44397.Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce, making a profit of 20 percent. What was the ratio of the mixture?
1:10
1:5
2:7
3:8
Explanation:
Let 1:k be the ratio in which Joseph mixed the two types of rice.
Then a sample of (1+k) ounces of the mixture should equal 1 ounce of rice of the first type, and k ounces of rice of the second type.
The rice of the first type costs 5 cents an ounce and that of the second type costs 6 cents an ounce. Hence, it cost him:
(1 ounce 5 cents per ounce) + (k ounces 6 cents per ounce) = 5+6k
Since he sold the mixture at 7 cents per ounce, he must have sold the net 1+k ounces of the mixture at 7(1+k).
Since he earned 20% profit doing this, 7(1+k) must be 20% more than 5+6k.
Hence, we have the equation
7(1+k)=$\left(1+\dfrac{20}{100}\right)$(5+6k)
7+7k=$\left(\dfrac{120}{100}\right)$(5+6k)
7+7k=$\dfrac{6}{5}$(5+6k)
7+7k=6+$\dfrac{36k}{5}$
1=$\dfrac{k}{5}$
k=5
Hence, the required ratio is 1:k= 1 : 5
Then a sample of (1+k) ounces of the mixture should equal 1 ounce of rice of the first type, and k ounces of rice of the second type.
The rice of the first type costs 5 cents an ounce and that of the second type costs 6 cents an ounce. Hence, it cost him:
(1 ounce 5 cents per ounce) + (k ounces 6 cents per ounce) = 5+6k
Since he sold the mixture at 7 cents per ounce, he must have sold the net 1+k ounces of the mixture at 7(1+k).
Since he earned 20% profit doing this, 7(1+k) must be 20% more than 5+6k.
Hence, we have the equation
7(1+k)=$\left(1+\dfrac{20}{100}\right)$(5+6k)
7+7k=$\left(\dfrac{120}{100}\right)$(5+6k)
7+7k=$\dfrac{6}{5}$(5+6k)
7+7k=6+$\dfrac{36k}{5}$
1=$\dfrac{k}{5}$
k=5
Hence, the required ratio is 1:k= 1 : 5
44398.A and B enter in to a partnership and A invests Rs. 10,000 in the partnership. At the end of 4 months he withdraws Rs.2000. At the end of another 5 months, he withdraws another Rs.3000.
If B receives Rs.9600 as his share of the total profit of Rs.19,100 for the year, how much did B invest in the company?
If B receives Rs.9600 as his share of the total profit of Rs.19,100 for the year, how much did B invest in the company?
Rs. 12,000
Rs. 8,000
Rs. 6,000
Rs. 96,000
Explanation:
The total profit for the year is 19100. Of this B gets Rs.9600. Therefore, A would get (19100−9600) =Rs. 9500
The partners split their profits in the ratio of their investments.
Therefore, the ratio of the investments of,
A:B=9500:9600=95:96.
A invested Rs.10000 initially for a period of 4 months. Then, he withdrew Rs.2000.
Hence, his investment has reduced to Rs.8000 (for the next 5 months).
Then he withdraws another Rs.3000. Hence, his investment will stand reduced to Rs.5000 during the last three months.
So, the amount of money that he had invested in the company on a money-month basis will be,
=4×10000+5×8000+3×5000
=40000+40000+15000
=95000
If A had 95000 money months invested in the company, B would have had 96,000 money months invested in the company (as the ratio of their investments is 95:96).
If B had 96,000 money-months invested in the company, he has essentially invested $\dfrac{96000}{12}$=Rs. 8000
The partners split their profits in the ratio of their investments.
Therefore, the ratio of the investments of,
A:B=9500:9600=95:96.
A invested Rs.10000 initially for a period of 4 months. Then, he withdrew Rs.2000.
Hence, his investment has reduced to Rs.8000 (for the next 5 months).
Then he withdraws another Rs.3000. Hence, his investment will stand reduced to Rs.5000 during the last three months.
So, the amount of money that he had invested in the company on a money-month basis will be,
=4×10000+5×8000+3×5000
=40000+40000+15000
=95000
If A had 95000 money months invested in the company, B would have had 96,000 money months invested in the company (as the ratio of their investments is 95:96).
If B had 96,000 money-months invested in the company, he has essentially invested $\dfrac{96000}{12}$=Rs. 8000
44399.Two alloys A and B are composed of two basic elements. The ratios of the compositions of the two basic elements in the two alloys are 5:3 and 1:2, respectively. A new alloy X is formed by mixing the two alloys A and B in the ratio 4:3. What is the ratio of the composition of the two basic elements in alloy X?
1:1
2:3
5:2
4:3
Explanation:
The new alloy X is formed from the two alloys A and B in the ratio 4:3.
Hence, 7 parts of the alloy contains 4 parts of alloy A and 3 parts of alloy B.
Let 7x ounces of alloy X contain 4x ounces of alloy A and 3x ounces of alloy B.
