Important Quant Miscellaneous Questions for IBPS PO Prelims 2018
Important Quant Miscellaneous Questions for IBPS PO Prelims 2018. Welcome to the www.letsstudytogether.co online Quant section. If you are preparing for IBPS PO/Clerk exams 201718, you will come across a section on Quantitative Aptitude Section. Here we are providing you “Quant Miscellaneous Questions” for IBPS PO Prelims based on the latest pattern for your daily practice.
This “Quant Miscellaneous Questions for IBPS PO Prelims 2018” is also important for other banking exams such as SBI PO, IBPS PO, IBPS Clerk, SBI Clerk, IBPS IBPS SO, SBI SO and other competitive exams
Important Quant Miscellaneous Questions for IBPS PO Prelims: Set25
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Question 1 of 10
1. Question
A starts a company and after 3 months B also joins the company. The initial investment of A and B is in the ratio of 3:2, respectively. A and B receives Rs. (x + 600) and Rs. (x – 360), respectively as profit after completion of one year of the company. Find the profit Share of B.
Correct
A. Rs. 960
The ratio of profit share of A and B is
A: B = 3 × 12: 2 × (12 – 3) = 2: 1According to question,
(x + 600)/(x – 360) = 2: 1
x + 600 = 2x – 720
3x – 2x = 600 + 720
x = 1320Therefore, profit share of B = 1320 – 360 = Rs. 960
Incorrect
A. Rs. 960
The ratio of profit share of A and B is
A: B = 3 × 12: 2 × (12 – 3) = 2: 1According to question,
(x + 600)/(x – 360) = 2: 1
x + 600 = 2x – 720
3x – 2x = 600 + 720
x = 1320Therefore, profit share of B = 1320 – 360 = Rs. 960

Question 2 of 10
2. Question
On the occasion of Rakshabandhan, there are (x + 10) boys and (x + 6) girls in a community hall, each girl tied the rakhi on the wrist of every boy, and every boy gave a chocolate to each girl as return gift, if total number of rakhi and chocolates used was 1344, then find the value of x?
Correct
C. 18
Number of rakhi used = (x + 10)(x + 6)
Number of chocolates used= (x + 6)(x + 10)
According to question,
(x + 10)(x + 6) + (x + 6)(x + 10) = 1344
x^{2}+ 16x + 60 + x^{2 }+ 16x + 60 = 1344
x^{2}+ 16x + 60 = 672
x^{2}+ 16x – 612 = 0
x^{2}+ 34x – 18x – 612 = 0
x(x+ 34) – 18(x + 34) = 0
(x + 34)(x – 18) = 0
x = 34 or x = 18
Incorrect
C. 18
Number of rakhi used = (x + 10)(x + 6)
Number of chocolates used= (x + 6)(x + 10)
According to question,
(x + 10)(x + 6) + (x + 6)(x + 10) = 1344
x^{2}+ 16x + 60 + x^{2 }+ 16x + 60 = 1344
x^{2}+ 16x + 60 = 672
x^{2}+ 16x – 612 = 0
x^{2}+ 34x – 18x – 612 = 0
x(x+ 34) – 18(x + 34) = 0
(x + 34)(x – 18) = 0
x = 34 or x = 18

Question 3 of 10
3. Question
Two numbers are in the ratio of 8:x, respectively. If 14 is added to the first number and 6 is subtracted from second number then ratio becomes 5:3, respectively. Find the value of x, if the difference between two numbers is 24. (Note: 8 ? x)
Correct
D. 6
Let, the two numbers be 8k and kx.
According to question,
8k – kx = 24
k(8 – x) = 24… (i)
And, (8k + 14): (kx – 6) = 5: 3
24k + 42 = 5kx – 30
k(24 – 5x) = 72
k(5x – 24) = 72
Using, equation (i), we get,
(24/(8 – x)) × (5x – 24) = 72
120x – 576 = 576 – 72x
120x + 72x = 1152
192x = 1152
x = 1152 ÷ 192 = 6
Incorrect
D. 6
Let, the two numbers be 8k and kx.
According to question,
8k – kx = 24
k(8 – x) = 24… (i)
And, (8k + 14): (kx – 6) = 5: 3
24k + 42 = 5kx – 30
k(24 – 5x) = 72
k(5x – 24) = 72
Using, equation (i), we get,
(24/(8 – x)) × (5x – 24) = 72
120x – 576 = 576 – 72x
120x + 72x = 1152
192x = 1152
x = 1152 ÷ 192 = 6

