Aptitude and Reasoning (Math)
Q. Find perimeter of a square, if its diagonal is 10 √ 2 cm.
A. 5 cm
B. 10 cm
C. 20 cm
Answer: (B) 10 cm
Explanation:
The diagonal along with two sides of the square will form a right angled triangle.
Those two sides will be considered as base and the height of the triangle
Now,
Base = Height = x ( Since, These are sides of square)
Now , applying Pythagoras theorem
Base2+Height2= Diagonal2
x2+x2=(10√2)2
2x2= 100x2
2x2= 200
x2= 200/2
x2= 100
x2= 10x10
x= 10 cm
A. 5 cm
B. 10 cm
C. 20 cm
Answer: (B) 10 cm
Explanation:
The diagonal along with two sides of the square will form a right angled triangle.
Those two sides will be considered as base and the height of the triangle
Now,
Base = Height = x ( Since, These are sides of square)
Now , applying Pythagoras theorem
Base2+Height2= Diagonal2
x2+x2=(10√2)2
2x2= 100x2
2x2= 200
x2= 200/2
x2= 100
x2= 10x10
x= 10 cm
Q. A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded to form a solid sphere. The radius of the sphere is:
A. 21 cm
B. 22 cm
C. 23 cm
Answer: (A) 21 cm
Explanation:
In this condition,
Volume of cone (LxBxH) is equal to the volume of Sphere(4/3 π r3) as it is melted and given an another shape.
Now,
4/3 π r3 = LxBxH
r3 = 3/4 x 7/22 x LxBxH
r3 = 3/4 x 7/22 x 49x33x24
r3 = 3x7x49x3x3
r3 = 21x21x21
r = 21 cm
A. 21 cm
B. 22 cm
C. 23 cm
Answer: (A) 21 cm
Explanation:
In this condition,
Volume of cone (LxBxH) is equal to the volume of Sphere(4/3 π r3) as it is melted and given an another shape.
Now,
4/3 π r3 = LxBxH
r3 = 3/4 x 7/22 x LxBxH
r3 = 3/4 x 7/22 x 49x33x24
r3 = 3x7x49x3x3
r3 = 21x21x21
r = 21 cm
Q. Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse.
A. 25 cm
B. 15 cm
C. 30 cm
Answer: (C) 30 cm
Explanation:
The given triangle is a right angled triangle
Base= 18 cm
Height= 24 cm
To find, length of Hypotenuse we will apply Pythagoras theorem,
Hypotenuse2= Base2+Height2
Hypotenuse2= 182+242
Hypotenuse2= 18x18+24x24
Hypotenuse2= 900
Hypotenuse2= 30x30
Hypotenuse= 30 cm
A. 25 cm
B. 15 cm
C. 30 cm
Answer: (C) 30 cm
Explanation:
The given triangle is a right angled triangle
Base= 18 cm
Height= 24 cm
To find, length of Hypotenuse we will apply Pythagoras theorem,
Hypotenuse2= Base2+Height2
Hypotenuse2= 182+242
Hypotenuse2= 18x18+24x24
Hypotenuse2= 900
Hypotenuse2= 30x30
Hypotenuse= 30 cm
Q. The decimal expansion of 22/7 is
1. Terminating
2. Non-terminating and repeating
3. Non-terminating and Non-repeating
Answer: (2) Non-terminating and repeating
Explanation:
22/7 = 3.14285714286..
Therefore, correct answer is Non-terminating and repeating
1. Terminating
2. Non-terminating and repeating
3. Non-terminating and Non-repeating
Answer: (2) Non-terminating and repeating
Explanation:
22/7 = 3.14285714286..
Therefore, correct answer is Non-terminating and repeating
Q. For some integer n, the odd integer is represented in the form of:
1. n
2. n + 1
3. 2n + 1
Answer: (3) 2n + 1
Explanation:
For any integer n, e.g : 1, 2, 3, 4, 5.....
2n is always even number.
so, 2n + 1 will always give an odd number.
Therefore, correct answer is 2n + 1.
1. n
2. n + 1
3. 2n + 1
Answer: (3) 2n + 1
Explanation:
For any integer n, e.g : 1, 2, 3, 4, 5.....
2n is always even number.
so, 2n + 1 will always give an odd number.
Therefore, correct answer is 2n + 1.
