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A man purchased a TV and fridge. If the price of TV is 150% of price of fridge then price of fridge is what percentage of the total cost of TV and fridge?
Let the cost of the fridge = 100x
The price of the TV = 150x (Since the cost of TV is 150% of the cost of the fridge)
The price of the fridge $$\ \frac{\ 100x}{100x+150x}$$ times the price of TV and fridge
The price of the fridge =40% of the price of TV and fridge
B is the correct answer.
If a labour output function for laundry service is described by the following equation: $$O = (a^\circ - 4L^\circ)^6$$, where L denotes Labour and O denotes output. Then, output ..........
In question it is mentioned L denotes labor.
Here $$a^0$$ and $$L^0$$ are both constants, since they are not being mentioned as variables.
Also, the equation does not have "L" term in it
So, the output is independent of Labour L.
C is the correct answer.
Assume that the taxes on petrol is 125% of the price of petrol per litre as received by the retailer minus the taxes. If in the last week, the petrol prices per litre as received by the retailer minus the taxes was 35, 34, 35.5, 37, 37.5 and 38, the average amount of tax collected per litre of petrol is
Sum of all (Petrol price per litre - taxes) in the last week = 35+34+35.5+37+37.5+38
= 217
Taxes on petrol = 1.25 times (the price of petrol - taxes)
2.25* taxes on petrol = 1.25 times the price of petrol
The price of petrol = 1.8 times the taxes on petrol.
1.8 times the taxes on petrol - taxes on petrol = 217
0.8 times the taxes on petrol = 217
Taxes on petrol = $$\ \frac{217}{0.8}$$ = 271.25
The average amount of tax collected = $$\ \frac{271.25}{6}$$
= 45.2
A is the correct answer.
If the price of Sugar increases by 20%, and Salman intends to spend only an additional 5% on Sugar, then find out the percentage decrease in his sugar consumption.
Let the initial price and expenditure of sugar be 100p, 100e respectively.
Consumption = $$\ \frac{\ Expenditure\ }{\Pr ice}$$
=$$\ \frac{\ e\ }{p}$$
The increased price of sugar = 120p
Increased expenditure on sugar = 105e
Changed consumption = $$\ \frac{105\ e\ }{120p}$$
=0.875 $$\ \frac{\ e\ }{p}$$
Percentage decrease in consumption = $$\ \frac{(0.875-1)}{1}$$*100
=12.5
B is the correct answer.
If 20% of x = y, then y% of 20 is same as
It is given 20% of x = y
x=5y
We have to find y% of 20 = $$\ \frac{\ y}{100}\times\ 20$$
=$$\ \frac{\ y}{5}$$
=$$\ \frac{\ x}{25}$$
= 4% of x
A batsman was having 32 runs per innings as his average after 15th innings. His average increased by 2 runs after 16th inning. Then what was his score in the 16th inning?
This can be answered either by using averages concept or alligations concept.
Alligations :$$p=\frac{\left(p1q1+p2q2\right)}{q1+q2}$$ where q1 and q2 are the number of innings in two group's , p1 and p2 are there respective averages(average runs) , p is the overall average.
Here q1=15 , q2=1 , p1=32 , p= 34 (since increased by 2 runs)
On solving the equation you get p2=64
Three candidates "A", "B", "C" participated in an election. "A" gets 40% of the votes more than "B". "C" gets 20% votes more than "B". "A" also overtakes "C" by 4000 votes. If 90% voters voted and no invalid or illegal votes were cast, then what will be the number of voters in the voting list?
Let 100x be the total numbers of voters in the voters list. => 90x voters voted
A=1.4B (since A received 40% more votes than B)
C=1.2B
Given A-C = 4000
1.4B-1.2B=4000 => B=20000 => A=24000 and C = 28000
So total voter who voted =A+B+C= 72000 which is 90% of the total list.
So when 90%= 72000 , 100%= 80000.
Find the value of 1% of 1% of 25% of 1000 is
$$\frac{1}{100}\cdot\frac{1}{100}\cdot\frac{25}{100}\cdot1000\ =0.025$$
A and B rent a pasture for 10 months; A puts in 80 cows for 7 months. How many can B put in for the remaining 3 months, if he pays half as much again as A?
