34730.Jake left point A for point B. 2 hours and 15 minutes later, Paul left A for B and arrived at B at the same time as Jake. Had both of them started simultaneously from A and B travelling towards each other, they would have met in 120 minutes. How much time (hours) did it take for the slower one to travel from A to B if the ratio of speeds of the faster to slower is 3:1?
202.5 minutes
205.4 minutes
208.5 minutes
201.5 minutes
Explanation:
It seems there is some problem with this question.
Let the distance between A and B is D km. As Paul is faster, take the speeds of Jake and Paul
are s and 3s kmph.
As the speeds are in the ratio of 1 : 3, times taken by them should be 3 : 1.
Take the times taken by them are 3x , x. But We know that 3x - x = 2 hour 15 min. So 2x = 9/4 hours, x = 9/8 hours.
So time taken by the slower one (Jake) takes 3x time = 3 x 9/8 = 27/8 hours = 202.5 minutes.
It seems there is some problem with this question.
Let the distance between A and B is D km. As Paul is faster, take the speeds of Jake and Paul
are s and 3s kmph.
As the speeds are in the ratio of 1 : 3, times taken by them should be 3 : 1.
Take the times taken by them are 3x , x. But We know that 3x - x = 2 hour 15 min. So 2x = 9/4 hours, x = 9/8 hours.
So time taken by the slower one (Jake) takes 3x time = 3 x 9/8 = 27/8 hours = 202.5 minutes.
34731.A completes a work in 2 days, B in 4 days, C in 9 and D in 18 days. They form group of two such that difference is maximum between them to complete the work. What is difference in the number of days they complete that work?
14/5 days
14/3 days
16/2 days
18/4 days
Explanation:
If C and D form a pair and A and B form a pair the difference is maximum.
Now C and D together can complete the work = 9 $\times$ 189+18 = 6 days.
A and B together can complete the work = 2$\times$ 42+4 = 4/3 days.
Difference = 6 - 4/3 = 14/3 days.
If C and D form a pair and A and B form a pair the difference is maximum.
Now C and D together can complete the work = 9 $\times$ 189+18 = 6 days.
A and B together can complete the work = 2$\times$ 42+4 = 4/3 days.
Difference = 6 - 4/3 = 14/3 days.
34732.How many 4 digit numbers contain number 2.
3170
3172
3174
3168
Explanation:
Total number of 4 digit numbers are 9000 (between 1000 and 9999).
We find the numbers without any two in them. So total numbers are 8 x 9 x 9 x 9 = 5832
So numbers with number two in them = 9000 - 5832 = 3168
Total number of 4 digit numbers are 9000 (between 1000 and 9999).
We find the numbers without any two in them. So total numbers are 8 x 9 x 9 x 9 = 5832
So numbers with number two in them = 9000 - 5832 = 3168
34733.How many three digit numbers abc are formed where at least two of the three digits are same.
260
285
252
253
Explanation:
Total 3 digit numbers = 9 x 10 x 10 = 900
Total number of 3 digit numbers without repetition = 9 x 9 x 8 = 648
So number of three digit numbers with at least one digit repeats = 900 - 648=252.
Total 3 digit numbers = 9 x 10 x 10 = 900
Total number of 3 digit numbers without repetition = 9 x 9 x 8 = 648
So number of three digit numbers with at least one digit repeats = 900 - 648=252.
34734.How many kgs of wheat costing Rs.24/- per kg must be mixed with 30 kgs of wheat costing Rs.18.40/- per kg so that 15% profit can be obtained by selling the mixture at Rs.23/- per kg?
15
16
18
12
Explanation:
S.P. of 1 kg mixture = Rs.23. Gain = 15%.
C.P. of 1 kg mixture = Rs.[(100/115) x 23] = Rs.20
Let the quantity of wheat costing Rs.24 is x kgs.
Using weighted average rule = x $\times$ 24+30 $\times$ 18.4x+30=20
Solving we get x = 12
S.P. of 1 kg mixture = Rs.23. Gain = 15%.
C.P. of 1 kg mixture = Rs.[(100/115) x 23] = Rs.20
Let the quantity of wheat costing Rs.24 is x kgs.
Using weighted average rule = x $\times$ 24+30 $\times$ 18.4x+30=20
Solving we get x = 12
34735.What is the next number of the following sequence 7, 14, 55, 110, ....?