Q. Which of the following is not irrational?
A. (3 + √7)
B. (3 – √7)
C . (3 + √7) (3 – √7)
Answer: (C) (3 + √7) (3 – √7)
Explanation:
(3 + √7) (3 – √7)
= 32 – (√7)2
= 9 – 7 = 2 (which is rational)
[By a2 – b2 = (a – b) (a + b)]
Therefore, correct answer is (3 + √7) (3 – √7)
A. (3 + √7)
B. (3 – √7)
C . (3 + √7) (3 – √7)
Answer: (C) (3 + √7) (3 – √7)
Explanation:
(3 + √7) (3 – √7)
= 32 – (√7)2
= 9 – 7 = 2 (which is rational)
[By a2 – b2 = (a – b) (a + b)]
Therefore, correct answer is (3 + √7) (3 – √7)
Q. The addition of a rational number and an irrational number is equal to:
A. rational number
B. Irrational number
C . Both
Answer: (B) Irrational number
Explanation:
Always Irrational number.
Therefore, correct answer is Irrational number.
A. rational number
B. Irrational number
C . Both
Answer: (B) Irrational number
Explanation:
Always Irrational number.
Therefore, correct answer is Irrational number.
Q. A number x is represented as p/q where p and q are integer, q is not 0 (zero) then a is?
A. rational number
B. Irrational number
C . Whole number
Answer: (B) rational number
Explanation:
Always rational number, e.g: 2/3, 0/5 are rational numbers
Therefore, correct answer is rational number.
A. rational number
B. Irrational number
C . Whole number
Answer: (B) rational number
Explanation:
Always rational number, e.g: 2/3, 0/5 are rational numbers
Therefore, correct answer is rational number.
Q. The average of 5 consecutive odd numbers a, b, c, d and e is 45. What is the product of b and d ?
A. 2020
B. 2021
C. 1021
Answer: (B) 2021
Explanation:
Let the first odd number be n,
∴ a= n
b= n + 2
c= n + 4
d= n + 6
e= n + 8
A/c,
n + 4 = 45
n = 45 - 4
n = 41
∴ a = 41
b= 41 + 2= 43
c= 41+ 4= 45
d= 41 + 6= 47
e= 41 + 8= 49
∴ b x d = 43 x 47=2021
A. 2020
B. 2021
C. 1021
Answer: (B) 2021
Explanation:
Let the first odd number be n,
∴ a= n
b= n + 2
c= n + 4
d= n + 6
e= n + 8
A/c,
| n + (n +2) + (n + 4) + (n + 6) + (n + 8) | = 45 |
| 5 |
| n + n +2 + n + 4 + n + 6 + n + 8 | = 45 |
| 5 |
| n + n +2 + n + 4 + n + 6 + n + 8 | = 45 |
| 5 |
| 5n + 20 | = 45 |
| 5 |
| 5 (n + 4) | = 45 |
| 5 |
n + 4 = 45
n = 45 - 4
n = 41
∴ a = 41
b= 41 + 2= 43
c= 41+ 4= 45
d= 41 + 6= 47
e= 41 + 8= 49
∴ b x d = 43 x 47=2021
Tap A can fill a tank in 6 hours and tap B can empty the same tank in 10 hours. If both taps are opened together, then how much time will be taken to fill the tank?
A. 15 hours
B. 20 hours
C. 25 hours
Answer: ( A) 15 hours
Explanation:
Tap A can fill in 6 hours = 1 tank
Again,
Tap B can empty in 10 hours = 1 tank
Now,
In 1 hour tank will filled by A and B
Now,
1 tank, fills by A and B = 1x 15 hours
= 15 hours
A. 15 hours
B. 20 hours
C. 25 hours
Answer: ( A) 15 hours
Explanation:
Tap A can fill in 6 hours = 1 tank
| Tap A can fill in 1 hours = | 1 | part |
| 6 |
Again,
Tap B can empty in 10 hours = 1 tank
| Tap B can empty in 1 hours = | 1 | part |
| 10 |
Now,
In 1 hour tank will filled by A and B
| = | 1 | - | 1 | part |
| 6 | 10 |
| = | 1x5 - 1x3 | part |
| 30 |
| = | 2 | part |
| 30 |
| = | 1 | part |
| 15 |
Now,
| 1 | part fills by A and B = 1 hours |
| 15 |
1 tank, fills by A and B = 1x 15 hours
= 15 hours
What will come at the place ? In the given series?
BZA, DYC, EXE, ? , JVI.
(A) HAG
(B) HGJ
(C) HWG
(D) HYG
Answer: C
BZA, DYC, EXE, ? , JVI.
(A) HAG
(B) HGJ
(C) HWG
(D) HYG
Answer: C
If the total ages of Iqbal and Shikhar is 12 years more than the total age of Shikhar and Charu. Charu is how many years younger than Iqbal?
