Quantitative Aptitude Questions (Inequality) for NIACL Assistant Mains 2018 Day-206

Dear Readers, NIACL is conducting Online Examination for the recruitment of Assistant. To enrich your preparation here we are providing new series of inequality  – Quantitative Aptitude Questions. Candidates those who are appearing for NIACL Assistant Mains Exams can practice these Quantitative Aptitude questions daily and make your preparation effective.

Practice Aptitude Questions (Inequality) Day-206

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Directions (Q1 – 5): In each of the following question, read the given statements and compare the two given quantities on its basis.

a) Quantity I > Quantity II

b) Quantity < quantity II

c) Quantity I≥ Quantity II

d) Quantity I ≤ Quantity II

e) Quantity I = Quantity II or cannot be established

 Quantity I:

The area of the rectangle field is 2704 sq.m and its length to breath is in the ratio of 4:1.what is the perimeter of the field

Quantity II

The radius and the height of the right circular cylinder are increased by 25% and 35% respectively. What will be the percentage increase in the curved area?

a) a

b) b

c) c

d) d

e) e

  1. Quantity I:

The ratio of amrit’s to amrita’s age is 7:5 and the sum of their ages is 72.what will be the total ages after 12 years?

Quantity II:

The average age of the whole class is 12.05years.the average age of the girls is 12.5 years and the average age of 45 boys is 11.75 years. What are the total number girls in the class?

a) a

b) b

c) c

d) d

e) e

  1. Quantity I:

7X2 – 34X-5 =0

Quantity II:

2Y2 – 3Y -14 =0

a) a

b) b

c) c

d) d

e) e

  1. Quantity I:

Find the probability of taking a red card or black queen from the deck of 52 cards?

Quantity II:

What is the probability of forming a committee consists of 3 people out of 5 men and 4 women with minimum one woman?

a) a

b) b

c) c

d) d

e) e

  1. Quantity I:

The difference between SI and CI for 2 years is Rs.226.08 and rate of interest is 12%.Find the principal invested on CI and SI

Quantity II:

Rs.5000 is invested by Sindhu on CI for 3 years at the rate of 12%. Find the interest earned by her if amount is compounded half yearly.

a) a

b) b

c) c

d) d

e) e

Directions (6-10): Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.

6) Quantity I:

Distance, if a man covers a distance in 22 hours, he covers first half at 15 km/hr and 2nd half at 18 km/hr
Quantity II:

Distance, if a man covers a distance of three equal parts in 20 hours. He covers first part at 10 km/hr, 2nd at 15 km/hr and 3rd at 30 km/hr

A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established

7) Quantity I: Days in which B can complete work alone, if A and B can complete work in 40 days, B and C in 20 days and C and A in 30 days.
Quantity II:
Days in which B can complete work alone, if A and B can complete work in 24 days and A is 50% more efficient than B.
A) Quantity I > Quantity II
B) Quantity I ≥ Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥ Quantity I
E) Quantity I = Quantity II or Relation cannot be established

8) Quantity I: If the S.I on a sum of money for 2 years at 10% per annum is Rs. 100, what is the C.I on the same at the same rate and for the same time?

Quantity II: What principal will be amount to Rs.48250 at compound interest in 2 years, the rate of interest for 1st& 2nd year being  6% and 8% respectively?

A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established

9) Quantity I: If 25 % of (P – Q) = 15% of (P + Q), then what percent of P is Q?

Quantity II: If (a+b / a-b = 8/7) and a is not equal to zero what percentage (to the nearest integer) of  a +7b is a -7b?

A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established

10) Quantity I: Harini scored 20% marks and failed by 10 marks, Stenika scored 30% marks and obtained 25 marks more than those  required to pass. The pass percentage is.

Quantity II: 2 numbers are smaller than a third number by 40% and 45% respectively. How much percent is the second number less than the first.

A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established

Answers:

1. Answer: A

Quantity I :

4×2 = 2704

X= 26

Perimeter = 2( 26+104) =260m

Quantity II :

Required change = 2π(1.25r*1.35h) – 2πrh / (2πrh) *100 =68.75%

Quantity I>Quantity II

2. Answer: A

Quantity I :

54+42 = 96 years

Quantity II:

12.05 *(x+45)= 12.5 *x +11.75*45

=>x=30

Quantity I> Quantity II

3. Answer: E

X = 5, -1/7

Y = 7/2 , -4/2

Quantity I = Quantity II or cannot be established

4. Answer: B

Quantity I:

= 13/52 + 4/52 = 17/52

Quantity II:

