Algebra: Aptitude Questions with Answers
Algebra is a vast topic and you will benefit from reviewing the Algebra Concepts first:
Let’s go through Algebra questions now!
Q. 1 Amelia ran a lemonade stand for two weeks, from Sunday to Saturday, and earned $276 in total profit. On hot days, she charged 25% more for each cup of lemonade compared to regular days. The cost for making each cup of lemonade was 75 cents, and she sold 32 cups every day. During this period, there were 3 hot days. What was the selling price per cup on the hot days?
A) $ \$1.50 $
B) $ \$1.88 $
C) $ \$2.25 $
D) $ \$2.50 $
Check Solution
Ans: B
Amelia operated the stand Monday through Friday (5 days) over a two week period, thus she operated the stand for total of 10 days.
7 regular days –> sales = 7*32*x = 224x;
3 hot days –> sales = 3*32*(1.25x) = 120x;
Total sales = 224x+120x = 344x.
Total cost = 10*32*0.75 = 240.
Profit = 344x – 240 = 276 –> x=1.5.
1.25x=~1.88.
Q. 2 If $( x^2 + 6x + 9 = 0 )$, find the value of $( x )$.
A) -3
B) 3
C) 0
D) None
Check Solution
Ans: A) -3
The given equation can be rewritten as $( (x + 3)^2 = 0 )$.
Therefore, $( x = -3 )$.
Q. 3 If $( y = 2x + 3 )$ and $( x = 4 )$, what is the value of $( y^2 – x^2 )$?
A) 125
B) 100
C) 95
D) 85
Check Solution
Ans: C) 105
Substitute $( x = 4 )$ into the equation for $( y )$:
$[y = 2 \cdot 4 + 3 = 8 + 3 = 11]$
Now, $( y^2 – x^2 = 11^2 – 4^2 = 121 – 16 = 105 )$.
Q. 4 If $( x + y = 5 )$ and $( x^2 + y^2 = 13 )$, find the value of $( xy )$.
A) 6
B) 8
C) 10
D) 12
Check Solution
Ans: A) 6
We know that $( (x + y)^2 = x^2 + y^2 + 2xy )$.
Substitute values:
$[5^2 = 13 + 2xy]$
$[25 = 13 + 2xy]$
$[2xy = 12 \Rightarrow xy = 6]$
Q. 5 If $( a + b = 7 )$ and $( ab = 10 )$, what is the value of $( a^3 + b^3 )$?
A) 133
B) 157
C) 175
D) 191
Check Solution
Ans: A) 133
We know that $( a^3 + b^3 = (a + b)(a^2 – ab + b^2) )$.
$[a^2 + b^2 = (a + b)^2 – 2ab = 49 – 20 = 29]$
So, $( a^3 + b^3 = 7(29 – 10) = 7 \cdot 19 = 133 )$.
Q. 6 Solve for $( x ): ( 2^{x+2} = 64 )$.
A) 3
B) 4
C) 5
D) 6
Check Solution
Ans: B) 4
Rewrite ( 64 ) as a power of ( 2 ):
$[64 = 2^6]$
So, $( 2^{x+2} = 2^6 )$.
Therefore, $( x + 2 = 6 )$ and $( x = 4 )$.
Q. 7 If $( (x – 1)(x + 1) = 12 )$, find the value of $( x^2 )$.
A) 11
B) 12
C) 13
D) 14
Check Solution
Ans: C) 13.
Expand the left side:
$[x^2 – 1 = 12]$
So, $( x^2 = 13 )$.
Q. 8 If $( x^2 + 4x + k = 0 )$ has real roots, what is the range of values for $( k )$?
A) $( k \leq 4 )$
B) $( k \geq 4 )$
C) $( k \leq -4 )$
D) $( k \geq -4 )$
Check Solution
Ans: A) $( k \leq 4 )$
For real roots, the discriminant $( b^2 – 4ac \geq 0 )$:
Here, $( a = 1, b = 4, c = k )$.
$[4^2 – 4 \cdot 1 \cdot k \geq 0]$
$[16 – 4k \geq 0]$
$[k \leq 4]$
Q. 9 If $( 3x + y = 7 )$ and $( x – y = 5 )$, find the value of $( x + y )$.
A) 6
B) 7
C) 8
D) 9
Check Solution
Ans: C) 8
Solve the second equation for ( y ):
$[y = x – 5]$
Substitute into the first equation:
$[3x + (x – 5) = 7]$
$[4x – 5 = 7 \Rightarrow 4x = 12 \Rightarrow x = 3]$
Substitute $( x = 3 )$ back into $( y = x – 5 )$:
$[y = -2]$
Q. 10 If $( x = 2 )$ is a root of $( x^2 + px + 4 = 0 )$, what is the value of $( p )$?
A) -6
B) 4
C) 2
D) -4
Check Solution
Ans: D) -4
Substitute $( x = 2 )$ into the equation:
$[2^2 + 2p + 4 = 0]$
$[4 + 2p + 4 = 0 \Rightarrow 2p = -8 \Rightarrow p = -4]$
Q. 11 Solve for x, given the equation: 13x – 10 = 2x + 9
A) 1.7
B) 0.7
C) 2.7
D) 3.2
Check Solution
Ans: A
We can move variables on one side and numbers on other side.
13x – 2x = 10 + 9
11x = 19
Q. 12 Solve for x, given the equation: 13x – 12 = 6x + 7
A) 1.7
B) 2.7
C) 3.7
D) 4.2
Check Solution
Ans: B
We can move variables on one side and numbers on other side.
13x – 6x = 12 + 7
7x = 19
Q. 13 Solve for x, given the set of equations: 2x + 3y = 2; and 8x + 8y = 13
A) 1.9
B) 2.9
C) 3.9
D) 4.4
Check Solution
Ans: B
We should aim to make the coefficient of y same. For that we need to take LCM of coefficient of y in both the equations. LCM = 24
Now multiply equation 1 with 8 and equation 2 with 3 to get the equations as: 16x + 24y = 16 and 24x + 24y = 39. Subtract the equations to solve for x now.
Q. 14 Alex has a 10-year age advantage over Bernad. Five years earlier, Alex’s age was double that of Bernad’s. What is Bernad’s current age?
A) 5
B) 10
C) 12
D) 15
Check Solution
Ans: D
Let’s assume Bernad’s present age to be x. Then, Alex’s present age would be x + 10.
Five years ago, Bernad’s age would be x – 5, and Alex’s age would be (x + 10) – 5 = x + 5.
Given that 5 years ago, Alex was twice as old as Bernad. So, we can write:
x + 5 = 2(x – 5)
Solving for x:
x + 5 = 2x – 10
x = 15
Therefore, Bernad’s present age is 15 years
Q. 15 Marbles need to be distributed into five separate holes in a wall, ensuring that each hole has at least one marble and that the number of marbles in each hole is not less than the hole immediately to its right. If the third hole from the left must have 3 marbles, what is the minimum number of marbles necessary to meet all conditions?
A) 5
B) 9
C) 10
D) 11
Check Solution
Ans: D
since the middle slot is 3… the next two slots shud be 3+3 =6
2 more slots lefts..they can contain 1 each
Q. 16 Three years ago, the age of Dog X was four times that of Dog Y. Four years from now, the age of Dog X will be three times the age of Dog Y. What is the current age difference between Dog X and Dog Y?
A) 36
B) 42
C) 54
D) 60
Check Solution
Ans: B
A – 3 = 4(B – 3) –> A – 4B = -9………….1
A + 4 = 3(B + 4) –> A – 3B = 8………….2
(2) – (1) –> B = 17
A – B = 59 – 17 = 42
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