Showing posts with label eLitmus Aptitude Question. Show all posts

Thursday, December 28, 2017

Elitmus Previous Years Questions On Number System

SET 1

Number System 01

01.

If n is any odd number greater than 1, then n(n2 - 1) is always divisible by?

(a) 96

(b) 48

(d) None of these

02.

If a number 774958A96B is to be divisible by 8 and 9, the respective values of A and B will be

(a) 7 and 8

(b) 8 and 0

(d) None of these

03.

Three consecutive positive even numbers are such that thrice the first number exceeds double the third by 2, the third number is?

(a) 10

(b) 14

(d) 12

04.

Three bells chime at intervals of 18 min, 24 min and 32 min respectively. At a certain time, they begin to together, What length of time will elapse before they chime together again?

(a) 2 h and 24 min

(b) 4h and 48 min

(d) 5h

05.

Two positive integers differ by 4 and sum of their reciprocals is 10/21. Then, one of the numbers is?

(a) 3

(b) 1

(d) 21

06.

(56-1) is divisible by?

(a) 13

(b) 31

(d) None of these

07.

The remainder obtained when a prime number greater that 6 is divided by 6 is

(a) 1 or 3

(b) 1 or 5

(d) 4 or 5

08.

For the product n(n+1)(2n+1) , which one of the following is not necessarily true?

(a) It is even

(b) Divisible by 3

(d) Never divisible by 237

09.

72 Hens Rs …96.7.. Then what does each hen cost, where two digits in place of “…. , …..”are not visible written in illegible hand-writing?

(a) Rs 3.23

(b) Rs 5.11

(d) Rs 7.22

10.

Which is the least number that must be subtracted from 1856 so that the remainder when divided by 7, 12, 16, is 4?

(a) 137

(b) 1361

(d) 172

Answers : Find Detailed Solutions at the end of the the page.

1. C

2. B

3. B

4. B

5. A

6. B

7. B

8. D

9. C

10. D

SET 2

Number System 02

01.

The number of common terms in the two sequences 17, 21, 25, ….. , 417 and 16, 21, 26, …,466 is ?

(a) 19

(b) 20

(d) 22

02.

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed ??

(a) 374

(b) 500

(d) 376

03.

What is the number of distinct terms in the expansion of (a + b + c)20 ?

(a) 231

(b) 253

(d) 210

04.

What are the last two digits of 72008 ?

(a) 21

(b) 61

(d) 41

05.

The integers 1, 2, 3, … 40 are written on blackboard. The following operation is then repeated 39 times. In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end ?

(a) 820

(b) 821

(d) 819

06.

An intelligence agency decides on a code of 2 digits selected from 0, 1, 2, …, 9. But on the slip on which the code is hand written allows confusion between top and bottom, because there are indistinguishable. Thus, for example, the code 91 could be confused with 16. How many codes are there such that there is no possibility of any confusion?

(a) 25

(b) 75

(d) None of these

07.

A young girl counted in the following way on the fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8 and thumb 9 and then back to the index finger for 10, middle finder for 11 and so on. She counted up to 1994. She ended on her ?

(a) thumb

(b) index finger

(d) ring finger

08.

Let U(n+1)= 2Un + 1, ( n=0, 1, 2, …………) U0=0 then U(10) would be nearest to ?

(a) 1023

(b) 2047

(d) 8195

09.

The product of all integers from 1 to 100 will have the following numbers of zeros at the end ?

(a) 20

(b) 24

(d) 22

10.

The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is ?

(a) 26

(b) 18

(d) None of these

Answers : Find Detailed Solutions at the end of the the page.

1. B

2. D

3. A

4. C

5. C

6. C

7. B

8. A

9. B

10. A

Number System Solution 01

Both the sequences ( 17, 21, 25 ……………… and (16, 21, 26….. are arithmetic progression with a common difference of 4 and 5 respectively.

In both the sequence first common term is 21.

Hence a new arithmetic sequence containing the common terms of both the series can be formed with a common difference of LCM of ( 4, 5) is 20

New sequence will be 21, 41, 61, ….401

nth term = a + (n-1)d

401= 21 + (n-1)20

n-1=19

Hence, n=20

(b)

The number required is greater than 999 and less than and equal to 4000.

Now out off our digits, 0, 1, 2, 3, 4.

To form a number greater than 999 and less than 4000.

