121.
ÃÂÃÂ is equal to :
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Solution:
$${0.34\overline {67} + 0.13\overline {33} }$$ = $$\frac{3467 - 34}{9900}$$ + $$\frac{1333 - 13}{9900}$$ = $$\frac{3433 + 1320}{9900}$$ = $$\frac{4753}{9900}$$ = $$\frac{4801 - 48}{9900}$$ = $${0.48\overline {01} }$$
122.
The value of + + ÃÂÃÂÃÂÃÂ correct to 4 decimal places is :
(A) 0.3075
(B) 0.3082
(C) 0.3083
(D) 0.3085
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Solution:
$$\eqalign{ & = \frac{1}{4} + \frac{1}{{4 \times 5}} + \frac{1}{{4 \times 5 \times 6}} \cr & = \frac{1}{4}\left( {1 + \frac{1}{5} + \frac{1}{{30}}} \right) \cr & = \frac{1}{4}\left( {\frac{{30 + 6 + 1}}{{30}}} \right) \cr & = \frac{1}{4} \times \frac{{37}}{{30}} \cr & = \frac{{37}}{{120}} \cr & = 0.3083 \cr} $$
123.
32.4 ÃÂÃÂÃÂÃÂÃÂÃÂ 11.5 ÃÂÃÂÃÂÃÂÃÂÃÂ 8.5 = ?
(A) 3149.5
(B) 3129.1
(C) 3167.1
(D) 3162.5
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Solution:
324 × 115 × 85 = 3167100 Sum of decimal place = 3 ∴ 32.4 × 11.5 × 8.5 = 3167.1
124.
The value of ÃÂÃÂ ÃÂÃÂ is closest to:
(A) 0.006
(B) 0.06
(C) 0.6
(D) 6
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Solution:
$$\eqalign{ & \frac{{489.1375 \times 0.0483 \times 1.956}}{{0.0873 \times 92.581 \times 99.749}} \approx \frac{{489 \times 0.05 \times 2}}{{0.09 \times 93 \times 100}} \cr & = \frac{{489}}{{9 \times 93 \times 10}} \cr & = \frac{{163}}{{279}} \times \frac{1}{{10}} \cr & = \frac{{0.58}}{{10}} \cr & = 0.058 \approx 0.06 \cr} $$
125.
Which of the following are in descending order of their values ?
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Solution:
Converting each of the given fractions in to decimal form, we get $$\frac{5}{9}$$ = 0.55 $$\frac{7}{11}$$ = 0.63 $$\frac{8}{15}$$ = 0.533 $$\frac{11}{17}$$ = 0.647 Clearly, 0.647 > 0.63 > 0.55 > 0.533 So, $$\frac{11}{17}$$ > $$\frac{7}{11}$$ > $$\frac{5}{9}$$ > $$\frac{8}{15}$$
126.
3927 + 5526 ÃÂÃÂÃÂ÷ 12.5 = ?
(A) 750.24
(B) 756.24
(C) 4369.08
(D) 4369.24
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Solution:
Given expression : $$\eqalign{ & = 3927 + \frac{{5526}}{{12.5}} \cr & = 3927 + \frac{{55260}}{{125}} \cr} $$ = 3927 + 442.08 = 4369.08
127.
10.0001 + 9.9999 - 8.9995 = ?
(A) 9.0005
(B) 10.9995
(C) 11.0001
(D) 11.0005
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Solution:
10.0001 + 9.9999 - 8.9995 = 20.0000 - 8.9995 = 11.0005
128.
+ 2 - = ÃÂÃÂ +
(A) 2
(B) 8
(C) 512
(D) 324
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Solution:
$$\eqalign{ & 2\frac{{1.5}}{5} + 2\frac{1}{6} - 1\frac{{3.5}}{{15}} = \frac{{{x^{\frac{1}{3}}}}}{4} + 1\frac{7}{{30}} \cr & \Rightarrow \frac{{11.5}}{5} + \frac{{13}}{6} - \frac{{18.5}}{{15}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr} $$ ⇒ L.C.M. of 5, 6 and 15 is 30 $$\eqalign{ & \Rightarrow \frac{{69 + 65 - 37}}{{30}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr & \Rightarrow \frac{{97}}{{30}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr & \Rightarrow \frac{{{x^{\frac{1}{3}}}}}{4} = \frac{{97}}{{30}} - \frac{{37}}{{30}} \cr & \Rightarrow {x^{\frac{1}{3}}} = \frac{{60}}{{30}} \times 4 \cr & \Rightarrow {x^{\frac{1}{3}}} = 8 \cr & \Rightarrow x = {\left( 8 \right)^3} \cr & \Rightarrow x = 512 \cr} $$ Hence, the number is 512
129.
Solve this : ÃÂÃÂÃÂÃÂ ÃÂÃÂÃÂÃÂ = ?
(A) 0.8
(B) 1.45
(C) 2.40
(D) 3.25
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Solution:
= $$\frac{0.0203 × 2.92}{0.0073 × 14.5 × 0.7}$$ = $$\frac{203 × 292}{73 × 145 × 7}$$ = $$\frac{4}{5}$$ = 0.8
130.
Solve : 1576 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 45.02 + 23.99 ÃÂÃÂÃÂÃÂÃÂàÃÂÃÂÃÂà= ?
(A) 340
(B) 420
(C) 380
(D) 460
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Solution:
Given expression : = 1576 ÷ 45.02 + 23.99 × $$\sqrt {255} $$ = 1575 ÷ 45 + 24 × $$\sqrt {256} $$ = 35 + 24 × 16 = 35 + 384 = 419 $$ \approx $$ 420