\,\,=1.21+2.12-3.33=0\\{{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr=0\end{array}$
Question 5
If l × m= 3l +2m, then 2 ×3 + 3×4 is equal to:
A
29
B
27
C
39
D
37
Question 5 Explanation: 
We are given by the information
l × m = 3l +2m
$ \displaystyle \begin{array}{l}\because l\times m=3l+2m\\2\times 3+3\times 4\\=3\times 2+2\times 3+3\times 3+2\times 4\\=6+6+9+8=29\end{array}$
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