Practice Quantitative Aptitude Questions For IBPS 2017 Exams (Miscellaneous& Quadratic Equation):
Dear Readers, Important Practice Aptitude Questions for IBPS Exams 2017 was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this material.
Directions (Q. 1-10):
1). A rectangular pool 20 metres wide and 60 metres long issurrounded by a walkway of uniform width. If the total areaof the walkway is 516 square metres, how wide, in metres, isthe walkway?
Then, (20 + 2x)(60 + 2x) = 516 +1200
= 1200 +120x + 40x + 4x^2 = 516 +1200
= 4x^2 +160x - 516 = 0
= x^2 + 40x -129 = 0
= x^2 + 43x - 3x -129 = 0
= x = 3, x = -43
Width of walkway = 3 metres
Answer is: C
2). Priya was married 8 years ago. Today her age is1 2/7 timesto that at the time of marriage. At present her daughter’sage is1/6thof her age. What was her daughter’s age 3 yearsago?
Priya’s present age = (x + 8) years
x + 8 = 9x/7
= 7x + 56 = 9x
= 2x = 56
x =28
Priya’s present age = 28 + 8 = 36 years
Her daughter’s age 3 years ago
= 36 × 1/6-3
= 6 - 3 =3 years
Answer is: C
3). Udaya brought 25 kg of rice at Rs. 32 per kg and 15 kg ofrice at Rs. 36 per kg. what profit did he get when he mixedthe two varieties together and sold it at Rs. 40.20 per kg?
= [(25 × 32) + (15 × 36)] = (800 + 540)
= 1340
S.P of 40 kg of mixture
= (40 × 40.2)
Profit = (1608 -1340)
= 268
Profit% = 268/1340 × 100
= 20%
Answer is: D
4). Three containers P, Q and R are having mixtures of milkand water in the ratio 1 : 5, 3 : 5 and 5 : 7, respectively. If thecapacities of the containers are in the ratio 5 : 4 : 5, then findthe ratio of the milk to the water if the mixtures of all thethree containers are mixed together.
5 × 1/6 : 4 × 3/8 : 5 × 5/12
= 5/6 : 3/2 : 25/12
and the ratio of water in the containers are,
5 × 5/6 : 4 × 5/8 : 5 × 7/12
= 25/6 : 5/2 : 35/12
Ratio of mixture of milk and water in the containers
= [1/6 × 5 + 3/8 × 4 + 5/12 × 5] : [5/6 × 5 + 5/8 × 4 + 7/12 × 5] = 106 : 230
= 53 : 115
Answer is: C
5). A contract is to be completed in 46 days and 117 men wereset to work, each working 8 hours a day. After 33 days, 4/7of the work is completed. How many additional men may beemployed so that the work may be completed in time, eachman now working 9 hours a day?
117 men were supposed to finish the whole work in46 × 8 = 368 hours.
But 117 men completed4/7of the work in 33 × 8= 264 hours
117 men could complete the work in 462 hours.
Now (117 + x) men are supposed to do 3/7of the work,working 9 hours a day, in 13 × 9 = 117 hours, so as tofinish the work in time.
i.e. (117 + x) men are supposed to complete the wholework in 117 × 7/3= 273 hours.
(117 + x) × 273 = 117 × 462
(117 + x) × 7 = 3 × 462
x + 117 = 3 × 66 = 198
x = 81
Required number of additional men to finish thework in time = 81
Answer is: B
Directions (Q.6-10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answera) if x< y
b) if x> y
c) if x ≤ y
d) if x≥ y
e) if x = y, or relationship between x and y cannot be established.
6). I. 5x=7y+21
II. 11x + 4y+ 109=0
x=-7
y=-8
x>y
Answer is: B
7). I. 6x^2-49x+99=0
II. 5y^2+17y+14=0
(3x-11) (2x-9)=0
x=11/3, 9/2
5y^2+17y+14=>
(5y+7) (y+2)=>
y=-2,-7/5
x>y
Answer is: B
8). I. 2x^2 - 11x + 12=0
II. 2y^2 - 17y + 36=0
x=3/2,4
2y^2-17y+36=>
y=4, 9/2
Answer is: C
9). I. 5x^2=19x-12
II. 5y^2+11y=12
x=3,4/5
5y^2+11y=12
y = 4/5,-3
Answer is: D
10). I. x^3=(1331)^1/3
II. 2y^2-21y+55=0
2y^2-21y+55= 0
(2y-11) (y-5)= 0
y=5,11/2
x>y
Answer is: B
