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.

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showing 1-10 of 187 comments,   sorted newest to the oldest
Anil
2015-04-09 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
(0) (0) Reply
PRASHANT SHRIVASTAVA
2016-03-16 13:38:39
6) Suppose students $=100$

$10×95=950\\ 20×90=1800$

$950+1800+70x=80×100\\ x=75$
(0) (0) Reply
PRASHANT SHRIVASTAVA
2016-03-16 13:33:18
4) $d= t×10$
$d=(t-2)×15$

$t=6$ hrs
$d=60$

$60=5×s$
speed $=12$ km/hr
(0) (0) Reply
PRASHANT SHRIVASTAVA
2016-03-16 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
(0) (0) Reply
PRASHANT SHRIVASTAVA
2016-03-16 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
(0) (0) Reply
PRASHANT SHRIVASTAVA
2016-03-16 13:10:15
1) Volume, V $=\dfrac{22}{7}×12×12\left(\dfrac{4}{3}×12×\dfrac{1}{2}+h\right)=3312×\dfrac{22}{7}$
$12×12\left(\dfrac{4}{3}×12×\dfrac{1}{2}+h\right)=3312\\ \dfrac{4}{3}×12×\dfrac{1}{2}+h=23\\ 8+h=23\\ h=23-8=15\text{ m}$
(0) (0) Reply
Ritesh Singh
2016-02-04 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
(0) (0) Reply
Ashutosh R
2015-08-11 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.
(0) (0) Reply
Yash
2015-07-31 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
(0) (0) Reply
Titas
2015-07-28 21:30:00
$4.$ Since we do not know at what time Shyam actually sets on his journey, we will consider any one of the time limits provided (either $12$ p.m or $2$ p.m) to be taking $x$ hours from the time he started, that is we have to consider either duration to be $x$ (or $y$ or $z$ or whatever you wish to take) hours. Now, if we consider this for $2$ p.m, then by time = distance/speed, we have

$x=\dfrac{d}{10}~~\cdots(1)$

Now, $(x-2)=\dfrac{d}{15}$
$\Rightarrow 15x-30=d\\ \Rightarrow 15x=30+d\\ \Rightarrow x=\dfrac{d+30}{15}~~\cdots(2)$

From $(1)$ and $(2)$, we have
$\dfrac{d+30}{15}=\dfrac{d}{10}\\ \Rightarrow 10d+300=15d\\ \Rightarrow 5d=300\\ \Rightarrow d=60\text{ km}$

Now,
$x-1=\dfrac{d}{10}-1\\ x-1=\dfrac{60}{10}-1=5$
[We are talking about $1$ p.m and we have chosen the time duration of up to to $2$ p.m to be $x$, so up to $1$ p.m,it is $(x-1)$ hours]

So,the time taken up to $1$ p.m is $5$ hours.
Therefore, speed = distance/time $=\dfrac{60}{5}$ km/hr
$=12$ km/hr
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