Practice Algebraic expressions - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   Directions :
In each of the following questions two equations I and II are given. You have to solve both the equations and give answer
  1. if a < b
  2. if a ≤ b
  3. if a ≥ b
  4. if a = b
  5. if a > b
I. $a^2$ – 5a + 6 = 0
II. $b^2$ – 3b + 2 = 0

(a)

(b)

(c)

(d)

(e)

Explanation:

For eqn 1, the roots (a) will be 2, ×3. As – 2 – 3 = 6 (ac) and (– 2) + (– 3) = – 5

(b). Similarly, for eqn II, the roots (b) will be 2, 1.

Q-2)   Directions :
In each question below one or more equation(s) is /are given. On the basis of these, you have to find out the relationship between p and q. Give answer
  1. if p = q
  2. if p > q
  3. if p < q
  4. if p ≤ q
  5. if p ≥ q
I. 2p (p – 4) = 8 (p + 5)
II. $q^2$ + 12 + 7q

(a)

(b)

(c)

(d)

(e)

Explanation:

I. 2p (p – 4) = 8 (p +5)

or $p^2$ –8p–20 =0

or (p + 2) (p –10) = 0

⇒ p = –2, or + 10

II. $q^2$ + 7q + 12 =0

(q + 3) (q +4) =0

q = –3 or –4

p > q

Q-3)   Directions :
In each of the following questions two equations I and II are given. You have to solve both the equations and give answer
  1. if a < b
  2. if a ≤ b
  3. if a ≥ b
  4. if a = b
  5. if a > b
I. $2a^2$ + 5a + 3 = 0
II. $2b^2$ – 5b + 3 = 0

(a)

(b)

(c)

(d)

(e)

Explanation:

a = –3/2 & –1; b = $3/2$ & 1

Q-4)   Directions :
In each of the following questions two equations I and II are given. You have to solve both the equations and give answer
  1. if a < b
  2. if a ≤ b
  3. if a ≥ b
  4. if a = b
  5. if a > b
I. 2a + 3b = 31
II. 3a = 2b + 1

(a)

(b)

(c)

(d)

(e)

Explanation:

2a + 3b = 31 ... (i)

3a – 2b = 1 ..(ii)

Multiply (i) by 2 and (ii) by 3 and then adding

(i) and (ii), we get a = $65/13$ = 5 . Putting the value of 'a'

in any equation, we get b = 7.

Hence, b > a or a < b.

Q-5)   Directions :
In each question below one or more equation(s) is /are given. On the basis of these, you have to find out the relationship between p and q. Give answer
  1. if p = q
  2. if p > q
  3. if p < q
  4. if p ≤ q
  5. if p ≥ q
I. $2p^2$ = 23p – 63
II. 2q ($q^{– 8}$) = $q^{–36}$

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $2p^2$ = 23p – 63

or, $2p^2$ – 23p + 63 = 0

II. 2q ($q ^{–8}$) = $q^{ –36}$

or, (2p – 9) (p – 7) = 0

or, $q^{ –7} × q^{36} = 1/2$

∴p = $9/2$ or 7

or, $q^{29} = 1/2$

∴q = $(1/2)^{1/29}$

Hence, p > q

Q-6)   Directions :
In each question below one or more equation(s) is /are given. On the basis of these, you have to find out the relationship between p and q. Give answer
  1. if p = q
  2. if p > q
  3. if p < q
  4. if p ≤ q
  5. if p ≥ q
I. p ($p^{–1}$) = ($p^{–1}$)
II. $q^2$ = $4q^{–1}$

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $p(p^{– 1}) = p^{–1}$

II. $q^2 = 4(q ^{–1})$

or, p × $1/p=1/p$

∴p = 1

or, $q^3$ = 4

∴q = $(4)^{1/3}$ > 1

Hence, q > p

Q-7)   Directions :
In each of the following questions there are two equations. Solve them and give answer
  1. if P < Q
  2. if P > Q
  3. if P ≤ Q
  4. if P ≥ Q
  5. if P = Q
I. $P^2$ + 3P – 4 = 0
II. $3Q^2$ – 10Q + 8 = 0

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $P^2$ + 3P – 4 = 0

$P^2$ + 4P – P – 4 = 0

⇒P(P + 4) –1(P + 4) = 0

⇒P = 1, – 4

II. $3Q^2$ – 10Q + 8 = 0

$3Q^2$ – 6Q – 4Q + 8 = 0

3Q (Q – 2) – 4 (Q – 2) = 0

(3Q – 4) (Q – 2) = 0

⇒Q = $4/3$, 2

∴Q > P

Q-8)   Directions :
In each of the following questions, one or two equation(s) is/are given. On their basis you have to determine the relation between x and y and then give answer.
  1. if x < y
  2. if x > y
  3. if x ≤ y
  4. if x ≥ y
  5. if x = y
I.$x^2$ + 3x + 2 = 0
II.$2y^2$ = 5y