A. 11 years
B. 12 years
C. 15 years
D. None of the above
Answer: D. 12 years
Solution:
Let the age of Iqbal be x
Let the age of Shikhar be y
Let the age of Charu be z
Then, according to question,
(x+y) – (y+z) = 12
⇒x+y-y-z = 12
⇒x-z = 12
Thus, Charu is 12 years younger than Iqbal
A. 11 years
B. 12 years
C. 15 years
D. None of the above
Answer: D. 12 years
Solution:
Let the age of Iqbal be x
Let the age of Shikhar be y
Let the age of Charu be z
Then, according to question,
(x+y) – (y+z) = 12
⇒x+y-y-z = 12
⇒x-z = 12
Thus, Charu is 12 years younger than Iqbal
The present age of Aradhana and Aadrika is in the ratio 3:4. 5 years back, the ratio of their ages was 2:3. What is the present age of Aradhana?
1. 12 years
2. 15 years
3. 20 years
Answer: (2) 15 years
Solution:
Let the present age of Aradhana be 3x
Let the present age of Aadrika be 4x
5 years back, Aradhana’s age = (3x-5) years
5 years back, Aadrika’s age = (4x-5)
According to the question, (3x-5) : (4x-5) = 2:3
⇒(3x-5) ÷ (4x-5) = 2/3
⇒3(3x-5) = 2(4x-5)
⇒9x-15 = 8x-10
⇒x = 5
Therefore, Aradhana’s current age = 3×5 = 15 years
1. 12 years
2. 15 years
3. 20 years
Answer: (2) 15 years
Solution:
Let the present age of Aradhana be 3x
Let the present age of Aadrika be 4x
5 years back, Aradhana’s age = (3x-5) years
5 years back, Aadrika’s age = (4x-5)
According to the question, (3x-5) : (4x-5) = 2:3
⇒(3x-5) ÷ (4x-5) = 2/3
⇒3(3x-5) = 2(4x-5)
⇒9x-15 = 8x-10
⇒x = 5
Therefore, Aradhana’s current age = 3×5 = 15 years
A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded to form a solid sphere. The radius of the sphere is:
A. 21 cm
B. 22 cm
C. 23 cm
Answer: (A) 21 cm
Explanation:
In this condition,
Volume of cone (LxBxH) is equal to the volume of Sphere(4/3 π r3) as it is melted and given an another shape.
Now,
4/3 π r3 = LxBxH
r3 = 3/4 x 7/22 x LxBxH
r3 = 3/4 x 7/22 x 49x33x24
r3 = 3x7x49x3x3
r3 = 21x21x21
r = 21 cm
A. 21 cm
B. 22 cm
C. 23 cm
Answer: (A) 21 cm
Explanation:
In this condition,
Volume of cone (LxBxH) is equal to the volume of Sphere(4/3 π r3) as it is melted and given an another shape.
Now,
4/3 π r3 = LxBxH
r3 = 3/4 x 7/22 x LxBxH
r3 = 3/4 x 7/22 x 49x33x24
r3 = 3x7x49x3x3
r3 = 21x21x21
r = 21 cm
Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse.
