Set 1 :
1-> 15% of 750 = 112.50. Multiply it with 4 .i.e 112.50*4 = 450 .this is the s.i.
2-> 2000/5 = 400 is the simple interest p.a .Now determine what sum has 400 as its 8%
0.08x = 400 => x = 5000 .
3-> Not sure..Is total annual interest = 56,000 ? in that case the amount at 15% would be much bigger.Please confirm ..
4 -> let sum be x.
0.05x*3 + 0.06x*4 + 0.08x*5 = 3950 => 0.79x = 3950 => x = 5000.
5-> let sum be x.
5*(0.03x) = 900 => 0.03x = 180 => x = 6000.
Set 2 :
6 -> S.I for 2 years = 125,
so simple interest per annum = 125/2 = 62.50
this is obtained at 10% rate of ineterest on a sum.Let this sum be x.
then 62.50 = 0.10x => x = 625 .
Now, find S.I on 625 for 3 years at the rate of 12% p.a. which is = 12% of 625 * 3 = 225.
7 -> 12 1/2 % of 6400 = 800
Now, let period be x
so, 800 * x = 2000 => x = 2.5 years
8 -> 6000 * x/100 * 3 1/3 = 3000 => x = 3000 * 1/200 = 15%
9 -> S.I on sum = 5600-4000 = 1600 for 2 years .
so, s.i for 1 year = 800.
Now, 4000 * x/100 = 800 => x = 20%.
10 ->Ques. not clear.
SET - 2 :-
10. A lent Rs. 25000 to B for 4 years and Rs. 40,000 to C for 3 1/2 years and got Rs. 24,000 S.I from both B and C. Find the rate PCPA
Sol :- SI = PTR/100
Let x be the rate of Interest (R)
(25000*4*x)/100 + (40000*7/2*x)/100 = 24000
1000x+1400x = 24000
2400x=24000
x = 10%
This seems to be a relatively easy one but the solutions seems to be a little tricky... Can one of u be a lamb and post the 'working out' of this problem...thanks in advance
What are values of X, Y and Z if
X + Y + 2Z = 8
2X - Y + 3Z = 13
Kindly post the steps too
This seems to be a relatively easy one but the solutions seems to be a little tricky... Can one of u be a lamb and post the 'working out' of this problem...thanks in advance
What are values of X, Y and Z if
X + Y + 2Z = 8
2X - Y + 3Z = 13
Kindly post the steps too
Sol :- X + Y + 2Z = 8 ------> A
2X - Y + 3Z = 13 -------> B
Now A+B is
3X+5Z = 21 -------> C
As we know that 21 is a multiple of 3. And X has coefficient as 3. So Z should be either 0 or multiple of 3.
Put Z = 0 in C we get X = 7, Substitute both the values in A
We get Y = 1.
These values satisfy equation B also.
therefore X = 7, Y = 1 and Z = 0 are the values.
Now lets check when Z = 3. We get X = 2, substituting these values in A we get Y = 0
Now these values are satisfied by B also.
So here values are X = 2, Y = 0 and Z = 3.
There will be multiple answers.
Lets start with Time and Distance Concepts .......
This Topic covers following concepts:
-Relationship between time, speed and distance
-Important formulae to solve the related problems
-Average speed and relative speed
TIME: TIME is a component to compare the durations of events and the intervals between them, and to quantify the motions of objects.
SPEED: The speed of a body can be defined as the distance covered by it in Unit time. This statement gives birth to three forumlae.
DISTANCE: The numerical description of "How far two objects are" is called distance. It may also refer to the length between two points that may be present or may get created.
1) SPEED = DISTANCE/TIME
When distance is constant,
This also proves SPEED IS PROPORTIONATE WITH 1/TIME.
If you notice the distance traveled in a unit time with different speed, greater the speed in a time unit; greater will be the distance covered under that time span.
2) TIME = DISTANCE/SPEED
When Time is constant,
Thus, DISTANCE IS PROPORTIONATE TO SPEED.
Example: Time taken in a particular journey can be changed only in two cases.
Case-1: Change in speed of object -
Case-2: Change in distance covered -
3) DISTANCE = TIME * SPEED
When Speed is constant,
Thus, DISTANCE IS PROPORTIONATE TO TIME.
so by multiple you mean only two right?
