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Quantitative Aptitude Questions and Answers with Explanation
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In this section, you can find numerous aptitude questions with answers and explanation. The quantitative aptitude questions with answers mentioned above covers various categories and extremely helpful for competitive exams. All the answers are explained in detail with very detailed answer descriptions.
The quantitative aptitude questions mentioned above also contain aptitude questions asked for various placement exams and competitive exams. These will help students who are preparing for any type of competitive examinations.
Quantities aptitude questions given here are extremely useful for all kind of competitive exams like Common Aptitude Test (CAT),MAT, GMAT, IBPS Exam, CSAT, CLAT , Bank Competitive Exams, ICET, UPSC Competitive Exams, CLAT, SSC Competitive Exams, SNAP Test, KPSC, XAT, GRE, Defense Competitive Exams, L.I.C/ G. I.C Competitive Exams , Railway Competitive Exam, TNPSC, University Grants Commission (UGC), Career Aptitude Test (IT Companies) and etc., Government Exams etc.
Comments(188)
tarun2017-03-05 14:58:16
why don't you put reasoning and verbal ability questions. I like this page very much, because of aptitude questions. Why don't you put reasoning and verbal questions.
Anil2015-04-09 13:59:29
1) A water tank is hemispherical below and cylindrical at the top. If the radius is $12$ m and capacity is $3312\pi$ cubic metre, the height of the cylindrical portion in metres is:
2) If three metallic spheres of radii $6$ cms, $8$ cms and $10$ cms are melted to from a single sphere, the diameter of the new sphere will be:
3) Sixty men can build a wall in $40$ days, but though they begin the work together, $5$ men quit every ten days. The time needed to build the wall is:
4) Shyam is travelling on his cycle and has calculated to reach point 'A' at $2$ PM If he travels at $10$ kmph. he will reach there at $12$ noon if he travels at $15$ kmph. At what speed must he travel to reach point 'A' at $1$ PM:
5) An aeroplane travelling at $700$ kmph in level flight drops a bomb from a height of $1000$ metres on a target. The time taken from releasing the bomb to hitting the target is nearest to the figure:
6) In a History examination, the average for the entire class was $80$ marks. If $10\%$ of the students scored $95$ marks and $20\%$ scored $90$ marks. What was the average marks of the remaining students of the class
2) If three metallic spheres of radii $6$ cms, $8$ cms and $10$ cms are melted to from a single sphere, the diameter of the new sphere will be:
3) Sixty men can build a wall in $40$ days, but though they begin the work together, $5$ men quit every ten days. The time needed to build the wall is:
4) Shyam is travelling on his cycle and has calculated to reach point 'A' at $2$ PM If he travels at $10$ kmph. he will reach there at $12$ noon if he travels at $15$ kmph. At what speed must he travel to reach point 'A' at $1$ PM:
5) An aeroplane travelling at $700$ kmph in level flight drops a bomb from a height of $1000$ metres on a target. The time taken from releasing the bomb to hitting the target is nearest to the figure:
6) In a History examination, the average for the entire class was $80$ marks. If $10\%$ of the students scored $95$ marks and $20\%$ scored $90$ marks. What was the average marks of the remaining students of the class
