Question 1
The average weight of 50 men was 72.The weight of the 46th person was 68 which has been wrongly
entered as 46.Find the corrected average weight.
a) 73 b) 78 c) 75 d) none of these.
Answer : d) none of these.
Solution :
Average weight of 50 men = 72.
Sum of weights of these 50 men = 72 x 50.
Since the weight 68 is wrongly entered as 46.
Amount of weight decreased in the sum = 68 - 46 = 22.
Therefore, the corrected new sum = 72 x 50 + 22.
And, required average of 50 men = (72 x 50 + 22) / 50 = 3622 / 50 = 72.44
Hence the answer is option d.
Question 2
In an entrance test, a candidate's point were wrongly taken as 42 instead of 24. Because of that, the
average score for the exam got increased by 3/4. Find the number
of candidates who have attended the exam.
a) 24 b) 12 c) 36 d) 48
Answer : a) 24.
Solution :
Let there be N candidates in the exam.
And, let X be incorrect average score.
Incorrect total marks = X x N
Since, the points 2.4 is wrongly entered as 4.2.
Corrected total marks = (X x N) - (42 - 24) = NX - 18 ....(1)
Since the average increased by 3/4, then the correct average = X - 3/4.
Correct average = (X - 3/4)N = NX - 3N/4 ....(2)
Equating (1) and (2), we get,
NX - 18 = NX - 3N/4
3N = 18 x 4
N = 18 x 4/3 = 24.
Hence, the number of candidates is 24.
Question 3
The average of 30 non-negative numbers is 30. Out of these numbers, the average of 10 numbers is 28
and that of the other 18 numbers is 32. The average of the
remaining numbers is:
a) 44 b) 22 c) 11 d) none of these.
Answer : b) 22.
Solution :
Average of 30 numbers = 30.
Sum of 30 numbers = 30 x 30 = 900
Average of 10 numbers = 28
Sum of theses 10 numbers = 28 x 10 = 280.
Average of 18 numbers = 32
Sum of these 18 numbers = 18 x 32 = 576.
Now, number of remaining numbers = 30 - 10 - 18 = 2.
We have to find the average of these two numbers.
Sum of these remaining numbers = 900 - (280 + 576) = 44.
Average of remaining numbers = 44/2 = 22.
Question 1
Two cans A and B contains milk worth Rs.7 per litre and Rs.9 per litre respectively. If the contents of A
and B are transferred to another can C in the ratio 3 : 7 then the cost per litre of the mixture in can C is:
a) Rs.9.40 b) Rs.10.10 c) Rs.9.40 d) Rs.8.40
Answer : d) Rs.8.40
Solution :
Cost of 1 litre of A = Rs.7 = cheaper quantity.
Cost of 1 litre of B = Rs.9 = dearer quantity.
Let the mean price be Rs.X.
Applying the rule of alligation,
Therefore, (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c) = (9 - X) : (X - 7)
Given ratio = 3 / 7 = 9 - X / X - 7
63 – 7X = 3X - 21
X = 8.4.
Hence, the required mean price = Rs.8.40.
Question 2
Two qualities of rice at Rs.63 per kg and Rs.67.50 per kg are mixed with another quality of rice in the
ratio 2:2:3. The final mixture sold at Rs.76.50 per kg then the rate of third quality rice per kg was:
a) Rs.87.50 b) Rs.91.50 c) Rs.81.50 d) Rs.99.50
Answer : b) Rs.91.50
Solution :
Given that, first and 2nd varieties are mixed in equal proportion.(2:2)
Their average price = Rs.(63 + 67.50) / 2 = Rs.65.25
The new mixture is formed by mixing two varieties, one at Rs.65.25 per kg and the other at Rs.X per kg
in the ratio 4:3. (note that, 2:2:3 becomes 4:3).
Given that, mean price of new mixture = Rs.76.50
We have to find X.
Applying the rule of alligation,
Therefore, (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c) =
X - 76.50 / 11.25 = 4/3
3X – 229.50 = 45
X = 91.50
Hence, the required price of 3rd quality rice = Rs.91.50.
Question 3
In what ratio, a liquid A of cost Rs.31 per litre should be mixed with liquid B of cost Rs.36 per litre, so
that the cost of the liquid of mixture is Rs.32.35 per litre?
a) 2:1 b) 3:1 c) 3:2 d) 4:3
Answer : b) 3:1
Solution :
Cost of 1 litre of liquid A = Rs.31 = cost of cheaper quantity.
Cost of 1 litre of liquid B = Rs.36 = cost of dearer quantity.
Given, mean price = Rs. 32.25
Applying the rule of alligation,
Therefore, required ratio = (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c)
= 3.75 : 1.25 = 3:1.
Hence, the answer is option b.
Question 1
The dimension of a rectangular shaped wall was X x Y. If the breadth is increased by 40% and the length
is decreased by 30% then the % of area of the new wall compared with previous one is:
a) 88% b) 98% c)68% d)78%
Answer : b) 98%
Solution :
Dimensions of the wall = X x Y.
Without loss of generality, assume that, X is length and Y is breadth.
Area of original wall = XY unit2.
