Ques:- 1) In a club there are certain no. of males and females. If 15 females are absent then no. of males will be twice that of females. If 45 males are absent then female strength will be 5 times that of males. Find no. of males actually present.
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Recent Answer : :
let the third no be x ,
then the first no is 3x ,
second no is 2*3x ie 6x
average is (3x+6x+x )/3 = 20
10x/3 = 20
x= 60/10 = 6
third no is 6 , second is 6x = 36, first no is 3x = 18
largest no is 36 i.e second no
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Recent Answer : :
7days
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Recent Answer : :
2*3(15+4)=114
114*1/2=57 rs
Ques:- if a car starts from A towards B with some velocity due to some problem in the engine afetr travelling 30 km .if the car goes 4/5 th of its of actual velociteythe car reaches B 45 min later to the actual time.if the car engine fails after travelling 45km the car reaches the destination B 36min late to the actual time ,wat is the initial velocity of car and what is the distance between A and B on km.
Recent Answer : :
20&130
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Recent Answer : :
I think it’s 32 not 37
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24 days
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Ques:- The sum of the first four primes is
A. 10
B. 11
C. 16
D. 17
Recent Answer : :
the sum of first 4 primes numbers =2+3+5+7 =>17
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Options:
A) EDRIRL
B) DCQHQK
C) ESJFME
D) DEQJQM
Ans: A)
Ques:- In a class of 60 students, the number of boys and girls participating in the annual sports is in the ratio 3 : 2 respectively. The number of girls not participating in the sports is 5 more than the number of boys not participating in the sports. If the number of boys participating in the sports is 15, then how many girls are there in the class? (1) Data inadequate (2) 20 (3) 25 (4) 30 (5) None of these
Recent Answer : :
let no. of boys not participating be x
then the no. of girls not participating = x+5
no. of boys : girls participating = 3:2
given no.of boys participating = 15
therefore, the ratio is now 15:y(say)
then 3:2 = 15 : x
on solving 3/2 =15/y ie.., 3y =30 we get y =10
hence no. of girls participating =10
therefore total no of students paricipating = 15+10=25
total no of students in class =60 given
hence no. of students not participating = 60-25=35
therefore x+(x+5)=35
2x=30
x=15
therfore no of girls not participating =15+5=20
therefore total no of girls in class = no of girls
participating + no of girls not participating
=10+20
=30 is the answer
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A. Rs. 9471
B. Rs. 12, 628
C. Rs. 18,040
D. Rs. 18,942
Recent Answer : :
B
Ques:- A garrison of 2000 men has provisions for 54 days. At the end of 15 days, a reinforcement arrives, and it is now found that the provisions will last only for 20 days more. What is the reinforcement?
Recent Answer : :
1900
Ques:- A cow is tethered in the middle of a field with a 14 feetlong rope. If the cow grazes 100 sq feet per dayapproximately. What time will be taken by the cow to grazethe grazeable area of the field?
Recent Answer : :
( d ) 6 days
Let the number of males be given the name M.
Let the number of females be given the name F.
If 15 females are absent, then M will be twice that of
present females.
This means that M = 2 * (F – 15)
M = 2 * F – 30.
or 2 * F – M = 30.
Now if in addition to the 15 females being absent, we also
have 45 males being absent,
then this gives the equation,
(F – 15) = 5 * (M – 45)
which simplifies to
F – 15 = 5 * M – 225
5 * M – F = 210
Pulling the equations together, we get
5 * M – F = 210
-M + 2 * F = 30
Multiply the first equation by 2, and keep the second
equation as is.
10 * M – 2 * F = 420
– M + 2 * F = 30
Add the equations.
9 * M = 450
M = 50
Verify answer.
Calculate F
from – M + 2 * F = 30
-50 + 2 * F = 30
2 * F = 30 + 50
F = 40.
If 15 females are absent, then number of males will be twice
that of females.
40 – 15 = 25.
50 = 2 * 25. Confirmed.
If also 45 males were absent, then female strength would be
5 times that of males.
Female strength is 25 due to the 15 females being absent.
50 – 45 = 5.
25 = 5 * 5. Confirmed.