UNITS OF MEASUREMENT:
TIME can be measured in:
Hours
Minutes
Seconds
Some conversions related to TIME are as follows:
1 Hour = 60 Minutes = 60 X (60 Seconds) = 3600 Seconds
DISTANCE can be measured in various units. some important are:
Miles
Kilometers
Meters
Yards
Feet
Some conversions related to DISTANCE are as follows:
1 Kilometer = 1000 Meters
1 Kilometer = 0.6214 Miles
=1 Mile = 1.609 Kilometer
= 8 Kilometer = 5 Miles
1 Yard = 3 feet
SPEED IS MEASURED IN TERMS OF DISTANCE COVERED / TIME. SOME IMPORTANT UNITS ARE:
Miles/hour
Kilometers/hour
Meters/second
Some conversions related to SPEED are as follows:
To convert kilometer per hour to meter per second: Kilometer/Hour =
5 Meters/18 seconds = 5/18
To convert meters per second to Kilometer to hour: Meters/Second =
18 kilometers/ 5 Seconds = 18/5
ajayjr Saysso by multiple you mean only two right?
No many more can come.. Like X = 12, Y= 2 and Z = -3 is also possible.
AVERAGE SPEED = TOTAL DISTANCE TRAVELED/ TOTAL TIME TAKEN
Means,If an object travels D1, D2, D3,......... Dn distances, with speeds S1, S2, S3,.......Sn in T1,T2,T3.........Tn times respectively, then Average Speed is
= (D1+D2+D3+......+Dn)/(T1+T2+T3+.....+Tn)
= (S1T1+S2T2+......SnTn)/(T1+T2+T3+.....+Tn)
1)If an object travels a distance D1 and D2 at a speed of S1 and S2 Miles per Hour, respectively, in time T1 and T2 then the total time taken 'T' is given by:
T = T1+T2 = D1/S1 + D2/S2
The total distance covered 'D' is given by:
D= D1+D2 = S1T1 + S2T2
2)If a certain distance D, from point X to point Y is traveled at 'a' km/hr and the same distance is covered i.e. from Y to X at 'b' km/hr, then the average speed during the whole journey can be calculated as:
Average Speed = km/hr
Also, if T1 and T2 are the times taken to travel from X to Y and from Y to X respectively, the distance D from X to Y can be calculated as:
D = (t1+t2) km
D = (t1-t2) km
D = (a-b) km
3)While a certain distance 'D' is covered, if A man changes his speed in the ratio X : Y, then the ratio of time taken becomes Y : X.
4)If an object travels a distance D from X to Y with Speed 'a' in time T1 and travels back from Y to X with m/n of the usual speed 'a', then the change in time taken to travel the same distance can be calculated as:
Change in Time = (n/m - 1)* T1 ; for n>m
Change in Time = (1-n/m)* T1 ; for m.n
5) IF two objects, say A and B, start at the same time in opposite directions from two different points (P and Q) and arrive at the opposite points in 'a' and 'b' hours respectively after having met.
Then the ratio of their speed is
A's Speed/ B's Speed = Root b/ Root a
So Buddy what this infers is that for a three variable problem we need three equations right?
ajayjr SaysSo Buddy what this infers is that for a three variable problem we need three equations right?
No when there are 2 equations we can get value of a single variable if the ratio of other two variables of both the equations is same. like a1/a2=b1/b2. Otherwise we definitely need 3 equations to have unique solution.
ur definitely not a zero in Quant...Example for the above mentioned logic plz
***VERY IMPORTANT ***
Its imperative that all aspirants get themselves acquainted with the meanings and difference between terms like :-
Indeterminate
Undefined
Inconsistent
Independant
Expect atleast two problems with one or more of these as options!!
Mixing English with Quants now :banghead::banghead::banghead:
Its imperative that all aspirants get themselves acquainted with the meanings and difference between terms like :-
Indeterminate
Undefined
Inconsistent
Independant
Expect atleast two problems with one or more of these as options!!
