Test: Algebra - 2 - CAT MCQ

# Test: Algebra - 2 - CAT MCQ

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## 25 Questions MCQ Test - Test: Algebra - 2

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Test: Algebra - 2 - Question 1

### If a1, a2, ... are in A.P., then,  is equal to (2019)

Detailed Solution for Test: Algebra - 2 - Question 1

As a1, a2, a3 ... are in A.P.
then,

{Where d = Common difference}.

Test: Algebra - 2 - Question 2

### Let x and y be positive real numbers such that log5 (x + y) + log5 (x – y) = 3, and log2 y – log2 x = 1 – log2 3. Then xy equals (2019)

Detailed Solution for Test: Algebra - 2 - Question 2

Given, log5(x + y) + log5(x – y) = 3
⇒ log5[(x + y)(x – y)] = 3
⇒ (x + y)(x – y) = 53 = 125
⇒ x2 – y2 = 125 ...(i)
And log2y – log2x = 1 – log2 3.

Then if x = 3k, then y = 2k
Now putting this value of x and y in (i), we get
(3k)2 – (2k)2 = 125
⇒ 5k2 = 125, ∴ k = 5
∴ x × y = 3k × 2k = 6 × 25 = 150

Test: Algebra - 2 - Question 3

### If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be (2019)

Detailed Solution for Test: Algebra - 2 - Question 3

We can transform each of the options for ‘n’ years.
(997)214 + 3 ≡ (p – 3)2n – 1 + 3
(997)15 – 3 ≡ (p – 3)n – 3
(1003)15 + 6 ≡ (p + 3)n + 6
(1003)215 – 3 ≡ (p + 3)2n – 3
As per the condition, in one year, the population ‘p’ becomes ‘3 + 2p’
Putting the value of n = 1 in each option, and checking to get 3 + 2p,
we have (997)214 + 3 ≡ p ≠ 3 + 2p
(1003)15 + 6 ≡ p + 9 ≠ 3 + 2p
(1003)215 – 3 ≡ (p + 3)2n – 3 = 3 + 2p
(997)15 – 3 ≡ (p – 3) – 3 ≡ p – 6

Test: Algebra - 2 - Question 4

If (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280, then what is the value of 1 + 2 + 3 + ... + n?

(2019)

Detailed Solution for Test: Algebra - 2 - Question 4

The sequence (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280 is in A.P. with first term (a) = 2n + 1
common difference (d) = 2 and last term (l) = 2n + 47,
Let ‘m’ be the number of terms in this sequence, then
l = a + (m – 1)d
2n + 47 = (2n + 1) + (m – 1) (2) ⇒ m = 24
Now, (2n + 1) + (2n + 3) + (2n + 5) +....+ (2n + 47) = 5280

⇒ 24(2n + 1 + 23) = 48(n + 12) = 5280
⇒ 48(n + 12) = 5280
∴ n = 98 Now, 1 + 2 + 3 + ... +

Test: Algebra - 2 - Question 5

Let A be a real number. Then the roots of the equation x2 – 4x – log2A = 0 are real and distinct if and only if

(2019)

Detailed Solution for Test: Algebra - 2 - Question 5

We know that the quadratic equation ax2 + bx + c = 0 has real and distinct roots, if b2 – 4ac > 0 Hence for real and distinct roots of
x2 – 4x – log2 A = 0, D > 0
∴ (–4)2 – 4 × 1 × (–log2A) > 0
⇒ 16 + 4 log2A > 0 ⇒ log2A > – 4

Test: Algebra - 2 - Question 6

If x is a real number, thenis a real number if and only if

Detailed Solution for Test: Algebra - 2 - Question 6

The given expression will be real only if

⇒ 4x – x2 ≥ 3 ⇒ x2 – 4x + 3 ≤ 0
⇒ (x – 1) (x – 3) ≤ 0 ⇒ 1 ≤ x ≤ 3

Test: Algebra - 2 - Question 7

Let a1, a2, ... be integers such that a1 – a2 + a3 – a4 + ... + (–1)n–1. an = n, for all n ≥ 1.
Then a51 + a52 + . . . + a1023 equals

(2019)

