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- if a < b
- if a ≤ b
- if a ≥ b
- if a = b
- if a > b
I. $a^2$ – 5a + 6 = 0
II. $b^2$ – 3b + 2 = 0
(a)
(b)
(c)
(d)
(e)
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- if p = q
- if p > q
- if p < q
- if p ≤ q
- if p ≥ q
I. 2p (p – 4) = 8 (p + 5)
II. $q^2$ + 12 + 7q
(a)
(b)
(c)
(d)
(e)
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- if a < b
- if a ≤ b
- if a ≥ b
- if a = b
- if a > b
I. $2a^2$ + 5a + 3 = 0
II. $2b^2$ – 5b + 3 = 0
(a)
(b)
(c)
(d)
(e)
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- if a < b
- if a ≤ b
- if a ≥ b
- if a = b
- if a > b
I. 2a + 3b = 31
II. 3a = 2b + 1
(a)
(b)
(c)
(d)
(e)
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- if p = q
- if p > q
- if p < q
- if p ≤ q
- if p ≥ q
I. $2p^2$ = 23p – 63
II. 2q ($q^{– 8}$) = $q^{–36}$
(a)
(b)
(c)
(d)
(e)
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- if p = q
- if p > q
- if p < q
- if p ≤ q
- if p ≥ q
I. p ($p^{–1}$) = ($p^{–1}$)
II. $q^2$ = $4q^{–1}$
(a)
(b)
(c)
(d)
(e)
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- if P < Q
- if P > Q
- if P ≤ Q
- if P ≥ Q
- if P = Q
I. $P^2$ + 3P – 4 = 0
II. $3Q^2$ – 10Q + 8 = 0
(a)
(b)
(c)
(d)
(e)
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.- if x < y
- if x > y
- if x ≤ y
- if x ≥ y
- if x = y
I.$x^2$ + 3x + 2 = 0
II.$2y^2$ = 5y
(a)
(b)
(c)
(d)
(e)
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.- if x < y
- if x > y
- if x ≤ y
- if x ≥ y
- if x = y
I.$ x^2$ – 5x + 6 = 0
II.$y^2$ + y – 6 = 0
(a)
(b)
(c)
(d)
(e)
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- if x < y
- if x > y
- if x =y
- if x ≥ y
- if x ≤ y
I. $x^2$ – 6x = 7
II. $2y^2$ + 13y + 15 = 0
(a)
(b)
(c)
(d)
(e)
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- if x < y
- if x > y
- if x =y
- if x ≥ y
- if x ≤ y
I. $10x^2$ – 7x + 1 = 0
II. $35y^2$ – 12y + 1 = 0
(a)
(b)
(c)
(d)
(e)
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 - if p = q
- if p > q
- if q > p
- if p ≥ q
- if p ≥ q
(i) $4p^2$ = 16 (ii) $q^2$ – 10q + 25 = 0
(a)
(b)
(c)
(d)
(e)
(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 - if p = q
- if p > q
- if q > p
- if p ≥ q
- if p ≥ q
(i) $4p^2$ – 5p + 1 = 0 (ii) $q^2$ – 2q + 1 = 0
(a)
(b)
(c)
(d)
(e)
(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.- if p = q
- if p > q
- if q > p
- if p ≥ q
- if p ≥ q
I. $6q^2+1/2=7/2a$
II. $12p^2$ + 2 = 10p
(a)
(b)
(c)
(d)
(e)
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.- if p = q
- if p > q
- if q > p
- if p ≥ q
- if p ≥ q
I. $2p^2$ + 40 = 18p
II. $q^2$ = 13q – 42
(a)
(b)
(c)
(d)
(e)
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.- if p = q
- if p > q
- if q > p
- if p ≥ q, and
- if q ≥ p
I. $q^2$ + q = 2
II. $p^2$ + 7p +10 = 0
(a)
(b)
(c)
(d)
(e)
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- if p is greater than q
- if p is smaller than q
- if p is equal to q
- if p is either equal to or greater than q
- 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)
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- if a is greater than b
- if a is smaller than b
- if a is equal to b
- if a is either equal to or greater than b
- if a is either equal to or smaller than b
I. 3a + 2b = 14
II. a + 4b – 13 = 0
(a)
(b)
(c)
(d)
(e)
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- if a is greater than b
- if a is smaller than b
- if a is equal to b
- if a is either equal to or greater than b
- if a is either equal to or smaller than b
I. $√2304$ a
II. $b^2$ = 2304
(a)
(b)
(c)
(d)
(e)
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- if p is greater than q
- if p is smaller than q
- if p is equal to q
- if p is either equal to or greater than q
- 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)
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