The digit at thousands place can be selected in 3 ways( 0 and 4 cannot be taken )

The digit at hundreds place can be selected in 5 ways.

The digit at tens place can be selected in 5 ways.

Total required number of ways= 3 x 5 x 5 x 5 = 375 ways

Since, 4000 is also one of the required number.

Therefore, total number of ways=375 + 1

(d)

According to question, if two numbers say a and b are erased and replaced by a new number a + b – 1, then in every repetition, the number of integers gets reduced by 1 and consequently at the last repetition there will be only one number left.

Whatever may be our selection of two numbers a and b. In any and every repetition, the final number so arrived will not changes.

Now, the sum of integers from 1 to 40= n(n + 1)/2 = 820

As, discussed above the sum of integers of the first, second, third ……… repetitions will be 819, 818, 817, ………….. so on respectively. Therefore, after 39 operations there will be only 1 number left and that will be 820 – 39=781

(c)

Thumb Finger :1, 9, 17, ….

Index Finger: 2, 8, 10, 16, 18, ….

Middle Finger: 3, 7, 11, 16, 19, ….

Ring Finger : 4, 6, 12, 14, 20, ……

Little Finger : 5, 13, 21, ….

Numbers on thumb forms an AP with common difference=8

Numbers on middle forms an AP with common difference=4

1993 will be on the thumb.

Hence, 1994 will be on the index finger.

U(n+1)= 2Un + 1, ( n=0, 1, 2, …………) U0=0

Put

n=0, U1=1

n=1, U2=3

n=2, U3=7

n=4, U5=31

Seeing this pattern we can conclude i.e. U(n)= 2n -1

Hence U(10)= (2)10 - 1= 1023

Every combination of 5 and 2 will give one zero, and number of zero in the product of any number is decided by the number of 2 and 5, whichever is less.

Hence, this problem can be solved by determining the number of 2 and 5 between 1 to 100.

Clearly, there are 20 numbers which are divisible by 5. Besides, there are four numbers 25, 50, 75, and 100 which will have one addition 5. Hence, number of zeroes in the product of all the numbers from 1 to 100 is 21.

(b)

10.

There are 50 odd numbers less than 100 which are not divisible by 2. Out of these 50 there are 17 number which are divisible by 3.

Out of remaining there are 7 numbers which are divisible by 5.

Hence, numbers which are not divisible by 2, 3, 5 = 50 – 17 – 7 = 26

Number System Solution 02

We have use put different values of n ( odd numbers ) greater than 1.

i.e. n=3, 5, 7, 9

When n=3 n(n2 - 1)= 24

When n=5 n(n2 - 1)= 120

When n=7 n(n2 - 1)=336

using options we find that all the numbers are divisible by 24

(c)

Numbers are divisible by 8 if the number formed by the last three digits is evenly divisible by 8.

Numbers are divisible by 9 if the sum of all the individual digit is divisible by 9.

The number 774958A96B is divisible by 8 if 96B is divisible by 8. And 96B is divisible by 8 if B is either 0 and 8. Now to make the same number divisible by 9 sum of all the digits should be divisible by 9. Hence, (55 + A + B) is divisible by 9 if (A + B) is either 0 or 8

i.e. either A = 0 or B = 8 or A = 8 or B = 0

Since, the number is divisible by both A and B. Hence, A and B may take either values i.e. 8 and 0

(b)

Let the three even number are ( x – 2), x, (x + 2)

Then, 3(x – 2) – 2(x + 2) = 2

3x – 6 – 2x – 4 = 2 i.e. x=12

Hence, the third number is (12 + 2) = 14

(b)

We have to take the L.C.M of 18, 24 and 32 i.e. 288 min.

Hence, bells will chime together again after 4 Hours and 48 Minutes.

(b)

(56 - 1) = (53)2 - (1)2 = (125)2 - (1)2

=(125 + 1)(125 – 1) = 126 x 124=31 x 2 x 2 x 126

Here we can easily conclude. ( 56-1) is divisible by 31

(b)

We have take some prime number greater than 6 i.e. 7, 11, 13, 17, 19, 23, 29,31, 37, 41

Now we have divide the numbers by 6. The remainder is always either 1 or 5.

(b)

We have to check for each option separately by taking the different values of n.

Option (a) : Check for n= 3, 4, 5, 6…

Option (b) : Check for n= 3, 4, 5, 7…

Option (c) : Divisible

Option (d) : For n= 237 : n(n + 1)(2n + 1) is divisible.

