141.
(A) 0.1
(B) 10
(C) 102
(D) 103
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Solution:
$$\eqalign{ & {\text{According to the question,}} \cr & \sqrt {\frac{{{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}}}{{{{\left( {0.003} \right)}^2} + {{\left( {0.021} \right)}^2} + {{\left( {0.0065} \right)}^2}}}} \cr & = \sqrt {\frac{{0.0009 + 0.0441 + 0.004225}}{{0.000009 + 0.000441 + 0.00004225}}} \cr & = \sqrt {\frac{{0.049225}}{{0.00049225}}} \cr & = \sqrt {100} \cr & = 10 \cr} $$
142.
ÃÂÃÂ ÃÂÃÂ is equal to = ?
(A) 2
(B) 2.5
(C) 3
(D) 3.5
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Solution:
$$\eqalign{ & {\text{Given expression,}} \cr & \frac{{28 + 14 + 7 + 4 + 2 + 1}}{{28}} \cr & = \frac{{56}}{{28}} \cr & = 2 \cr} $$
143.
ÃÂÃÂ ÃÂÃÂ is equal to = ?
(A) a - b
(B) b - a
(C) 1
(D) 0
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Solution:
$$\eqalign{ & \frac{1}{{1 + {2^{a - b}}}} + \frac{1}{{1 + {2^{b - a}}}} \cr & = \frac{1}{{1 + {2^{a - b}}}} + \frac{1}{{1 + {2^{ - \left( {a - b} \right)}}}} \cr & = \frac{1}{{1 + {2^{a - b}}}} + \frac{{{2^{a - b}}}}{{{2^{a - b}} + 1}} \cr & = \frac{{1 + {2^{a - b}}}}{{1 + {2^{a - b}}}} \cr & = 1 \cr} $$
144.
The value of 3 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 18 of 3 ÃÂÃÂÃÂÃÂÃÂà6 + 21 ÃÂÃÂÃÂÃÂÃÂà6 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 18 - 3 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 2 + 3 - 3 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 9 of 3 ÃÂÃÂÃÂÃÂÃÂà9 is:
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Solution:
$$\eqalign{ & 3 \div 18{\text{ of }}3 \times 6 + 21 \times 6 \div 18 - 3 \div 2 + 3 - 3 \div 9{\text{ of }}3 \times 9 \cr & = \frac{3}{{54}} \times 6 + \frac{{21 \times 6}}{{18}} - \frac{3}{2} + 3 - \frac{3}{{27}} \times 9 \cr & = \frac{1}{3} + 7 - \frac{3}{2} + 3 - 1 \cr & = 9 + \frac{1}{3} - \frac{3}{2} \cr & = \frac{{54 + 2 - 9}}{6} \cr & = \frac{{47}}{6} \cr} $$
145.
1 - [5 - {2 + (- 5 + 6 - 2) 2}] is equal to:
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Solution:
1 - [5 - {2 + (- 5 + 6 - 2) 2}] = 1 - [5 - {2 + (- 1) 2}] = 1 - [5 - {2 - 2}] = 1 - [5 - 0] = 1 - 5 = -4
146.
Find the value of
(A) -16
(B)
(C) 16
(D) 16.2
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Solution:
$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {248 + \sqrt {52 + \sqrt {144} } } \cr & \Rightarrow \sqrt {248 + \sqrt {52 + 12} } \cr & \Rightarrow \sqrt {248 + \sqrt {64} } \cr & \Rightarrow \sqrt {256} \cr & \Rightarrow \pm 16 \cr} $$
147.
A teacher wants to arrange his students in an equal number of rows and columns. If there are 1369 students, the number of students in the last row are ?
(A) 37
(B) 33
(C) 63
(D) 47
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Solution:
Let the number of students in a row be 'x' According to question ⇒ x × x = 1369 ⇒ x2 = 1369 ⇒ x = 37
148.
ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ is equal to = ?
(A) 12
(B) 11
(C) 10
(D) 15
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Solution:
According to question $$\root 3 \of {{{\left( {333} \right)}^3} + {{\left( {333} \right)}^3} + {{\left( {334} \right)}^3} - 3 \times 333 \times 333 \times 334} $$ As we know that a3 + b3 + c3 - 3abc = $$\frac{1}{2}$$ (a + b + c)[(a - b)2 + (b - c)2 + (c - a)2] = $$\frac{1}{2}$$ (333 + 333 + 334)[(333 - 333)2 + (333 - 334)2 + (334 - 333)2] = $$\frac{1}{2}$$ × 1000[0 + 1 + 1] = $$\frac{1}{2}$$ × 1000 × 2 = $$\root 3 \of {1000} $$ = 10
149.
If a + 2b = 6 and ab = 4, then what is
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Solution:
$$\eqalign{ & \frac{2}{a} + \frac{1}{b} \cr & = \frac{{2b + a}}{{ab}} \cr & = \frac{6}{4} \cr & = \frac{3}{2} \cr} $$
150.
Let ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ When y is added to x, the result is ÃÂÃÂ What is the value of y?
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Solution:
$$\eqalign{ & x = \frac{{5\frac{3}{4} - \frac{3}{7} \times 15\frac{3}{4} + 2\frac{2}{{35}} \div 1\frac{{11}}{{25}}}}{{\frac{3}{4} \div 5\frac{1}{4} + 5\frac{3}{5} \div 3\frac{4}{{15}}}} \cr & x = \frac{{\frac{{23}}{4} - \frac{3}{7} \times \frac{{63}}{4} + \frac{{72}}{{35}} \div \frac{{36}}{{25}}}}{{\frac{3}{4} \div \frac{{21}}{4} + \frac{{28}}{5} \div \frac{{49}}{{15}}}} \cr & x = \frac{{\frac{{23}}{4} - \frac{{27}}{4} + \frac{{10}}{7}}}{{\frac{1}{7} + \frac{{12}}{7}}} \cr & x = \frac{{ - \frac{4}{4} + \frac{{10}}{7}}}{{\frac{{13}}{7}}} \cr & x = \frac{{\frac{{ - 7 + 10}}{7}}}{{\frac{{13}}{7}}} \cr & x = \frac{3}{{13}} \cr & x + y = \frac{7}{{13}} \cr & y = \frac{7}{{13}} - \frac{3}{{13}} \cr & y = \frac{4}{{13}} \cr} $$