Select Exam

Frequestly asked Aptitude question in elitmus test with solution

Question 1 ]
How many 3 digit numbers are there tens digit place is more than hundreds digit place and units place less than hundreds digit place?
        A.     120
        B.     100
        C.     130
        D.     90

Explanation :
we have numbers {0,1,2,3,4,5,6,7,8,9}
now we would decide the cases on behalf of value of hundred place digit. and accordingly choose unit's digit and ten's digit.
value of hundred place digit must be chosen in such a manner that there should exist a greater digit for ten's place and a smaller digit for unit place.
thus we can have values of hundred's digit from 1,2,3,4,5,6,7,8 only
now starting from hundred's digit=8 :
Hundred's digit possible values for ten's digit possible values for unit digit total ways
8 9 0,1,2,3,4,5,6,7 1*1*8=8
7 8,9 0,1,2,3,4,5,6 1*2*7=14
6 7,8,9 0,1,2,3,4,5 1*3*6=18
5 6,7,8,9 0,1,2,3,4 1*4*5=20
4 5,6,7,8,9 0,1,2,3 1*5*4=20
3 4,5,6,7,8,9 0.1.2 1*6*3=18
2 3,4,5,6,7,8,9 0,1 1*7*2=14
1 2,3,4,5,6,7,8,9 0 1*8*1=8
so, total no of cases=8+14+18+20+20+18+14+8 =120
Hence, total no. of possible 3 digit numbers = 120


Question 2]
if log xy2 =a, log x3y =b,find log y/log x = ?
        A.     3a-b/2b-a
        B.     3a-b/2b+a
        C.     3a+b/2b-a
        D.     3a-b/2b

Explanation :
Given, log xy2 = a,
=> log x + log y^2 = a
=> logx + 2logy = a
=> logx = a - 2logy ----------(1)
=> logy = a - logx/2 ----------(2)

Similarly,
logx3 + logy = b
=> 3logx + logy = b -----------(3)
=> 3(a-2logy) + logy = b -------(from 1)
=> 3a - 6logy + logy = b
=> 3a - b = 5logy
=> logy = 3a - b/5.

Putting value of logy in eq(3),
=> 3logx+a-logx/2=b
=> 6logx+a-logx=2b
=> 5logx=2b-a
=> logx=2b-a/5
Hence, logy/logx=3a-b/2b-a.


Question 3 ]
What is the difference between highest ever increasing number and second highest ever increasing number? (eg. 19,12345,189..etc)
        A.     1
        B.     10
        C.     100000
        D.     1,111,11,111

Explanation :
Highest ever increasing number will be 123456789,
and the second highest number like this will be 012345678
Hence, their difference is 123456789 - 012345678 = 111111111.


Question 4 ]
Maximum value of N for {(1008)-(1111122)} is divisible by 3N. What will be maximum value for N?
        A.     6
        B.     2
        C.     1
        D.     0

Explanation :
Given, [(100)4]2-(111112)2
=> a2-b2=(a+b)(a-b) = 98888888 * 100111112.
Here, sum of digits i.e 9+9+8+8+8+8+8+8=66 which is divisible by 3 but not 9.
Hence, the value of 'n'should be 1.


Question 5]
In a party men drink only beer,women drink only wine,sons drink only orange juice & daughters drink only apple juice.There are 1.05L of beers, 1.05L of wine 3.5L of orange juice and 4.9L of apple juice. If every one has the same drinking capacity then what can be the minimum number of people present at the party?
        A.     30
        B.     32
        C.     35
        D.     None of these

Explanation :
Take HCF of 1.05,1.05,3.5,4.9 which is 0.35.
Then divide each quantity by HCF to get the number people drinking that particular beverage i.e 3+3+10+14=30.


Question 6 ]
In a class exactly 10% students get 70 marks, exactly 25% get 80 marks, exactly 15% get 85 marks, exactly 25% get 90 marks and remaining get 95 marks.what is the minimum number of students in the class if the avg. of sum of marks is an even no.
        A.     20
        B.     80
        C.     100
        D.     200

Explanation :
Let total no. of students be x.
=> [x(10*70 + 25*80 + 15*85 + 25* 90 + 25*90)/100]/ x = 86
Here, the average is always 86 and does not depend upon the no. of students.
Hence the answer is the minimum value that is 20.


