Quizsummary
0 of 10 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
Information
No. of Questions 
10 
Time 
10 min 
Medium 
english 
Marks 
10 
Positive Marks 
+1 
Negative Marks 
0 
INSTRUCTION : All question carry 1 mark and there is only single option correct for every questionMark review to see the question later 
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Time has elapsed
You have reached 0 of 0 points, (0)
Average score 

Your score 

Categories
 easy 0%
 medium 0%
 tough 0%

your performance is poor!!! 🙁 🙁
Attemt more quiz to enhance your score, good luck

your performance is below average!!!!! 🙁
Attempt more quiz to enhance your score, good luck

your performance is average!! good going 🙂
Attempt more quiz to enhance your score, good luck

nice!! your score is above average 🙂 🙂
Attemt more quiz to enhance your score, good luck

your are scoring good :)) 😀
Attemp more quiz to score 100%, good luck

wow you are scoring very good :)) 😀
attempt more quiz to score 100%, good luck

voila!!!! you made it.. you are scoring 100%:)) 😀
attempt more quiz to keep the same, good luck
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 Answered
 Review

Question 1 of 10
1. Question
1 pointsA sum of money is to be distributed among A, B,C, D in the proportion of 5:2:4:3. If C gets Rs. 1000 more than D, what is B’s share?
Correct
Solution:
Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x−3x=1000
⇒x=1000
B’s share =Rs.2x=Rs.(2×1000)= Rs. 2000Incorrect
Solution:
Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x−3x=1000
⇒x=1000
B’s share =Rs.2x=Rs.(2×1000)= Rs. 2000 
Question 2 of 10
2. Question
1 pointsA sum of Rs.312 was divided among 100 boys and girls in such a way that the boy gets Rs.3.60 and each girl Rs.2.40 the number of girls is?
Correct
Solution:
Step (i) Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x+y=100
Step (ii) A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.6x+2.4y=312
Step (iii)
Solving (i) and (ii)
3.6 x+3.6 y=360 multiplying by 3.6
3.6 x + 2.4 y = 312 1.20 y = 48 y = 40 The number of girls is 40
Incorrect
Solution:
Step (i) Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x+y=100
Step (ii) A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.6x+2.4y=312
Step (iii)
Solving (i) and (ii)
3.6 x+3.6 y=360 multiplying by 3.6
3.6 x + 2.4 y = 312 1.20 y = 48 y = 40 The number of girls is 40

Question 3 of 10
3. Question
1 pointsIn a mixture 60 litres, the ratio of milk and water 2:1. If the this ratio is to be 1:2, then the quantity of water to be further added in liters is:
Correct
Solution:
Quantity of milk = 60×23=40 litres
Quantity of water in it =(60−40) litres =20 litres.
New ratio =1:2
Let, the quantity of water to be added further be x litres.
Then milk : water = 40/(20+x)
Now, 40/(20+x)=1/2
⇒20+x =80
=60
Incorrect
Solution:
Quantity of milk = 60×23=40 litres
Quantity of water in it =(60−40) litres =20 litres.
New ratio =1:2
Let, the quantity of water to be added further be x litres.
Then milk : water = 40/(20+x)
Now, 40/(20+x)=1/2
⇒20+x =80
=60

Question 4 of 10
4. Question
1 pointsA sum of Rs. 36.90 is made up of 180 coins which are either 10 paise coins or 25 p coins. The number of 10 p coins is:
Correct
Solution:
Step (i) Total number of coins = 180
Let x be number of 10p coins and y be number of 25p coins
x+y=180 ——–(i)
Step (ii) Given 10p coins and 25p coins make the sum = Rs. 36.90
10x/100+25y/100=36.9
⇒10x+25y=3690—–(ii)
Step (iii)
Solving (i) and (ii),
10x+10y=180010x+25y=3690=
15y=1890
y=126
Substitute y value in equation (i)
x=180–126=54
Number of 10p coins = 54
Incorrect
Solution:
Step (i) Total number of coins = 180
Let x be number of 10p coins and y be number of 25p coins
x+y=180 ——–(i)
Step (ii) Given 10p coins and 25p coins make the sum = Rs. 36.90
10x/100+25y/100=36.9
⇒10x+25y=3690—–(ii)
Step (iii)
Solving (i) and (ii),
10x+10y=180010x+25y=3690=
15y=1890
y=126
Substitute y value in equation (i)
x=180–126=54
Number of 10p coins = 54

Question 5 of 10
5. Question
1 pointsThe ratio of the number of boys and girls in a college is 7:8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?
Correct
Solution:
Option(C) is correctOriginally, let the number of boys and girls in the college be
7 x and8 x respectively.Their increased number is (120% of7 x ) and (110 % of8 x ).⇒⇒(120/100×7x) and (110/100×8x)
⇒ 42 x/ 5 and 44 x/ 5 So, the required ratio = (42x/5 : 44x/5)
=21:22
Incorrect
Solution:
Option(C) is correctOriginally, let the number of boys and girls in the college be
7 x and8 x respectively.Their increased number is (120% of7 x ) and (110 % of8 x ).⇒⇒(120/100×7x) and (110/100×8x)
⇒ 42 x/ 5 and 44 x/ 5 So, the required ratio = (42x/5 : 44x/5)
=21:22

