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Numbers & Algebra - Numbers and Algebra MCQ

31:  

Two new charity organizations C1 and C2 were formed, with x members each, on January 1, 2012. On first day of each, subsequent month, in C1, number of members increases by a certain number a, while in C2, number of members increases in such a way that ratio of the number of members in a month to the preceding month bear a ratio equal to b. On May 1, 2012, both organizations had the same number of members. If a = 20x, then b will be

A.

2

B.

3

C.

2.5

D.

3.5

 
 

Option: B

Explanation :

Number of members in C1 on May 1, 2012 = x + 4ya
Number of members in C2 on May 1, 2012= xb4
x + 4a = xb4 and a = 20x
∴ x(b4 - 81)
As x   ≠ 0, b4 - 81 = 0
∴ b = 3

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32:  

Which of the following is true?

A.

√2+√5=√7

B.

√2+√5 ≤ √7

C.

√2+√5<√7

D.

√2+√5 > √7

 
 

Option: D

Explanation :

(√2+√5)² = 2 + 2√10 +5 =7+2√10
(√7)² = 7
∴ √2+√5> 7
Alternatively:
√2 > √1 and √5 > √4
∴ √2+√5 > 2+1 
=> √2 + √5 > 3 and √7 < √9 = 3
=> √7 < 3. 
so √2 + √5>√7

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33:  

What is the least number which on being divided by 5, 6, 8, 9, 12 leaves in each case a remainder 1 but when divided by 13 leaves no remainder?

A.

2987

B.

3601

C.

3600

D.

2986

 
 

Option: B

Explanation :

L.C.M. of 5, 6, 8, 9 and 12 is 360
Required number = 360 K + 1 = (13 x 27 + 9) K + 1
= (13 x 27) K + (9K + 1)
Now this number must be divisible by 13 K = 10 and required number = 3601.

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34:  

Average marks of 15 students in a class is 145, maximum marks being 150. If two lowest scores are removed, the average increases by 5. Also, two lowest scores are consecutive multiples of 9. What is lowest score in the.class?

A.

126

B.

117

C.

108

D.

None of these

 
 

Option: C

Explanation :

Total marks of 15 students = 15 x 145 = 2175.
Average marks of 15 students (excluding two lowest scores) = 13 x (145 + 5) = 1950.
∴ Total of two lowest scores = 2175 1950 = 225
Given that two scores are consecutive multiples of 9 * 9x + 9x + 9 = 225 18x = 216
=> 18/x = 216 => x = 216/18
Lowest score = 9x = (216/18) x 9 = 108

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35:  

A gardener had a number of shrubs to plant in horizontal rows. At first he tried to plant 5 shrubs in each row, then 6, then 8 and then 12, but had always 1 left. On trying 13, in one row he had none left. What is the smallest number of shrubs that he could have had?

A.

481

B.

477

C.

468

D.

121

 
 

Option: A

Explanation :

Number is 120K + 1= ((13 * 9 + 3) K + 1) 
= 13 * 9K + 3 K + 1, which is divisible by 13.
3K + 1 is divisible by 13.
 ∴  K = 4. Number = 481

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