A. 25 cm
B. 15 cm
C. 30 cm
Answer: (C) 30 cm
Explanation:
The given triangle is a right angled triangle
Base= 18 cm
Height= 24 cm
To find, length of Hypotenuse we will apply Pythagoras theorem,
Hypotenuse2= Base2+Height2
Hypotenuse2= 182+242
Hypotenuse2= 18x18+24x24
Hypotenuse2= 900
Hypotenuse2= 30x30
Hypotenuse= 30 cm
A. 25 cm
B. 15 cm
C. 30 cm
Answer: (C) 30 cm
Explanation:
The given triangle is a right angled triangle
Base= 18 cm
Height= 24 cm
To find, length of Hypotenuse we will apply Pythagoras theorem,
Hypotenuse2= Base2+Height2
Hypotenuse2= 182+242
Hypotenuse2= 18x18+24x24
Hypotenuse2= 900
Hypotenuse2= 30x30
Hypotenuse= 30 cm
The ratio of speed of boat in still water to speed of stream is 8 : 1. It takes 4 hours by boat to cover 54 km in downstream & 42 km in upstream. Find the downstream speed of boat:
A. 25 km/h
B. 26 km/h
C. 27 km/h
D. 17km/h
Answer: C. 27 km/h
Explanation:
Ratio from speed of boat in still water to speed of stream = 8 : 1
Let
The speed of boat in still water be 8x km/h
Speed of stream be x km/h
We know that,
Time= Distance/Speed
Time required for downstream
= 54/(8x+x)
= (6/x) hrs
Time required for upstream= 42/(8x-x)
= (6/x) hrs
A/Q
4 = (6/x) + (6/x)
Since, Total required time is 4 hrs
x = 3
Speed of the boat in downstream
= (8x+x)=9×3= 27 km/h
A. 25 km/h
B. 26 km/h
C. 27 km/h
D. 17km/h
Answer: C. 27 km/h
Explanation:
Ratio from speed of boat in still water to speed of stream = 8 : 1
Let
The speed of boat in still water be 8x km/h
Speed of stream be x km/h
We know that,
Time= Distance/Speed
Time required for downstream
= 54/(8x+x)
= (6/x) hrs
Time required for upstream= 42/(8x-x)
= (6/x) hrs
A/Q
4 = (6/x) + (6/x)
Since, Total required time is 4 hrs
x = 3
Speed of the boat in downstream
= (8x+x)=9×3= 27 km/h
Manoj gave 60% of his salary to his wife and invested rest amount in mutual funds. His wife spends 30% amount on grocery and 20% on rent. From remaining amount, she purchased gold worth Rs. 18000. Find salary of Manoj :
A. Rs. 50,0000
B. Rs. 60,000
C. Rs. 70,000
D. Rs. 75000
Answer: B. Rs. 60,000
Explanation:
Let the salary be Rs. X
Amount Manoj gives his wife= 60% of x
= Rs. (3x/5)
Again,
From her 100% amount , she spends for gold= (100% -30%-20%)
Which is 50%
A/Q
50% of her money = Rs 18000
(50/100) × (3x/5) = 18000
x = Rs. 60,000
A. Rs. 50,0000
B. Rs. 60,000
C. Rs. 70,000
D. Rs. 75000
Answer: B. Rs. 60,000
Explanation:
Let the salary be Rs. X
Amount Manoj gives his wife= 60% of x
= Rs. (3x/5)
Again,
From her 100% amount , she spends for gold= (100% -30%-20%)
Which is 50%
A/Q
50% of her money = Rs 18000
(50/100) × (3x/5) = 18000
x = Rs. 60,000
A can do a work in 36 days while B can do the same in 48 days. If A work for ‘x’ days while B work for ‘x+2’ days then one-third of the work is complete. Find the value of x :
A. 8
B. 6
C. 7
D. 10
Answer: B. 6
Explanation:
In 36 days A completes = 1 work
In one day A can complete= (1/36) part of the work
In x days A can complete = (x/36) part of the work
Again,
in (x+2) days B completes = (x+2/48) parts
A/Q
(x/36) + (x+2/48) = 1/3 part
(4x+3x+6/144) = 1/3 part
Therefore,
x = 6
A. 8
B. 6
C. 7
D. 10
Answer: B. 6
Explanation:
In 36 days A completes = 1 work
In one day A can complete= (1/36) part of the work
In x days A can complete = (x/36) part of the work
Again,
in (x+2) days B completes = (x+2/48) parts
A/Q
(x/36) + (x+2/48) = 1/3 part
(4x+3x+6/144) = 1/3 part
Therefore,
x = 6
A shopkeeper marked the price of an article by 40% above cost price and gave a discount of Rs. 224. On the final amount, he charged 10% tax. In the whole transaction, he earned Rs. 158.6. Find cost price of the article :
A. Rs 750
B. Rs 760
C. Rs 770
D. Rs 780
Answer: A. Rs 750
Explanation:
Let the CP be x
He made 40% profit
So, SP= (140/100) × x
= Rs 1.4x
Discount Rs 224
So,
SP after discount = Rs(1.4x - 224)
But ,
Tax on SP is 10%
Final SP = (110/100) × (1.4x-224)
= Rs 1.1(1.4x-224)
A/Q,
1.1(4x-224) - x = 158.6
