RRB Mathematics - Decimal and Fraction
Exam Duration: 45 Mins Total Questions : 30
The fraction \(101\frac { 27 }{ 100000 } \)in decimal form is:
- (a)
.01027
- (b)
.10127
- (c)
101.00027
- (d)
101.000027
\(101\frac { 27 }{ 100000 } =101+\frac { 27 }{ 100000 } =101+.00027=101.00027.\)
Which of the follwing fractions is less than\(\frac { 7 }{ 8 } \)and greater than \(\frac { 1 }{ 3 } \)?
- (a)
\(\frac { 1 }{ 4 } \)
- (b)
\(\frac { 23 }{ 24 } \)
- (c)
\(\frac { 11 }{ 12 } \)
- (d)
\(\frac { 17 }{ 24 } \)
\(\frac { 7 }{ 8 } \)=0.875, \(\frac { 1 }{ 3 } \)=0.333, \(\frac { 1 }{ 4 } \)=0.25, \(\frac { 23 }{ 24 } \)=0.958, \(\frac { 11 }{ 12 } \)=0.916, \(\frac { 17 }{ 24 } \)=0.708
Clearly, 0.708 lies between 0.333 and 0.875
\(\therefore \) \(\frac { 17 }{ 24 } \)lies between \(\frac { 1 }{ 3 } \) and \(\frac { 7 }{ 8 } \)
337.62+8.591+34.4=?
- (a)
370.611
- (b)
380.511
- (c)
380.611
- (d)
426.97
48.95-32.006=?
- (a)
16.089
- (b)
16.35
- (c)
16.89
- (d)
16.944
832.58-242.31=779.84-?
- (a)
179.57
- (b)
199.57
- (c)
295.05
- (d)
None of these
Let 832.58-242.31=779.84-x
Then x=(779.84+242.31)-832.58=1022.15-832.58=189.57
Which of the following equal to \(3.14\times{ 10 }^{ 6 }\)
- (a)
314
- (b)
3140
- (c)
3140000
- (d)
None of these
\(3.14\times{ 10 }^{ 6 }\)=3.140000x1000000=3140000
16.02\(\times \)0.001=?
- (a)
0.001602
- (b)
0.01602
- (c)
0.1602
- (d)
1.6021
1602 x 1 = 1602. Sum of decimal places = 5
\(\therefore \) 16.02 x 0.001 = 0.01602
0.014 x 0.014 = ?
- (a)
0.000196
- (b)
0.00196
- (c)
19.6
- (d)
196
14 x 14 = 196.
Sum of decimal places = 6
\(\therefore \) 0.014 x 0.014 = 0.000196
\(\left( .00625\quad of\quad \frac { 23 }{ 5 } \right) \)when expressed as a vulgar, fraction, equals :
- (a)
\(\frac { 23 }{ 80 } \)
- (b)
\(\frac { 23 }{ 800 } \)
- (c)
\(\frac { 23 }{ 8000 } \)
- (d)
\(\frac { 125 }{ 23 } \)
\(\left( .00625\quad of\quad \frac { 23 }{ 5 } \right) =\left( \frac { 625 }{ 100000 } \times \frac { 23 }{ 5 } \right) =\frac { 23 }{ 800 } \)
.0.4\(\times\)? = .000016.
- (a)
0.0004
- (b)
0.04
- (c)
4
- (d)
None of these
Let .04\(\times\)x = .000016. Then, x = \({.000016\over.04} = {.0016\over4} = .0004.\)
\({.009\over?}=.01\)
- (a)
.0009
- (b)
.09
- (c)
.9
- (d)
9
Let \({.009\over x} = .01.\) Then, \(x={.009\over.01}={.9\over1}=.9.\)
The price pf commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was rs.4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
- (a)
2010
- (b)
2011
- (c)
2012
- (d)
2013
Suppose commodity X will cost 40 paise more than Y sfter z years. Then, (4.20+0.40z) - (6.30+0.15z) = 0.40
\(\Leftrightarrow \) 0.25z=0.40 + 2.10 \(\Leftrightarrow \) z=\(\frac { 2.50 }{ 0.25 } =\frac { 250 }{ 25 } =10\).