Question 7 ]
x is the 9 digit largest number(digits are 1 to 9 with out repeat) which is divisible by 11 and y second largest number which also divisible by 11 then difference between x and y ?
        A.     198
        B.     192
        C.     188
        D.     None of these

Explanation :
As we know "sum of digit at even place and sum of digit at odd places must be adjusted such that their difference is 0 or divisible by 11.
we have 9 digit without repetition, so sum of digit from 1 to 9 = 45
Let sum of digit at odd place will be 'w' and sum of even place will be 'Z'.
W + Z= 45
W - Z = 11
2W = 56
So W = 28 and Z = 17
Now the digits at odd place are 2, 4, 6, 7, 9
Sum of digit at odd place is 2+4+6+7+9 = 28
The digits at even place are 1,3,5,8
Sum of digits at even place is 1+3+5+8 =17
So the largest no will be find by put no. of odd and even places in increasing order accordingly
1st largest no is 987563412 and 2nd largest no is 987563214
Hence, the required Difference between them X-Y = 198.


Question 8]
There are 10 pair of socks in a cupboard from which 4 individual socks are picked at random. The probability that there is at least one pair is:
        A.     195/323
        B.     99/323
        C.     198/323
        D.     185/323

Explanation :
The required probability is 1-p(drawing four socks which ar different).
Since, p(drawing four socks different) (20/20)*(18/19)*(16/18)*(14/17) => 224/323.
So, Probability is 1-224/323 = 99/323.


Question 9]
Two squares are chosen on a chessboard at random. What is the probability that they have a side in common?
        A.     1/18
        B.     64/4032
        C.     63/64
        D.     1/9

Explanation :
In 64 squares, there are:
(1) 4 at-corner squares, each has ONLY 2 squares each having a side in common with...
(2) 6*4 = 24 side squares, each has ONLY 3 squares such that each has a side in common with...
(3) 6*6 = 36 inner squares, each has 4 squares such that each has a side in common with...
So we have the calculation:
P = (4/64)*(2/63) + (24/64)*(3/63)+ (36/64)*(4/63)
P = 1/18.

Question 10]
Batting average of 40 innings of a player is 52. His highest score exceeds lowest score by 172 runs.Two innings excluded,average of remaining 38 innings is 40,What is highest score?
        A.     340
        B.     236
        C.     336
        D.     366

Explanation :
Consider highest and lowest score be H and L
Total score in 40 innings =40*52
Total score in 38 innings =38*40
Remaning are lowest and highest scores i.e
=> 40 * 52 - 38 * 40 = H + L = 560 -------(i)
Given that H - L = 172 ------(ii)
By solving eq(i) and eq(ii), we get, H=366.


Question 11 ]
A number is of form abbbbbba , where a is not equal to b, and neither a nor b equals 0(zero). eg. 12222221 . What is the probability that such possible numbers are divisible by 3.
        A.     4/3
        B.     2/3
        C.     1/2
        D.     1/3

Explanation :
According to the question,
'a' can be any of the 9 numbers from (1,2,3,...,9).
'b' can be any of the remaining 8 numbers.
Total: 9×8 = 72 i.e there can be 72 possible numbers.
If the number is divisible by 3 i.e a+b+b+b+b+b+b+a is divisible by 3.
=> 2a + 6b is divisible by 3.
=> 2a is divisible by 3 (because 6b is always divisible by 3).
=> a can be (3,6,9)

Thus, if the number is divisible by 3,
a can be any of the 3 numbers from (3,6,9).
b can be any of the 8 remaining numbers.
Total: 3×8 = 24 i.e there can be 24 possible numbers which are divisible by 3.
The required probability => 24/72 = 1/3.


Question 12 ]
In boring a well 60 m depth, the cost is Rs.4/– for the first meter. The cost of each successive meter increases by Rs.3/–.What is the cost of boring the entire well?
        A.     5414
        B.     4520
        C.     5550
        D.     None of these

Explanation :
According to the question, cost per meter increases as: 4,7,10,13...
They are in A.P. with common difference = 3.
60th term (as well is 60 m) would be= a+(n-1)d=> 4+(59*3)=181
summation Sn= (n/2)(a+l)
=> 30(4 + 181) = 5550.


