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DIRECTIONS: Read the following given below and answer the questions that follow.
i) Eleven students Anwar, Baham, Chetan, Dayal, Eshwar, Farooq, Gajendra, Hariom, Inayat,
Jatin and Kishore are sitting on a bench in a lecture room facing towards the teacher.
ii) Dayal is towards the left of Farooq and second to the right of Chetan.
iii) Eshwar is second to the left of Anwar and sitting on one end.
iv) Jatin is the neighbourer of Anwar and Baham and is third to the left of
Gajendra.
v) Hariom is next to left of Dayal and is third to the right of Inayat.
1) Which two students are sitting in two ends?
a) Kishore and Dayal b) Eshwar and Dayal
c) Eshwar and Farooq d) Kishore and Farooq
2) Which group of students are sittting just next to the right of Gajendra?
a) Chetan, Hariom, Dayal, Eshwar b) Chetan, Hariom, Inayat, Baham
c) Chetan, Hariom, Inayat, Dayal d) none of the above
3) Who is sitting just in the middle?
a) Inayat b) Chetan c) Baham d) Jatin
4) Which of the five given statements is unnecessary?
a) (i) and (ii) b) (i) c) (iii) d) All four are necesssary
5) Which of the following is correct?
a) Eshwar and Anwar are the closest neighbourers of Jatin
b) Gajendar, Inayat and Baham are sitting to the left of Chetan
c) Hariom is in the middle of the line
d) Anwar, Kishore and Eshwar are to the right of Jatin
Ques. A man sold two cows for Rs. 210 at a total profit of 5 %. He sold one cow at a loss
of 10% and another at a profit of 10%. What is the price of each cow?
- 50, 100
- 100, 50
- 150, 100
- 50, 150
Ans :50 and 150
Solution : after profit, price is 210
so cost price of 2 cows = 200
x+y=200
selling price of 2 cows =210
0.9x + 1.1y =210 //1 cow sold at 10 % loss and another is at 10% profit
solving both equations x=50 and y=150
so one cow is of x = Rs. 50
and Other is of 200-x = (200-50) = Rs. 150
Ques. In a class there are less than 500 students . when it is divided by 3 it gives a
whole number. Similarly when it is divided by 4,5 or 7 gives a whole number. find the
no. of students in the class
- 430
- 420
- 410
- 440
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?
Ans: 14/3 days.
Sol: 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 = 9×189+18 = 6 days.
A and B together can complete the work = 2×42+4 = 4/3 days.
Difference = 6 – 4/3 = 14/3 days.
How many 4 digit numbers contain number 2.
a. 3170
b. 3172
c. 3174
d. 3168
Ans: D
Sol:
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
How many three digit numbers abc are formed where at least two of the three digits
are same.
Ans: 252
Sol:
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
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? Ans: 12
Sol:
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×24+30×18.4x+30=20 Solving we get x = 12
What is the next number of the following sequence 7, 14, 55, 110, ….?
Ans: 121
Sol:
Next number = Previous number + Reverse of previous number So
7 ,7+7=14, 14+41 = 55, 55+55 = 110, 110+011 = 121
How many numbers are divisible by 4 between 1 to 100
Ans: 24
Sol: 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.
(11111011)2= ()8
Ans: 373
Sol: 11111011)2=(251)10=(373)8 or
You can group 3 binary digits from right hand side and write their equivalent octal form.
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?
Ans: 714 / 80000
Sol:
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
161?85?65?89 = 100, then use + or – in place of ? and take + as m,- as n then find
value of m-n.
Ans: -1
Sol:
161 – 85 – 65 + 89 = 100
so m’s =1, n’s = 2 => (m – n)= – 1
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?
Ans: 60
Sol: 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.
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
Ans: 31
Sol:
Let the total no of participants including Rahul = x Excluding rahul=(x-1)
15(x−1)+56(x−1) = x
31x – 31=30x
Total no. of participants x =31
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.
Ans:
Sol:
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) = ⇒7c65c2⇒710=70
For (5, 3) = ⇒7c55c3⇒2110=210
For (4, 4) = ⇒7c45c4⇒355=175 So Total ways = 70+210+175=455
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.
Ans: 256
Sol:
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
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.
Ans: 2375
Sol:
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
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?
Ans: 2772
Sol:
We can select only one woman, and remaining 6 from men. So 12C6×3C1
= 2772
Tennis players take part in a tournament. Every player plays twice with each of his
opponents. How many games are to be played?
Ans: 210
Sol:
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.
Find the unit digit of product of the prime number up to 50 .
Ans: 0
Sol: No need to write all the primes upto 50. There are two primes 2, 5 gives unit digit of
So the entire product has unit digit 0.
If [x^(1/3)] – [x^(1/9)] = 60 then find the value of x.
Ans: 49
Sol:
Let t = x1/9
So,
t3−t=60
Therefore, (t-1) x t x (t + 1) = 60 =3 x 4 x 5. therefore, t = x1/9 =4.
hence, x = 49
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?
Ans: 18
Sol:
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