Now, alloy A is formed of the two basic elements mentioned in the ratio 5:3.
Hence, 4x ounces of alloy A contains
$\left(\dfrac{5}{5+3}\right)4x=\dfrac{5x}{2}$ ounces of first basic element
and $\dfrac{3}{5+3}4x=\dfrac{3x}{2}$ ounces of the second basic element.
Also, alloy B is formed of the two basic elements mentioned in the ratio 1:2.
Hence, let the 3x ounces of alloy A contain $\left(\dfrac{1}{1+2}\right)$3x=x ounces of the second basic
element. ounces of the first basic element and $\left(\dfrac{2}{1+2}\right)$3x=2x ounces of the second basic element.
Then the total compositions of the two basic elements in the 7x ounces of alloy X would contain
$\dfrac{5x}{2}$ ounces (from A) + x ounces (from B) = $\dfrac{7x}{2}$ ounces of first basic element,
and basic elements in alloy X is $\dfrac{3x}{2}$(from A) + 2x(from B) = $\dfrac{7x}{2}$ ounces of the second basic element.
Hence, the composition of the two basic elements in alloy X
is $\dfrac{7x}{2}:\dfrac{7x}{2}$=1:1
The new alloy X is formed from the two alloys A and B in the ratio 4:3.
Hence, 7 parts of the alloy contains 4 parts of alloy A and 3 parts of alloy B.
Let 7x ounces of alloy X contain 4x ounces of alloy A and 3x ounces of alloy B.
Now, alloy A is formed of the two basic elements mentioned in the ratio 5:3.
Hence, 4x ounces of alloy A contains
$\left(\dfrac{5}{5+3}\right)4x=\dfrac{5x}{2}$ ounces of first basic element
and $\dfrac{3}{5+3}4x=\dfrac{3x}{2}$ ounces of the second basic element.
Also, alloy B is formed of the two basic elements mentioned in the ratio 1:2.
Hence, let the 3x ounces of alloy A contain $\left(\dfrac{1}{1+2}\right)$3x=x ounces of the second basic
element. ounces of the first basic element and $\left(\dfrac{2}{1+2}\right)$3x=2x ounces of the second basic element.
Then the total compositions of the two basic elements in the 7x ounces of alloy X would contain
$\dfrac{5x}{2}$ ounces (from A) + x ounces (from B) = $\dfrac{7x}{2}$ ounces of first basic element,
and basic elements in alloy X is $\dfrac{3x}{2}$(from A) + 2x(from B) = $\dfrac{7x}{2}$ ounces of the second basic element.
Hence, the composition of the two basic elements in alloy X
is $\dfrac{7x}{2}:\dfrac{7x}{2}$=1:1
44400.A bus and a truck are available to cross a jungle. The speed of the truck is thrice that of the bus. The capacity of the truck is 50 persons and that of bus is 30 persons. The average occupancy of the bus is twice that of the truck. The tickets for the bus and the truck cost Re 1 and Re 1.50 respectively. What is the ratio of the average rupee collection of the truck to that of the bus in a day? Assume there is no wastage time between trips and the occupancy of the bus/truck is defined as the ratio of the actual number of persons boarding it and its capacity.
15:4
15:2
15:8
15:7
Explanation:
Average Rupee collection = Speed× capacity × Occupancy × Ticket rate
ratio of average Rupee collection of truck to that of boat= product of above rates
⇒(3×50×1×1.5):(1×30×2×1)=15:4
Average Rupee collection = Speed× capacity × Occupancy × Ticket rate
ratio of average Rupee collection of truck to that of boat= product of above rates
⇒(3×50×1×1.5):(1×30×2×1)=15:4
44401.If N>1000, what could be the least value of N?
1249
1023
1202
1246
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
As, N>1000, the least value of N is when k=2
⇒N= 1246
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
As, N>1000, the least value of N is when k=2
⇒N= 1246
44402.Manish, Rahul and Bharti have some stones with each of the. Five times the number of stones with Rahul equals seven times the number of stones with Manish while five times the number of stones with Manish equals seven times the number of stones with Bharti. What is the minimum number of stones that can be there with all three of them put together?
113
109
93
97
Explanation:
Let the stones with Manish, rahul and Bharti be m,r and b respectively.
Given, 5r=7m and 5m=7b
⇒25r=35m and 35m=49b
⇒25r=35m=49b=k
⇒$\dfrac{r}{49}=\dfrac{m}{35}=\dfrac{b}{25}$
The least possible integral values for r,m,b will be r=49, m=35 and b=25
⇒Total=49+35+25=109
Let the stones with Manish, rahul and Bharti be m,r and b respectively.
Given, 5r=7m and 5m=7b
⇒25r=35m and 35m=49b
⇒25r=35m=49b=k
⇒$\dfrac{r}{49}=\dfrac{m}{35}=\dfrac{b}{25}$
The least possible integral values for r,m,b will be r=49, m=35 and b=25
⇒Total=49+35+25=109
44403.The ratio of marks obtained by Vinod and Basu is 6:5. If the combined average of their percentage is 68.75 and their sum of the marks is 275, find the total marks for which exam was conducted.
