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Arithmetic Aptitude :: Simplification

Home > Arithmetic Aptitude > Simplification > General Questions

71. 1(1!)+2(2!)+3(3!)....2012(2012!) = ?

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Answer: Option A

Explanation:

1(1!)=1 => 2!-1
1(1!)+2(2!)=1+4=5 => 3!-1
1(1!)+2(2!)+3(3!)=1+4+18=23 => 4!-1
........................
.......................
1(1!)+2(2!)+3(3!)+........+2012(2012!)=2013!-1

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72. If X^Y denotes X raised to the power of Y, find out last two digits of (2957^3661)+(3081^3643)

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Answer: Option C

Explanation:

(2957^4)^3660 *2957 +(3081)^3643
last 2 digits of (57)^4 is 01
so (01)^3660 yields last two digits as 01
{since the unit's digit is always 1 and tens digit is obtained by 0*0=0}
so 2957*01 gives last 2 digits 57
similarly (81)^3643 gives ten's digit as the unit digit of 8*3 ie 4
so 57 +41=98

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73. if f(x) is a function for all real numbers x holds good for maximum of 2x+4 and 12+3x. Then what is the value of x so that f(x) is equal to 2x+4.

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Answer: Option B

Explanation:

maximum (2x+4 ,12+3x)
and ques says maximum is 2x+4
if x= -9 2x+4 = -14 and 12+3x= -15 maximum is -14
Value of x = -9

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74. If ab3133ab is divisible by 12.What is the minimum value of (a+b)?

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Answer: Option C

Explanation:

The Trick is that we should check ab3133ab for its divisibility by 3



A number is divisible by 3 only when the sum of all the digits of that number is divisible by 3 so,



2a+2b+3+1+3+3 = 2a+2b+10 = 2(a+b)+10



Going by options only 4 and 7 satisfy



Then if we put 4 then also it's not divisible by 12 so the final answer is 7

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75. A number lying between 1000 and 2000 is such that on division by 2, 3, 4,5,6,7 and 8 Leavers remainders 1, 2, 3, 4, 5,6and 7 respectively. The number is

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Answer: Option B

Explanation:

LCM(2,3,4,5,6,7,8)= 1680
2-1=3-2=4-3=...=8-7=1
So, the number = 1680-1 = 1679

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76. Simplify the below

5y-(4x+3y)-6x

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Answer: Option D

Explanation:

5y-(4x+3y)-6x
= 5y-4x-3y-6x
= 2y-10x

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77. (2/9+2/3)*2 1/4 /(5/6*5/12)

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Answer: Option C

Explanation:

2/9+2/3=8/9
2 1/4=9/4
25/72
8/9*9/4*25/72=1

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78. How many 1/15 are there in 243/5 ?

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Answer: Option D

Explanation:

Solution : Just simply divide 243/5 by 1/15



The answer is 729.

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79. If x+y+z=9 and both y and z are positive integers greater than zero, then the maximum value x can take is?

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Answer: Option B

Explanation:

Acc. to the question y and z are positive integers greater than zero so taking the lowest possible value for them is 1 so putting y=z=1 in equation x+y+z=9 we get x value as 7 which is the maximum possible value of x.

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80. [ 2/9 + 2/3 ]*2 1/4+5/6*5/12

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Answer: Option C

Explanation:

[2+6 /9]*9/4 divide 5/6 * 5/12
8/9 *9/4 divide 5/6 *5/12
8/9 * 9/4 * 6/5 * 5/12=1

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