Hello and welcome to exampundit. Here is a Full Length Numerical Ability Sectional Test for RBI Assistant Mains 2017. The set contains 40 questions with timer and explanation.
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The following Set of Numerical Ability Sectional Test is for RBI Assistant Mains examinations.
Total Time: 30 minutes (RBI Assistant Mains Time)
No. of Questions: 40
Level: Moderate
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- Question 1 of 40
1. Question
Directions: Read the following table and information carefully and answer below questions.
A company manufactures six products-A, B, C,D,E and F. A part of each product is exported and the remaining units are sold inside the country. The table shows the units produced, the percentage of units exported. The ratio of sold and unsold units of the part of products sold inside the country along with the tax required to be paid by the company for respective products is mentioned in the table. Now answer the following questions as per the information provided.
Product Units produced % exported Sold: Unsold % of tax on sold A 763992 17 7:9 11 B 86432 28 11:2 13 C 97367 41 3:2 21 D 654372 37 1:11 17 E 643721 47 9:2 23 F 43986 63 5:8 19 Questions: What is the difference between the in-country tax paid by D and A?
CorrectSolution:
A=763992×0.83x(7/16)x0.11=30516
D=654372×0.63x(1/12)x0.17=5840
Difference=30516-5840=₹24676
IncorrectSolution:
A=763992×0.83x(7/16)x0.11=30516
D=654372×0.63x(1/12)x0.17=5840
Difference=30516-5840=₹24676
- Question 2 of 40
2. Question
Directions: Read the following table and information carefully and answer below questions.
A company manufactures six products-A, B, C,D,E and F. A part of each product is exported and the remaining units are sold inside the country. The table shows the units produced, the percentage of units exported. The ratio of sold and unsold units of the part of products sold inside the country along with the tax required to be paid by the company for respective products is mentioned in the table. Now answer the following questions as per the information provided.
Product Units produced % exported Sold: Unsold % of tax on sold A 763992 17 7:9 11 B 86432 28 11:2 13 C 97367 41 3:2 21 D 654372 37 1:11 17 E 643721 47 9:2 23 F 43986 63 5:8 19 Questions: Highest Tax on in country sold units is paid by which of the following:
CorrectSolution:
A=₹30516
F=43986×0.37x(5/13)x0.19=₹1189
D=₹5840
C=97367×0.59x(3/5)x0.21=₹7238
Thus A has highest tax
IncorrectSolution:
A=₹30516
F=43986×0.37x(5/13)x0.19=₹1189
D=₹5840
C=97367×0.59x(3/5)x0.21=₹7238
Thus A has highest tax
- Question 3 of 40
3. Question
Directions: Read the following table and information carefully and answer below questions.
A company manufactures six products-A, B, C,D,E and F. A part of each product is exported and the remaining units are sold inside the country. The table shows the units produced, the percentage of units exported. The ratio of sold and unsold units of the part of products sold inside the country along with the tax required to be paid by the company for respective products is mentioned in the table. Now answer the following questions as per the information provided.
Product Units produced % exported Sold: Unsold % of tax on sold A 763992 17 7:9 11 B 86432 28 11:2 13 C 97367 41 3:2 21 D 654372 37 1:11 17 E 643721 47 9:2 23 F 43986 63 5:8 19 Questions: If product E needs to pay 31 % tax on exported units. Find the total amount of tax paid by the company for product E in the whole year of 1990?
CorrectSolution:
Total tax=tax on In-country sold units + Tax on export units
=(643721×0.53x{9/11}x0.23)+(0.47×643721)
= ₹366750
IncorrectSolution:
Total tax=tax on In-country sold units + Tax on export units
=(643721×0.53x{9/11}x0.23)+(0.47×643721)
= ₹366750
- Question 4 of 40
4. Question
Directions: Read the following table and information carefully and answer below questions.
A company manufactures six products-A, B, C,D,E and F. A part of each product is exported and the remaining units are sold inside the country. The table shows the units produced, the percentage of units exported. The ratio of sold and unsold units of the part of products sold inside the country along with the tax required to be paid by the company for respective products is mentioned in the table. Now answer the following questions as per the information provided.
Product Units produced % exported Sold: Unsold % of tax on sold A 763992 17 7:9 11 B 86432 28 11:2 13 C 97367 41 3:2 21 D 654372 37 1:11 17 E 643721 47 9:2 23 F 43986 63 5:8 19 Questions: If the warehouse of the company stores only the in-country saleable unsold units of each product, then what is the relation between the current warehouse storage of product A and product C.
CorrectSolution:
Unsold A stored in warehouse=763992×0.83×9/16=356688
Unsold C stored in warehouse=97367×0.59×2/5=7650
7650/356688 x 100 =2.14 %
Hence option b holds true
IncorrectSolution:
Unsold A stored in warehouse=763992×0.83×9/16=356688
Unsold C stored in warehouse=97367×0.59×2/5=7650
7650/356688 x 100 =2.14 %
Hence option b holds true
- Question 5 of 40
5. Question
Directions: Read the following table and information carefully and answer below questions.
A company manufactures six products-A, B, C,D,E and F. A part of each product is exported and the remaining units are sold inside the country. The table shows the units produced, the percentage of units exported. The ratio of sold and unsold units of the part of products sold inside the country along with the tax required to be paid by the company for respective products is mentioned in the table. Now answer the following questions as per the information provided.
Product Units produced % exported Sold: Unsold % of tax on sold A 763992 17 7:9 11 B 86432 28 11:2 13 C 97367 41 3:2 21 D 654372 37 1:11 17 E 643721 47 9:2 23 F 43986 63 5:8 19 Questions: If the government imposes an agricultural tax of 17% on the unsold units of the in-country saleable units of product D only, then the agricultural tax paid by the company for D is:
CorrectSolution: 654372×0.63x(11/12)x0.17= ₹64243
IncorrectSolution: 654372×0.63x(11/12)x0.17= ₹64243
- Question 6 of 40
6. Question
A blind school has 3433 visually handicapped students. 80 to 85 percent of the students are good musicians and 30 to 40 percent of the students are good actors. Each student has excellence in at least one of the skills of music and acting. What is the absolute difference between the minimum and maximum possible numbers of students who are both great musicians and actors?
