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In this series of quizzes,we will be providing most important questions on Arithmetic (Percentage, Profit Loss,TSD,Time and Work,etc) which are usually asked in the exams.In Prelims Examination,usually 10-15 questions are asked from this topic.
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Question 1 of 10
1. Question
In a cricket team of 11 players, one player weighing 72 kg got injured and was replaced by a new player thereby increasing the average weight of the entire team by 100 gm. What is the weight of the new player?
Correct
Let the original weight of the new player is W kg and the average weight of the original 11 players is n kg.
Hence total weight of the 11 players = 11n kg
Now according to the question, (11n-72 + W)/11=n+0.1
=> 11n – 72 + W = 11n + 1.1
=> W = 72 + 1.1
=> W = 73.1 kg
Incorrect
Let the original weight of the new player is W kg and the average weight of the original 11 players is n kg.
Hence total weight of the 11 players = 11n kg
Now according to the question, (11n-72 + W)/11=n+0.1
=> 11n – 72 + W = 11n + 1.1
=> W = 72 + 1.1
=> W = 73.1 kg
Question 2 of 10
2. Question
The area of a circular field is 9856 cm^{2}. If a person crosses the field along the diameter, calculate the distance that he will travel (in cm)
Correct
πr^{2} = 9856
=> (22/7) r^{2}=9856
=> r^{2}=9856*7/22
=> r= 56
=> So, the distance that he has to travel = 2*56 = 112
Incorrect
πr^{2} = 9856
=> (22/7) r^{2}=9856
=> r^{2}=9856*7/22
=> r= 56
=> So, the distance that he has to travel = 2*56 = 112
Question 3 of 10
3. Question
If P, Q and R started a business by initial investments of Rs.10000, Rs.15000 and Rs.20000 respectively and after 6 months P, Q and R additionally invested Rs.10000, Rs.8000 and Rs.6000 respectively, then what is the share of Q in the profit of Rs.144000 after 1 year?
Correct
Investment of P = (10000 x 6) +(20000 x 6) = (30000 x6)
Investment of Q = (15000 x 6) +(23000 x 6) = (38000 x6)
Investment of R = (20000 x 6) +(26000 x 6) = (46000 x6)
Ratio of investments of P, Q and R
= (30000 x6): (38000 x6): (46000 x6)
= 30: 38: 46 = 15: 19: 23
Share of Q = (19/57) x 144000 = Rs.48000.
Incorrect
Investment of P = (10000 x 6) +(20000 x 6) = (30000 x6)
Investment of Q = (15000 x 6) +(23000 x 6) = (38000 x6)
Investment of R = (20000 x 6) +(26000 x 6) = (46000 x6)
Ratio of investments of P, Q and R
= (30000 x6): (38000 x6): (46000 x6)
= 30: 38: 46 = 15: 19: 23
Share of Q = (19/57) x 144000 = Rs.48000.
Question 4 of 10
4. Question
An election was contested by two parties A and B. A got 18 percentage points more votes in comparison to B and B got 82000 votes. Find by what margin did A win the election?
Correct
Let the votes got by A = x%
Votes got by B = x-18%
And x+x-18% = 100%
2x = 118%
x = 59%
Votes obtained by A = 59%
Votes obtained by B = 59-18 = 41%
41/100x = 82000
x = 200000 votes
Margin = 18% of total votes
= 18*200000/100
= 36000 votes
Incorrect
Let the votes got by A = x%
Votes got by B = x-18%
And x+x-18% = 100%
2x = 118%
x = 59%
Votes obtained by A = 59%
Votes obtained by B = 59-18 = 41%
41/100x = 82000
x = 200000 votes
Margin = 18% of total votes
= 18*200000/100
= 36000 votes
Question 5 of 10
5. Question
A watch was sold at a loss of 15%. Had it been sold for Rs.300 more, the seller would have gained 15% instead. What is the no-profit, no-loss selling price of the watch?
Correct
Initial price = x
15% loss price = 85x/100
According to question, (85x/100)+300 = x (1+15/100)= 115x/100
→ 300 = (115x/100)-(85x/100)
→ 300 = 30x/100
→ x = 300*100/30=1000
Hence the correct answer is option B.
Incorrect
Initial price = x
15% loss price = 85x/100
According to question, (85x/100)+300 = x (1+15/100)= 115x/100
→ 300 = (115x/100)-(85x/100)
→ 300 = 30x/100
→ x = 300*100/30=1000
Hence the correct answer is option B.
Question 6 of 10
6. Question
There are two mixtures of milk and water A and B respectively. The ratio of milk and water in these two mixtures is 7 : 12 and 2 : 5 respectively. If both the mixtures are mixed in a way such that the ratio of water to milk in resulting mixture is 7 : 3 then find out the ratio in which first mixture A is mixed with the second mixture B?
