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) $\frac{(6868............\;6)}{4034\;times}$

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:

.-. 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: $If\;P:Q=\;\;1\frac13\;:\;1\frac1{2\;}\;and\;Q\;:\;R=\frac12:\frac13\;then\;P\;:\;R\;\;=$

= 24:27:18
=8:9:6
Required P : R = 8 :-6 (or) 4:3

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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

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PART-2 MATHEMATICS

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PART-2
MATHEMATICS