Wipro Aptitude Questions with Solutions

 Wipro Aptitude Questions with Solutions


1.The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
A) 1
B) 2
C) 3
D) 4
Answer: B) 2
Explanation:
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
=>ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.



2.The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
A) 123
B) 127
C) 235
D) 305
Answer: B) 127
Explanation:
Required number = H.C.F. of (1657 – 6) and (2037 – 5)
= H.C.F. of 1651 and 2032 = 127.



3.Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
A) 40
B) 80
C) 120
D) 200
Answer: A) 40
Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.


4.The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above condition is
A) 4
B) 6
C) 8
D) 12
Answer: A) 4
Explanation:Let the required numbers be 33a and 33b.
Then 33a +33b= 528   =>   a+b = 16.


Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).herefore, Required numbers are  ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)The number of such pairs is 4



5.The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 100, then their difference is
A) 10
B) 46
C) 70
D) 90
Answer: A) 10
Explanation:
Let the numbers be x and (100-x).
Then, x(100−x)=5*495
=>  x^2−100x+2475=0         x^2-100x+2475=0
=>  (x-55) (x-45) = 0
=>  x = 55 or x = 45
The numbers are 45 and 55
Required difference = (55-45) = 10


6.The H.C.F and L.C.M of two numbers are  84 and 21 respectively.  If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is
A) 12
B) 48
C) 84
D) 108
Answer: C) 84
Explanation:
Let the numbers be x and 4x. Then,  x×4x=84×21 ⇔ x2=84×214⇔ x=21x×4x=84×21 ⇔ x2=84×214⇔ x=21
Hence Larger Number = 4x = 84


7.The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275 , then the other is
A) 279
B) 283
C) 308
D) 318
Answer: C) 308
Explanation:
Other number = (11×7700) =275 ×          x= 308.



8.Two number , both greater than 27,have hcf 27 and lcm 162 The sum of the number is
a)1
b)162
c)135
d)81
Answer:C)135
Explanation:
H.C.F*l.C.M= product of two no.
162*27=4374
now find out the factors of 4374
ANd find out the suitavle factors.I.e 54*81=4374 which have h.cf=27 and L.c.m-162.
therefore 54+81=135


9.Find the greatest number, which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
A.127
b. 132
c. 114
d. 108
Answer:A
Explanation: Hence, make the dividend completely divisible by the divisor. This is possible, if we subtract remainder from the dividend.
Therefore,
1657 – 6 = 1651
2037 – 5 = 2032
H.C.F. of 1651 and 2032 is 127.



10.From a pack of 52 playing cards, 4 cards are removed at random. In how many ways can the 1st place and 3rd place cards be drawn out such that both are black ?
A: 64,974
B: 62,252
C: 69,447
D:1,592,500
Answer:
Explanation: There are 26 black cards to choose from on the first draw.
Now the second card can either be black or red.
If black, then there are 25 to choose from, and then for
the third card there will then be 24 black cards to choose
from.If red, then there are 26 reds to choose from, and then
for the third card there are then 25 black cards to choose from.
In either case, there are 49 cards left to from on the
4th draw.
So the number of ways we can do this is
26*(25*24 + 26*25)*49 = 1,592,500


11.The traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 5 : 20 : 00 hours, then find the time at which they will change simultaneously.
A. 5 : 28 : 00 hrs
b. 5 : 30 : 00 hrs
c. 5 : 38 : 00 hrs
d. 5 : 40 : 00 hrs
Correct Option: (c)
Explanation:
Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.
Therefore, find the L.C.M. of 40, 72 and 108.
L.C.M. of 40, 72 and 108 = 1080
The traffic lights will change again after 1080 seconds = 18 min
The next simultaneous change takes place at 5 : 38 : 00 hrs.



12. A five-digit number is formed by using digits 1,2,3,4 and 5 without repetition. What is the probability that the number is divisible by 4?
A)1/5
B)2/5
C)3/5
D)4/5
Answer:A
Solution:
A number divisible by 4 formed using the digits 1,2,3,4 and 5 has to have the last two digits 12 or 24 or 32 or 52.
In each of these cases, the five digits number can be formed using the remaining 3 digits in 3! = 6 ways.
A number divisible by 4 can be formed in 6×4=24 ways.
Total number that can be formed using the digits 1,2,3,4 and 5 without repetition
=5!=120
Required probability = 24/120 = 1/5



13.A rectangular courtyard 4.55 meters long and 5.25 meters wide is saved exactly with square tiles of same size. Find the largest size of the tile used for this purpose?
A. 25 cm
b. 45 cm
c. 21 cm
d. 35 cm
Correct Option: (d)
Explanation:
Here, we are asked to find the largest size of tile. Therefore, calculate H.C.F.
Step 1: Covert numbers without decimal places i.e 455 cm and 525 cm
Step 2: Find the H.C.F. of 455 and 525
H.C.F. of 455 and 525 = 35 cm
Hence, the largest size of the tile is 35 cm.