Question 13]
If the perimeter of a rhombus is 52 cm and one of its diagonals is 24 cm, then what is the area of the rhombus?
        A.     120 sq.cm
        B.     115 sq.cm
        C.     105 sq.cm
        D.     None of these

Explanation :
Perimeter of rhombus = 4a (with side a)
4a=52 then a=13,
lets length of one diagonal p=24cm,
lenght of other diagonal is q;
area of rhombus = pq/2
Using pythogorus theorem a^2= (p/2)^2 + (q/2)^2
We will get q=10.
Then, area = 10*24/2 = 120 sq.cm


Question 14] If 1/x + 1/y = 1/z where x = 12 and y,z positive integer, how many possible value of y and z are there?
        A.     4
        B.     7
        C.     12
        D.     Infinite

Explanation :
We can take the several values of y in 1/12+1/y such that
=> 1/12 + 1/y = 1/z.
=> 12y/12 + y = z.
In this equation, we put positive integer of y and and on many positive integer value of y we will get the many positive value of 'z' so infinite solutions.


Question 15] If a>b>c>d>e>0 and the avg of a & e is 7 and b & d is 4.5.How many possible combination will satisfy the above condition?
        A.     8
        B.     5
        C.     6
        D.     9

Explanation :
According to the question,
a + e = 14.
So, (13,1) (12,2) (11,3) (10,4) (9,5). Between a and e, 3 digits should be there minimum and a>e. So, we can't take other options.
Similarly,
b + d = 9 can be (7,2) (6,3) min 1 digit gap and b>d but (11,4) (10,3) (9,4) do not valid for the above pair of b and d.
(1 2 3 7 13)
(1 2 4 7 13)
(1 2 5 7 13)
(1 2 6 7 13)
(1 3 4 6 13)
(1 3 5 6 13)
(2 3 5 6 12)
(2 3 4 6 12)
Hence, there are total 8 such combinations.


Question 16] Fresh milk contains 90% water while condensed milk contains 20% water. How much of condensed milk can be obtained from 72 liter of fresh milk.
        A.     7.2 liter
        B.     8 liter
        C.     9 liter
        D.     14.4 liter

Explanation :
According to the question, fresh milk contain 72*10/100=7lit( pure milk)
Given, condensed milk contain 20% water & 80% pure milk
Hence,80% of condensed milk=7.2
Then, 100 % ---->condensed milk= 9 liter


Question 17] A five digit number is formed by using digits 1,3,5,7 and 9. without repeating any of them. what is the sum all such possible numbers?
        A.     6666600
        B.     6666660
        C.     6666666
        D.     none of these

Explanation :
According to the formula,
Sum of all such possible numbers=(Sum of all digits)*(n-1)!*(1111....n terms)
=> (1+3+5+7+9)*(5-1)!*(11111)
=> 25*4!*11111
=> 25*24*11111
=> 600*11111
=> 6666600.


Question 18]
ABC is scalene triangle. Its angle A is kept constant, rest sides are tripled in length such that AC'=3AC and AB'=3AB. how much is the area increased in triangle AB'C' from ABC?
        A.     800%
        B.     850%
        C.     200%
        D.     None of these

Explanation :
Let's assume triangle(ABC) having sides 3,4,5 and angle A = 90 degree.
Hence, Area(ABC) = 1/2 * 3* 4 = 6.
Similarly, assume triangle AB'C' where AB' = 3AB, AC' = 3AC' (A/Q)
Hence Area(AB'C) = 1/2 * 9 * 12 = 54.
Difference = 54-6 = 48.
Hencde, the required % Change = 48 * 100/ 6 = 800%.



Question 19 ] How many five digit numbers can be formed by using the digits 0,1,2,3,4,5 such that the number is divisible by 4?
        A.     100
        B.     150
        C.     144
        D.     120

Explanation :
Divisible by 4 means last two digit should be always divided by 4.
Unit place = (04,20,40,12,24,32,52)
_ _ _04
_ _ _ 20
_ _ _40
Total no=4x3x2x3=72 --(i)
_ _ _12
_ _ _24
_ _ _32
_ _ _52
Total no = 3x3x2x4=72 ---(ii)
Hence,the required number is = 72 + 72 = 144.