CorrectSolution:
Total Students = 2001
% of students who are musicians = 80% to 85% of 3433 = 2746 to 2918
So, % of students who are good in acting alone =100%-85% to 100%-80% = 15% to 20% = 515 to 687
% of students who are good actors = 30% to 45% = 1030 to 1545
So, minimum = 1030-687=201
and maximum = 1545-515=499
and 499-201=298IncorrectSolution:
Total Students = 2001
% of students who are musicians = 80% to 85% of 3433 = 2746 to 2918
So, % of students who are good in acting alone =100%-85% to 100%-80% = 15% to 20% = 515 to 687
% of students who are good actors = 30% to 45% = 1030 to 1545
So, minimum = 1030-687=201
and maximum = 1545-515=499
and 499-201=298 - Question 7 of 40
7. Question
20 college students are waiting in a queue for discount coupons of KFC. Each coupon costs ₹5. Ten of the students have the ₹5 coin and others have currency notes of ₹10. Determine in how many ways can all the 20 students stand in the queue such that the person at the coupon counter will not face any issue in paying back changes provided that the person at the counter does not have any changes.
CorrectSolution:
For the coupon seller to have change always, the person with ₹5 should come before the person with ₹ 10
We can also say that first person must have ₹5 and 20th person must have ₹10
→ let us take this case 5_ _ _(10 places), 10 _ _ _ (10 places), no of cases – 1Take the first 10 numbers as one part and next 10 as second part
now take one 10 into the first part, this can be done in ^{9}C_{1 }ways because 1st number should be 5 always
and the 5 which is replaced by 10 can be arranged in ^{9}C_{1} ways in 2nd partone 10 is replaced, no of cases: ^{9}C_{1}x^{9}C_{1} = 81
now replace two 10’s , 5 5 _ _ _ _ _ _ _ _ ; ……..10 —– ^{8}C^{2} ways
5 10 5 _ _ _ _ _ _ _ ; …… 10 —– ^{7}C_{1} ways
same will be the cases for two 5’s to be arranged in the second part
two 10’s are replaced, no of cases – (^{8}C_{2} + ^{7}C_{1})x(^{8}C_{2} + ^{7}C_{1}) = 1225
replace three 10’s, 5 5 5 _ _ _ _ _ _ _; ……… 10 ——- ^{7}C_{3} ways
5 10 5 5 _ _ _ _ _ _ ; ……… 10 ——– ^{6}C_{2} ways
5 10 5 10 5 _ _ _ _ _ ; …….. 10 ——- ^{5}C_{1 }ways
three 10’s are replaced, no of cases – (^{7}C_{3} + ^{6}C_{2} + ^{5}C_{1})x(^{7}C_{3 }+ ^{6}C_{2} + ^{5}C_{1}) = 3025
four 10’s replaced , no of cases : (^{6}C_{4} + ^{5}C_{3} + ^{4}C_{2} + ^{3}C_{1})^{ 2} = 1156five 10’s replaced , no of cases : (^{5}C_{5 }+ ^{4}C_{4} + ^{3}C_{3} + ^{2}C_{2} + ^{1}C_{1}) ^{2} = 25
max only five 10’s can come into first part
so total number of cases = 1+81+1225+3025+1156+25= 5513IncorrectSolution:
For the coupon seller to have change always, the person with ₹5 should come before the person with ₹ 10
We can also say that first person must have ₹5 and 20th person must have ₹10
→ let us take this case 5_ _ _(10 places), 10 _ _ _ (10 places), no of cases – 1Take the first 10 numbers as one part and next 10 as second part
now take one 10 into the first part, this can be done in ^{9}C_{1 }ways because 1st number should be 5 always
and the 5 which is replaced by 10 can be arranged in ^{9}C_{1} ways in 2nd partone 10 is replaced, no of cases: ^{9}C_{1}x^{9}C_{1} = 81
now replace two 10’s , 5 5 _ _ _ _ _ _ _ _ ; ……..10 —– ^{8}C^{2} ways
5 10 5 _ _ _ _ _ _ _ ; …… 10 —– ^{7}C_{1} ways
same will be the cases for two 5’s to be arranged in the second part
two 10’s are replaced, no of cases – (^{8}C_{2} + ^{7}C_{1})x(^{8}C_{2} + ^{7}C_{1}) = 1225
replace three 10’s, 5 5 5 _ _ _ _ _ _ _; ……… 10 ——- ^{7}C_{3} ways
5 10 5 5 _ _ _ _ _ _ ; ……… 10 ——– ^{6}C_{2} ways
5 10 5 10 5 _ _ _ _ _ ; …….. 10 ——- ^{5}C_{1 }ways
three 10’s are replaced, no of cases – (^{7}C_{3} + ^{6}C_{2} + ^{5}C_{1})x(^{7}C_{3 }+ ^{6}C_{2} + ^{5}C_{1}) = 3025
four 10’s replaced , no of cases : (^{6}C_{4} + ^{5}C_{3} + ^{4}C_{2} + ^{3}C_{1})^{ 2} = 1156five 10’s replaced , no of cases : (^{5}C_{5 }+ ^{4}C_{4} + ^{3}C_{3} + ^{2}C_{2} + ^{1}C_{1}) ^{2} = 25
max only five 10’s can come into first part
so total number of cases = 1+81+1225+3025+1156+25= 5513 - Question 8 of 40
8. Question
Arshi is a shopaholic. She went to a bag shop to buy a handmade wallet. She took ₹ 15 to the shop in the form of one rupee notes and 20 paise coins. After returning from the shop after buying the wallet, she was left with as many one rupee notes as she originally had 20 paise coins and as many 20 paise coins as she had originally one rupee notes. The total amount also reduced by two-third. What was the cost of the wallet?