14.John, Smith and Kate start at same time, same point and in same direction to run around a circular ground. John completes a round in 250 seconds, Smith in 300 seconds and Kate in 150 seconds. Find after what time will they meet again at the starting point?
A. 30 min
b. 25 min
c. 20 min
d. 15 min
Correct Option: (b)
Explanation:
L.C.M. of 250, 300 and 150 = 1500 sec
Dividing 1500 by 60 we get 25, which mean 25 minutes.
John, Smith and Kate meet after 25 minutes.




15.Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
A) 4
B) 10
C) 15
D) 16
Answer: D) 16
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes,they will together (30/2)+1=16 times



16.The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is
A) 1677
B) 1683
C) 2523
D) 3363
Answer: B) 1683
Explanation:
L.C.M of 5, 6, 7, 8 = 840
Therefore, Required Number is of the form 840k+3.    =(93*9 +3)k + 3 = (93*9*k) + (3k +3)
Least value of k for which (840k+3) is divisible by 9 is k = 2
Therefore, Required  Number = (840 x 2+3)=1683



17.A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in Width. Find the least number of square tiles of equal size required to cover the entire floor of the room.
A) 107
B) 117
C) 127
D) 137
Answer: B) 117
Explanation:
Let us calculate both the length and width of the room in centimeters.
Length = 6 meters and 24 centimeters = 624 cm
width = 4 meters and 32 centimeters = 432 cm
As we want the least number of square tiles required, it means the length of each square tile should be as large as possible.
Further,the length of each square tile should be a factor of both the length and width of the room.
Hence, the length of each square tile will be equal to the HCF of the length and width of the room = HCF of 624 and 432 = 48
Thus, the number of square tiles required = (624 x 432 ) / (48 x 48) = 13 x 9 = 117



18.Find the largest number of four digits exactly divisible by 12,15,18 and 27.
A)9720
B)6580
C)4578
D)1023
Answer:A
Explanation:
The Largest number of four digits is 9999.
Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.
On dividing 9999 by 540,we get 279 as remainder .
Required number = (9999-279) = 9720.



19.Find the least number which when divided by 5,6,7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder .
A)1683
B)1982
C)2001
D)none
Ans:A
Sol. L.C.M. of 5,6,7,8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 X 2 + 3)=1683



20.L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :
A) -2
B) -1
C) 1
D) 2
Answer:C
Solution:  161 = 7*23    x=23   y= 7        Therefore, 3y –x = -2


21. The number of prime factors of (3 x 5)^12 * (2 x 7)^10  * (10)^25 is:
A): 47
B): 60
C): 72
D): None of these
Correct Answer : D
Explanation: The equation can be factorize as = 3^12 * 5^12 * 2^10 * 7^10 * 2^25 * 5^ 25
total no of prime factor = 12 + 12 + 10 + 10 + 25 + 25 = 94



22 What least value must be assigned to * so that the number 63576*2 is divisible by 8?
A): 1
B): 2
C): 3
D): 4
Correct Answer :C
Explanation: The test for divisibility by 8 is that the last 3 digits of the number in question have to be divisible by 8.

So, 6*2 has to be divisible by 8.
I know 512 is divisible by 8.
Also 592 is divisible by 8.
So, 632 is divisible by 8.
So * is 3.



23 The smallest number, which is a perfect square and contains 7936 as a factor is:
A): 251664
B): 231564
C): 246016
D): 346016
Correct Answer :C
Explanation:
7936 => 2^2 * 2^2 * 2^2 * 2^2 * 31^1
To make it as a perfect square, we have to multiply 7936 with 31…
Hence the reqd no. is 7936*31 = 246016



24 In a division problem, the divisor is ten times the quotient and five times the remainder. If remainder is 46, the number will be:
A): 3360
B): 5336
C): 1616
D): 2051
Correct Answer :B
Explanation:
Divisor = (5 x 46) = 230

10 x Quotient = 230
Dividend = (Divisor x Quotient) + Remainder
= (230 x 23) + 46
= 5290 + 46
= 5336.


25  If a number is exactly divisible by 85, then what will be the remainder when the same number is divided by 17?
A): 3
B): 1
C): 4
D): 0
Correct Answer : D
Explanation: number=divisor*quotient+remainder
so 17*5+0;
remainder is 0;
divisor is 17;
quotient is 5;


26 The least perfect square number which is exactly divisible by 3, 4, 7, 10 and 12 is:
A): 8100
B): 17600
C): 44100
D): None of these
Correct Answer : C


27 (x power n+y power n) is divisible by (x-y):
A): for all values of n
B): only for even values of n
C): only for odd values of n
D): for no values of n
Correct Answer : D
Explanation: for no values of n



28. P is an integer. P is greater than 883.If P -7 is a multiple of 11, then the largest number that will always divide (P+4)(P+15) is
A)242
B)343
C)321
D)none
Answer:A
Explanation: p-7= 11*a (as it is multiple of 11)
p=11*(a+7)
so (p+4)(p+15)= (11a+7+4)(11a+7+15);
= (11a+11)(11a+22);
=11*11(a+1)(a+2);
=121*2
=242



29.The greatest number that will divide 63, 138 and 228 so as to leave the same remainder in each case:
A): 15
B): 20
C): 35
D): 40
Correct Answer :A
Explanation: The greatest number = H.C.F of (138-63), (228-138), (228-63)
H.C.F of 75, 90, 165 = 15.
15 is the greatest number.