Correct
Let’s suppose m kg of mixture A is mixed with n kg of mixture B.
Hence the total volume of the milk in resulting mixture = (7m/19+2n/7) liter
Total volume of the resulting mixture = (m + n) liter
According to the question,
=> 10 (7m/19+2n/7) = 3m + 3n
By solving the above equation we will get,
=> m/n=19/91
Incorrect
Let’s suppose m kg of mixture A is mixed with n kg of mixture B.
Hence the total volume of the milk in resulting mixture = (7m/19+2n/7) liter
Total volume of the resulting mixture = (m + n) liter
According to the question,
=> 10 (7m/19+2n/7) = 3m + 3n
By solving the above equation we will get,
=> m/n=19/91
Question 7 of 10
7. Question
Ratio of population of Banglore to Chennai is 5 : 9 and the ratio of number of males in Banglore and Chennai is 7 : 15. What is the ratio of number of females in Chennai to number of females in Banglore?
Correct
let the population of Banglore be 5n.
According to the question; “Ratio of population of Banglore to Chennai is 5 : 9” hence population of Chennai will be 9n.
Also according to the question; Number of males in Banglore = 7m
Number of males in Chennai = 15m
As we know that;
Total population = Number of Male + Number of females
Let number of females in Banglore and Chennai be p and q respectively.
Therefore; 5/9 = (7m+p)/(5m+q)
From this equation we can’t determine the ratio of p : q.
Incorrect
let the population of Banglore be 5n.
According to the question; “Ratio of population of Banglore to Chennai is 5 : 9” hence population of Chennai will be 9n.
Also according to the question; Number of males in Banglore = 7m
Number of males in Chennai = 15m
As we know that;
Total population = Number of Male + Number of females
Let number of females in Banglore and Chennai be p and q respectively.
Therefore; 5/9 = (7m+p)/(5m+q)
From this equation we can’t determine the ratio of p : q.
Question 8 of 10
8. Question
A sum of Rs. 2000 is lent in two parts on simple interest, one at 8% per annum and other at 9.2% per annum for 7 years and the interest obtained is Rs. 1204. What are the two parts respectively?
Correct
Let the two parts lent be x and (2000-x)
Simple interest = 1204 = (x*7*8)/100 + [(2000-x) *7*9.2]/100
120400 = 56x + 128800 – 64.4x
8.4x = 8400
x = 1000
So, the two parts are Rs 1000 and Rs 1000.
Incorrect
Let the two parts lent be x and (2000-x)
Simple interest = 1204 = (x*7*8)/100 + [(2000-x) *7*9.2]/100
120400 = 56x + 128800 – 64.4x
8.4x = 8400
x = 1000
So, the two parts are Rs 1000 and Rs 1000.
Question 9 of 10
9. Question
A man going to his office at a speed of 40 kmph, he reaches the office 15 minutes late and while going at a speed of 56 kmph, he reaches the office late by five minutes. At what speed should the man travel to reach the office on time?
Correct
Let ‘d’ km be the distance the man travels to reach his office and ‘t’ hrs be the usual time taken to reach office on time.
Therefore d = 40(t + 1/4) = 56(t + 1/12)
t = 1/3 hr
Therefore d = 40(1/3 + 1/4)
d = 40(7/12) = 70/3 km.
Therefore usual speed of the man = distance/time = (70/3)/(1/3) = 70 kmph
Hence the correct option is (4).
Incorrect
Let ‘d’ km be the distance the man travels to reach his office and ‘t’ hrs be the usual time taken to reach office on time.
Therefore d = 40(t + 1/4) = 56(t + 1/12)
t = 1/3 hr
Therefore d = 40(1/3 + 1/4)
d = 40(7/12) = 70/3 km.
Therefore usual speed of the man = distance/time = (70/3)/(1/3) = 70 kmph
Hence the correct option is (4).
Question 10 of 10
10. Question
One pipe A can fill the tank completely in 24 minutes and another pipe B can fill it completely in 30 minutes. A 3^{rd} pipe which is drain pipe empties the tank completely in 12 minutes. The inlet pipes are opened for 10 minutes and then drain pipe is also opened. Find the time taken to empty the water tank.
Correct
=> Let the tank is emptied after x minutes.
Means, The third drain pipe is opened only for x minutes
=> x/12 – (x+10)/24 – (x+10)/30 = 0
=> 10x – 5(x+10) – 4 (x+10) / 120 = 0
=> 10x – 5x – 50 – 4x – 40 = 0
=> x = 90
Incorrect
=> Let the tank is emptied after x minutes.
Means, The third drain pipe is opened only for x minutes
=> x/12 – (x+10)/24 – (x+10)/30 = 0
=> 10x – 5(x+10) – 4 (x+10) / 120 = 0
=> 10x – 5x – 50 – 4x – 40 = 0
=> x = 90