Algebraic Expressions Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 1 EXERCISES
The following question based on Algebraic Expressions topic of quantitative aptitude
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) if p ≤ q
(b) if p ≥ q
(c) if p = q
(d) if p < q
e) if p > q
The correct answers to the above question in:
Answer: (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
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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) if a is either equal to or greater than b
b) if a is either equal to or smaller than b
c) if a is greater than b
d) if a is equal to b
e) if a is smaller than b
Answer »Answer: (d)
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
Question : 2
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) if p ≥ q and
b) if q ≥ p
c) if p = q
d) if q > p
e) if p > q
Answer »Answer: (d)
(i) $4p^2$2 = 16; p = $√4$ = 2
(ii) $q^2 $ – l0q + 25 = 0 ⇒ (q – 5)(q – 5) = 0
or, q = 5 ∴ q > p
Question : 3
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) if p ≤ q
b) if p ≥ q
c) if p = q
d) if p < q
e) if p > q
Answer »Answer: (d)
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
Question : 4
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) if a is either equal to or greater than b
b) if a is either equal to or smaller than b
c) if a is greater than b
d) if a is equal to b
e) if a is smaller than b
Answer »Answer: (c)
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
Question : 5
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) if p ≥ q, and
b) if q ≥ p
c) if p = q
d) if q > p
e) if p > q
Answer »Answer: (b)
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
Question : 6
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) if p is either equal to or greater than q
b) if p is either equal to or smaller than q.
c) if p is greater than q
d) if p is equal to q.
e) if p is smaller than q
Answer »Answer: (d)
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
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