Quantitative Aptitude - 45 | IBPS | SBI

1. The perimeter of a rhombus is 52 cm and one of its diagonal is 10cm. Find its area in (cm^2).

a)130

b)120

c)150

d)140

e)110

2.A cow is tied in a corner of a rectangular field with dimensions 55cm x 65cm. The length of the rope is 28cm. Find the area(cm^2) which the cow can graze.

a)414

b)313

c)515

d)616

e)717

3.Find the perimeter of the square whose area is 15 cm^2 more than the area of a circle of radius 7 cm.

a)42 cm

b)22 cm

c)52 cm

d)48 cm

e)50 cm

4.The age of a man is four times that of his son. 15 years ago, the man was 7 times as old as his son. What will be the age of his son after 12 years?

a)30 years

b)42 years

c)52 years

d)36 years

e)62 years

5.Rancho has Rs 31,000 in the form of Rs 100, Rs 500, Rs 1000 notes. The ratio of the number of these notes is 7:5:3. What is the amount of Rs 500 notes?

a)Rs 2500

b)Rs 1500

c)Rs12500

d)Rs 10500

e)Rs 5000

6.The price of an article increases by 20% every year. If the difference between the price at the end of the third year and fourth year is Rs 144, what was the price at the end of the second year?

a)Rs 420

b)Rs 500

c)Rs 600

d)Rs 650

e)Data inadequate

7.The profit on selling an article for Rs 550 is thrice the loss on selling it for Rs 350. What is the CP of the article?

a)Rs 200

b)Rs 300

c)Rs 400

d)Rs 500

e)Rs 600

8.The CI and SI on a certain sum for 2 years are Rs 1653.64 and Rs 1560 respectively. Find the sum and rate of interest.

a)Rs 7800, 10%

b)Rs 7800, 15%
c)Rs 6500, 10%

d)Rs 6500, 12%

e)Rs 7800, 12%

9.A trader claims to sell rice at CP but mixes stones and earn 25% profit. What percentage of stones is mixed by the trader?

a)16 2/3%

b)20%

c)25%

d)12.5%

e)22%

10.Farhan worked for 2 hours and then Raju joined him. After 4 more hours, Farhan stopped, and Raju took another 20 hours to complete the remaining part of the job. If both of them work together they can complete the job in 18 hours. How many hours will Farhan and Raju take to complete the work, if they work separately?

a)24 hours, 36 hours

b)36 hours, 24 hours

c)57 hours, 24 hours

d)54 hours, 27 hours

e)20 hours, 35 hours

Answers

1.(b)
Given, perimeter of rhombus = 52cm
Hence, side = 52/4 = 13cm
Let D1 and D2 be the diagonals of a rhombus
D1 = 10cm
D2/2 = (13^2 -5^2)^1/2 =(144)^1/2 = 12
D2 = 12 x 2 =24cm
Area of a rhombus = 1/2 x D1 x D2
=1/2 x 24 x 10 =120

2.(d)
Area grazed = area of sector with radius 28 cm
(pi r^2)/4 = 1/4 x 22/7 x 28 x 28 = 616 cm^2

3.(c)
Area of the circle = 22/7 x 7 x7 = 154 cm^2
Now, area of square = 154 + 15 = 169 cm^2
Side of square = 13cm
Hence perimeter = 13 x 4 = 52 cm

4.(b)
Let the present age of the man be x years and that of his son be y years.
Then x = 4y...(1)
According to the question, 15 years ago
(x -15) = 7(y -15)
Solving we get, x -7y = -90...(2)
Substituting 1 in 2
We get 4y -7y = -90
Hence y =30 years
Son's age after 12 years = 30 +12 =42 years
 
5.(c)
Let x be the number
Ratio = 7x : 5x : 3x
(100 x 7x) + (500 x 5x) +(1000 x 3x) = 31000
x = 5
Hence Rs 500 notes = 500 x 5 x 5 = Rs 12500

6.(c)
Let the initial price be x.
Price(first year) = x
Price second year=1.2x
Price third year = 1.44x
Price fourth year = 1.728x
Now 1.728x -1.44x =144
Solving, we get x =500
Hence, price at the end of the second year = 500 x 120/100 = Rs 600

7.(c)
Let CP be x
According to the question:
550 - x = 3(x -350)
x =400

8.(d)
Let the sum be x and rate be r%.
Then, the difference of SI and CI = r% of SI for two years
r/100 x 1560/2 = 1653.60 -1560
r =12%
Sum = (1560 x 100)/(2 x 12) = Rs 6500

9.(b)
Let there be x kg of stones in 100 kg rice
then, x/(100 -x) x 100 = 25
x =20kg
Percentage of stones mixed = 20/100 x 100 = 20%

10.(d)
Number of hours Farhan worked = 2 + 4 = 6 hoursRaju worked for (4 +20) = 24 hours
Let the per hour work of Farhan be 1/F
And that of Raju = 1/R
Then, 1/F +1/R = 1/18
or
18/F +18/R =1...(1)
For completing the work
6/F +24/R =1...(2)
Equating (1) and (2), we get F =2R
Substituting value of F in 1, we get R =27 hours, F = 54 hours