Mixing English with Quants now :banghead::banghead::banghead:
ajayjr Saysur definitely not a zero in Quant...Example for the above mentioned logic plz
See when two equations are
2x+3y-z=0
4x+6y+4z=6
Here we have unique solution for z. Because 2/4 = 3/6
And this is possible if a1/a2=b1/b2 or a1/a2=c1/c2 or b1/b2=c1/c2
If all the three ratios are same like a1/a2=b1/b2=c1/c2 then we don't get any solution because both the equations get cancelled.
hello!
i needed help with a pattern in number system.
the question types are:
1)find the maximum value of n such that 50! is perfectly divisible by 2520^n?
here the largest prime factor(i.e 7) was taken as restricting factor..
2)) find the maximum value of n such that 50! is perfectly divisible by 12600^n?
here instead of the biggest prime no.(i.e 7), 3 was taken as restricting prime factor.
so,
can somebody plz help me with this concept? i know that the longer way would be to calculate it for all the prime factors and see which one will be the smallest...but thus can be really lengthy (like in case of 504^n if we take 2 as the prime factor).. so can sumbody plz tell dat how is the limiting factor considered?
thanks..
hello!
i needed help with a pattern in number system.
the question types are:
1)find the maximum value of n such that 50! is perfectly divisible by 2520^n?
here the largest prime factor(i.e 7) was taken as restricting factor..
2)) find the maximum value of n such that 50! is perfectly divisible by 12600^n?
here instead of the biggest prime no.(i.e 7), 3 was taken as restricting prime factor.
so,
can somebody plz help me with this concept? i know that the longer way would be to calculate it for all the prime factors and see which one will be the smallest...but thus can be really lengthy (like in case of 504^n if we take 2 as the prime factor).. so can sumbody plz tell dat how is the limiting factor considered?
thanks..
According to my knowledge that is the only method that we follow. I follow the same method. If we are quick that is not a long method.We can do it quickly.
SET - 1 On Time and Distance :-
1. A man rows 750m in 675 seconds against the stream and returns in 7 1/2 minutes. His rowing speed in still water is:
a) 3 km/hr
b) 4 km/hr
c) 5 km?hr
d) 6 km/hr
e) 7 km/hr
2. A train 150 m long is running at a speed of 90 km/hr. Time taken by the train to cross a tree is.
a) 3 sec
b) 4 sec
c) 6 sec
d) 8 sec
e) 5 sec
3. Two trains A and B of equal lengths of 200 meters each running in opposite direction cross each other in 16 seconds. What is the speed of train A?
a) 40 km/hr
b) Data inadequate
c) 90 km/hr
d) 80 km/hr
e) None of these
4. Sunil left for city 'x' from city 'y' at 5.50 a.m. He traveled at the speed of 80km/hr for 2 hours 15 minutes. After that the speed was reduced to 60km/hr. If the distance between the two cities is 350 kms, at what time did sunil reach the city 'y'?
a) 9.50 am
b) 10.35 am
c) 9.55 am
d) 10.05 am
e) None of these
5. Find the speed of train in km/hr whose length is 200m and crosses a platform of length 240m in 22 seconds.
a) 60
b) 48
c) 53
d) Data inadequate
e) None of these
SET - 2 On Time and Distance :-
1. Two cars start at the same time from A and B which are 120 km apart. If the two cars travel in opposite direction they meet after one hour and if they travel in same direction ( from A towards B) then A meets B after 6 hours. What is the speed of the car starting from A?
a) 70 kmph
b) 120 kmph
c) 60 kmph
d) Data Inadequate
e) None of these
2. With a uniform speed a car covers a distance in 8 hours. Were the speed increased by 4 km/hr the same distance could be covered in 7 1/2 hours. What is the distance covered?
a) 640 km
b) 480 km
c) 420 km
d) Cannot be determined
e) None of these
3. Starting with the initial speed of 30 km/hr, the speed is increased by 4km/hr every two hours. How many hours will it take to cover a distance of 288 km?
a) 4
b) 6
c) 12
d) 8
e) None of these
4. A boat running downstream cover a distance of 16kms in 2 hours, while for covering the same distance upstream it takes 4 hours. What is the speed of the boat in still water?
a) 4kmph
b) 6kmph
c) 8kmph
d) Data inadequate
e) None of these
5. The speed of a car increases by 2km after every hour. If the distance traveled in the first hour was 35km, what was the total distance traveled in 12 hours?
a) 522 km
b) 456 km
c) 556 km
d) 482 km
e) None of these