PRASHANT SHRIVASTAVA2016-03-16 13:38:39
6) Suppose students $=100$
$10×95=950\\
20×90=1800$
$950+1800+70x=80×100\\
x=75$
$10×95=950\\
20×90=1800$
$950+1800+70x=80×100\\
x=75$
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PRASHANT SHRIVASTAVA2016-03-16 13:33:18
4) $d= t×10$
$d=(t-2)×15$
$t=6$ hrs
$d=60$
$60=5×s$
speed $=12$ km/hr
$d=(t-2)×15$
$t=6$ hrs
$d=60$
$60=5×s$
speed $=12$ km/hr
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PRASHANT SHRIVASTAVA2016-03-16 13:27:12
3) Total man days $=2400$
$60×10+55×10+50×10+45×10=2100$ man days
$\dfrac{300}{40}=7.5$ days
Total days $=10+10+10+10+7.5=47.5$ days
$60×10+55×10+50×10+45×10=2100$ man days
$\dfrac{300}{40}=7.5$ days
Total days $=10+10+10+10+7.5=47.5$ days
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PRASHANT SHRIVASTAVA2016-03-16 13:16:24
2) $\dfrac{4}{3}×\dfrac{22}{7}×r×r×r=\dfrac{4}{3}×\dfrac{22}{7}×(216+512+1000)$
$r×r×r=1728$
$r=12$ cm
$r×r×r=1728$
$r=12$ cm
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PRASHANT SHRIVASTAVA2016-03-16 13:10:15
1) Volume, V $=\dfrac{22}{7}×12×12\left(\dfrac{4}{3}×12×\dfrac{1}{2}+h\right)=3312×\dfrac{22}{7}$
$12×12\left(\dfrac{4}{3}×12×\dfrac{1}{2}+h\right)=3312\\
\dfrac{4}{3}×12×\dfrac{1}{2}+h=23\\
8+h=23\\
h=23-8=15\text{ m}$
$12×12\left(\dfrac{4}{3}×12×\dfrac{1}{2}+h\right)=3312\\
\dfrac{4}{3}×12×\dfrac{1}{2}+h=23\\
8+h=23\\
h=23-8=15\text{ m}$
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Ritesh Singh2016-02-04 05:02:52
we will have to take up physics formulae for 5th one
$s=ut+0.5at^2$
here $s=1000$
$u=0$ (initial speed)
$a=g=9.8$ m/s2
So time can be calculated as $14.14$ sec
$s=ut+0.5at^2$
here $s=1000$
$u=0$ (initial speed)
$a=g=9.8$ m/s2
So time can be calculated as $14.14$ sec
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Ashutosh R2015-08-11 12:47:55
Answer for Q2:
If three metallic spheres melted down and formed a new sphere, volume of all three sphere will be equivalent to the volume of new sphere and final equation will be
$(r_1)^3+(r_2)^3+(r_3)^3=R^3$
here,
$6^3+8^3+10^3=R^3\\
\Rightarrow 216+512+1000=R^3\\
\Rightarrow R=1728^{1/3}\\
\Rightarrow R=12\text{ cm}$
Hence, diameter of the new sphere will be $(2R)$
i.e $24$ cm.
If three metallic spheres melted down and formed a new sphere, volume of all three sphere will be equivalent to the volume of new sphere and final equation will be
$(r_1)^3+(r_2)^3+(r_3)^3=R^3$
here,
$6^3+8^3+10^3=R^3\\
\Rightarrow 216+512+1000=R^3\\
\Rightarrow R=1728^{1/3}\\
\Rightarrow R=12\text{ cm}$
Hence, diameter of the new sphere will be $(2R)$
i.e $24$ cm.
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Yash2015-07-31 05:52:28
4. Since distance is same,
$s_1×t_1=s_2×t_2\quad$($s_1=10$ kmph, $s_2=15$ kmph)
$10(14-t)=15(12-t)\quad$ ($2$ o'clock on $24$ hrs clock is $14,$ while $12$ is $12$)
$140-10t=180-15t\quad$ ($t$ is the start time)
$5t=40$
$t=8$ hrs, means he had started journey at $8$ am
$d=s_1(14-t)\\
d=10(14-8)\\
d=60\text{ km}$
Time to reach destination $=13-8=5$ hrs
Speed $=\dfrac{d}{t}=\dfrac{60}{5}=12$ kmph
$s_1×t_1=s_2×t_2\quad$($s_1=10$ kmph, $s_2=15$ kmph)
$10(14-t)=15(12-t)\quad$ ($2$ o'clock on $24$ hrs clock is $14,$ while $12$ is $12$)
$140-10t=180-15t\quad$ ($t$ is the start time)
$5t=40$
$t=8$ hrs, means he had started journey at $8$ am
$d=s_1(14-t)\\
d=10(14-8)\\
d=60\text{ km}$
Time to reach destination $=13-8=5$ hrs
Speed $=\dfrac{d}{t}=\dfrac{60}{5}=12$ kmph
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