Question 4 of 10
4. Question
The average salary of all the employees in a bank is Rs. 800 per day. The average salary of clerks and managers is Rs.500 per day and Rs. 1250 per day, respectively. Find the number of clerks if there are total 20 employees in the branch. (Employees = Clerks + Managers)
Correct
C. 12
Let, the number of clerk be ‘x’
So, number of managers be (20 – x).
According to question,
(20 – x) × 1250 + x × 500 = 800 × 20
25000 – 1250x + 500x = 16000
750x = 25000 – 16000
750x = 9000
x = 9000 ÷ 750
x = 12
Incorrect
C. 12
Let, the number of clerk be ‘x’
So, number of managers be (20 – x).
According to question,
(20 – x) × 1250 + x × 500 = 800 × 20
25000 – 1250x + 500x = 16000
750x = 25000 – 16000
750x = 9000
x = 9000 ÷ 750
x = 12

Question 5 of 10
5. Question
The profit earned on selling an article for Rs. (x + 200) is twice the loss incurred on selling the same article for Rs. (x – 250). Find the value of ‘x’, if to earn a profit of 20%, it must be sell for Rs. 1140.
Correct
D. 1050
Cost price of article = 1140 ÷ 1.2 = Rs. 950
According to question,
(x + 200) – 950 = 2(950 – (x – 250))
x – 750 = 2(950 – x + 250)
x = 2400 – 2x + 750
3x = 3150
x = 3150 ÷ 3
x = 1050
Incorrect
D. 1050
Cost price of article = 1140 ÷ 1.2 = Rs. 950
According to question,
(x + 200) – 950 = 2(950 – (x – 250))
x – 750 = 2(950 – x + 250)
x = 2400 – 2x + 750
3x = 3150
x = 3150 ÷ 3
x = 1050

Question 6 of 10
6. Question
The sum of perimeter of a rectangle(R) and a square(S) is 210 cm. The length and breadth of rectangle(R) is 40% more and 20% less than the side of square(S), what is the length of diagonal of rectangle(R)?
Correct
B. 5√65 cm
Let, the side of square be ‘s’ cm.
So, length of rectangle = 1.4s cm
And, breadth of rectangle = 0.8s cm
According to question,
2(1.4s + 0.8s) + 4s = 210
4.4s+ 4s = 210
8.4s= 210
s = 210 ÷ 8.4
s = 25 cm
So, length of rectangle = 1.4 × 25 = 35 cm
And breadth of rectangle = 0.8 × 25 = 20 cm
So, diagonal of rectangle = √(35)^{2 }+ (20)^{2 }= √1625 = 5√65 cm
Incorrect
B. 5√65 cm
Let, the side of square be ‘s’ cm.
So, length of rectangle = 1.4s cm
And, breadth of rectangle = 0.8s cm
According to question,
2(1.4s + 0.8s) + 4s = 210
4.4s+ 4s = 210
8.4s= 210
s = 210 ÷ 8.4
s = 25 cm
So, length of rectangle = 1.4 × 25 = 35 cm
And breadth of rectangle = 0.8 × 25 = 20 cm
So, diagonal of rectangle = √(35)^{2 }+ (20)^{2 }= √1625 = 5√65 cm