(d)

We have to check for each option.

(a) 3.23 x 72 = 232.56

(b) 5.11 x 72 = 367.92

(d) 7.22 x 72 =519.84

Option (c) is nearest to Rs. …..96.7…

(c)

10.

Firstly, we have to take LCM of 7, 12, 16 = 336

If we divide 1856 by 336, then remainder is 176. Since it is given that remainder in this condition is 4. Hence, the lease number to be subtracted = (176 – 4)172.

(d)

Elitmus Previous Years Questions on Permutations and Combination

Permutations And Combinations 01

01.Five-digit numbers are formed using only 0, 1, 2, 3, 4 exactly once. What is the difference between the greatest and smallest numbers that can be formed?

(a) 19800

(b) 41976

(c) 32976

(d) None of these

02.In how many ways can Eight Directors, Vice-Chairman and Chairman of a firm be seated at a round table, if the Chairman has to sit between the Vice-Chairman and a Director?

(a) 9! x 2

(b) 2 x 8!

(c) 2 x 7!

(d) None of these

03. A man has 9 friends: 4 boys and 5 girls. In how many ways can he invite them, if there have to be 3 exactly girls in the invitees?

(a) 320

(b) 160

(c) 80

(d) 200

04.

Boxes numbered 1, 2, 3, 4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?

(a) 8

(b) 10

(c) 15

(d) 22

05.

A, B, C, D are four towns, any three of which are non-collinear. Then, the number of ways to construct three roads each joining a pair of towns so that the roads do not form a triangle is?

(a) 7

(b) 8

(c) 9

(d) 24

06.

If a 4-digit number is formed with digits 1, 2, 3 and 5. What is the probability that the number is divisible by 25, if repetition of digits is not allowed?

(a) 1/12

(b) 1/4

(c) 1/6

(d) None of these

07. A five digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?

(a) 6666600

(b) 6666660

(c) 6666666

(d) None of these

08.139 persons have signed for an elimination tournament. All players are to be paired up for the first round, but because 139 is an odd number one player gets a bye, which promotes him to the second round, without actually playing in the first round. The pairing continues on the next round, with a bye to any player left over. If the schedule is planned so that a minimum number of matches is required to determine the champion, the number of matches which must be played is

(a) 136

(b) 137

(c) 138

(d) 139

09.A box contains 6 red balls, 7 green balls and 5 blue balls. Each ball is of different size. The probability that the red ball selected is the smallest red ball?

(a) 1/18

(b) 1/3

(c) 1/6

(d) 2/3

10. A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

(a) 3

(b) 4

(c) 5

(d) 6

Answers :

1. C

2. B

3. B

4. D

5. D

6. A

7. A

8. C

9. C

10. D

Permutation And Combinations 02

01. In a chess competition involving some boys and girls of a school, every student had to play exactly one game with the every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is?

(a) 200

(b) 216

(c) 235

(d) 256

02.

A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is?

(a) 12 x 81

(b) 16 x 192

(c) 20 x 125

(d) 24 x 216

03.

An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …. 9 such that the first digit of the code is non-zero. The code, handwritten on a slip, can however potentially create confusion, when read upside down- for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?

(a) 80

(b) 78

(c) 71

(d) 69

04.

How many numbers can be made with digits 0, 7, 8 which are greater than 0 and less than a million?

(a) 496

(b) 486

(c) 1084

(d) 728

05.

In how many ways is it possible to choose a white square and a black square on a chess-board so that the squares must not lie in the same row or column?

(a) 56

(b) 896

(c) 60

(d) 768

06.

How many four-letter computer passwords can be formed using only the symmetric letters. (no repetition allowed) (Symmetric letters :- A, H, I, M, O, T, U, V, W, X AND Z)

(a) 7920

(b) 330

(c) 14640

(d) 419430

07.

How many three-letter computer password can be formed with at least one symmetric letter ? (Symmetric letters: A, H, I, M, O, T, U, V, W, X and Z)

(a) 990

(b) 2730

(c) 12870

(d) 1560000

08.

For a scholarship, at the most n candidates out of 2n+1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is ?

(a) 3

(b) 4

(c) 6

(d) 5

09.

Ten points are marked on a straight line and 11 points are marked on another straight line. How many triangles can be constructed with vertices from among the above points?

(a) 495

(b) 550

(c) 1045

(d) 2475

10.

How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit at unit’s place must be greater than the in the ten’s place?

