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Arithmetic : Level 3 Test -1

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Question 1
A report consists of 20 sheets each of 55 lines and each such line consist of 65 characters. This report is retyped into sheets each of 65 lines such that each line consists of 70 characters. The approximate percentage reduction in number of sheets is closest to
A
20
B
5
C
30
D
35
Question 1 Explanation:
Total volume of first report = (20 x 55 x 65) = 71500
Let n be the number of pages.
Therefore, the new volume will be = (65 x 70 x n) = 4550n
As the volume remains same, we can say that
71500 = 4550n
n = 15.71 = 16 sheets.
Clearly, the sheet reduced by 4 and percentage reduction in number of sheets =(4/20) x 100 = 20%.
Question 2
The rate of increase of the price of sugar is observed to be two percent more than the inflation rate expressed in percentage. The price of sugar, on January 1, 1994, is Rs. 20 per kg. The inflation rate for the years 1994 and 1995 are expected to be 8% each. The expected price of sugar on January 1, 1996 would be
A
23.60
B
24.00
C
24.20
D
24.60
Question 2 Explanation:
Inflation rate for the year 1994 & 1995 = 8% each
The rate of increase in price of sugar will be = 10%.
As, price of sugar on Jan 1, 1994 is Rs 20
Then, it will be Rs 22 on Jan 1, 1995 and Rs. 24.20 on Jan 1,
1996.
Question 3
The number of votes not cast for the Praja Party increased by 25% in the National General Election over those not cast for it in the previous Assembly Polls, and the Praja Party lost by a majority twice as large as that by which it had won the Assembly Polls. If a total 2,60,000 people voted each time. How many voted for the Praja Party in the Assembly Elections.
A
1,10,000
B
1,50,000
C
1,40,000
D
1,20,000
Question 3 Explanation:
Let the number of votes in the previous polls = n
So, the number of votes not cast for the party would be = 1.25n.
Total numbers of votes are 260000
Therefore, the number of votes cast in the two polls would be = (260000 – x) and (260000 – 1.25x)
Majority of votes in the previous polls = (votes cast) – (votes not cast) = (260000 – x) – x = (260000 – 2x).
Majority of votes lost in this year’s polls
= (votes not cast) – (votes cast) =1.25x – (260000 – 1.25x)
= (2.5x – 260000). According to the question
(Margin of loss this year) = 2 x (Margin of victory last year).
(2.5x – 260000) = 2(260000 – 2x).
Solving this equation we get,
x = 120000.
The number of votes not casted for the party in the previous assembly polls = 12000.
So, the number of votes casted for the party = (260000 – 120000) = 140000.
Question 4
2/5 of the voters promise to vote for P and the rest promised to vote for Q. Of these, on the last day 15% of the voters went back of their promise to vote for P and 25% of voters went back of their promise to vote for Q, and P lost by 2 votes. Then the total number of voters is
A
100
B
110
C
90
D
95
Question 4 Explanation:
Let the total number of votes = 100
2/5 of 100 = 40
Initially, there were 40 of these, who promised to vote for P
Number of people who votes for Q = 60
On the last day,
Number of people who broke their promise, who were with P= (15% of 40) = 6 voters
Number of people who broke their promise, who were with Q = (25% of 60) = 15 voters
So the number of votes P have = (40 – 6 + 15) = 49 votes
The number of votes Q have = (60 – 15 + 6) = 51 votes.
Hence, margin of victory for Q = (51 – 49) = 2,
Hence 2% of total = 2 ; Hence the total number of voters is 2
But we are already given by the condition that P losses by 2 votes hence the total number of voters are 100
Question 5
Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?
A
2 : 3
B
4 : 3
C
3 : 2
D
3 : 4
Question 5 Explanation:
Let the ratio of content = p: q
Quantity of A in the mixture = 5p/6 + 1q/4
Quantity of B in the mixture = 1p/6 + 3q/4
Therefore, the required mixture = 1:1
{5p/6 + 1q/4} / {1p/6 + 3q/4} = 1/1
By solving,
We get p/q = 3/4
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