It is given that B pays 1.5 times that of A
Let B put N number of cows for 3 month
$$\frac{Money\ paid\ by\ B}{Money\ paid\ by\ A}=\ \frac{3}{2}\ =\ \frac{3\times\ N}{80\times\ 7}$$
$$N=280$$
The average of 11 numbers is 10.9. If the average of the first six numbers is 10.5 and that of the last six numbers is 11.4, then the middle number is :
Let $$a_1,\ a_2.....\ a_{11}$$ be the 11 number
$$a_1+a_2+.....\ +a_{11}\ =\ 11\times\ 10.9\ =\ 119.9$$ .... (I)
Average of first six is 10.5 thus $$a_1+a_2+.....\ +a_6\ =\ 6\times\ 10.5\ =\ 63$$ ....(II)
Average of last six is 10.5 thus $$a_6+a_7+.....\ +a_11\ =\ 6\times\ 11.4\ =\ 68.4$$ ....(III)
(II)+(III)-(I) gives $$a_6\ =\ 11.5$$
The monthly salaries of A and B together amount to 40,000. A spends 85% of his salary and B, 95% of his salary. If now their savings are the same, then the salary of A is
Let a and b represent salaries of A and B respectively
then $$a+b=40000$$...(I)
After spending 85% of his salary, savings of A will be = (0.15)a
After spending 95% of his salary, savings of A will be = (0.05)b
Their saving are equal, therefore (0.15)a = (0.05b) or 3a=b ...(II)
Solving eqn (I)and (II) we get a = 10,000
Average of five numbers is 61. If the average of first and third number is 69 and the average of second and fourth number is 69, what is the fifth number ?
Let the five numbers be a, b, c, d, e in the given order.
a+b+c+d+e = 61*5=305 ----- Eq 1
a+c = 69*2 = 138
b+d = 69*2 = 138
From eq 1: e=305-(138+138) =305-276 =29
B is the correct answer.
A certain amount was to be distributed among A, B and C in the ratio 2 :3 :4 respectively, but was erroneously distributed in the ratio 7:2:5 respectively. As a result of this, B got Rs.40 less. What is the amount ?
Let the amount be. Rs x
B's share in the ratio 2:3:4 = $$\ \frac{\ 3}{9}$$ of x=$$\ \frac{\ x}{3}$$
B's share in the ratio 7:2:5 = $$\ \frac{\ 2}{14}$$ of x = $$\ \frac{\ x}{7}$$ , therefore
$$\ \frac{\ x}{3}$$-$$\ \frac{\ x}{7}$$= 40
$$\ \frac{\ 4x}{21}$$ = 40
x = (40×21)/4 = Rs 210
A is the correct answer.
Last year there were 610 boys in a school. The number decreased by 20 percent this year. How many girls are there in the school if the number of girls is 175 percent of the total number of boys in the school this year?
Number of boys in the school last year = 610
Number of boys in the school in the present year = 610*0.8 (It is given that the boys decreased by 20% this year)
= 488
Number of girls in the class = 1.75* Number of boys
=1.75*488
=854
A is the correct answer.
Rs.73,689/- are divided between A and B in the ratio 4 :7. What is the difference between thrice the share of A and twice the share of B?
Rs.73,689/- are divided between A and B in the ratio 4 :7.
Share of A = $$\ \frac{\ 4}{11}\times\ 73689$$ /- = 26796
Share of B = $$\ \frac{\ 7}{11}\times\ 73689$$ /- = 46893
Now, we have to find the difference between thrice the share of A and twice the share of B
= 3*26796 - 2*46893 =Rs. 13398
The average marks in English subject of a class of 24 students is 56. If the marks of three students were misread as 44, 45 and 61 of the actual marks 48, 59 and 67 respectively, then what would be the correct average?
Given,
Sum of the marks misread = 44+45+61 = 150
Actual marks of the students = 48, 59 and 67
Actual marks sum = 48+59+67 = 174
Difference = 174-150 = 24
Cporrect average = 56+$$\ \frac{\ 24}{24}$$
= 57
D is the correct answer.
The respective ratio between the present age of Manisha and Deepali is 5 : X. Manisha is 9 years younger than Parineeta. Parineeta's age after 9 years will be 33 years. The difference between Deepali's and Manisha's age is same as the present age of Parineeta. What will come in place of X?
Let the present age of Manisha be 5a
The present age of Deepali = Xa
The present age of Praneeta = 5a+9
Praneeta's age after 9 years = 5a+18=33
5a=15
a=3
The difference between Deepali's and Manisha's age is the same as the present age of Parineeta
Xa-5a=5a+9
3X-15=15+9
3X=39 ==> X=13
D is the correct answer.