Therefore,
CP(x) = Rs 750
A. Rs 750
B. Rs 760
C. Rs 770
D. Rs 780
Answer: A. Rs 750
Explanation:
Let the CP be x
He made 40% profit
So, SP= (140/100) × x
= Rs 1.4x
Discount Rs 224
So,
SP after discount = Rs(1.4x - 224)
But ,
Tax on SP is 10%
Final SP = (110/100) × (1.4x-224)
= Rs 1.1(1.4x-224)
A/Q,
1.1(4x-224) - x = 158.6
Therefore,
CP(x) = Rs 750
X liters of milk is taken out and replaced with water from a container having 240 liters milk. Now, 20% of the mixture is taken out and replaced with water. In final mixture, the difference in quantity of milk & water is 128 liters. Find X :
A. 12
B. 10
C. 9
D. 11
Answer: B. 10
Explanation:
Initial Quantity of milk in the container = 240 L
Quantity of milk when x L is taken out = (240-x) L
Quantity of water added= x L
Again,
when 20% mixture is taken out,
So, Quantity of milk taken out = 20% of (240-x)
= (240-x)/5 L
So, Quality of water added there after = (240-x)/5 L
Quantity of milk left in the container= (240-x) - (24-x)/5 L
Total quantity of water in the container now = x+(240-x)/5 L
A/Q,
Difference of Milk and Water = {(240-x)-(240-x)/5} - {x+(240-x)/5)}
128= {(240-x)-(240-x)/5} - {x+(240-x)/5)}
x = 10
A. 12
B. 10
C. 9
D. 11
Answer: B. 10
Explanation:
Initial Quantity of milk in the container = 240 L
Quantity of milk when x L is taken out = (240-x) L
Quantity of water added= x L
Again,
when 20% mixture is taken out,
So, Quantity of milk taken out = 20% of (240-x)
= (240-x)/5 L
So, Quality of water added there after = (240-x)/5 L
Quantity of milk left in the container= (240-x) - (24-x)/5 L
Total quantity of water in the container now = x+(240-x)/5 L
A/Q,
Difference of Milk and Water = {(240-x)-(240-x)/5} - {x+(240-x)/5)}
128= {(240-x)-(240-x)/5} - {x+(240-x)/5)}
x = 10
In the following question, by using which mathematical operators will the expression become correct?
14 ? 2 ? 4 ? 6 ? 4
A. ×, ÷, > and ×
B. ÷, ×, > and ×
C. ÷, +, = and ×
D. ÷, +, > and ×
Answer: The correct option is B
Explanation:
14 ÷ 2 × 4 > 6 × 4
7 × 4 > 24
28 > 24
14 ? 2 ? 4 ? 6 ? 4
A. ×, ÷, > and ×
B. ÷, ×, > and ×
C. ÷, +, = and ×
D. ÷, +, > and ×
Answer: The correct option is B
Explanation:
14 ÷ 2 × 4 > 6 × 4
7 × 4 > 24
28 > 24
If 72 x 96 = 6927, 58 x 87 = 7885, then 79 x 86 = ?
A. 6897
B. 9768
C. 6927
D. 7885
Answer: The correct option is A. 6897
Explanation:
72 x 96
→ 96 x 72
→ 69 x 27
→ 6927
Again,
58 x 87
→ 87 x 58
→ 78 x 85
→ 7885
Similarly,
79 x 86
→ 86 x 79
→ 68 x 97
→ 6897
A. 6897
B. 9768
C. 6927
D. 7885
Answer: The correct option is A. 6897
Explanation:
72 x 96
→ 96 x 72
→ 69 x 27
→ 6927
Again,
58 x 87
→ 87 x 58
→ 78 x 85
→ 7885
Similarly,
79 x 86
→ 86 x 79
→ 68 x 97
→ 6897
Price of a diamond is directly proportional to the square of its weight. If the diamond break into 4 pieces by mistake, the ratio of their weight becomes 1 : 2 : 3 : 4, therefore loss will arise Rs 1,40,000. Find original price of the diamond?
A. Rs 240000
B. Rs 200000
C. Rs 220000
D. Rs 180000
Answer: The correct option is B. 200000
Explanation:
Let the weight of the 4 pieces be x, 2x, 3x and 4x
∴ Total weight of the diamond = x + 2x + 3x + 4x
= 10x
∴ Its original price= (10x)²
= 100x² .....(i)
Now,
Present prce of diamond
= (x)² + (2x)² + (3x)² + (4x)²
= 30x²
∴ Loss= 100x² - 30x²
=70x²
A/q
70x² = Rs 140000
∴ x² = Rs 2000
Now,
Putting the value of x² in eq(i) ,
Original price= 100 × 2000
= Rs 200000
A. Rs 240000
B. Rs 200000
C. Rs 220000
D. Rs 180000
Answer: The correct option is B. 200000
Explanation:
Let the weight of the 4 pieces be x, 2x, 3x and 4x
∴ Total weight of the diamond = x + 2x + 3x + 4x
= 10x
∴ Its original price= (10x)²
= 100x² .....(i)
Now,
Present prce of diamond
= (x)² + (2x)² + (3x)² + (4x)²
= 30x²
∴ Loss= 100x² - 30x²
=70x²
A/q
70x² = Rs 140000
∴ x² = Rs 2000
Now,
Putting the value of x² in eq(i) ,
Original price= 100 × 2000
= Rs 200000
A 240 m long train crosses a platform twice its length in 2 min what is the speed of the train ?
