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Numbers & Algebra - Numbers and Algebra MCQ

11:  

HCF and LCM of two numbers is given. It is possible to ind out the two numbers uniquely if
I. either sum or difference between the two numbers is known.
II. HCF of two numbers = LCM of two numbers.
III. (LCM/HCF)= Prime number.

 

A.

I and II only

B.

II  only

C.

II and III only

D.

I, II and III

 
 

Option: D

Explanation :

Let HCF be h and LCM be l.
I.  Let, numbers be ah and bh. Then abh = and (a + b)h = m
=> (a — b)h = n
Using these ah and bh can be uniquely determined. Thus, I is true II.
II.  If HCF = LCM, then two numbers are equal and same as HCF or LCM. Thus, II is true.
III. LCM/HCF = prime i.e. l/h =P Then one of the numbers is equal to h and other is equal to E. Thus, III is true.

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12:  

A number when divided by sum of 555 and 445 gives two times their diference as quotient and 30 as remainder. The number is

A.

1220

B.

1250

C.

22030

D.

220030

 
 

Option: D

Explanation :

Number = (555+445)*(555-445)*2+30
= (555+445)*2*110+30
= 220000+30 = 220030

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13:  

 What is the remainder when 2050 x 2071 x 2095 is divided by 23?

A.

17

B.

12

C.

6

D.

3

 
 

Option: C

Explanation :

2050 x 2071 x 2095
=(23x89+3)x(23x90+1)x(23x91+2)
=(23[89 x 90+ 89 + 90 x 3]+ 3)x(23 x 91+2)
=(23 *K+ 3)* (23 * 91 + 2) )
=(23 *[K* 91 + K*  2 + 91 * 3]+ 6)
= 23 * L + 6 (where K and L are constants).
Hence remainder is 6.

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Sakshi singh said: (10:38pm on Friday 21st July 2017)
but sir remainder 17 are also come .if we calculate it well, some quick arithmetic as followed by using multiples of 23 like 230,2300 etc. 2300-230=2070 is clearly a multiple of 23.(2070-2050)*(2070-2071)*(2070-2095)=20*(-1)*(-25) by 23500/23remainder is come 17.

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14:  

 HCF of y3 - 4y and 4y (y3 + 8) is

A.

y(y + 2)

B.

y - 2

C.

2 (y + 2)

D.

(y + 2)(y2- 2y + 4)

 
 

Option: A

Explanation :

y3-4y = y(y2-4)
= y(y-2)(y+2)
4y(y3+8)
=4y(y+2)(y2-2y+4)
∴ HCF = y(y + 2).

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15:  

A ball is dropped from a height of 12 m and it rebounds 1/2 of the distance it falls. If it continues to fall and rebound in this way, how far will it travel before coming to rest?

A.

36 m

B.

30 m

C.

48 m

D.

60 m

 
 

Option: A

Explanation :

Total distance travelled by the ball before coming to rest
= 12 + 2 (6 + 3 + 1.5 +...to?)
= 12 + 2. ( 6/ (1-1/2))
= 12 + 2(12)
= 36m

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