Profit earned by an organization is distributed among officers and clerks where the individual amount received by them is in the ratio of 5 : 3. If the number of officers is 45 and the number of clerks is 80 and the amount received by each officer is Rs.25,000, what was the total amount of profit earned?
Profit received by each officer= Rs.25000
Therefore, Profit received by each clerk= $$\ \frac{\ 3}{5}$$×25000 = Rs.15000
Therefore, Total earned profit
= Rs. (45×25000+80×15000)
= Rs.(1125000+1200000)
= Rs.23.25 lakh
D is the correct answer.
Kajal spends 55% of her monthly income on grocery, clothes and education in the ratio of 4 : 2 : 5 respectively. If the amount spent on clothes is Rs.5540/-, what is Kajal’s monthly income?
Let the monthly income of Kajal be 100k
It is given that she spent 55% of the monthly income on grocery, clothes and education in the ratio of 4 : 2 : 5 respectively
i.e she spent 55k on grocery, clothes and education in the ratio of 4 : 2 : 5 respectively.
Amount spent on clothes = $$\ \frac{\ 2}{11}\times\ 55k$$
10k = 5540 (given)
So 100k = 55400
Monthly income = 100k = 55400
A is the correct answer.
Mrs. X spends Rs. 535 in purchasing some shirts and ties for her husband. If shirts cost Rs. 43 each and the ties cost Rs.21 each, then what is the ratio of the shirts to the ties, that are purchased ?
Given,
Cost of each shirt = Rs. 43
Cost of each tie = Rs. 21
Let x, y be the number of shirts and ties purchased
x*43+ y*21 = 535
y = $$\ \frac{\ 535-43x}{21}$$
when x = 10, y is an integer i.e y =5
ratio of shirts to ties purchased = 10:5 =2:1
B is the correct answer.
In an examination, Raman scored 25 marks less than Rohit. Rohit scored 45 more marks than Sonia. Rohan scored 75 marks which is 10 more than Sonia. Ravi's score is 50 less than, maximum marks of the test. What approximate percentage of marks did Ravi score in the examination, if he gets 34 marks more than Raman ?
Marks obtained by Rohan =75
Marks obtained by Sonia = 65
Marks obtained by Rohit = 65+45=110
Marks obtained by Raman = 110-25 = 85
Marks obtained by Ravi = 85+34 = 119
Maximum Marks = 119+50 = 169
Required percentage = $$\ \frac{\ 119}{169}\times\ 100$$
=70%
B is the correct answer.
Average weight of 19 men is 74 kgs, and the average weight of 38 women is 63 kgs. What is the average weight (rounded off to the nearest integer) of all the men and the women together?
The average weight of 19 men is 74 kgs
The average weight of 38 women is 63 kgs
The average weight of all the men and women together = $$\ \frac{\ 74\times\ 19+63\times\ 38}{19+38}$$
=$$\ \frac{\ 3800}{57}$$
= 66.66 kg
=approx 67 kg
D is the correct answer.
Mr Giridhar spends 50% of his monthly income on household items and out of the remaining he spends 50% on transport, 25% on entertainment, 10% on sports and the remaining amount of Rs. 900 is saved. What is Mr Giridhar's monthly income?
Let the total monthly income of Mr Giridhar be 100g
The proportion of income he spends on household items = 50% i.e 50g
Of the remaining 50%, the proportion of income he spends on transport, entertainment and sports are 50%, 25%, 10%
15% of 50g = Amount saved
7.5g = 900(given)
100g = 12000
The monthly income of Mr Giridhar = Rs.12000
In a team there are 240 members (males and females). Two-third of them are males. Fifteen percent of males are graduates. Remaining males are nongraduates. Three-fourth of the females are graduates. Remaining females are non-graduates
What is the difference between the number of females who are non-graduates and the number of males who are graduates?
Total members :240
Males = 2/3 (240) =160
so Females = 240-160 =80
Graduate Males = 160(0.15) =24
Non graduate males = 160-24 = 136
Graduate Females = (0.75) 80 = 60
So non graduate females = 80-60 =20
Now difference between the number of females who are non-graduates and the number of males who are graduates : 24-20 =4
In a team there are 240 members (males and females). Two-third of them are males. Fifteen percent of males are graduates. Remaining males are nongraduates. Three-fourth of the females are graduates. Remaining females are non-graduates
What is the sum of the number of females who are graduates and the number of males who are non-graduates?