A. 6m/s
B. 8m/s
C. 10m/s
D. 12m/s
Answer: The correct option is A. 6 m/s
Explanation:
Here, Length of train = 240 m
∴ Length of platform = 2× 240 m
=480 m
Distance to cross the platform=240+480
=720 m
Now,
Required speed = Distance/time
=720/120 m/sec
= 6 m/s
A. 6m/s
B. 8m/s
C. 10m/s
D. 12m/s
Answer: The correct option is A. 6 m/s
Explanation:
Here, Length of train = 240 m
∴ Length of platform = 2× 240 m
=480 m
Distance to cross the platform=240+480
=720 m
Now,
Required speed = Distance/time
=720/120 m/sec
= 6 m/s
Length of the bridge, which a train 130 metres long and travelling at 45km/h can cross in 30 seconds is :
A. 200 metres
B. 245 metres
C. 225 metres
Answer: The correct option is B
Explanation:
Length of the bridge
= Distance traveled by the train in 30 sec - length of train
= speed × time - 130 metres
Length of the bridge
= 45×1000/60×60 (m/sec) × 30 sec - 130 metres
= 245 metres
A. 200 metres
B. 245 metres
C. 225 metres
Answer: The correct option is B
Explanation:
Length of the bridge
= Distance traveled by the train in 30 sec - length of train
= speed × time - 130 metres
Length of the bridge
= 45×1000/60×60 (m/sec) × 30 sec - 130 metres
= 245 metres
A shopkeeper mixed two varieties of rice at Rs. 24/kg and Rs. X/kg in the ratio 2:3 respectively and sold the mixture at Rs. 29.88/kg at 20% profit. Find the value of X :
A. 25.5
B. 27
C. 30
Answer: The correct option is A
Explanation:
Here , profit % = 20%
SP of 1 kg mixed rice = Rs. 29.88
Therefore, CP of 1 kg mixed rice
= (100/120)×29.88
= Rs. 24.9
Again,
Respective ratio = 2:3
let, Rs. 24/ kg rice be 2A kg
Therefore, Its CP = 2A×24 = Rs. 48A
and Rs. X/kg rice be 3A kg
Therefore, its CP= 3A×X= Rs. 3AX
So,
Total CP of 5A kg mixed rice = Rs. (48A + 3AX)
CP of 1 kg mixed rice = (48A+3AX)/5A
=(48+3X)/5
A/Q, (48+3X)/5 = 24.9
Therefore, X = 25.5
A. 25.5
B. 27
C. 30
Answer: The correct option is A
Explanation:
Here , profit % = 20%
SP of 1 kg mixed rice = Rs. 29.88
Therefore, CP of 1 kg mixed rice
= (100/120)×29.88
= Rs. 24.9
Again,
Respective ratio = 2:3
let, Rs. 24/ kg rice be 2A kg
Therefore, Its CP = 2A×24 = Rs. 48A
and Rs. X/kg rice be 3A kg
Therefore, its CP= 3A×X= Rs. 3AX
So,
Total CP of 5A kg mixed rice = Rs. (48A + 3AX)
CP of 1 kg mixed rice = (48A+3AX)/5A
=(48+3X)/5
A/Q, (48+3X)/5 = 24.9
Therefore, X = 25.5
Ankur , Bhanu and Chatur can finish an assignment in their company together in 20 days. They started the assignment together and Ankur left it after first 6 days. After next 4 days, Bhanu also left the assignment. Then Chatur completed the remaining three fifth of the assignment in 72 days. How many days would Bhanu alone take to finish the whole assignment?
