# COGNIZANT Numbers Questions with Solutions – Merittrack

1. What will be largest number which divides 37, 59 and 74 leaving remainder 2, 3 and 4 respectively?

A)4
B)5
C)3
D)7
Answer:D
Explanation: Let the divisor be X.
Using the standard identity,
Dividend = (Divisor × Quotient) + Reminder
We have,
37 = (X × Q1) + 2
⇒ (X × Q1) = 35 [Where Q1 is a quotient]
Similarly,
59 = (X × Q2) + 3
⇒ (X × Q2) = 56 [Where Q2 is a quotient]
74 = (X × Q3) + 4
⇒ (X × Q3) = 70 [Where Q3 is a quotient]
Now, since X is the largest number satisfying the given condition, X will be the HCF of 35, 56 and 70.
HCF [35, 56, 70] = 7
Hence, the largest number which satisfies the given requirements is 7.

1. What largest number of four digits is exactly divisible by 89?

A) 9768
B) 9988
C) 9968
D) 8888
Answer:C
Explanation: As we know that the largest 4 digit no. is 9999, & hence in order to find the largest 4 digit no. divisible by 89 we will divide 9999 and check if it will completely divide the no or we will get some remainder.
⇒ On dividing 9999 by 89 we get 112 as quotient and 31 as remainder
∴ The largest 4 digit no. divisible by 89 will be 9999 – 31 = 9968

1. When n is divided by 4, the remainder is 3. The remainder when 2n is divided by 4 is:

A)1
B)2
C)3
D)6
Answer:B
Explanation: Given, when n is divided by 4, the remainder is 3.
Let the quotient be x.
Thus, n = 4x + 3
Now, 2n = 8x + 6
When, 2n is divided by 4
8x will be completely divisible 4.
Now, when 6 is divided by 4 the remainder is 2.
Thus, the remainder when 2n is divided by 4 is 2

1. What least value must be given to ‘a’ so that the number 91875a6 is divisible by 8?

A)5
B)6
C)3
D)4
Answer:C
Explanation: As per the standard divisibility rule of 8, for any number to be divisible by 8, the last three digits should be divisible by 8. According to this rule, we have “5a6” should be divisible by 8. We try to get all the multiples of 8 between 500 and 600, and then obtain the least number ending with 6. Hence, we have:
63 × 8 = 504
64 × 8 = 512
65 × 8 = 520
66 × 8 = 528
67 × 8 = 536
68 × 8 = 544
69 × 8 = 552
We do not need to go beyond this point because we have already obtained the number we need, and that number is 536. Hence, the least value of a is 3.

1. Find the least multiple of 23, which when divided by 18, 21 and 24 leaves the remainder 7, 10 and 13 respectively.

A) 3013
B)3105
C)2967
D)3059
Answer:A
Explanation: When the number is divided by 18, it leaves a remainder of 7.
When the number is divided by 21, it leaves a remainder of 10.
When the number is divided by 24, it leaves a remainder of 13.
Observing the constant difference of 11 between divisor and remainder:
We can conclude that if 11 is subtracted from the LCM of 18, 21 and 24, then it will leave a remainder of 7 when divided by 18, a remainder of 10 when divided by 21 and a remainder of 13 when divided by 24.
When 11 is subtracted from any multiple of LCM, it will give the same result.
Now,
18 = 2 × 3 × 3
21 = 3 × 7
24 = 2 × 2 × 2 × 3
∴ LCM of 18, 21 and 24 = 504
So, any number of the form (504n – 11) will give the required remainders, when n is an integer.
But, this number also has to be divisible by 23.
Using trial and error method, for n = 6:
We find that: 504n – 11 = 504 × 6 – 11 = 3024 – 11 = 3013

1. How many times 4 is written in numbers from 1 to 60?

A)62
B)61
C)16
D)17
Answer:C
Explanation: We can write 4 at unit place of 4, 14, 24, 34, 44, 54
Also we can wite at ten’s place 40, 41, 42, 43, 44, 45, 46, 47, 48, 49
∴ total = 16 times

1. The least number of the following which on dividing 343000, gives the result a perfect cube, is:

A)5
B)7
C)8
D)9
Answer:C
Explanation: 343000 = 23 × 53 × 73
As we can see, 343000 is already a perfect cube. It will only give a perfect cube on dividing by any one of its perfect cube factors.
Since the smallest perfect cube factor apart from 1 is 8, it should be divided by 8 to get a perfect cube.