New length (after decreasing 30%) = X – 30X/100 = 70X/100
New breadth (after increasing 40%) = Y = Y + 40Y/100 = 140Y/100.
New area = 70X/100 x 140Y/100 = 49XY/50.
Therefore, required % = (new area / original area ) x 100 = ((49XY/50) / XY) x 100= 98%.
Hence, the answer is 98%.
Question 2
If the area of a square shaped field decreases by 36%, then the % of each side of the field decreases by:
a) 15% b) 12% c) 20% d) 19%
Answer : c) 20%
Solution :
Let us assume that the original area of the field is 100 unit2.
Then, its original sides = sqrt(100) = 10 units.
% decrease in area = 36%
New area = 100 - 36 = 64 unit2.
Therefore, new side of the field = sqrt(64) = 8 units.
i.r., decrease on 10 units = (10 - 8) = 2 units.
Decreasing % = (units of decreasing / original units) x 100 = 2 x 100/10 = 20%
Hence, the answer is 20%.
Question 3
The dimension of a black board is L m x B m. If l and b decreases by 20% and 40% respectively, then the
area of the board before alteration exceeds the area of new one by:
a) 48% b) 52% c) 39% d) 62%
Answer : b) 52%
Solution :
Let length = L m and breadth = B m.
Original area = (LB) m2.
New length (after decreasing 20%) = (100 - 20)% of L= 80% of L = 80L /100 m = 4L / 5 m.
New breadth (after decreasing 40%) = (100 - 40)% of B = 60% of B = 60B / 100 = 3B / 5 m.
New area = 4L/5 x 3B/5 = 12(LB)/25 m2.
Difference in area = original area - new area = Lb - 12(LB)/25 = 13(LB)/25 m2.
Therefore, required % = difference / original area x 100 = [(13(LB)/25) / (LB) ] x 100 = 13 x 4 = 52%.
Question 1
A fruit seller has 16 apples and cost of each apple is Rs.12. Three persons A,B and C decided to buy that
apples. If A and B gives Rs.84 and Rs.48 to the seller for some apples, then how many apples C can buy?
a) 8 b) 11 c) 5 d) 6
Answer : c) 5
Solution :
Given that the seller has 16 apples and he sells each apple at Rs.12
Then total cost of 16 apples = 16 x 12 = Rs.192
A and B buys for Rs.84 and Rs.48.
Then the cost of remaining apples = Rs.(192 - (84+48))
= Rs.60.
TC can buy for Rs.60
Since each apple costs Rs.12, then C can buy 60/12 = 5 apples.
Hence the answer is 5.
Question 2
Thomas bought X number of sports goods for Rs.9000. If each item was cheaper by Rs.30 then with the
same amount he could have bought 50 more items than X. Find the number of items bought by Thomas
?
a) 100 b) 150 c) 75 d) 125
Answer : a) 100
Solution :
Transaction I
Thomas have bought X items for Rs.9000.
Then the cost of each item = Rs.9000/X ...(1)
Transaction II
If each item was cheaper by Rs.30, he could have bought 50 more items.
Therefore, with Rs.30 discount, the amount of each item = Rs.9000 / X + 50 ...(2)
We know that the cost of each item in transaction II will be lesser than that of transaction I by Rs. 30.
i.e (1) - (2) = 30
Since he bought 50 more items when cost of each item is less by Rs.30 then we have
9000/X - 9000/X + 50 = 30
1/X - 1/X + 50 = 30/9000
50 / X(X + 50) = 1/300
X(X + 50) = 15000
X^2 + 50X = 15000
X^2 + 50X - 15000 = 0
(X+150)(X-100) = 0
X = -150 or X = 100
X cannot be a negative value. Therefore, X = 100.
Hence the answer is 100.
Question 3
On Republic day, chocolates were to be distributed among 350 kids in a play school. But 140 kids were
absent on that day and each kid present got 3 chocolates more. Find the total number of chocolates
bought for the distribution.
a) 3200 b) 2750 c) 2950 d) 1575
Answer : d) 1575
Solution :
Let the total number of chocolates be X.
Total number of kids in the school(when everyone is present) = 350.
If everyone is present, number of chocolates per kid = X/350 chocolates ...(1)
But, 140 kids are absent.
Therefore, total number of kids present = 350 - 140 = 210
Therefore, number of chocolates per present kid = X/210 chocolates ...(2)
210 kids were present and each kid gets 3 extra.
i.e (2)-(1)=3
Or X/210 - X/350=3
350X - 210X / (210 x 350)=3
140X / 210 x 350=3
X = 3 x 210 x 350 / 140 = 1575
Then the required answer is 1575.
Question 4
8 friends planned to go to hotel for dinner and to share the bill amount equally. If one of them have
forget to bring the wallet, then what will be the extra amount contribute by each to pay the bill of
Rs.1904 ?
a) Rs.28 b) Rs.54 c) Rs.34 d) Rs.38
Answer : c) Rs.34
Solution :
Given that the bill amount = Rs.1904
Actual share of each = Rs.1904/8 = Rs.238
If one of 8 is left, then the sharing amount = Rs.1904/7 = Rs.272
Then the extra amount given by each = Rs.(272 - 238) = Rs.34.
Hence the answer is Rs.34