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Sample Question Paper Quantitative 03

01.

Let X be the set of integers {9, 15, 21, 27, ……375}. Y denotes a subset of X, such that the sum of no two elements of Y is 384. Find the maximum number of elements in Y

(a) 29

(b) 30

(c) 31

(d) 32

02.

The digits of a three-digit number A are written in the reverse order to form another three digit number B. If B > A and B – A is perfectly divisible by 7, then which of the following is necessarily true ?

(a) 100 < A < 299

(b) 106 < A < 305

(c) 112 < A < 311

(d) 118 < A < 317

03.

If a1 = 1 and an+1 – 3an + 2 = 4n for every positive integer n, then a100 equals ?

(a) 3^{99} – 200

(b) 3^{99 }+ 200

(c) 3^{100 }– 200

(d) 3^{100 }+ 200

04.

The total number of integer pairs (x, y) satisfying the equation x + y = xy is/are

(a) 0

(b) 1

(c) 2

(d) None of these

05.

What is the number of distinct triangles with integral values sides and perimeter as 14 ?

(a) 6

(b) 5

(c) 4

(d) 3

06.

(a) 27/14

(b) 21/13

(c) 49/27

(d) 256/147

07.

The number of common roots between the two equations x^{3 }+ 3x^{2 }+ 4x + 5 = 0, x^{3 }+ 2x^{2 }+ 7x + 3 = 0

(a) 0

(b) 1

(c) 2

(d) 3

08.

If both a and b belong to the set {1, 2, 3, 4} then the number of equations of the form ax^{2 }+ bx + 1 = 0 having real roots is

(a) 10

(b) 7

(c) 6

(d) 12

09.

In a Public Sector Unit (PSU), there are 45600 employees. When the PSU offered a Voluntary Retirement Scheme(VRS), 40% of the employees applied for the VRS. After scrutinizing, PSU has rejected 15% of the applications. But only 9120 employees took the retirement through the scheme. What percentage of the total number of employees did not take retirement even though their applications are not rejected ?

(a) 25%

(b) 14%

(c) 24%

(d) 12.75%

10.

If three-digit number is eleven times the two-digit number formed by using the hundred’s and the unit’s digit of the three-digit number respectively, in the ten’s and unit’s place of the two-digit number. If the difference between the digit in ten’s place and the digit in hundred’s place is 1, then what is the digit in the unit’s place ?

In a triangle,there is a property..that is sum of any two sides length of triangle is greater than the third side.
That’s why here perimeter is 14,hence possible sides are (4,4,6),(4,5,5),(6,5,3),(6,6,2)

In a triangle,there is a property..that is sum of any two sides length of triangle is greater than the third side.
That’s why here perimeter is 14,hence possible sides are (4,4,6),(4,5,5),(6,5,3),(6,6,2)

I think 4th answer is only one pair as (0,0) is not a integer pair.

Sample Question Paper Quantitative 03 is very tough

questions label will like this or moderated .

5 no solution pzz discribe

In a triangle,there is a property..that is sum of any two sides length of triangle is greater than the third side.

That’s why here perimeter is 14,hence possible sides are (4,4,6),(4,5,5),(6,5,3),(6,6,2)

In question no. 2

z and X can take the value 8 and 1 respectively

i.e 891-198=693 i.e divisible by 7

hence a) 100<A<299 is correct

In a triangle,there is a property..that is sum of any two sides length of triangle is greater than the third side.

That’s why here perimeter is 14,hence possible sides are (4,4,6),(4,5,5),(6,5,3),(6,6,2)

3)

an+1–3an+2=4nan+1–3an+2=4n

a2−3a1+2=4∗1a2−3a1+2=4∗1

a2=5=32−2∗2a2=5=32−2∗2

a3−3a2+2=4∗2a3−3a2+2=4∗2

a3=21=33−2∗3a3=21=33−2∗3

Hence we can generalize an=3n−2∗nan=3n−2∗n

a100=3100−2∗100a100=3100−2∗100

Option C

3)

an+1–3an+2=4nan+1–3an+2=4n

a2−3a1+2=4∗1a2−3a1+2=4∗1

a2=5=32−2∗2a2=5=32−2∗2

a3−3a2+2=4∗2a3−3a2+2=4∗2

a3=21=33−2∗3a3=21=33−2∗3

Hence we can generalize an=3n−2∗nan=3n−2∗n

a100=3100−2∗100a100=3100−2∗100

Option C