CorrectSolution:
Let number of one rupee notes=X
Number of 20 paise coins=Y
Arshi started with (100X+20Y) and came back with (100Y and 20A) paise
Also, 100Y+20X=(1/3) (100X+20Y)
=>X=7Y
By hit and trial method,
Put Y=1=>X=7=>total ₹ 7.2 is less
Put Y=2=>X=14=>Total=₹ 14.4
This is correct.
Hence she spent=(2/3)x14.4=₹ 9.60=cost of wallet
IncorrectSolution:
Let number of one rupee notes=X
Number of 20 paise coins=Y
Arshi started with (100X+20Y) and came back with (100Y and 20A) paise
Also, 100Y+20X=(1/3) (100X+20Y)
=>X=7Y
By hit and trial method,
Put Y=1=>X=7=>total ₹ 7.2 is less
Put Y=2=>X=14=>Total=₹ 14.4
This is correct.
Hence she spent=(2/3)x14.4=₹ 9.60=cost of wallet
- Question 9 of 40
9. Question
In an island of the Maldives, the natives have a peculiar process of determining their average earnings and expenditures. According to an old tradition, the average monthly earnings had to be calculated on the basis of 14 months in the calendar year, while the average monthly expenditure was to be calculated on the basis of 9 months in the year. This weird system of calculation always resulted in the natives underestimating their savings because there occurs an underestimation of their earning. The expenditure per month gets overestimated. Now keeping the above points in view try to answer the below questions:
Mr. Ghosh comes to his native island from Africa and makes his native community comprising of 173 families to calculate their average earning and the average expenditure on the basis of 12 months per the calendar year. The average estimated earning in his community according to the old system is 77 fasios per month. Assuming there are no other changes, what will be the percentage change in savings of the 173 families?
CorrectSolution:
Average Monthly earning (old system) = 77
So, Total Income of 173 families/14 = 77
So, Total Income of 173 families = 77×14
Now,Average Monthly earning of 173 families (new system) =
Total Income/12 = 77×14/12 = 89.83 fasios
But, we do not know the average monthly expenditure in either system.
Nor do we know the savings.
So, the required answer cannot be determined.IncorrectSolution:
Average Monthly earning (old system) = 77
So, Total Income of 173 families/14 = 77
So, Total Income of 173 families = 77×14
Now,Average Monthly earning of 173 families (new system) =
Total Income/12 = 77×14/12 = 89.83 fasios
But, we do not know the average monthly expenditure in either system.
Nor do we know the savings.
So, the required answer cannot be determined. - Question 10 of 40
10. Question
In an island of the Maldives, the natives have a peculiar process of determining their average earnings and expenditures. According to an old tradition, the average monthly earnings had to be calculated on the basis of 14 months in the calendar year, while the average monthly expenditure was to be calculated on the basis of 9 months in the year. This weird system of calculation always resulted in the natives underestimating their savings because there occurs an underestimation of their earning. The expenditure per month gets overestimated. Now keeping the above points in view try to answer the below questions:
In the previous question, the average estimated monthly expenditure is 21 fasios per month for the island. Determine the percentage change in the estimated savings of the 173 families.
CorrectSolution:
Average Monthly Income (old system) = 77
So, Total Income of 173 families/14 = 77
So, Total Income of 173 families = 77×14
Now, Average Monthly Income of 173 families (new system) =Total Income/12 = 77×14/12 = 89.83 fasios
Now, Average Monthly expenditure (old system) = 21
So Average Monthly Expenditure (new system) = 21×9/12 = 15.75Total Savings (old system) = 77-21=56
Total Savings (new system) = 89.83-15.75 = 74.08
%change = (74.08-56)/56 x100 = 32.3%IncorrectSolution:
Average Monthly Income (old system) = 77
So, Total Income of 173 families/14 = 77
So, Total Income of 173 families = 77×14
Now, Average Monthly Income of 173 families (new system) =Total Income/12 = 77×14/12 = 89.83 fasios
Now, Average Monthly expenditure (old system) = 21
So Average Monthly Expenditure (new system) = 21×9/12 = 15.75Total Savings (old system) = 77-21=56
Total Savings (new system) = 89.83-15.75 = 74.08
%change = (74.08-56)/56 x100 = 32.3% - Question 11 of 40
11. Question
Siddharth spends 30% of his income on his father’s medication, 20% on agricultural property and 10% on daughter’s education. The corresponding percentages for Akarsh are 40%. 25% and 13%. Given below are two conditions in 2 statements.
A.Siddharth spends more on agriculture than Akarsh.
B. Akarsh spends more on daughter’s education than Siddharth.We can determine who spends more on Father’s medication with the help of which of the above statements?
CorrectSolution:
Let Siddharth’s and Akarsh’s total income be 100x and 100y respectively.
So the amount spent by Siddharth on father, agriculture and daughter will be 30x, 20x and 10x respectively.
The corresponding amounts spent by Akarsh will be 40y, 25y and 13y.
We need to find relation between 30x and 40y i.e. we need the ratio 3x/4y
Now let us consider statement A alone:
We get 20x>25y
So 4x>5y i.e. x>1.25y
If we take x=1.3y, then 3x/4y=3.9x/4y<1
But if x=2y, then 3x/4y=6y/4y>1
So statement A alone is not sufficient.Consider statement B alone.
It says 13y>10x
So x<1.3y
For x=1.29y, 3x/4y = 3×1.29y/4y<1
Thus even for the highest value of x, we get 3x<4y.
For any value of x lesser than 1.29, 3x is surely going to be less than 4y.
