44084.If 0.75 : x :: 5 : 8, then x is equal to:
1.12
1.2
1.25
1.30
Explanation:
$(x \times 5) = (0.75 \times 8) $=> x =$ \dfrac{6}{5} = 1.20$
$(x \times 5) = (0.75 \times 8) $=> x =$ \dfrac{6}{5} = 1.20$
44085.Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit s salary?
Rs. 17,000
Rs. 20,000
Rs. 25,500
Rs. 38,000
Explanation:
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then,$\dfrac{2x+4000}{3x+4000}=\dfrac{40}{57}$
=> 57 (2x + 4000) = 40 (3x+4000)
=> 6x = 68,000
=> 3x = 34000
Sumit s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then,$\dfrac{2x+4000}{3x+4000}=\dfrac{40}{57}$
=> 57 (2x + 4000) = 40 (3x+4000)
=> 6x = 68,000
=> 3x = 34000
Sumit s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.
44086.The 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?
8 : 9
17 : 18
21 : 22
Cannot be determined
Explanation:
Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
=>$ ( \dfrac{120}{100} \times 7x) and ( \dfrac{110}{100} \times 8x)$
=> $\dfrac{42x}{5} and \dfrac{44x}{5}$
The required ratio =$ (\dfrac{42x}{5} : \dfrac{44x}{5})$
Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
=>$ ( \dfrac{120}{100} \times 7x) and ( \dfrac{110}{100} \times 8x)$
=> $\dfrac{42x}{5} and \dfrac{44x}{5}$
The required ratio =$ (\dfrac{42x}{5} : \dfrac{44x}{5})$
44087.Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:
27
33
49
55
Explanation:
Let the numbers be 3x and 5x.
Then, $\dfrac{3x-9}{5x-9} = {12}{23}$
=> 23(3x - 9) = 12(5x - 9)
=> 9x = 99
=> x = 11.
ஃ The smaller number = (3 x 11) = 33.
Let the numbers be 3x and 5x.
Then, $\dfrac{3x-9}{5x-9} = {12}{23}$
=> 23(3x - 9) = 12(5x - 9)
=> 9x = 99
=> x = 11.
ஃ The smaller number = (3 x 11) = 33.
44088.In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
50
100
150
200
Explanation:
Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.
Then, sum of their values = Rs.$(\dfrac{25x}{100} + \dfrac{10 \times2x}{100} + \dfrac{5 \times3x}{100}) = Rs.\dfrac{60x}{100}$
ஃ $\dfrac{60x}{100} = 30 $<=> x = $\dfrac{30 \times100}{60} = 50.$
Hence, the number of 5 p coins = (3 x 50) = 150.
Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.
Then, sum of their values = Rs.$(\dfrac{25x}{100} + \dfrac{10 \times2x}{100} + \dfrac{5 \times3x}{100}) = Rs.\dfrac{60x}{100}$
ஃ $\dfrac{60x}{100} = 30 $<=> x = $\dfrac{30 \times100}{60} = 50.$
Hence, the number of 5 p coins = (3 x 50) = 150.
44407.In the previous question, what is the sum of the number of biscuits hidden by the last 2 men?
142
144
180
178
Explanation:
Suppose N=5x+1
A took (x+1) biscuit.
Now 4x is of the form 5y+1 then x must be in the form 5z+4
⇒4(5z+4)=5y+1
⇒y=4z+3 and x=5z+4
The ratio of number of biscuits that A and B took is
[(5z+4)+1]:[(4z+3)+1]=5:4
So, we can say that any two successive persons A, B, C and D take coins in the ratio of 5:4
Let the number of biscuits that A, B, C and D took be a, b, c and d respectively.
a:b=b:c=c:d=5:4
a:b:c:d=125:100:80:64
⇒a=125k
⇒x=125k−1 and N=5x+1=625k−4
⇒N<100, then k=1
⇒N=621
⇒621=(5×124)+3
4×124=(5×99)+1
4×99=(5×79)+1
4×79=(5×63)+1
The number of biscuits hidden by 3rd and the 4th men is 79+1=80 and 63+1=64
i.e. A total of 144.
Suppose N=5x+1
A took (x+1) biscuit.
Now 4x is of the form 5y+1 then x must be in the form 5z+4
⇒4(5z+4)=5y+1
⇒y=4z+3 and x=5z+4
The ratio of number of biscuits that A and B took is
[(5z+4)+1]:[(4z+3)+1]=5:4
So, we can say that any two successive persons A, B, C and D take coins in the ratio of 5:4
Let the number of biscuits that A, B, C and D took be a, b, c and d respectively.
a:b=b:c=c:d=5:4
a:b:c:d=125:100:80:64
⇒a=125k
⇒x=125k−1 and N=5x+1=625k−4
⇒N<100, then k=1
⇒N=621
⇒621=(5×124)+3
4×124=(5×99)+1
4×99=(5×79)+1
4×79=(5×63)+1
The number of biscuits hidden by 3rd and the 4th men is 79+1=80 and 63+1=64
i.e. A total of 144.
44408.If 15% of x = 20% of y, then x:y is ___?
4:3
5:4
4:5
3:4
Explanation:
Given 15% of x = 20% of y
=> 15x = 20y
=> x/y = 20/15
=> x : y = 4 : 3
Given 15% of x = 20% of y
=> 15x = 20y
=> x/y = 20/15
=> x : y = 4 : 3
44409.82.5% of 2360?
1960
1954
1947
1931
Explanation:
Given 82.5% of 2360
$\\dfrac{82.5}{100}$ x 2360 = 0.825 x 2360 = 1947
Given 82.5% of 2360
$\\dfrac{82.5}{100}$ x 2360 = 0.825 x 2360 = 1947
44410.In a city of engineers of 4800, the number of female engineers are 3/8th of them. The number of female engineers is what percentage of the number of male engineers?
76
60
56
52
Explanation:
Total engineers = 4800
Number of female engineers = 3/8 x 4800 = 1800
Then, number of male engineers = 4800 - 1800 = 3000
Now, required is female engineers is what percentage of the number of male engineers
=> 1800 is what % of 3000
=> p x 3000/100 = 1800
=> 30p = 1800
p = 60%
Hence, female engineers is 60 percentage of the number of male engineers.
Total engineers = 4800
Number of female engineers = 3/8 x 4800 = 1800
Then, number of male engineers = 4800 - 1800 = 3000
Now, required is female engineers is what percentage of the number of male engineers
=> 1800 is what % of 3000
=> p x 3000/100 = 1800
=> 30p = 1800
p = 60%
Hence, female engineers is 60 percentage of the number of male engineers.
44411.In a class 60% of the students pass in English and 45% pass in Hindi. If 25% of them pass in both subjects, what percentage of the students fail in both the subjects?
20%
25%
30%
15%
Explanation:
Percentage of students passed in English = 60%
Percentage of students passed in Hindi = 45%
Percentage of students passed in both subjects = 25%
ATQ,
Percentage of students passed in only English = 60 - 25 = 35%
Percentage of students passed in only Hindi = 45 - 25 = 20%
Therefore, Percentage of students failed = 100 - (35 + 25 + 20)
= 100 - 80 = 20%.
Percentage of students passed in English = 60%
Percentage of students passed in Hindi = 45%
Percentage of students passed in both subjects = 25%
ATQ,
Percentage of students passed in only English = 60 - 25 = 35%
Percentage of students passed in only Hindi = 45 - 25 = 20%
Therefore, Percentage of students failed = 100 - (35 + 25 + 20)
= 100 - 80 = 20%.