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $x^2$ + 3x + 2 = 0

or, $x^2$ + 2x + x + 2 = 0

or, (x + 2)(x + 1) = 0

or, x = – 2, – 1

II. $2y^2$ = 5y

or, $2y^2$ – 5y = 0

or, y(2y – 5) = 0

or, y = 0, $5/2$

Hence, y > x

Q-9)   Directions :
In each of the following questions, one or two equation(s) is/are given. On their basis you have to determine the relation between x and y and then give answer.
  1. if x < y
  2. if x > y
  3. if x ≤ y
  4. if x ≥ y
  5. if x = y
I.$ x^2$ – 5x + 6 = 0
II.$y^2$ + y – 6 = 0

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $x^2$ – 5x + 6 = 0

or, $x^2$ – 3x – 2x + 6 = 0

or, x(x – 3) – 2(x – 3) = 0

or, (x – 3)(x – 2) = 0

or, x = 2, 3

II. $y^2$ + y – 6 = 0

or, $y^2$ + 3y – 2y – 6 = 0

or, y(y + 3) – 2 (y + 3) = 0

or, (y + 3) (y – 2) = 0

or, y = 2, –3

Hence, x ≥ y

Q-10)   Directions :
In each of these questions two equations are given. You have to solve these equations and Give answer
  1. if x < y
  2. if x > y
  3. if x =y
  4. if x ≥ y
  5. if x ≤ y
I. $x^2$ – 6x = 7
II. $2y^2$ + 13y + 15 = 0

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $x^2$ – 6x = 7

or, $x^2$ – 6x – 7 = 0

or, (x – 7) (x + 1) = 0

or, x = 7, – 1

II. $2y^2$ + 13y + 15 = 0

or, $2y^2$ + 3y + 10y + 15 = 0

or, (2y + 3) (y + 5) = 0 or, y =${-3}/2,$-5

Hence, x > y

Q-11)   Directions :
In each of these questions two equations are given. You have to solve these equations and Give answer
  1. if x < y
  2. if x > y
  3. if x =y
  4. if x ≥ y
  5. if x ≤ y
I. $10x^2$ – 7x + 1 = 0
II. $35y^2$ – 12y + 1 = 0

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $10x^2$ – 7x + 1 = 0

or, $10x^2$ – 5x – 2x + 1 = 0

or, (2x – 1) (5x – 1) = 0

or, x = 1/2, 1/5

II. $35y^2$ – 12y + 1= 0

or, $35y^2$ – 7y – 5y + 1 = 0

or, (5y – 1) (7y – 1) = 0

or, y = $1/5 , 1/7$

Hence, x ≥ y

Q-12)   Directions :
In each question one or more equation(s) is (are) provided. On the basis of these you have
  1. if p = q
  2. if p > q
  3. if q > p
  4. if p ≥ q
  5. if p ≥ q
(i) $4p^2$ = 16 (ii) $q^2$ – 10q + 25 = 0

(a)

(b)

(c)

(d)

(e)

Explanation:

(i) $4p^2$2 = 16; p = $√4$ = 2

(ii) $q^2 $ – l0q + 25 = 0 ⇒ (q – 5)(q – 5) = 0

or, q = 5 ∴ q > p

Q-13)   Directions :
In each question one or more equation(s) is (are) provided. On the basis of these you have
  1. if p = q
  2. if p > q
  3. if q > p
  4. if p ≥ q
  5. if p ≥ q
(i) $4p^2$ – 5p + 1 = 0 (ii) $q^2$ – 2q + 1 = 0

(a)

(b)

(c)

(d)

(e)

Explanation:

(i) $4p^2$ – 5p + 1 = 0 or, $4p^2$ – 4p – p + i = 0

or, 4p(p – 1) – 1(p – 1) = 0

or, (4p – 1)(p – 1) = 0

or, p = 1 and p = $1/4$

(ii) $q^2$–2q +1 = 0

⇒(q – 1) (q –1) = 0

or, q = 1

∴q ≥ p

Q-14)   Directions :
In each question below one or more equation(s) is/are provided. On the basis of these, you have to find out relation between p and q.
  1. if p = q
  2. if p > q
  3. if q > p
  4. if p ≥ q
  5. if p ≥ q
I. $6q^2+1/2=7/2a$
II. $12p^2$ + 2 = 10p

