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The sum of 3 numbers is 285. Ratio between 2^{nd} and 3^{rd} numbers is 6 ∶ 5. Ratio between 1^{st} and 2^{nd} numbers is 3 ∶ 7. The 3^{rd} number is?
Given:
The sum of 3 numbers is 285. Ratio between 2nd and 3rd numbers is 6 ∶ 5.
Ratio between 1st and 2nd numbers is 3 ∶ 7.
Calculation:
Let the 1^{st} number be x
Let the 2^{nd} number be y
Let the 3^{rd} number be z
The sum of 3^{rd} number is 285
According to the question
Ratio of 2^{nd} and 3^{rd} number
⇒ y : z = 6 : 5
Ratio of 1^{th} and 2^{nd} number
⇒ x : y = 3 : 7
Ratio of x : y : z
⇒ (x × y) : (y × y) : (y × z) = (3 × 6) : (7 × 6) : (7 × 5)
⇒ x : y : z = 18 : 42 : 35
The 3^{rd} number z = 285 × 35 /(18 + 42 + 35)
⇒ z = 285 × 35 / 95 = 105
∴ The 3rd number is 105
The ratio of average of expenses on food, clothing and miscellaneous items to the average of expenses on savings and rent is
(108 + 36 + 72) ∶ (54 + 90)
= 216 ∶ 144
= 3 ∶ 2
If a^{2} + b^{2} + c^{2}  ab  bc  ca = 0, then a ∶ b ∶ c is:
Given :
Equation is a^{2} + b^{2} + c^{2}  ab  bc  ca = 0
Solution :
Here we have 3 variables and only one equation .
Note To solve these type of question we assume the value of any 2 variables.
Let assume that b = 1 and C = 1
⇒ a^{2 }+ 1 + 1  a  1  a = 0
⇒ a^{2} + 1  2a = 0
⇒ ( a  1 )^{2} = 0 [ ∵ ( A+B )^{2} = A^{2} + B^{2} + 2AB ]
Now a = 1
Now we have a = 1 , b = 1 and c = 1
So a : b : c = 1 : 1 : 1
Hence the correct answer is "1 : 1 : 1".
A man has 25 paise, 50 paise and 1 Rupee coins. There are 220 coins in all and the total amount is 160. If there are thrice as many 1 Rupee coins as there are 25 paise coins, then what is the number of 50 paise coins?
Given:
Total coin = 220
Total money = Rs. 160
There are thrice as many 1 Rupee coins as there are 25 paise coins.
Concept used:
Ratio method is used.
Calculation:
Let the 25 paise coins be 'x'
So, one rupees coins = 3x
50 paise coins = 220 – x – (3x) = 220 – (4x)
According to the questions,
3x + [(220 – 4x)/2] + x/4 =160
⇒ (12x + 440 – 8x + x)/4 = 160
⇒ 5x + 440 = 640
⇒ 5x = 200
⇒ x = 40
So, 50 paise coins = 220 – (4x) = 220 – (4 × 40) = 60
∴ The number of 50 paise coin is 60.
If a : b = 3 : 2, b : c = 2 : 1, c : d = 1/3 : 1/7 and d : e = 1/4 : 1/5 find a : b : c : d : e.
Given :
(i) a : b = 3 : 2,
(ii) b : c= 2 : 1,
(iii) c : d = 1/3 : 1/7,
(iv) d : e = 1/4 : 1/5
Calculations:
To solve these type of questions fill the blank by side values and then multiply all the ratios
Shortcut Trick:
a : b = 3 : 2 now check which options satisfied this ratio
1) a : b = 100 : 75 = 4 : 3 not equal to 3 : 2
2) a : b = 100 : 30 = 10 : 3 not equal to 3 : 2
3) a : b = 105 : 70 = 21 : 14 = 3 : 2 which is equal to 3 : 2
4) a : b = 105 : 35 = 21 : 7 = 3 : 1 which is not equal to 3 : 2
Hence option c is correct option
u : v = 4 : 7 and v : w = 9 : 7. If u = 72, then what is the value of w?
Given:
u : v = 4 : 7 and v : w = 9 : 7
Concept Used: In this type of question, number can be calculated by using the below formulae
Formula Used: if u ∶ v = a ∶ b, then u × b = v × a.
Calculation:
u : v = 4 : 7 and v : w = 9 : 7
To make ratio v equal in both cases
We have to multiply the 1st ratio by 9 and 2nd ratio by 7
u : v = 9 × 4 : 9 × 7 = 36 : 63 (i)
v : w = 9 × 7 : 7 × 7 = 63 : 49 (ii)
Form (i) and (ii), we can see that the ratio v is equal in both cases
So, Equating the ratios we get,
u ∶ v ∶ w = 36 ∶ 63 ∶ 49
⇒ u ∶ w = 36 ∶ 49
When u = 72,
⇒ w = 49 × 72/36 = 98
∴ Value of w is 98
A diamond weighing 35 grams cost Rs. 12,250 is cut down into two pieces having weights in the ratio of 5 ∶ 2. If the price directly proportional to the square of the weight then find the loss incurred.
Given:
New ratio is 5 ∶ 2.
The price of diamond before broken is Rs. 12250.
Concept used:
Using the concept simple ratio.
Calculation:
Let the weight of each piece of diamond be 5x and 2x.
