Practice Quantitative Aptitude Questions For IBPS 2017 Exams (Application Problems & 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. 6-10): In each of these questions two equations (I) and (II) are given. You have to solve both the equations and give answer
Answer With Explanation:
4 bag + 7 box = 21 × 11 = 231
12 bag + 12 box = 556
Total cost of 18 bags and 18 boxes
= 556 × 3 / 2
= 834
Answer is: b)
Then
Harish brother’s age = x + 3
Harish mother’s age = x + 26
Harish sister’s age = (x + 3) + 4 = x + 7
Harish Father’s age = (x + 7) + 28 = x + 35
Age Harish father when Harish brother was born = x + 35 – (x + 3) = 32
Answer is: d)
36 × 18 × 3 = a^3 + 8a^3
a^3 = 216
a = 6 cm
Surface area of the smaller cube = 6 × 6^2
= 216 cm^2
Answer is: e)
Work done by Y in 1 hour = 1/10
Work done by Z in 1 hour = 1/12
Work done by X,Y and Z in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Y and Z in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the machines were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Y an Z to complete the pending work = (23/60) / (11/60) = 23/11
This is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm
Answer is: c)
n(s) = 24C3 = 2024
Number of ways to select one ball of each colour
= n (E) = 4C1 × 8C1 × 12C1 = 4 × 8 × 12 = 384
P (E) = 384 / 2024
= 48 / 253
Answer is: a)
or x(x – 11) – 71(x – 11) = 0
or (x – 11) (x – 71) = 0
x = 11, 71
II. y^2 = 5041
y = +71, -71
No relation can be established between x and y.
Answer is: e)
or 3x(2x – 3) – 4(2x – 3) = 0
or (2x – 3) (3x – 4) = 0
x = 3/2, 4/3
II. 7y^2 – 7y – 6y + 6 = 0
or 7y(y – 1) – 6(y – 1) = 0
or (7y – 6) (y – 1) = 0
y = 1, 6/7
x > y
Answer is: a)
or 3x(2x – 5) – 16(2x – 5) = 0
or (3x – 16) (2x – 5) = 0
x = 16/3, 5/2
II. 2y^2 – 4y – 5y + 10 = 0
or 2y(y – 2) – 5(y – 2) = 0
or (y – 2) (2y – 5) = 0
y = 2, 5/2
x ≥ y
Answer is: b)
or 2x(x + 1) – 1(x + 1) = 0
or (2x – 1) (x + 1) = 0
x = –1, 1/2
II. 2y^2 + 3y + 10y + 15 = 0
or y(2y + 3) + 5(2y + 3) = 0
or (y + 5) (2y + 3) = 0
y = –5, – 3/2
x > y
Answer is: a)
or x(x + 4) + 8(x + 4) = 0
or (x + 4) (x + 8) = 0
x = –4, –8
II. 2y^2 + 6y + 9y + 27 = 0
or 2y(y + 3) + 9(y + 3) = 0
or (2y + 9) (y + 3) = 0
y = – 9/2, –3
No relation can be established between x and y.
Answer is: e)
This post was last modified on December 26, 2019 4:48 pm