(a) 54

(b) 60

(c) 17

(d) 2 x 4!

Answers :Find Detailed Solutions at the end of the page

1. A

2. A

3. D

4. D

5. D

6. A

7. C

8. A

9. C

10. B

Set 1 Detailed Solutions
1.
Greatest five digit number : 43210
Smallest five digit number : 10234
Difference = 43210-10234= 32976
(c)
2. We consider vice-chairman and the chairman as 1 Unit. Now, 9 persons can be arranged along a circular table in 8! ways. And vice-chairman and chairman can be arranged in 2 different ways. Hence required number of ways=2 x 8!
(b)
3. 3 Girls can be selected out of 5 girls in 5C3 ways.
Since number of boys to be invited is not given, hence out of 4 boys, he can invite them (2)4 ways.
Hence required number of ways is = 5C3 x (2)4 = 160
(b)
4.Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls=25=32.
Two adjacent boxes with blue can be got in 4 ways.
i.e. (1, 2) (2, 3) (3, 4) (4, 5)
Three adjacent boxes with blue can be got in 3 ways
i.e. (1, 2, 3), (2, 3, 4) and (3, 4, 5)
Four adjacent boxes with blue can be be got in 2 ways.
i.e. (1234) , (2345)
and five boxes with blue can be got in 1 way.
Hence total number of ways of filling the boxes such that adjacent boxes have blue
(4 + 3 + 2 + 1)=10
(d)
5. To construct 2 roads, three towns can be selected out of 4 in 4 x 3 x 2 = 24 ways. Now, if the third road goes from the third town to the first town, a triangle is formed and if it goes to the fourth town, a triangle is not formed. So there are 24 ways to form a triangle and 24 ways of avoiding a triangle.
(d)
6. Total number of 4 digit numbers that can be formed = 4!. If the number is divisible by 25, then the last two digit are 25. So the first two digits can be arranged in 2! ways.
Hence required probability = 2!/4! = 1/12
(a)
7. Keeping one digit in fixed position, other four can be arranged in 4! ways= 24 ways. Thus each of the 5 digits will occur in each of the five place 4! times. Hence the sum of digits in each position is 24(1 + 3 + 5 + 7 + 9) = 600. So, the sum of all numbers
=600(1+10+100+1000+10000)=6666600
(a)
8. Required number of matches played will be (139 – 1)=138
(c)
9. Required probability is = 1/6
(c)
10. Let the number of students in the front row be x and the number of rows be n.
Hence, number of students in the next rows would be (x – 3) , (x – 6), (x – 9),……. and so on.
Now we have to check for each value of n=3, 4, 5, 6
Firstly take n=3
x + (x – 3) + (x – 6) =630
» 3x = 639 i.e. x = 213 ( Thus, n=3 is possible)
Likewise if n=4
x + (x – 3) + (x – 6)+(x – 9) =630
» 4x – 18 = 630 i.e. x = 162 ( Thus, n=4 is possible)
Likewise if n=5
x + (x – 3) + (x – 6)+(x – 9) +( x -12) =630
» 5x – 30 = 630 i.e. x = 132 ( Thus, n=5 is possible)
Likewise if n=6
x + (x – 3) + (x – 6)+(x – 9)+(x -12)+(x – 15)=630
» 6x – 45 = 630 i.e. x =112.5 (NOT AN INTEGER) ( Thus, n=6 is not possible)

(d)