1. The number of times 2 is used while writing the numbers from 1 to 100 is:

A)19
B)18
C)21
D)20
Answer:D
Explanation: In the unit’s place, starting with 2, 12 ….92, we have 10 2s used.
Also in 20 – 29, we have 2, 10 times in the ten’s place.
Hence, total no. of 2s used = 10 + 10 = 20

1. The sum of two numbers is 20 and their product is 80. What will be the sum of their reciprocals?

A)1
B)2
C) ¼
D) ½
Answer:
Explanation: Let the two numbers be a and b.
Given, a + b = 20 and ab = 80.
Sum of their reciprocals:
1/a+1/b=a+b/ab=20/80=1/4

10.The product of two positive fractions is 21/40 and their quotient is 56/15. The greater fraction is

A)7/6
B)7/5
C)4/5
D)5/8
Answer:
Explanation: Let’s assume that the two fractions are X & Y
∴product of two fractions = X× Y = 21/40   ……………………………..(i)
& Quotient of two fractions = X/Y = 56/15    ……………………………(ii)
⇒ Y = 15X/56
Putting the value of Y in equation (i)
X × (15X/56) = 21/40
⇒ X2 = 49/25
⇒ X = 7/5
& Y = (15/56) × (7/5) = 3/8
∴The greater fraction is = 7/5

1. If 13 + 23 + 33 +…… + 103 = 3025, then find the value of 23 + 43 + 63 + ….. + 203

A) 6050
B) 9075
C) 12100
D) 24200
Answer:D
Explanation: Given, 13 + 23 + 33 +…… + 103 = 3025
23 + 43 + 63 + ….. + 203
= 23 (13 + 23 + 33 +…… + 103)
= 8 × 3025
= 24200

1. In an examination, 35% of the candidates failed in Mathematics and 25% in English. If 10% failed in both Mathematics and English, then what is the percentage of students who passed in both the subjects?

A) 50
B)55
C)57
D)60
Answer:
Explanation: Let circle M represent students who failed in maths
& Let circle E represent students who failed in English
Unshaded part represents students who passed in exam
Now, % of students failed in maths =n(M) = 35% of students failed in English=n(E)= 25% of students failed in both maths &English =n(M∩ E)= 10
We know that, % of students who failed in either one or both subjects:
n(M ⋃E)= n(M)+ n(E) – n(M ⋂E)
⇒ n(M U E)= 35 + 25– 10
⇒ n(M U E)= 50
∴ % of students who passed in both subjects = 100 –(% of students who passed in any one subject)
⇒% of students who passed in both subjects = 100 – 50 = 50

1. A, B, C start running at the same time and at the same point in the same direction in a circular stadium. A completes a round in 252 s, B in 300 s and C in 198 s. After what time will they meet again at the starting point?

A) 1155 mins
B) 1550 mins
C) 1150 mins
D) 1515 mins

Answer:A
Explanation: The time at which they will meet again at the starting point will be the LCM of 252, 300 and 198 s.Hence,
LCM [252,300,198] = 69300 s = 69300/60 mins = 1155 mins

14.A rectangular courtyard of a temple 2.56 m long and 6.25 m wide is to be paved exactly with square marble stones, all of the same size. What is the largest size of the marble stone which could be used for the purpose?

A)2m
B)1m
C)1cm
D)2cm
Answer:C
Explanation: If the courtyard is to be paved exactly with square stones,
The square stones must be such that each side of the courtyard is a whole multiple of each side of the stone.Thus, one side of the stone must be a factor of 2.56m
The other side must be a factor of 6.25m
As both sides of a square are equal,
∴ the value of each side of the square stone = HCF of the 2 given dimensions.
∴ HCF of 2.56 and 6.25 = 0.01m.
Thus, the maximum side of each square stone should be 1cm

1. Consider those numbers between 100 and 1000 such that when each number is divided by 6,7 and 11 it leaves 5 as remainder in each case. What is the sum of the numbers?