Question 20 ] If logN base3+logN base9,sum is whole number,then how many number possible for N between 100 to 100?
        A.     1
        B.     20
        C.     111
        D.     None of these

Explanation :
We can write logN base3=(logN baseX/log3 baseX);
and logN base9=(logN baseX/log9 baseX);
Now, logN base3 + logN base9= ( logN baseX / log3 baseX + logN baseX / log9 baseX)
= logN baseX (1/log3 baseX + 1/2log3 baseX) bcz (3 square =9)
= logN baseX ( 3log3 baseX / 2log3 baseX *log3 baseX) by simple addition
= logN base X ( 3/2log3 base X)
= logN base X *3 / log9 base X
= 3 * logN base 9.
So, N should be 81 because it is a 9 square and it also ies between 10 to 100.



Other Useful Questions for elitmus [Try these questions too]

1. In a division sum, the divisor is 10 times the quotient and 5 times
the remainder. If the remainder is 46, the dividend is:
(1) 4236 (2) 4306
(3) 4336 (4) 5336

2. If 1.5 x= 0.04 y, then the value of (y-x) (y+x) is:
(1) 730/77 (2) 73/77
(3) 7.3/77 (4) 703/77

3. An employee may claim Rs. 7.00 for each km when he travels by taxi and
Rs. 6.00 for each km if he drives his own car. If in one week he claimed
Rs. 595 for traveling km. How many kms did he travel by taxi?
(1) 55 (2) 65
(3) 62 (4) 70

4. The square root of 3 + “5 is :
(1) “3 /2 + 1/”2 (2) “3 /2 - 1/”2
(3) “5 /2 - 1/”2 (4) “(5/2) + “(1/2)

5. The mean temperature of Monday to Wednesday was 370C and of Tuesday to
Thursday was 340C, if the temperature on Thursday was 4/5th that of
Monday, then what was the temperature on Thursday?
(1) 36.50C (2) 360C
(3) 35.50C (4) 340C

6. A certain number of two digits is three times the sum of its digits.
If 45 be added to it, the digits are reversed. The number is:
(1) 72 (2) 32
(3) 27 (4) 23

7. Three years ago the average age of A and B was 18 years. While C
joining them now, the average becomes 22 years. How old (in years) is C
now?
(1) 24 (2) 27
(3) 28 (4) 30

8. If 2^(2x-1) = 8^(3-x), then the value of x is:
(1) -1 (2) -2 
(3) 2 (4) 3

9. A man’s basic pay for a 40 hours’ week is Rs. 200. Overtimes is paid
at 25% above the basic rate. In a certain week, he worked overtime and
his total was Rs. 300. He therefore, worked for a total of (in hours):
(1) 52 (2) 56
(3) 58 (4) 62

10. On a Rs. 10, 000 payment order, a person has choice between 3
successive discounts of 10%, 10% and 30% and 3 successive discounts of
40%, 5% and 5%. By choosing the better one he can save (in Rupees):
(1) 200 (2) 255
(3) 400 (4) 433

11. Rs. 600 are divided among A, B, C so that Rs. 40 more than 2/5 th of
A’s share, Rs. 20 more that 2/7 th of B’s share and Rs. 10 more than 9/17
th of C’s may all be equal. What is A’s share (in Rupees)?
(1) 150 (2) 170
(3) 200 (4) 280

12. A, B, C started a business with their investment in the ratio 1 : 3 :
5. After 4 months, A invested the same amount as before and B as well as
C withdrew half of their investments. The ratio of their profits at the
end of the year was:
(1) 5 : 6 : 10 (2) 6 : 5 :10
(3) 10 : 5 : 6 (4) 4 : 3 : 5

13. If 9 men working 71/2 hours a day can finish a piece of work in 20
days, then how many days will be taken by 12 men, working 6 hours a day
to finish the work? It is being given that 2 men of latter type work as
much as 3 men of the former type?
(1) 91/2
(2) 11
(3) 121/2
(4) 13

14. Three pipes A, B and C can fill a cistern in 6 hours. After working
at it together for 2 hours, C is closed and A and B can fill the
remaining part in 7 hours. The number of hours taken by C alone to fill
the cistern is:
(1) 12 (2) 14
(3) 16 (4) 18

No comments:

Post a Comment