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# Decimal Fraction Questions

Home > Quantitative Aptitude > Decimal Fraction > General Questions
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An internet recently hired 8 new network,in advertisement 20 network already employed.All new network cam from university A.In addition 75%of computer advertisement came from same university A.what fraction of original 20 network advertisement came from same university A?

A$$\frac{13}{20}$$

B$$\frac{11}{20}$$

C13

D11

Answer: Option A

Explanation:

8 new n/w,20 is already employed
Total n/w = 20+8 = 28 n/w
so that 75% came from same university then $$28 \times \frac{75}{100} = 21$$
we all know 8 n/w is recently hired then 21-8 = 13
=> the fraction is = $$\frac{13}{20}$$

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Arrange the fractions $$\frac{3}{5},\frac{4}{7},\frac{8}{9} and \frac{9}{11}$$ in their descending order.

A$$\frac{2}{9}>\frac{6}{11}>\frac{4}{5}>\frac{3}{7}$$.

B$$\frac{8}{9}>\frac{6}{11}>\frac{4}{5}>\frac{3}{7}$$.

C$$\frac{8}{9}>\frac{9}{11}>\frac{4}{5}>\frac{3}{7}$$.

D$$\frac{8}{9}>\frac{9}{11}>\frac{3}{5}>\frac{4}{7}$$.

Answer: Option D

Explanation:

clearly, $$\frac{3}{5}=0.6,\frac{4}{7}=0.571,\frac{8}{9}=0.88,\frac{9}{11}=0.818$$.
now, $$0.88>0.818>0.6>0.571$$.
so,$$\frac{8}{9}>\frac{9}{11}>\frac{3}{5}>\frac{4}{7}$$.

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what value will replace the question mark in the following equations? (i) $$5172.49+378.352+?=9318.678$$
(ii) $$?-7328.96=5169.38$$

A12496.43,3768.798

B3767.836, 12498.34

C1249.34,3767.836

D4394.12,3767.836

Answer: Option B

Explanation:

(i) $$5172.49+378.352+x=9318.678$$
then, $$x=9318.678-(5172+378.352)=9318.678-5550.842$$
=3767.836
(ii) $$?-7328.96=5169.38$$
let $$x-7328.96=51.69.38$$
then,$$x=5169.38+7328.96=12498.34$$

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Evaluate:
(i) 35+.07
(ii) 2.5 +0.0005
(iii) 136.09+43.9

A(i) 500 (ii) 3.1 (iii) 5000

B(i)500 (ii)5000 (iii)3.1

C(i) 3.1 (ii)5000 (iii)500

D(i) 5000 (ii) 500 (iii) 3.1`

Answer: Option B

Explanation:

(i) $$\frac{35}{.07}=\frac{35\times100}{.07\times100}=\frac{3500}{7}=500$$
(ii) $$\frac{2.5}{0.0005}=\frac{2.5\times10000}{.0005\times10000}=\frac{25000}{5}=5000$$
(iii) $$\frac{136.09}{43.9}=\frac{136.09\times10}{43.9\times10}=\frac{1360.9}{439}=3.1$$

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Simplify: $$\frac{0.05\times0.05\times0.05+0.04\times0.04\times0.04}{0.05\times0.05-0.05\times0.04+0.04\times0.04}$$

A0.9

B0.06

C0.09

D0.009

Answer: Option C

Explanation:

Given expression = $$\left(\frac{a^3+b^3}{a^2-ab+b^2}\right)$$, where a=0.05, b=0.04
=(a+b)=(0.05+0.04)=0.09

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Which of the following has fraction in ascending order?