A) 462
B)929
C) 1386
D) 1396
Answer:C
Explanation: As the number leaves same remained in each case,
∴ required numbers = 5 + (common multiples of 6, 7 and 11)
∴ Consider the LCM of 6, 7 and 11.
LCM = 462
⇒ One of the required numbers = 462 + 5 = 467.
Next and last common multiple of 6, 7 and 11 in the given range = 462 × 2 = 924
⇒ Second required number = 924 + 5 = 929
∴ Sum of required numbers = 467 + 929 = 1396

1. Ram distributes his pens among 5 people A, B, C, D and E in the ratio 1/2 : 1/3 : 1/4 : 1/5 : 1/6, then what will be the minimum number of pens Ram has?

A) 84
B)87
C)80
D)77
Answer:B
Explanation: Given ratio is 1/2 : 1/3 : 1/4 : 1/5 : 1/6.
Converting the fractions into whole numbers by multiplying by the LCM of the denominators of each fraction.
LCM of 2, 3, 4, 5, 6 = 60
Ratio becomes, 12×60:13×60:14×60:15×60:16×6012×60:13×60:14×60:15×60:16×60
Ratio of number of pens received = 30 : 20 : 15 : 12 : 10
Each ratio is the minimum pens which a person receives.
Thus the minimum number of pens = 30 + 20 + 15 + 12 + 10 = 87

1. Sukhiram plants 12544 mango trees in his garden and arranges them so that there are as many rows as there are mango trees in each row. Find the number of rows.

A)108
B)112
C)118
D)122
Answer:B
Explanation: Given, Sukhiram plants 12544 mango trees in his garden and arranges them so that there are as many rows as there are mango trees in each row.
Let the number of rows be a.
∴ Number of mangoes in each row = a
Total number of mangoes = a × a = a2
∴ a2 = 12544 = 72 × 28
⇒ a = 7 × 24 = 112

1. Three men step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. The minimum distance each should cover so that all can cover the distance in complete steps is

A)9630 cm
B) 9360 cm
C) 6930 cm
D) 6950 cm
Answer:C
Explanation: Minimum distance to be covered by all men to cover the same distance in complete steps = LCM (63, 70, 77)
= 7 × LCM (9, 10, 11)
= 7 × 9 × 10 × 11
= 6930 cm

1. What approximate value will come in place of question mark in the following question?

23.001 × 18.999 × 7.998 =?

A) 3500
B) 2500
C) 1500
D) 7500

Answer:
Explanation: In this type of question, we are expected to calculate Approximate value (not exact value), so we can replace the given numbers by their nearest perfect places which makes the calculation easy.
Let 23.001 ≈ 23, 18.999 ≈ 19 and 7.998 ≈ 8
Now, the given expression:
23.001 × 18.999 × 7.998 =?
⇒ ? ≈ 23 × 19 × 8
⇒ ? ≈ 437 × 8
⇒ ? ≈ 3496 ≈ 3500
Hence, the required approximate value in place of question mark is 3500.

20.The mean temperature of Monday to Wednesday was 40°C and of Tuesday was 36°C. The average of Tuesday to Thursday was 44°C. If the temperature of Thursday was 5/4th that of Monday, the temperature of Thursdays was

A) 60.5°C
B) 64°C
C) 65°C
D) 60°C
Answer:D
Explanation: Let the temp. on Monday be M, Wednesday be W, and Thursday be Th. We have,
(M + 36 + W)/3 = 40
⇒ M + W = 120 – 36 = 84°C
⇒ W = 84 – M
Also, given
(36 + W + Th)/3 = 44
⇒ W + Th = 132 – 36 = 96°C
It is given that Th = 5M/4
⇒ W + 5M/4 = 96
⇒ W = 96 – 5M/4
84 – M = 96 – 5M/4
⇒ M/4 = 12
⇒ M = 48°C
⇒ Th = 5 × 48/4 = 60°C