Total members = 240
Number of males = $$\ \frac{\ 2}{3}\times\ 240$$ = 160
Number of males who are non-graduates = $$\ \frac{\ 85}{100}\times\ 160$$ = 136
Number of females = $$\ \frac{\ 1}{3}\times\ 240$$ = 80
Number of females who are graduates = $$\ \frac{\ 3}{4}\times\ 80$$ = 60
Sum of male non graduates and female graduates = 136+60=196
D is the correct answer.
In a team there are 240 members (males and females). Two-third of them are males. Fifteen percent of males are graduates. Remaining males are nongraduates. Three-fourth of the females are graduates. Remaining females are non-graduates
What is the ratio between the total number of males and the number of females who are non-graduates?
Total members in the team = 240
Males = $$\ \frac{\ 2}{3}\times\ 240$$ = 160
Females = $$\ \frac{\ 1}{3}\times\ 240$$ = 80
Number of females who are non graduates = $$\ \frac{\ 1}{4}\times\ 80$$ = 20
ratio between the total number of males and the number of females who are non-graduates
= 160:20
=8:1
B is the correct answer.
Ramsukh bhai sells rasgulla(a favourite Indian sweets) at Rs. 15 per kg. A rasgulla is made up of flour and sugar in the ratio 5 : 3. The ratio of price of sugar and flour is 7 : 3 (per kg). Thus he earns $$66\frac{2}{3}$$% profit. What is the cost price of sugar?
SP of 1kg rasgulla= Rs.15
Profit= $$\frac{200}{3}$$%
CP of 1 kg rasgulla = ($$\ \frac{\ 15\times100}{\ \frac{\ 500}{3}}$$) = Rs. 9
CP of 1 kg of sugar= 7x
CP of 1 kg of floor= 3x
3x 7x
9
7x-9 9-3x
$$\ \frac{\ 7x-9}{9-3x}\ =\ \ \frac{\ 5}{3}$$
x=2
CP of sugar= 7x= 14 per kg
D is the correct answer.
A reduction of 20% in the price of sugar enables a person to purchase 6 kg more for Rs. 240. What is the original price per kg of sugar?
Let the initial price of 100 kg Sugar was Rs. 100.
Now, a 20% decrease in rate then rate of 100 kg sugar would be Rs. 80.
So, Rs. 80 = 100 kg sugar.
Rs. 100 = 100× $$\ \ \frac{100}{80}$$ = 125 Kg sugar.
Increment in sugar = 25 kg.
% increment = $$\ \ \frac{125-100}{100}\times\ 100$$ = 25%.
25% increment = 6 kg of sugar
total sugar initially = 6*4 =24 kg
Original price of the sugar = $$\ \ \frac{240}{24}$$
= Rs. 10 per kg
A is the correct answer.
The speed of scooter, car and train are in the ratio of 1 : 4 : 16. If all of them cover equal distance then the ratio of time taken/velocity for each of the vehicle is:
Let the distance covered be 16 km, and the velocities be 1kmph, 4kmph and 16kmph.
Then the time taken: 16h, 4h, and 1h respectively
Ratio of time and velocity for each vehicle: $$\frac{16}{1}:\frac{4}{4}:\frac{1}{16}$$
=$$16:1:\frac{1}{16}$$
=256:16:1.
A, B, C and D purchased a restaurant for Rs. 56 lakhs. The contribution of B, C and D together is 460% that of A, alone. The contribution of A, C and D together is 366.66% that of B’s contribution and the contribution of C is 40% that of A, B and D together. The amount contributed by D is:
A+B+C+D = 56 -- Eq 1
B+C+D = 4.6A -- Eq 2
A+C+D = $$\ \ \frac{11B}{3}$$ --- Eq 3
A+B+D = 2.5C --Eq 4
Using Eq 1 and 2, we get A = 10
Using Eq 1 and 4, we get C = 16
Using Eq 1 and 3, we get B = 12
On substituting the values of A, B, C in Eq 1, we get D = 18
D is the correct answer.
The salary of Raju and Ram is 20% and 30% less than the salary of Saroj respectively. By what percent is the salary of Raju more than the salary of Ram?
