Exploring Algebraic Identities MCQs for UPSC Prelims
135 practice questions covering Exploring Algebraic Identities, organised into 2 topics.
135 questions include a written explanation.
Pick a topic below, or try the samples first.
Two square tiles have sides $n+4$ centimetres and $n-4$ centimetres. The larger tile's area is $128\ \text{cm}^2$ more than the smaller tile's area. What is the value of $n$?
A$8$
B$12$
C$16$
D$4$
Show answer and explanation
Correct answer: A - $8$
The difference of areas is $(n+4)^2-(n-4)^2=(2n)(8)=16n$, so $16n=128$ and $n=8$. Checking, the tiles are $12\ \text{cm}$ and $4\ \text{cm}$: $144-16=128$. Taking $n=16$ gives a difference of $256$, $n=12$ gives $192$, and $n=4$ makes the smaller tile vanish altogether.
Q2
hard
A trinomial is squared and the result is $9x^2+4y^2+16z^2-12xy-16yz+24zx$. Which trinomial was squared?
A$3x-2y+4z$
B$9x-4y+16z$
C$3x+2y+4z$
D$3x-2y-4z$
Show answer and explanation
Correct answer: A - $3x-2y+4z$
The square roots of the three square terms are $3x$, $2y$ and $4z$, and the signs must make $xy$ and $yz$ negative while $zx$ stays positive, which happens when only $2y$ is negative. At $x=y=z=1$ the expression is $25$, so the bracket must be $5$, and $3-2+4=5$. The other brackets give $9$, $-3$ and $21$, whose squares are $81$, $9$ and $441$.
Q3
easy
When $\left(\frac{3}{4}s+8t\right)^2$ is expanded using $(a+b)^2=a^2+2ab+b^2$, what is the middle term?
A$\frac{3}{2}st$
B$6st$
C$12st$
D$24st$
Show answer and explanation
Correct answer: C - $12st$
The middle term is $2ab$ with $a=\frac{3}{4}s$ and $b=8t$, so it is $2\times\frac{3}{4}\times 8\,st=12st$. Using $ab$ without doubling gives $6st$, doubling twice gives $24st$, and $\frac{3}{2}st$ comes from doubling only the fraction and forgetting the $8$. Checking with $s=4$ and $t=1$, the square is $121$ and $9+48+64=121$.
Q4
hard
A rectangle has area $x^2-11x+28$ square units, and one of its sides has length $x-4$ units. What is the length of the other side?
A$x-24$
B$x+7$
C$x-7$
D$x-4$
Show answer and explanation
Correct answer: C - $x-7$
Factorising, the constants must multiply to $28$ and add to $-11$, so $x^2-11x+28=(x-4)(x-7)$ and the other side is $x-7$. Checking at $x=10$: the area is $100-110+28=18$, one side is $6$, and $18\div 6=3=10-7$. A side of $x+7$ would give $102$ at $x=10$, and $x-4$ or $x-24$ do not pair with $x-4$ to give the right constant term.
Q5
medium
Assertion (A): Shridharacharya's method gives $45^2=50\times 40+5=2005$. Reason (R): for any numbers $a$ and $b$, $a^2=(a+b)(a-b)+b^2$.
ABoth A and R are true, and R is the correct explanation of A
BBoth A and R are true, but R is not the correct explanation of A
CA is true, but R is false
DA is false, but R is true
Show answer and explanation
Correct answer: D - A is false, but R is true
(R) is a correct rearrangement of the difference of squares. (A) applies it with the wrong correction: the amount added must be $b^2=5^2=25$, not $5$, so the method gives $2000+25=2025$, and $45\times 45$ is indeed $2025$, not $2005$. So the reason stands while the assertion does not.
Q6
easy
Simplify $\frac{x^2-9}{x+3}$, where $x\neq -3$.
A$x-3$
B$x+3$
C$x^2-3$
D$\frac{x-3}{x+3}$
Show answer and explanation
Correct answer: A - $x-3$
The numerator is a difference of squares, so $x^2-9=(x+3)(x-3)$ and the common factor $x+3$ cancels, leaving $x-3$. At $x=2$ the original is $\frac{4-9}{5}=-1$, and $2-3=-1$. The other answers give $5$, $1$ and $-\frac{1}{5}$ at $x=2$, because they cancel the wrong factor or cancel only part of it.
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