So statement B is sufficient.IncorrectSolution:
Let Siddharth’s and Akarsh’s total income be 100x and 100y respectively.
So the amount spent by Siddharth on father, agriculture and daughter will be 30x, 20x and 10x respectively.
The corresponding amounts spent by Akarsh will be 40y, 25y and 13y.
We need to find relation between 30x and 40y i.e. we need the ratio 3x/4y
Now let us consider statement A alone:
We get 20x>25y
So 4x>5y i.e. x>1.25y
If we take x=1.3y, then 3x/4y=3.9x/4y<1
But if x=2y, then 3x/4y=6y/4y>1
So statement A alone is not sufficient.Consider statement B alone.
It says 13y>10x
So x<1.3y
For x=1.29y, 3x/4y = 3×1.29y/4y<1
Thus even for the highest value of x, we get 3x<4y.
For any value of x lesser than 1.29, 3x is surely going to be less than 4y.
So statement B is sufficient. - Question 12 of 40
12. Question
A pump can fill or empty an oil reservoir. The capacity of the oil reservoir is 3600m^{3}. The filling capacity of the pump connected to the reservoir is 10m^{3}/min less than its emptying capacity.If the pump takes 12 minutes extra to fill the reservoir than to empty it, determine the emptying capacity of the pump?
CorrectSolution:
Let x m^{3}/min be the filling capacity of the pump.
Therefore the emptying capacity=(x+10) m^{3}/min
The time taken to fill the reservoir=3600/x
Time taken to empty the reservoir=3600(x+10)
It takes 12 more minutes to fill the reservoir than to empty it
{3600/x } – {3600/(x+10) } =12
=>3600x+36000-3600x=12(x^{2}+10x)
=>36000=12(x^{2}+10x)
=>3000= x^{2}+10x
=> x^{2}+10x-3000=0
=>x=-60 or 50
Taking positive value of x=50
Emptying capacity =50+10=60 m^{3}/min
IncorrectSolution:
Let x m^{3}/min be the filling capacity of the pump.
Therefore the emptying capacity=(x+10) m^{3}/min
The time taken to fill the reservoir=3600/x
Time taken to empty the reservoir=3600(x+10)
It takes 12 more minutes to fill the reservoir than to empty it
{3600/x } – {3600/(x+10) } =12
=>3600x+36000-3600x=12(x^{2}+10x)
=>36000=12(x^{2}+10x)
=>3000= x^{2}+10x
=> x^{2}+10x-3000=0
=>x=-60 or 50
Taking positive value of x=50
Emptying capacity =50+10=60 m^{3}/min
- Question 13 of 40
13. Question
Ronnie starts driving from the market to the Star’s lab in his car. Meanwhile, Dr. Stein also drives from the market one hour after Ronnie and overtakes Ronnie after covering 30% of the distance from market to the lab. Both of them continue the drive even after the overtaking. On reaching Star’s lab, Dr. Stein reverses his car and meets Ronnie’s car, after covering 23 1/3 of the distance between Star’s lab and the market. In how much time did Dr. Stein cover the distance between market to the lab? ( in hours)
CorrectSolution:
Let the distance from market to Star’s lab be x
When Dr. Stein overtakes Ronnie for the first time, both of them cover= 3x/10
When Dr. Stein meets Ronnie after that, Dr. Stein covers (7x/10)+(7x/30) =28x/30
And Ronnie covers= (23x/30)-(9x/30)=14x/30
Therefore, Dr. Stein is twice as fast as Ronnie.
Dr.Stein starts one hour after Ronnie and catches up in 1 hour.
Thus, Dr. Stein covers 0.3x (30 % of x) in one hour or x in 10/3 hr or 3 1/3 h.
IncorrectSolution:
Let the distance from market to Star’s lab be x
When Dr. Stein overtakes Ronnie for the first time, both of them cover= 3x/10
When Dr. Stein meets Ronnie after that, Dr. Stein covers (7x/10)+(7x/30) =28x/30
And Ronnie covers= (23x/30)-(9x/30)=14x/30
Therefore, Dr. Stein is twice as fast as Ronnie.
Dr.Stein starts one hour after Ronnie and catches up in 1 hour.
Thus, Dr. Stein covers 0.3x (30 % of x) in one hour or x in 10/3 hr or 3 1/3 h.
- Question 14 of 40
14. Question
Tanmay spent less than ₹ 17500 to buy one quintal each of pomegranate, orange and guava. Which one of the following statements is sufficient ot determine which one of the three fruits bought was the costliest ?
A. 2 kg pomegranate and 1 kg guava cost less than 1 kg pomegranate and 2 kg guava.
B. 1 kg pomegranate and 2 kg orange together cost the same as 1 kg orange and 2 kg guava.
CorrectSolution: From A alone, 2P+G<1+2G=>P<G
We got no idea about olive, so it is not sufficient.
From B alone, P+2P=O+2G => P+N=2G
This also is not alone.
Using both, As P<G and P+O=2G=> G<O
Hence both are required.
IncorrectSolution: From A alone, 2P+G<1+2G=>P<G
We got no idea about olive, so it is not sufficient.
From B alone, P+2P=O+2G => P+N=2G
This also is not alone.
Using both, As P<G and P+O=2G=> G<O
Hence both are required.
- Question 15 of 40
15. Question
Nitu got an order for 480 khaddar blouses. She bought 12 charkhas and appointed some spinners to do it. However, many did not come on the working day. As a result, each of those who did come had to spin 32 more blouses than originally assigned with equal distribution of work. How many spinners had been appointed earlier and how many did not come?
CorrectSolution:
Solving by options,
Option a->
12 charkhas need to make 480 blouses.
Each charkha=40 blouses.
Due to the absence of few spinners, now each spinner had to make 40+32=72 blouses.