Total weight of diamond = 5x + 2x = 7x
Cost of the diamond = (7x)^{2} = 49x^{2}
Cost of first piece = (5x)^{2} = 25x^{2}
Cost of second piece = (2x)^{2} = 4x^{2}
Total cost of diamond after weight = 25x^{2} + 4x^{2} = 29x^{2}
⇒ New cost = 29x^{2}
According to the question,
49x^{2} = 12250
⇒ x^{2} = 250
The loss in the price of diamond after broken = 49x^{2} – 29x^{2}
⇒ The required loss = 20x^{2}
⇒ 250 × 20
⇒ Rs. 5000
∴ The loss in the price of diamond after broken is Rs. 5000.
In a bag, there are coins of 5ps, 10ps, and 25ps in a ratio of 3 : 2 : 1. If there are Rs. 60 in all, how many 5ps coins are there?
Given:
5p : 10p : 25p = 3 : 2 : 1 = 3x : 2x : x
Concept:
1 Rupee = 100 paise
Calculation:
60 Rupees = 60 × 100 = 6000 paise
⇒ 5 × 3x + 10 × 2x + 25 × 1x = 6000
⇒ 15x + 20x + 25x = 6000
⇒ 60x = 6000
⇒ x = 100
∴ Number of 5 paise coins = 3x = 3 × 100 = 300
Rs.750 are divided among A, B and C in such a manner that A : B is 5 : 2 and B : C is 7 : 13. What is A’s share?
Given:
Total rupees = Rs 750
Calculation:
A : B = 5 : 2
B : C = 7 : 13
A : B : C = 5 × 7 : 2 × 7 : 2 × 13 = 35 : 14 : 26
Total Sum = 750
⇒ 35 x + 14x + 26x = 750
⇒ x = 10
So, A's share = 35 × 10 = Rs 350
∴ The required answer is Rs 350
In a family, the age of father, mother, son, and grandson are A, B, C, and D respectively. Given that A  B = 3, B + C = 78, C + D = 33 and the average age of the family is 34 years, then (B  C) is:
Given:
A  B = 3, B + C = 78, C + D = 33 and the average age of the family is 34 years
Formula used:
Total age = Average age × total number of peoples
Calculation:
Total age of the family = 4 × 34 = 136 years
A + B + C + D = 136
⇒ A + B = 136 – 33 (∵ C + D = 33)
A + B = 103 years  (i)
A – B = 3 years (Given)  (ii)
From (i) and (ii)
2A = 106 years
∴ A = 53 years
B = 103 – 53 = 50 years ( from i )
Also C + B = 78 (Given)
⇒ C = 78 – 50 = 28 years
∴ B – C = 50 – 28 = 22 years
The monthly salaries of A and B are in the ratio 5 : 6. If both of them get a salary increment of Rs. 2000, the new ratio becomes 11 : 13. What is the new monthly salary of A?
Given:
Ratio of salaries of A and B = 5 : 6
New ratio = 11 : 13
Salary increment of each = Rs. 2000
Calculations:
Let the unit of ratio be x
A’s salary = 5x
B’s salary = 6x
After increment,
⇒ (5x + 2000)/(6x + 2000) = 11/13
⇒ 13(5x + 2000) = 11(6x + 2000)
⇒ 65x + 26000 = 66x + 22000
⇒ x = Rs. 4000
⇒ A’s new month salary = 5x + 2000 = 5 × 4000 + 2000
⇒ A’s new month salary = Rs. 22000
∴ The new monthly salary of A is Rs. 22000.
Rahul has a bag which contains Rs. 1, 50 paisa, and 25 paisa coins and the ratio of number of coins is 1 ∶ 1/2 ∶ 1/3. If Rahul has a total amount of Rs 1120, then find the total value of 25 paisa coins.
Given:
Total amount = Rs 1120
Ratio of coins of Rs 1, 50 paisa, and 25 paisa = 1 ∶ 1/2 ∶ 1/3
Formula Used:
Value of coins = Number of coins × per coin value
Calculation:
We know that
Rs 1 = 100 paisa
Ratio of value per coin = 100 ∶ 50 ∶ 25
⇒ 4 ∶ 2 ∶ 1
Ratio of number of coins = 1 ∶ 1/2 ∶ 1/3
⇒ 6 ∶ 3 ∶ 2
Total value of coins = 24 ∶ 6 ∶ 2
⇒ 12 ∶ 3 ∶ 1 = 16 units
⇒ 16 units = 1120
⇒ 1 units = 70
Value of 25 paisa coins = 1 units
⇒ 1 × 70
⇒ Rs. 70
∴ The total number of 25 paisa coins is Rs. 70.
A man divided an amount between his sons in the ratio of their ages. The sons received Rs. 54000 and Rs. 48000. If one son is 5 years older than the other, find the age of the younger son.
Let the son’s ages be ‘x’ years and ‘y’ years
Given, x  y = 5
⇒ x = y + 5
Now,
Ratio of their shares = ratio of their ages
⇒ 54000 ∶ 48000 = x ∶ y
⇒ 9 ∶ 8 = (y + 5) ∶ y
⇒ 9y = 8y + 40
⇒ y = 40
∴ The younger son’s age is 40 years
The number of students in 3 classes is in the ratio 2 : 3 : 4. If 12 students are increased in each class this ratio changes to 8 : 11 : 14. The total number of students in the three classes in the beginning was
Let the number of students in the classes be 2x, 3x and 4x respectively;
Total students = 2x + 3x + 4x = 9x
According to the question,
or, 24x + 96 = 22x + 132
or, 2x = 132 − 96 or,
x = 36/2 = 18
Hence,Original number of students,
9x = 9 × 18
= 162
A box has 210 coins of denominations onerupee and fifty paise only. The ratio of their respective values is 13 : 11. The number of onerupee coin is
Respective ratio of the Number of coins;
= 13 : 11 × 2 = 13 : 22
Hence, Number of 1 rupee coins;
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