Detailed Solution for Test: Algebra - 2 - Question 7

For n = 1 , a1 = 1
For n = 2 , a1 – a2 = 2 ⇒ a2 = –1
For n = 3 , a1 – a2 + a3 = 3 ⇒ a3 = 1
For n = 4 , a1 – a2 + a3 – a4 = 4 ⇒ a4 = –1
From the above shown pattern, we conclude that each odd term = 1 and each even term = –1
⇒ a51 + a52 + ... + a1022 + a1023
= (a51 + a52 + ....+ a1022) + a1023
= 0 + 1 = 1

Test: Algebra - 2 - Question 8

Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is

(2018)

Detailed Solution for Test: Algebra - 2 - Question 8

Since, 5x, 16y, 12z are in AP.
∴ 32y = 5x + 12z …(1)
∵ x, y, z are in GP
∴ y2 = xz ...(2)
Squaring both sides of (1), we get
1024y2 = 25x2 + 144z2 + 120xz
⇒ 1024xz = 25x2 + 144z2 + 120xz
⇒ 25x2+144z2 - 904xz = 0
⇒ 25x2 - 900xz - 4xz + 144 z2 = 0
⇒ 25x(x - 36z) - 4z(x - 36z) = 0
⇒ (25x - 4z) (x - 36z) = 0

[r is the common ratio]

But  because x, y, z > 0 and x < y < z
∴ common ratio = 5 / 2

Test: Algebra - 2 - Question 9

If x is a positive quantity such that 2x = 3log 52, then × is equal to

(2018)

Detailed Solution for Test: Algebra - 2 - Question 9

2x = 3log52
Taking logarithms to base 5 on both sides, we have
x (log52) = log52 . log53
x = log53 = log5

Test: Algebra - 2 - Question 10

Given that X2018Y2017 = 1/2 and X2016Y2019 = 8, the value of x2 + y3 is

(2018)

Detailed Solution for Test: Algebra - 2 - Question 10

x2018 y2017 = 1/2 and x2016 y2019 = 8

⇒ y4035 = 24035

Test: Algebra - 2 - Question 11

If log1281 = p, thenis equal to

(2018)

Detailed Solution for Test: Algebra - 2 - Question 11

log1281 = p ⇒ log1234 = p

Test: Algebra - 2 - Question 12

If log2(5 + log3 a) = 3 and log5(4a + 12 + log2 b) = 3, then a + b is equal to

(2018)

Detailed Solution for Test: Algebra - 2 - Question 12

5 + log3 a = 23 = 8 ⇒ log3a = 3  ⇒ a = 27 Similarly, 4a + 12 + log2b = 53 = 125
Since a = 27, 4(27) + 12 + log2b = 125
⇒ log2b = 5  ⇒ b = 32.
∴ a + b = 27 + 32 = 59

Test: Algebra - 2 - Question 13

Let a1, a2...., a2n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ... + a3n = 1830, then what is the smallest positive integer m such that m(a1 + a2 + ... + an) > 1830?

(2017)

Detailed Solution for Test: Algebra - 2 - Question 13

a1 = 3, a2 = 7, d = 4

n(6n + 1) = 610
6n2 + n - 610 = 0
solved, n = 10
Now, m(a1 + a2 + ... an) > 1830

m > 8.7 → m > 9

Test: Algebra - 2 - Question 14

If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

(2017)

Detailed Solution for Test: Algebra - 2 - Question 14

(a + 6d)2 = (a + 2d)(a + 16d)

Test: Algebra - 2 - Question 15

If 92x–1 – 81x–1 = 1944, then x is

(2017)

Detailed Solution for Test: Algebra - 2 - Question 15

92x–1 – 81x–1 = 1944

4x – 4 = 5 → x = 9/4

Test: Algebra - 2 - Question 16

The value of log 0.008 √5 + log √3 81 - 7 is equal to

(2017)

Detailed Solution for Test: Algebra - 2 - Question 16

Test: Algebra - 2 - Question 17

If x + 1 = x2 and x > 0, then 2x4 is

(2017)

Detailed Solution for Test: Algebra - 2 - Question 17

x2 – x – 1 = 0
(∵ x > 0)

Test: Algebra - 2 - Question 18

Two positive real numbers, a and b, are expressed as the sum of m positive real numbers and n positive real numbers respectively as follows:
a = s1 + s2 +…+ sm and b = t1 + t2 +…+ tn
If [a] = [s1] + [s2] +…+ [sm] + 4 and [b] = [t1 ] + [t2 ] +…+ [tn] + 3,
Where [x] denotes the greatest integer less than or equal to x, what is the minimum possible value of m + n?