Question 6 of 10
6. Question
1 pointsIf 0.75 : x :: 5 : 8, then x is equal to:
Correct
x×5 = 0.75×8
x= 1.2
Incorrect
x×5 = 0.75×8
x= 1.2

Question 7 of 10
7. Question
1 pointsIf 20 men or 24 women or 40 boys can do a job in 12 days working for 8 hours a day, how many men working with 6 women and 2 boys take to do a job four times as big working for 5 hours a day for 12 days?
Correct
Let’s try solving this Problem using ratio approach.
Amount of work done by 20 men = 24 women = 40 boys or 1 man = 1.2 woman = 2 boys.
Let us, therefore, find out the amount of men required, if only men were working on the job, to complete the new job under the new conditions and then make adjustments for the women and children working with the men.
The man hours required to complete the new job = 4 times the man hours required to complete the old job. (As the new job is 4 times as big as the old job)
Let n be the number of men required.
$n\times 5\times 12=20\times 8\times 12\times 4$= or n=128
i.e. 128 men working on the job will be able to complete the given job.
However, the problem states that 6 women and 2 boys are working on the job.
6 women $=\dfrac{6}{1.2}= 5 men and 2 boys = 1 man.
∴ The equivalent of 5+1=6 men are already working.
Thus, final number of men working,
=128−6=122 men
Incorrect
Let’s try solving this Problem using ratio approach.
Amount of work done by 20 men = 24 women = 40 boys or 1 man = 1.2 woman = 2 boys.
Let us, therefore, find out the amount of men required, if only men were working on the job, to complete the new job under the new conditions and then make adjustments for the women and children working with the men.
The man hours required to complete the new job = 4 times the man hours required to complete the old job. (As the new job is 4 times as big as the old job)
Let n be the number of men required.
$n\times 5\times 12=20\times 8\times 12\times 4$= or n=128
i.e. 128 men working on the job will be able to complete the given job.
However, the problem states that 6 women and 2 boys are working on the job.
6 women $=\dfrac{6}{1.2}= 5 men and 2 boys = 1 man.
∴ The equivalent of 5+1=6 men are already working.
Thus, final number of men working,
=128−6=122 men

Question 8 of 10
8. Question
1 pointsAn outgoing batch of students wants to gift a PA system worth Rs 4,200 to their school. If the teachers, offer to pay 50% more than the students and an external benefactor gives three times the teacher’s contribution, then how much should the teachers donate?
Correct
The ratio of the share students : teacher: benefactor=1:1.5:4.5
So the proportion to teacher’s share = 1.5/7
Hence, the teachers would donate 1.5/7×4200=Rs 900
Incorrect
The ratio of the share students : teacher: benefactor=1:1.5:4.5
So the proportion to teacher’s share = 1.5/7
Hence, the teachers would donate 1.5/7×4200=Rs 900

Question 9 of 10
9. Question
1 pointsTwo cogged wheels of which one has 32 cogs and other 54 cogs, work into each other. If the latter turns 80 times in three quarters of a minute, how often does the other turn in 8 seconds?
Correct
Less Cogs ⇒ more turns and less time ⇒ less turns
Cogs Time Turns A 54 45 80 B 32 8 ? Number of turns required= 80×54/32×8/45=24 times
Incorrect
Less Cogs ⇒ more turns and less time ⇒ less turns
Cogs Time Turns A 54 45 80 B 32 8 ? Number of turns required= 80×54/32×8/45=24 times

Question 10 of 10
10. Question
1 pointsTwo men, sitting on the dining table. One man has 7 eggs and other had 5 eggs. A third man passing by requested them to share their food in return for money. The three of them shared the eggs equally and the third traveller paid the other two a total of Rs 24. Find the difference between the amounts received by first two men?
Correct
Solution:
As the two men had a total of 12 (=7+5) eggs, also they agreed to share eggs in such a manner that all three of them got an equal number of eggs. So each of them must have got 4(=12/3) eggs.
The first and the second man must have given 3(=7−4) eggs and 1(=5−4) egg to the third man.
Hence, the ratio of their share is 3:1
Now, money has to be distributed in the ratio of their contribution. As total money paid by the stranger is Rs. 24.
Hence the first and the second man get Rs 18(=24×3/(3+1)) and Rs 6(=24×13+1)
6 ( = 24 × 1/( 3 + 1) ) respectively.The 1st man gets Rs 12(=18−6)more than the second.
Incorrect
Solution:
As the two men had a total of 12 (=7+5) eggs, also they agreed to share eggs in such a manner that all three of them got an equal number of eggs. So each of them must have got 4(=12/3) eggs.
The first and the second man must have given 3(=7−4) eggs and 1(=5−4) egg to the third man.
Hence, the ratio of their share is 3:1
Now, money has to be distributed in the ratio of their contribution. As total money paid by the stranger is Rs. 24.
Hence the first and the second man get Rs 18(=24×3/(3+1)) and Rs 6(=24×13+1)
6 ( = 24 × 1/( 3 + 1) ) respectively.The 1st man gets Rs 12(=18−6)more than the second.