Let the salary of Saroj be 100x
The salary of Raju = 80x
The salary of Ram = 70x
the percent by which the salary of Raju more than the salary of Ram = $$\ \frac{\ 80x-70x}{70x}\times\ 100$$
=$$\ \frac{\ 1}{7}\times\ 100$$
=14.28%
D is the correct answer.
The radius of a wire is decreased to one - third and its volume remains the same. The new length is how many times the original length?
We know that wire is in the shape of a cylinder.
So, the volume of wire = $$\pi\ r^2h$$
$$\pi\ r^2h_1$$ = $$\pi\ (\frac{r}{3})^2h_2$$
Cancelling π on both sides,
$$\ r^2h_1$$ = $$\frac{\ r}{9}^{^2}h_2$$
$$9h_1$$ = $$h_2$$
length is increased by 9 times
D is the correct answer.
A 4 cm cube is cut into 1cm cubes. What is the percentage increase in the surface area after such cutting?
Initial surface area of the cube = 6$$(side)^2$$
= 6$$(4)^2$$
= 96$$cm^2$$
No. of new cubes = $$\ \ \frac{\ Volume\ of\ older\ cube}{Volume\ of\ 1\ new\ cube}$$
=$$\ \ \frac{\ 4\times\ 4\times\ 4}{1\times\ 1\times\ 1}$$
=64 cubes
Newer surface area of 64 cubes = 64*6$$(1)^2$$ = 384$$cm^2$$
Percentage increase in surface area = $$\ \ \frac{384-96}{96}$$ *100
=$$\ \ \frac{28800}{96}$$
= 300%
B is the correct answer
The following pie chart shows the hourly distribution in a day (in degrees) of all the major activities of a student. Moreover, a day has 24 hours.
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The percentage of time which he spends in school is:
The time spent in school corresponds to $105^0$$
Percentage of time spent in school = $$\ \frac{\ 105}{360}\times\ 100$$
= Approx 30%
B is the correct answer.
The following pie chart shows the hourly distribution in a day (in degrees) of all the major activities of a student. Moreover, a day has 24 hours.
How much time (in per cent) does he spend in games in comparison to sleeping?
Time spent on games corresponds to $$30^0$$ and on sleeping to $$120^0$$
Required ratio = $$\ \frac{\ 30}{120}\times\ 100$$
=25%
C is the correct answer.
The following pie chart shows the hourly distribution in a day (in degrees) of all the major activities of a student. Moreover, a day has 24 hours.
If he spends the time in games equal to the home work and remains constant in other activities, then the percentage decrease in time of sleeping:
The time spent in playing games and in doing homework corresponds to 30° and 45° respectively.
If the time spent in playing games is equal to that spent on doing homework, then there should be an increase of 15° in the time spent in games.
If this 15 ° is obtained by decreasing the time spent in sleeping (with the other activities remaining constant),
The percentage decrease in the time spent in sleeping = $$\ \frac{\ 15^0}{120^0}\times\ 100$$
= $$12.5^0$$
B is the correct answer.
The following pie chart shows the hourly distribution in a day (in degrees) of all the major activities of a student. Moreover, a day has 24 hours.
What is the difference in time (in hours) spent in school and in home work?
School corresponds to $$105^0$$ and homework corresponds to $$45^0$$
Difference = $$105^0$$ - $$45^0$$
=$$60^0$$
= $$\ \frac{\ 60}{360}\times\ 24$$
= 4 hrs
C is the correct answer.
The following pie chart shows the hourly distribution in a day (in degrees) of all the major activities of a student. Moreover, a day has 24 hours.
If he spends $$\frac{1}{3}rd$$ time of homework in Mathematics then the number of hours he spends in rest of the subjects in home work:
The total number of hours = 24
Time spent on homework = $$45^0$$
=$$\ \frac{\ 45}{360}\times\ 24$$ = 3 hrs
It is given that she spent $$\frac{1}{3}rd$$ time of homework in Mathematics.
The time she spent on Mathematics = $$\ \frac{\ 1}{3}\times\ 3$$
=1 hr
The time she spent on the rest of the subjects = 3-1 = 2 hr
B is the correct answer.
In a retail outlet, the average revenue was Rs.10,000 per day over a 30 day period. During this period, the average revenue on weekends (total 8 days) was Rs. 20,000 per day. What was the average daily revenue on weekdays?
