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NMMS Exam 2017 Previous Year Question Paper with Answers And Complete Analysis With Key For Practice Test Paper

PART-2
MATHEMATICS

Q). if $\frac{154}{69}=$ is expressed as $a+\frac1{b+\frac1{c+\frac1d}}$
then the value of a + b + c- d =
A)10
B)5
C)4
D)12
E)

View Answer

C)4
Explanation: $a+\frac1{b+{\displaystyle\frac1{c+{\displaystyle\frac1d}}}}\;=\frac{154}{69}=2+\frac{16}{69}$

= $2+\frac1{\displaystyle\frac{69}{16}}$

= $2+\frac1{4+{\displaystyle\frac5{16}}}$

= $2+\frac1{4+{\displaystyle\frac1{\displaystyle\frac{16}5}}}$

= $2+\frac1{4+{\displaystyle\frac1{3+{\displaystyle\frac15}}}}$

a+b+c-d=2+4+3-5
=4

Q). if $3^a=4,4^b\;=\;5,\;5^C\;=\;6,\;6^d\;=\;7,\;7^e\;=\;8,\;8^f\;=\;9$
then the value of product of abcdef
A)2
B)8
C)16
D)4

View Answer

A)2
Explanation: $\begin{array}{l}3^a=4=5^\frac1b=6^\frac1{bc}=7^\frac1{bcd}=8^\frac1{bcde}=9^\frac1{bcdef}\\\end{array}$

= $\begin{array}{l}3^a=9^\frac1{bcdef}=\left(3^2\right)^\frac1{bcdef}=3^\frac2{bcdef}\\\end{array}$

3^a= $\begin{array}{l}3^\frac2{bcdef}\\\end{array}$

Bases are equal, so powers must be equal.
a =2/bcdef
so abcdef= 2

Q). If xyz + xy + xz + yz + x + y + z = 384, then the value of x + y + z = , if x, y, z are positive integers.
A)18
B)24
C)20
D)15

View Answer

C)20
Explanation: This cyclic expression can be factorised first. As we notice individual power of all 3 variables in every term, is not more than 1. So there are chances that its each factor contains just one ofthe3 variables,
xyz + xy + zx + yz+x + y + z = 384
. => xy (z + 1) + (x + y) (z + 1) + z = 384
Add 1 on both sides
=> xy(z + 1)+(x + y) (z +1)+(z + 1) = 384 +1
=> (z + 1) (xy + x + y +1 ) = 385
=>(z+1) {x(y+1) + l(y+1)} =385
=> (z + 1) (x + 1) (y + 1) = 385
=> (x +1 ), (y + 1) and (z + 1)
hold either of these values 5 or 7 or 11.
=>x,y,z hold either of the values 4,6 or 10.
Hence x + y + z = 4 + 6+10 = 20

Q). The value of
$\frac1{20}+\frac1{30}+\frac1{42}+\frac1{56}+\frac1{72}+\frac1{90}+\frac1{110}+\frac1{132}\;$
A) 1/6
B) 1/24
C)1/42
D) 1/56

View Answer

D) 1/56
Explanation: $\frac1{20}+\frac1{30}+\frac1{42}+\frac1{56}+\frac1{72}+\frac1{90}+\frac1{110}+\frac1{132}\;$

Each term in the given series can be 1 written as $\frac{1}{n(n+1)}$

$\frac1{4x5}+\frac{\displaystyle1}{\displaystyle5x6}+\frac{\displaystyle1}{\displaystyle6x7}+\frac1{7x8}+\frac{\displaystyle1}{\displaystyle8x9}+\frac1{9x10}+\frac1{\;10x11\;}+\frac1{11x12}$

The given sum can be expressed as
${\textstyle\sum_1^8}\frac1{k\left(k+1\right)}={\textstyle\sum_{k=1}^{k=8}}\;\left(\frac1k-\frac1{k+1}\right)=\frac11$

$\frac17-\frac1{7+1}=\frac17-\frac18=\frac{8-7}{56}=\frac1{56}$

Q). $\frac{(66666...........6)^2}{2017\;times}+\frac{\displaystyle(88888.....8)}{\displaystyle2017\;times}$
A) $\frac{(444............\;4)}{4034\;times}$
B) $\frac{(555............\;5)}{2017\;times}$
C) $\frac{(666............\;6)}{2017\;times}$
D) $\frac{(6868............\;6)}{4034\;times}$