Set 2 Detailed Solutions
1. Let there be m boys and n girls.
Then nC2 = 45 » n(n – 1)=90» n =10
mC2 = 190 » m(m – 1) = 380 » m=20
Number of games played between one boy and one girl
= 10C1 x 20C1= 10 x 20=200
(a)
2. Any of the 4 colours can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The stripe can again be colored in 3 ways. ( We can repeat the colour of the first stripe, but not use the colour of the second stripe).
Similarly, there are 3 ways to colour each of the remaining stripes.
» The number of ways the flag can be coloured is
4 x (3)5 = (12)(3)4
(a)
3. The available digits are 0, 1, 2 …………… 9. The first digit can be chosen in 9 ways ( 0 not acceptable ), the second digit can be accepted in 9 ways ( digits repetition not allowed). Thus the code can be made in 9 x 9=81 ways.
Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways.
Total number of ways confusion can arise = 4 x 3=12
Thus, required answer= 81-12=69
(d)
4.
Number of ways for selecting single digit=2
Number of ways for selecting two digit= 2 x 3=6
Number of ways for selecting three digit= 2 x 3 x 3=18
Number of ways for selecting four digit= 2 x 3 x 3 x 3=54
Number of ways for selecting five digit=2 x 3 x 3 x 3 x 3=162
Number of ways for selecting six digit= 2 x 3 x 3 x 3 x 3 x 3=486
Hence, total number of ways = (2 + 6 + 18 + 54 + 162 + 486)=728
(d)
5. There are 32 black and 32 white square on a chess-board then number of ways in choosing one white and one black square on the chess.
32C1 x 32C1= 32 x 32=1024
Number of ways in which square lies in the same row
White square=4
Black square=4
Number of rows=8
4C1 x 4C1 x 8 = 128
»Number of ways in which square lies in the same column= 128
Total number in which square lie on the same row or same column= 128 + 128=256.
(d)
6. Ist place of the four letter password can be filled in 11 ways.
IInd place of four letter password can be filled in 10 ways.
IIIrd place of four letter password can be filled in 9 ways.
IVth place of four letter password can be filled in 8 ways.
Hence, required number of ways= 11 x 10 x 9 x 8=7920 ways
(a)
7. Three letter password from 26 letters can be selected in 26 x 25 x 24 ways. Three letter password from 15 asymmetric letters can be selected in 15 x 14 x 13 ways.
Hence, three letter password with at least one symmetric letter can be made in (26 x 25 x 24)-(15 x 14 x 13)=12870 ways.
(c)
8. At lease one candidate out of (2n + 1) candidates can be selected in (2n+1 – 1) ways.
» 22n+1 – 1 = 63 » 22n+1 = 64 = (2)6» n=2.5
Since n cannot be a fraction. Hence n=3.
(a)
9. Required number of triangles formed
10C2 x 11 + 11C2 x 10= 45 x 11 + 55 x 10 = 1045
(c)
10. The digit in the unit’s place should be greater than that in the ten’s place. Hence, if digit 5 occupies the unit place then remaining four digits need not to follow any order.
Hence required number of ways = 4!
However, if digit 4 occupies the unit place then 5 cannot occupies the ten’s positions. Hence, digits at the ten’s place will be one among 1, 2 or 3. This can happen in 3 ways. The remaining 3 digits can be filled in the remaining three places in 3! ways. Hence in all we have (3 x 3!) numbers ending in 4.
Similarly, if we have 3 in the unit’s place, the ten’s place can be either 1 or 2. This can happen in 2 ways. The remaining 3 digits can be arranged in the remaining 3 places in 3! ways. Hence we will have (2 x 3!) numbers ending in 3. Similarly, we can find that there will be 3! numbers ending in 2 and no number ending with 1. Hence total number of numbers
= 4! + (3) x 3! + (2)3! + 3!
= 4! + 6 x 3! = 24 + (6 x 6) = 60

(b)

Elitmus Previous Year Questions on Alligations and Mixtures

SET 1

Allegations Set 01

01.

Two containers of equal capacity are full of a mixture of oil and water. In the first, the ratio of oil to water is 4: 7 and in the second it is 7: 11. Now both the mixtures are mixed in a bigger container. What is the resulting ratio of oil to water?

(a) 149: 247

(b) 247: 149

(d) 241: 143

02.

Two vessels contain spirit and water mixed respectively in the ratio of 1: 3 and 3: 5. Find the ratio in which these are to be mixed to get a new mixture in which the ratio of spirit to water is 1: 2?

(a) 2: 1

(b) 3: 1

(d) 1: 3

03.

The price of a pen and pencil is Rs 35. The pen was sold at a 20% profit and the pencil at a 10% loss. If in the transaction a man gains Rs 4, how much is cost price of the pen?

(a) Rs 10

(b) Rs 25

(d) None of these

04.

A person purchased a cupboard and a cot for Rs 18,000. He sold the cupboard at a profit of 20% and the cot at a profit of 30%. If his total profit was 25.83%, find the cost price of the cupboard?

(a) Rs 10,500

(b) Rs 12,000

(d) Rs 10,000

05.

A vessel is full of a mixture of kerosene and petrol in which there is 18% kerosene. Eight litres are drawn off and then the vessel is filled with petrol. If the kerosene is now 15%, how much does the vessel hold?

(a) 40 litres

(b) 32 litres

(d) 48 litres

06.

