| s.no |
question |
Answer |
time |
| 1) |
A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes? |
Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr
Distance covered in 6 minutes =[(1/60)*6] km = (1/10)km = 100 m.
Distance between the thief and policeman = (200 – 100),
m = 100 m.
|
5min |
| 2) |
A fast typist can type some matter in 2 hours and a slow typist can type the same in
3 hours. If both type combinely, in how much time will they finish? |
The fast typist's work done in 1 hr = 1/2,
The slow typist's work done in 1 hr = 1/3,
If they work combinely, work done in 1 hr = 1/2+1/3 = 5/6,
So, the work will be completed in 6/5 hours. i.e., 1+1/5 ,
hours = 1hr 12 min,
|
5min |
| 3) |
Gavaskar's average in his first 50 innings was 50. After the 51st innings, his
average was 51. How many runs did he score in his 51st innings. (supposing that he
lost his wicket in his 51st innings) |
Total score after 50 innings = 50*50 = 2500
Total score after 51 innings = 51*51 = 2601
So, runs made in the 51st innings = 2601-2500 = 101
If he had not lost his wicket in his 51st innings, he would have scored an unbeaten 50 in his 51st innings
. |
5min |
| 4) |
A rectangular plate with length 8 inches, breadth 11 inches and thickness 2 inches
is available. What is the length of the circular rod with diameter 8 inches and equal to
the volume of the rectangular plate |
Volume of the circular rod (cylinder) = Volume of the
rectangular plate
(22/7)*4*4*h = 8*11*2
h = 7/2 = 3.5
|
5min |
| 5) |
What is the sum of all numbers between 100 and 1000 which are divisible by 14 ? |
The number closest to 100 which is greater than 100 and
divisible by 14 is 112, which is the first term of the series which has to be summed.
The number closest to 1000 which is less than 1000 and divisible by 14 is 994, which is the last term of the series.
112 + 126 + .... + 994 = 14(8+9+ ... + 71) = 35392
|
5min |
| 6) |
If s(a) denotes square root of a, find the value of s(12+s(12+s(12+ ...... upto
infinity |
Let x = s(12+s(12+s(12+.....
We can write x = s(12+x). i.e., x^2 = 12 + x. Solving this quadratic equation,
we get x = -3 or x=4. Sum cannot be -ve and hence sum = 4
|
5min |
| 7) |
A cylindrical container has a radius of eight inches with a height of three inches.
Compute how many inches should be added to either the radius or height to give the
same increase in volume? |
Let x be the amount of increase. The volume will increase by
the same amount if the radius increased or the height is increased.
So, the effect on increasing height is equal to the effect on increasing the
radius.
i.e., (22/7)*8*8*(3+x) = (22/7)*(8+x)*(8+x)*3
Solving the quadratic equation we get the x = 0 or 16/3. The possible increase
would be by 16/3 inches
|
5min |
| 8) |
A man took a loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was |
Principal = Rs. (100×5400 / 12×3)
= Rs. 15,000. |
5min |
| 9) |
The price of a T.V set worth Rs.20,000 is to be paid in 20 instalments of Rs. 1000 each. If the rate of interest be 6% per annum, and the first instalments be paid at the time of purchase, then the value of the last instalments covering the interest as well will be |
Monet paid in cash= Rs. 1000.
Balance amount = Rs. (20000 - 1000)
= Rs. 19000. |
5min |
| 10 |
A lent Rs.5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs.2200 in all from both of them as interest. The rate of interest per annum is |
Let the rate be R% p.a.
Then,(5000xRx2/100) + (3000xRx4/100)=100R+120R= 2200
=R=(2200/220)
Rate =10%. |
5min |
| 11) |
A money lender finds that dues to a fall in the annual rate of interest from 8% to 7x3/4%, his yearly income diminishes by Rs.61.50. His capital is |
Let the capital be Rs. x.
Then= (x × 8×1/100) - (x × 31/4×1/100)
= 61.50.
=32x - 31x =6150×4
‹=›x= 24600. |
5min |
| 12) |
A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9. p.c.p.a in 5 years. What is the sum? |
Principal = Rs. (100×4016.25 / 9×5)
= Rs. (401625/45)
= Rs. 8925. |
5min |
APTITUDE