55822.The cost of diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant got Rs.70,000 less. Find the original price of the diamond.
Rs.1.4 lakh
Rs.2 lakh
Rs.1 lakh
Rs.2.5 lakh
Explanation:
Let the original weight of the diamond be 10x. Hence, its original price will be k(100x2 ), where k is a constant. The weights of the pieces after breaking are x, 2x, 3x and 4x. Therefore, their prices will be kx2, 4kx2 , 9kx2 and 16kx2 . So the total price of the pieces = (1 + 4 + 9 + 16) kx2 = 30kx2 .
Hence, the difference in the price of the original diamond and its pieces = 100kx 2 – 30kx 2 = 70kx 2 = 70000. Hence, kx2 = 000 and the original price = 100 kx2= 100 × 1000 = 100000 = Rs.1 lakh.
Let the original weight of the diamond be 10x. Hence, its original price will be k(100x2 ), where k is a constant. The weights of the pieces after breaking are x, 2x, 3x and 4x. Therefore, their prices will be kx2, 4kx2 , 9kx2 and 16kx2 . So the total price of the pieces = (1 + 4 + 9 + 16) kx2 = 30kx2 .
Hence, the difference in the price of the original diamond and its pieces = 100kx 2 – 30kx 2 = 70kx 2 = 70000. Hence, kx2 = 000 and the original price = 100 kx2= 100 × 1000 = 100000 = Rs.1 lakh.
55823.A student gets an aggregate of 60% marks in five subject in the ratio 10 : 9 : 8 : 7 : 6. If the passing marks are 50% of the maximum marks and each subject has the same maximum marks, in how many subjects did he pass the examination?
2
3
4
5
Explanation:
Let his marks be 10, 9, 8, 7 and 6 in the five subjects. Hence, totally he has scored 40 marks. This constitutes only 60% of the total marks. Hence, total marks 40/0.6 = 66.7 or 67 approx. , which is the maximum marks in all 5 subjects.
Since the total marks in each subject is the same, hence maximum marks in each subject will be 67/5 = 13 approx. Out of this 50% is the passing marks . In other words to pass in a subject, he needs to score 6.5 marks. We can see that only in 1 subject, he scored less than this viz. 6. Hence, he passed in 4 subject.
Let his marks be 10, 9, 8, 7 and 6 in the five subjects. Hence, totally he has scored 40 marks. This constitutes only 60% of the total marks. Hence, total marks 40/0.6 = 66.7 or 67 approx. , which is the maximum marks in all 5 subjects.
Since the total marks in each subject is the same, hence maximum marks in each subject will be 67/5 = 13 approx. Out of this 50% is the passing marks . In other words to pass in a subject, he needs to score 6.5 marks. We can see that only in 1 subject, he scored less than this viz. 6. Hence, he passed in 4 subject.
55824.The number of candidates writing three different entrance exams is in the ratio 4:5:6. There is a proposal to increase these numbers of candidates by 40%, 60% and 85% respectively. What will be the ratio of increased numbers?
14:15:16
12:15:19
13:19:21
56 : 80 : 111
Explanation:
Given ratio of number of candidates is 4:5:6
Let the number of candidates for 3 exams be 4k, 5k and 6k respectively.
After increasing, number of candidates become (140% of 4k), (160% of 5k) & (185% of 6k)
That is, (140x4k)/100, (160x5k)/100 and (185x6k)/100
= 56k/10, 80k/10 and 111k/10
Now, the required new ratio is: 56k/100 : 80k/10 : 111k/10
= 56 : 80 : 111
Given ratio of number of candidates is 4:5:6
Let the number of candidates for 3 exams be 4k, 5k and 6k respectively.
After increasing, number of candidates become (140% of 4k), (160% of 5k) & (185% of 6k)
That is, (140x4k)/100, (160x5k)/100 and (185x6k)/100
= 56k/10, 80k/10 and 111k/10
Now, the required new ratio is: 56k/100 : 80k/10 : 111k/10
= 56 : 80 : 111
55825.The salary of two friends Ramu and Raju are in the ratio 4:5. If the salary of each one increases by Rs.6000, then the new ratio becomes 48:55. What is Raju s present salary?
Rs.10500
Rs.10500
Rs.11500
Rs.12500
Explanation:
Ratio their salary is 4:5
Let the original salary of Ramu and Raju be 4k and 5k respectively.