Total revenue of the 30-day period = Rs.10000 x 30 days = Rs.300000
Total revenue of the weekends (8 days) = Rs.20000 x 8 days = Rs. 160000
Total revenue of weekdays = (300000-160000) = Rs. 140000
Avg. daily revenue on weekdays = 140000/22 = Rs.6363.63 $$\approx\ $$ Rs.6364
A bakery opened with its daily supply of 40 dozen rolls. Half of the rolls were sold by noon, and 60% of the remaining rolls were sold between noon and closing time. How many dozen rolls were left unsold?
They started with 40 dozen rolls. Half of them (=20 dozen) were sold by noon.
40% of the remaining rolls (half of 40 dozen) were left unsold, which is 0.4 x 20 = 8 dozen rolls
Sonali invests 15% of her monthly salary in insurance policies. She spends 55% of her monthly salary in shopping and on household expenses. She saves the remaining amount of Rs. 12,750. What is Sonali’s monthly income?
Let Sonali's monthly income be 100x
Now Expenditure on insurance policies = 15x
Expenditure on shopping and household expenses = 55x
Therefore remaining amount = 100x-55x-15x = 30x
Which is equal to 12,750
Now therefore 30x=12,750
x = 425
Therefore total income = 100x = 42,500
In an examination, out of 480 students 85% of the girls and 70% of the boys passed. How many boys appeared in the examination if total pass percentage was 75%?
Total number of students = 480
Percentage of total students passed
= 75 % of total student = 75×480/100=360student
Now, using the condition from the question,
Let the number of boys be x
Then, 70% of x + 85% of (480 - x) = 360
=> 70×x/100+85×(480−x)/100=360
=> 70x - 85x + 40800 = 36000
=> 40800 - 36000 = 85x - 70x
=> 4800 = 15x
=> x=4800/15=320
There are 320 boys who have appeared for the examination.
An agent sells goods of value of Rs. 15,000. The commission which he receives at the rate of $$12\frac{1}{2}\%$$ is
The commission amount will be :
15,000 (12.5/100)
= 15,000(25/200)
= Rs 1875
An alloy of gold and silver weighs 50 gms. It contains 80% gold. How much gold should be added to the alloy so that percentage of gold is increased to 90?
Gold in 50g of alloy
$$=80\times\frac{50}{100}=40g$$
Let x g gold must be added.
Now, according to the question,
$$40+x=\frac{90}{100}\left(50+x\right)$$
$$x=50g$$
Thus, 50 g of gold must be added to it to get 90%.
Thus, the correct option is A.
Weekly incomes of two persons are in the ratio of 7 : 3 and their weekly expenses are in the ratio of 5 : 2. If each of them saves Rs. 300 per week, then the weekly income of the first person is
Let the incomes of two persons are 7x and 3x
Expenditure of first person = 7x - 300
Expenditure of second person = 3x - 300
According to the question,
7x−300/3x−300=5/2
=> 14x-600 = 15x-1500
x = 900
Income of first person
= 7x = 7×900=RS.6300
Wheat is now being sold at Rs. 27 per kg. During last month its cost was Rs. 24 per kg. Find by how much per cent a family reduces its consumption so as to keep the expenditure fixed.
Old price of rice = Rs.24/k
New price of rice = Rs.27 per/kg
Therefore, Increase in price of rice = New price - Old price = 27 - 24 = Rs.3/kg
Therefore, Increased percentage in price
= 3/24×100=12.5 %
Now, reduction in consumption, so as to keep the expenditure fixed.
According to the formula,
$$\frac{a}{100+a}\times\ 100$$
= $$\frac{12.5}{100+12.5}\times\ 100$$
=11.1 %
A 14.4 kg gas cylinder runs for 104 hours when the smaller burner on the gas stove is fully opened while it runs for 80 hours when the larger burner on the gas stove is fully opened. Which of these values are closest to the percentage difference in the usage of gas per hour, of the smaller burner over the larger burner?