\(\therefore \) X will cost 40 paise more than Y 10 years after 2001 i.e., in 2011.
When 0.232323.... is converted into a fraction, then the result is;
- (a)
\(1\over5\)
- (b)
\(2\over9\)
- (c)
\(23\over99\)
- (d)
\(23-\over100\)
0.232323...........\(0.\bar{23}={23\over99}\)
The least among the following is:
- (a)
0.2
- (b)
1 \(\div\) 0.2
- (c)
\(0.\overline{2}\)
- (d)
\((0.2)^{2}\)
\(1\div0.2={1\over0.2}={10\over2}=5; 0.\overline{2}=0.222......;(0.2)^{2}=0.04.\)
\(0.04<0.2<0.22......<5.\)
Since 0.04 is the least, so \((0.2)^{2}\) is the least.
If 213 \(\times\)16 = 3408, then 1.6 \(\times\) 21.3 is equal to :
- (a)
0.3408
- (b)
3.408
- (c)
34.08
- (d)
340.8
\(1.6\times 21.3=\left( \frac { 16 }{ 10 } \times \frac { 213 }{ 10 } \right) =\left( \frac { 16\times 213 }{ 100 } \right) \frac { 3408 }{ 100 } =34.08\)
When 52416 is divided by 312, the quotient is 168. What will be the quotient when 52.416 is divided by 0.0168 ?
- (a)
3.12
- (b)
312
- (c)
3120
- (d)
None of these
\({96.54-89.63\over96.54+89.63}\div{965.4-896.3\over9.654+8.963}=?\)
- (a)
\(10^{-2}\)
- (b)
\(10^{-1}\)
- (c)
10
- (d)
None of these
The value of \({1\over4}+{1\over4\times5}+{1\over4\times5\times6}\) correct to 4 decimal places is :
- (a)
0.3075
- (b)
0.3082
- (c)
0.3083
- (d)
0.3085
\(\frac { 1 }{ 4 } +\frac { 1 }{ 4\times 5 } +\frac { 1 }{ 4\times 5\times 6 } =\frac { 1 }{ 4 } \left( 1+\frac { 1 }{ 5 } +\frac { 1 }{ 30 } \right) =\frac { 1 }{ 4 } \left( \frac { 30+6+1 }{ 30 } \right) =\frac { 1 }{ 4 } \times \frac { 37 }{ 30 } =\frac { 37 }{ 120 } =0.3083\)
The sum of the first 20 terms of the series \({1\over5\times6}+{1\over6\times7}+{1\over7\times8}+.....is :\)
- (a)
0.16
- (b)
1.6
- (c)
16
- (d)
None of these
The value of \(3.6 \times0.48\times2.50\over0.12\times0.09\times0.5\) is :
- (a)
80
- (b)
800
- (c)
8000
- (d)
80000
\(\frac { 3.6\times 0.48\times 2.50 }{ 0.12\times 0.09\times 0.5 } =\frac { 36\times 48\times 250 }{ 12\times 9\times 5 } =800\)
\({5\times 1.6 - 2 \times 1.4 \over 1.3 } = ?\)
- (a)
0.4
- (b)
1.2
- (c)
1.4
- (d)
4
Give expression = \(\frac { 8-2.8 }{ 1.3 } =\frac { 5.2 }{ 1.3 } =\frac { 52 }{ 13 } =4\)
The value of \(({8.6\times 5.3 + 8.6 \times 4.7 \over 4.3 \times 9.7 - 4.3 \times 8.7})\) is :
- (a)
3.3
- (b)
6.847
- (c)
13.9
- (d)
20
Given expression=\(\frac { 8.6\times \left( 5.3+4.7 \right) }{ 4.3\times \left( 9.7-8.7 \right) } =\frac { 8.6\times 10 }{ 4.3\times 1 } =20\)
Evalute : \((2.39)^{2} - (1.61)^{2} \over 2.39 \ - \ 1.61\)
- (a)
2
- (b)
4
- (c)