A. 15 days
B. 30 days
C. 60 days
D. None of above
Answer: The correct option is C
Explanation:
Together three of them can completes in 20 days= 1 work
so, in 6 days they together can finish
=(1/20)x6 Part of the work
=3/10 part of the work
∴ Part of work left to be completed
= 1-(3/10)= (7/10) part of the work
Again,
Chatur completes in 72 days= (3/5) part of the work
∴ Bhanu and Chatur together finishes in 4 days
=(7/10)-(3/5) = 1/10 part of the work
Chatur finishes in 72 days = 3/5 part of the work
∴ Chatur can finish in 4 days
= (3/5)x(4/72)
= 1/30 part of the work
∴ Part of work Bhanu can finish in 4 days
= (1/10)-(1/30)=1/15 part of the work
Now,
Bhanu can finish 1/15 part in= 4 days
∴ Bhanu can finish the complete work
=4x(15/1)= in 60 days
A. 15 days
B. 30 days
C. 60 days
D. None of above
Answer: The correct option is C
Explanation:
Together three of them can completes in 20 days= 1 work
so, in 6 days they together can finish
=(1/20)x6 Part of the work
=3/10 part of the work
∴ Part of work left to be completed
= 1-(3/10)= (7/10) part of the work
Again,
Chatur completes in 72 days= (3/5) part of the work
∴ Bhanu and Chatur together finishes in 4 days
=(7/10)-(3/5) = 1/10 part of the work
Chatur finishes in 72 days = 3/5 part of the work
∴ Chatur can finish in 4 days
= (3/5)x(4/72)
= 1/30 part of the work
∴ Part of work Bhanu can finish in 4 days
= (1/10)-(1/30)=1/15 part of the work
Now,
Bhanu can finish 1/15 part in= 4 days
∴ Bhanu can finish the complete work
=4x(15/1)= in 60 days
An alloy of aluminium, copper and Iron contains 85% aluminium, 8% copper and 7% iron. A second alloy of aluminium and iron melted with the first and the mixture then contains 75% aluminium, 5% copper and 20% iron. Find the percentage of aluminium in the second alloy ?
A. 49.4%
B. 58.33%
C. 53.75%
D. None of above
Answer: The correct option is B
Explanation:
For First Alloy:
Let the Mass of first alloy be x
Percentage of Aluminium= 85%
∴ Mass of Aluminium=85x/100 (unit)
Percentage of Coopper=8%
Mass of Copper= 8x/100 (unit)
For Second Alloy:
Let its mass be y
Percentage of Aluminium be p%
∴ Mass of Aluminium= py/100 (unit)
For the Mixed Alloy:
Mass of copper in first alloy = Mass of coper in Mixed alloy
8x/100 = 5(x+y)/100
x/y = 5/3
∴ x= 5 and y= 3
Again,
Mass of aluminium in first alloy + Mass of aluminium in second alloy
=85x/100 + py/100 (unit)
Mass of aluminium in mixed alloy = 75(x+y)/100
A/Q
85x/100 + py/100 = 75(x+y)/100
Putting x=5 and y=3 in the above equation,
p = 58.33%
A. 49.4%
B. 58.33%
C. 53.75%
D. None of above
Answer: The correct option is B
Explanation:
For First Alloy:
Let the Mass of first alloy be x
Percentage of Aluminium= 85%
∴ Mass of Aluminium=85x/100 (unit)
Percentage of Coopper=8%
Mass of Copper= 8x/100 (unit)
For Second Alloy:
Let its mass be y
Percentage of Aluminium be p%
∴ Mass of Aluminium= py/100 (unit)
For the Mixed Alloy:
Mass of copper in first alloy = Mass of coper in Mixed alloy
8x/100 = 5(x+y)/100
x/y = 5/3
∴ x= 5 and y= 3
Again,
Mass of aluminium in first alloy + Mass of aluminium in second alloy
=85x/100 + py/100 (unit)
Mass of aluminium in mixed alloy = 75(x+y)/100
A/Q
85x/100 + py/100 = 75(x+y)/100
Putting x=5 and y=3 in the above equation,
p = 58.33%
If a person lends a certain amount at an interest, compounded annually for 2 years at a rate of 20% p.a. instead of 25% p.a. Then the interest payable will be less by :
A. 24%
B. 19.25%
C. 21.77%
D. None of above
Answer: The correct option is C
Explanation:
Let the Principal amount be Rs 100
n= 2
r= 20%
∴ CI = CA - P
CI = Rs 44
Again,P= Rs 100
n= 2
r= 25%
CI = CA - P
= Rs 56.25
Now,
Amount is less by = Rs 56.25 - Rs 44
= Rs 12.25
Less % = (12.25/56.25)x100
= 21.77%
A. 24%
B. 19.25%
