Quantitative Aptitude Miscellaneous Questions for IBPS Clerk : Set-7

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Quantitative Aptitude Miscellaneous Questions for IBPS Clerk. Welcome to the www.letsstudytogether.co online Quant section. If you are preparing for IBPS PO/Clerk 2017 exam, you will come across a section on Quantitative Aptitude Section. Here we are providing you “Quantitative Aptitude Miscellaneous Questions” for IBPS Clerk based on the latest pattern for your daily practice.

This “Quantitative Aptitude Miscellaneous Questions” is also important for other banking exams such as IBPS PO, IBPS Clerk, SBI Clerk, IBPS  IBPS SO, SBI SO and other competitive exams.

Quantitative Aptitude Miscellaneous Questions for IBPS Clerk : Set-7


1.Virat spends one-third more than what he saves every month. Disha spends 50% more than what Virat spends every month. Monthly salaries of Virat and Disha are in the ratio 9:10. Disha’s savings is what percentage of her
spending?
a. 24.0%
b. 25.5%
c. 29.6%
d. 27.4%
e. 28.3%

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C. 29.6%

Let Virat’s salary = 9y and Disha’s salary is 10y
Let Virat’s savingis x
So, Virat’s spending = 4x/3
Therefore, Virat’s salary = x + 4x/3 = 7x/3
And, Disha’s salary = (7x/3) * (10/9) =70x/27
Disha’s spending = (4x/3)*(1.5) = 2x
Disha’s saving = 70x/27 – 2x = 16x/27
Therefore, required percentage = 100*(16x/27)/2x = 29.6%

2.Modi started a business with an investment of Rs. 48000. After “t” months, Yogi joined him with an investment of Rs. 48000, and after another “t” months Chandu joined them both with an investment of Rs. 24000. If at the end of one year of business, the profit share of Modi was equal to the sum of the profit shares of Yogi and Chandu, find the value of “t”?
a. 2
b. 3
c. 4
d. 5
e. None of the above

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B. 3

Modi Yogi Chandu
 Investment 4800 4800 2400
Time Period for Investment(Month) 12 12-t 12-2t

Let percentage profit is X
According to question,
4800*(X/100))*12 = (4800*(X/100)*(12-t)) + (2400*(X/100))*(12-2t)
=> 576*X = 576*X – 48*X*t + 288*X – 48*X*t
=> 96*t = 288
=> t = 3

3.If the letters of the word GARDEN are rearranged in all possible ways and all the words are written in the order of a dictionary, then what would be the rank of the word GARDEN?
a. 427
b. 361
c. 379
d. 373
e. Cannot be determined

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C. 379

Let’s arrange the letters of the given word in alphabetical manner.
A, D, E, G, N, R
Number of words starting with A = 5! = 120
Number of words starting with D = 5! = 120
Number of words starting with E = 5! = 120
GARDEN is the first word starting with GAR
So, Number of words starting with G which come before GARDEN = Number of words starting with GAD + Number of words starting with GAE + Number of words starting with GAN = 3! + 3! + 3! = 18
Therefore, rank of GARDEN in the dictionary = 120+120+120+18+1 = 379

4.Pipe A can fill an empty Tank with water in 24 hours. Pipe B can fill the empty Tank with a mixture that contains 24% pure alcohol, in 16 hours. If both the pipes are opened alternately for one hour, starting from pipe A, what will be the alcohol concentration in the tank when it gets filled?
a. 12%
b. 13%
c. 14%
d. 15%
e. 16%

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C. 14%

Let the volume of the tank = 48 litre
Water filled by Pipe A in one hour = 48/24 = 2 litre
24% pure Alcohol filled by Pipe B in one hour = 48/16 = 3 litre
Volume filled in 2 hours = 2+3 = 5 litre
Total Volume filled in 18 hours = 45 litre
In the 19th hour Pipe A fills another 2 litres in the tank
So, remaining volume = 48 – 45 – 2 = 1 liter
The remaining volume will be filled by Pipe B in (1/3) hour
Quantity of Water in the tank = 10*2 = 20 litre
Quantity of mixture in the tank that contains 24% alcohol = 9*3+3*(1/3) = 28 litres
Total volume of alcohol in the tank = 28*(0.24) = 6.72 litres
Therefore, concentration of alcohol in the tank = 100*6.72/48 = 14%

5.A trader mixed two varieties of wheat whose cost prices are Rs.30 per kg and Rs.38 per kg. He sold the mixture at the rate of Rs.40 per kg and earned 25% profit. What quantity of the costlier variety of wheat was mixed with 12 kg of the cheaper variety of wheat?
a. 3 kg
b. 4 kg
c. 6 kg
d. 5 kg
e. 8 kg

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B. 4 kg

S.P. = Rs. 40 and profit = 25%
∴ C.P. per kg of the mixture = Rs. 32
Let x kg of the costlier wheat be used.
∴ 32(x + 12) = 38x + 30(12)
∴ 32x + 384 = 38x + 360
∴ 6x = 24 i.e. x = 4
Thus, 4 kg of the costlier wheat is used.

