Q.1 Ramesh covers two-third of a certain distance at 4 km/h and the remaining at 5 km/h. if he takes 42 minutes in all, the distance is:
a) 2.5 km
b) 3 km
c) 4 km
d) 4.6 km
e) none of these
Q.2 A train at 48 km/h Completely crosses another train having half its length and traveling in opposite direction at 42 km/h, in 12 seconds. it also a railway platform in 45 seconds. Find the length of plate form.
a) 560m
b) 400m
c) 600m
d) 450m
e) none of these
Q.3 The ratio of the age of a man and his wife is 4:3. After 4 years this ratio will be 9:7. if the at the time of marriage the ratio was 5:3., then how many years ago were they married:
a) 4
b) 8
c) 12
d) 16
e) none of these
Q.4 The batting average for 40 innings of cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. if these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the players is :
a) 165 runs
b) 170 runs
c) 172 runs
d) 174 runs
e) none of these
Q.5 The average age of 5 members of a committee is the same as it was 3 years ago, because an old member has been replaced by a new member. The difference the age of old and new member is :
a) 2 years
b) 4 years
c) 8 years
d) 15 years
e) none of these
Q.6 Two third of the first number is equal to cube of the second number. If the second number is equal to 12% of 50. What is the sum of the first and second number?
a) 330
b) 360
c) 390
d) 372
e) none of these
Q.7 The digit in units place of a number is equal to the digit in tens place of half of that number and the digit in the tens place of that number is less than the digit in units place of half of the number by 1. If the sum of the digits of the number is seven, then what is the number?
a) 52
b) 16
c) 34
d) 36
e) none of these
Q.8 If the numerator of a fraction is increased by 2 and denominator is increased by 3, the fraction becomes 7/9 and if numerator as well as denominator are decreased by 1, the fraction becomes 4/5. What is the original fraction?
a) 13/16
b) 9/11
c) 5/6
d) 17/21
e) none of these.
Q.9 Walking at 3/4 of his usual speed, a man is late by 2*1/2 hr. The usual time would have been:
a) 7*1/2hrs.
b) 3*1/2 hrs.
c) 3*1/4 hrs.
d) 7/8 hrs.
e) none of these
Q.10 If a students walks from his house to school at 5 kmph, he is late by 30 min. However, if he walks at 6 kmph, he is late by 5 min. only. The distance of his school from his house is:
a) 2.5 km
b) 3.6 km
c) 5.5 km
d) 12.5 km
e) none of these
Answers:
Q.1 Option : B
Total Distance = x,
Now, 2/3*x/4 + 1/3*x /5= 42/60
----> 10x+4x/60 = 42/60
---> 14x = 42
x = 3 km.
Q.2 Option : B
Exp : let length of 1st train= x and length of 2nd train = x/2
then, Relative speed = Lt1 + Lt2/time
----> 48+42 = x + x/12
---> 90km/h = 3x/12*2
---> 90*5*2*12/18*3=x
----> x = 200m
Now, St = Lt + Lp/ 45
---> 48*5*45/18 = 200 + Lp
----> 600 = 200 + Lp
----> Lp = 400m
Q.3 Option : C
Exp : 4x+4/3x+4 = 9/7
or 28x + 28 = 27x + 36
or x = 8
2nd conditions, let y yr ago they got married
32-y/24-y = 5/3
96 -3y= 120-5y or y = 12.
Q.4 Option : D
Exp : N A S
40 50 2000
38 48 1824
Now, Let Lowest score = x and Highest= x+ 172
x + x +172 = 2000 - 1824 = 176 or, x = 2
So, Highest score = 2+ 172 = 174.
Q.5 Option : D
Exp : Increase during 3 years= (3*5)years =15 yr.
So, the difference b/w ages of old and new member is 15 years.
Q.6 Option : 1
Exp : The second No. = 12% of 50 = 12/100*50= 6
Let the First no. be x. Then,
2/3*x =6^3
---> x= 6*6*6*3/2 = 324
Req. Sum = 324 + 6 = 330
Q.7 Option : A
Exp : Let 1/2 of the no. = 10x + y
and the no. = 10v + w
From the given conditions,
w = x and v = y - 1
thus the no. = 10 (y-1) + x
therefore, 2(10x + y) = 10(y-1)+x
=> 8y - 19x = 10
v + w = 7 .................... (i)
=> y - 1 + x = 7
therefore, x + y =8 ....................(ii)
after solving equation (i) and (ii), we get
x= 2 and y = 6
hence from equation (a)
No. = 10 (y - 1) + x= 52.
Q.8 Option : C
Exp : Let the numerator and denominator be x and y respectively
Then, x+2/y+3 = 7/9
or, 9(x+2) = 7(y+3)
=> 9x - 7y = 3 .......... (i)
Again, x - 1/y - 1 =4/5
=> 5x - 4y = 1 .......... (ii)
Solving (i) and (ii) we get, x = 5, y= 6
Req. fraction = 5/6.
Q.9. Option : A
Exp: Suppose, usual time = T. If Usual Speed changes to 3/4 then time changes to 4/3T
therefore, 4/3T - T= 2*1/2
which is, => T/3 =5/2
=>T = 15/2 =7*1/2 hr.
Q.10. Option : D
Exp : Time Difference =30-5 = 25min.
therefore, T1- T2 =T . Now, let distance = x.
So, x/5 - x/6 = 25/60 h
=> 6x - 5x /30 = 25/60
=> x = 25/2 = 12.5 kmph
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