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Quant 5th NOV

1. What is the range of numbers between 400 and 700 that are divisible by 2 and 3?

2. Col A: x^2 + y^2
Col B: (x + 1)^2 * (y + 1)^2

3. Given a series p1, p2, p3, p4................ If p1 = 1 & pn = 24*p(n-1)+8, then
Col A: The remainder when p66 is divided by 6
Col B: 4
(Here 1, 2, 3, 4, 66, n & n-1 are suffixes)

4. If 1pen & 1stapler costs 10$ and 3pens & 2staplers costs 25$, then
Col A: Cost of 1pen
Col B: 4.20$

5. Given a circle in the coordinate system with centre at (3, 4), find the radius of the circle?

6. Given average of a & b as 10. When 'c' is added to a & b, then the average is 10(again).
Col A: Value of 'c'
Col B: 10

7. If -1 < src="http://www.drrajusgre.com/nov_img1.JPG">
Col A: a^2 + b^2 + c^2
Col B: f^2

1 comments:

Anonymous said...

NOV 5th

1)Ans:294
if a number is divisible by bothb2 & 3 then it is divisible by 6
so range of numbers divisible by 6 between 400 to 700

400/6=66.66 So 67*6=402
700/6=116.66 So 116*6=696

range is 696-402=294

2)Ans: A < B
so stupid question to give Explanation

3)Ans: A < B
Col A:P1=1
P2=24*1+8=32 remainder of 32/6=2
P3=24*32+8=776 remainder of 776/6=2
P4=24*776+8=18632 remainder of 18632/6=2
......................................
....................
Remainder of P66/6=2
Col B:4
A < B
ANY MORE EASY APPROACHES

4)Ans: A > B
Col A:
pen=x stapler=y
1pen &1stapler=10$
x+y=10 ----------------Eq(1)
3pens&2staplers=25$
3x+2y=25 -----------------Eq(2)
Eq(2)-2*Eq(1)
3x+2y=25
-2x-2y=20
---------------------
x=5
cost of 1pen=5$
Col B:4.20$


5)Data insufficient
we cannot determine the radius only from centre
it can be any number in range of 0 < r < infinity

6) Ans: A = B
Col A:
Given (a+b)/2=10 So a+b=20
then (a+b+c)/3=10
a+b+c=30 but a+b=20 So c=30-20=10
Col B:10

7)Ans:0.8x
Since given x is -ve number between 0 and -1
sqrt(x) is an imaginary number
so 0.8x has greater value

8)Ans: A = B
length of side of triangle with sides a&b be say x
since it is a right angled triangle
x=sqrt(a^2+b^2)
Since the other triangle is also a right angled triangle
f=sqrt(x^2+c^2)=sqrt(a^2+b^2+c^2)
Col A:(a^2+b^2+c^2)
Col B:f^2=(a^2+b^2+c^2)

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