6.Three dices are thrown one after the other. Find in how many ways the dices can be thrown such that the sum of the three numbers displayed by the dices is more than 3 but less than 16.
a. 11 ways
b. 211 ways
c. 205 ways
d. 5 ways
e. 198 ways

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C. 205 ways

Total number of ways in which the three dices can be thrown = 6C1*6C1*6C1 = 6*6*6 = 216 ways
Possibility that the sum of numbers is less than 3 is never possible.
Possibilities that the sum of numbers is 3:
(1,1,1) = 1 way
Possibilities that the sum of numbers is 18:
(6,6,6) = 1 way
Possibilities that the sum of numbers is 17:
(6,6,5), (6,5,6), (5,6,6) = 3 ways
Possibilities that the sum of numbers is 16:
(5,5,6), (5,6,5), (6,5,5) = 3 ways
(6,4,6), (6,6,4), (4,6,6) = 3 ways
Total number of ways such that the sum of the numbers is not, more than 3 or less than 16 = 1 + 1 + 3 + 3 + 3 = 11
Therefore, the total number of ways such that the sum of the numbers is more than 3 but less than 16:
216 – 11 = 205 ways.

7.A ball is dropped from a height a 200m. After ricocheting with the ground it only bounces back 75% of its height from which it drops, repeatedly, till it comes to a stop. Find the distance travelled by the ball.
a. 1600m
b. 350m
c. 800m
d. 1400m
e. 600m

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D. 1400m

After each collision with the ground the ball bounces back 75% or 3/4th
of its height from which it drops. So, it forms an infinite geometric progression with the initial term = 200m and common ratio = 3/4.
But, when the ball rises to a certain height it also drops that amount height to touch the ground.
So, the value of the infinite G.P. has to be multiplied by 2.
But also, the ball does not rise back to its initial 200m height.
So, 200m has to be subtracted from the derived value.
Therefore, the distance travelled by the ball will be:
2*{200/(1-3/4)} – 200 = 2*800 – 200 = 1400m.

8.An examination of Mathematics in class X consisted of total 120 questions. The marking pattern for each correct answer, each wrong answer and each unanswered question in the exam is 1 mark, -1/2 mark and -1/4 mark respectively. If Rishabh scored total 50 marks in the examination, then what is the maximum number of questions that Rishabh could have answered wrong in the examination?
a. 35
b. 40
c. 45
d. 50
e. 55

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C. 45

Let number of correct answered questions = C
Let number of wrong answered questions = W
Let number of Unanswered questions = U
Thus, C+W+ U= 120 ………….. (1)
Also, C – (1/2)W – (1/4)U = 50 .…………. (2)
Þ 4C -2W -U = 200 .………….(3)
Adding (1) and (3), we get
5C -W = 320
W = 5C- 320
When C = 64, 65, 66…..
Values of W will be 0, 5, 10, 15………..
When C = 73, W = 45
and when C = 74, W = 50 which is not possible.
Therefore, maximum possible value of W = 45

9.Worker-A can dig 12m2 area in 2 days, working 8 hours a day while worker B can dig 14m2 in 3 days, working 6 hours a day. In a square park of area 3136m2 four equal sized largest possible circular digging job is planned. If both the workers are employed together to do the job and they both work 9 hours a day and A left the job 52 days after start, on which day after the start the job got completed?
a. 300
b. 301
c. 302
d. 303
e. 304

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C. 302

10.Arnold and Brandon jointly start a business firm. Arnold is a working partner while Brandon is a sleeping partner. The investment in the business of Arnold is Rs. 10000 while that of Brandon is Rs. 12000. 12.5% of the profit goes to Arnold for his managerial services, while rest is divided between them proportionate to their capitals. What is the profit share of Arnold in the business if the firm makes a profit of Rs. 1760?
a. Rs. 220
b. Rs. 540
c. Rs. 700
d. Rs. 920
e. None of these

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D. Rs. 920

The amount which Arnold receives for managing business = 12.5% of Rs. 1760 = (12.5/100)*1760 = Rs.220
The amount left to be divided in the ratio of 10000: 12000 i.e. 5: 6 will be 1760 – 220 = Rs. 1540
Therefore, Arnold’s share = (5/11)*1540 = Rs. 700
Therefore, total profit share of Arnold = 700 + 220 = Rs. 920


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