150
200
400
None
Explanation:
Let Vinod marks be 6x and Basu s is 5x. Therefore, the sum of the marks=6x+5x=11x
But the sum of the marks is given as 275=11x. We get x=25 therefore, Vinod’s marks is 6x=150 and Basu’s marks = 5x=125.
Therefore, the combined average of their marks % = $\dfrac{150+125}{2}$=137.5
If the total mark of the exam is 100 then their combined average of their percentage is 68.75
Therefore, if their combined average of their percentage is 137.5 then the total marks would be $\dfrac{137.5}{68.75}$×100=200
Let Vinod marks be 6x and Basu s is 5x. Therefore, the sum of the marks=6x+5x=11x
But the sum of the marks is given as 275=11x. We get x=25 therefore, Vinod’s marks is 6x=150 and Basu’s marks = 5x=125.
Therefore, the combined average of their marks % = $\dfrac{150+125}{2}$=137.5
If the total mark of the exam is 100 then their combined average of their percentage is 68.75
Therefore, if their combined average of their percentage is 137.5 then the total marks would be $\dfrac{137.5}{68.75}$×100=200
44404.A fort has provisions for 60 days. If after 15 days 500 men strengthen them and the food lasts 40 days longer, how many men are there in the fort?
3500
4000
6000
None
Explanation:
Let there be x men in the beginning so that after 15 days the food for them is left for 45 days.
After adding 500 men the food lasts for only 40 days.
Now (x+500) men can have the same food for 40 days.
Therefore by equating the amount of food we get,
45x=(x+500)$\times$ 40
5x=20,000
x=4000
Therefore there are 4,000 men in the fort.
After adding 500 men the food lasts for only 40 days.
Now (x+500) men can have the same food for 40 days.
Therefore by equating the amount of food we get,
45x=(x+500)$\times$ 40
5x=20,000
x=4000
Therefore there are 4,000 men in the fort.
44405.In a pocket of A, the ratio of Rs.1 coins, 50p coins and 25p coins can be expressed by three consecutive odd prime numbers that are in ascending order. The total value of coins in the bag is Rs 58. If the number of Rs.1, 50p, 25p coins are reversed, find the new total value of coins in the pocket of A?
Rs 68
Rs 43
Rs 75
Rs 82
Explanation:
Since the ratio of the number of Rs. 1, 50p and 25p coins can be represented by 3 consecutive odd numbers that are prime in ascending order, the only possibility for the ratio is 3:5:7.
Let the number of Re1, 50p and 25p coins be 3k,5k and 7k respectively.
Hence, total value of coins in paise
⇒100×3k+50×5k+25×7k
=725k
=5800
⇒k=8.
If the number of coins of Rs. 1,50p and 25p is reversed, the total value of coins in the Bag (in paise)
=100×7k+50×5k+25×3k=1025k (In above we find the value of k).
⇒8200p=Rs. 82.
Since the ratio of the number of Rs. 1, 50p and 25p coins can be represented by 3 consecutive odd numbers that are prime in ascending order, the only possibility for the ratio is 3:5:7.
Let the number of Re1, 50p and 25p coins be 3k,5k and 7k respectively.
Hence, total value of coins in paise
⇒100×3k+50×5k+25×7k
=725k
=5800
⇒k=8.
If the number of coins of Rs. 1,50p and 25p is reversed, the total value of coins in the Bag (in paise)
=100×7k+50×5k+25×3k=1025k (In above we find the value of k).
⇒8200p=Rs. 82.
44406.If N<1000, how many biscuits were left after the fourth man took his share?
624
621
252
257
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+3andx=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
After the fourth man took his share (5×63+1), the biscuits lefts is 4×63= 252
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+3andx=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
After the fourth man took his share (5×63+1), the biscuits lefts is 4×63= 252
- Ratio and Proportion Test 4
- Percentage Test 2
- Percentage Test 3
- Percentage Test 4
- Percentage Test 5
- Percentage Test 6
- Ratio and Proportion Test 1
- Ratio and Proportion Test 2
- Ratio and Proportion Test 3
- Percentage Test 1
- Average Test 1
- Average Test 2
- Average Test 3
- Interest Test 1
- Interest Test 2
- Interest Test 3
- Interest Test 4
- Interest Test 5
- Time and Distance Test 3
- Number system Test 2
- Partnership Test 1
- Partnership Test 2
- Profit,Loss and Discount Test 1
- Profit,Loss and Discount Test 2
- Profit,Loss and Discount Test 3
- Time and Distance Test 1
- Time and Distance Test 2
- Number System Test 1
- Time and Work Test 1
- Time and Work Test 2
- Time and Work Test 3
- Time and Work Test 4
- Time and Work Test 5
- Time and Work Test 6
- Time and Work Test 7