9C3 = 84

= 4C1*5C2 + 4C2*5C1 +4C3*5C0

= 74

= 74/84

Quantity I<Quantity II

5. Answer: A

Quantity I:

226.08 = PR2/1002

=>P= 15700

Quantity II:

Amount = 5000 (1+(12/2)/100)6

=> 7093

Quantity I > Quantity II

6. Answer: A

Quantity I:
(2*15*18)/33 * 22 = 360 km
Quantity II:

LCM of 10, 15 and 30 = 30

Let us take distance covered by each part be 30 km

Total time = 30/10 + 30/15 +30/30 = 6 hrs

Average speed = Total distance/ Total speed

= (30+30+30)/6 = 90/6 = 15
Average speed will become 15 km/hr
So distance 15*20 = 300 km
Quantity I> Quantity II

7. Answer: C

Quantity I:
A + B = 40……..3
B + C = 20………6   ……..(LCM = 120)
C+ A = 30……….4
Total = 2 (A+B+C) = 3+6+4 = 13
So A+B+C = 13/2
(A+B+C) – (B+C) = 13/2 – 4 = 5/2
So B can complete work in 120/(5/2) = 48 days
Quantity II:

Efficiency A …….. B   = 3 : 2
So days = 2 …. 3
LCM of 2 and 3 is 6
A = 2………6/2 = 3
B = 3………6/3 = 2
Total A+B = 3+2 = 5
So 6/5 == 24
So 1 == 20
So 3 == 60
So Quantity II > Quantity I

8. Answer: C

Quantity I:

SI=PTR/100

P=100*SI/TR

=100*100/20

=500

Amount = (p*(1+R/100)n)

=500(1+10/100)2

= 500(110/100)2

=500(11/10)2

=500*11/10*11/10

=60500/100

=RS 605

CI=Rs (605-500)

=Rs 105

Quantity II:

Let Rs.P be the required sum.

48250 = p(1 + 6/100)(1 + 8/100)

48250=P(106/100)(108/100)

48250=P*1.06*1.08

48250=1.144P

P=48250/1.144

P=Rs 42147

Hence the required principal is Rs 42147

Quantity II > Quantity I
9. Answer: C

Quantity I:

25% of (P – Q) = 15% of (P +Q)

25 /100 (P – Q) = 15/100 (P +Q)

25(p-Q) = 15(P+Q)
5(P-Q) = 3(P+Q)
5P – 5Q = 3P +3Q

2P = 8Q

P / Q = 4

p =4Q

Required percentage = (Q/P* 100)

= (Q /4Q *100)

=25%

Quantity II:

Dividing both the numerator and the denominator of the given equation

a+b/a – b = 8/7

(a/b + 1) / (a/b – 1) = 8/7

Cross multiplying this equation yields

7a/b + 7 = 8a/b -8

8a/b – 7a/b = 8+7

a/b = 15

now the percentage of a +7b the expression a -7b is

a-7b/a+7b * 100

dividing both the numerator and the denominator of the expression be b

= a/b – 7 / a/b +7 *100

           = 15-7/15+7 *100

           =8/22 *100

           =36%

Quantity II > Quantity I
10. Answer: A

Quantity I:

Let total marks = x

Then (20% of x) + 10 = (30% of x) -25

(20/100*x)+10 = (30/100*x) – 25

(20x/100)+10 = 30x /100 – 25

10x/100 = 35

X = 350

So passing marks = (20% of 350)+10

=(20/100*350)+10

=80

Pass percentage = 80/350 * 100

= 22.8%

Quantity II:

Let the 3rd number be x

Then 1st number = 60% of x

= 60/100*x

=3x/5

2nd number = 55% of x

=55/100*x

=11/20x

Difference = (3x/5 – 11x/20)

= 12x – 11x/20

=x/20

Required percentage = (x/20*5/3x*100)

=8.33%

Quantity I > Quantity II

Daily Practice Test Schedule | Good Luck

Topic Daily Publishing Time
Daily News Papers & Editorials 8.00 AM
Current Affairs Quiz 9.00 AM
Quantitative Aptitude “20-20” 11.00 AM
Vocabulary (Based on The Hindu) 12.00 PM
General Awareness “20-20” 1.00 PM
English Language “20-20” 2.00 PM
Reasoning Puzzles & Seating 4.00 PM
Daily Current Affairs Updates 5.00 PM
Data Interpretation / Application Sums (Topic Wise) 6.00 PM
Reasoning Ability “20-20” 7.00 PM
English Language (New Pattern Questions) 8.00 PM

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