30.Find the largest number, smaller than the smallest four-digit number, which when divided by 4,5,6and 7 leaves a remainder 2 in each case:
A): 422
B): 842
C): 12723
D): None of these
Correct Answer : B
Explanation: Take LCM of 4,5,6,7. It is 420
BUt the no must leave remainder 2 in each case, so the no is of the form: 420k + 2.
The smallest 4-digit no is 1000. So keeping k=0,1,2,3….
We get that the largest no smaller than the smallest 4 -digit no is 842



31,What is the highest power of 5 that divides 90 x 80 x 70 x 60 x 50 x 40 x 30 x 20 x 10?
A): 10
B): 12
C): 14
D): None of these
Correct Answer : A
Explanation: Take LCM of Each Number
90/5=5*2*3*3——————>here we will get one 5
80/5=5*2*2*2*2—————>here we will get one 5
70/5=5*2*7————–___—->here we will get one 5
60/5=5*2*2*3——————>here we will get one 5
50/5=5*5*2___——————>here we will get Two 5^2
40/5=5*2*2*2——————>here we will get one 5
30/5=5*2*3———————>here we will get one 5
20/5=5*2*2———————>here we will get one 5
10/5=5*2————————>here we will get one 5
Here we will get one 5 in each number instead of 50(5*5*2)
So answer is 5^10



32.If a and b are natural numbers and a-b is divisible by 3, then a3-b3 is divisible by:
A): 3 but not by 9
B): 9
C): 6
D): 27
Correct Answer : B
Explanation: If a − b is divisible by 3, then a − b = 3k, for some integer k
(a − b)² = (3k)²
a² − 2ab + b² = 9k²
a³ − b³ = (a−b) (a² + ab + b²)
= (a−b) (a² − 2ab + b² + 3ab)
= 3k (9k + 3ab)
= 3k * 3 (3k + ab)
= 9 k(3k+ab)
Since k(3k+ab) is an integer, then 9k(3k+ab) is divisible by 9



33.What is the greatest positive power of 5 that divides 30! exactly?
A): 5
B): 6
C): 7
D): 8
Correct Answer : C
Explanation:  The question is, how many powers of 5 are in the factors of 30! (that’s 30factorial, for those above)…
Only the numbers 5, 10, 15, 20, 25, and 30 have divisors of 5. And 25 is divisible by 5^2.
So the answer is 5*5*5*5*(5^2)*5 = 5^7.



34.What is the smallest four-digit number which when divided by 6, leaves a remainder of 5 and when divided by 5 leaves a remainder of 3?
A): 1043
B): 1073
C): 1103
D): None of these
Correct Answer : D
Explanation: remainder when  m is divided by 5  = 2
Smallest m is 2.
Hence, N = 1001 + 6 * 2 = 1013.



35.P is an integer. P>883. If P-7 is a multiple of 11, then the largest number that will always divide (P+4) (P+15) is:
A): 11
B): 121
C): 242
D): None of these
Correct Answer : C
Explanation: Given P is an integer>883.
P-7 is a multiple of 11=>there exist a positive integer a such that
P-7=11 a=>P=11 a+7
(P+4)(P+15)=(11 a+7+4)(11 a+7+15)

=(11 a+11)(11 a+22)
=121(a+1)(a+2)
As a is a positive integer therefore (a+1)(a+2) is divisible by 2.Hence (P+4)(P+15) is divisible by 121*2=242



36.Let C be a positive integer such that C + 7 is divisible by 5. The smallest positive integer n (>2) such that C + n^2 is divisible by 5 is:
A): 4
B): 5
C): 3
D): Does not exist
Correct Answer : D
Explanation: c + n^2 is divisible by 5 if and only if c and n^2 are both divisible by 5.
But, if c is divisible by 5 then c + 5 will not be divisible by 5.
So, option(d ) is correct.



37.Four bells begin to toll together and then each one at intervals of 6 s, 7 s, 8 s and 9 s respectively.
The number of times they will toll together in the next 2 hr is:
A): 14 times
B): 15 times
C): 13 times
D): 11 times
Correct Answer : A
Explanation: first we to find the L.C.M. of 6, 7, 8 and 9.
Prime factorization of 6 = 2*3
Prime factorization of 7 = 7
Prime factorization of 8 = 2*2*2