Gas usage per hour of the larger burner = $$\ \frac{\ 14.4}{\ \ 80}$$
Gas usage per hour of the smaller burner = $$\ \frac{\ 14.4}{\ \ 104}$$
The required percentage = $$\ \dfrac{\ \ \frac{14.4\ }{80}-\ \frac{\ 14.4}{104}}{\ \frac{14.4\ }{80}}$$ = 1- $$\ \frac{\ \ 80}{\ 104}$$ = $$\ \frac{\ \ 24}{\ 104}$$ = 23.07
The average of nine numbers is M and the average of three of these is P. If the average of remaining numbers is N, then
This can be solved easily by using allegations
$$p=\frac{\left(p1q1+p2q2\right)}{q1+q2}$$
Here p = M , p1=P , p2=N
q1=3 , q2=6
$$\ \therefore\ M=\frac{\left(3P+6N\right)}{3+6}$$
On solving 3M = P+2N
The average of 5 consecutive numbers is n. If the next two numbers are also included the average will
Average of 5 consecutive number = $$\frac{\left(n\right)+\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+\left(n+4\right)}{5}$$ = n+2
If 2 more consecutive number are added then its average = $$\frac{\left(n\right)+\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+\left(n+4\right)\ +\left(n+5\right)+\left(n+6\right)}{7}$$ = n+3
Hence average is increased by 1
Three friends had a dinner at a restaurant. When the bill was received Amita paid $$\dfrac{2}{3}$$ as much as Veena paid and Veena paid $$\dfrac{1}{2}$$ as much as Tanya paid. What fraction of the bill did Veena pay?
Let the amount Tanya paid be 6x
Amount paid by Veena is 3x
Then Amita paid by 2x
Required fraction $$\frac{3x}{6x+3x+2x}$$ = $$\frac{3}{11}$$
The numbers of students studying Physics, Chemistry and Zoology in a college were in the ratio 4 : 3 : 5 respectively. If the number in these three disciplines increased by 50%, 25% and 10% respectively in the next year, then what was the new respective ratio?
Let the number of student in Physics, Chemistry and Zoology be 4x, 3x and 5x respectively
New number of Physics student = 4x*(1+$$\frac{50}{100}$$) = 6x
New number of Chemistry student = 3x*(1+ $$\frac{25}{100}$$) = $$\frac{15x}{4}$$
New number of Zoology student = 5x * (1 + $$\frac{10}{100}$$) = $$\frac{11x}{2}$$
Required ratio is $$6$$ : $$\frac{15}{4}$$ : $$\frac{11}{2}$$
Multiplying overall by 4, answer is 24:15:22
The current birth rate per thousand is 32, whereas corresponding death rate is 11 per thousand. The growth rate in terms of population increase in per cent is given by
For 1000 people, 32 person are born and 11 die.
Overall increase in population = 32-11 = 21
%increase in popoulation = $$\frac{21}{1000}\times\ 100$$% = 2.1%
SNAP Averages, Ratio and Proportion is one of the most important arithmetic topics in the Quantitative, Data Interpretation & Data Sufficiency section of the SNAP 2026 Exam. Questions from this topic test a candidate's understanding of averages, ratios, proportions, partnerships, mixtures, and comparative quantitative relationships. These concepts form the foundation for several advanced arithmetic topics and are frequently used in aptitude-based problem solving. Practicing topic-wise questions and solving problems from SNAP Previous Papers is an effective way to strengthen concepts and improve accuracy. Additionally, attempting SNAP Mock Tests will help candidates understand the latest exam pattern and the types of questions that can be expected in the actual exam. To gain a better understanding of all the topics covered in the exam, candidates should also refer to the SNAP Syllabus.
Yes, Averages, Ratio and Proportion are important topics in SNAP Quantitative Aptitude. These concepts are frequently tested directly and also serve as the basis for solving questions from mixtures, partnerships, percentages, and other arithmetic topics.
SNAP does not prescribe a fixed number of questions from this topic. However, candidates can expect direct and application-based questions involving averages, ratios, proportions, and related arithmetic concepts.
SNAP may include questions on simple and weighted averages, ratio comparisons, direct and inverse proportions, partnerships, mixtures, and real-life quantitative applications involving proportional relationships.
Start by understanding the fundamental concepts and formulas related to averages, ratios, and proportions. Practice a variety of problems, solve questions from SNAP Previous Papers, and regularly attempt SNAP Mock Tests to improve speed, accuracy, and problem-solving skills.
The difficulty level is generally easy to moderate. Most questions are concept-based and can be solved quickly with a strong understanding of the underlying arithmetic principles and sufficient practice.
Cracku's SNAP Averages, Ratio and Proportion Questions are designed according to the latest SNAP exam pattern and difficulty level. They provide topic-wise practice questions, detailed solutions, shortcut methods, and performance analysis to help aspirants strengthen their concepts and improve their scores.
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