6
- (d)
8
Give expression= \(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ a\quad b } =\frac { \left( a+b \right) \left( a-b \right) }{ \left( a-b \right) } =(a+b)=(2.39+1.61)=4\)
simplify : \(5.32 \times 56 + 5.32 \times 44 \over (7.66)^2 - (2.34)^2\)
- (a)
7.2
- (b)
8.5
- (c)
10
- (d)
12
Given expression= \(\frac { 5.32\times (56+44) }{ (7.66+2.34)(7.66-2.34) } =\frac { 5.32\times 100 }{ 10\times 5.32 } =10\)
The expression (11.98 x 11.98 + 11.98 x X + 0.02 x 0.02) will be a perfect square for X equal to
- (a)
0.02
- (b)
0.2
- (c)
0.04
- (d)
0.4
Given expression= \(\left( 11.98 \right) ^{ 2 }+{ \left( 0.02 \right) }^{ 2 }+11.98\times x\)
For the given expression to be a perfect square, we must have
11.98 \(\times \) x=2 \(\times \) 11.98 \(\times \) 0.02 or x=0.04.
The value of \((0.137 + 0.098)^2 - (0.137 - 0.098)^2\over 0.137 \times 0.098\) is
- (a)
0.039
- (b)
0.235
- (c)
0.25
- (d)
4
Given expression=\(\frac { { (a }+{ b })^{ 2 }-(a-b)^{ 2 } }{ ab } =\frac { 4ab }{ ab } =4\)
\(({10.3\times 10.3 \times 10.3 + 1\over 10.3\times 10.3 - 10.3 + 1})\) is equal to :
- (a)
9.3
- (b)
10.3
- (c)
11.3
- (d)
12.3
\([{8(3.75)^3 + 1\over (7.5)^2 - 6.5 }]\) is equal to :
- (a)
\({9\over 5}\)
- (b)
2.75
- (c)
4.75
- (d)
8.5
Given expression \(=\frac { \left( { 2\times 3.75) }^{ 3 }+({ 1 } \right) ^{ 3 } }{ \left( { 7.5 } \right) ^{ 2 }-\left( 7.5\times 1 \right) +\left( { 1 } \right) ^{ 2 } } =\frac { \left( 7.5 \right) ^{ 3 }+\left( { 1 } \right) ^{ 3 } }{ \left( { 7.5 } \right) ^{ 2 }-(7.5\times 1)+\left( 1 \right) ^{ 2 } } \)
\(=\left( \frac { { a }^{ 3 }+{ b }^{ 3 } }{ { a }^{ 2 }-ab+{ b }^{ 2 } } \right) =(a+b)=\quad (7.5+1)=8.5\)
The value of \((2.3)^3 - .027 \over (2.3)^2+.69+.09\) is
- (a)
0
- (b)
1.6
- (c)
2
- (d)
3.4
Given expression\(=\frac { { \left( 2.3 \right) }^{ 3 }-{ \left( 0.3 \right) }^{ 3 } }{ \left( 2.3 \right) ^{ 2 }+(2.3\times 0.3)+\left( 0.3 \right) ^{ 2 } } =\left( \frac { { a }^{ 3 }-{ b }^{ 3 } }{ { a }^{ 2 }+ab+{ b }^{ 2 } } \right) \)
The value of \((0.06)^2 + (0.47)^2+ (0.079)^2 \over (0.006)^2+(0.047)^2+(0.0079)^2\) is :
- (a)
0.1
- (b)
10
- (c)
100
- (d)
1000
Given expression \(=\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ \left( { \frac { a }{ 10 } } \right) ^{ 2 }+\left( \frac { b }{ 10 } \right) ^{ 2 }+\left( \frac { c }{ 10 } \right) ^{ 2 } } \), where a=0.6, b=0.47 and c=0.079.
\(=\frac { 100\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }{ \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) } =100\)