Dear Readers, Find the aptitude test questions to crack latest bank exams. We regularly provide 20 aptitude test questions daily for students. Aspirants practice these questions on a regular basis to improve your score in aptitude section. Aspirants preparing for the exams can make use of this “20-20” Quantitative Aptitude Questions. Here we have started New Series of Practice Materials specially for Crack NIACL Assistant Prelims 2018. Aspirants those who are preparing for the exams can use this “20-20” Quantitative Aptitude Questions.
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Directions (6 – 10): what value should come in place of question mark (?) in the following questions?
[(2464 ÷ 8 ÷ 14) + (28 × 7)^{1/2}] = (?)^{2}
Correct
Answer: a)
((2464/8)/14) + (196)^{1/2} = x^{2}
(308/14) + 14 = x^{2}
X^{2} = 22 + 14 = 36
X = 6
Incorrect
Answer: a)
((2464/8)/14) + (196)^{1/2} = x^{2}
(308/14) + 14 = x^{2}
X^{2} = 22 + 14 = 36
X = 6
Question 11 of 20
11. Question
C is 40% less efficient than A. A and B together can finish a piece of work in 10 days. B and C together can do it in 15 days. In how many days can A alone finish the same piece of work?
Correct
Answer: c)
The efficiency ratio of A and C = 100 : 60 = 5 : 3
The day ratio of A and B = 3 : 5 (3x, 5x)
(A + B)’s one day work = 1/10
(B + C)’s one day work = 1/15
(A – C)’s one day work = (1/10) – (1/15) = 1/30
According to the question,
(1/3x) – (1/5x) = 1/30
2x/(3x*5x) = 1/30
2/15x = 1/30
60 = 15x
X = 4
A alone can finish a work in = 3x = 12 days
Incorrect
Answer: c)
The efficiency ratio of A and C = 100 : 60 = 5 : 3
The day ratio of A and B = 3 : 5 (3x, 5x)
(A + B)’s one day work = 1/10
(B + C)’s one day work = 1/15
(A – C)’s one day work = (1/10) – (1/15) = 1/30
According to the question,
(1/3x) – (1/5x) = 1/30
2x/(3x*5x) = 1/30
2/15x = 1/30
60 = 15x
X = 4
A alone can finish a work in = 3x = 12 days
Question 12 of 20
12. Question
The diameter of a smaller circle and larger circle are the side of a square and the diagonal of the square respectively. If the side of the square is 16 cm, then find the ratio of the areas of the smaller circle and the larger circle?
Correct
Answer: b)
The diameter of the smaller circle = Side of the square = 16 cm
The radius of the smaller circle(r) = 8 cm
The diameter of the larger circle = Diagonal of the square = √2*Side = 16√2
The radius of the larger circle = 8√2
The ratio of area of smaller circle and area of larger circle
= > 8^{2 }: (8√2)^{2}
= > 64 : 64*2 = 1 : 2
Incorrect
Answer: b)
The diameter of the smaller circle = Side of the square = 16 cm
The radius of the smaller circle(r) = 8 cm
The diameter of the larger circle = Diagonal of the square = √2*Side = 16√2
The radius of the larger circle = 8√2
The ratio of area of smaller circle and area of larger circle
= > 8^{2 }: (8√2)^{2}
= > 64 : 64*2 = 1 : 2
Question 13 of 20
13. Question
The ratio of present ages of son and daughter of a lady is 7 : 8 and the age of lady is sum total ages of her daughter and son. If the ratio of ages of son after 3 years and the age of daughter 6 years ago is 4 : 3 then what will be the ratio of the present age of lady and the age of her son after four years?
Correct
Answer: d)
The ratio of present ages of son and daughter of a lady = 7 : 8 (7x, 8x)
The present age of lady = 7x + 8x = 15x
According to the question,
(7x + 3) / (8x – 6) = 4/3
21x + 9 = 32x – 24
11x = 33
x = 3
Son’s present age = 21years
Daughter’s present age = 24 years
Present age of lady = 45 years
Required ratio = 45: (21 + 4) = 45 : 25 = 9 : 5
Incorrect
Answer: d)
The ratio of present ages of son and daughter of a lady = 7 : 8 (7x, 8x)
The present age of lady = 7x + 8x = 15x
According to the question,
(7x + 3) / (8x – 6) = 4/3
21x + 9 = 32x – 24
11x = 33
x = 3
Son’s present age = 21years
Daughter’s present age = 24 years
Present age of lady = 45 years
Required ratio = 45: (21 + 4) = 45 : 25 = 9 : 5
Question 14 of 20
14. Question
A shopkeeper offers two successive discounts of 10 % and 15 % on an article and is still able to earn 53 % profit. If he offers only 20 % discount, then what will be his profit percentage?
Correct
Answer: b)
Let us assume the marked price of the article be Rs. 100,
Sp= 100*(90/100)*(85/100) = Rs. 76.5
(153/100)*Cp = 76.5
Cp = 76.5*(100/153) = Rs. 50
If the shopkeeper offers only 20 % discount, the profit % will be,
In a bag, there are 6 red balls and 4 green balls. Three balls are picked at random. What is the probability that two balls are red and one ball is green colour?
Correct
Answer: c)
Total probability n(S) = 10C_{3}
Required probability n(E) = 6C_{2} and 4C_{1}
P(E) = n(E)/n(S) = (6C_{2} and 4C_{1})/10C_{3} = 1/2
Incorrect
Answer: c)
Total probability n(S) = 10C_{3}
Required probability n(E) = 6C_{2} and 4C_{1}
P(E) = n(E)/n(S) = (6C_{2} and 4C_{1})/10C_{3} = 1/2
Question 16 of 20
16. Question
Directions (16 – 20): Study the following table carefully and answer the questions given below.