121
128
126
124
Explanation:
Next number = Previous number + Reverse of previous number
So
7 ,7+7=14, 14+41 = 55, 55+55 = 110, 110+011 = 121
Next number = Previous number + Reverse of previous number
So
7 ,7+7=14, 14+41 = 55, 55+55 = 110, 110+011 = 121
34736.How many numbers are divisible by 4 between 1 to 100
22
24
26
28
Explanation:
There are 25 numbers which are divisible by 4 till 100. (100/4 = 25). But we should not
consider 100 as we are asked to find the numbers between 1 to 100 which are divisible by 4.
So answer is 24.
There are 25 numbers which are divisible by 4 till 100. (100/4 = 25). But we should not
consider 100 as we are asked to find the numbers between 1 to 100 which are divisible by 4.
So answer is 24.
34737.$\left(11111011\right)_{2}$ = $\left(\right)_{8}$
453
358
373
473
Explanation:
$\left(11111011\right)_{2}$ =$ \left(251\right)_{10}$= $\left(373\right)_{10}$
or
You can group 3 binary digits from right hand side and write their equivalent octal form.
$\left(11111011\right)_{2}$ =$ \left(251\right)_{10}$= $\left(373\right)_{10}$
or
You can group 3 binary digits from right hand side and write their equivalent octal form.
34738.There are 1000 junior and 800 senior students in a class.And there are 60 sibling pairs where each pair has 1 junior and 1 senior. One student is chosen from senior and 1 from junior randomly.What is the probability that the two selected students are from a sibling pair?
845 / 50000
714 / 80000
741 / 40000
854 / 50000
Explanation:
Junior students = 1000
Senior students = 800
60 sibling pair = 2 x 60 = 120 student
One student chosen from senior = 800C1
=800
One student chosen from junior=1000C1=1000
Therefore, one student chosen from senior and one student chosen from junior n(s) = 800 x 1000=800000
Two selected students are from a sibling pair n(E)=120C2=7140
therefore,P(E) = n(E) / n(S)=7140/800000
= 714/80000
Junior students = 1000
Senior students = 800
60 sibling pair = 2 x 60 = 120 student
One student chosen from senior = 800C1
=800
One student chosen from junior=1000C1=1000
Therefore, one student chosen from senior and one student chosen from junior n(s) = 800 x 1000=800000
Two selected students are from a sibling pair n(E)=120C2=7140
therefore,P(E) = n(E) / n(S)=7140/800000
= 714/80000
34739.161?85?65?89 = 100, then use + or - in place of ? and take + as m,- as n then find value of m-n.?
-1
1
0
none of these
Explanation:
161 - 85 - 65 + 89 = 100
so m`s =1, n`s = 2 => (m - n)= - 1
161 - 85 - 65 + 89 = 100
so m`s =1, n`s = 2 => (m - n)= - 1
34740.In a cycle race there are 5 persons named as J,K,L,M,N participated for 5 positions so that in how many number of ways can M finishes always before N?
40
80
30
60
Explanation:
Total number of ways in which 5 persons can finish is 5! = 120 (there are no ties)
Now in half of these ways M can finish before N.
Total number of ways in which 5 persons can finish is 5! = 120 (there are no ties)
Now in half of these ways M can finish before N.
34741.Rahul took a part in cycling game where 1/5 ahead of him and 5/6 behind him excluding him. Then total number of participants are
35
31
36
37
Explanation:
Let the total no of participants including Rahul = x
Excluding rahul=(x-1)
$\dfrac{1}{5}\left(x-1\right)$ + $\dfrac{5}{6}\left(x-1\right)$= x
31x - 31=30x
Total no. of participants x =31
Let the total no of participants including Rahul = x
Excluding rahul=(x-1)
$\dfrac{1}{5}\left(x-1\right)$ + $\dfrac{5}{6}\left(x-1\right)$= x
31x - 31=30x
Total no. of participants x =31
34742.If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans. In how many ways can we remove 8 cans so that atleast 1 blue can and 1 red can remains in the refrigerator.
455
486
475
445
Explanation:
Possible ways to draw 8 balls from the refrigerator which contains atleast 1 blue and 1 red can
after the drawing are (6,2) (5,3) (4,4).
