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

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NMMS Exam Online Mock Test


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}
3a=\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}

View Answer
A) \frac{(444............\;4)}{4034\;times}
Explanation: \frac{(66666...........6)^2}{2017\;TIMES}+\frac{\displaystyle(88888.....8)}{\displaystyle2017\;TIMES}
\{6+6.10+6.100++6.10^{n-1}\}
6\left(\frac{10^{n-1}}{10-1}\right)=\frac69\left(10^{n-1}\right)=\frac23\left(10^{n-1}\right)
(6666 ……….6)2
\left[\frac23\left(10^{n-1}\right)\right]^2=\frac49\left(10^{n-1}\right)
Similarly we can write
88888………8 { n times}
\{8+8.10+8.100++8.10^{n-1}\}
8\left(\frac{10^{n-1}}{10-1}\right)=\frac89\left(10^{n-1}\right)
So we get from given
(6666 ……….6)2 + 8888…..8
\frac49\left(10^{n-1}\right)^2=\frac89\left(10^{n-1}\right)
\frac49\left(10^{n-1}\right)\left[10^{n-1}+2\right]=\frac49\left(10^{2n-1}\right)\;
\frac12x\frac1{2017}\left(444...4\right)
\frac{(444............\;4)}{4034\;times}

Q). Simplify 16-\left[\begin{array}{l}9\frac12+\{8-(6-\overline{4-2}\;\;\;)\}\\\end{array}\right]
A) 7
B) 5
C) 2 ½
D) 6

View Answer
C) 2 ½
Explanation: 16\;-\left[\frac{19}2+{8-(6-2)}\right]
16\;-\left[\frac{19}2+{8-(4)}\right]
16\;-\left[\frac{19}2+\frac{(4)}1\right]
16\;-\left[\frac{19+8}2\right]
16\;-\left[\frac{27}2\right]=\frac{32-27}2=\frac52=2\frac12

Q). HCF of 1560, 1755 and 2925 is
A) 168
B) 195
C) 156
D) 242

View Answer
B) 195
Explanation: HCF of 1560, 1755 and 2925 is 195.

Q). 14% of a number is 35, then the number is
A) 250
B) 165
C)174
D)182

View Answer
A) 250
Explanation: Let the number be ‘x’.
14% of ‘x’ = 35
(14/100) x =35
X= \frac{35x100}{14}=2.5x100=250

Q). If a : b = 3 : 5 and b : c = 4 : 7, then a: b : c is
A) 16:14:18
B) 12 : 20 : 35
C) 12: 14 : 25
D) 12:25: 20

View Answer
B) 12 : 20 : 35
Explanation: nmms 2017
.-. a : b : c = 12 : 20 : 35

Q). If P:Q=1 \frac13 :1 \frac1{2} and Q:R=\frac12:\frac13 then P : R=
A) 1 : 5
B) 2 : 3
C) 4 : 3
D) 5 : 6

View Answer
C) 4 : 3
Explanation: If\;P:Q=\;\;1\frac13\;:\;1\frac1{2\;}\;and\;Q\;:\;R=\frac12:\frac13\;then\;P\;:\;R\;\;=
nmms 2017
= 24:27:18
=8:9:6
Required P : R = 8 :-6 (or) 4:3

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