C. 21.77%
D. None of above
Answer: The correct option is C
Explanation:
Let the Principal amount be Rs 100
n= 2
r= 20%
∴ CI = CA - P
CI = Rs 44
Again,P= Rs 100
n= 2
r= 25%
CI = CA - P
= Rs 56.25
Now,
Amount is less by = Rs 56.25 - Rs 44
= Rs 12.25
Less % = (12.25/56.25)x100
= 21.77%
What least number must be added to 1056, so that the sum is completely divisible by 23 :
A. 2
B. 3
C. 4
Answer: The correct option is A. 2
Explanation:
After dividing 1056 by 23 remainder is 21
So, required number= 23-21 = 2
A. 2
B. 3
C. 4
Answer: The correct option is A. 2
Explanation:
After dividing 1056 by 23 remainder is 21
So, required number= 23-21 = 2
What is the unit digit in 63741793 × 625317 × 341491:
A. 0
B. 2
C. 5
Answer: The correct option is A. 0
Explanation:
Unit digit in 63741793
= Unit digit in 41793
=Unit digit in 42×896+1
= Unit digit in (42)896 × 4
= Unit digit in (16)896 × 4
=Unit digit in 6×4
=4
Again,
Unit digit in 625317=5 and 1 in 341491
Required digit=Unit digit in 4×5×1= 0
A. 0
B. 2
C. 5
Answer: The correct option is A. 0
Explanation:
Unit digit in 63741793
= Unit digit in 41793
=Unit digit in 42×896+1
= Unit digit in (42)896 × 4
= Unit digit in (16)896 × 4
=Unit digit in 6×4
=4
Again,
Unit digit in 625317=5 and 1 in 341491
Required digit=Unit digit in 4×5×1= 0
Shyam and Ram entered into a partnership with investment in the ratio 3 : 2 respectively. Among them, Ram is the working partner for which he gets 10% of the profit and Shyam is the sleeping partner. If at the end of one year they earned a profit of Rs. 75000 out of which 5% goes to the charity. Find the share of Ram in the profit.
A. Rs 32000
B. Rs. 33000
C. Rs 31000
D. Rs. 35000
Answer: correct option is (B)
Explanation:
Amount goes to Charity
= (5/100)x75000
= Rs. 3750
Amount Ram gets for work
= (10/100)x75000
= Rs 7500
Remaining profit= Rs (75000-3750-7500)
= Rs 63750
Shares of Ram in remaining profit
=(2/5)x63750
= Rs 25500
Now,
Total share of Ram in profit
= Rs (7500 + 25500)
= Rs 33,000
A. Rs 32000
B. Rs. 33000
C. Rs 31000
D. Rs. 35000
Answer: correct option is (B)
Explanation:
Amount goes to Charity
= (5/100)x75000
= Rs. 3750
Amount Ram gets for work
= (10/100)x75000
= Rs 7500
Remaining profit= Rs (75000-3750-7500)
= Rs 63750
Shares of Ram in remaining profit
=(2/5)x63750
= Rs 25500
Now,
Total share of Ram in profit
= Rs (7500 + 25500)
= Rs 33,000
The speed of current is 5 km/h. What will be the respective downstream speed of a boy rowing a boat , If one third of the distance covered going downstream in a certain time is equal to the distance covered going upstream in the same time.
A. 15km/h , 5 km/h
B. 20 km/h , 10 km/h
C. 25 km/h , 20 km/h
D.None of the above
Answer: Correct option is A
Explanation:
Given speed of the current
= 5 km/h
Let, the speed of the boat in still water be x km/h
Now,
For Downstream
Time = t , Relative speed of the Boat= (x+5) km/h
Distance = Speed x Time
∴ Distance= (x+5)t (km)
Again,
For Upstream
Time= t, Relative speed of the Boat= (x-5) km/h
Distance=(x-5)t (km)
A/Q
(x+5)t/3 = (x-5)t
x= 10 km/h
∴ Downstream speed= (10+5)km/h = 15 km/h and Upstream speed=(10-5) km/h = 5 km/h
A. 15km/h , 5 km/h
B. 20 km/h , 10 km/h
C. 25 km/h , 20 km/h
D.None of the above
Answer: Correct option is A
Explanation:
Given speed of the current
= 5 km/h
Let, the speed of the boat in still water be x km/h
Now,
For Downstream
Time = t , Relative speed of the Boat= (x+5) km/h
Distance = Speed x Time
∴ Distance= (x+5)t (km)
Again,
For Upstream
Time= t, Relative speed of the Boat= (x-5) km/h
Distance=(x-5)t (km)
A/Q
(x+5)t/3 = (x-5)t
x= 10 km/h
∴ Downstream speed= (10+5)km/h = 15 km/h and Upstream speed=(10-5) km/h = 5 km/h
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