A$$\frac{2}{5},\frac{3}{5},\frac{1}{3},\frac{4}{7},\frac{5}{6},\frac{6}{7}$$

B$$\frac{1}{3},\frac{2}{5},\frac{4}{7},\frac{3}{5},\frac{5}{6},\frac{6}{7}$$

C$$\frac{1}{3},\frac{2}{5},\frac{3}{5},\frac{4}{7},\frac{5}{6},\frac{6}{7}$$

D$$\frac{1}{3},\frac{2}{5},\frac{3}{5},\frac{5}{6},\frac{4}{7},\frac{6}{7}$$

Answer: Option B

Explanation:

Converting each of the given fractions into decimal form, we get
$$\frac{1}{3}=0.33,\frac{2}{5}=0.4,\frac{4}{7}=0.57,\frac{3}{5}=0.6,\frac{5}{6}=,0.82,\frac{6}{7}=0.857$$
Clearly $$0.33<0.4<0.57<0.6<0.82<0.857.$$
so, $$\frac{1}{3},\frac{2}{5},\frac{4}{7},\frac{3}{5},\frac{5}{6},\frac{6}{7}$$

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Which of the following has fraction in ascending order?

A$$\frac{8}{9},\frac{9}{11},\frac{7}{9},\frac{2}{3},\frac{3}{5}$$

B$$\frac{2}{3},\frac{3}{5},\frac{7}{9},\frac{9}{11},\frac{8}{9}$$

C$$\frac{3}{5},\frac{2}{3},\frac{9}{11},\frac{7}{9},\frac{8}{9}$$

D$$\frac{3}{5},\frac{2}{3},\frac{7}{9},\frac{9}{11},\frac{8}{9}$$

Answer: Option D

Explanation:

Converting each of the given fractions into decimal form, we get
$$\frac{2}{3}=0.66,\frac{3}{5}=0.6,\frac{7}{9}=0.77,\frac{9}{11}=0.81,\frac{8}{9}=0.88$$
Clearly $$0.6<0.66<0.77<0.81<0.88.$$
so, $$\frac{3}{5}<\frac{2}{3}<\frac{7}{9}<\frac{9}{11}<\frac{8}{9}$$

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Which of the following are descending order of their value?

A$$\frac{11}{17},\frac{7}{11},\frac{5}{9},\frac{8}{15}$$

B$$\frac{5}{9},\frac{7}{11},\frac{8}{15},\frac{11}{17}$$

C$$\frac{5}{9},\frac{8}{15},\frac{11}{17},\frac{7}{11}$$

D$$\frac{11}{17},\frac{7}{11},\frac{8}{15},\frac{5}{9}$$

Answer: Option A

Explanation:

Converting each of the given fractions into 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 $$0.647>0.63>0.55>0.533<0.88.$$
so, $$\frac{11}{17}>\frac{7}{11}>\frac{5}{9}>\frac{8}{15}$$

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What is the diffrence between the biggest and the smallest fraction among $$\frac{2}{3},\frac{3}{4},\frac{4}{5} and \frac{5}{6}$$ ?

A$$\frac{1}{30}$$

B$$\frac{1}{6}$$

C$$\frac{1}{12}$$

D$$\frac{1}{20}$$

Answer: Option B

Explanation:

Converting each of the given fractions into decimal form, we get
$$\frac{2}{3}=0.66,\frac{3}{4}=0.75,\frac{4}{5}=0.8,\frac{5}{6}=0.8333,\frac{8}{9}=0.88$$
Clearly $$0.8333>0.8>0.75>0.66$$
so, $$\frac{5}{6}>\frac{4}{5}>\frac{3}{4}>\frac{2}{3}$$
therefore, required diffrence =$$\left(\frac{5}{6}-\frac{2}{3}\right)=\frac{1}{6}$$

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Which of the following fraction is greater then $$\frac{3}{4}$$ and less then $$\frac{5}{6}$$ ?

A$$\frac{9}{10}$$

B$$\frac{1}{2}$$

C$$\frac{2}{3}$$

D$$\frac{4}{5}$$

Answer: Option D

Explanation:

$$\frac{3}{4}=0.75,\frac{5}{6}=0.8333,\frac{1}{2}=0.5,\frac{2}{3}=0.66,\frac{4}{5}=0.8,\frac{9}{10}=0.9$$
Clearly, 0.8 lies between 0.75 and 0.8333
therefore, $$\frac{4}{5}$$ lies between $$\frac{3}{4} and\frac{5}{6}$$

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