According to option, 12 are total spinners and 4 were absent hence 8 were spinning.
8 spinners made=8×72 >480 hence the option does not hold true.
Option c->
10 spinners did 480
One spinner did 48.
Remaining 6 spinners need to spin 48+32=130 blouse each
Tota 6 spinners will spin= 130x 6= 480 which satisfies the question
Hence answer is option C
IncorrectSolution:
Solving by options,
Option a->
12 charkhas need to make 480 blouses.
Each charkha=40 blouses.
Due to the absence of few spinners, now each spinner had to make 40+32=72 blouses.
According to option, 12 are total spinners and 4 were absent hence 8 were spinning.
8 spinners made=8×72 >480 hence the option does not hold true.
Option c->
10 spinners did 480
One spinner did 48.
Remaining 6 spinners need to spin 48+32=130 blouse each
Tota 6 spinners will spin= 130x 6= 480 which satisfies the question
Hence answer is option C
- Question 16 of 40
16. Question
Directions: Read the following information carefully and answer the questions.
A total of 3549237 athletes participated in an international marathon to be organized worldwide. The pie chart shows the distribution of athletes from different states. The table shows the ratio of athletes from rural and urban areas of the different places. It also shows the ratio of persons who live in pucca and kaccha houses in the rural areas. The information on the percentage of graduates and the persons with computer literacy from different places is also denoted in the table. Now keeping the data in view, solve the following questions.
City Rural : Urban Pucca : Kaccha % of Graduates % of Computer Literates Agra 7 : 6 3 : 7 31 67 Noida 3 : 2 13 : 6 79 89 Pune 5 : 11 7 : 19 13 21 Thane 19 : 21 9 : 4 49 39 Goa 9 : 13 11 : 19 23 37 Nagpur 13 : 17 21 : 4 56 24 Questions: Which place has the highest number of graduates?
CorrectSolution:
Agra=(3549237/360)x56x0.31=552104×0.31=171152
Noida=9859x129x0.79=1271811×0.79=1004730
Pune=9859x67x0.13=660553×0.13=85872
Thane=9859x29x0.49=28591×0.49=140097
Goa=9859x15x0.23=147885×0.23=34014
Noida has highest.
IncorrectSolution:
Agra=(3549237/360)x56x0.31=552104×0.31=171152
Noida=9859x129x0.79=1271811×0.79=1004730
Pune=9859x67x0.13=660553×0.13=85872
Thane=9859x29x0.49=28591×0.49=140097
Goa=9859x15x0.23=147885×0.23=34014
Noida has highest.
- Question 17 of 40
17. Question
Directions: Read the following information carefully and answer the questions.
A total of 3549237 athletes participated in an international marathon to be organized worldwide. The pie chart shows the distribution of athletes from different states. The table shows the ratio of athletes from rural and urban areas of the different places. It also shows the ratio of persons who live in pucca and kaccha houses in the rural areas. The information on the percentage of graduates and the persons with computer literacy from different places is also denoted in the table. Now keeping the data in view, solve the following questions.
City Rural : Urban Pucca : Kaccha % of Graduates % of Computer Literates Agra 7 : 6 3 : 7 31 67 Noida 3 : 2 13 : 6 79 89 Pune 5 : 11 7 : 19 13 21 Thane 19 : 21 9 : 4 49 39 Goa 9 : 13 11 : 19 23 37 Nagpur 13 : 17 21 : 4 56 24 Questions: Which place has the least number of computer literates for the following places?
CorrectSolution:
Agra=552104×0.67=369901
Noida=1271811×0.89=1131912
Pune=660553×0.21=138716
Goa=147885×0.37=54717
Goa has least computer literate
IncorrectSolution:
Agra=552104×0.67=369901
Noida=1271811×0.89=1131912
Pune=660553×0.21=138716
Goa=147885×0.37=54717
Goa has least computer literate
- Question 18 of 40
18. Question
Directions: Read the following information carefully and answer the questions.
A total of 3549237 athletes participated in an international marathon to be organized worldwide. The pie chart shows the distribution of athletes from different states. The table shows the ratio of athletes from rural and urban areas of the different places. It also shows the ratio of persons who live in pucca and kaccha houses in the rural areas. The information on the percentage of graduates and the persons with computer literacy from different places is also denoted in the table. Now keeping the data in view, solve the following questions.
City Rural : Urban Pucca : Kaccha % of Graduates % of Computer Literates Agra 7 : 6 3 : 7 31 67 Noida 3 : 2 13 : 6 79 89 Pune 5 : 11 7 : 19 13 21 Thane 19 : 21 9 : 4 49 39 Goa 9 : 13 11 : 19 23 37 Nagpur 13 : 17 21 : 4 56 24 Questions: The number of rural persons living in rural pucca houses of Nagpur is what percent more or less than the number of urban persons from Nagpur?
CorrectSolutions:
Nagpur pucca rural=9859×64 x(13/30)x(21/25)=630976 x(13/30)x(21/25)=259675
Nagpur urban=630976×17/30=357553
Answer=(357553-259675)/357553 x100= 27.3 % less
IncorrectSolutions:
Nagpur pucca rural=9859×64 x(13/30)x(21/25)=630976 x(13/30)x(21/25)=259675
Nagpur urban=630976×17/30=357553
Answer=(357553-259675)/357553 x100= 27.3 % less
- Question 19 of 40
19. Question
Directions: Read the following information carefully and answer the questions.
A total of 3549237 athletes participated in an international marathon to be organized worldwide. The pie chart shows the distribution of athletes from different states. The table shows the ratio of athletes from rural and urban areas of the different places. It also shows the ratio of persons who live in pucca and kaccha houses in the rural areas. The information on the percentage of graduates and the persons with computer literacy from different places is also denoted in the table. Now keeping the data in view, solve the following questions.
