Any High School level book can do the job,but i prefer Arun Sharma(CAT), the concepts of P&C; and Probability are both explained very well, along with a lot of examples. You need not do the Level 3 exercises, they are more suited to CAT Applicants. But the concepts can be grasped very well if you work with all the available examples.
hi puys,
Please suggest me some good book for GMAT math section.( The main bottleneck in my math prep is probability. )
Thanks a lot sausi007. i will search for the book by Arun Sharma. If you have soft copy of the book, please do send me.
A company has 2 types of machines, type R and type S operating @ constant rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R are used?
3
4
6
9
12
2. If the speed of X meters [er second equivalent to the speed of Y kilometre per hour, what is Y in Terms of X?
5x/18
6X/5
18X/5
60X
3600000X
Ans 1.
R type machine takes 36 hrs to do a certain job. In 1 hr, it will do 1/36th of the job.
S type machine takes 18 hrs to do a certain job. In 1 hr, it will do 1/18th of the job.
Its mentioned that the company used the same number of each type of machine to do the job in 2 hours,so lets take x number of machien from R type is used, hence same number x of S types is also used.
Also, machines are working are constant rate.
So, final euation is,
Both types of machine will do (x/36+x/18 )th of work in 1 hr.So, in 2 hr , they will complete the work,
=> 2 (x/36+x/18 )= 1
=> x = 36/6= 6
C is the answer.
Ans 2.
x m/s = y km/hr
=> x (m/s) = (y * 1000)/(3600) (m/s)
=> x = 5y/18
=> y = 18x/5
C is the answer.
If Y Intercept of line L is 4, and the slope of line L is negetive , whic of the following could be X intercept of L
1) -1
2) 0
3) 6
Two similar triangle with side s & Side S
The areas of triangle with base Side S triangle is twice the area of side s Triangle
In terms of s, S is????????
My take: 3) 6
&
S=4s
If Y Intercept of line L is 4, and the slope of line L is negetive , whic of the following could be X intercept of L
1) -1
2) 0
3) 6
Two similar triangle with side s & Side S
The areas of triangle with base Side S triangle is twice the area of side s Triangle
In terms of s, S is????????
My take: 3) 6
&
S=4s
1. Y intercept of 4 means that the line cuts the Y axis at 4, or, at x=0, y=4.
Now since the slope of the line is -ve, it can only cut the x axis at values of x>0, hence the only answer, 6
2. Similar triangles have the property that the ratio of their Areas is eaqual to the square of the ratio of their sides, thus considering the area of the smaller being a and the bigger being A, we get,
(s/S)^2 = a/A
now we know that A=2a
(s^2/S^2)=1/2
LOL, i found the correct answer while posting !!
Perils of accessing PG from office at the fore.
So finally we get the correct answer S=root2.s
1St is right But 2nd is wromg........Its root2s...
PLz Expalain Both
Hi Sausi,
Nice to know the formula about similar triangles ...
Otherwise, we could have solved it this way, as both are similar triangles,
so,we can have one right angled triangle with s=3,h=4,hyp=5 and the other triangle to have area as 2 times of smaller one is to have sides as ,
S=3 root 2, h'=4 root 2,hyp'=5 root 2.
So, S=s root 2.
1. Y intercept of 4 means that the line cuts the Y axis at 4, or, at x=0, y=4.
Now since the slope of the line is -ve, it can only cut the x axis at values of x>0, hence the only answer, 6
2. Similar triangles have the property that the ratio of their Areas is eaqual to the square of the ratio of their sides, thus considering the area of the smaller being a and the bigger being A, we get,
(s/S)^2 = a/A
now we know that A=2a
(s^2/S^2)=1/2
LOL, i found the correct answer while posting !!
Perils of accessing PG from office at the fore.
So finally we get the correct answer S=root2.s
Hi Lads,
If we go by the simple formula - y=mx+b by substituting, x =0 and y =4, we get b =4. Now we need where y = 0 as we have to find out x intercept, so, 0=mx+4 =>-mx =4 , as we know m being the slope, which is negative , x has to be positive to be equal to 4. Therefore x =6 is the answer. In fact, this also gives the slope as m =-2/3.
Thanks,
N
If y(u - c) = 0 and j(u - k) = 0, which of the following must be true, assuming c
(A) yj (B) yj > 0
(C) yj = 0
(D) j = 0
(E) y = 0
from the first equation ,
either y = 0 or u - c = 0 which implies = u = c
from the second equation ,
either j = 0 or u = k
the above 2 results imply that u = c = k which is not the case , so either of the 2 ie y or j has to be zero ,
would go for option
c) ie yj = 0
Please let me know the answer as well .
thanks,.
N
If y(u - c) = 0 and j(u - k) = 0, which of the following must be true, assuming c
(A) yj (B) yj > 0
(C) yj = 0
(D) j = 0
(E) y = 0
The answer is (c)
The function g(x) is defined for integers x such that if x is even, g(x) = x/2 and if x is odd, g(x) = x + 5. Given that g(g(g(g(g(x))))) = 19, how many possible values for x would satisfy this equation?
A. 1
B. 5
C. 7
D. 8
E. 11
The function g(x) is defined for integers x such that if x is even, g(x) = x/2 and if x is odd, g(x) = x + 5. Given that g(g(g(g(g(x))))) = 19, how many possible values for x would satisfy this equation?
A. 1
B. 5
C. 7
D. 8
E. 11
The function g(x) is defined for integers x such that if x is even, g(x) = x/2 and if x is odd, g(x) = x + 5. Given that g(g(g(g(g(x))))) = 19, how many possible values for x would satisfy this equation?
A. 1
B. 5
C. 7
D. 8
E. 11
Op A....the only possible value of x for which g(g(g(g(g(x))))) = 19 hold is
x=122
A real gud question buddy....but the thing is i solve the fn g(g(g(g(g(x))))) = 19
and then made the conclusion of only one possible value...although solving the fn was
not that tough but m sure there is some gud way...
if u know some quick fix trick plsss share...
How many roots does this equation have?
sq rt (x^2 + 1) + sq rt (x^2 + 2) =2
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
I would go with E (4 roots)
How many roots does this equation have?
sq rt (x^2 + 1) + sq rt (x^2 + 2) =2
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
x^2 >=0, so sq rt (x^2 + 1) + sq rt (x^2 + 2) >= 1 + sq rt 2 , which is larger than 2. Hence no (real) solutions for x.
Although complex numbers solutions can be obtained, these are not part of regular GMAT.
Ans A (0)