ElitmusZone » Sample Question Paper Quantitative 4

Sample Question Paper Quantitative 4

01.

Let P = {2, 3, 4, ………. 100} and Q = {101, 102, 103, ….. 200}. How many elements of Q are there such that they do not have any element of P as a factor ?

(a) 20

(b) 24

(c) 23

(d) 21

02.

What is the sum of all the 2-digit numbers which leave a remainder of 6 when divided by 8 ?

(a) 612

(b) 594

(c) 324

(d) 872

03.

Find the reminder of 2^{1040} divided by 131.

(a) 1

(b) 3

(c) 5

(d) 7

04.

p is a prime and m is a positive integer. How many solutions exist for the equation p^{6} – p = (m^{2} + m + 6) (p – 1) ?

(a) 0

(b) 1

(c) 2

(d) Infinite

05.

A rectangle is drawn such that none of its sides has length greater than ‘a’. All lengths less than ‘a’ are equally likely. The chance that the rectangle has its diagonal greater than ‘a’ is (in terms of %).

(a) 29.3%

(b) 21.5%

(c) 66.66%

(d) 33.33%

06.

If x is a real number, [x] is greatest integer less than or equal to x, then 3|x|+2-[x]=0. Will the above equation have any real roots ?

(a) Yes

(b) No

(c) real root for x < 0

(d) real roots for x>0

07.

A student is asked to form numbers between 3000 and 9000 with digits 2, 3, 5, 7 and 9. If no digit is to be repeated, in how many ways can the student do so ?

(a) 24

(b) 120

(c) 60

(d) 72

08.

A car A starts from a point P towards another point Q. Another car B starts (also from P) 1H after the first car and overtakes it after covering 30% of the distance PQ. After that, the cars continues. On reaching Q, car B reverses and meets car A, after covering 70/3 of the distance QP. Find the time taken by car B to cover the distance PQ ( in hours) .

(a) 3

(b) 4

(c) 5

(d) 10/3

09.

Find the complete set of values that satisfy the relations ||x| – 3| < 2 and ||x| – 2| < 3.

(a) (-5, 5)

(b) (-5, -1) U (1, 5)

(c) (1, 5)

(d) (-1, 1)

10.

When asked for his taxi number, the driver replied, “If you divide the number of my taxi by 2, 3, 4, 5, 6 each time you will find a remainder of 1, But if you divide it by 11, the remainder is zero. You will also not find any other driver with a taxi having lower number who can say the same”. What is the taxi number ?

Solution of Question 3.
(2^1040 )/131 gives remainder ?
a/c to fermat’s rule..
(a)^p-1/p=1
when hcf(a,p)=1;
then first we have to arrange to this is in fermat’s rule.
then we write, {(2^(131-1))8}/131
then (1)^8/131=1;
hence remainder is equals to 1.

Solution of Question 4
put p=2
we will get
64-2=m^2+m+6
m^2 +m-56=0
m=7 or -8
m=-8 is not possible,becoz given dat m is +ve integer
only m=7 satisfy
let me check
2^6-2=(7^2+7+6)(2-1)
62=62
only one solution is possible

Sorry question number 3

Solution of Question 3.

(2^1040 )/131 gives remainder ?

a/c to fermat’s rule..

(a)^p-1/p=1

when hcf(a,p)=1;

then first we have to arrange to this is in fermat’s rule.

then we write, {(2^(131-1))8}/131

then (1)^8/131=1;

hence remainder is equals to 1.

Solution of Question 4

put p=2

we will get

64-2=m^2+m+6

m^2 +m-56=0

m=7 or -8

m=-8 is not possible,becoz given dat m is +ve integer

only m=7 satisfy

let me check

2^6-2=(7^2+7+6)(2-1)

62=62

only one solution is possible

Q3-

(2^1040 )/131 gives remainder ?

Apply Euler’s

E(131) = 130 //Prime Number E(P) = P-1

1040 mod 130 = 0

New Question –

2^0 mod 131

1 mod 131 = 1 (Ans)

Q.6 > x is real no

[x] is greatest no

then equation is 3|x|+2-[x]=0

is not a Quadratic equation (ax^2+bx+c) so we can not find roots.

No(Ans)

Can someone plzz xplain the last part of Q8.

p6−p=(m2+m++6)(p−1)

Now take p common from L.H.S and the equation becomes…

p(p5−1)=(m2+m+6)(p−1)

We can expand p5−1

as follows

p5–1=(p−1)(p4+p3+p2+p)

So equation becomes..

p(p−1)(p4+p3+p2+p+1)

=(m2+m+6)(p−1)

Divide both the side by (p−1)

and equation becomes….

p(p4+p3+p2+p+1)

=(m2+m+6)

Now(m2+m+6)

will always be even number for every natural number of m.

And p4+p3+p2+p

is an even number because sum of two prime numbers is even.

Thus (p4+p3+p2+p+1)

is odd.

And as a result p(p4+p3+p2+p+1)

is also odd because product of a prime number and odd number is odd.

So L.H.S is odd while R.H.S is even. It means we have NO SOLUTION.

But WAIT !!!!

What will happen if value of p is 2. 2 is the only even prime number.

If value of p is 2 the our L.H.S will be even because product of a even number and odd number is even.

Equation becomes…..

2∗(24+23+22+2+1

) =(m2+m+6)

62=(m2+m+6)

m2+m−56=0

m=7

or m=−8

Value of m cannot be negative as it is given that m is a positive integer.

So only for one prime number this equation holds true when p = 2 and m = 7.

Thus this equation has only ONE SOLUTION.