City Rural : Urban Pucca : Kaccha % of Graduates % of Computer Literates Agra 7 : 6 3 : 7 31 67 Noida 3 : 2 13 : 6 79 89 Pune 5 : 11 7 : 19 13 21 Thane 19 : 21 9 : 4 49 39 Goa 9 : 13 11 : 19 23 37 Nagpur 13 : 17 21 : 4 56 24 Questions: If 17% of graduates from Pune shift to Thane, what is the new number of graduates in Thane?
CorrectSolution:
Graduates shifting from Pune= 85872×0.17=14598
New number of graduates in Thane=140097+14598=154695
IncorrectSolution:
Graduates shifting from Pune= 85872×0.17=14598
New number of graduates in Thane=140097+14598=154695
- Question 20 of 40
20. Question
Directions: Read the following information carefully and answer the questions.
A total of 3549237 athletes participated in an international marathon to be organized worldwide. The pie chart shows the distribution of athletes from different states. The table shows the ratio of athletes from rural and urban areas of the different places. It also shows the ratio of persons who live in pucca and kaccha houses in the rural areas. The information on the percentage of graduates and the persons with computer literacy from different places is also denoted in the table. Now keeping the data in view, solve the following questions.
City Rural : Urban Pucca : Kaccha % of Graduates % of Computer Literates Agra 7 : 6 3 : 7 31 67 Noida 3 : 2 13 : 6 79 89 Pune 5 : 11 7 : 19 13 21 Thane 19 : 21 9 : 4 49 39 Goa 9 : 13 11 : 19 23 37 Nagpur 13 : 17 21 : 4 56 24 Questions: If another place Chennai has 37.7% less computer literates than Noida. Find the number of computer literates in Chennai.
CorrectSolution:
Number of computer literates in Noida=1131912
100-37.7=62.3%
Hence Number of computer literates in Chennai= 1131912×0.623=705181
IncorrectSolution:
Number of computer literates in Noida=1131912
100-37.7=62.3%
Hence Number of computer literates in Chennai= 1131912×0.623=705181
- Question 21 of 40
21. Question
What will come in the place of (?)?
7560.6÷4×11.2= (?)×8.8+111.11
CorrectIncorrect - Question 22 of 40
22. Question
What will come in the place of (?)?
1015/29÷65/871×517.5/207=(?)
CorrectIncorrect - Question 23 of 40
23. Question
What will come in the place of (?)?
(23.32)^2+5099.32-712.65×32.31=(?)
CorrectIncorrect - Question 24 of 40
24. Question
What will come in the place of (?)?
√5476×√13689÷√3442.25=(?)
CorrectIncorrect - Question 25 of 40
25. Question
What will come in the place of (?)?
66% of 800-281.71= (?)-23% of 789
CorrectIncorrect - Question 26 of 40
26. Question
What should come in place of question mark in the following questions?
16.01 21.51 ? 4829.16 309074.74
CorrectLOGIC-
*1³+5.5, *2³+6.5, *3³+7.5 and so on
IncorrectLOGIC-
*1³+5.5, *2³+6.5, *3³+7.5 and so on
- Question 27 of 40
27. Question
What should come in place of question mark in the following questions?
243 171 ? 63 27 3
CorrectLOGIC-
-72, -60, -48, -36, -24
IncorrectLOGIC-
-72, -60, -48, -36, -24
- Question 28 of 40
28. Question
What should come in place of question mark in the following questions?
306 ? 182 132 90
CorrectLogic –
306=18*17
240=16*15
182=14*13
132=12*11 and so on
IncorrectLogic –
306=18*17
240=16*15
182=14*13
132=12*11 and so on
- Question 29 of 40
29. Question
Each of the two mobile phones i.e., iPhone and OPPO were sold at same price . A profit of 10% was made on iPhone while a loss of 30% incurred on OPPO. Find the total gain or loss?
CorrectSolution:
As SP is same so SP1=SP2
So, total SP=2x assuming SP to be x
CP of iPhone=(100/110) of x (10% profit)
CP of oppo= (100/70) of x ( 30% loss)
Total CP= [(100/110) of x]+[(100/70)of x]=(180/77) 0f x
Profit/loss%=SP-CP/CP*100=[2x-(180x/77)/(180x/77)]*100= -(13/90)*100
Since negative sign. So its loss which is approximately 14%
IncorrectSolution:
As SP is same so SP1=SP2
So, total SP=2x assuming SP to be x
CP of iPhone=(100/110) of x (10% profit)
CP of oppo= (100/70) of x ( 30% loss)
Total CP= [(100/110) of x]+[(100/70)of x]=(180/77) 0f x
Profit/loss%=SP-CP/CP*100=[2x-(180x/77)/(180x/77)]*100= -(13/90)*100
Since negative sign. So its loss which is approximately 14%
- Question 30 of 40
30. Question
A man bought 10 jam bottles of equal size. If he sells jam at Rs.50 per kg, he losses Rs.200. While selling it at Rs.60 per kg he gains Rs.150 on the whole. Find the amount of honey in each bottle?
CorrectSolution:
Let the amount of jam = x kg
ATQ, 50x+200=60x
So, x=35 kg
Hence each bottle has 3.5 kg
IncorrectSolution:
Let the amount of jam = x kg
ATQ, 50x+200=60x
So, x=35 kg
Hence each bottle has 3.5 kg
- Question 31 of 40
31. Question
A can do a piece of work in 20 days. B can do half the work in 12 days. C can do 1/3 of work in 5 days. If they all started working together but A leaves the task 2 days before completion and C leaves the work 4 days before completion. B works alone until the task is finished. Find the total time (approximate value) to complete the task.
