QUANS - 11
1.
A student divided a number by 2/3 instead of
multiplying the same by 3/2. Calculate the percentage of error in his
calculation.
(a)
0% (b) 25% (c) 44.44%
(d) 125%
Let the number be x. Then
the actual result should be x
* 3/2 =
3x/2.
Due to the mistake of the student
the result arrived was x / 2/3 =
3x/2.
Since both the results are same
there is no error in percentage.
Ans: (a)
2.
A man bought a Horse and Cart. If he sells the
horse at 10% loss and the cart at 20% profit he will not lose anything. If he
sells the horse at 5% loss and the cart at 5% gain then he will lose Rs 10 in
the bargain. Find the amount paid by him for the purchase of the horse and the
cart?
Let the price of the horse be H and that of
the cart C.
Then we have an equation: H + C = 0.9H + 1.2C (Horse was sold for 10%
loss and Cart 20% profit) Solving the
two we get 0.2C – 0.1H = 0 -----------(i)
0.95H + 1.05C = H + C – 10 (Horse was sold for 5% loss and Cart for 5%
profit)
Solving the above we get 0.05C – 0.05H = -10--------------(ii)
Equating (i) and (ii) 0.2C – 0.1H = 0
0.05C – 0.05H = -10 we get
Ans: H = Rs
400.00 and C =Rs 200.00
3. During
a vacation time I had visited my cousin’s home town and had a wonderful time.
We would go for jogging in the mornings and on some days instead played Tennis
in the afternoons. There were days when we were too lazy to do anything and as
such stayed at home. There were 14 days when we either went for jogging or
played Tennis. There were 12 mornings and 18 evenings when we did nothing or
stayed at home. How many days did I stay at my cousin’s place?
Let A, B and C represent the days when we
went for jogging, or played Tennis or stayed at home. We now have the following
information.
B + C = 12, A + C = 18, and A + B = 14
Adding all the three, we get
2( A+B+C) = 44 A + B + C = 22
A = 22 – 12
= 10, B = 22 – 18 = 4, C =
22 – 14 = 8
Thus I
stayed for 22 days at my cousin’s place.
4.
A stamp collector has the habit of arranging and
re-arranging his stamps now and then. In one such case when he arranged the
stamps in pairs one stamp was left out. When arranged in groups of three, four,
five and six in each case one stamp remained. However, when arranged in group
of seven thre was no stamp remaining. How many stamps he had with him?
The number should be an unknown figure say
‘x’ and the number of stamps is arrived at
X (LCM of 2,3,4,5,and 6) + 1 x * 60 + 1 = 60x
+ 1.
Now substituting value for ‘x’ from 1
onwards we observe that 5 fits in nicely. Applying the value of 5 for ‘x’
The total number of stamps comes to 5*60+1
= 301 which is divisible by 7 without any reminder and fulfils the oter
conditions in the question. Hence,
Ans:
301
5.
A truck when travelling at a speed of 30
km/hr reaches the destination one hour
earlier and when travels at 20 km/hr is late by one hour. At what speed the
truck should go to reach the place on time?
Distance / speed = time Let D be the distance travelled by the
truck. Then we have an equation
D/20 – D/30 = 2 . Solving, we
get the distance as 120 km.
At a speed of 30 km/hr the truck will take
120/30 = 4 hours and is one hour early.
Hence the actual time to travel should be 5
hours and applying this, we get the speed as
120 / 5 = 24 km/hr.
Ans: 24 km/hr.
6.
A lady visits a restaurant and after having her
food, pays half the amount she had towards the bill plus $1 as tip. She then
visits the nearby mall and purchases some articles and pays half the amount she
is left towards cost of purchases and on coming outside saw a beggar and gave
him $2. Finally she visits the nearby bookshop purchased some books and paid
half the money she is left with and after coming out saw another beggar and
gave him $3. She was now left with $1 only. How much money she had originally?
This question is best Answered going from
backwards.
She was left with $1 and prior to that had
given the beggar $3. Totalling it comes to $4 which represents half the money
after her paying for the books. Hence, while entering the book shop she was
having $4 * 2 = $8
Earlier to this while leaving the mall she
was having this much of money with her and prior to that had given another
beggar $2. Totalling these two amounts comes to $10 which incidentally is half
the amount left with her after paying for the goods at the mall.
Hence, before entering the mall she was
having $10 * 2 = $20 with her.
This was the sum she was having while
leaving the restaurant where she gave a tip of $1. Together the sum works out
to $ 21 which again was half the amount she had originally.
Hence the amount she had originally was $21
* 2 = $ 42.
Ans:
$42
7.
A tree on the first day grows half its height, on
the second day the tree grows 1/3 of its height the previous day, on the third
day ¼ th of the height on the second day, on the fourth day 1/5th of
the height on the third day and so on.
Find on which day the tree will be 100 times its original height?
Let the initial height of the tree be ‘x’
On the first day the height of the tree
will be x * ½ = 3/2 x
On the second day the height will be 3/2 *1/3 = ½ + 3/2 = 2x
On the 3rd day the height will be 2 * ¼ = ½
+ 2x = 5/2 x and so on.
In other words the tree grows at the rate
of ½ its original height every day.
Every two days the tree grows its original
height of ‘x’. Going on this basis, we observe
The tree will be 100 times its original
height on the 198th day.
Ans:
198 days.
8.
The length of the side of a square is (x+2) and
the side of an equilateral triangle is 2x. If the perimeter of the Triangle and
the Square are equal, then what is the value of ‘ x ‘?
We have the following equation from the
given information.
2x * 3 = 4(x+2) 6x = 4x + 8 2x = 8
x = 4
Ans:
x = 4.
9.
Find the next term in the series: 6, 24, 60, 120, 210, ?
The series goes in this format : 1*2*3=6, 2*3*4=24,
3*4*5=60, 4*5*6=120, etc
The next series is 6*7*8=336
Ans:
336
10.
Find the next term in the series: 1, 5, 13, 25, ?
The series follows the pattern
(n-1)2 + n2 Hence
the answer is (5-1)2 + 52 = 41.
Ans: 41
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