After increasing Rs.6000, the ratio becomes 48:55
That is,
(4k+6000)/(5k+6000) = 48/55
55(4k+6000) = 48(5k+6000)
220k+330000 = 240k+288000
20k= 42000
We have to find the original salary of Raju; that is, 5k.
If 20k = 42000 then 5k = 10500.
Hence the required answer is Rs.10500
Ratio their salary is 4:5
Let the original salary of Ramu and Raju be 4k and 5k respectively.
After increasing Rs.6000, the ratio becomes 48:55
That is,
(4k+6000)/(5k+6000) = 48/55
55(4k+6000) = 48(5k+6000)
220k+330000 = 240k+288000
20k= 42000
We have to find the original salary of Raju; that is, 5k.
If 20k = 42000 then 5k = 10500.
Hence the required answer is Rs.10500
55826.The ratio of salary of two persons X and Y is 5:8. If the salary of X increases by 60% and that of Y decreases by 35% then the new ratio of their salaries become 40:27. What is X s salary?
Rs.15000
Rs.12000
Rs.19500
data inadequate.
Explanation:
Ratio of salary of X and Y is 5:8
Let the original salary of X and Y be Rs.5k and Rs.8k respectively.
After increasing 60%, new salary of X = 160% of 5k = 160x5k/100 = 80k/10 ...(1)
After decreasing 35%, new salary of Y = (100-35)% of 8k = 65% of 8k = 52k/10 ...(2)
Given that, new ratio is 40:27
That is, 80k/10 : 52k/10 = 40/27
This does not give the value of k; so that we cannot find X s exact salary.
Hence the answer is data inadequate.
Ratio of salary of X and Y is 5:8
Let the original salary of X and Y be Rs.5k and Rs.8k respectively.
After increasing 60%, new salary of X = 160% of 5k = 160x5k/100 = 80k/10 ...(1)
After decreasing 35%, new salary of Y = (100-35)% of 8k = 65% of 8k = 52k/10 ...(2)
Given that, new ratio is 40:27
That is, 80k/10 : 52k/10 = 40/27
This does not give the value of k; so that we cannot find X s exact salary.
Hence the answer is data inadequate.
55827.Out of three positive numbers, the ratio of the first and the second numbers is 3 : 4 that of the second and the third numbers is 5 : 6 if the product of the second and the third numbers is 4320. What is the sum of three numbers?
C)185
C)185
177
165
160
Explanation:
Ratio of 1st and 2nd numbers = 3 : 4
Ratio of 2nd and 3rd numbers = 5 : 6
Let the 2nd number = 5x, third number = 6x
Product of 2nd and 3rd numbers = 4320
5x × 6x = 4320
x2 = 144
x = 12
2nd number = 60, 3rd number = 72,
1st number = 60 /4 × 3 = 45
Sum of three numbers = 60 + 72 + 45 = 177
Ratio of 1st and 2nd numbers = 3 : 4
Ratio of 2nd and 3rd numbers = 5 : 6
Let the 2nd number = 5x, third number = 6x
Product of 2nd and 3rd numbers = 4320
5x × 6x = 4320
x2 = 144
x = 12
2nd number = 60, 3rd number = 72,
1st number = 60 /4 × 3 = 45
Sum of three numbers = 60 + 72 + 45 = 177
55828.If P/Q = 5, then find the value of $\dfrac{P^{2}+Q^{2}}{P^{2}−Q^{2}}$.