CorrectSolution:
Let the no. of days be denoted as ‘d’
ATQ
d-2/20+d/24+d-4/15=1
solving we get d=8.63 nearly equal to 8.6 days
IncorrectSolution:
Let the no. of days be denoted as ‘d’
ATQ
d-2/20+d/24+d-4/15=1
solving we get d=8.63 nearly equal to 8.6 days
- Question 32 of 40
32. Question
Amit borrowed Rs.50,000 from two money lenders. For the first loan he paid 24%pa as simple interest and for second loan he paid 28%pa. If the total interest he paid at the end of the year be Rs.12560, how much amount did he borrow from each lender?
CorrectSolution:
Let the money received at 24% be Rs.x
Then money received at 28% be Rs. 50000-x
Then,[x*24/200*1]+[(50000-x)]*1*28/100=12560
6x/25+350000-7x/25=12560
Or, x=36000
So, at 24% amount =36000
thus, at 28% amount is 14000IncorrectSolution:
Let the money received at 24% be Rs.x
Then money received at 28% be Rs. 50000-x
Then,[x*24/200*1]+[(50000-x)]*1*28/100=12560
6x/25+350000-7x/25=12560
Or, x=36000
So, at 24% amount =36000
thus, at 28% amount is 14000 - Question 33 of 40
33. Question
If a sum of money grows to 144/121 times when invested for two years in a scheme where interest is compounded annually, how long will the same sum of money take to treble if invested at the same rate of interest in a scheme where interest is computed using simple interest method?
CorrectSolution:
The sum of money grows to 144/121 times in 2 years.
If P is the principal invested, then it has grown to 144/121 P in two years when invested in compound interest.
In compound interest, if a sum is invested for two years, the amount is found using the following formula
A=(1+R/100)² P in this case.
=>(1+R/100)²=144/121 =>(1+R/100)²=(12/11)²=>R=100/11
If r =100/11% , then in simple interest the time it will take for a sum of money to treble is found out as follows:
Let P be the principal invested. Therefore, if the principal trebles = 3P, the remaining 2P has come on account of simple interest.
Simple Interest =PNR/100 , where P is the simple interest, R is the rate of interest and ‘N’ is the number of years the principal was invested.
Therefore, 2P =PN×100/11×100 => 2 =N/11 or N = 22 yearsIncorrectSolution:
The sum of money grows to 144/121 times in 2 years.
If P is the principal invested, then it has grown to 144/121 P in two years when invested in compound interest.
In compound interest, if a sum is invested for two years, the amount is found using the following formula
A=(1+R/100)² P in this case.
=>(1+R/100)²=144/121 =>(1+R/100)²=(12/11)²=>R=100/11
If r =100/11% , then in simple interest the time it will take for a sum of money to treble is found out as follows:
Let P be the principal invested. Therefore, if the principal trebles = 3P, the remaining 2P has come on account of simple interest.
Simple Interest =PNR/100 , where P is the simple interest, R is the rate of interest and ‘N’ is the number of years the principal was invested.
Therefore, 2P =PN×100/11×100 => 2 =N/11 or N = 22 years - Question 34 of 40
34. Question
Two friends A and B leave City P and City Q simultaneously and travel towards Q and P at constant speeds. They meet at a point in between the two cities and then proceed to their respective destinations in 54 minutes and 24 minutes respectively. How long did B take to cover the entire journey between City Q and City P?
CorrectSolution:
Let us assume Car A travels at a speed of a and Car B travels at a speed of b. Further, let us assume that they meet after t minutes.
Distance traveled by car A before meeting car B = a * t. Likewise distance traveled by car B before meeting car A = b * t.
Distance traveled by car A after meeting car B = a * 54. Distance traveled by car B after meeting car A = 24 * b.
Distance traveled by car A after crossing car B = distance traveled by car B before crossing car A (and vice versa).
=> at = 54b ———- (1)
and bt = 24a ——– (2)
Multiplying equations 1 and 2
we have ab * t2 = 54 * 24 * ab
=> t2 = 54 * 24
=> t = 36
So, both cars would have traveled 36 minutes prior to crossing each other. Or, B would have taken 36 + 24 = 60 minutes to travel the whole distance.IncorrectSolution:
Let us assume Car A travels at a speed of a and Car B travels at a speed of b. Further, let us assume that they meet after t minutes.
Distance traveled by car A before meeting car B = a * t. Likewise distance traveled by car B before meeting car A = b * t.
Distance traveled by car A after meeting car B = a * 54. Distance traveled by car B after meeting car A = 24 * b.
Distance traveled by car A after crossing car B = distance traveled by car B before crossing car A (and vice versa).
=> at = 54b ———- (1)
and bt = 24a ——– (2)
Multiplying equations 1 and 2
we have ab * t2 = 54 * 24 * ab
=> t2 = 54 * 24
=> t = 36
So, both cars would have traveled 36 minutes prior to crossing each other. Or, B would have taken 36 + 24 = 60 minutes to travel the whole distance. - Question 35 of 40
35. Question
A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
CorrectSolution:
Let x and (12-x) litres of milk be mixed from the first and second container respectively
Amount of milk in x litres of the first container = .75x
Amount of water in x litres of the first container = .25x
Amount of milk in (12-x) litres of the second container = .5(12-x)
Amount of water in (12-x) litres of the second container = .5(12-x)
Ratio of water to milk = [.25x + .5(12-x)] : [.75x + .5(12-x)]=3:5 ⇒(.25x+6−.5x)/(.75x+6−.5x)=3/5
⇒(6−.25x)/(.25x+6)=3/5
⇒30−1.25x=.75x+18
⇒2x=12⇒x=6
Since x = 6, 12-x = 12-6 = 6 Hence 6 and 6 litres of milk should mixed from the first and second container respectivelyIncorrectSolution:
Let x and (12-x) litres of milk be mixed from the first and second container respectively
Amount of milk in x litres of the first container = .75x
Amount of water in x litres of the first container = .25x
Amount of milk in (12-x) litres of the second container = .5(12-x)
Amount of water in (12-x) litres of the second container = .5(12-x)
Ratio of water to milk = [.25x + .5(12-x)] : [.75x + .5(12-x)]=3:5 ⇒(.25x+6−.5x)/(.75x+6−.5x)=3/5
⇒(6−.25x)/(.25x+6)=3/5
⇒30−1.25x=.75x+18
⇒2x=12⇒x=6
Since x = 6, 12-x = 12-6 = 6 Hence 6 and 6 litres of milk should mixed from the first and second container respectively - Question 36 of 40
36. Question
A group of people decided to cut 128 trees in a certain number of days. For the first 4 days, they were able to achieve their planned per day target. However, for the remaining days, the group was able to cut 4 more trees daily than planned. In this way, the group had cut 144 trees one day before the planned finish date. What was the number of trees the group was planning to cut per day?