The following table shows the amount invested by six different companies during six different months (in lakhs)
Find the difference between the investment made by company A and E together in March and company B and D together in January (in lakhs)?
Correct
Answer: a)
Investment made by A and E in March = 52 + 83 = 135 lakhs
Investment made by B and D in January = 28 + 32 = 60 lakhs
Hence, required difference = 135 – 60 = 75 lakhs
Incorrect
Answer: a)
Investment made by A and E in March = 52 + 83 = 135 lakhs
Investment made by B and D in January = 28 + 32 = 60 lakhs
Hence, required difference = 135 – 60 = 75 lakhs
Question 17 of 20
17. Question
Directions (16 – 20): Study the following table carefully and answer the questions given below.
The following table shows the amount invested by six different companies during six different months (in lakhs)
Find the sum of the amount invested by company B and D in all the given months together (in lakhs)?
Correct
Answer: c)
The amount invested by company B in all the given months
= > 28 + 40 + 38 + 45 + 20 + 95 = 266 lakhs
The amount invested by company D in all the given months
= > 32 + 55 + 72 + 90 + 25 + 80 = 354 lakhs
Required sum = 266 + 354 = 620 lakhs
Incorrect
Answer: c)
The amount invested by company B in all the given months
= > 28 + 40 + 38 + 45 + 20 + 95 = 266 lakhs
The amount invested by company D in all the given months
= > 32 + 55 + 72 + 90 + 25 + 80 = 354 lakhs
Required sum = 266 + 354 = 620 lakhs
Question 18 of 20
18. Question
Directions (16 – 20): Study the following table carefully and answer the questions given below.
The following table shows the amount invested by six different companies during six different months (in lakhs)
In which month, the investment made by these companies were maximum?
Correct
Answer: b)
The total investment of companies,
In January = 25 + 28 + 50 + 32 + 24 + 35 = 194 lakhs
In February = 45 + 40 + 60 + 55 + 28 + 38 = 266 lakhs
In March = 52 + 38 + 46 + 72 + 83 + 65 = 356 lakhs
In April = 80 + 45 + 70 + 90 + 65 + 76 = 426 lakhs
In May = 15 + 20 + 47 + 25 + 13 + 30 = 150 lakhs
In June = 62 + 95 + 27 + 80 + 50 + 20 = 334 lakhs
April month has the maximum investments.
Incorrect
Answer: b)
The total investment of companies,
In January = 25 + 28 + 50 + 32 + 24 + 35 = 194 lakhs
In February = 45 + 40 + 60 + 55 + 28 + 38 = 266 lakhs
In March = 52 + 38 + 46 + 72 + 83 + 65 = 356 lakhs
In April = 80 + 45 + 70 + 90 + 65 + 76 = 426 lakhs
In May = 15 + 20 + 47 + 25 + 13 + 30 = 150 lakhs
In June = 62 + 95 + 27 + 80 + 50 + 20 = 334 lakhs
April month has the maximum investments.
Question 19 of 20
19. Question
Directions (16 – 20): Study the following table carefully and answer the questions given below.
The following table shows the amount invested by six different companies during six different months (in lakhs)
The investment made by company A in June is approximately what percentage less than of the investment made by company D in the same month?
11) C is 40% less efficient than A. A and B together can finish a piece of work in 10 days. B and C together can do it in 15 days. In how many days can A alone finish the same piece of work?
a) 15 days
b) 9 days
c) 12 days
d) 18 days
e) None of these
12) The diameter of a smaller circle and larger circle are the side of a square and the diagonal of the square respectively. If the side of the square is 16 cm, then find the ratio of the areas of the smaller circle and the larger circle?
a) 5 : 11
b) 1 : 2
c) 3 : 7
d) 7 : 13
e) None of these
13) The ratio of present ages of son and daughter of a lady is 7 : 8 and the age of lady is sum total ages of her daughter and son. If the ratio of ages of son after 3 years and the age of daughter 6 years ago is 4 : 3 then what will be the ratio of the present age of lady and the age of her son after four years?
a) 4 : 7
b) 5 : 3
c) 6 : 11
d) 9 : 5
e) None of these
14) A shopkeeper offers two successive discounts of 10 % and 15 % on an article and is still able to earn 53 % profit. If he offers only 20 % discount, then what will be his profit percentage?
a) 54 %
b) 60 %
c) 58 %
d) 66 %
e) None of these
15) In a bag, there are 6 red balls and 4 green balls. Three balls are picked at random. What is the probability that two balls are red and one ball is green colour?
a) 3/7
b) 5/8
c) 1/2
d) 11/19
e) None of these
Directions (16 – 20): Study the following table carefully and answer the questions given below.
The following table shows the amount invested by six different companies during six different months (in lakhs)
16) Find the difference between the investment made by company A and E together in March and company B and D together in January (in lakhs)?
a) 75 lakhs
b) 65 lakhs
c) 50 lakhs
d) 85 lakhs
e) None of these
17) Find the sum of the amount invested by company B and D in all the given months together (in lakhs)?
a) 670 lakhs
b) 560 lakhs
c) 620 lakhs
d) 590 lakhs
e) None of these
18) In which month, the investment made by these companies were maximum?
a) March
b) April
c) May
d) June
e) None of these
19) The investment made by company A in June is approximately what percentage less than of the investment made by company D in the same month?
a) 23 %
b) 15 %
c) 45 %
d) 5 %
e) 50 %
20) Find the average investment made by company F during all the months (in lakhs)?