Q1. Solve for x.
3x – 8 = 2x – 9
(a) 12
(b) –2
(c) –1
(d) 1
Q2. Two positive numbers x and y are inversely proportional. If x increases by 10%, then y decreases by
(a) 10%
(b) 2/10%
(c) 100/11%
(d) 10/11%
Q3. Solve for x:
x/2+2/3=1/3
(a) 2/3
(b) –2/3
(c) 1
(d) –4/3
Q4. Solve for x :
x/21+1/7=1/3
(a) 5
(b) – 5
(c) 4
(d) – 4
Q5. Solve for x :
3x/2–2=x/2+3
(a) 5
(b) – 5
(c) –7.5
(d) – 4
Q6. If x is an integer, then (x + 1)⁴ – (x – 1)⁴ is always divisible by
(a) 6
(b) 8
(c) 9
(d) 12
Q7. The expression x² – y² + x + y – z² + 2yz – z has one factor, which is
(a) y – x + z
(b) x – y + z + 1
(c) x + y – z + 1
(d) x – y – z + 1
Q8. A factor common to x² + 7x + 10 and x² – 3x – 10 is
(a) x – 5
(b) x + 5
(c) x + 2
(d) x – 2
Q9. The sum of three consecutive integers is 99. Find the smallest integer.
(a) 33
(b) 34
(c) 32
(d) 35
Q10. The sum of two positive numbers is 63. If one number x is double the other, then the equation is
(a) x/(x–63)=2
(b) (63–x)/x=2
(c) (x–63)/x=2
(d) x/(63–x)=2
Solutions
S1. Ans.(c)
Sol.
3x – 8 = 2x – 9
⇒ 3x – 2x = – 9 + 8
⇒ x = –1
S2. Ans.(c)
Sol.
As x is inversely proportional to y, x = k/y ⇒ y = k/x
Let X and Y be the new values of x and y respectively.
∴ X = x + 10% of x
= x+10/100 of x
= x+1/10 x = 11/10 x
Now, Y = k/X = k/(11/10 x)
= 10/11 (k/x) = 10/11 y
Decrease in y = y –10/11 y = 1/11 y
∴ Percentage decrease = (1/11 y)/y×100 = 100/11%
S3. Ans.(b)
Sol.
x/2+2/3=1/3
⇒ x/2 = 1/3–2/3
⇒ x/2 = –1/3
⇒ x = –2/3
S4. Ans.(c)
Sol.
x/21+1/7 = 1/3
⇒ x/21 = 1/3–1/7
⇒ x/21 = (7–3)/21
⇒ x/21 = 4/21
⇒ x = 4
S5. Ans.(a)
Sol.
3x/2–x = x/2+3
⇒ 3x/2–x/2 = 3 + 2
⇒ 2x/2 = 5
⇒ x = 5
S6. Ans.(b)
Sol.
(x+1)^4–(x –1)^4
= [(x+1)^2 ]^2–[(x–1)^2 ]^2
= [(x+1)^2+(x–1)^2 ][(x+1)^2–(x–1)^2 ]
= [2 (x^2+1)][4×x×1]
= 8x(x^2+1)
Therefore, the given expression is always divisible by 8.
S7. Ans.(b)
Sol.
The given expression is
x^2–y^2+x+y –z^2+2yz –z
= x^2–(y^2+z^2 –2yz)+x+y –z
= x^2–(y–z)^2+x+y –z
= (x+y–z)(x–y+z)+(x+y –z)
= (x+y –z)(x –y+z+1)
∴ x – y +z + 1 is a factor of the given expression.
S8. Ans.(c)
Sol.
x² + 7x + 10 = x² + 5x + 2x + 10
= x (x + 5) + 2 (x + 5)
= (x + 5) (x + 2)
x² – 3x – 10 = x² – 5x + 2x – 10
= x (x – 5) + 2 (x – 5)
= (x – 5) (x + 2)
Hence, the common factor x + 2.
S9. Ans.(c)
Sol.
Let the smallest integer be x.
∴ Two other integers are
x + 1 and x + 2
x + x + 1 + x + 2 = 99
⇒ 3x + 3 = 99
⇒ 3x = 96
⇒ x = 32
Three integers are 32, 33 and 34.
Hence, the smallest integer = 32.
S10. Ans.(d)
Sol.
One number = x
∴ Another number = x/2
According to the question,
x+x/2 = 63
⇒ x/2 = 63 –x
⇒ x/(63–x) = 2