Question 7 of 10
7. Question
A bag contains 36 balls of 3 different colours Magenta, Khaki, and Violet. The probability of drawing a Khaki ball is 1/4, and the ratio of number of Magenta balls to the number of Violet ball is 5:4. If 2 balls are drawn from the bag randomly then what will be the probability that two balls are of same colour?
Correct
A. 23/70
Probability of drawing a Khaki ball = 1/4,
So, number of Khaki balls = (1/4) × 36 = 9
Therefore, number of balls of Magenta and Violet colours = 36 – 9 = 27
So, number of Magenta balls = (5/9) × 27 = 15
And, number of Violet balls = 27 – 15 = 12
Therefore, required probability = (^{9}C_{2} + ^{15}C_{2} + ^{12}C_{2})/^{36}C_{2}= (9 × 4 + 15 × 7 + 6 × 11)/(18 × 35) = 23/70
Incorrect
A. 23/70
Probability of drawing a Khaki ball = 1/4,
So, number of Khaki balls = (1/4) × 36 = 9
Therefore, number of balls of Magenta and Violet colours = 36 – 9 = 27
So, number of Magenta balls = (5/9) × 27 = 15
And, number of Violet balls = 27 – 15 = 12
Therefore, required probability = (^{9}C_{2} + ^{15}C_{2} + ^{12}C_{2})/^{36}C_{2}= (9 × 4 + 15 × 7 + 6 × 11)/(18 × 35) = 23/70

Question 8 of 10
8. Question
Sita and Gita together can complete a work in 18 days. Sita starts the work and completes onethird of work in ‘x’ days and leaves the work. The remaining work is done by Gita in ‘x’ days. Find the time taken by Rita to complete the same work who is 20% more efficient than Sita.
Correct
D. 45 days
Time taken by Sita to complete the work = 3x days
Time taken by Gita to complete the work = (3/2)x days
According to question,
1/3x + 2/3x = 1/18
3/3x = 1/18
x = 18
So, time taken by Sita to complete the work = 54 days
Therefore, time taken by Rita to complete the work = 54/1.2 = 45 days.
Incorrect
D. 45 days
Time taken by Sita to complete the work = 3x days
Time taken by Gita to complete the work = (3/2)x days
According to question,
1/3x + 2/3x = 1/18
3/3x = 1/18
x = 18
So, time taken by Sita to complete the work = 54 days
Therefore, time taken by Rita to complete the work = 54/1.2 = 45 days.

Question 9 of 10
9. Question
Ram and Shyam invested in a scheme offering simple interest for same period of time. If interest obtained by Ram and Shyam at rate of 8% per annum and 12% per annum, respectively is Rs. 2232 and Rs. 2700, respectively. Find the amount invested by Shyam if amount invested by Ram is Rs. 1800 more than amount invested by Shyam.
Correct
A.
Let, amount invested by Shyam is Rs. S, and time of investment is ‘t’ years.
So, amount invested by Ram is Rs. (S + 1800).
According to question,
S × 12% × t = 2700
Also, (S + 1800) × 8% × t = 2232
From both equations, we get,
2700/12S = 2232/8(S + 1800)
225/S = 279/(S + 1800)
225S + 405000 = 279S
54S = 405000
S = 405000/54
S = 7500
Incorrect
A.
Let, amount invested by Shyam is Rs. S, and time of investment is ‘t’ years.
So, amount invested by Ram is Rs. (S + 1800).
According to question,
S × 12% × t = 2700
Also, (S + 1800) × 8% × t = 2232
From both equations, we get,
2700/12S = 2232/8(S + 1800)
225/S = 279/(S + 1800)
225S + 405000 = 279S
54S = 405000
S = 405000/54
S = 7500

Question 10 of 10
10. Question
A train of length (x + 30) m can cross a man standing on the platform in 12 sec and, it can pass a man who is moving at speed of 18 Km/hour in same direction that of train, in 15 sec, find the length of train.
Correct
C. 300 m
Let, the speed of train is ‘s’ m/s
According to question,
(x + 30)/12 = s
And, (x + 30)/15 = s – 18 × (5/18)
Therefore, comparing both equations, we get,
(x + 30)/12 = (x + 30)/15 + 18 × (5/18)
(x + 30)/12 = (x + 30 + 75)/15
5x + 150 = 4x + 420
5x – 4x = 420 – 150
x = 270
Therefore, length of train = 270 + 30 = 300 m
Incorrect
C. 300 m
Let, the speed of train is ‘s’ m/s
According to question,
(x + 30)/12 = s
And, (x + 30)/15 = s – 18 × (5/18)
Therefore, comparing both equations, we get,
(x + 30)/12 = (x + 30)/15 + 18 × (5/18)
(x + 30)/12 = (x + 30 + 75)/15
5x + 150 = 4x + 420
5x – 4x = 420 – 150
x = 270
Therefore, length of train = 270 + 30 = 300 m
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