(2016)

Detailed Solution for Test: Algebra - 2 - Question 18

If a positive number a is expressed as the sum of two positive numbers s1 and s2 then [a] could be at the most 1 more than [s1] + [s2], i.e., the fractional parts of s1 and s2 together, can provide at most 1.
Similarly, the fractional parts of s1, s2, s3, s4, s5 can together, provide at most 4.
Conversely, if [a] is 4 more than [s1] + [s2] + [s3] + [s4] + [sm], then m has to be at least 5.
Similarly, the least value of n is 4.
∴ (m + n)min = 5 + 4 = 9

Test: Algebra - 2 - Question 19

P1, P2, P3, ..., P11 are 11 friends. The number of balls with P1 through P11 in that order is in an Arithmetic Progression. If the sum of the number of balls with P1, P3, P5, P7, P9 and P11 is 72, what is the number of balls with P1, P6 and P11 put together?

(2014)

Detailed Solution for Test: Algebra - 2 - Question 19

Let the number of balls with
Pi = ai (i = 1 to 11)
a1 + a3 + a5 ...... + a11 = 6 (a6) = 72.
As a6 would be the arithmetic mean of these 11 numbers and
2(a6) = (a1 + a11)
= (a2 + a10)
= (a3 + a9)
= (a4 + a8)
= (a5 + a7)
∴ a1 + a6 + a11 = 3 (a6)
= 36

Test: Algebra - 2 - Question 20

If log32, log3(2x – 5) and log3  are in Arithmetic Progression, then x is equal to

(2014)

Detailed Solution for Test: Algebra - 2 - Question 20

According to question,

Let 2x = a

⇒ a = 4 or 8
∴ x = 2 or 3
Hence, 2x – 5 = –1,
when x = 2, which is not possible.
∴ x = 3.
Or
All numbers are in AP,
2 log3 (2x – 5) = log3 2 + log3 (2x – 7/2)
Suppose 2x = t
2 log3 (t – 5) = log3 2 + log3 (t – 7/2)
→ log3 (t – 5)2 = log3 (2t – 7)
→ (t – 5)2 = 2t – 7
→ t2 – 10t + 25 = 2t – 7
→ t2 – 12t + 32 = 0
→ t = 4.8
2x = 4.8
∴ x = 2, 3.

Test: Algebra - 2 - Question 21

A ray of light along the line  gets reflected on the x-axis to become a ray along the line

(2014)

Detailed Solution for Test: Algebra - 2 - Question 21

Take –y in place of y
Then equation is

Test: Algebra - 2 - Question 22

If  where p ≤ n, then the maximum value of X for n = 8 is :

(2014)

Detailed Solution for Test: Algebra - 2 - Question 22

X = (log10 1 + log10 2 + ... + log10 n) – (log10 1 + log10 2 + .... + log10 p) – (log10 1 + log10 2 + .... + log10 (n – p))
⇒ X = log10 n! – log10 p! –  log10 (n – p)!
log (m × n) = log m + log n

X is maximum when   is maximum.

Test: Algebra - 2 - Question 23

If x + y = 1, then what is the value of (x3 + y3 + 3xy)?

(2012)

Detailed Solution for Test: Algebra - 2 - Question 23

Given that x + y = 1
⇒ x + y – 1 = 0
⇒ x3 + y3 – 1 = – 3xy
(a3 + b3  + c3 = 3abc if a + b + c = 0)
⇒ x3 + y3 + 3xy = 1

Test: Algebra - 2 - Question 24

If log165 = m and log53 = n, then what is the value of log36 in terms of ‘m’ and ‘n’?

(2011)

Detailed Solution for Test: Algebra - 2 - Question 24

...(i)
...(ii)
From equations (i) and (ii), we get

Let log36 be equal to k; therefore,

Test: Algebra - 2 - Question 25

If a = b2 = c3 = d4 then the value of loga (abcd) would be :

(2010)

Detailed Solution for Test: Algebra - 2 - Question 25

Let a = b2 = c3 = d4

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