Two solutions of 90% and 97% purity are mixed resulting in 21 litres of the mixture of 94% purity. How much is the quantity of the first solution in the resulting mixture?

(a) 15 litres

(b) 12 litres

(d) 6 litres

07.

In what ratio should water be mixed with soda costing Rs 12 per litre so as to make a profit of 25% by selling the diluted liquid at Rs 13.75 per litre?

(a) 10: 1

(b) 11: 1

(d) 12: 1

08.

A sum of Rs 36.90 is made up of 90 coins that are either 20 paise coins or 50 paise coins. Find out how many 20 paise coins are there in the total amount?

(a) 47

(b) 43

(d) 63

09.

A bonus of Rs 9,85,000 was divided among 300 workers of a factory. Each male workers gets 5000 rupees and each female worker gets 2500 gets. Find the number of male workers in the factory?

(a) 253

(b) 47

(d) 206

10.

In what proportion must water be mixed with milk so as to gain 20% by selling the mixture at the cost price of the milk? ( Assume that water is freely available )

(a) 1: 4

(b) 1: 5

(d) 1: 12

Answers: Find Detailed Solutions at the end of the page.

1. A

2. C

3. B

4. C

5. D

6. C

7. C

8. C

9. C

10. B

SET 2

Allegations Set 02

01.

A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture?

(a) 10 gals

(b) 8.5 gals

(d) 8.33 gals

02.

4oo students took a mock exam in Delhi, 60% of the boys and 80% of the girls cleared the cut off in the examination. If the total percentage of students qualifying is 65%, how many girls appeared in the examination?

(a) 100

(b) 120

(d) 300

03.

What will be the ratio of petrol and kerosene in the final solution formed by mixing petrol and kerosene that are present in three vessels of equal capacity in ratios 4: 1, 5: 2 and 6: 1 respectively?

(a) 166: 22

(b) 83: 22

(d) None of these

04.

A dishonest grocer professes to sell pure butter at cost price, but he mixes it with adulterated fat and thereby gains 25%. Find the percentage of adulterated fat in the mixture assuming that adulterated fat is freely available?

(a) 20%

(b) 25%

(d) 40%

05.

In a Singapore zoo, there are deers and there are ducks. If the heads are counted, there are 180, while the legs are 448. What will be the number of deer in the zoo

(a) 136

(b) 68

(d) 22

06.

Two vessels contain a mixture of spirit and water. In the first vessel and ratio of spirit to water is 8: 3 and in the second vessel the ratio is 5: 1. A 35-litre cask is filled from these vessels so as to contain a mixture of spirit and water in the ratio of 4: 1. How many litres is taken from the first vessel?

(a) 11 litres

(b) 22 litres

(d) 17.5 litres

07.

In what ratio should water be mixed with soda costing Rs 12 per litre so as to make a profit of 25% by selling the diluted liquid at Rs 13.75 per litre?

(a) 10:1

(b) 11:1

(d) 12:1

08.

A 20 percent gain is made by selling the mixture of two types of ghee at Rs 480 per kg. If the type costing 610 per kg was mixed with 126 kg of the other, how many kilograms of the former was mixed?

(a) 138 kg

(b) 34.5 kg

(d) Cannot be determined

09.

A bartender stole champagne from a bottle that contained 50% of spirit and he replaced what he had stolen with champagne having 20% spirit. The bottle then contained only 25% spirit. How much of the bottle did steal ?

(a) 80%

(b) 83.33%

(d) 88.88%

10.

A mixture of 20 litres of brandy and water contains 10% water. How much water should be added to it to increase the percentage of water to 25%?

(a) 2 litres

(b) 3 litres

(d) 4 litres

Answers:

1. D

2. A

3. B

4. A

5. C

6. A

7. C

8. D

9. B

10. D

Amcat & Elitmus Previously asked question

Pages

Thursday, December 28, 2017

Elitmus Previous Years Questions On Number System

Elitmus Previous Years Questions On Number System

SET 1

Number System 01

SET 2

Number System 02

Number System Solution 01

Elitmus Previous Years Questions on Permutations and Combination

Elitmus Previous Years Questions on Permutations and Combination

Permutations And Combinations 01

Permutation And Combinations 02

Elitmus Previous Year Questions on Alligations and Mixtures

Elitmus Previous Year Questions on Alligations and Mixtures

SET 1

Allegations Set 01

SET 2

Allegations Set 02

Pages

Pages