For (6, 2) $\Rightarrow$ 7c6 $\times$ 5c2 $\Rightarrow$ 7 $\times$ 10=70
For (5, 3) $\Rightarrow$ 7c5 $\times$ 5c3 $\Rightarrow$ 21$\times$ 10=210
For (4, 4) $\Rightarrow$ 7c4 $\times$ 5c4 $\Rightarrow$ 35 $\times$ 5=175
So Total ways = 70+210+175=455
Possible ways to draw 8 balls from the refrigerator which contains atleast 1 blue and 1 red can
after the drawing are (6,2) (5,3) (4,4).
For (6, 2) $\Rightarrow$ 7c6 $\times$ 5c2 $\Rightarrow$ 7 $\times$ 10=70
For (5, 3) $\Rightarrow$ 7c5 $\times$ 5c3 $\Rightarrow$ 21$\times$ 10=210
For (4, 4) $\Rightarrow$ 7c4 $\times$ 5c4 $\Rightarrow$ 35 $\times$ 5=175
So Total ways = 70+210+175=455
34743.There are 16 people, they divide into four groups, now from those four groups select a team of three members,such that no two members in the team should belong to same group.
268
286
265
256
Explanation:
We can select any three of the 4 groups in 4C3
ways. Now from each of these groups we can select 1 person in 4 ways.
So total ways = 4 x 4 x 4 x 4 = 256
We can select any three of the 4 groups in 4C3
ways. Now from each of these groups we can select 1 person in 4 ways.
So total ways = 4 x 4 x 4 x 4 = 256
34744.How many five digit numbers are there such that two left most digits are even and remaining are odd and digit 4 should not be repeated.
2578
2365
2475
2375
Explanation:
We have
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
But 44 is one case we have to omit. So total ways for leftmost two digits are 4 x 5 - 1 = 19
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}
So total ways = 19 x 5 x 5 x 5 = 2375
We have
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
But 44 is one case we have to omit. So total ways for leftmost two digits are 4 x 5 - 1 = 19
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}
So total ways = 19 x 5 x 5 x 5 = 2375
34745.7 people have to be selected from 12 men and 3 women, Such that no two women can come together. In how many ways we can select them?
2562
2786
2896
2772
Explanation:
We can select only one woman, and remaining 6 from men.
So 12C6$\times$3C1
= 2772
We can select only one woman, and remaining 6 from men.
So 12C6$\times$3C1
= 2772
34746.Tennis players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?
210
250
240
220
Explanation:
We can select two teams out of 15 in 15C2 ways. So each team plays with other team once.
Now to play two games, we have to conduct 15C2 x 2 = 210 games.
We can select two teams out of 15 in 15C2 ways. So each team plays with other team once.
Now to play two games, we have to conduct 15C2 x 2 = 210 games.
34747.Find the unit digit of product of the prime number up to 50 .
0
1
10
none of these
Explanation:
No need to write all the primes upto 50. There are two primes 2, 5 gives unit digit of 0.
So the entire product has unit digit 0.
No need to write all the primes upto 50. There are two primes 2, 5 gives unit digit of 0.
So the entire product has unit digit 0.
34748.If [$x^{1/3}$] - [$x^{1/9}$] = 60 then find the value of x.
$4^8$
$6^8$
$4^9$
$5^8$
Explanation:
Let t = $x^{1/9}$
So,
t3-t=60
Therefore, (t-1) x t x (t + 1) = 60 =3 x 4 x 5.
therefore, t = $x^{1/9}$ =4.
hence, x = $4^9$
Let t = $x^{1/9}$
So,
t3-t=60
Therefore, (t-1) x t x (t + 1) = 60 =3 x 4 x 5.
therefore, t = $x^{1/9}$ =4.
hence, x = $4^9$
34749.A family X went for a vacation. Unfortunately it rained for 13 days when they were there.But whenever it rained in the mornings, they had clear afternoons and vice versa. In all they enjoyed 11 mornings and 12 afternoons. How many days did they stay there totally?
18
16
14
12
Explanation:
Total they enjoyed on 11 mornings and 12 afternoons = 23 half days
It rained for 13 days. So 13 half days.
So total days = (13 + 23) / 2 = 18
Total they enjoyed on 11 mornings and 12 afternoons = 23 half days
It rained for 13 days. So 13 half days.
So total days = (13 + 23) / 2 = 18