22/21
26/24
65/63
50/47
Explanation:
Given that P/Q = 5 ⇒$\dfrac{P^{2}+Q^{2}}{P^{2}−Q^{2}}$=$\dfrac{Q^2[(P^{2}/Q^{2})+1]} {Q^2[(P^{2}/Q^{2})−1]}$
(∵ By taking Q2 common in both numerator and denominator)
=$\dfrac{[(P^{2}/Q^{2})+1]}{[(P^{2}/Q^{2})−1]}$
(∵ By cancelling Q2 term in both numerator and denominator)
=$\dfrac{[(P/Q)^{2}+1]} {[(P/Q)^{2}−1]}$
=$\dfrac{[(5)^2+1]}{[(5)^2−1]}$
(∵ P/Q = 5)
=$\dfrac{[(25)+1]}{[(25)−1]}$=26/24
Given that P/Q = 5 ⇒$\dfrac{P^{2}+Q^{2}}{P^{2}−Q^{2}}$=$\dfrac{Q^2[(P^{2}/Q^{2})+1]} {Q^2[(P^{2}/Q^{2})−1]}$
(∵ By taking Q2 common in both numerator and denominator)
=$\dfrac{[(P^{2}/Q^{2})+1]}{[(P^{2}/Q^{2})−1]}$
(∵ By cancelling Q2 term in both numerator and denominator)
=$\dfrac{[(P/Q)^{2}+1]} {[(P/Q)^{2}−1]}$
=$\dfrac{[(5)^2+1]}{[(5)^2−1]}$
(∵ P/Q = 5)
=$\dfrac{[(25)+1]}{[(25)−1]}$=26/24
55829.In the income statement of Asha and Ravenna, the ratio of their income in the year 2017 was 5 : 4. The ratio of Asha’s income in the year 2018 to that in 2017 is 3 : 5 and the ratio of Ravenna’s income in the year 2018 to that in 2017 is 3 : 2. If Rs. 10242 is the sum of the income of Asha and Ravenna in the year 2018, then find the income of Ravenna in the year 2017?
Rs. 1138
Rs. 2776
Rs. 2420
Rs. 4552
Explanation:
Let the income of Asha in 2018 and 2017 be 3x and 5x respectively.
Let the income of Ravenna in 2018 and 2017 be 3y and 2y respectively
Since, the ratio of their income in the year 2017 was 5 : 4
5x : 2y = 5 : 4
2x = y
The sum of their incomes in 2018 is Rs. 10242
3x + 3y = 10, 242
9x = 10, 242
x = 1,138 and y = 2276
Ravenna’s income for the year 2017 = 2y = Rs. 4552
Let the income of Asha in 2018 and 2017 be 3x and 5x respectively.
Let the income of Ravenna in 2018 and 2017 be 3y and 2y respectively
Since, the ratio of their income in the year 2017 was 5 : 4
5x : 2y = 5 : 4
2x = y
The sum of their incomes in 2018 is Rs. 10242
3x + 3y = 10, 242
9x = 10, 242
x = 1,138 and y = 2276
Ravenna’s income for the year 2017 = 2y = Rs. 4552
55830.If 2/3 of A=30% of B=0.5 of C. Then A : B : C is ?
9 : 20 : 12
9 : 20 : 14
2 : 10 : 5
12 : 25 : 15
Explanation:
⇒2A/3=3B/10=C/2=k (k is a constant)
⇒A=3K/2,B=10K/3,C=2K
⇒ A : B : C = 3K/2:10K/3:2K
⇒ A : B : C = 3K/2:10K/3:2K
⇒ A : B : C = 9 : 20 : 12
⇒2A/3=3B/10=C/2=k (k is a constant)
⇒A=3K/2,B=10K/3,C=2K
⇒ A : B : C = 3K/2:10K/3:2K
⇒ A : B : C = 3K/2:10K/3:2K
⇒ A : B : C = 9 : 20 : 12
55831.The average score in an examination taken by 52 students of a class is 85. If the scores of the best 5 performers are not considered, the average score of the remaining students falls by 2. If, none of the first five highest scorers is not below 80 and if each of the 5 top scorers had distinct integral scores, find the maximum possible score of the topper.
108
109
177
193
Explanation:
Let the score of the topper be T.
Total score of the 52 students = 52 × 85 = 4420
Total score of the remaining 47 students after scores of the best five performers are removed = 47 × 83 = 3901
Total score of the top five students = 4420 – 3901 = 519
T + (total score of the next 4 top scores) = 519
T is the maximum when the total score of the next 4 top scorers is minimum.
Total score of the next 4 top scorers has a minimum value of 80 + 81 + 82 + 83 = 326 (since all the top 5 scores are distinct) and the least is 80.
T has a maximum value of 519 – 326 = 193
Let the score of the topper be T.
Total score of the 52 students = 52 × 85 = 4420
Total score of the remaining 47 students after scores of the best five performers are removed = 47 × 83 = 3901
Total score of the top five students = 4420 – 3901 = 519
T + (total score of the next 4 top scores) = 519
T is the maximum when the total score of the next 4 top scorers is minimum.
Total score of the next 4 top scorers has a minimum value of 80 + 81 + 82 + 83 = 326 (since all the top 5 scores are distinct) and the least is 80.
T has a maximum value of 519 – 326 = 193