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $6q^2+1/2=7/2q$

or,$12q^2$ - 7q + 1 = 0.

or $(q-{1/4})(q-{1/3})$=0

∴ q = $1/4$ pr $1/3$

II. $12p^2$ + 2 = 10p

or, $6p^2$ – 5p + 1 = 0

or,$(p-{1/3})(p-{1/2})$=0

∴ p=${1/3}$ or $1/2$

Hence, p ≥ q

Q-15)   Directions :
In each question below one or more equation(s) is/are provided. On the basis of these, you have to find out relation between p and q.
  1. if p = q
  2. if p > q
  3. if q > p
  4. if p ≥ q
  5. if p ≥ q
I. $2p^2$ + 40 = 18p
II. $q^2$ = 13q – 42

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $2p^2$ + 40 = 18p

or, $p^2$ – 9p + 20 = 0

or, (p – 4) (p – 5) = 0

∴ p = 4 or 5

II. $q^2$ = 13q – 42

or, $q^2$ – 13q + 42 = 0

or, (q – 7) (q – 6) = 0

∴ q = 6 or 7

Hence, q > p

Q-16)   Directions:
In each question, one/two equations are provided. On the basis of these you have to find out the relation between p and q.
  1. if p = q
  2. if p > q
  3. if q > p
  4. if p ≥ q, and
  5. if q ≥ p
I. $q^2$ + q = 2
II. $p^2$ + 7p +10 = 0

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $q^2$ + q = 2

or, $q^2$ + q – 2 = 0

II. $p^2$ + 7p + 10 = 0

or, $p^2$ + 5p + 2p + 10 = 0

or, (q + 2) (q – 1) = 0

or, (q + 5)(p + 2) = 0

∴ q = – 2 or 1

∴p = – 5 or – 2

Hence, q ≥ p

Q-17)   Directions :
For the two given equations I and II give answer
  1. if p is greater than q
  2. if p is smaller than q
  3. if p is equal to q
  4. if p is either equal to or greater than q
  5. if p is either equal to or smaller than q.
I. p = $√4/√9$
II. $9q^2$ – 12q + 4 =0

(a)

(b)

(c)

(d)

(e)

Explanation:

I. p = $√4/√9$=$2/3$

II. $9q^2$ – 12q + 4 = 0

or, $9q^2$ – 6q – 6q + 4 = 0

or, 3q(3q – 2) – 2(3q – 2) = 0

or, (3q – 2)(3q – 2) = 0

or, q = $2/3$

Therefore, p = q

Q-18)   Directions :
For the two given equations I and II, give answer
  1. if a is greater than b
  2. if a is smaller than b
  3. if a is equal to b
  4. if a is either equal to or greater than b
  5. if a is either equal to or smaller than b
I. 3a + 2b = 14
II. a + 4b – 13 = 0

(a)

(b)

(c)

(d)

(e)

Explanation:

I. 3a + 2b = 14

II. a + 4b = 13

Subtract equation I from equation II after multiplying II by 3.

We get 3a + 12b – 3a – 2b = 39 – 14

⇒10b = 25

⇒b = 2.5

Put value of b in equation II. We set a + 4 × 2.5 = 13.

Therefore, a = 3. Thus, a > b

Q-19)   Directions :
For the two given equations I and II, give answer
  1. if a is greater than b
  2. if a is smaller than b
  3. if a is equal to b
  4. if a is either equal to or greater than b
  5. if a is either equal to or smaller than b
I. $√2304$ a
II. $b^2$ = 2304

(a)

(b)

(c)

(d)

(e)

Explanation:

From I :

If $√{2304}$= a

then a = ± 48

(Do not consider – 48 as value of a)

Again,

From II :

If $b^2$ = 2304 then b = ± 48

Hence a = b

Q-20)   Directions :
For the two given equations I and II give answer
  1. if p is greater than q
  2. if p is smaller than q
  3. if p is equal to q
  4. if p is either equal to or greater than q
  5. if p is either equal to or smaller than q.
I.$3p^2$ + 15p = –18
II. $q^2$ + 7q + 12 =0

(a)

(b)

(c)

(d)

(e)

Explanation:

I. $3p^2$ + 15p + 18 = 0

or,$3p^2$ + 9p + 6p + 18 = 0

or, 3p (p + 3) + 6(p + 3) = 0

or, (3p + 6)( p + 3) = 0

or, p = ${-6}/3,-3$= – 2, – 3

II.$q^2$ + 7q + 12 = 0

or, $q^2$ + 4q + 3q + 12 = 0

or, q(q + 4) + 3(q + 4) = 0

or, (q + 3)(q + 4) = 0

or, q = – 3, –4

Therefore, p ≥ q