CorrectSolution:
Let the original target be x trees/day and the number of planned days be y.
=) y=128/x
Also, as per the question
(y-3-1)×(x+4) + 4x = 144
(y-4)×(x+4) + 4x = 144
xy – 4x + 4y – 16 + 4x = 144
x×128/x+4×128/x=160
512/x=32
x = 16IncorrectSolution:
Let the original target be x trees/day and the number of planned days be y.
=) y=128/x
Also, as per the question
(y-3-1)×(x+4) + 4x = 144
(y-4)×(x+4) + 4x = 144
xy – 4x + 4y – 16 + 4x = 144
x×128/x+4×128/x=160
512/x=32
x = 16 - Question 37 of 40
37. Question
In a class, if 50% of the boys were girls, then there would be 50% more girls than boys. What percentage of the overall class is girls?
CorrectSolution:
ATQ, the number of girls should be 1.5 * the number of boys.
When 50% of the boys are taken as girls, let the number of boys = x
Number of girls = 1.5x
total number of students = 2.5x
Original number of boys = 2x (50% of boys = x)
Original number of girls = 0.5x
therefore, Girls form 20% of the overall class.IncorrectSolution:
ATQ, the number of girls should be 1.5 * the number of boys.
When 50% of the boys are taken as girls, let the number of boys = x
Number of girls = 1.5x
total number of students = 2.5x
Original number of boys = 2x (50% of boys = x)
Original number of girls = 0.5x
therefore, Girls form 20% of the overall class. - Question 38 of 40
38. Question
In an ice-cream factory, 30% of total Vanilla-flavored ice-cream produced is mixed with 40% of total Banana-flavored ice-cream produced, in a ratio of 1:1. Find the overall percentage of ice-cream produced (approximately) which is mixed with each other. (Ice-creams of only these 2 flavors are produced in the factory)
CorrectSolution:
Let the total Vanilla-flavored ice-cream produced and total Banana-flavored ice-cream produced be x and y respectively
Given,30% of x = 40% of y (Since ratio is 1:1)………….(1)
=> x/y=4/3
=> x/y+1=4/3+1(On adding 1 to both sides of the equation)
=> (x+y)/y=7/3
=> y/(x+y)=3/7
Total ice-cream mixed = 0.3x + 0.4y = 0.4y + 0.4y = 0.8y (from (1))
% of overall ice-cream mixed = 0.8y/(x+y)×100=0.8×3/7×100 = 34.28%~34%IncorrectSolution:
Let the total Vanilla-flavored ice-cream produced and total Banana-flavored ice-cream produced be x and y respectively
Given,30% of x = 40% of y (Since ratio is 1:1)………….(1)
=> x/y=4/3
=> x/y+1=4/3+1(On adding 1 to both sides of the equation)
=> (x+y)/y=7/3
=> y/(x+y)=3/7
Total ice-cream mixed = 0.3x + 0.4y = 0.4y + 0.4y = 0.8y (from (1))
% of overall ice-cream mixed = 0.8y/(x+y)×100=0.8×3/7×100 = 34.28%~34% - Question 39 of 40
39. Question
Two persons A and B start moving at each other from point P and Q respectively which are 1400 Km apart. Speed of A is 50 Km/hr and that of B is 20 Km/hr. How far is A from Q when he meets B for the 22nd time?
CorrectSolution:
Total distance travelled by both of them for 22nd meeting = 1400 + 21 x 2 x 1400 = 43 x 1400
Distance travelled by each will be in proportion of their speed:-
Therefore, distance travelled by A = 50/(50 + 20) x 43 x 1400 = 43000
Now, for every odd multiple of 1400, A will be at Q and for every even multiple of 1400 A will be at P. So, at 42000 Km (1400 x 30, even multiple) A will beat P. So at their 22 meeting, A will be 1000 Km from P, therefore, 400 Km from Q.IncorrectSolution:
Total distance travelled by both of them for 22nd meeting = 1400 + 21 x 2 x 1400 = 43 x 1400
Distance travelled by each will be in proportion of their speed:-
Therefore, distance travelled by A = 50/(50 + 20) x 43 x 1400 = 43000
Now, for every odd multiple of 1400, A will be at Q and for every even multiple of 1400 A will be at P. So, at 42000 Km (1400 x 30, even multiple) A will beat P. So at their 22 meeting, A will be 1000 Km from P, therefore, 400 Km from Q. - Question 40 of 40
40. Question
In an examination the candidate was asked to attempt 50 questions where all the questions were mandatory.1 mark is given for each correct answer and 25% of the correct answer’s mark is deducted if answer is incorrect. Nisha appeared the exam and scored 27.5 marks. How many incorrect answers she gave?
CorrectSolution:
Let x be the no of correct answers so (50-x) will be the number of wrong ans.
x-(50-x)/4=27.5
=>x=32..So no of incorrect answers=(50-32)=18IncorrectSolution:
Let x be the no of correct answers so (50-x) will be the number of wrong ans.
x-(50-x)/4=27.5
=>x=32..So no of incorrect answers=(50-32)=18
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