BACK
PERCENTAGESEXAMPLE PROBLEMS:
22. A school has only three classes which contain 40,50,60 students
respectively . The pass percent of these classes are 10, 20 and 10
respectively . Then find the pass percent in the school.
SOLUTION:
Number of passed candidates =
10/100*40+20/100 *50+10/100 * 60
=4+10+6
=20
Total students in school = 40+50+60 =150
So required percentage = 20/150 *100
= 40 /3
=13 1/3 %
23. There are 600 boys in a hostel . Each plays either hockey or football
or both .If 75% play hockey and 45 % play football ,Find how many
play both?
SOLUTION:
n(A)=75/100 *600
=450
n(B) = 45/100 *600
= 270
n(A^B)=n(A) + n(B) – n(AUB)
=450 + 270 -600
=120
So 120 boys play both the games.
24.A bag contains 600 coins of 25p denomination and 1200 coins of 50p
denomination. If 12% of 25p coins and 24 % of 50p coins are removed, Find the percentage of money removed from the bag ?
SOLUTION:
Total money = (600 * 25/100 +1200 *50/100)
=Rs 750
25p coins removed = 12/100 *600
=72
50p coins removed = 24/100 *1200
=288
So money removed =72 *1/4 +288 *1/2
= Rs 162
So required percentage=162/750 *100
=21 .6%
25. P is six times as large as Q.Find the percent that Q is less than P?
SOLUTION:
Given that P= 6Q
So Q is less than P by 5Q.
Required percentage= 5Q/P*100 %
=5/6 * 100 %
=83 1/3%
26.For a sphere of radius 10 cm ,the numerical value of surface area is what percent of the numerical value of its volume?
SOLUTION:
Surface area = 4 *22/7 *r2
= 3/r(4/3 * 22/7 * r3)
=3/r * VOLUME
Where r = 10 cm
So we have S= 3/10 V
=3/10 *100 % of V
= 30 % of V
So surface area is 30 % of Volume.
27. A reduction of 21 % in the price of wheat enables a person to buy 10 .5
kg more for Rs 100.What is the reduced price per kg.
SOLUTION:
Let the original price = Rs x/kg
Reduced price =79/100x /kg
==> 100/(79x/100)-100/x =10.5
==> 10000/79x-100/x=10.5
==> 10000-7900=10.5 * 79 x
==> x= 2100/10.5 *79
So required price = Rs (79/100 *2100/10.5 *79) /kg
= Rs 2 per kg.
28.The length of a rectangle is increased by 60 % .By what percent would
the width have to be decreased to maintain the same area?
SOLUTION:
Let the length =l,Breadth= b.
Let the required decrease in breadth be x %
then 160/100 l *(100-x)/100 b=lb
160(100-x)=100 *100
or 100-x =10000/160
=125/2
so x = 100-125/2 =75/2 =37.5
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