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# Simplification Questions

Home > Quantitative Aptitude > Simplification > General Questions
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Choose the correct option.

Find min value of f(x) : |-5-x| + |2-x|+|6-x|+ |10-x|; where x is an integer

A19

B17

C23

D10

Explanation:

Note: | a | = | ?a |
| ?5 ? x | = | ?(?5 ? x) | = | 5 + x | and
| 2 ? x | = | ?(2 ? x) | = | ?2 + x | = | x ? 2 |
=> | ?5 ? x | + | 2 ? x | + | 6 ? x | + | 10 ? x |
= | 5 + x | + | x ? 2 | + | 6 ? x | + | 10 ? x |
Also | a | + | b | ? | a + b | for all a and b
and | c | + | d | ? | c + d | for all c and d
and so:
| a | + | b | + | c | + | d | ? | a + b | + | c + d | ? | a + b + c + d |
| a | + | b | + | c | + | d | ? | a + b + c + d |

Therefore
| ?5 ? x | + | 2 ? x | + | 6 ? x | + | 10 ? x |
= | 5 + x | + | x ? 2 | + | 6 ? x | + | 10 ? x |
? | 5 + x + x ? 2 + 6 ? x + 10 ? x | = | 19 | = 19

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Factorize y^2 + 8y - 48 = 0

A(y-12) (y-4)

B(y+12) (y-4)

C(y-12) (y+4)

DNone of these

Explanation:

Solution:  y^2+ 8y-48=0

y^2+12y-4y-48=0

y(y+12)-4(y+12)=0

(y-4)(y+12)

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How many 1's are there in the binary form of 8*1024 + 3*64 + 3

A4

B5

C3

D6

Explanation:

8*1024 + 3*64 + 3 = 8387
= 10000011000011

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In the following, which is the closest approximation of (50.2*0.49)/199.8 ?

A0.5

B0.112

C0.125

D0.25

Explanation:

(50.2*0.49)/199.8 ~ (50*0.5)/200 = 0.125

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If A=2/3(B-C) and c=1/2(A+B) and A+B+C=3000, then find C ?

A1000

B100

C1100

D1001

Explanation:

C=1/2(A+B)
A+B+(1/2(A+B))=3000
3/2(A+B)=3000
A+B=2000
C=1/2(A+B)
C=1000

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Assume that f(1)=0 and f(m+n)=f(m)+f(n)+4(9mn-1). For all natural numbers (Integers>0)m and n. What is the value of f(17)?

A5436

B4831

C5508

D4832

Explanation:

f(2) = f(1+1)=f(1)+f(1)+4(9*1*1 - 1 )= 0+0+4*8 = 32
f(4) = f(2+2)=f(2)+f(2)+4(9*2*2 - 1 )= 32+32+4*35 = 204
f(8) = f(4+4)=f(4)+f(4)+4(9*4*4 - 1 )= 204+204+4*143 = 980
f(16) = f(8+8)=f(8)+f(8)+4(9*8*8 - 1 )= 980+980+4*575 = 4260
f(17)= f(1+16)= f(16)+f(1)+4(9*16*1 -1)= 4260+0+ 4*143= 4832

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In this question, A^B means A raised to the power B. Let f(X)=1+X+X^2+....X^6. The remainder when f(X^7) is divided by f(X) is

A0

B6

C7

DNone of these

Explanation:

f(x)=1+x+x^2+x^3+.........+x^6
When we, put x = 0 in the given functions, then we will have remainder as 1.
f(0) = 1.
1/1 remainder = 0.
For x =1, remainder also will be 0.
But for x >1 (x = 2,3,4.......)
Remainder would be 7.

You can find remainder for x > 1 through this formula,
f(x) = 1+x+x^2+.....+x^(n).
for x > 1,
then the remainder when f(x^(n+1)) is divided by f(x) is (n+1).
n = maximum power of the f(x).

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(p/q-q/p)=21/10. Then find 4p/q+4q/p?

A58/3

B58/5

C68/5

DNone of these

Explanation:

let us take p/q=x and q/p=1/x
x-1/x=21/10
then by solving we get a quadratic equation like 10x^2-10-21x=0
10x^2-25x+4x-10=0
then by taking factors we get two values for x,x=-2/5 and x=5/2
by substituting x=5/2
4(5/2)+4(2/5)=58/5

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A certain function f satisfies the equation f(x)+2*f(6-x)=x for all real numbers x. The value of f(1) is

A1

B2

C3

DCannot be determined

Explanation:

Put x =1 => f(1)+2*f(6-1) = 1 => f(1) + 2*f(5) = 1
Put x = 5 => f(5)+2*f(6-5) = 5 => f(5) + 2*f(1) = 5
Put f(5) = 5 - 2*f(1) in the first equation
=> f(1) + 2*(5 - 2*f(1)) = 1
=> f(1) + 10 - 4f(1) = 1
=> f(1) = 3

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Find the number of zeroes in the expression 15*32*25*22*40*75*98*112*125

A12

B9

C14

D7

Explanation:

Maximum power of 5 in the above expression can be calculated like this. Count all the powers of 5 in the above expression. So number of zeroes are 9.

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